Chapter 11

CHAPTER 11

PROBLEM 11.1 Determine the modulus of resilience for each of the following metals: (a ) Stainless steel AISI 302 (annealed):

E = 190 GPa σ Y = 260 MPa

(b) Stainless steel AISI 302 (cold-rolled):

E = 190 GPa σ Y = 520 MPa

(c) Malleable cast iron:

E = 165 GPa σ Y = 230 MPa

SOLUTION (a)

E = 190 × 109 Pa, σ Y = 260 × 106 Pa uY =

σ Y2 2E

=

(260 × 106 ) 2 = 177.9 × 103 N ⋅ m/m3 (2)(190 × 109 ) uY = 177.9 kJ/m3 

(b)

E = 190 × 109 Pa, σ Y = 520 × 106 Pa uY =

σ Y2 2E

=

(520 × 106 ) 2 = 712 × 103 N ⋅ m/m3 9 (2)(190 × 10 ) uY = 712 kJ/m3 

(c)

E = 165 × 109 Pa, σ Y = 230 × 106 Pa uY =

σ Y2 2E

=

(230 × 106 )2 = 160.3 × 103 N ⋅ m/m3 9 (2)(165 × 10 ) uY = 160.3 kJ/m3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.2 Determine the modulus of resilience for each of the following alloys: (a ) Titanium:

E = 16.5 × 106 psi σ Y = 120 ksi

(b) Magnesium:

E = 6.5 × 106 psi

σ Y = 29 ksi

(c) Cupronickel (annealed):

E = 20 × 106 psi

σ Y = 16 ksi

SOLUTION (a)

E = 16.5 × 106 psi, σ Y = 120 × 103 psi uY =

(b)

=

2E

(120 × 103 ) 2 (2)(16.5 × 106 )

uY = 436 in ⋅ lb/in 3 

E = 6.5 × 106 psi, σ Y = 29 × 103 psi uY =

(c)

σ Y2

σ Y2 2E

=

(29 × 103 ) 2 (2)(6.5 × 106 )

uY = 64.7 in ⋅ lb/in 3 

E = 20 × 106 psi, σ Y = 16 × 103 psi uY =

σ Y2 2E

=

(16 × 103 ) 2 (2)(20 × 106 )

uY = 6.40 in ⋅ lb/in 3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.3 Determine the modulus of resilience for each of the following grades of structural steel: (a ) ASTM A709 Grade 50: σ Y = 50 ksi (b) ASTM A913 Grade 65: σ Y = 65 ksi (c) ASTM A709 Grade 100: σ Y = 100 ksi

SOLUTION E = 29 × 106 psi for all three steels given.

Structural steel: (a)

σ Y = 50 ksi = 50 × 103 psi uY =

(b)

2E

=

(50 × 103 ) 2 (2)(29 × 106 )

uY = 43.1 in ⋅ lb/in 3 

σ Y = 65 ksi = 65 × 103 psi uY =

(c)

σ Y2

σ Y2 2E

=

(65 × 106 ) 2 (2)(29 × 106 )

uY = 72.8 in ⋅ lb/in 3 

σ Y = 100 ksi = 100 × 103 psi uY =

σ Y2 2E

=

(100 × 103 ) 2 (2)(29 × 106 )

uY = 172.4 in ⋅ lb/in 3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.4 Determine the modulus of resilience for each of the following aluminum alloys: (a ) 1100-H14:

E = 70 GPa σ Y = 55 MPa

(b) 2014-T6:

E = 72 GPa σ Y = 220 MPa

(c) 6061-T6:

E = 69 GPa

σ Y = 150 MPa

SOLUTION Aluminum alloys: (a)

E = 70 × 109 Pa σ Y = 55 × 106 Pa uY =

σ Y2 2E

=

(55 × 106 ) 2 = 21.6 × 103 N ⋅ m/m3 9 (2)(70 × 10 ) uY = 21.6 kJ/m3 

(b)

E = 72 × 109 Pa σ Y = 220 × 106 Pa uY =

σ Y2 2E

=

(220 × 106 ) 2 = 336 × 103 N ⋅ m/m3 9 (2)(72 × 10 ) uY = 336 kJ/m3 

(c)

E = 69 × 109 Pa σ Y = 150 × 106 Pa uY =

σ Y2 2E

=

(150 × 106 ) 2 = 163.0 × 103 N ⋅ m/m3 (2)(69 × 109 ) uY = 163.0 kJ/m3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.5 The stress-strain diagram shown has been drawn from data obtained during a tensile test of an aluminum alloy. Using E = 72 GPa, determine (a) the modulus of resilience of the alloy, (b) the modulus of toughness of the alloy.

SOLUTION (a)

σ Y = Eε Y uY =

σ Y2 2E

=

1 2 1 Eε Y = (72 × 109 )(0.006) 2 2 2

uY = 1296 kJ/m3 

= 1296 × 103 N ⋅ m/m3

(b)

Modulus of toughness = total area under the stress-strain curve

The average ordinate of the stress-strain curve is 500 MPa = 500 × 106 N/m 2.

The area under the curve is

A = (500 × 106 )(0.18) = 90 × 106 N/m 2.

modulus of toughness = 90 × 106 J/m3 = 90 MJ/m3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.6 The stress-strain diagram shown has been drawn from data obtained during a tensile test of a specimen of structural steel. Using E = 29 × 106 psi, determine (a) the modulus of resilience of the steel, (b) the modulus of toughness of the steel.

SOLUTION

(a)

σ Y = Eε Y uY =

σ Y2 2E

=

1 2 1 Eε Y = (29 × 106 )(0.002) 2 2 2 uY = 58.0 in ⋅ lb/in 3 

(b)

Modulus of toughness = total area under the stress-strain curve

A 1 = (57)(0.25 − 0.002) = 14.14 kips/in 2 = 14.14 in ⋅ kip/in 3 A2 =

A3 =

1 (28)(0.25 − 0.021) = 3.21 kips/in 2 2 = 3.21 in ⋅ kip/in 3 2 (20)(0.25 − 0.075) = 2.33 kips/in 2 3 = 2.33 in ⋅ kip/in 3

modulus of toughness = uY + A1 + A2 + A3 modulus of toughness ≈ 20 in ⋅ kip/in 3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.7 The load-deformation diagram shown has been drawn from data obtained during a tensile test of a 0.875-in.diameter rod of an aluminum alloy. Knowing that the deformation was measured using a 15-in. gage length, determine (a) the modulus of resilience of the alloy, (b) the modulus of toughness of the alloy.

SOLUTION Volume of stressed material involved in the measurement: V= =

(a)

π 4

π 4

d 2L (0.875)2 (15) = 9.0198 in 3

Modulus of resilience. PY = 30 kips, δ Y = 0.104 in. UY =

1 1 PY δ Y − (30)(0.104) = 1.56 in ⋅ kip = 1560 in ⋅ lb 2 2

modulus of resilience = uY = (b)

UY 1560 = V 9.0198

uY = 173.0 in ⋅ lb/in 3 

Modulus of toughness. A1 = (30)(1.85 − 0.104) = 52.38 kip ⋅ in = 52380 in ⋅ lb 1 (5)(1.85 − 0.104) = 4.365 kip ⋅ in = 4365 in ⋅ lb 2 2 A3 = (4)(1.85 − 0.104) = 4.656 kip ⋅ in = 4656 in ⋅ lb 3 U = U Y + A1 + A2 + A3 = 62961 in ⋅ lb

A2 =

U 62961 = V 9.0198

modulus of toughness = 6980 in ⋅ lb/in 3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.8 The load-deformation diagram shown has been drawn from data obtained during a tensile test of structural steel. Knowing that the cross-sectional area of the specimen is 250 mm 2 and that the deformation was measured using a 500-mm gage length, determine (a) the modulus of resilience of the steel, (b) the modulus of toughness of the steel.

SOLUTION Assuming that yielding occurs at P = 62.5 kN and δ = 0.6 mm, UY =

1 (62.5 × 103 )(0.6 × 10−3 ) = 18.75 N ⋅ m = 18.75 J 2

Volume of stressed material:

V = AL = (250)(500) = 125 × 103 mm3 = 125 × 10−6 m3

uY =

UY 18.75 = = 150 × 103 V 125 × 10−6 uY = 150 kJ/m3 

A1 = (62.5 × 106 )(96 × 10−3 ) = 6 × 103 N ⋅ m = 6 × 103 J

Total energy:

A2 =

1 (28 × 103 )(96 − 8.6) × 10−3 = 1.22 × 103 N ⋅ m = 1.22 × 103 J 2

A3 =

2 (15 × 106 )(61 × 10−3 ) = 0.61 × 103 N ⋅ m = 0.61 × 103 J 3

U = U Y + A1 + A2 + A3 = 7.85 × 103 J U 7.85 × 103 = = 63 × 106 J/m3 V 125 × 10−6

modulus of toughness = 63 MJ/m3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.9 Using E = 29 × 106 psi, determine (a) the strain energy of the steel rod ABC when P = 8 kips, (b) the corresponding strain energy density in portions AB and BC of the rod.

SOLUTION P = 8 kips, E = 29 × 103 ksi A=

P π 2 σ2 d , V = AL, σ = , u = 4 A 2E

U = uV

Portion

d(in.)

A(in2)

L(in.)

V(in3)

σ (ksi)

AB

0.625

24

0.3608

7.363

26.08

BC

0.75

36

0.4418

15.904

18.11

(b)

U (in ⋅ kip)

−3

86.32 × 10−3

5.65 × 10−3

89.92 × 10−3

11.72 × 10

176.24 × 10−3

Σ

(a)

u (in ⋅ kip/in 3 )

U = 176.2 × 10−3 in ⋅ kip

U = 176.2 in ⋅ lb 

In AB : u = 11.72 × 10−3 in ⋅ kip/in 3

u AB = 11.72 in ⋅ lb/in 3 

In BC : u = 5.65 × 10−3 in ⋅ kip/in 3

uBC = 5.65 in ⋅ lb/in 3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.10 Using E = 200 GPa, determine (a) the strain energy of the steel rod ABC when P = 25 kN, (b) the corresponding strain-energy density in portions AB and BC of the rod.

SOLUTION AAB =

π 4

(20) 2 = 314.16 mm 2 = 314.16 × 10−6 m 2

ABC =

π 4

(16) 2 = 201.06 mm 2 = 201.06 × 10−6 m 2

P = 25 × 103 N U = =

(a) (b)

P 2L 2 EA

(25 × 103 ) 2 (1.2) (25 × 103 ) 2 (0.8) + (2)(200 × 109 )(314.16 × 10−6 ) (2)(200 × 109 )(201.06 × 10−6 )

U = 5.968 + 6.213 = 12.18 N ⋅ m

σ AB = u AB =

σ BC = uBC =

U = 12.18 J 

P 25 × 103 = = 79.58 × 106 Pa AAB 314.16 × 10−6 2 σ AB

2E

=

(79.58 × 106 ) 2 = 15.83 × 103 9 (2)(200 × 10 )

u AB = 15.83 kJ/m3 

P 25 × 103 = = 124.28 × 106 Pa AAB 201.16 × 10−6 2 σ BC

2E

=

(124.28 × 106 ) 2 = 38.6 × 103 (2)(200 × 109 )

uBC = 38.6 kJ/m3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.11 A 30-in. length of aluminum pipe of cross-sectional area 1.85 in 2 is welded to a fixed support A and to a rigid cap B. The steel rod EF, of 0.75-in. diameter, is welded to cap B. Knowing that the modulus of elasticity is 29 × 106 psi for the steel and 10.6 × 106 psi for the aluminum, determine (a) the total strain energy of the system when P = 8 kips, (b) the corresponding strain-energy density of the pipe CD and in the rod EF.

SOLUTION Member EF carries a force P = 8000 lb in tension while member CD carries 8000 lb in compression. Area of member EF: A = (a)

π 4

d2 =

π 4

(0.75)2 = 0.4418 in 2

Strain energy. CD :

U CD =

P2 L ( −8000) 2 (30) = = 48.95 in ⋅ lb 2 EA (2)(10.6 × 106 )(1.85)

EF :

U EF =

P2 L (8000) 2 (48) = = 119.89 in ⋅ lb 2 EA (2)(29 × 106 )(0.4418)

Total: U = U CD + U EF = 168.8 in ⋅ lb

(b)

U = 168.8 in ⋅ lb. 

Strain energy density. CD :

σ =−

8000 = −4324 psi, 1.85

u=

σ2 2E

=

(−4324) 2 (2)(10.6 × 106 ) u = 0.882 in ⋅ lb/in 3 

EF :

σ=

8000 = 18108 psi, 0.4418

u=

σ2 2E

=

(18108) 2 (2)(29 × 106 ) u = 5.65 in ⋅ lb/in 3 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.12 Rod AB is made of a steel for which the yield strength is σ Y = 450 MPa and E = 200 GPa; rod BC is made of an aluminum alloy for which σ Y = 280 MPa and E = 73 GPa. Determine the maximum strain energy that can be acquired by the composite rod ABC without causing any permanent deformations.

SOLUTION AAB =

ABC =

π 4

π 4

(10) 2 = 78.54 mm 2 = 78.54 × 10−6 m 2 (14)2 = 153.94 mm 2 = 153.94 × 10−6 m 2

Pall = σ Y A for each portion. AB :

Pall = (450 × 106 )(78.54 × 10−6 ) = 35.343 × 103 N

BC :

Pall = (280 × 106 )(153.94 × 10−6 ) = 43.103 × 103 N

Use the smaller value.

P = 35.343 × 103 N

U=

P 2 LBC P 2 LAB (35.343 × 103 ) 2 (1.2) + = 2 E AB AAB 2 EBC ABC (2)(200 × 109 )(78.54 × 10−6 ) +

(35.343 × 103 ) 2 (1.6) (2)(73 × 109 )(153.94 × 10−6 ) U = 136.6 J 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.13 A single 6-mm-diameter steel pin B is used to connect the steel strip DE to two aluminum strips, each of 20-mm width and 5-mm thickness. The modulus of elasticity is 200 GPa for the steel and 70 GPa for the aluminum. Knowing that for the pin at B the allowable shearing stress is τ all = 85 MPa, determine, for the loading shown, the maximum strain energy that can be acquired by the assembled strips.

SOLUTION Apin =

π 4

d2 =

π 4

(6) 2 = 28.274 mm 2 = 28.274 × 10−6 m 2

τ all = 85 × 106 Pa Double shear:

P = 2 Aτ = (2)(28.274 × 10−6 )(85 × 106 ) = 4.8066 × 103 N

For strips AB, DB, BE,

A = (20)(5) = 100 mm 2 = 100 × 10−6 m 2 1 FAB = FDB = P = 2.4033 × 103 N 2 U AB = U DB = U BE =

Total:

2 FAB LAB (2.4033 × 103 )(0.5) = = 0.2063 J 2 Ea AAB (2)(70 × 109 )(100 × 10−6 )

2 FBE LBE (4.8066 × 103 ) 2 (1.25 − 0.5) = = 0.4332 J 2 Es ABE (2)(200 × 109 )(100 × 10−6 )

U = U AB + U DB + U BE = 0.846 J

U = 0.846 J 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.14 Rod BC is made of a steel for which the yield strength is σ Y = 300 MPa and the modulus of elasticity is E = 200 GPa. Knowing that a strain energy of 10 J must be acquired by the rod when the axial load P is applied, determine the diameter of the rod for which the factor of safety with respect to permanent deformation is six. U Y = ALuY A=

π 4

d2

SOLUTION For factor of safety of six on the energy, U Y = (6)(10) = 60 J uY =

σ Y2

(300 × 106 )2 (2)(200 × 109 )

=

2E

= 225 × 103 J/m3 U Y = ALuY A=

UY 60 = LuY (1.8)(225 × 103 ) = 148.148 × 10−6 m 2

A= d=

π 4

d2

4A

π

=

(4)(148.148 × 10−6 )

π −3

= 13.73 × 10 m d = 13.73 mm 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.15 The assembly ABC is made of a steel for which E = 200 GPa and σ Y = 320 MPa. Knowing that a strain energy of 5 J must be acquired by the assembly as the axial load P is applied, determine the factor of safety with respect to permanent deformation when (a) x = 300 mm, (b) x = 600 mm.

SOLUTION

σ Y = 320 MPa = 320 × 106 Pa, AAB =

4

2 d AB =

π

π 4

(12) 2 = 113.097 mm 2 = 113.097 × 10−6 m 2

π

2 d BC = (18) 2 = 254.47 mm 2 = 254.47 × 10−6 m 2 4 4 = AAB

ABC = Amin

π

E = 200 GPa = 200 × 109 Pa

Force at yielding or allowable axial force. P = PY = σ Y Amin = (320 × 106 )(113.097 × 10−6 ) = 36.191 × 103 N

(a)

x = 300 mm:

LAB = 0.300 m,

U Y = U AB + U BC = =

LBC = 0.600 m

P LAB P LBC P 2  LAB L  + = + BC   ABC  2EAAB 2 EABC 2E  AAB 2

2

(36.191 × 103 ) 2  0.300 0.600  + 9  −6 −6  (2)(200 × 10 )  113.097 × 10 254.97 × 10 

= (3.2745 × 10−3 )(2652.6 + 2353.2) = 16.392 J

(b)

Applied energy:

U = 5J

Factor of safety:

UY 16.392 = U 5

x = 600 mm: UY =

LAB = 0.600 m,

F .S . = 3.28  LBC = 0.300 m

36.191 × 103  0.600 0.300  + 9  −6 −6  (2)(200 × 10 ) 113.097 × 10 254.97 × 10 

= (3.2745 × 10−3 )(5305.2 + 1176.6) = 21.225 J

Factor of safety:

UY 21.225 = U 5

F .S . = 4.25 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.16 Using E = 10.6 × 106 psi, determine by approximate means the maximum strain energy that can be acquired by the aluminum rod shown if the allowable normal stress is σ all = 22 ksi.

SOLUTION Amin =

π

(1.5) 2 = 1.7671 in 2

4

σ all = 22000 psi Pall = σ all Amin = 38877 lb U=



P 2dx P 2 = 2 EA 2 E

dx

πd

2

=

4

2 P 2 dx π E d2



Use Simpson’s rule to compute the integral. h = 1.5 in. Section

d(in.)

1/d 2 (in −2 )

multiplier

m (1/d 2 ) (in −2 )

1

1.50

0.4444

1

0.4444

2

2.10

0.22675

4

0.9070

3

2.55

0.15379

2

0.3076

4

2.85

0.12311

4

0.4924

5

3.00

0.11111

1

0.1111

Σ

2.2625

ò

B A

dx d2

U=

=

æ 1 ö h 1.5 Σ m çç 2 ÷÷÷ = (2.2625) = 1.13125 in-1 3 3 èç d ø

(2)(38877) 2 (1.13125) π (10.6 × 106 )

U = 102.7 in ⋅ lb 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.17 Show by integration that the strain energy of the tapered rod AB is U=

P2 L 4 EAmin

where Amin is the cross-sectional area at end B.

SOLUTION Radius: r =

cx L

Amin = π c 2

A = π r2 = U=



2L L

π c2 2

x2

L P dx P 2 = 2 EA 2 E 2



2L L

L2 dx π c 2 x2

2L

=

P 2 L2  1  −  2 Eπ c 2  x  L =

P 2 L2  1 1 +  − 2 EAmin  2 L L 

U=

P2 L  4 EAmin

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.18 In the truss shown, all members are made of the same material and have the uniform crosssectional area indicated. Determine the strain energy of the truss when the load P is applied.

SOLUTION Equilibrium of joint C. Solving the force triangle, P = 0.577 P tan 60° P = = 1.155P sin 60°

FBC = FCD

Member CB: Member CD:

U CD =

Total strain energy:

U BC =

2 FBC LBC (0.577 P) 2 l P 2l = = 0.1667 EA 2EA 2EA

2 FCD LCD (1.155P) 2 (2l ) P 2l = = 1.3333 EA 2EA 2EA

U = U BC + U CD U = 1.500

P 2l  EA

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.19 In the truss shown, all members are made of the same material and have the uniform cross-sectional area indicated. Determine the strain energy of the truss when the load P is applied.

SOLUTION 2

LBC = LCD =

5 1  l2 +  l  = l 2 2  

Joint C. (equilibrium)  Fx = 0 :

−

2 2 FBC − FCD = 0 5 5

FCD = − FBC

 Fy = 0 : FBC =

5 P 2

1 1 FBC − FCD − P = 0 5 5 FCD = −

5 P 2

Strain energy. F 2L 1  2 2 FBC LBC + FCD LCD  = 2 EA 2EA  2 2   1  5   5   5   5   P  l +  P  l =  2 EA  2   2   2   2    

U =

U = 1.398

P 2l  EA

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.20 In the truss shown, all members are made of the same material and have the uniform cross-sectional areas indicated. Determine the strain energy of the truss when the load P is applied.

SOLUTION Geometry.

β = tan −1

1 = 26.57° 2

lBD = 2l , lBC = lCD =

l/2 = 1.118l sin 26.57°

Equilibrium of joint C. From the force triangle, P/2 FBC = FCD = = 1.118P sin 26.57° Equilibrium of joint D.  Fx = 0: FBD − (1.118P) cos 26.57° = 0 FBD = − P

Strain energy. F 2L U =  i i = U BC + U BD + U CD 2 AE 1  (1.118P)2 (1.118l ) (− P) 2 (2l ) (1.118P) 2 (1.118l )  = + +   2E  2A A A  = [1.3974 + 1 + 1.3974]

P 2l P 2l = 3.795 2 AE 2 AE U = 1.898

P 2l  AE

    

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.21 In the truss shown, all members are made of the same material and have the uniform cross-sectional area indicated. Determine the strain energy of the truss when the load P is applied.

SOLUTION Equilibrium of joint C. +

FCD = −

2 P 3

 Fx = 0: − FBC − FCD = 0 FBC =

1 P 3

 Fy = 0: −

3 FCD − P = 0 2 1 2

Equilibrium of joint D.

 Fy = 0: FBD +

Strain energy. Member

F

BC CD BD

1 3 −

P

2 3

P

P

U =

3 FCD = 0 2

FBD = P

1 F 2L 1 F 2L =  A 2 EA 2E

L

A

F2L/A

l

A

1 2 P l/A 3

2l

A

8 2 P l/A 3

3l

A

3P 2l/A

4.732P2l/A

Σ

U =

1  P 2l  4.732   2 E  A 

U = 2.37

P 2l  EA

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.22 Each member of the truss shown is made of steel and has the crosssectional area shown. Using E = 29 × 106 psi, determine the strain energy of the truss for the loading shown.

SOLUTION 7.52 + 42 = 8.5 ft = 102 in.

LBC =

LCD = 7.5 ft = 90 in. ABC = 3 in 2 ,

ACD = 4 in 2

E = 29000 ksi

Equilibrium at joint C.

 Fy = 0:

4 FBC − 24 = 0 8.5

 Fx = 0: − FCD −

FBC = 51 kips

7.5 (51 kips) + 20 kips = 0 8.5 FCD = −25 kips

Strain energy. U = =

Fi 2 Li F2 L F2 L = BC BC + CD CD 2 EAi 2 EABC 2 EACD

(51) 2 (102) (25) 2 (90) + (2)(29000)(3) (2)(29000)(4)

= 1.5247 + 0.2425 U = 1.767 kip ⋅ in 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.23 Each member of the truss shown is made of aluminum and has the cross-sectional area shown. Using E = 72 GPa, determine the strain energy of the truss for loading shown.

SOLUTION

Lengths of members: LBC = (3.22 + 2.42 )1/ 2 = 4 m LCD = (12 + 2.42 )1/ 2 = 2.6 m E = 72 GPa = 72 × 109 Pa

Forces in kN. Equilibrium of truss.

 M B = 0: (30)(2.4) − (80)(3.2) + Dy (2.2) = 0 D y = 83.636 kN ↑

 Fy = 0: Dy − By − 80 = 0 83.636 − B y − 80 = 0

B y = 3.636 kN ↓

Member forces. FBC = B y

4m  4  = (3.636 kN)   = 6.061 kN 2.4 m  2.4 

FCD = − Dy

Strain energy. U =



U = U BC + U CD =



2.6 m  2.6  = −(83.636 kN)   = −90.606 kN 2.4 m  2.4 

Fi Li 2 AE

2 FBC LBC F2 L (6.061 × 103 ) 2 (4) (90.606 × 103 ) 2 (2.6) + CD CD = + 2 EABC 2EACD (2)(72 × 109 )(2 × 10−3 ) (2)(72 × 109 )(2.5 × 10−3 )

= 0.510 J + 59.290 J 

U = 59.8 J 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.24 Taking into account only the effect of normal stresses, determine the strain energy of the prismatic beam AB for the loading shown.

SOLUTION v  M K = 0: − M − (wv)   = 0 2

1 M = − wv 2 2

U=



L

0

M2 1 dv = 2 EI 2 EI

w2 = 8EI =



L

0

w2 L5 40EI



2

L

0

1 2  wv  dv 2 

w2 v 5 v dv = 8 EI 5

L

4

0

U =

w2 L5  40EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.25 Taking into account only the effect of normal stresses, determine the strain energy of the prismatic beam AB for the loading shown.

SOLUTION L  M B = 0 : − RAL + (wL)   = 0 2

Bending moment:

M = RA x −

U=



L

0

=



L

0

wL 2

1 2 w wx = ( Lx − x 2 ) 2 2

M2 w2 dx = 2 EI 8EI

w2 = 8EI

RA =



L

0

( Lx − x 2 ) 2 dx

w2  L2 x3 2 Lx 4 x5  − +  ( L x − 2 Lx + x )dx =  8 EI  3 4 5 2 2

3

L

4

0

2 5

w L 1 1 1  − + 8EI  3 2 5  U =

w2 L5  240 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.26 Taking into account only the effect of normal stresses, determine the strain energy of the prismatic beam AB for the loading shown.

SOLUTION

A to D:

ΣM B = 0: −RA L − M 0 = 0

RA =

M0 ↓ L

ΣM A = 0:

RB L − M 0 = 0

RB =

M0 ↑ L

ΣM J = 0:

M0x +M =0 L

M0x L

M =− U AD =

D to B:

0

M 02 M 2 dx = 2 EI 2 EIL2

ΣM K = 0: −M + M= U DB =

Total:



a



a

0

x 2 dx =

M 02 a 3 6 EIL2

M 0v L

M 0v L



b

0

M 02 M 2 dv = 2 EI 2 EIL2

U = U AD + U DB



b

0

v 2 dv =

M 02 b3 6 EIL2 U=

M 02 (a3 + b3 ) 6 EIL2



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.27 Taking into account only the effect of normal stresses, determine the strain energy of the prismatic beam AB for the loading shown.

SOLUTION Symmetric beam and loading:

RA = RB ΣFy = 0: RA + RB − 2 P = 0

Over portion AD,

M = RA x = Px U AD =

Over portion DE,

RA = RB = P



a

0

M2 P2 dx = 2 EI 2 EI

M = Pa

U DE =



a

0

P 2 x3 x dx = 2 EI 3

a

2

= 0

P 2 a3 6 EI

P 2 a 2 ( L − 2a ) 2 EI

Over portion EB, By symmetry, U EB = U AD =

Total:

P 2 a3 6 EI

U = U AD + U DE + U EB

U=

P2a2 (3L − 4a )  6 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.28 Using E = 200 GPa, determine the strain energy due to bending for the steel beam and loading shown. (Ignore the effect of shearing stresses.)

SOLUTION Over portion AC,

M= U AC =

1 Px 2



L/2

0

M2 P2 dx = 2 EI 8EI

P 2 x3 = 8EI 3

By symmetry,

U CB = U AC =

L/2

= 0



L/2

0

x 2 dx

P 2 L3 192 EI

P 2 L3 192 EI P 2 L3 96 EI

Total:

U = U AC + U CB =

Data:

P = 180 × 103 N, L = 4.8 m, E = 200 × 109 Pa I = 178 × 106 mm 4 = 178 × 10−6 m 4 U=

(180 × 103 )2 (4.8)3 = 1048 N ⋅ m (96)(200 × 109 )(178 × 10−6 )

U = 1048 J 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.29 Using E = 200 GPa, determine the strain energy due to bending for the steel beam and loading shown. (Ignore the effect of shearing stresses.)

SOLUTION M = Px

Over portion AD,

U AD =



a

0

M2 1 dx = 2 EI 2 EI

P 2 x3 = 2 EI 3

a

= 0



a

0

P 2 a3 6 EI M = Pa

Over portion DE, U DE =

( Pa)2 a P 2a 3 = 2EI 2EI

U EB = U AD =

By symmetry,

( Px) 2 dx

P 2a3 6 EI

U = U AD + U DE + U EB =

Data:

5 P 2a 3 6 EI

P = 80 × 103 N, a = 1.6 m, E = 200 × 109 Pa I = 163 × 106 mm 4 = 163 × 10−6 m 4 U =

5 (80 × 103 ) 2 (1.6)3 = 670 N ⋅ m 6 (200 × 109 )(163 × 10−6 )

U = 670 J 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.30 Using E = 29 × 106 psi, determine the strain energy due to bending for the steel beam and loading shown. (Ignore the effect of shearing stresses.)

SOLUTION Equilibrium.

 M B = 0:

(4 kip)(24 in.) − C (96 in.) = 0 C = 1 kip ↓

Bending moment diagram. (0 ≤ x ≤ 24 in.)

Over AB:

M = −4 x kip ⋅ in (0 ≤ v ≤ 96 in.)

Over BC:

M = −1v kip ⋅ in U = U AB + U CB =

Strain energy. Data:



24

0

I = 16.4 in 4 ,

W6 × 9,

2 M AB dx + 2 EI



96

0

2 M CB dv 2 EI

E = 29 × 103 ksi

96 1  24 1 16 1   (−4 x) 2 dx + ( −v)2 dv  = (24)3 + (96)3    0 2 EI  0 2 3 3 EI    184320 184320 = = = 0.38755 in ⋅ kip EI (29 × 103 )(16.4)

U=





U = 388 in ⋅ lb 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.31 Using E = 29 × 106 psi, determine the strain energy due to bending for the steel beam and loading shown. (Ignore the effect of shearing stresses.)

SOLUTION Over A to B:

M = − Px U AB =

Over B to C:

0

M 2 dx P 2 = 2 EI 2 EI



a

0

x 2 dx =

P 2 a3 6 EI

− M = Pa = constant U BC =

By symmetry,



a

M 2b P 2 a 2b = 2 EI 2 EI

U CD = U AB =

P 2 a3 6 EI P 2 a 2 (2a + 3b) 6 EI

Total:

U = U AB + U BC + U CD =

Data:

P = 2 × 103 lb, a = 15 in., b = 60 in. 1 I = (1.5)(3)3 = 3.375 in 4 12 U=

(2 × 103 )2 (15) 2 [(2) (15) + (3) (60)] (6) (29 × 106 ) (3.375)

U = 322 in ⋅ lb 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.32 Assuming that the prismatic beam AB has a rectangular cross section, show that for the given loading the maximum value of the strain-energy density in the beam is umax = 15

U V

where U is the strain energy of the beam and V is its volume.

SOLUTION ΣM K = 0: −M − ( wv)

v =0 2

1 M = − wv 2 2 U=



L 0

M2 1 dv = 2 EI 2 EI

w2 = 8EI



0

0

1 2  wv  dv 2 

w2 x 5 v dv = 8EI 5

L

4

0

2 5

=

w L 40 EI

M max =

1 2 wL 2

umax =

L



2

L

2 σ max

2E

σ max = =

2 M max c2

2 EI 2

(

=

M max c I 1 4

w2 L4 c 2 2 EI 2

=

w2 L4 c 2 8 EI 2

)

L 121 bd 3 1 1 U LI = 2 = = Lbd = V 2 15 15 umax 5c 5 ( d2 ) umax = 15

U  V

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.33 The ship at A has just started to drill for oil on the ocean floor at a depth of 5000 ft. The steel drill pipe has an outer diameter of 8 in. and a uniform wall thickness of 0.5 in. Knowing that the top of the drill pipe rotates through two complete revolutions before the drill bit at B starts to operate and using G = 11.2 × 106 psi, determine the maximum strain energy acquired by the drill pipe.

SOLUTION

ϕ = (2) (2π ) = 4π rad L = 5000 ft = 60 × 103 in. d co = o = 4 in. ci = co − t = 3.5 in. 2 J=

π

(c

2 TL ϕ= GJ

4 o

)

− ci4 = 166.406 in 4 T=

GJ ϕ L 2

U=

T 2 L  GJ ϕ  L GJ ϕ 2 = =  2GJ  L  2GJ 2L

U=

(11.2 × 106 )(166.406)(4π ) 2 (2)(60 × 103 )

U = 2.45 × 106 in ⋅ lb 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.34 Rod AC is made of aluminum and is subjected to a torque T applied at C. Knowing that G = 73 GPa and that portion BC of the rod is hollow and has an inner diameter of 16 mm, determine the strain energy of the rod for a maximum shearing stress of 120 MPa.

SOLUTION do = 12 mm 2 d ci = i = 8 mm 2

co =

J AB =

π 2

co4 =

π 2

(12) 4 = 32.572 × 103 mm 4 = 32.572 × 10−9 m 4

J BC =

π

(c 2

4 o

)

− ci4 =

π 2

(124 − 84 ) = 26.138 × 103 mm 4 = 26.138 × 10−9 m 4

τ all =

Total:

Tc J min

T=

J minτ all (26.138 × 10 −9 )(120 × 106 ) = = 261.38 N ⋅ m c 12 × 10−3

U AB =

T 2 LAB (261.38) 2 (400 × 10−3 ) = = 5.747 J 2GJ AB (2)(73 × 109 )(32.572 × 10−9 )

U BC =

T 2 LBC (261.38)2 (500 × 10−3 ) = = 8.951 J 2GJ BC (2)(73 × 109 )(26.138 × 10−9 )

U = U AB + U BC

U = 14.70 J 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.35 Show by integration that the strain energy in the tapered rod AB is U=

7 T 2L 78 GJ min

where J min is the polar moment of inertia of the rod at end B.

SOLUTION r= J=

cx L

π

r4 =

π c4

2 L4 2 L T 2 dx = U= L 2GJ 2



=

T 2 L4 2GJ min



2L L

x 4 , J min =



2L L

π 2

c4

T 2 dx  π c4 4  2G  x  4 2 L 

dx x4 2L

T 2 L4  1  = − 2GJ min  3x3  L U=

T 2 L4 2GJ min

 1 1  + 3   − 3 3L   3(2 L)

U=

7 T 2L  48 GJ min

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.36 The state of stress shown occurs in a machine component made of a grade of steel for which σ Y = 65 ksi. Using the maximum-distortionenergy criterion, determine the factor of safety associated with the yield strength when (a) σ y = +16 ksi, (b) σ y = −16 ksi.

SOLUTION 1 2 = 4 ksi σx −σz 8 − 0 = 2 2 = 4 ksi τ xz = 14 ksi

σ ave = (0 + 8)

2

 σ −σz  2 + τ xz R=  x  2   = 42 + 142 = 14.56 ksi σ a = σ ave + R = 18.56 σ b = σ ave − R = −10.56 σc = σ y  σ  (σ a − σ b ) 2 + (σ b − σ c ) 2 + (σ c − σ a )2 = 2  Y   F .S . 

(a)

2

σ c = σ y = 16 ksi  65  (18.56 + 10.56) 2 + (−10.56 − 16)2 + (16 − 18.56)2 = 2    F .S .  8450 847.97 + 705.43 + 6.55 = ( F .S .) 2

(b)

2

F .S . = 2.33 

σ c = σ y = −16 ksi  65  (18.56 + 10.56) 2 + (−10.56 + 16) 2 + (−16 − 18.56) 2 = 2    F .S .  8450 847.97 + 29.59 + 1194.39 = ( F .S .) 2

2

F .S . = 2.02 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.37 The state of stress shown occurs in a machine component made of a grade of steel for which σ Y = 65 ksi. Using the maximum-distortionenergy criterion, determine the range of values of σ y for which the factor of safety associated with the yield strength is equal to or larger than 2.2.

SOLUTION 1 2 = 4 ksi σx −σz 8 − 0 = 2 2 = 4 ksi τ xz = 14 ksi

σ ave = (0 + 8)

2

 σ −σz  2 R=  x  + τ xz 2   = 42 + 142 = 14.56 ksi σ a = σ ave + R = 18.56 ksi σ b = σ ave − R = −10.56 ksi σc = σ y

 σ  (σ a − σ b ) 2 + (σ b − σ c ) 2 + (σ c − σ a ) 2 = 2  Y   F .S .   65  (18.56 + 10.56) 2 + (−10.56 − σ y )2 + (σ y − 18.56) 2 = 2    2.2 

2

2

847.97 + (111.51 + 21.12σ y + σ y2 ) + (σ y2 − 37.12σ y + 344.47) = 1745.87 2σ y2 − 16σ y − 441.92 = 0 16 ± 162 + (4)(2)(441.92) (2)(2) = 4 ± 15.39 σ y = 19.39 ksi, −11.39 ksi

σy =

−11.39 ksi ≤ σ y ≤ 19.39 ksi 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.38 The state of stress shown occurs in a machine component made of a brass for which σ Y = 160 MPa. Using the maximum-distortion-energy criterion, determine the range of values of σ z for which yield does not occur.

SOLUTION 1 2 = 60 MPa

σ ave = (100 + 20) σx −σ y 2

τ xy

100 − 20 2 = 40 MPa = 75 MPa =

2

 σx −σy  2 R=   + τ xy 2   = 402 + 752 = 85 MPa σ a = σ ave + R = 145 MPa σ b = σ ave − R = −25 MPa σc = σ z (σ a − σ b ) 2 + (σ b − σ c ) 2 + (σ c − σ a ) 2 = 2σ Y2 (145 + 25)2 + ( −25 − σ z ) 2 + (σ z − 145) 2 = (2)(160) 2 28900 + (625 + 50σ z + σ z2 ) + (σ z2 − 290σ z + 21025) = 51200 2σ z2 − 240σ z − 650 = 0 240 ± 2402 + (4)(2)(650) = 60 ± 62.65 (2)(2) σ z = 122.65 MPa, −2.65 MPa

σz =

−2.65 MPa < σ z < 122.65 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.39 The state of stress shown occurs in a machine component made of a brass for which σ Y = 160 MPa. Using the maximum-distortion-energy criterion, determine whether yield occurs when (a) σ z = +45 MPa, (b) σ z = −45 MPa.

SOLUTION 1 2 = 60 MPa

σ ave = (100 + 20) σx −σ y 2

τ xy

100 − 20 2 = 40 MPa = 75 MPa =

2

 σx −σy  2 R=   + τ xy 2   = 402 + 752 = 85 MPa σ a = σ ave + R = 145 MPa

σ b = σ ave − R = −25 MPa σc = σ z ?

(σ a − σ b ) 2 + (σ b − σ c ) 2 + (σ c − σ a ) 2 < 2σ Y2

(a)

σ c = σ z = +45 MPa ?

(145 + 25) 2 + (−25 − 45) 2 + (45 − 145)2 < 2(160) 2 = 51200 28900 + 4900 + 10000 = 43800 < 51200

(b)

(No yield.) 

σ c = σ z = −45 MPa ?

(145 + 25) 2 + (−25 + 45)2 + ( −45 − 145) 2 < 51200 28900 + 400 + 36100 = 65400 > 51200

(Yield occurs.) 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.40 Determine the strain energy of the prismatic beam AB, taking into account the effect of both normal and shearing stresses.

SOLUTION Reactions:

RA =

M0 M ↓, RB = 0 ↑ L L

Shear:

V =−

Bending moment:

M=

M0 L

M0 v L

For bending, L

L M 02 M2 dv = v 2 dv 2 0 2 EI 2 EIL M L3 M 2 L = 0 2 = 0 6 EI 6 EIL

U1 =



τ xy =

3V  y2  1 − 2  2 A c 

0



For shear,

2

2 τ xy

 9M 02 y2  1 − 2  = 8G (bd ) 2 L2 c  

9V 2 u= = 2G 8GA2 L



 

U 2 = udv =

=

0

9 M 02 2 2

8Gbd L



1 d 2

c=

 y2 y4  1 − 2 2 + 4  c c   c L c  9M 02 b y2 y4  1 2 − + u b dy dx =   dy dx −c 8Gb 2 d 2 L2 0 − c  c 2 c 4 

 

c

L 0

9M 02 2 y3 1 y5  + = dx  y −  3 c 2 5 c 4  8Gbd 2 L2  −c



L 0

4 2   2c − 3 c + 5 c  dx  

9 M 02

6 M 02 c 3 M 02  16  = = c L   5 Gbd 2 L 5 GbdL 8Gbd 2 L2  15 

Total: with

U = U1 + U 2 = I= U=

M 02 L 3 M 02 + 6 EI 5 GbdL

1 3 bd 12 2 M 02 L Ebd 3

+

3 M 02 5 GbdL

U=

2 M 02 L  3 E d 2  1+ ⋅ ⋅ 2   3  Ebd  10 G L 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.41* A vibration isolation support is made by bonding a rod A, of radius R1, and a tube B, of inner radius R2, to a hollow rubber cylinder. Denoting by G the modulus of rigidity of the rubber, determine the strain energy of the hollow rubber cylinder for the loading shown.

SOLUTION ΣFx = 0: −τ (2π rL) + Q = 0

τ= Q2 2G 8π 2 r 2 L2 G Q2 U = u dV = 2 2 8π GL u=

τ2

=



=

Q 2π rL

Q2 4π GL2

L

R2

0

R1

 



dV Q2 = r 2 8π 2 GL2

dr Q2 dx = r 4π GL2



L 0

L

R2

0

R1

 

(

R

2π r dr dx r2

)

lnr R2 dx 1

U=

R Q2 ln 2  4π GL R1

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.42 The cylindrical block E has a speed v0 = 16 ft/s when it strikes squarely the yoke BD that is attached to the 78 -in.-diameter rods AB and CD. Knowing that the rods are made of a steel for which σ Y = 50 ksi and E = 29 × 106 psi, determine the weight of block E for which the factor of safety is five with respect to permanent deformation of the rods.

SOLUTION At the onset of yielding, the force in each rod is F = σ Y A.

Corresponding strain energy: U AB =

2 FAB LAB σ Y2 A2 L σ Y2 AL = = 2 EAAB 2 EA 2E

U CD = U AB =

σ Y2 AL 2E

U m = U AB + U CD =

σ Y2 AL

E 1W 2 1  U m =  mv02  ( F .S .) =  v0  ( F .S .) 2  2 g 

Solving for W, Data:

W=

2 gU m v02 ( F .S .)

=

2 gσ Y2 AL v02 ( F .S .) E

g = 32.17 ft/ sec2 = 386 in/ sec2 , A=

π 4

d2 =

π 7

σ Y = 50 × 103 psi,

2

E = 29 × 106 psi

= 0.60132 in 2   48

L = 3.5 ft = 42 in.

F .S . = 5

v0 = 16 ft/sec = 192 in/sec W=

(2)(386)(50 × 103 )2 (0.60132)(42) (192)2 (5)(29 × 106 )

W = 9.12 lb 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.43 The 18-lb cylindrical block E has a horizontal velocity v0 when squarely the yoke BD that is attached to the 78 -in.-diameter and CD. Knowing that the rods are made of a steel for which σ Y and E = 29 × 106 psi, determine the maximum allowable speed rods are not to be permanently deformed.

it strikes rods AB = 50 ksi v0 if the

SOLUTION At the onset of yielding, the force in each rod is F = σ Y A.

Corresponding strain energy: U AB =

2 FAB LAB σ 2 A2 L σ Y2 AL = Y = 2EAAB 2EA 2E

U CD =

2 σ 2 AL FCD LCD = Y 2EACD 2E

Total:

Solving for v02 ,

Um =

1 2 1W 2 mv0 = v0 2 2 g

v02 =

2 gU m 2 gσ Y2 AL = W EW

g = 32.17 ft/s 2 = 386 in/s 2

π

v0 =

d2 =

π 7

E

σ Y = 50 × 103 psi

2

2   = 0.60132 in , 4 48 L = 3.5 ft = 42 in.

A=

σ Y2 AL

2 gσ Y2 AL EW

v0 =

Data:

U m = U AB + U CD =

E = 29 × 106 psi W = 18 lb

(2)(386)(50 × 103 ) 2 (0.60132)(42) = 305.6 in/sec (29 × 106 )(18) v0 = 25.5 ft/sec 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.44 Collar D is released from rest in the position shown and is stopped by a small plate attached at end C of the vertical rod ABC. Determine the mass of the collar for which the maximum normal stress in portion BC is 125 MPa.

SOLUTION

σ m = 125 × 106 Pa π 2

Portion BC:

ABC =

4

(9) = 63.617 mm 2 = 63.617 × 10−6 m 2

Pm = σ m ABC = 7952 N

Corresponding strain energy: U BC =

π

(12)2 = 113.907 mm 2 = 113.907 × 10−6 m 2 4 P2 L (7952) 2 (4) = m AB = = 10.574 J 2 E AB AAB (2)(105 × 109 )(113.907 × 10−6 )

AAB = U AB

Pm2 LBC (7952) 2 (2.5) = = 17.750 J 2 EBC ABC (2)(70 × 109 )(63.617 × 10−6 )

U m = U BC + U AB = 28.324 J

Corresponding elongation Δ m : 1 Pm Δ m = U m 2 2U m (2)(28.324) Δm = = = 7.12 × 10−3 m Pm 7952

Falling distance: Work of weight = U m

h = 0.6 + 7.12 × 10−3 = 0.60712 m Wh = mgh = U m m=

Um 28.324 = gh (9.81)(0.60712)

m = 4.76 kg 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.45 Solve Prob. 11.46, assuming that both portions of rod ABC are made of aluminum. PROBLEM 11.44 Collar D is released from rest in the position shown and is stopped by a small plate attached at end C of the vertical rod ABC. Determine the mass of the collar for which the maximum normal stress in portion BC is 125 MPa.

SOLUTION

σ m = 125 × 106 Pa π 2

Portion BC:

ABC =

4

(9) = 63.617 mm 2 = 63.617 × 10−6 m 2

Pm = σ m ABC = 7952 N

Corresponding strain energy: U BC =

π

(12)2 = 113.907 mm 2 = 113.907 × 10−6 m 2 4 P2 L (7952)2 (4) = m AB = = 15.861 J 2 EAAB (2)(70 × 109 )(113.907 × 10−6 )

AAB = U AB

Pm2 LBC (7952) 2 (2.5) = = 17.750 J 2 EABC (2)(70 × 109 )(63.617 × 10−6 )

U m = U BC + U AB = 33.611 J

Total: Corresponding elongation Δ m :

1 Pm Δ m = U m 2 2U m (2)(33.611) Δm = = = 8.45 × 10−3 m Pm 7952

Falling distance: Work of weight = U m .

h = 0.6 + Δ m = 0.60845 m Wh = mgh = U m m=

Um 33.611 = gh (9.81)(0.60845)

m = 5.63 kg 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.46 The 48-kg collar G is released from rest in the position shown and is stopped by plate BDF that is attached to the 20-mm-diameter steel rod CD and to the 15-mm-diameter steel rods AB and EF. Knowing that for the grade of steel used σ all = 180 MPa and E = 200 GPa, determine the largest allowable distance h.

SOLUTION Let Δ m be the maximum elongation.

Δm =

σ AB L E

=

σ CD L E

=

σ EF L E

σ AB = σ CD = σ EF = 180 MPa = 180 × 106 Pa L = 2.5 m

For each rod, Rod CD:

ACD = U CD =

Rods AB and EF:

π 4

Δm =

(180 × 106 )(2.5) = 0.00225 m 200 × 109

U =

2 Fm2 L ( EAΔ m /L) 2 L EAΔ m = = 2EA 2 EA 2L

(20) 2 = 314.16 mm 2 = 314.16 × 10−6 m 2

(200 × 109 )(314.16 × 10−6 )(0.00225)2 = 63.617 J (2)(2.5)

AAB = AEF = U AB = U EF =

Total strain energy:

E = 200 × 109 Pa

π 4

(15) 2 = 176.71 mm 2 = 176.71 × 10−6 m 2

(200 × 109 )(176.71 × 10−6 )(0.00225) 2 = 35.674 J (2)(2.5)

U m = U AB + U CD + U EF = 134.97 J

Work of falling collar: U m = mg (h + Δ m ) = (48)(9.81)(h + Δ m )

Equating,

(48)(9.81)(h + Δ m ) = 134.97

h + Δ m = 0.28662 m

h = 0.28662 − 0.00225 = 0.285 m

h = 285 mm 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.47 Solve Prob. 11.46, assuming that the 20-mm-diameter steel rod CD is replaced by a 20-mm-diameter rod made of an aluminum alloy for which σ all = 150 MPa and E = 75 GPa. PROBLEM 11.46 The 48-kg collar G is released from rest in the position shown and is stopped by plate BDF that is attached to the 20-mm-diameter steel rod CD and to the 15-mm-diameter steel rods AB and EF. Knowing that for the grade of steel used σ all = 180 MPa and E = 200 GPa, determine the largest allowable distance h.

SOLUTION Let Δ m be the maximum elongation.

L = 2.5 m

Δm =

If rod CD yields,

E AB

=

σ CD L

σ CD = 150 × 106 Pa, Δm =

If rods AB and EF yield,

σ AB L

ECD

=

σ EF L EEF

ECD = 75 × 109 Pa

(150 × 106 )(2.5) = 0.005 m 75 × 109

σ AB = σ EF = 180 × 106 Pa E AB = EEF = 200 × 109 Pa Δm =

(180 × 106 )(2.5) = 0.00225 m 200 × 109

The smaller value of Δ m governs.

2 F 2 L ( EAΔ m /L) 2 L EAΔ m = = 2EA 2 EA 2L

For each rod,

U =

Rod CD:

ACD = U CD =

Rods AB and EF:

Δ m = 0.00225 m

π 4

(20) 2 = 314.16 mm 2 = 314.16 × 10−6

(75 × 109 )(314.16 × 10−6 )(0.00225) 2 = 23.857 J (2)(2.5)

AAB = AEF = U AB = U EF =

Total strain energy:

π 4

(15) 2 = 176.71 mm 2 = 176.71 × 10−6 m 2

(200 × 109 )(176.71 × 10−6 )(0.00225) 2 = 35.674 J (2)(2.5)

U m = U AB + U CD + U EF = 95.205 J

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.47 (Continued)

Work of falling collar: U m = mg (h + Δ m ) = (48)(9.81)(h + Δ m )

Equating,

(48)(9.81)(h + Δ m ) = 95.205 h + Δ m = 0.20218 m

h = 0.20218 − 0.00225 = 0.19993 m h = 200 mm 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.48 The steel beam AB is struck squarely at its midpoint C by a 45-kg block moving horizontally with a speed v0 = 2 m/s. Using E = 200 GPa, determine (a) the equivalent static load, (b) the maximum normal stress in the beam, (c) the maximum deflection of the midpoint C of the beam.

SOLUTION From Appendix C, for W150 × 13.5, I x = 6.83 × 106 mm 4 = 6.83 × 106 m 4 S x = 91.1 × 103 mm3 = 91.1 × 10−6 m3

Kinetic energy: T =

1 1 m v02 = (45)(2)2 = 90 J 2 2

From Appendix D, Case 4: ym =

PL3 , 48EI

M max =

PL 4

Principle of work and energy. U =

(a)

1 P 2 L3 Pm ym = m = T 2 96EI

Equivalent static load. Pm =

96EIT = L3

(96)(200 × 109 )(6.83 × 10−6 )(90) = 20.907 × 103 N (3.0)3 Pm = 21.0 kN 

(b)

Maximum normal stress.

σm =

M max P L (20.907 × 103 )(3.0) = m = = 172.1 × 106 Pa S 4S (4)(91.1 × 10−6 )

σ m = 172.1 MPa  (c)

Maximum deflection. ym =

2U (2)(90) = = 8.61 × 10−3 m Pm 20.907 × 103

ym = 8.61 mm 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.49 Solve Prob. 11.48, assuming that the W150 × 13.5 rolled-steel beam is rotated by 90° about its longitudinal axis so that its web is vertical. PROBLEM 11.48 The steel beam AB is struck squarely at its midpoint C by a 45-kg block moving horizontally with a speed v0 = 2 m/s. Using E = 200 GPa, determine (a) the equivalent static load, (b) the maximum normal stress in the beam, (c) the maximum deflection of the midpoint C of the beam.

SOLUTION From Appendix C, for W150 × 13.5, I y = 0.916 × 106 mm 4 = 0.916 × 10−6 m 4 S y = 18.2 × 103 mm3 = 18.2 × 10−6 m3

Kinetic energy: T =

1 1 m v02 = (45)(2)2 = 90 J 2 2

From Appendix D, Case 4: ym =

PL3 , 48EI

M max =

PL 4

Principle of work and energy. U =

(a)

1 P 2 L3 Pm ym = m = T 2 96EI

Equivalent static load. Pm =

96EIT = L3

(96)(200 × 109 )(0.916 × 10−6 )(90) = 7.657 × 103 N (3.0)3 Pm = 7.66 kN 

(b)

Maximum normal stress.

σm =

M max P L (7.657 × 103 )(3.0) = m = = 316 × 106 Pa S 4S (4)(18.2 × 10−6 )

σ m = 316 MPa  (c)

Maximum deflection. ym =

2U (2)(90) = = 23.5 × 10−3 m Pm 7.657 × 103

ym = 23.5 mm 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.50 A 25-lb block C moving horizontally with a velocity v0 hits the post AB squarely as shown. Using E = 29 × 106 psi, determine the largest speed v0 for which the maximum normal stress in the pipe does not exceed 18 ksi.

SOLUTION I x = 21.4 in 4 ,

W5 × 16: Maximum stress:

S x = 8.55 in 3

σ m = 18 ksi

Maximum bending moment: M m = σ m S x = (18 ksi)(8.55 in 3 ) = 153.9 kip ⋅ in

Equivalent force:

Pm L = M m

M m 153.9 kip ⋅ in = = 1.71 kips = 1710 lb L 90 in.

Pm =

From Appendix D, ym = Um =

T =

Kinetic energy: T =

Equating,

Pm L3 (1710)(90)3 = = 0.66956 in. 3EI (3)(29 × 106 )(21.4) 1 1 Pm ym = (1710)(0.66956) = 572.48 in ⋅ lb 2 2 = 47.706 ft ⋅ lb

1W 2 v0 2 g

25 v02 = 0.3882v02 ft ⋅ lb (2)(32.2) T = Um

0.3882v02 = 47.706

Maximum speed.

v0 = 11.09 ft/s 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.51 Solve Prob. 11.50, assuming that the post AB has rotated 90° about its longitudinal axis. PROBLEM 11.50 A 25-lb block C moving horizontally with a velocity v0 hits the post AB squarely as shown. Using E = 29 × 106 psi, determine the largest speed v0 for which the maximum normal stress in the pipe does not exceed 18 ksi.

SOLUTION I y = 7.51 in 4 ,

W5 × 16: Maximum stress:

S y = 3.00 in 3

σ m = 18 ksi

Maximum bending moment: M m = σ m S y = (18 ksi)(3.00 in 3 ) = 54.0 kip ⋅ in

Equivalent force: Pm =

Pm L = M m

Mm 54.0 kip ⋅ in = = 0.600 kips = 600 lb L 90 in.

From Appendix D, ym = Um =

T =

Kinetic energy: T =

Equating,

(600)(90)3 Pm L3 = = 0.66945 in. 3EI (3)(29 × 106 )(7.51) 1 1 Pm ym = (600)(0.66945) = 200.83 in ⋅ lb 2 2 = 16.736 ft ⋅ lb

1W 2 v0 2 g

25 v02 = 0.3882v02 ft ⋅ lb (2)(32.2) T = Um

0.3882v02 = 16.736

Maximum speed.

v0 = 6.57 ft/s 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.52 The 2-kg block D is dropped from the position shown onto the end of a 16-mmdiameter rod. Knowing that E = 200 GPa, determine (a) the maximum deflection of end A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the rod.

SOLUTION I=

π d

4

π  16 

4

3 4   =   = 3.2170 × 10 mm 42 4 2 

= 3.2170 × 10−9 m 4 c=

d = 8 mm = 8 × 10−3 m LAB = 0.6 m 2

Appendix D, Case 1: Pm L3AB M m = Pm LAB 3EI 3EI (3)(200 × 109 )(3.217 × 10−9 ) = 8.9361 × 103 ym Pm = 3 ym = (0.6)3 LAB

ym =

Um =

Work of dropped weight:

1 1 Pm ym = (8.9361 × 103 ) ym2 = 4.4681 × 103 ym2 2 2

mg (h + ym ) = (2)(9.81)(0.040 + ym ) = 0.7848 + 19.62 ym

Equating work and energy, 0.7848 + 19.62 ym = 4.4681 × 103 ym2 ym2 − 4.3911 × 10−3 ym − 175.645 × 10−6 = 0

(a)

ym =

{

1 4.3911 × 10−3 + (4.3911 × 10−3 ) 2 + (4)(175.645 × 10−6 ) 2

= 15.629 × 10−3 m

} ym = 15.63 mm 

Pm = (8.9361 × 103 )(15.629 × 10−3 ) = 139.66 N

(b)

M m = − Pm LAB = −(139.66)(0.6)

(c)

σm =

|M m | c (83.8)(8 × 10−3 ) = = 208 × 106 Pa −9 I 3.2170 × 10

|M m | = 83.8 N ⋅ m 

σ m = 208 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.53 The 2-kg block D is dropped from the position shown onto the end of a 16-mm-diameter rod. Knowing that E = 200 GPa, determine (a) the maximum deflection of end A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the rod.

SOLUTION

π d

4

π  16 

4

3 4   =   = 3.2170 × 10 mm 42 4 2 

I=

= 3.2170 × 10−9 m 4 c=

Over AB:

d = 8 mm = 8 × 10−3 m a = 0.6 m 2

M = − Pm x U AB = =



a 0

M m = − Pm a

Pm2 x 2 P 2 a3 dx = m 2 EI 6 EI

(0.6)3 Pm2 (6)(200 × 109 )(3.2170 × 10−9 )

= 55.953 × 10−6 Pm2

By symmetry of bending moment diagram, U BC = U AB = 55.953 × 10−6 Pm2 U m = U AB + U BC = 111.906 × 10−6 Pm2 1 Pm ym = U m = 111.906 × 10−6 Pm2 2 Pm = 4.4681 × 103 ym Um =

Work of dropped weight:

1 Pm ym = 2.2340 × 103 ym2 2

mg (h + ym ) = (2)(9.81)(0.040 + ym ) = 0.7848 + 19.62 ym

Equating work and energy, 0.7848 + 19.62 ym = 2.2340 × 103 ym2 ym2 − 8.7825 × 10−3 ym − 351.298 × 10−6 = 0

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.53 (Continued)

(a)

ym =

{

1 8.7825 × 10−3 + (8.7282 × 10−3 ) 2 + (4)(351.298 × 10−6 ) 2

= 23.636 × 10−3 m = 23.6 mm

} ym = 23.6 mm 

Pm = (4.4681 × 103 )(23.636 × 10−6 ) = 105.61 N

(b)

M m = −(105.61)(0.6) = −64.4 N ⋅ m

(c)

σm =



|M m | = 64.4 N ⋅ m 

|M m | c (64.4)(8 × 10−3 ) = = 157.6 × 106 Pa −9 I 3.2170 × 10

σ m = 157.6 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.54 The 45-lb block D is dropped from a height h = 0.6 ft onto the steel beam AB. Knowing that E = 29 × 106 psi, determine (a) the maximum deflection at point E, (b) the maximum normal stress in the beam.

SOLUTION S5 × 10 rolled steel shape: I x = 12.3 in 4

S x = 4.90 in 4

a = 2 ft = 24 in.

b = 4 ft = 48 in.

L = 6 ft = 72 in.

h = 0.6 ft = 7.2 in.

RA =

Pmb ↑, L

RB =

Pm a ↑, L

Mm = ME =

Pm ab L

From Appendix D, Case 5, yE =

Pma 2b2 Pm (24) 2 (48) 2 = = 17.2245 × 10−6 Pm 3EIL (3)(29 × 106 )(12.3)(72)

Pm = 58057 yE 1 Pm ym = 29028 yE2 2

Total strain energy:

Um =

Work of falling weight:

W (h + yE ) = 45(7.2 + yE ) = 324 + 45 yE 324 + 45 yE = 29028 yE2

Equating work and energy:

yE2 − 1.5502 × 10−3 yE − 11.1615 × 10−3 = 0

(a)

yE =

1 1.5502 × 10−3 + (1.5502 × 10−3 ) 2 − (4)(−11.1615 × 10−3 )   2 

Deflection at E. (b)

yE = 0.1061 in. 

Pm = (58057)(0.1064) = 6.179 × 103 lb Mm =

(6.179 × 103 )(24)(48) = 98.86 × 103 lb ⋅ in 72

σm =

98.86 × 103 Mm = = 20.2 × 103 psi 4.90 Sx

Maximum stress.

σ m = 20.2 ksi 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.55 Solve Prob. 11.54, assuming that a W4 × 13 rolled-steel shape is used for beam AB. PROBLEM 11.54 The 45-lb block D is dropped from a height h = 0.6 ft onto the steel beam AB. Knowing that E = 29 × 106 psi, determine (a) the maximum deflection at point E, (b) the maximum normal stress in the beam.

SOLUTION W4 × 13 rolled steel shape:

RA =

I x = 11.3 in 4

S x = 5.46 in 3

a = 2 ft = 24 in.

b = 4 ft = 48 in.

L = 6 ft = 72 in.

h = 0.6 ft = 7.2 in.

Pmb ↑, L

RB =

Pma ↑, L

Mm = ME =

Pm ab L

From Appendix D, Case 5, yE =

Pma 2b2 Pm (24) 2 (48) 2 = = 18.7488 × 10−6 Pm 3EIL (3)(29 × 106 )(11.3)(72)

Pm = 53337 yE 1 Pm ym = 26668 yE2 2

Total strain energy:

Um =

Work of falling weight:

W (h + yE ) = 45(7.2 + yE ) = 324 + 45 yE

Equating work and energy,

324 + 45 yE = 26668 yE2

yE2 − 1.6874 × 10−3 yE = 12.1494 × 10−3 = 0

(a)

yE =

1 1.6874 × 10−3 + (1.6874 × 10−3 ) 2 − (4)(−12.1494 × 10−3 )   2 

Deflection at E. (b)

yE = 0.1111 in. 

Pm = (53337)(0.1111) = 5.926 × 103 lb Mm =

(5.926 × 103 )(24)(48) = 94.81 × 103 lb ⋅ in 72

σm =

94.81 × 103 Mm = = 17.36 × 103 psi 5.46 Sx

Maximum stress.

σ m = 17.36 ksi 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.56 A block of weight W is dropped from a height h onto the horizontal beam AB and hits it at point D. (a) Show that the maximum deflection ym at point D can be expressed as  2h  ym = yst 1 + 1 +   yst  

where yst represents the deflection at D caused by a static load W applied at that point and where the quantity in parenthesis is referred to as the impact factor. (b) Compute the impact factor for the beam and the impact of Prob. 11.52. PROBLEM 11.52 The 2-kg block D is dropped from the position shown onto the end of a 16-mm-diameter rod. Knowing that E = 200 GPa, determine (a) the maximum deflection of end A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the rod.

SOLUTION Work of falling weight:

Work = W (h + ym ) U=

Strain energy:

1 1 Pym = kym2 2 2

where k is the spring constant for a load applied at point D. Equating work and energy, W ( h + ym ) = ym2 −

1 2 kym 2

2W 2W ym − h=0 k k

ym2 − 2 yst ym − 2 yst h = 0

(a)

ym =

2 yst + 4 yst2 + 8 yst h 2

where

yst =

W k  2h  ym = yst 1 + 1 +    yst  

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.56 (Continued)

For Problem 11.52,

W = mg = (2)(9.81) = 19.62 N 9

E = 200 × 10 Pa

I=

π  16 

4

3 4 −9 4   = 3.217 × 10 mm = 3.217 × 10 m 4 2 

L = 0.6 m h = 40 mm = 40 × 10−3 m

Using Appendix D, Case 1,

yst =

WL3 3EI

yst =

(19.62)(0.6)3 = 2.196 × 10−3 m (3)(200 × 109 )(3.217 × 10−9 )

2h (2)(40 × 10−3 ) = = 36.44 yst 2.196 × 10−3

(b)

Impact factor.

= 1 + 1 + 36.44

Impact factor = 7.12 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.57 A block of weight W is dropped from a height h onto the horizontal beam AB and hits point D. (a) Denoting by ym the exact value of the maximum deflection at D and by ym′ the value obtained by neglecting the effect of this deflection on the change in potential energy of the block, show that the absolute value of the relative error is ( ym′ − ym )/ym , never exceeding ym′ /2h. (b) Check the result obtained in part a by solving part a of Prob. 11.52 without taking ym into account when determining the change in potential energy of the load, and comparing the answer obtained in this way with the exact answer to that problem. PROBLEM 11.52 The 2-kg block D is dropped from the position shown onto the end of a 16-mm-diameter rod. Knowing that E = 200 GPa, determine (a) the maximum deflection of end A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the rod.

SOLUTION U=

1 1 Pm ym = kym2 2 2

where k is the spring constant for a load at point D. exact: Work = W ( h + ym )

Work of falling weight:

approximate : Work ≈ Wh 1 2 kym = W (h + ym ) (1) exact 2 1 2 (2) approximate kym′ = Wh 2

Equating work and energy,

where ym′ is the approximate value for ym . Subtracting,

1 k ym2 − ym′2 = Wym 2

(

)

ym2 − ym′2 = ( ym − ym′ )( ym + ym′ ) =

Relative error:

2W ym k

ym − ym′ 2W = ym k ( ym + y m )

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.57 (Continued)

2W ym′2 = k h

But

=

from Eq. (2).

ym − ym′ ym′2 y′ = < m ′ ym h ( ym + y m ) 2 h

(a)

Relative error

(b)

From the solution to Problem 11.52,

ym = 15.63 mm

Approximate solution:

W = mg = (2)(9.81) = 19.62 N E = 200 × 109 Pa I=

π d

4

4

π  10  3 4   =   = 3.217 × 10 mm 42 4 2  = 3.217 × 10−9 m 4

L = 0.6 m, h = 40 mm

= 40 × 10−3 m k=

3EI (3)(200 × 109 )(3.217 × 10−9 ) = L3 (0.6)3

= 8.936 × 103 N/m ym′2 =

2Wh (2)(19.62)(40 × 10−3 ) = k 8.936 × 103 = 175.65 × 10−6 m 2

ym′ = 13.25 × 10−3 m = 13.25 mm

Relative error: 

=

15.63 − 13.25 15.63

relatives error = 0.152  ym′ = 0.166  2h

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.58 Using the method of work and energy, determine the deflection at point D caused by the load P.

SOLUTION ΣM A = 0: Pa + RB L = 0 Pa L

RB = −

Over portion DA:

M = − Px U DA =

Over portion AB:

M =− U AB =

Total:





a 0

M2 P2 dx = 2 EI 2 EI



a 0

x 2dx =

P 2 a3 6 EI

Pav L L

0

M2 P2a2 dv = 2 EI 2 EIL2

U = U DA + U AB = 1 Pδ D = U 2



L 0

v 2dv =

Pa 2 L 6 EI

P 2 a 2 (a + L) 6 EI

δD =

2U P

δD =

Pa 2 ( a + L) ↓  3EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.59 Using the method of work and energy, determine the deflection at point D caused by the load P.

SOLUTION Pb Pa , RB = L L

Reactions:

RA =

Over AD:

M = RA x =

Pbx L

M2 P 2b2 dx = 0 2 EI 2 EIL2 2 2 3 P b a = 6 EIL2

U AD =

Over DB:



a

M = RB v =

Total:



b

U = U AB + U BC = 1 Pδ D = U 2

0

x 2dx

Pav L

M2 P2 a2 dv = 0 2 EI 2 EIL2 P 2 a 2 b3 = 6 EIL2

U DB =



a

δD =



b 0

v 2dv

P 2 a 2 b 2 (a + b) P 2a 2b 2 = 6 EIL 6 EIL2

2U P

δD =

Pa 2 b 2 ↓  3EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.60 Using the method of work and energy, determine the slope at point D caused by the couple M 0 .

SOLUTION RB =

Reaction:

M0 ↑ L

Bending moment diagram.

Strain energy:

Over 0 ≤ x ≤ a:

M = M0

Over 0 ≤ v ≤ L:

M =

U = U DA + U AB = =



0

M 02 dy + 2 EI



L

0

1  M0 2 EI  L

2

 v  dv 

 1 1 L3  M 2  L  M 02  a + 2  = 0  + a  2 EI L 3  2 EI  3  

1 M Dθ D = U : 2

θD =

a

M 0v L

M0 EI

1 M2  L  M 0θ D = 0  + a  2 2EI  3  L   3 + a  

θD =

M0 ( L + 3a ) 3EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.61 Using the method of work and energy, determine the slope at point D caused by the couple M 0 .

SOLUTION M0 ↓ L

RA =

Reactions:

M =−

Over portion AD: U AD = = M=

Over portion DB:

U DB = =

M 02 M2 dx = 0 2 EI 2 EIL2 M 0 a3



a

1 M 0θ D = U 2

a

0

M0 ↑ L

M0x L

x 2 dx

6 EIL2 M 0v L M 02 M2 dv = 0 2 EI 2 EIL2 2 M0 b



b



b

0

v 2 dv

6 EIL2

U = U AD + U DB =

Total:



RB =

θD =

2U M0

M 02 (a3 + b3 ) 6 EIL2

θD =

M 0 ( a 3 + b3 ) 3EIL2



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.62 Using the method of work and energy, determine the deflection at point C caused by the load P.

SOLUTION Symmetric beam and loading: RA = RB =

From A to C ,

1 P 2 1 Px 2

M = RA x = U AC = =



a 0

M dx + 2 EI

2

P 8EI



a 0



2a a

x 2 dx +

M2 dx 4 EI

P2 16 EI



2a a

x 2 dx

P 2 a3 P2 3 P 2 a3 (2a)3 − a3  = +  16 EI 24 EI 48EI 

By symmetry, Total:

U CB = U AB =

3 P 2 a3 16 EI

U = U AB + U BC = 1 Pδ C = U 2

δC =

3 P 2 a3 8 EI 2U P

δC =

3Pa3 ↓  4 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.63 Using the method of work and energy, determine the deflection at point C caused by the load P.

SOLUTION Bending moment:

M = − Pv

Over AB: M2 P2 L 2 dv = v dv L/ 2 4 EI 4 EI L/ 2 3 P 2  3  L   7 P 2 L3 = L −    = 12 EI   2   96 EI

U AB =

Over BC:

U BC = =

Total:



L



L/2 0



M2 P2 dv = 2 EI 2 EI



L/ 2 0

v 2 dv

1 P 2 L3 48 EI

U = U AB + U BC = 1 Pδ C = 0 2

δC =

3 P 2 L3 32 EI 2U P

δC =

3PL3 ↓  16 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.64 Using the method of work and energy, determine the slope at point A caused by the couple M 0 .

SOLUTION RB =

M0 L

M = RB v =

Over AC:

U AC = U AC =

Over CB:

M2 dv L/ 2 2(2 EI ) L

M 02 4 EIL2



L L/2

v 2 dv

=

 3  L 3  L −    12 EIL2   2  

=

7 M 02 L 96 EI

U CB = U CB =

Total:



M0 v L

M 02



L/ 2 0

M2 dv 2 EI

M 02 2 EIL2



L/2 0

v 2 dv =

U = U AC + U CB = 1 M 0θ A = U 2

θA =

1 M 02 L 48 EI

3 M 02 L 32 EI 2U M0

θA =

3M 0 L 16 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.65 Using the method of work and energy, determine the slope at point D caused by the couple M 0 .

SOLUTION RA =

Reactions:

M0 ↓, L

RB =

M0 ↑ L

Bending moment diagram. L , 2 L Over 0 < v < , 2 Over 0 < x <

U = U AD + U BD =

Strain energy:

U= =



L/2

0

(−

M0 L

x

)

0

2

2(2 EI )

M 02 L 



dx +



L/2

0

L/2

(

2 M AD dx + 2(2 EI )

M0 L

v

)

2 EI

2

dv =



L/2

0

M 0x L M v M = 0 L

M =−

2 M BD dv 2 EI

M 0  1  ( L / 2)3 M 0  1  ( L / 2)3 + EIL  4  3 EIL  2  3

1 1  3 1 M 02 L + = = EI  96 48  96 EI 32 EI

1 M 0θ D = U : 2

M 02 L

1 1 M 02 L M 0θ D = 2 32 EI

θD =

1 M 0L 16 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.66 Torques of the same magnitude T are applied to the steel shafts AB and CD. Using the method of work and energy, determine the length L of the hollow portion of shaft CD for which the angle of twist C is equal to 1.25 times the angle of twist at A.

SOLUTION T is the same for each shaft.

θC = 1.25θ A

1 Tθ A 2

U AB =

U CD =

and

1 T θC 2

Then U CD = 1.25 U AB Shaft AB:

(1)

LAB = 60 in. U AB =

U CE =

Shaft portion ED: U ED =

Shaft CD:

π 2

LCE = L,

co = 2 in.,

(c

π

4 o

)

− ci4 =

π 2

c4 =

π 2

(2)4 = 25.133 in 4

2

ci = 1.5 in.

(24 − 1.54 ) = 17.1806 in 4

T 2 LCE T 2L T 2L = = 0.058205 2GJ CE (2G)(17.1806) 2G LED = 60 − L,

J DE = J AB = 25.133 in 4

T 2 LED T 2 (60 − L) T2 T 2L = = 2.3873 − 0.039789 2GJ ED 2G (25.138) 2G 2G

U CE + U ED = 2.3873

Using Eq. (1), 2.3873

J AB =

T 2 LAB T 2 (60) T2 = = 2.3873 2GJ AB (2G )(25.133) 2G

Shaft portion CE: J CE =

c = 2 in.

T2 T 2L + 0.018416 2G 2GJ

T2 T 2L T2 + 0.018416 = (1.25)(2.3873) 2G 2G 2G

0.018416 L = (0.25)(2.3873)

L = 32.4 in. 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.67 The 20-mm-diameter steel rod BC is attached to the lever AB and to the fixed support C. The uniform steel lever is 10 mm thick and 30 mm deep. Using the method of work and energy, determine the deflection of point A when L = 600 mm. Use E = 200 GPa and G = 77.2 GPa.

SOLUTION Member AB. (Bending) I=

1 (10)(30)3 = 22.5 × 103 mm 4 12 = 22.5 × 10−9 m 4

a = 500 mm = 0.500 m M B = Pa = (450)(0.500) = 225 N ⋅ m M = Px U AB = =



a 0

M dx = 2 EI



a 0

P 2 x 2 dx P 2 a 3 = 2 EI 6 EI

2

(450) (0.500)3 = 0.9375 J (6)(200 × 109 )(22.5 × 10−9 )

Member BC. (Torsion) T = M B = 225 N ⋅ m J=

π

c=

1 d = 10 mm 2

c 4 = 15.708 × 103 mm 4 = 15.708 × 10−9 m 4

2 L = 600 mm = 0.600 m U BC =

Total:

T 2L (225)2 (0.600) = = 12.5242 J 2GJ (2)(77.2 × 109 )(15.708 × 10−9 )

U = U AB + U BC = 0.9375 + 12.5242 = 13.4617 J 1 Pδ A = U 2

δA =

2U (2)(13.4617) = = 59.8 × 10−3 m P 450

δ A = 59.8 mm ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.68 The 20-mm-diameter steel rod BC is attached to the lever AB and to the fixed support C. The uniform steel lever is 10 mm thick and 30 mm deep. Using the method of work and energy, determine the length L of the rod BC for which the deflection at point A is 40 mm. Use E = 200 GPa and G = 77.2 GPa.

SOLUTION Member AB. (Bending) I=

1 (10)(30)3 = 22.5 × 103 mm 4 12 = 22.5 × 10−9 m 4

a = 500 mm = 0.500 m M B = Pa = (450)(0.500) = 225 N ⋅ m M = Px U AB = =

Member BC. (Torsion)



a 0

M2 dx = 2 EI



a 0

P 2 x 2 dx P 2 a3 = 2 EI 6 EI

(450) 2 (0.500)3 = 0.9375 J (6)(200 × 109 )(22.5 × 10−9 )

T = M B = 225 N ⋅ m c= J= U BC =

1 d = 10 mm 2

π 2

c 4 = 15.708 × 103 mm 4 = 15.708 × 10−9 m 4

T 2L (225) 2 L = = 20.874 L 2GJ (2)(77.2 × 109 )(15.708 × 10−9 )

1 Pδ A = U AB + U BC 2 1 (450)(40 × 10−3 ) = 0.9375 + 20.874 L 2

L = 0.386 m = 386 mm 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.69 Two solid steel shafts are connected by the gears shown. Using the method of work and energy, determine the angle through which end D rotates when T = 820 N ⋅ m. Use G = 77.2 GPa.

SOLUTION Shaft CD:

T = 820 N ⋅ m J =

π 2

c4 =

π 2

c=

(0.020) 4 = 251.33 × 10−9 m 4

G = 77.2 × 109 Pa, U CD =

Equilibrium of shafts :

T 2L (820) 2 (0.60) = = 10.397 J 2GJ (2)(77.2 × 109 )(251.33 × 10−9 )

rB 100 mm TCD = (820 N ⋅ m) = 1366.67 N ⋅ m rC 60 mm

T = 1366.67 N ⋅ m J =

π 2

c4 =

π 2

U AB =

c=

1 d = 25 mm = 0.025 m 2

(0.025) 4 = 613.59 × 10−6 m 4

G = 77.2 × 109 Pa,

Total strain energy:

L = 0.60 m

TCD T = AB rC rB TAB =

Shaft AB:

1 d = 20 mm = 0.020 m 2

L = 0.40 m

T 2L (1366.67)2 (0.40) = = 7.886 J 2GJ (2)(77.2 × 109 )(613.59 × 10−9 )

U = U CD + U AB = 18.283 J 1 Tθ = U 2

θ =

2U (2)(18.283) = = 0.04459 rad TA 820

θ = 2.55° 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.70 The thin-walled hollow cylindrical member AB has a noncircular cross section of nonuniform thickness. Using the expression given in Eq. (3.53) of Sec. 3.13, and the expression for the strain-energy density given in Eq. (11.19), show that the angle of twist of member AB is

φ=

TL 4 2 G

ds t

where ds is an element of the center line of the wall cross section and  is the area enclosed by that center line.

SOLUTION From Eq. (3.53),

τ=

T 2t 

Strain energy density: u= U= =

Work of torque:

τ2 2G

 

L 0

L 0

=

T2 8Gt 2 Ꮽ 2

ut ds dx T2 8GᏭ2

ds T 2L dx = t 8GᏭ 2

1 T 2L = Tϕ = 2 8GᏭ 2

ds t

ds t

ϕ=

TL 4GᏭ 2

ds  t

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.71 Each member of the truss shown has a uniform cross-sectional area A. Using the method of work and energy, determine the vertical deflection of the point of application of the load P.

SOLUTION Reactions:

C =

3 P →, 4

Ax =

3 P ←, 4

Ay = P ↑

Members AB and AC are zero force members. FBD = − P FCD =

3 P 4

Equilibrium of joint A. Using the force triangle, 2

25 2 3  2 FAD = P2 +  P  = P 16 4  . 5 FAD = P (tension) 4 U =

Strain energy:

Fi 2 Li 2 EA

Fi

Li

Fi 2 Li

AD

5 + P 4

5  4

1.9531P 2

BD

−P



P 2

CD

3 − P 4

3  4

0.4219P 2  = 3.375P 2

U =

3.375P 2 2 EA

1 Py = U 2 B

Deflection at B.

yB =

2U P yB = 3.375

P  EA

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.72 Each member of the truss shown is made of steel and has a crosssectional area 400 mm 2. Using E = 200 GPa, determine the deflection of D caused by the 16-kN load.

SOLUTION

Equilibrium of entire truss.  M A = 0: D = 0  Fx = 0:

Ax = 0

 Fy = 0:

Ay = 16 kN ↑

Equilibrium of joint A. From the force triangle, FACE F 16 kN = AB = 17 15 8 FAB = 30 kN (compression) FACE = 34 kN

(tension)

FDE = 30 kN

(comperssion)

By symmetry, FBCD = 34 kN

(tension)

Equilibrium of joint B. ΣFy = 0: − −

8 FACE − FBE = 0 17

8 (34 kN) − FBE = 0 FBE = −16 kN 17

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.72 (Continued)

A = 400 mm 2 = 400 × 10−6 m 2

Members:

E = 200 GPa = 200 × 109 Pa EA = (200 × 109 )(400 × 10−6 ) = 80 × 106 N U =Σ

Strain energy:

Fi 2 Li 1 = ΣFi 2 Li 2EA 2 EA

Fi 2 Li

Fi (kN)

Li (m)

AB

−30

1.5

1350

DE

−30

1.5

1350

ACE

+34

1.7

1965.2

BCD

+34

1.7

1965.2

BE

−16

0.8

204.8

 = 6835.2 (kN) 2 ⋅ m

= 6.8352 × 109 N 2 ⋅ m U =

6.8352 × 109 = 42.72 N ⋅ m (2)(80 × 106 )

Principle of work and energy: 1 Pδ = U ( P = 16 kN = 16 × 103 N) 2 1 (16 × 103 ) δ = 42.72 δ = 5.34 × 10−3 m 2

Deflection of point D.

δ = 5.34 mm ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.73 Each member of the truss shown is made of steel and has a crosssectional area of 5 in2. Using E = 29 × 106 psi, determine the vertical deflection of point B caused by the 20-kip load.

SOLUTION RA = RB = 10 kips ↑ LAC = LCD = 6 ft = 72 in. LBC = 2.5 ft = 30 in. 62 + 2.52 = 6.5 ft = 78 in.

LAB = LBC =

Equilibrium of joint A. ↑ +ΣFy = 0: ΣFx = 0:

2.5 FAB + 10 = 0 6.5

FAB = −26 kips

6 FAB + FAB = 0 6.5

FAC = 24 kips

Equilibrium of joint C. FBC = 0, FCD = 24 kips FBD = FAB = −26 kips

By symmetry,

U =Σ

Strain energy: Member

F 2L 1 =  F 2L 2 EA 2 EA

F (kips)

L (in.)

F 2 L (kip 2 ⋅ in)

AB

−26

78

52728

AC

24

72

41472

BC

0

30

0

CD

24

72

41472

BD

−26

78

52728

Σ

188400 U =

188400 = 0.64966 kip ⋅ in (2) (29 × 103 ) (5)

Vertical deflection of point B. 1 Pδ B = U 2 2U (2)(0.64966) δB = = P 20

δ B = 0.0650 in. ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.74 Each member of the truss shown is made of steel and has a uniform cross-sectional area of 5 in2. Using E = 29 × 106 psi, determine the vertical deflection of joint C caused by the application of the 15-kip load.

SOLUTION Members BD and AE are zero force members. M A = 0: 2.5 RE − (12) (15) = 0

For entire truss,

RE = 72 kips

FED = − RE = −72 kips

For equilibrium of joint E, Joint C:

ΣFy = 0: −

2.5 FCD − 15 = 0 6.5

ΣFx = 0: −

6 FCD − FBC = 0 6.5

ΣFx = 0:

Joint D:

FCD = −39 kips FBC = 36 kips Σ Fx = 0:

Joint B:

− FAB + FBC = 0

6 ( FAD + 39) = 0 6.5 = 39 kips

72 − FAD

Um = Σ

Strain energy: Member

FAB = 36 kips

F 2L 1 = ΣF 2 L 2 EA 2 EA

F (kips)

L (in.)

F 2 L (kip 2 ⋅ in)

AB

36

72

93312

BC

36

72

93312

CD

−39

78

118638

DE

−72

72

373248

BD

0

30

0

AE

0

30

0

AD

39

78

118638

Σ

Data:

3

797148

E = 29 × 10 ksi A = 5 in Um =

2

797148 = 2.7488 kip ⋅ in (2) (29 × 103 ) (5)

1 Pm Δ m = U m 2

Δm =

2U m (2) (2.7488) = Pm 15

Δ m = 0.366 in. ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.75 Each member of the truss shown is made of steel; the cross-sectional area of member BC is 800 mm 2 and for all other members the crosssectional area is 400 mm2. Using E = 200 GPa, determine the deflection of point D caused by the 60-kN load.

SOLUTION ΣM A = 0: 2.4 RD − (0.5)(60) = 0

Entire truss:

ΣFy = 0: 12.5 −

Joint D:

0.5 FCD = 0 1.3

ΣFx = 0: 60 − FBD −

ΣFy = 0: −

FCD = 32.5 kN

1.2 FCD = 0 FBD = 30 kN 1.3

1.2 FAB = 0 1.3

FAB = 32.5 kN

0.5 FAB + FBC = 0 1.3

FBC = 12.5 kN

ΣFx = 0: 30 −

Joint B:

1.2 (32.5) = 0 1.3 F 2L 1 F 2L U =Σ = Σ 2 EA 2 E A

ΣFx = 0: −FAC +

Joint C:

Member

RD = 12.5 kN

F (kN)

FAC = 30 kN

L (m)

A (10−6 m 2 )

F 2 L/A (N 2 /m)

CD

32.5

1.3

400

3.4328 × 1012

BD

30

1.2

400

2.7 × 1012

AB

32.5

1.3

400

3.4328 × 1012

BC

12.5

0.5

800

0.0977 × 1012

AC

30

1.2

400

2.7 × 1012 12.3633 × 1012

Σ E = 200 × 109 Pa U=

12.3633 × 1012 = 30.908 J (2)(200 × 109 )

Work-energy: 1 PΔ = U 2

Δ=

2U (2)(30.908) = = 1.030 × 10−3 m P 60 × 103 Δ = 1.030 mm → 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.76 The steel rod BC has a 24-mm diameter and the steel cable ABDCA has a 12-mm diameter. Using E = 200 GPa, determine the deflection of joint D caused by the 12-kN load.

SOLUTION FAB = FBD = FDC = FCA

Owing to symmetry,

U AB = U BD = U DC = U CA U = 4U BD + U BC = 4

2 2 FBD LBD FBC LBC + 2 EABD 2 EABC

Let P be the load at D. 1 Pδ D = U 2

δD =

2U P

=4

Joint B:

Σ Fy = 0:

Joint D:

FBD 2 LBD FBC 2 LBC + EABD P EABC P ΣFx = 0: 4 FBC + (2) FBD = 0 5 8 4 FBC = − FBD = − P 5 3

3 2 FBD − P = 0 5 5 FBD = P 6 2

2

 5  PLBD  4  PLBC P  25 LBD 16 LBC  δD = 4  +  =  +   6  EABD  3  EABC E  9 ABD 9 ABC 

Data:

P = 12 × 103 N

E = 200 × 109 Pa

LBD = 600 × 10−3 m

ABD =

LBC = 960 × 10−3 m

ABC =

δD =

12 × 103 200 × 109

π 4

π 4

(12) 2 = 113.097 mm 2 = 113.097 × 10−6 m 2 (24) 2 = 452.39 mm 2 = 452.39 × 10−6 m 2

600 × 10−3 16 960 × 10−3   25 −3 + ⋅  ⋅  = 1.111 × 10 m −6 9 452.39 × 10−6   9 113.097 × 10

δ D = 1.111 mm ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.77 Using the information in Appendix D, compute the work of the loads as they are applied to the beam (a) if the load P is applied first, (b) if the couple M is applied first.

SOLUTION From Appendix D, Case 1, yAP =

PL3 3EI

θAP = −

yAM =

M 0 L2 2 EI

θAM =

PL2 2 EI

From Appendix D, Case 3,

(a)

M0L EI

First P, then M0. U = A1 + A2 + A3 =

1 1 Py AP + Py AM + M 0θ AM 2 2 U=

(b)

P 2 L3 PM 0 L2 M 02 L + +  6 EI 2 EI 2 EI

First M0, then P. U = A4 + A5 + A6 =

1 1 Py AP + M 0θ AP + M 0θ AM 2 2 U=

P 2 L3 M 0 PL2 M 02 L + +  6 EI 2 EI 2 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.78 Using the information in Appendix D, compute the work of the loads as they are applied to the beam (a) if the load P is applied first, (b) if the couple M is applied first.

SOLUTION Appendix D, Cases 1 and 3,

(a)

P( L/2)3 PL3 = 3EI 24 EI 2 M ( L/2) M L2 = 0 = 0 2 EI 8 EI

P( L/2) 2 PL2 = 2 EI 8EI M L = 0 EI

yBP =

θCP =

yBM

θ BM

First P, then M0. U = A1 + A2 + A3 =

1 1 PyBP + PyBM + M 0θCM 2 2

U=

(b)

P 2 L3 PM 0 L2 M 02 L + + 48EI 8EI 2 EI



First M0, then P. U = A4 + A5 + A6 =

1 1 PyBP + M 0θCP + M 0θCM 2 2

U=

P 2 L3 M 0 PL2 M 02 L + + 48EI 8EI 2 EI









PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.79 Using the information in Appendix D, compute the work of the loads as they are applied to the beam (a) if the load P is applied first, (b) if the couple M is applied first.

SOLUTION From Appendix D, Case 4, ↓ yC =

PL3 48EI

θB = −

PL2 16 EI

From Appendix D, Case 7, ↓ yC =

M0 M L2 [( L/2)3 − L2 ( L/2)] = − 0 6 EIL 16 EI

θB = (a)

M0L 3EI

First P, then M0. U = A1 + A2 + A3

(b)

=

1 1 PyCP + PyCM + M 0θ BM 2 2

=

P 2 L3 PM 0 L2 M 02 L − + 96 EI 16 EI 6 EI



First M0, then P. U = A4 + A5 + A6 =

1 1 PyCP + M 0θ BP + M 0θ BM 2 2

=

P 2 L3 M 0 PL2 M 02 L − + 96 EI 16 EI 6 EI





PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.80 For the beam and loading shown, (a) compute the work of the loads as they are applied successively to the beam, using the information provided in Appendix D, (b) compute the strain energy of the beam by the method of Sec. 11.4 and show that it is equal to the work obtained in part a.

SOLUTION (a)

Label the forces PD and PE . Using Appendix D, Case 5, 3L L P a 2 b 2 PE ( 4 ) ( 4 ) 3 PE L3 = E = = 3EIL 3EIL 256 EI PE ( L4 )  2  L 2   L   L 3  Pb 7 PE L3  L −      −    = = E [( L2 − b 2 ) x − x3 ] = 6 EIL 6 EIL   4    4   4   768 EI  2

δ EE δ DE

Likewise,

δ DD =

2

3 PD L3 256 EI

and δ ED =

7 PD L3 768 EI

Let PD be applied first. U = A1 + A2 + A3 1 1 PDδ DD + PDδ DE + PE δ EE 2 2 2 3 3 PD L 7 PD PE L3 3 PE2 L3 = + + 512 EI 768 EI 512 EI

U=

with

PD = PE = P U=

1 P 2 L3 48 EI

U=

P 2 L3  48EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.80 (Continued)

(b)

RA = RB = P

Reactions: Over portion AD:

U AD =



L/ 4 0

Over portion DE: Over portion EB: Total:

L   0 < x < 4  : M = Px   M2 P2 dx = 2 EI 2 EI PL M= 4



L/ 4 0

3

x 2 dx =

U DE =

P2 1  L  1 P 2 L3 ⋅   = 2 EI 3  4  384 EI M 2 ( L2 ) 2 EI

By symmetry, U EB = U AD =

2

+

P 2 L2 1 L P 2 L3 ⋅ ⋅ = 2 EI 16 2 64 EI

1 P 2 L3 384 EI

1 1  P 2 L3  1 U = U AB + U DE + U EB =  + +   384 64 384  EI

U=

1 P 2 L3  48 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.81 For the beam and loading shown, (a) compute the work of the loads as they are applied successively to the beam, using the information provided in Appendix D, (b) compute the strain energy of the beam by the method of Sec. 11.4 and show that it is equal to the work obtained in part a.

SOLUTION (a)

Label the forces PB and PC . Using Appendix D, Case 1,

δ BB =

PB ( L/2)3 1 PB L3 = 3EI 24 EI 1 PB L3 L PB ( L/2) 2 + 24 3EI 2 2 EI 1  P L3 5 PB L3  1 = +  B = 48 EI  24 16  EI

L 2

δ CB = δ BB + θ B =

δ CC =

1 PC L3 3 EI

δ BC =

2 3 PC P   L   L   5 PC L3 (3Lx 2 − x3 ) = C  3L   −    = 6 EI 6 EI   2   2   48 EI

Apply PB first, then PC . U = A1+ A2 + A3 U = =

with

PB = PC = P,

1 1 PB δ BB + PBδ BC + PCδ CC 2 2 1 PB L3 5 PB PC L3 1 PC2 L3 + + 48 EI 48 EI 6 EI

5 1  P 2 L3  1 U = + +   48 48 6  EI

U=

7 P 2 L3  24 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.81 (Continued)

(b)

Over AB:

L L   M = Pv + P  v −  = P  2v −  2 2   U AB =

Over BC:



M2 P2 dv = L/2 2 EI 2 EI L



1   2 4v − 2 Lv + L2  dv  L/2  4  2

=

P2 2 EI

3 2  4  L   1 2 L  1 2 L  3   L −    − 2L ⋅  L −    + L  L −   3 2  2   2    2   4    

=

P2 2 EI

2 3  7 3 3 3 1 3  13 P L L L L − + =   4 8  48 EI 6

M = Pv

U BC =



L/2 0

M2 P2 dv = 2 EI 2 EI



L/2 0

P2 1  L  ⋅   v dv = 2 EI 3  2  =

Total:

2 3  13 1  P L +  U = U AB + U BC =   48 48  EI

3

2

P 2 L3 48EI U=

7 P 2 L3  24 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.82 For the beam and loading shown, (a) compute the work of the loads as they are applied successively to the beam, using the information provided in Appendix D, (b) compute the strain energy of the beam by the method of Sec.11.4 and show that it is equal to the work obtained in part a.

SOLUTION (a)

Label the applied couples M A and M B . Apply M A at point A first. Note that M B = 0 during this phase. From Appendix D,

θ AA = U1 =

M AL EI 1 M 2 AL M Bθ AA = 2 2EI

Now apply M B at point B. Note that M A remains constant during this second phase. From Appendix D,

θ BB =

M B ( L/2) M L = B 2 EI EI

Since the curvature of portion AB does not change as M B is applied,

θ AB = θ BB = U2 = =

M BL 2EI

1 M Bθ BB + M Aθ AB 2 M B2 L M AM B + 4 EI 2EI

Total strain energy: U = U1 + U 2 =

Set

M A2 L M B 2 L M AM B L + + 2EI 4EI 2EI

M A = M B = M 0. U =

5M 0 2 L  4EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.82 (Continued)

(b)

Bending moment diagram. L  Over portion AB:  0 < x <  2 

M = −M 0 U AB =



L /2

0

M 02 M 2L dx = 0 2 EI 4 EI

L  Over portion BC:  < x < L  2 

M = −2M 0



U BC =

M 2L  (2M 0 )2 dx = 0 L/ 2 2 EI EI



L

Total strain energy:  

U = U AB + U BC  U =

5M 0 2 L  4EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.83 For the prismatic beam shown, determine the deflection of point D.

SOLUTION Add force Q at point D. M2 dx 0 2 EI L M ∂M ∂U 1 dx = = δD = 0 EI ∂ Q EI ∂Q U=



L





L 0

M

∂M dx ∂Q

∂M =0 ∂Q

Over portion AD:

L  0 < x < 2   

M = − Px,

Over portion DB:

L   2 < x < L  

L  ∂M L   = − x −  M = − Px − Q  x −  , ∂ 2 2 Q   

Set Q = 0.

δD =

1 EI



=

P EI



=

P EI

3 2  1 3 1  L   L  1 2 L 1  L    L −   −  L +  3 2   2  2 2 2  2    3

L/2 0

L

(− Px)(0)dx +

1 EI



  L  (− Px)  −  x −   dx L/2 2    L

 2 L   x − x  dx 2 

L/2 

3  1 1 1 1  PL = − − +   3 24 4 16  EI

δD =

5 PL3 ↓  48EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.84 For the prismatic beam shown, determine the deflection of point D.

SOLUTION Add force Q at point D. M2 dx 0 2 EI ∂U 1 = δD = ∂ Q EI U=



L



L 0

L  0 < x < 2   

1 M = − wx 2 2

Over portion DB:

L   2 < x < L  

1 L  M = − wx 2 − Q  x −  2 2 

Set Q = 0.

δD = =



w 2 EI

w = 2 EI =

L/2 

1 2 1  − 2 wx  (0) dx + EI  

0



∂M dx ∂Q

∂M =0 ∂Q

Over portion AD:

1 EI

M



L

∂M L  = − x −  2 ∂Q 

L   1 2    − 2 wx   −  x − 2   dx   

L/2 

 3 L 2  x − x  dx L/2  2  L

 1 4 1  L 4  L  1 3  L  1  L 3   L −   −  L +     4 2   2 3  2  3  2    4

1  1 1 1 1  wL4 17 wL4 − +  =  − 2  4 64 6 48  EI 384 EI

δ D = 0.0443

wL4 ↓  EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.85 For the prismatic beam shown, determine the slope at point D.

SOLUTION Add couple M 0 at point D. U=



L 0

θD =

M2 dx 2 EI

∂U = ∂ M0

L  0 < x < 2   

M = − Px

Over portion DB:

L   2 < x < L  

M = − Px − M 0

Set M 0 = 0.

θD =

P = EI

L



L/2



P xdx = L/2 EI

(− Px)(0)dx +

L

2  1 1  PL = −   2 8  EI

L 0

M ∂M 1 dx = EI ∂ M 0 EI



L 0

M

∂M dx ∂ M0

∂M =0 ∂ M0

Over portion AD:

1 EI



1 EI



L L/2

∂M = −1 ∂ M0 (− Px)(−1)dx

 1 2 1  L 2   L −    2  2    2

θD =

3PL2 8EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.86 For the prismatic beam shown, determine the slope at point D.

SOLUTION Add couple M 0 at point D. L

M2 dx 2 EI

U=



θD =

∂U = ∂ M0

0



L 0

M ∂M 1 dx = EI ∂ M 0 EI



L 0

M

∂M dx ∂ M0

∂M =0 ∂ M0

Over portion AD:

L  0 < x < 2   

1 M = − wx 2 2

Over portion DB:

L   2 < x < L  

1 M = − wx 2 − M 0 2

Set M 0 = 0.

θD =

∂M = −1 ∂ M0

1 L  1 2  1 2 − wx  (0)dx +   − wx  (−1)dx L/2  2 EI L/2  2   3  w L 2 w 1 3 1 L = x dx =  L −    2 EI L/2 2 EI  3 3  2   1 EI



L





=

1  1  wL3 1 −  6  8  EI

θD =

7 wL3 48 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.87 For the prismatic beam shown, determine the deflection at point D.

SOLUTION Add force Q at point D. Reactions:

1 1 RA = − wL − Q 8 2 5 1 RE = wL − Q 8 2 U = U AD + UDE + U EB ; δ D =



L/2

M2 dx , 2 EI

Over portion AD:

U AD =

L  0 ≤ x ≤ 2   

1 1 M = RA x = − wLx − Qx, 8 2 1 ∂M =− x 2 ∂Q

Set Q = 0.

0

∂U ∂Q

1 M = − wLx 8

∂ U AD = ∂Q



L/2 0

M ∂M dx = EI ∂ Q



L/2  0

wLx  1  wL −  − x  dx = EI EI 8 2 16    =

Over portion DE: L  0 ≤ v ≤ 2   

UDE =



L/2 0



L/2 0

x 2 dx

wL4 384 EI

M2 dv 2 EI

M = RE v −

wL  L 1 1 v +  = wL(v − L) − Qv 2  4 8 2

∂M 1 =− v ∂Q 2

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.87 (Continued)

Set Q = 0.

M =−

∂U = ∂Q =

Over portion EB: L  0 ≤ u ≤ 2   



wL ( L − v) 8 L/2 0



L/2 0

δD =



L/2 0

( L − v)vdv =

wL 16 EI

 Lv 2 v3  −   3  2

L/2 0

wL4  1 1  wL4 − =   16 EI  8 24  192 EI

1 M = − wu 2 2

∂ UEB = ∂Q

M ∂M wL dv = 16 EI EI ∂ Q

∂M =0 ∂Q

UEB =



L/2 0

M2 du 2 EI

M ∂M du = 0 EI ∂ Q

∂ UAD ∂ UDE ∂ UEB wL4 wL4 + + = + +0 384 EI 192 EI ∂Q ∂Q ∂Q

δD =

wL4 ↑  128EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.88 For the prismatic beam shown, determine the deflection at point D.

SOLUTION Change force at D from P to Q. 1 (Q − P) ↑, 2

Reactions:

RA =

Strain energy:

U = U AD + U DE + U EB

RE =

1 (3P + Q) ↑ 2

Deflection of point D (formula).

δD = Portion AD:

∂U ∂U AD ∂U DE ∂U EB = + + ∂Q ∂Q ∂Q ∂Q U AD =



L/2 0

M = RA x =

M2 dx 2 EI

1 (Q − P) x 2

∂M 1 = x With Q = P, M = 0. ∂Q 2 L/2 M ∂M ∂U AD = dx = 0 0 ∂Q EI ∂Q



Portion DE:

U DE =

M2 dx L/ 2 2 EI



L

L 1 L 1 1   M = RA x − Q  x −  = (Q − P) x − Q  x −  = Q( L − x) − Px 2 2 2 2 2   1 ∂M = ( L − x) 2 ∂Q

With

Q = P,

1 M = − P(2 x − L). 2

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.88 (Continued)

∂U DE = ∂Q

M ∂M P L dx = − (2 x − L)( L − x) dx L / 2 EI ∂Q 4 EI L / 2 P L =− (3Lx − 2 x 2 − L2 )dx 4 EI L / 2



L





L

P 3 2 2 3 2  =−  Lx − x − L x  4 EI  2 3  L/2 =−

Portion EB:

PL3  3 2 PL3  3  1   2  1  1   − − 1 −    +    +  = − 4 EI  2 3 96 EI  2  4   3  8  2 

U EB = M = − Pv

Deflection of point D.



L/2

0

M2 dv 2 EI ∂U EB ∂M =0 = ∂Q ∂Q

δD = 0 −



L/ 2

0

PL3 +0 96EI

M ∂M dv = 0 EI ∂Q

δD =

PL3 ↑  96EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.89 For the prismatic beam shown, determine the slope at point D.

SOLUTION Add couple M 0 at point D. Reactions:

RA =

M 0 wL , − 8 L

RE = −

U = UAD + UDE + UEB ;

Over portion AD:

U AD =



L/2 0

M2 dx, 2 EI

Set M 0 = 0.

M =−

∂ UAD = ∂ M0



M 0 5wL + 8 L ∂U θD = ∂ M0 M = RA x =

wLx 8 L/2 0

M ∂M dx = EI ∂ M 0



L/2 

UDE =



L/2 0



L/2 0

x 2 dx = −

wL3 192 EI

M v wL wL  L v+ =− 0 + (v − L)  2  4 L 8

∂M v =− ∂ M0 L

1 M = − wL( L − v) 8

∂ UDE w = ∂ M 0 8 EI Over portion EB:

w 8 EI

M2 dv 2EI

M = RE v −

Set M 0 = 0.

wLx  x  −   dx  8 EI  L 

0

=−

Over portion DE:

∂M x = ∂ M0 L

M 0 x wLx − 8 L

UEB =

∂ UEB = ∂ M0 θD =

 

L/2 0 L/2 0



L/2 0

w ( L − v)vdv = 8EI

L/2

 Lv 2 v3  wL3 − =   2 3  96 EI  0

M2 1 dv M = − wu2 2 EI 2

∂M =0 ∂ M0

M ∂M dv = 0 EI ∂ M 0

∂ UAD ∂ UDE ∂ UEB wL3 wL3 + + =− + +0 ∂ M0 ∂ M0 ∂ M0 192 EI 96 EI

θD =

wL3 192 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.90 For the prismatic beam shown, determine the slope at point D.

SOLUTION Add counterclockwise couple M 0 at point D. Reactions:  M E = 0: − AL + P

L L − P + M0 = 0 2 2

M0 ↑ L

A=

L L   M A = 0: EL − P  L +  − P + M 0 = 0 2 2   M E = 2P − 0 ↑ L U = U AD + U DE + U EB

Strain energy:

Slope at point D (formula).

θD =

∂U ∂U AD ∂U DE ∂U EB = + + ∂M 0 ∂M 0 ∂M 0 ∂M 0 L  0 ≤ x ≤ 2   

Portion AD: UAD = M=



L/2 0

M2 dx 2 EI x ∂M = ∂M 0 L

M0 x L

Set M 0 = 0 so that M = 0. ∂U AD 1 = ∂M 0 EI

Portion DE:

L   2 ≤ x ≤ L  

U DE =



L/2

0

M

∂M dx = 0 ∂M 0

M2 dx L/ 2 2 EI



L

L L  x   M = Ax − P  x −  − M 0 = M 0  − 1 − P  x −  2 2  L  

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.90 (Continued)

∂M x = − 1; ∂M 0 L ∂U DE 1 = ∂M 0 EI



L

L/2

M

L  Set M 0 = 0 so that M = − P  x −  . 2  ∂M P dx = ∂M 0 EI

L  x  x − 1 −  dx  L/2  2  L



L

3 x2 L  −  dx  x − L/2 2 L 2   L L P  3 x 2 x3 Lx = − −  EI  2 2 L/2 3L L/ 2 2  =

=

P EI



L

   L/2   L

PL2  3 3 1 1 1 1  PL3 − + =  − − + EI  4 16 3 24 2 4  48 EI L  0 ≤ v ≤ 2   

Portion EB: M = − Pv

Slope at point D.

∂M =0 ∂M 0

θD = 0 +

U EB = ∂U EB 1 = ∂M 0 EI



L/2

0

M



L/2

0

M2 dx 2 EI

∂M dv = 0 ∂M 0

PL2 +0 48 EI

θD =

PL2 48 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.91 For the beam and loading shown, determine the slope at end A. Use E = 200 GPa.

SOLUTION Add couple M A at point A. Units:

Forces in kN; lengths in m.

E = 200 × 109 Pa, I = 163 × 106 mm 4 = 163 × 10−6 m 4 EI = (200 × 109 )(163 × 10−6 ) = 32.6 × 106 N ⋅ m 2 = 32600 kN ⋅ m 2

Reactions:

RA = 80 −

MA 4.8

RB = 80 + 2.4

U = UAB + UBC =  0

Over AB: Set M A = 0.

=

Over BC: Set M A = 0.

2 M2 2.4 M dx +  0 dv 2EI 2 EI

M = M A + RA x = M A + 80 x − ∂U AB 1 = ∂M A EI



2.4 0

MA 4.8

MA x 4.8

θA =

∂U ∂UAB ∂UBC = + ∂M A ∂M A ∂M A

∂M x   = 1− 4.8  ∂M A 

x  1  (80 x) 1 −  dx = EI  4.8 



2.4 0

(80 x − 16.6667 x 2 )dx

1  1 1 2 3  153.6 (80)   (2.4) − (16.6667)   (2.4)  = EI  EI 2  3 

M = RB v = 80v +

MA v, 4.8

∂U BC 1 = ∂M A EI =

θA =



2.4 0

1 ∂M = v 4.8 ∂M A

16.6667  1  (80v)  v  dv = 4.8 EI  



2.4 0

v 2 dv

(16.6667)(2.4)3 76.8 = 3EI EI 1 230.4 {153.6 + 76.8} = EI 32600

θA = 7. 07 × 10−3 rad



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.92 For the beam and loading shown, determine the slope at end C. Use E = 29 × 106 psi.

SOLUTION

Units:

Forces in kips; lengths in ft.

E = 29 × 103 ksi

I = 291 in 4

EI = (29 × 103 )(291) = 8.439 × 106 kip ⋅ in 2 = 58.604 × 103 kip ⋅ ft 2

Add clockwise couple M 0 at end C. U=

Strain energy:



M2 1 dx = M 2 dx 2 EI 2 EI



θC =

Slope at C (formula).

∂U 1 ∂M = M dx ∂M 0 EI ∂M 0



Bending moment diagram. Portion BC: (0 ≤ x ≤ 2 ft) M = −4 x − M 0 ∂M = −1; ∂M 0



2

0

Set M 0 = 0.

∂M M dx = ∂M 0



2

0

2

4 (−4 x)(−1)dx = x 2 = 8 kip ⋅ ft 2 2 0

Portion AB: (2 ft ≤ x ≤ 8ft) M = −8( x − 2) − 4 x − M 0 ∂M = −1; ∂M 0



8

2

M

∂M dx = ∂M 0

Set M 0 = 0. 8

 [−8( x − 2 − 4 x](−1)dx 2

8

=

Slope at end C.

θC =

8

8 4 ( x − 2) 2 + x 2 = (4)(6)2 + (2)(64 − 4) = 264 kip ⋅ ft 2 2 2 2 2

(calculated)

8 + 264 272 = EI 58.604 × 103

θC = 4.64 × 10−3 rad



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.93 For the beam and loading shown, determine the deflection at end C. Use E = 29 × 106 psi.

SOLUTION

Units:

Forces in kips; lengths in ft.

E = 29 × 103 ksi I = 291 in 4 EI = (29 × 103 )(291) = 8.439 × 106 kip ⋅ in 2 = 58.604 × 103 kip ⋅ ft 2

Let Q be the force at end C. It is later set equal to 4 kips. U=

Strain energy:



M2 1 dx = M 2 dx 2 EI 2 EI



Deflection at C (formula).

δC =

∂U 1 ∂M = M dx ∂Q EI ∂Q



Bending moment diagram. Portion BC: (0 ≤ x ≤ 2 ft) M = −Qx ∂M = −x Set Q = 4 kips. ∂Q 2 2 ∂M M dx = (−4 x)( − x)dx 0 0 ∂M o





2

=

4 3 x = 10.67 kip ⋅ ft 3 3 0

Portion AB: (2 ft ≤ x ≤ 8 ft) M = −8( x − 2) − Qx ∂M = −x ∂M 0

Set

Q = 4 kips.

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.93 (Continued)



6

2

M

∂M dx = ∂Q

8

 [−8( x − 2) − 4 x](− x)dx 2

=8



8

2

( x 2 − 2 x)dx + 4



8

2

x 2 dx

 x3 8 2 8 3 8 x x  = 8 − 16 +4 2 2 3 2 3 2   4096 64 1024 64 2048 32 = − − + + − 3 3 2 2 3 3 3 = 1536 kip ⋅ ft

Deflection at C.

δC =

(calculated) 1536 + 10.67 1546.67 = = 0.0264 ft EI 58.604 × 103

δ C = 0.317 in. ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.94 For the beam and loading shown, determine the deflection at point D. Use E = 200 GPa.

SOLUTION Units:

Forces in kN, lengths in m. E = 200 × 109 Pa = 200 × 106 kN/m 2 I = 51.2 × 106 mm 4 = 51.2 × 10−6 m 4 EI = (200 × 106 )(51.2 × 10−6 ) = 10240 kN ⋅ m 2

Let Q be the force applied at D. It will be set equal to 90 kN later. Reactions:  M B = 0: − 3.2 A + 2.6Q + (0.6)(90) = 0 A = 16.875 + 0.8125Q ↑  M A = 0: 3.2 B − 0.6 Q − (2.6)(90) = 0 B = 73.125 + 0.1875Q ↑

Strain energy: U=

1 2 EI



3.2

0

M 2 dx

Deflection at point D. (formula)

δD =

∂U 1 = ∂Q EI



3.2

0

M

∂M dx ∂Q

Over portion AD: (0 ≤ x ≤ 0.6 m) M = (16.875 + 0.8125 Q ) x ∂M = 0.8125 x; Set Q = 90 kN. ∂Q M = 90 x



0.6

0

∂M M = ∂Q



0.6

0

x3 (90 x)(0.8125 x) = 73.125 3

0.6

= 5.265kN ⋅ m3 0

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.94 (Continued) Over portion DE: (0.6 m ≤ x ≤ 2.6 m) M = (16.875 + 0.8125 Q ) x − Q( x − 0.6) = 16.875 x − 0.1875Qx + 0.6Q ∂M Set Q = 90 kN. = −0.1875 x + 0.6; ∂Q M = (0.6)(90) = 54 kN ⋅ m 2.6 2.6 ∂M M dx = (54)(−0.1875 x + 0.6)dx 0.6 0.6 ∂Q





x2 = −10.125 2

2.6 2.6

+ 32.4 x 0.6 = 32.4 kN ⋅ m3 0.6

Over portion ED: (2.6 m ≤ x ≤ 3.2 m; 0 ≤ v ≤ 0.6 m) M = Bv = (73.125 + 0.1875 Q )v ∂M = 0.1875 v Set Q = 90 kN. ∂Q M = 90 v 3.2 0.6 0.6 ∂M ∂M M dx = M dv = (90v)(0.1875 v)dv 2.6 0 0 ∂Q ∂Q





v3 = 16.875 3



0.6

= 1.215 kN ⋅ m3 0

Deflection at point D. (calculated)

δD =

5.265 + 32.4 + 1.215 38.88 = = 3.797 × 10−3 m 10240 EI

δ D = 3.80 mm ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.95 For the beam and loading shown, determine the deflection at point B. Use E = 200 GPa.

SOLUTION L M2 M2 dx + dx a 2 EI 0 2 EI ∂U δB = ∂P a M ∂M L M ∂M dx + dx = a EI ∂ P 0 EI ∂ P

U=



a





Portion AB:



(0 ≤ x ≤ a)

∂M =0 ∂P

1 M = − wx 2 2

 Portion BC:

a 0

M ∂M dx = 0 EI ∂ P

( a < x ≤ L) 1 M = − wx 2 − P( x − a) 2 ∂M = −( x − a) ∂P



L a

M ∂M w dx = EI ∂ P 2 EI



w 2 EI



=

L a L a

x 2 ( x − a)dx +

P EI

( x3 − ax 2 )dx +



P EI

L a



( x − a) 2 dx b

0

v 2 dv

 L4 aL3 a 4 a 4  Pb3 − +  − + 3 4 3  3EI  4 w  L4 aL3 a 4  Pb3 δB = 0 + +  − + 2 EI  4 3 12  3EI =

w 2 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.95 (Continued)

Data:

a = 0.6 m, b = 0.9 m, L = a + b = 1.5 m w = 5 × 103 N/m P = 4 × 103 N 1 I = (40)(80)3 = 1.70667 × 106 mm 4 12 = 1.70667 × 10−6 m 4 EI = (200 × 109 )(1.70667 × 10−6 ) = 341333 N ⋅ m 2

δB = 0 +

5 × 103  (1.5)4 (0.6)(1.5)3 (0.6)3  (4 × 103 )(0.9)3 − +  + (2)(341333)  4 3 12  (3)(341333)

= 7.25 × 10−3 m

δ B = 7.25 mm ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.96 For the beam and loading shown, determine the deflection at point B. Use E = 200 GPa.

SOLUTION Add force Q at point B. Units:

Forces in kN; lengths in m.

∂M =0 ∂Q

Over AB:

M = −8 x

Over BC:

1 M = −8(v + 1) − (18)v 2 − Qv 2

∂M = −v ∂Q

E = 200 × 109 Pa, I = 28.7 × 106 mm 4 = 28.7 × 10−6 m 4 EI = (200 × 109 )(28.7 × 10−6 ) = 5.74 × 106 N ⋅ m 2 = 5740 kN ⋅ m 2 M2 dx + 0 2EI 1

U=



δB =

1  ∂U =  ∂ Q EI 





1.5 0

1 0

M

M2 dv 2 EI

∂M dx + ∂Q



1.5 0

M

∂M  dv  ∂Q 

1.5   1 1.5 3 1  1  (9v + 8v 2 + 8v)dv −8(v + 1) − (18)v 2  (−v)dv  = 0 +  0  0 EI  EI 2   1 9 8 8 29.391 29.391 4 3 2 = = = 5.12 × 10−3 m  (1.5) + (1.5) + (1.5)  = EI  4 EI 3 2 5740 

=





δ B = 5.12 mm ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.97 For the beam and loading shown, determine the deflection at point C. Use E = 29 × 106 psi.

SOLUTION Units:

Forces in kips; lengths in ft.

E = 29 × 103 ksi

I = 57.5 in 4

EI = (29 × 103 )(57.5) = 1.6675 × 106 kip ⋅ in 2 = 11580 kip ⋅ ft 2

Add force Q at point C. 1 1 RA = 4 + Q ↓, RD = 12 − Q ↑ 2 2

Reactions:

U = UAC + UCD + UDB





1 1 M = RB v − 8(v + 3) = 12v − Q − 8v − 24 = 4v − 24 − Qv 2 2 ∂M 1 =− v Set Q = 0. ∂Q 2

Over CD: (0 < v < 3)



∂ U ∂ UAC ∂ UCD ∂ UDB = + + ∂Q ∂Q ∂Q ∂Q

1  ∂M 1  M = − 4 + Q x Set Q = 0. =− x ∂Q 2  2  ∂ UAC 1 3 2 3 2 (2)(3)3 18 1  (4 x)  x  dx = = = x dx = 3EI EI 0 EI 0 EI ∂Q 2 

Over AC: (0 < x < 3)

∂ UCD 1 = EI ∂Q

δC =

1 1  (24 − 4v)  v  dv = 0 EI 2  3

=

Over DB: (0 < u < 3)

M = −8u

δC =



3 0

(12v − 2v 2 )dv =

1 EI

 (3) 2 (3)3  − (2) (12)  2 3  

36 EI

∂M =0 ∂Q

∂ U DB =0 ∂Q

18 36 54 + +0= = 4.663 × 10−3 ft 11580 EI EI

δ C = 0.0560 in. ↑ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.98 For the beam and loading shown, determine the slope at end A. Use E = 200 GPa.

SOLUTION

Units:

Forces in kips; lengths in ft. E = 29 × 103 ksi

I = 57.5 in 4

EI = (29 × 103 )(57.5) = 1.6675 × 106 kip ⋅ in 2 = 11580 kip ⋅ ft 2

Add couple M A at end A. Reactions:

RA = −4 +

MA M , RB = 12 − A 6 6

θA =

U = UAD + UDB

Over AD: (0 < x < 6)

M = − M A + RA x = − M A − 4 x + ∂M x  = − 1 −  ∂M A 6  

∂ UA D 1 = ∂ M A EI



M = −8u

x 1  (4 x) 1 −  dx = 0 EI  6 6

θA =

∂M =0 ∂ MA

24 24 +0= =0 11580 EI

MA x 6

Set M A = 0.

=

Over DB: (0 < u < 3)

∂ UAD ∂ UDB ∂U = + ∂ MA ∂ MA ∂ MA



2  1  62 2 63  4 x − x 2  dx = (4) −   0 EI  3  2 3 3  6

24 EI

∂ U DB =0 ∂ MA θ A = 2.07 × 10−3 rad



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.99 For the truss and loading show, determine the horizontal and vertical deflection of joint C.

SOLUTION Add horizontal force Q at point C. From geometry, LBC = LCD =

5 l 2

Equilibrium of joint C.  Fx = 0: −  Fy = 0:

1

2 5

FBC +

FBC +

2 5

FCD + Q = 0

1

FCD − P = 0 5 5 Solving simultaneously, 5 5 5 5 FBC = P+ Q FCD = P− Q 2 4 2 4

Equilibrium of joint D.  Fx = 0: FBD +

 5 5  + P− Q = 0 4  5  2

2

Q 2 Fi 2 Li U = 2 EAi

FBD = − P +

Strain energy: Deflections.

Horizontal:

xC =

∂U 1 Fi Li ∂Fi =  ∂Q E Ai ∂Q

Vertical:

yC =

∂U 1 Fi Li ∂Fi =  ∂P E Ai ∂P

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.99 (Continued) In the table, Q is set equal to zero in the last two columns.

BC CD BD

Fi

Li

Ai

∂Fi /∂P

∂Fi /∂Q

Fi Li ∂Fi Ai ∂P

Fi Li ∂Fi Ai ∂Q

5 5 P+ Q 2 4

5 l 2

A

5 2

5 4

5 Pl 5 8 A

5 Pl 5 16 A

5 5 P− Q 2 4 1 −P + Q 2

5 l 2

A

5 2

5 Pl 5 8 A

−5 Pl 5 16 A

2l

2A

−1

5 4 1 + 2

−

5

  4 xC =

1 Fi Li ∂Fi Pl  =− 2 EA E Ai ∂Q

yC =

1 Fi Li ∂Fi  5  Pl 5 + 1  = E Ai ∂P  4  EA

Pl A  Pl 5 + 1  A

−

1 Pl 2 A

−

1 Pl 2 A xC =

Pl ←  2 EA

yC = 3.80

Pl ↓ EA

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.100 For the truss and loading shown, determine the horizontal and vertical deflection of joint C.

SOLUTION Add horizontal force Q at point C. From geometry,

LBC = LCD =

5  2

Equilibrium of joint C. 2 ( FBC + FCD ) − Q = 0  Fx = 0: 5  Fy = 0:

1 ( FBC − FCD ) − P = 0 5

Solving simultaneously, FBC =

U =

Strain energy: Deflections.

5 5 P+ Q 2 4

FCD = −

5 5 P+ Q 2 4

Fi 2 Li 2 EAi

Horizontal:

xC =

∂U 1 F L ∂F =  i i i ∂Q E Ai ∂Q

Vertical:

yC =

∂U 1 F L ∂F =  i i i ∂P E Ai ∂P

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.100 (Continued) In the table, Q is set equal to zero in the last two columns.

BC CD

−

Fi

Li

Ai

∂Fi / ∂P

∂Fi / ∂Q

Fi Li ∂Fi A ∂P

Fi Li ∂Fi A ∂Q

5 5 P+ Q 2 4

5  2

A

5 2

5 4

5 P 5 A 8

5 P 5 A 16

5  2

A

5 4

5 P 5 A 8



5 P 5 A 4

5 5 P+ Q 2 4

−

5 2

xC =

1 Fi Li ∂Fi  =0 E A ∂Q

yC =

1 Fi Li ∂Fi 5 P 5  = E Ai ∂P EA 4

−

5 P 5 A 16

0 xC = 0  yC = 2.80

P ↓ EA

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.101 Each member of the truss shown is made of steel and has a cross-sectional area of 500 mm 2 . Using E = 200 GPa, determine the vertical deflection of joint B.

SOLUTION Find the length of each member as shown. Add vertical force Q at joint B.

δB =

Joint C:

∂U ∂ F 2L 1 ∂F = Σ = ΣF L ∂ Q ∂ Q 2 EA EA ∂Q

ΣFy = 0:

4 FCB − 4.8 = 0 FCB = 6.0 kN 5

Joint B:

3 FCB + FCD = 0 FCD = −3.6 kN 5 4 4 ΣFx = 0: FAB + FBD − 3.6 = 0 5 5 3 3 ΣFy = 0: FAB − FBD − 4.8 − Q = 0 5 5

Solving simultaneously,

FAB = 6.25 + 0.8333 Q

ΣFx = 0:

kN

FBD = −1.75 − 0.8333 Q kN

Joint D:

ΣFy = 0:

3 FBD + FAD = 0 5

3 FAD = − FBD = 1.05 + 0.5 Q 5

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.101 (Continued)

Member

F (103 N)

AB

6.25 + 0.8333Q

AD

1.05 + 0.5Q

BD

−1.75 − 0.8333Q

∂ F/∂ Q 0.8333 0.5

L (m)

with Q = 0 F (∂ F/∂ Q ) L (103 N ⋅ m)

2.0

10.4167

2.4

1.26

−0.8333

2.0

2.9167

BC

6.0

0

1.5

0

CD

−3.6

0

2.5

0

Σ

14.593 1 Σ F (∂ F/∂ Q) L EA 14.593 × 103 =  (200 × 109 )(500 × 10−6 )

δB = 

= 145.9 × 10−6 m



 δ B = 0.1459 mm ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.102 Each member of the truss shown is made of steel and has a cross-sectional area of 500 mm 2 . Using E = 200 GPa, determine the horizontal deflection of joint B.

SOLUTION Find the length of each member as shown. Add horizontal force Q at joint B.

δB = Joint C: Joint B:

Solving simultaneously,

∂U ∂ F 2L 1 ∂F = Σ = ΣF L ∂ Q ∂ Q 2 EA EA ∂Q

ΣFy = 0:

4 FCB − 4.8 = 0 5

FCB = 6.0 kN

3 FCB + FCD = 0 FCD = −3.6 kN 5 4 4 ΣFx = 0: FAB + FBD − 3.6 − Q = 0 5 5 3 3 ΣFy = 0: FAB − FBD − 4.8 = 0 5 5 ΣFx = 0:

FAB = 6.25 + 0.625Q kN FBD = −1.75 + 0.625Q kN

Joint D:

ΣFy = 0:

3 FBD + FAD = 0 5

3 FAD = − FBD = 1.05 − 0.375Q 5

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.102 (Continued)

Member

F (103 N)

AB

6.25 + 0.625Q

AD

1.05 + 0.375Q

BD

−1.75 + 0.625Q

L (m)

F (∂ F/∂ Q ) L (103 N ⋅ m)

2.0

7.8125

−0.375

2.4

−0.9450

0.625

2.0

−2.1875

∂ F/∂ Q 0.625

BC

6.0

0

1.5

0

CD

−3.6

0

2.5

0

Σ

4.680 1 Σ F (∂ F/∂ Q) L EA 4.680 × 103  = (200 × 109 )(500 × 10−6 )

δB = 

= 46.8 × 10−6 m



δ B = 0.0468 mm ← 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.103 Each member of the truss shown is made of steel and has the crosssectional area shown. Using E = 29 × 106 psi, determine the vertical deflection of joint C.

SOLUTION Call the vertical load P. The vertical deflection of joint C is δ P .

δP =

∂ U ∂ F 2L 1 FL ∂ F = Σ = Σ ∂ P ∂ P 2 EA E A ∂ P

AC = 92 + 3.752 = 9.75 ft = 117 in.

Geometry:

BC = 52 + 3.752 = 6.25 ft = 75 in. 4 ft = 48 in., 5 ft = 60 in., 3.75 ft = 45 in.

Joint C:

ΣFx = 0: −

108 60 FAC − FBC = 0 117 75

ΣFy = 0: −

45 45 FAC − FBC − P = 0 117 75

Solving simultaneously,

FAC = 3.25P, FBC = −3.75 P

Joint B:

ΣFx = 0: − FAB −

60 FAC = 0 75

FAB = −3.00 P

Member

F

∂ F/∂ P

L (in.)

A (in 2 )

F (∂ F/∂ P) L/A

AB

−3.00P

−3.00

48

4

108.00P

AC

3.25P

3.25

117

2

617.91P

BC

−3.75P

−3.75

75

6

175.78P

Σ

901.69P

δP =

901.69 P (901.69)(7.5 × 103 ) = E 29 × 106

δ P = 0.233 in. ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.104 Each member of the truss shown is made of steel and has the crosssectional area shown. Using E = 29 × 106 psi, determine the horizontal deflection of joint C.

SOLUTION Call the vertical load P. Add horizontal load Q at joint C. The horizontal deflection of joint C is δ Q .

δQ = Geometry:

∂U ∂ F 2L 1 FL ∂ F = Σ = Σ ∂ Q ∂ Q 2 EA E A ∂ Q

AC = 92 + 3.752 = 9.75 ft = 117 in. BC = 52 + 3.752 = 6.25 ft = 75 in. 4 ft = 48 in. 5 ft = 60 in. 3.75 ft = 45 in.

Joint C:

ΣFx = 0: −

108 60 FAC − FBC + Q = 0 117 75

ΣFy = 0: −

45 45 FAC − FBC − P = 0 117 75

FAC = 3.25P + 2.4375Q

Solving simultaneously,

FBC = −3.75 P − 1.5625Q ΣFx = 0:

Joint B:

FAB =

4 FAC − FAB = 0 5

4 FBC = −3.00 P − 1.25Q 5 Q=0 F (∂ F/∂ Q ) L/A

L (in.)

A (in 2 )

−1.25

48

4

45.00P

3.25 P + 2.4375Q

2.4375

117

2

463.43P

−3.75P − 1.5625Q

−1.5625

75

6

73.24P

Member

F

AB

−3.00 P − 1.25Q

AC BC

∂ F/∂ Q

Σ

581.67P

δQ =

581.67 P (581.67)(7.5 × 103 ) = E 29 × 106

δ Q = 0.1504 in. → 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.105 Two rods AB and BC of the same flexural rigidity EI are welded together at B. For the loading shown, determine (a) the deflection of point C, (b) the slope of member BC at point C.

SOLUTION Add horizontal force Q and couple M C at C. ΣM A = 0: RC l + M C − ( P + Q )l = 0 MC l

RC = P + Q +

ΣFx = 0: P + Q + RAx = 0 M = RAx y = ( P + Q ) y,

Member AB:

U AB =



RAx = P + Q ←

∂M = y, ∂Q

∂M =0 ∂ MC

M2 dy 0 2EI l

Set Q = 0 and M C = 0.

∂ UAB 1 = ∂Q EI ∂ UAB 1 = ∂ M C EI Member BC:

 

∂M 1 dy = 0 ∂Q EI l ∂M M dx = 0 0 ∂ MA l

M

l



( Py )( y )dy =

0

1 Pl 3 3 EI

M   M = M C + RC x = M C +  P + Q + C  x l   ∂M ∂M x = x, =1− ∂Q ∂ MC l UBC =



M2 dx 0 2 EI l

Set Q = 0 and M C = 0.

∂ UBC 1 = ∂Q EI ∂U 1 = ∂ M A EI

 

∂M 1 l 1 Pl 3 dx = ( Px) x dx = 0 ∂Q EI 0 3 EI l l ∂M 1 x  ( Px) 1 −  dx M dx = 0 ∂ MA EI 0 l  l



M



= =

P EI



l

x2  x − 0 l 

  dx 

P  1 2 1 2  1 Pl 2  l − l = 3  6 EI EI  2

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.105 (Continued)

(a)

Deflection at C.

δC =

∂ UAB ∂ UBC + ∂Q ∂Q

δC

2 Pl 3 →  3EI

(b)

Slope at C.

θC =

∂ UAB ∂ UBC + ∂ MA ∂ MC

θC

Pl 2 6 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.106 A uniform rod of flexural rigidity EI is bent and loaded as shown. Determine (a) the horizontal deflection of point D, (b) the slope at point D.

SOLUTION Add couple M D at point D.

Reactions at A: Member AB:

RAy = 0,

RAx = P ←,

M A = M0

∂M = y, ∂P

M = M A + RA y = M D + Py U AB =



∂M =1 ∂ MD

M2 dy 0 2 EI l

Set M D = 0.

∂ UAB 1 = ∂P EI ∂ UAB 1 = ∂ M 0 EI Member BC:

 

l 0 l 0

M

∂M 1 dy = ∂P EI

M

∂M 1 dy = ∂ M0 EI



l

( Py ) y dy =

0



l 0

M = M A + RAl = MD + Pl UBC =



Pl 3 3 EI

( Py )(1) dy =

Pl 2 2 EI

∂M = l, ∂P

∂M =1 ∂ MD

M2 dx 0 2 EI l

Set M D = 0.

∂ UBC 1 = ∂P EI ∂ UBC 1 = ∂ MD EI

 

l 0 l 0

M

∂M 1 dx = ∂P EI

M

∂M 1 dx = ∂ MD EI



l 0

( Pl )(l ) dx =



l 0

Pl 3 EI

( Pl )(1) dx =

Pl 2 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.106 (Continued)

Member CD:

∂M =y ∂P

M = M D + Py UCD =



∂M =1 ∂ MD

M2 dy 0 2 EI l

Set M D = 0.

∂ UCD 1 = ∂P EI ∂ U CD 1 = ∂ M D EI (a)

(b)

 

l 0 l 0

M

∂M 1 dy = ∂P EI

M

∂M 1 dy = ∂ MD EI



l 0

( Py )( y ) dy =



l 0

Pl 3 3EI

( Py )(1) dy =

Pl 2 2 EI

Horizontal deflection of point D.

δP =

∂ UAB ∂ UBC ∂ UCD  1 1  Pl 3 + + =  +1+  3  EI ∂P ∂P ∂P 3

θD =

∂ UAB ∂ UBC ∂ UCD  1 1  Pl 2 + + =  +1+  ∂ MD ∂ MD ∂ MD  2 2  EI

δP =

5Pl 3 →  3EI

Slope at point D.

θD =

2 Pl 2 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.107 A uniform rod of flexural rigidity EI is bent and loaded as shown. Determine (a) the vertical deflection of point D, (b) the slope of BC at point C.

SOLUTION Add vertical force Q at point D and couple M C at point C.

Reactions at A:

RAx = P ←, RAy = Q ↓ M A = Ql + M C

Member AB:

∂M ∂M = l, =1 ∂Q ∂ MC

M = M A + RAx y = Ql + M C + Py ,

UAB =



M2 dy 0 2 EI l

Set Q = 0 and M C = 0.

∂ UAB 1 = ∂Q EI ∂ UAB 1 = ∂ M C EI

 

l

0 l

0

M

∂M 1 dy = ∂Q EI

M

dM 1 dy = ∂ MC EI

UBC =

1 EI



0



( Py )(l ) dy = l

0

Pl 3 2 EI

( Py )(1) dy =

∂M =x ∂Q

M = M C + Pl + Qx

Member BC:



l

Pl 2 2 EI

∂M =1 ∂ MC

M2 dx 0 2 EI l

Set Q = 0 and M C = 0.

∂ U BC 1 = ∂Q EI ∂ U BC 1 = ∂ M C EI

 

l

0 l

0

M M

∂M 2Q

dx =

1 EI



1 ∂M dx = ∂ MC EI

l

0

( Pl )( x) dx =



l

0

( Pl )(1) dx =

Pl 3 2 EI Pl 2 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.107 (Continued)

Member CD:

M = Py ∂UCD =0 ∂Q

(a)

∂M =0 ∂ MC

∂ UCD =0 ∂ MC

Vertical deflection of point D.

δD = (b)

∂M =0 ∂Q

3 ∂ UAB ∂ UBC ∂ UCD  1 1  Pl + + =  + + 0 ∂Q ∂Q ∂Q 2 2  EI

δD =

Pl 3 ↑  EI

Slope of BC at C.

θC =

2 ∂ UAB ∂ UBC ∂ UCD  1  Pl + + =  +1+ 0 ∂ MC ∂ MC ∂ MC  2  EI

θC =

3Pl 2 2 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.108 A uniform rod of flexural rigidity EI is bent and loaded as shown. Determine (a) the vertical deflection of point A, (b) the horizontal deflection of point A.

SOLUTION Add horizontal force Q at point A. 1 3 M = Pv + Qv Over AB: 2 2 3 ∂M 1 ∂M = v = v 2 ∂P 2 ∂Q UAB =



L

0

M2 dx 2 EI

Set Q = 0.

∂ UAB 1 = ∂P EI



L

0

M

∂M 1 dv = ∂P EI

L

0

1  1   2 Pv  2 v  dv   

3

1 PL 12 EI ∂ UAB 1 L ∂M 1 M dv = = 0 ∂Q ∂Q EI EI =



Over BC:

 

1  3 3 PL3 dv =  Pv  12 EI 2  2

L

0

∂M L  = − x − , ∂P 2 

L 3  M = −P  x −  + QL, 2 2  L M2 UBC = dx 0 2 EI

∂M 3 = L ∂Q 2



Set Q = 0.

∂ UBC 1 = ∂P EI ∂ UBC 1 = ∂Q EI (a)

 

L

0 L

0

1 ∂M M dx = ∂P EI

∂M −1 M dx = ∂Q EI



0 L

0

2

L P  L  P  x −  dx = x−  2 3EI  2 

3 L

= 0

1 PL3 12 EI L

2 3P  L  3  L  P x −  L dx x = − − =0  2   2  4 EI  2   0

Vertical deflection of point A.

δP = (b)



L

∂ UAB ∂ UBC + ∂P ∂P

δP =

PL3 ↓  6 EI

δ Q = 0.1443

PL3 →  EI

Horizontal deflection of point A.

δQ =

∂ UAB ∂ UBC 3 PL3 + = ∂Q ∂Q 12 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.109 For the beam and loading shown, and using Castigliano’s theorem, determine (a) the horizontal deflection of point B, (b) the vertical deflection of point B.

SOLUTION Add horizontal force Q at point B. Use polar coordinate ϕ . U=



π /2 0

M2 Rdϕ 2EI

Bending moment. ΣM J = 0: M − Pa − Qb = 0 M = Pa + Qb = PR sin ϕ + QR (1 − cos ϕ ) ∂M ∂M = R sin ϕ = R (1 − cos ϕ ) ∂P ∂Q Set Q = 0.

(a)

δQ = =

(b)

∂U 1 = ∂ Q EI



PR3 EI

(sin ϕ − sin ϕ cos ϕ )dϕ =



π /2 0

π /2 0

M

∂M 1 Rdϕ = ∂Q EI



π /2 0

PR sin ϕ R (1 − cos ϕ ) Rdϕ π /2

PR3 1 ( − cos ϕ − sin 2 ϕ ) EI 2 0

3

=

PR  π 1 2π 1 2  + sin 0   − cos + cos 0 − sin EI  2 2 2 2 

=

PR3  1  0 +1− + 0 EI  2 

δP = =

∂U 1 = ∂ P EI



PR3 EI

sin 2 ϕ dϕ =



π /2 0

π /2 0

M

δQ =

∂M 1 Rdϕ = ∂P EI PR3 EI



π /2 0



π /2 0

PR3 → 2 EI

PR sin ϕ R sin ϕ Rdϕ

1 (1 − cos 2ϕ )dϕ 2

π /2

PR3  1 PR3  1 π 1 1 1 1   ϕ − sin 2ϕ  = = ⋅ − ⋅ 0 − sin π + ⋅ sin 0    EI  2 2 EI  2 2 2 2 2 0 



=

PR3  π   − 0 − 0 + 0  EI  4 

δP =

π PR3 4 EI

↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.110 For the uniform rod and loading shown, and using Castigliano’s theorem, determine the deflection of point B.

SOLUTION Use polar coordinate ϕ . Calculate the bending moment M (ϕ ) using free body BJ. ΣM J = 0: Px − M = 0 M = Px = PR sin ϕ

Strain energy:

U= U=



0

π 0

M2 ds 2EI ( PR sin ϕ ) 2 ( Rdϕ ) 2 EI

=

P 2 R3 2 EI



=

P 2 R3 2 EI



=

P2 R2 2 EI

1  ϕ 2 

=

By Castigliano’s theorem,



L

δ=

π 0

sin 2 ϕ dϕ

π 1 − cos

2

0

π

− 0

2ϕ

dϕ

1 sin 2ϕ 4

π 0

   

2

πP R 4 EI

∂U ∂P

δ=

π PR3 2 EI

↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.111 Determine the reaction at the roller support and draw the bending moment diagram for the beam and loading shown.

SOLUTION Remove the support at A and add the reaction RA as a load. Bending moment:

M = RA x − M 0

Strain energy:

U=

1 2 EI



L

0

M 2 dx

Deflection at point A is zero. yA = =

1 ∂U ∂M M dx = ∂RA EI ∂RA



1 EI

Reaction at A. Bending moment:



L

0

( RA x − M 0 ) xdx =

1 EI

 L3 L2  − M 0  = 0  RA 3 2   RA =

3 M0 ↑  2 L

3  M = M 0  x − 1  2 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.112 Determine the reaction at the roller support and draw the bending moment diagram for the beam and loading shown.

SOLUTION Remove support B and add reaction RB as a load. U = U AC + U CB =

∂ U ∂ U AB yB = = ∂ RB ∂ RB Over AC:

M2 du + 0 2 EI ∂ U CB + =0 ∂ RB



L/ 2

0

M2 dv 2 EI

L ∂M  L  = u +  M = RB  u +  − Pu , ∂ 2 2 R    B

∂ U AB 1 = ∂ RB EI



L/ 2 

0

 L L   RB  u +  − Pu   u +  du 2 2   

=

RB EI



L/2 

2

L u +  dv −  0  2 3 R  L  P = B  L3 −    − 3EI   2   EI =

Over CB:



L/ 2

P EI



L/2

0

L  u  u +  du 2 

 1  L 3 L 1  L  2     + ⋅    2 2  2    3  2 

7 RB L3 5 PL3 − 24 EI 48 EI

∂M =v ∂ RB

M = RB v

∂ U CB 1 = ∂ RB EI



L/2

0

3

( RB v)vdv =

RB  L  1 RB L3 = 3EI  2  24 EI

1  RB L3 5 PL3  7 yB =  + − =0  48 EI  24 24  EI M C = RB

L 2

M A = RB L − P

L  5 1 = − PL 2  16 2 

RB =

5 P↑  16

MC =

5 PL  32

MA = −

3 PL  16

MB = 0 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.113 Determine the reaction at the roller support and draw the bending moment diagram for the beam and loading shown.

SOLUTION Remove the support at end A and add the reaction RA at end A as a load. The deflection y A at point A is zero. 1 L 2 Strain energy: U= M dx 2 EI 0 Deflection at A. 1 L ∂M ∂U yA = M dx = ∂RA EI 0 ∂RA Bending moment. 1 Portion AB: M = RA x − wx 2 2 1 L  M = RA x − wL  x −  Portion CB: 2 4 





yA =

1 ∂M 1 M dx = ∂RA EI EI

1 2  RA x − 2 wx  xdx   L 1 1 1   + RA x − wLx + wL2  xdx  /2 L 2 8 EI  





L/2 

0



=

1   EI 



L/2 

1 3 2  RA x − wx  dx + 2  

0

L /2

RA 3 w 4 x − x 3 8 0

+



1 1 2    2 2  RA x − wLx + wL x  dx  = 0 2 8  

L

L/2 

RA 3 wL 3 wL2 2 x − x + x 3 6 16

L

=0 L /2

R RA L wL wL wL   RA L3 wL L3 wL2 L2  − +  A L3 − + − + − =0 3 8 8 16  3 6 16   3 8 6 8 16 4  3

4

4

4

RA L3 1 1 1 1   1 + wL4  − − + + − =0 3 128 6 16 48 64   RA L3  41  − wL4  =0 3  384 

RA =

41 wL ↑  128

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.113 (Continued) Bending moment diagram. L  Portion AB:  0 ≤ x ≤  2  41 1 M = wLx − wx 2 128 2 dM  41  V = = w L − x dx  128  V = 0 at x = xm .

xm =

41 L  128

2

Mm =

1  41  wL2 2  128 

L  Portion BC:  ≤ x ≤ L  2  1 23 M = wL2 − wLx 8 128 7 MB = − wL2 128

M m = 0.0513 wL2 

M B = −0.0547 wL2 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.114 Determine the reaction at the roller support and draw the bending moment diagram for the beam and loading shown.

SOLUTION Remove support A and add reaction RA as a load. M2 dx 0 2 EI ∂U 1 = δA = ∂ RA EI U=

Portion AD:

(0 < x < a )



L

(a < x < L)

∂ U DB 1 = ∂ RA EI δA =



L

0

M

∂M dx = 0 ∂ RA

∂M =x ∂ RA

M = RA x

∂ U AD 1 = ∂ RA EI Portion DB:





a

0

( RA x) ( x) dx =

∂M =x ∂ RA

M = RA x − M 0 L a

RA a 3 3EI

( RA x − M 0 ) ( x) dx =

1 1 1 3 3 2 2   RA ( L − a ) − M 0 ( L − a )  2 EI  3 

∂ U AD ∂ U DB 1  1 3 1 3 1 3 1 2 2  + =  RA  a + L − a  − M 0 ( L − a )  = 0 3 3  2 ∂ RA ∂ RA EI   3  RA =

3 M 0 ( L2 − a 2 ) 2 L3

RA =

3 M 0b ( L + a) 2 L3

↑ 

MA = 0  M D − = RA a

M D− =

M D+ = M D− − M 0 M B = RA L − M 0

Bending moment diagram drawn to scale for a =

M D+ =

3 M 0 ab ( L + a)  2 L3

3 M 0 ab( L + a) − M0  2 L3

MB =

3 M 0 b( L + a ) − M0  2 L2

1 L. 3

By singularity functions, M = 3M 0 b( L + a) x/2 L3 − M 0  L − a 0  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.115 For the uniform beam and loading shown, determine the reaction at each support.

SOLUTION Remove support A and add reaction RA at point A as a load. The deflection yA at point A is zero. MB = 0 :

Statics:

L − M0 = 0 2 1 M RC = RA + 0 2 L  Fy = 0: RA + RB + RC = 0 RC L − RA

RB = − RA − RC U=

Strain energy:

yA =

Deflection:

1 2 EI



L/ 2

0

M 2 dx +

1 ∂U = ∂RA EI



L/ 2

0

M

1 2 EI



L

0

M 2 dv

1 ∂M dx + EI ∂RA



L

0

M

∂M dv = 0 ∂RA

Bending moment: L  Portion AB  0 ≤ x ≤  : M = RA x 2 

∂M = x ∂RA

M  1 M = RC v − M 0 =  RA + 0  v − M 0 L  2

Portion BC (0 ≤ v ≤ L) :

∂M 1 = v ∂RA 2

After dividing by EI,  1  M  1 RA + 0 v  − M 0   v  dυ = 0   0 0 2 L    2  L/2 L L 1 1 M0 L 2 1 RA x 2 dx + RA v 2 dv + v dv − M 0 vdv = 0 0 0 0 4 2 L 0 2



L/2

( RA x)( x) dx +





L 







RA L3 RA L3 M 0 L2 M 0 L2 + + − =0 24 12 6 4 RA L3 M 0 L2 − =0 8 12 RC =

1  2 M0  M0 + L 2  3 L 

RB = −

2 M0 4 M0 M + = −2 0 3 L 3 L L

RA =

2 M0 ↑  3 L

RC =

4 M0 ↑  3 L

RB = 2

M0 ↓  L

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.116 Determine the reaction at the roller support and draw the bending moment diagram for the beam and load shown.

SOLUTION Remove support A and add reaction RA as a load. M2 dx 0 2 EI ∂U 1 = δA = ∂ RA EI



U=

L  Portion AD:  0 < x <  3 

L



L/3

0

L

0

M

∂M dx ∂ RA

∂M =x ∂ RA

M = RA x

∂ U AD 1 = ∂ RA EI



M

∂M 1 dx = ∂ RA EI



L/3

0

( RA x) ( x) dx

3

=

RA  L  1 RA L3 = 3EI  3  81 EI

L  Portion DB:  < x < L  3 

L  M = RA x − P  x −  3  ∂M =x ∂ RA  L    RA − P  x −   x dx 3    R L 2 P L  2 L  = A x dx −  x − x  dx 3  EI L/3 EI L/3 

∂ U DB 1 = EI ∂ RA



L

L/3

M

1 ∂M dx = EI ∂ RA



=

RA 3EI



L

L/3



 3  L 3  P  1  3  L  3  L  2  L  2     L −    −  L −    L −    −  3   EI  3   3   6   3    

3 3  1 1  R L  1 1 1 1  PL = −  A − − − +   3 81 6 54  EI  3 81  EI

δA =

∂ U AD ∂ U DB  1 1 1  RA L3 14 PL3 + = + − − ∂ RA ∂ RA  81 3 81  EI 81 EI

1 RA L3 14 PL3 − 3 EI 81 EI =0 =

RA =

14 P↑  27

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.116 (Continued)

Bending moments:

By singularity functions,

 L  14 M D = RA   = PL  3  81

M D = 0.1728 PL 

4  2L  M B = RA L − P  = − PL  27  3 

M B = −0.1481 PL  M=

14 Px − P x − L/31  27

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.117 Three members of the same material and same cross-sectional area are used to support the load P. Determine the force in member BC.

SOLUTION Detach member BC at support C. Add reaction RC as a load. U =Σ

F 2L 2 EA

yC =

∂U FL ∂ F =Σ =0 EA ∂ RC ∂ RC

FBC = RC

Joint C:

ΣFx = 0: FBE sin ϕ − FBD sin ϕ = 0

Joint B:

FBE = FBD

ΣFy = 0: FBD cos ϕ + FBE cos ϕ + RB − P FBD = FBE =

P − RB 2cos ϕ

Member

F

BD

( P − RB )/2 cos ϕ

BE

( P − RB )/2 cos ϕ

BC

RB

∂ F/∂ RB −1/2 cos ϕ −1/2 cos ϕ 1

L

( FL/EA) (∂ F/∂ RB )

l/ cos ϕ

( RB − P)l/4 EA cos3ϕ

l/ cos ϕ

( RB − P)l/4 EA cos3ϕ

l

RB l/EA

yB = − Pl/2 EA cos3 ϕ + RB l/2 EA cos3 ϕ + RB l/EA = 0 RB =

P 1 + 2 cos3ϕ

FBC = RB

FBC =

P  1 + 2 cos3ϕ

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.118 Three members of the same material and same cross-sectional area are used to support the load P. Determine the force in member BC.

SOLUTION Detach member BC from its support at point C. Add reaction FC as a load. ΣFy = 0:

Joint B:

2 FBD + FC − P = 0 2

FBD = 2 P − 2 FC ΣFx = 0:

2 FBD + FBE = 0 2

FBE = − P + FC F 2R R = ΣF 2 2 EA 2 EA R ∂U ∂F = ΣF =0 δC = ∂ FC EA ∂ FC U =Σ

Member BC

F FC

BD

2 P − 2 FC

BE

− P + FC

∂ F/∂ FC 1

F (∂ F/∂ FC )

− 2

−2 P + 2 FC

1

Σ

FC − P + FC −3P + 4 FC

R (−3P + 4 FC ) = 0 EA 3 FC = P 4  FBC = FC

δC =

FBC =

3 P  4

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.119 Three members of the same material and same cross-sectional area are used to support the load P. Determine the force in member BC.

SOLUTION Detach member BC from support C. Add reaction FC as a load. F 2L 1 = ΣF 2 L 2 EA 2 EA 1 ∂U ∂F = ΣF L δC = ∂ FC EA ∂ FC U =Σ

Joint B:

ΣFy = 0: FC − P + ΣFx = 0: FBE +

3 FBD = 0 5

FBD =

4 FBD = 0 5

5 5 P − FC 3 3

4 4 FBE = − P + FC 3 3

Member

F

∂ F/∂ FC

L

F (∂ F/∂ FC ) L

BC

FC

1

3 l 4

3 FC l 4

BD

5 5 P − FC 3 3

BE

4 4 − P + FC 3 3

−

5 3

5 l 4

4 3

l

Σ

δC =

1  21  − Pl + 6 FC l  = 0  EA  4 

FC =

7 P 8

−

125 125 Pl + FC l 36 36 −

16 16 Pl + FC l 9 9

−

21 Pl + 6 FC l 4

FBC = FC

FBC =

7 P  8

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.120 Three members of the same material and same cross-sectional area are used to support the load P. Determine the force in member BC.

SOLUTION Cut member BC at end B and replace member force FBC by load FB acting on member BC at B.

δB =

∂U ∂ F 2L 1 ∂F L=0 = Σ = ΣF ∂ FB ∂ FB EA EA ∂ FB 3 FCD + FBC − P = 0 2

ΣFy = 0:

Joint C:

2

FCD =

3

2

P−

3

ΣFx = 0: FAC − FAC =

Member

1 3

P−

CD

1 FCD = 0 2

1 3

FB

∂ F/∂ FB

F FB

AC BC

FB

1 3 2 3

P− P−

1 3 2 3

1

FB

−

FB

−

F (∂ F/∂ FB ) L

L

1 3

l

FB l

l

1 1 − Pl + FB l 3 3

2

2

3

3

8 3

Pl +

8 3

FB l

1 8  4 8  − +  Pl +  +  FB l 3 3 3 3

Σ 1 3

δB = − + FB =



−

l

1 3

+

4 3

+

8 3 8 3

FBC = FB 

8  Pl  4 8  FB l + + =0   3  EA  3 3  EA P=

8+ 3 8+4 3

P = 0.652 P FBC = 0.652 P 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.121 Knowing that the eight members of the indeterminate truss shown have the same uniform cross-sectional area, determine the force in member AB.

SOLUTION Cut member AB at end A and replace member force FAB by load FA ← acting on member AB at end A.

δA =

∂U ∂ F 2L 1 ∂F L=0 = Σ = ΣF ∂ FA ∂ FA 2EA EA ∂ FA

Joint B:

5 FBD = − FA 4

Joint E:

ΣFy = 0: FBE − P +

FBE =

3 FA 4

3 FAE 5

5 5 P − FBE 3 3 5 5 = P − FA 3 4

FAE =

ΣFx = 0: −

4 FAE − FDE = 0 5

4 4 FDE = − FAE = − P + FA 5 3

Joint D:

ΣFy = 0: FAD +

3 FDB = 0 5

3 3 FAD = − FDB = − FA 5 4

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.121 (Continued)

Member

∂ F/∂ FA

F

L

F (∂ F/∂ FA ) L

1

l

FAl

−

3 4

3 l 4

27 FAl 64

5 5 P − FA 3 4

−

5 4

5 l 4

BD

5 − FA 4

−

5 4

5 l 4

125 FAl 64

BE

3 FA 4

3 4

3 l 4

27 FAl 64

DE

4 − P + FA 3

1

l

4 − Pl + FAl 3

AB

FA

AD

3 − FA 4

AE

−

125 125 Pl + FAl 48 64

−

Σ

63 27 Pl + FAl 16 4

1  63 27  − Pl + FAl  = 0  4 EA  16  7 FA = P 12

δA =



FAB = FA 

FAB =

7 P  12

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.122 Knowing that the eight members of the indeterminate truss shown have the same uniform cross-sectional area, determine the force in member AB.

SOLUTION Cut member AB at end A and replace member force FAB by load FA ← acting on member AB at end A.

δA = Joint B:

∂U ∂ F 2L 1 ∂F L=0 = Σ = ΣF ∂ FA ∂ FA 2 EA EA ∂ FA

ΣFx = 0: −FA −

4 FBD = 0 5

5 FBD = − FA 4

3 ΣFy = 0: −P − FBE − FBD = 0 5

Joint E:

ΣFy = 0: FBE +

ΣFx = 0: −

Joint D:

4 FAE − FDE = 0 5

ΣFy = 0: FAD + FAD =

3 FAE = 0 5

FBE = − P +

FAE =

3 FA 4

5 5 P − FA 3 4

4 FDE = − P + FA 3

3 FBD = 0 5

3 FA 4

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.122 (Continued)

Member

F

∂ F/∂ FA

L

AB

FA

1

l

AD

3 FA 4

3 4

3 l 4

AE

5 5 P − FA 3 4

−

5 4

5 l 4

BD

5 − FA 4

−

5 4

5 l 4

BE

−P +

3 FA 4

3 4

3 l 4

DE

4 − P + FA 3

1

l

F (∂ F/∂ FA ) L FAl 27 FAl 64 −

125 125 Pl + FAl 48 64 125 FAl 64 −

9 27 Pl + FAl 16 64 4 − Pl + FAl 3

9 27 − Pl + FAl 2 4

Σ 1  9 27  − Pl + FAl  = 0  EA  2 4  2 FA = P 3

δA =



FAB = FA 

FAB =

2 P  3

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.123 Rods AB and BC are made of a steel for which the yield strength is σ Y = 300 MPa and the modulus of elasticity is E = 200 GPa. Determine the maximum strain energy that can be acquired by the assembly without causing permanent deformation when the length a of rod AB is (a) 2 m, (b) 4 m.

SOLUTION AAB =

π 4

(12) 2 = 113.097 mm 2 = 113.097 × 10−6 m 2

ABC =

π 4

(8) 2 = 50.265 mm 2 = 50.265 × 10−6 m 2

P = σ Y Amin = (300 × 106 )(50.265 × 10−6 ) = 15.08 × 103 N U =Σ

(a)

a = 2 m, L − a = 5 − 2 = 3 m U=

(b)

P 2l P2 l = Σ 2 EA 2 E A

(15.08 × 103 ) 2  2 3  + = 44.0 N ⋅ m 9  −6 −6  (2)(200 × 10 ) 113.097 × 10 50.265 × 10 

U = 44.0 J 

a = 4 m L − a =1 m U=

(15.08 × 103 ) 2  4 1  = 31.4 N ⋅ m 9  −6 −6  (2)(200 × 10 ) 113.097 × 10 50.265 × 10 

U = 31.4 J 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.124 Assuming that the prismatic beam AB has a rectangular cross section, show that for the given loading the maximum value of the strain-energy density in the beam is umax =

45 U 8 V

where U is the strain energy of the beam and V is its volume.

SOLUTION L 1 = 0 RA = wL 2 2

M B = 0: −RA L + ( wL) M = RA x − U=



L

0

1 2 1 wL = w( Lx − x 2 ) 2 2

M w2 dx = 2 EI 8EI



L

0

( L2 x 2 − 2 Lx3 + x 4 )dx L

w2  L2 x3 2 Lx 4 x5  = − +   8EI  3 4 5 0 =

w2 L5 8EI

=

w2 L5 240 EI

M max =

σ max = umax =

1 1 1  3 − 2 + 5  

2 1  L L  1 2 w  L ⋅ −    = wL 2  2  2   8

M max c wL2 c = 8I I 2 σ max

2E

=

w2 L4 c 2 128EI 2

(

3 1 U 8LI 8L 12 bd = = 2 umax 15c 2 15 ( d2 )

=

)

8 8 Lbd = V 45 45 umax =

45 U  8 V

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.125 A 5-kg collar D moves along the uniform rod AB and has a speed v0 = 6 m/s when it strikes a small plate attached to end A of the rod. Using E = 200 GPa and knowing that the allowable stress in the rod is 250 MPa, determine the smallest diameter that can be used for the rod.

SOLUTION 1 2 1 mv0 = (5)(6) 2 = 90 J 2 2 2 P L ( Aσ max ) 2 L Um = m = 2 EA 2 EA 2 EU (2)(200 × 109 )(90) = 480 × 10−6 m 2 A= 2 m = 6 2 σ max L (250 × 10 ) (1.2) Um =

π 4

d2 = A

d=

4A

π

=

(4)(480 × 10−6 )

π

= 24.7 × 10−3 m

d = 24.7 mm 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.126 A 160-lb diver jumps from a height of 20 in. onto end C of a diving board having the uniform cross section shown. Assuming that the diver’s legs remain rigid and using E = 1.8 × 106 psi, determine (a) the maximum deflection at point C, (b) the maximum normal stress in the board, (c) the equivalent static load.

SOLUTION 1 (16)(2.65)3 = 24.813 in 4 12 L = 9.5 ft = 114 in. a = 2.5 ft = 30 in. 1 c = (2.65) = 1.325 in. 2 P L M =− m x a a M2 P 2 L2 a 2 P 2 L2 a U AB = dx = m 2 x dx = m 0 2 EI 6 EI 2 EIa 0 I=

Over portion AB:





M = − Pm v L M2 P2 U BC = dv = m 0 2 EI 2 EI

Over portion BC:





L

0

v 2 dv =

Pm2 L3 6 EI

Pm2 L2 (a + L) 6 EI 2U m Pm L2 (a + L) ym = = Pm 3EI

U = U AB + U BC =

Total:

1 Pm ym = U m 2

3EI (3)(1.8 × 106 )(24.813) ym = ym = 71.598 ym (114)2 (114 + 30) L ( a + L) 1 U m = Pm ym = 35.799 ym2 2 Pm =

2

= W ( h + ym ) = (160)(20 + ym ) = 3200 + 160 ym

Work of weight:

3200 + 160 ym = 35.799 ym2

Equating:

ym2 − 4.4694 ym − 89.388 = 0

{

1 4.4694 + 4.46942 + (4)(89.388) 2

(a)

ym =

(c)

Pm = (71.598)(11.95) = 856 lb

}

ym = 11.95 in.  Pm = 856 lb 

M m = −(856)(114) = 97535 lb ⋅ in

(b)

σm =

Mm c I

=

(97535)(1.325) = 5210 psi 24.813

σ m = 5.21 ksi 

 PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.127 A block of weight W is placed in contact with a beam at some given point D and released. Show that the resulting maximum deflection at point D is twice as large as the deflection due to a static load W applied at D.

SOLUTION Consider dropping the weight from a height h above the beam. The work done by the weight is Work = W (h + ym )

Strain energy:

U=

1 1 Pm ym = kym2 2 2

where k is the spring constant of the beam for loading at point D. Equating work and energy, Setting h = 0,

W ( h + ym ) = Wym =

1 2 kym , 2

1 2 kym . 2 ym =

2W . k

The static deflection at point D due to weight applied at D is

δ st =

W . k

ym = 2δ st 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.128 The 12-mm-diameter steel rod ABC has been bent into the shape shown. Knowing that E = 200 GPa and G = 77.2 GPa, determine the deflection of end C caused by the 150-N force.

SOLUTION

π

J=

c4 =

2

π  12 

4

3 4   = 2.0358 × 10 mm 2 2 

= 2.0358 × 10−9 m 4 1 I = J = 1.0179 × 10−9 m 4 2

Portion AB:

bending

M = − Px UAB, b =



LAB

0

M2 P2 dx = 2 EI 2 EI



LAB

0

x 2dx

P 2 L3AB (150)2 (200 × 10−3 )3 = 6 EI (6)(200 × 109 )(1.0179 × 10−9 ) = 0.14736 J =

torsion

T = PLBC T 2 LAB P 2 L2BC LAB = 2GJ 2GJ 2 (150) (200 × 10−3 ) 2 (200 × 10−3 ) = (2)(77.2 × 109 )(2.0358 × 10−9 ) = 0.57265 J

UAB , t =

Portion BC:

M = − Px P 2 L3BC M2 P 2 LBC 2 dx = x dx = 0 2 EI 3EI 0 6 EI 2 −3 3 (150) (200 × 10 ) = = 0.14736 J (6)(200 × 109 )(1.0179 × 109 )

UBC =

Total: Work-energy:



LBC



U = UAB, b + UAB, t + UBC = 0.86737 J 1 Pδ = U 2

δ=

2U (2)(0.86737) = 150 P = 11.57 × 10−3 m

δ = 11.57 mm ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.129 Two steel shafts, each of 0.75-in. diameter, are connected by the gears shown. Knowing that G = 11.2 × 106 psi and that shaft DF is fixed at F, determine the angle through which end A rotates when a 750-lb ⋅ in torque is applied at A. (Ignore the strain energy due to the bending of the shafts.)

SOLUTION Work-energy equation: 1 TAϕ A = U 2 2U ϕA = TA

Portion AB of shaft ABC: TAB = TA = 750 lb ⋅ in LAB = 5 + 6 = 11 in. J AB = U AB =

Portion BC of shaft ABC: Gear B: Gear E: Portion DE of shaft DEF: Portion EF of shaft DEF:

π d

4

π  0.75  =  = 31.063 × 10−3 in 4   22 2  2 

2 TAB LAB (750) 2 (11) = = 8.892 in ⋅ lb 2GJ AB (2)(11.2 × 106 )(31.063 × 10−3 )

U BC = 0 FBE =

TB TAB 750 = = = 250 lb rB rB 3

TE = rE FBE = (4)(250) = 1000 lb ⋅ in U DE = 0 TEF = TE = 1000 lb ⋅ in LEF = 8 in. U EF =

Total:

4

J EF =

π d

4

−3 4   = 31.063 × 10 in 22

2 TEF LEF (1000) 2 (8) = = 11.497 lb ⋅ in 2GJ EF (2)(11.2 × 106 )(31.063 × 10−3 )

U = U AB + U BC + U DE + U EF = 20.389 in ⋅ lb

ϕA =

2U (2)(20.389) = = 54.4 × 10−3 rad TA 750

ϕ A = 3.12° 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.130 Each member of the truss shown is made of steel and has a uniform crosssectional area of 3 in2. Using E = 29 ⋅ 106 psi, determine the vertical deflection of joint A caused by the application of the 24-kip load.

SOLUTION

Joint A:

Joint C: Strain energy:

4 FAC = 0 5

ΣFy = 0:

−16 −

Σ Fx = 0:

3 FAC + FAB = 0 5

FAB = 12 kips

ΣFy = 0:

4 FB − (20) = 0 5

FBC = 16 kips

U =Σ

Member

F 2L 1 = ΣF 2 L 2 EA 2 EA

FAC = −20 kips

E = 29 × 103 ksi A = 3 in.2 F 2 L (kip 2 ⋅ in)

F (kips)

L (in.)

AB

18

36

11664

AC

−30

60

54000

BC

24

48

27648

Σ U=

93312 93312 = 0.53628 kip ⋅ in (2) (29 × 103 ) (3)

1 PΔ = U 2

Δ=

2U (2) (0.53628) = P 24

Δ = 0.0447 in. ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.131 A disk of radius a has been welded to end B of the solid steel shaft AB. A cable is then wrapped around the disk and a vertical force P is applied to end C of the cable. Knowing that the radius of the shaft is r and neglecting the deformations of the disk and of the cable, show that the deflection of point C caused by the application of P is PL3  Ea 2  δC = 1 + 1.5 2  . 3EI  GL 

SOLUTION Torsion:

T = Pa Ut =

Bending:

T 2 L P2 a2 L = 2GJ 2GJ

M = Pv M 2 dv = 0 2 EI P 2 L3 = 6 EI

Ub =

Total:

Since J = 2 I ,



L



L

0

P 2 v 2 dv 2 EI

P 2 a 2 L P 2 L3 1 + = Pδ C 2GJ 6 EI 2 2 3 Pa L PL PL3  3EIa 2 + = δC = 1 + 3EI 3EI  GJ GJL2 U=

  

δC =

PL3  Ea 2 1 + 1.5 2 3EI  GL

   

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.132 Three rods, each of the same flexural rigidity EI, are welded to form the frame ABCD. For the loading shown, determine the angle formed by the frame at point D.

SOLUTION Add couple M 0 at point D. Statics:

M A = 0: M 0 + DL − PL = 0 D=P−

Strain energy:

M0 ↑ L

ΣFx = 0: − Ax + P = 0

Ax = P ←

ΣFy = 0: Ay + D = 0

Ay = P −

M0 ↓ L

U = UAB + UBC + UCD

∂U ∂ M0 ∂ U AB ∂ U BC ∂ U CD θD = + + ∂ M0 ∂ M0 ∂ M0

By Castigliano’s theorem, θ D =

Member AB:

∂M = 0 UAB = ∂ M0 L M ∂M ∂ UAB = dy = 0 0 EI ∂ M ∂ M0 0 M = Py



L

0

M2 dy 2 EI



Member BC:

M = M 0 + Dx = M 0 + Px −

Set M 0 = 0

M = Px

∂ UBC ∂ M0

∂M x =1− ∂M 0 L

M2 dx 0 2 EI L M ∂M P = dx = 0 EI ∂ M EI 0

UBC =

 

M0x L

L



L

0

x PL2  x 1 −  dy = L 6 EI 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.132 (Continued)

Member CD:

M = M0

Set M 0 = 0

M =0 U=



L

0

θD = 0 +

∂M =1 ∂ M0

M2 dy 2 EI PL2 +0 6 EI

∂U = ∂ M0



L

0

M ∂M dx = 0 EI ∂ M 0

θD =

PL2 6 EI



PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.133 The steel bar ABC has a square cross section of side 0.75 in. and is subjected to a 50-lb load P. Using E = 29 ⋅ 106 psi for the rod BD and the bar, determine the deflection of point C.

SOLUTION Assume member BD is a two-force member. ΣM A = 0: 10 FBD − (40)(50) = 0

Member ABC: Portion AB:

π

(0.2)2 = 31.416 × 10−3 in 2 4 F2 L (200) 2 (25) = BD BD = 2 EA (2)(29 × 106 )(31.416 × 10−3 ) = 0.5488 in ⋅ lb

ABD = U BD

FBD = 200 lb

I=

1 (0.75)(0.75)3 = 26.367 × 10−3 in 4 12

x = −150 x 10 10 M 2 1502 dx = = 0 2 EI 2 EI

M = −1500 UAB





10 0

x 2dx

(150) 2 (103 ) (2)(29 × 106 )(26.367 × 10−3 )(3) = 4.904 in ⋅ lb =

Portion BC:

M = −50v M2 502 30 2 dv = v dv 0 2 EI 2 EI 0 (50) 2 (30)3 = = 14.713 in ⋅ lb (2)(29 × 106 )(26.367 × 10 −3 )(3)

UBC =

Total:



30



U = U BD + U AB + U BD = 20.166 in ⋅ lb 1 Pδ C = U 2

δC =

2U (2)(20.166) = P 50

δ C = 0.807 in. ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 11.134 For the uniform beam and loading shown, determine the reaction at each support.

SOLUTION Remove support A and add reaction RA as a load. Σ M B = 0: −RA

L 1 2 − wL + RC L = 0 2 2

1 1 RA + wL 2 2

RC =

U = UAB + UBC =



∂ U ∂ UAB = δA = ∂ RA ∂ RA M = R A x,

Portion AB:

∂ UAB 1 = ∂ RA EI Portion BC:



L/2

0

M = RC v −

M2 dx + 0 2 EI ∂ UBC + =0 ∂ RA L/ 2



L

0

M2 dv 2 EI

∂M =x ∂ RA

∂M 1 M dx = ∂ RA EI



L/2

0

3

R L 1 RA L3 ( RA x) ( x) dx = A   = 3EI  3  24 EI

1 2 1 1 1 wv = RA v + wLv − wv 2 2 2 2 2

∂M 1 = v ∂ RA 2 ∂ UBC 1 = ∂ RA EI =

δA =



L 1

0

1 1 2  1   2 RAv + 2 w( Lv − v )   2 v  dv = 4 EI   

0

[ RAv 2 + w( Lv 2 − v3 )]dv

 L4 L4   R L3 wL4 1  L3 + w  −   = A +  RA 4 EI  3 4   12 EI 48 EI  3

∂ UAB ∂ UBC  1 1  R L3 wL4 + = +  A + =0 ∂ RA ∂ RA  24 12  EI 48EI

1 RA = − wL 6 RC =



L

RA =

1 1  1 − wL  + wL  2 6  2

RC =

1 wL ↓  6

5 wL ↑  12

ΣFy = 0: RA + RB + RC − wL = 0 1 5 − wL + RB + wL − wL = 0 6 12

RB =

3 wL ↑  4

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.