BeerMOM_ISM_C11.pdf

CHAPTER 11

PROBLEM 11.1 Determine the modulus of resilience for each of the following grades of structural steel: (a) ASTM A709 Grade 50:  Y  50 ksi (b) ASTM A913 Grade 65:  Y  65 ksi (c) ASTM A709 Grade 100:  Y  100 ksi

SOLUTION E  29  106 psi for all three steels given.

Structural steel: (a)

 Y  50 ksi  50  103 psi uY 

(b)

2E



(50  103 )2 (2)(29  106 )

uY  43.1 in.  lb/in 3 

 Y  65 ksi  65  103 psi uY 

(c)

 Y2

 Y2 2E



(65  106 ) 2 (2)(29  106 )

uY  72.8 in.  lb/in 3 

 Y  100 ksi  100  103 psi uY 

 Y2 2E



(100  103 ) 2 (2)(29  106 )

uY  172.4 in.  lb/in 3 

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PROBLEM 11.2 Determine the modulus of resilience for each of the following aluminum alloys: (a) 1100-H14:

E  70 GPa  Y  55 MPa

(b) 2014-T6:

E  72 GPa  Y  220 MPa

(c) 6061-T6:

E  69 GPa

 Y  150 MPa

SOLUTION Aluminum alloys: (a)

E  70  109 Pa  Y  55  106 Pa uY 

 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  Y  220  106 Pa uY 

 Y2 2E



(220  106 ) 2  336  103 N  m/m3 9 (2)(72  10 ) uY  336 kJ/m3 

(c)

E  69  109 Pa  Y  150  106 Pa uY 

 Y2 2E



(150  106 )2  163.0  103 N  m/m3 (2)(69  109 ) uY  163.0 kJ/m3 

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PROBLEM 11.3 Determine the modulus of resilience for each of the following metals: (a) Stainless steel AISI 302 (annealed):

E  190 GPa  Y  260 MPa

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

E  190 GPa  Y  520 MPa

(c) Malleable cast iron:

E  165 GPa  Y  230 MPa

SOLUTION (a)

E  190  109 Pa,  Y  260  106 Pa uY 

 Y2 2E



(260  106 ) 2  177.9  103 N  m/m3 (2)(190  109 ) uY  177.9 kJ/m3 

(b)

E  190  109 Pa,  Y  520  106 Pa uY 

 Y2 2E



(520  106 ) 2  712  103 N  m/m3 9 (2)(190  10 ) uY  712 kJ/m3 

(c)

E  165  109 Pa,  Y  230  106 Pa uY 

 Y2 2E



(230  106 ) 2  160.3  103 N  m/m3 9 (2)(165  10 ) uY  160.3 kJ/m3 

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PROBLEM 11.4 Determine the modulus of resilience for each of the following alloys: (a ) Titanium:

E  16.5  106 psi  Y  120 ksi

(b) Magnesium:

E  6.5  106 psi

(c) Cupronickel (annealed):

6

E  20  10 psi

 Y  29 ksi  Y  16 ksi

SOLUTION (a)

E  16.5  106 psi,  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,  Y  29  103 psi uY 

(c)

 Y2

 Y2 2E



(29  103 )2 (2)(6.5  106 )

uY  64.7 in.  lb/in 3 

E  20  106 psi,  Y  16  103 psi uY 

 Y2 2E



(16  103 ) 2 (2)(20  106 )

uY  6.40 in.  lb/in 3 

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␴ (ksi)

PROBLEM 11.5

100

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.

80 60 40 20 0

0.021 0.002

0.2

0.25

⑀

SOLUTION

(a)

 Y  E Y uY 

 Y2 2E



1 2 1 E 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 

 (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  20.0 in.  kip/in 3 

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␴ (MPa)

PROBLEM 11.6

600

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.

450

300

150

0.14

0.006

0.18

⑀

SOLUTION (a)

 Y  E Y uY 

 Y2 2E



1 2 1 E 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)  90.0  106 N/m 2.

modulus of toughness  90.0  106 J/m3  90.0 MJ/m3 

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PROBLEM 11.7

P (kN) P

400

400 mm d

300 200 100

P' 50

The load-deformation diagram shown has been drawn from data obtained during a tensile test of a specimen of an aluminum alloy. Knowing that the cross-sectional area of the specimen was 600 mm2 and that the deformation was measured using a 400-mm gage length, determine by approximate means (a) the modulus of resilience of the alloy, (b) the modulus of toughness of the alloy.

d (mm)

2.8

SOLUTION

   

P P  A 600  106 m 2

 L



 400 mm

Draw    curve:

(a)

Modules of resilience: (shaded area) uY 

1 (500 MPa)(0.007) 2 uY  1.750 MJ/m3 

(b)

Modules of toughness: (total area under    curve)  1.750 MJ/m3  (500 MPa)(0.125  0.007) 

1 (6.33  500)(0.125  0.007) 2

 1.750 MJ/m3  59 MJ/m3  10.46 MJ/m3  71.2 MJ/m3



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PROBLEM 11.8

P (kips) P 20 15 18 in. 10

d

5

P' 0.36

3.2

4

The load-deformation diagram shown has been drawn from data obtained during a tensile test of a 5 -in.-diameter rod of structural steel. Knowing 8 that the deformation was measured using an 18-in. gage length, determine by approximate means (a) the modulus of resilience of the steel, (b) the modulus of toughness of the steel.

d (in.)

0.025

SOLUTION 5 -in.-diameter rod: 3 A

   

 5

2

2    0.3068 in 4 8

P 12.5 kips   40 ksi A 0.3068 in 2

 L



0.025 in.  1.389  103 18 in.

 Y  40 ksi Draw  - curve:

(a)

Mod. of resilience: (shaded area) uY 

1 1  Y  Y  (40  103 psi)(1.389  103 ) 2 2 uY  28.0 in.  lb/in 3 

(b)

Mod. of toughness: (total area under  - curve)  28 in.  lb/in 2  (40 ksi)(0.02  0.0014) 

1 (40  65)(0.1778  0.02)  (62.5)(0.222  0.1778) 2

 28  744  9860  2760  13,390 in.  lb/in 2 Modulus of toughness  13.40 in.  kips/in 3 

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PROBLEM 11.9 C

3 ft

3 4

in.

5 8

in.

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.

B 2 ft A P

SOLUTION P  8 kips, E  29  103 ksi A

 4

d 2 , V  AL,  

2 P , u A 2E

U  uV Portion

d(in.)

A(in2)

L(in.)

V(in3)

 (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



(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

u BC  5.65 in.  lb/in 3 

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PROBLEM 11.10

20-mm diameter 16-mm diameter

B A

C P

1.2 m 2m

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.

0.8 m

SOLUTION AAB 

 4

(20) 2  314.16 mm 2  314.16  106 m 2

ABC 

 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

 AB  u AB 

 BC  uBC 

U  12.18 J 

P 25  103   79.58  106 Pa AAB 314.16  106 2  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  BC

2E



(124.28  106 ) 2  38.6  103 (2)(200  109 )

uBC  38.6 kJ/m3 

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PROBLEM 11.11 A B

E

F

D

P

C 30 in. 48 in.

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)

 4

d2 

 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 :

 

8000  4324 psi, 1.85

u

2 2E



(4324) 2 (2)(10.6  106 ) u  0.882 in.  lb/in 3 

EF :



8000  18,108 psi, 0.4418

u

2 2E



(18,108) 2 (2)(29  106 )

u  5.65 in.  lb/in 3 

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PROBLEM 11.12

0.5 m B A

C

20 mm

D

E P 1.25 m

5 mm

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  all  85 MPa, determine, for the loading shown, the maximum strain energy that can be acquired by the assembled strips.

SOLUTION Apin 

 4

d2 

 4

(6)2  28.274 mm 2  28.274  106 m 2

 all  85  106 Pa Double shear:

P  2 A  (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 

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PROBLEM 11.13 10-mm diameter B

A

6-mm diameter

a

C P 6m

Rods AB and BC are made of a steel for which the yield strength is 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 any permanent deformation when the length a of rod AB is (a) 2 m, (b) 4 m.

SOLUTION AAB  ABC 

 4

 4

(10) 2  78.54 mm 2  78.54  106 m 2 (6)2  28.274 mm 2  28.274  10 6 m 2

P   Y Amin  (300  106 )(28.274  106 )  8.4822  103 N U 

(a)

a  2m

U 

P2L 2EA

L  a  6  2  4m

(8.4822  103 ) 2 (2) (8.4822  103 )2 (4)  (2)(200  109 )(78.54  106 ) (2)(200  109 )(28.274  106 )

 4.5803  25.4466  30.0 N  m  30.0 J (b)

a  4m

U 



L  a  6  4  2m

(8.4822  103 ) 2 (4) (8.4822  103 )2 (2)  9 6 (2)(200  10 )(78.54  10 ) (2)(200  109 )(28.274  106 )

 9.1606  12.7233  21.9 N  m  21.9 J



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PROBLEM 11.14 B

C

P 1.8 m

Rod BC is made of a steel for which the yield strength is  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.

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

 Y2

(300  106 ) 2 (2)(200  109 )



2E

 225  103 J/m3 UY  ALuY A

UY 60  LuY (1.8)(225  103 )  148.148  106 m 2

A d

 4

d2

4A





(4)(148.148  106 )

 3

 13.73  10 m d  13.73 mm 

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PROBLEM 11.15 18-mm diameter C

B

12-mm diameter

A

The assembly ABC is made of a steel for which E  200 GPa and  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.

x

900 mm

P

SOLUTION

 Y  320 MPa  320  106 Pa, AAB 

4

2 d AB 



 4

(12) 2  113.097 mm 2  113.097  106 m 2



2 d BC  (18) 2  254.47 mm 2  254.47  106 m 2 4 4  AAB

ABC  Amin



E  200 GPa  200  109 Pa

Force at yielding or allowable axial force. P  PY   Y Amin  (320  106 )(113.097  106 )  36.191  103 N (a)

x  300 mm:

LAB  0.300 m,

UY  U AB  U BC  

LBC  0.600 m

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

2

(36.191  103 ) 2  0.300 0.600   6 6  9  (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 

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PROBLEM 11.16 A

Show by integration that the strain energy of the tapered rod AB is

2c

U c P

B

L

1 P2 L 4 EAmin

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

SOLUTION Radius: r 

cx L

Amin   c 2 A   r2 

 c2

x2 L2 2 L P 2dx P2  U L 2 EA 2E



2 2



2L L

L2 dx  c2 x2

2L



P L 2 E c 2

 1  x   L



P 2 L2 2 EAmin

1  1   2L  L   

U

P2 L  4 EAmin

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1.5 in.

PROBLEM 11.17

2.85 in. 2.55 in. 2.10 in.

P

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  all  22 ksi.

3 in.

A B 4 @ 1.5 in.  6 in.

SOLUTION Amin 

 4

(1.5)2  1.7671 in 2

 all  22,000 psi Pall   all Amin  38,877 lb U



P 2dx P 2  2 EA 2 E

dx

d

2



4

2 P 2 dx  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



2.2625

ò

B A

dx d

2

=

U

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

(2)(38877) 2 (1.13125)  (10.6  106 )

U  102.7 in.  lb 

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PROBLEM 11.18

B 1 2

l

1 2

l

A

P

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.

C D

A l

SOLUTION 2

LBC  LCD 

5 1  l   l  l 2 2 

 Fx  0:



2

Joint C. (equilibrium) 2 2 FBC  FCD  0 5 5

FCD   FBC  Fy  0 : FBC 

Strain energy.

1 1 FBC  FCD  P  0 5 5

5 P 2

FCD  

5 P 2

F 2L 1  2 2  FBC LBC  FCD LCD  2 EA 2 EA  2 2   1  5   5   5   5    P l  P l         2 EA  2   2   2   2    

U 

U  1.398

P 2l  EA

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l P A C

B

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.

A

308

D

SOLUTION Fy  0: 

3 FCD  P  0 2

Fx  0:  FBC  U 



1 FCD  0 2

FCD  

2 P 3

FBC 

1 P 3

F 2L 1 F 2L   2 EA 2 E A

Member

F

L

A

F2L/A

BC

1 P 3

l

A

1 2 P l /A 3

2l

A

8 2 P l /A 3

CD



2 P 3



U 

3P 2l /A 1  P 2l   3 2 E  A 

U  1.5

P 2l  EA

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PROBLEM 11.20

l P

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.

A C

B

A

A

30°

D

SOLUTION Equilibrium of joint C.

  Fy  0: 

3 FCD  P  0 2

FCD  

1 2

 Fx  0:  FBC  FCD  0 FBC 

2 P 3

1 P 3

Equilibrium of joint D.

 Fy  0: FBD 

Strain energy. Member

CD BD

1 3 

2 3

FBD  P

1 F 2L 1 F 2L   A 2 EA 2E

L

A

F2L/A

P

l

A

1 2 P l/A 3

P

2l

A

8 2 P l/A 3

3l

A

3P 2l/A

F

BC

U 

3 FCD  0 2

P

4.732P2l/A



U 

1  P 2l   4.732  2E  A 

U  2.37

P 2l  EA

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PROBLEM 11.21

P C 2A

B

2A A

3 4

l

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.

D

l

SOLUTION 3 3 FCD  FCB  0 5 5

Fx  0: FCB  FCD

Fy  0:  P  2 

4 FCD  0 5

5 FCB  FCD   P 8 3 FCD  0 5 3  FBD  FCD  0 5 3 5 3   P P 5 8 8

Fx  0:  FBD 

FBD

F 2L  2 EA 1  179   2 E  384

U 



1 F 2L  A 2E 2  Pl  A 

179 P 2l 768 EA

U  0.233

Member

F

L

A

F 2 L /A

CB

5  P 8

5 l 6

2A

125 2 P l /A 768

CD

5  P 8

5 l 6

2A

125 2 P l /A 768

BD

3 P 8

l

A

9 2 P l /A 64

P 2l  EA

179 2 P l /A 384

 

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PROBLEM 11.22

80 kN C

2500 mm2 2000 mm2

30 kN

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.

2.4 m D

B

2.2 m

1m

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 Dy  83.636 kN 

 Fy  0: Dy  By  80  0 83.636  By  80  0

By  3.636 kN 

Member forces. FBC  By

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   9 3 2 EABC 2 EACD (2)(72  10 )(2  10 ) (2)(72  109 )(2.5  103 )

 0.510 J  59.290 J 

U  59.8 J 

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PROBLEM 11.23

B 3 in2 2.5 ft

2 in2

C

24 kips

2.5 ft D

Each member of the truss shown is made of aluminum and has the cross-sectional area shown. Using E  10.5  106 psi, determine the strain energy of the truss for the loading shown.

40 kips

5 in2 6 ft

SOLUTION 62  2.52  6.5 ft  78 in.

LBC  LCD  Joint C:

6 6 FBC  FCD  24  0 6.5 6.5 2.5 2.5 Fy  0:  FBC  FCD  40  0 6.5 6.5

Fx  0: 

(1) (2)

Solving (1) and (2) simultaneously, FBC  65 kips FCD  39 kips Joint D Fy  0: FBD  U 

2.5 FCD  0 6.5

FBD  15 kips

F 2L 1 F 2L   2 EA 2 E A

Member

F (103 lb)

L(in.)

A(in 2 )

BC

65

78

3

109.85

BD

15

60

2

6.75

CD

39

78

5

23.73



F 2 L /A (109 lb 2 /in.)

140.33 U 

140.33  109  6682 lb  in.  6.68 kip  in.  (2)(10.5  106 )

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PROBLEM 11.24

w B A

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

L

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  8 EI 



L

0



2

L

0

1 2  2 wv  dv  

w2 v 5 v dv  8 EI 5

L

4

w2 L5 40EI

0

U 

w2 L5  40 EI

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PROBLEM 11.25

P D

A a

B

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

L

SOLUTION

M D  0: aP  LRB  0 RB  

aP aP   L L

Over portion AD: M   Px a

U AD   0

M2 1 a 2 2 P 2 x3 dx  P x dx   0 2 EI 2EI 2 EI 3

a

 0

P 2a3 6 EI

Over portion DB: M 

aP v L

L

U DB   0

M2 1 L a2P2 2 dv  v dv  2 EI 2 EI 0 L2

P 2a 2 L 2 P 2a 2 v3 v dv    2EIL2 0 2 EIL2 3

Total:

U  U AD  U DB

L

 0

P 2a 2 L 6EI U 

P 2a 2 (a  L)  6 EI

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P a

PROBLEM 11.26

P

D

E

a

A

B

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

L

SOLUTION Symmetric beam and loading:

RA  RB Fy  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

x 2 dx 

P 2 x3 2 EI 3

a

 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

P2 a2 (3L  4a )  6 EI

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PROBLEM 11.27

M0 A

B D a

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

b L

SOLUTION

A to D:

M B  0: RA L  M 0  0

RA 

M0  L

M A  0:

RB L  M 0  0

RB 

M0  L

M J  0:

M0x M 0 L

M  U AD  D to B:

a 0

M 02 M 2 dx  2 EI 2 EIL2

M K  0: M  M U DB 

Total:



M0x L



a 0

x 2 dx 

M 02 a3 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 02b3 6 EIL2 U

M 02 (a3  b3 ) 6 EIL2



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PROBLEM 11.28

8 kips D A

B S8 3 18.4 6 ft

Using E  29  106 psi, determine the strain energy due to bending for the steel beam and loading shown. (Neglect the effect of shearing stresses.)

3 ft

SOLUTION

M D  0:  RA L  aP  0 RA 

aP  L

M 

Over portion AD:

aP x L

L

U AD   0

Over portion DB:

2

M2 1 L  aP  dx    x  dx 2EI 2 EI 0  L 



P 2a 2 L 2  x dx 2EIL2 0



P 2a 2 L 6 EI

M   Pv a

U BD   0

M2 1 a 2 2 P 2a3 dv  P x dx   0 2 EI 2 EI 6EI P 2a 2 ( a  L) 6 EI

Total:

U  U AD  U DB 

Data:

P  8000 lb, L  6 ft  72 in.,

a  3 ft  36 in., E  29  106 psi

I  57.5 in 4 U 

(8000) 2 (36) 2 (72  36) (6)(29  106 )(57.5) U  895 in.  lb 

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2 kips A

B

C

PROBLEM 11.29

1.5 in.

2 kips D

D

3 in.

Using E  29  106 psi, determine the strain energy due to bending for the steel beam and loading shown. (Neglect the effect of shearing stresses.)

60 in. 15 in.

15 in.

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 

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PROBLEM 11.30

180 kN W360  64

C

A 2.4 m

B

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

2.4 m 4.8 m

SOLUTION Over portion AC,

M U AC 

1 Px 2



L/2 0

M2 P2 dx  2 EI 8 EI

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  1048J 

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80 kN D

W310  74

E

B

A 1.6 m

PROBLEM 11.31

80 kN

1.6 m

1.6 m

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

4.8 m

SOLUTION Over portion AD,

M  Px U AD  



a 0

M2 1 dx  2 EI 2 EI

P 2 x3 2 EI 3

a

 0



( Px)2 dx

M  Pa

( Pa)2 a P 2a3  2EI 2 EI

U EB  U AD 

By symmetry,

0

P 2 a3 6 EI

Over portion DE, U DE 

a

P 2a3 6EI

U  U AD  U DE  U EB  Data:

5 P 2a3 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 

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PROBLEM 11.32 w B

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

A

umax 

L

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 8 EI



L 0

( L2 x 2  2 Lx3  x 4 )dx L

w2  L2 x3 2 Lx 4 x5       8 EI  3 4 5 0 

w2 L5 8 EI



w2 L5 240 EI

M max 

 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  max

2E



w2 L4 c 2 128 EI 2



3 1 8 LI 8 L 12 bd U   2 umax 15c 2 15  d2 





8 8 Lbd  V 45 45 umax 

45 U  8 V

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TA 5 300 N · m

In the assembly shown, torques TA and TB are exerted on disks A and B, respectively. Knowing that both shafts are solid and made of aluminum (G  73 GPa), determine the total strain energy acquired by the assembly.

0.9 m

30 mm TB 5 400 N · m

PROBLEM 11.33

A

B 0.75 m

46 mm C

SOLUTION Over portion AB:

TAB  TA  300 N  m J AB 

 2

c4 

  30 

4

9 4 3 4    79.52  10 mm  79.52  10 m 2 2 

LAB  0.9 m U AB 

2 TAB LAB (300) 2 (0.9)  2GJ AB (2)(73  109 )(79.52  109 )

 6.977 J

Over portion BC:

TBC  TA  TB  300  400  700 N  m, LBC  0.75 m

J BC  U BC 

Total:

  46 

4

9 4 3 4    439.57  10 mm  439.57  10 m 2 2 

2 TBC LBC (700)2 (0.75)   5.726 J 2GJ BC (2)(73  109 )(439.57  109 )

U  U AB  U BC  6.977  5.726  12.70 J



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PROBLEM 11.34 The design specifications for the steel shaft AB require that the shaft acquire a strain energy of 400 in.  lb as the 25-kip  in. torque is applied. Using G  11.2  106 psi, determine (a) the largest inner diameter of the shaft that can be used, (b) the corresponding maximum shearing stress in the shaft.

36 in. B

A

2.5 in. 25 kip · in.

SOLUTION U  400 in.  lb T  25 kip  in.  25  103 lb  in.

L  48 in. U 

J  J 

But (a)

di4  d 04 

32



 2.54 

T 2L 2GJ

(25  103 ) 2 (48) T 2L   3.3482 in 4 2GU (2)(11.2  106 )(400) 4   d 0 

4  4 d   d 0  di4     i    2  2   2   32





J

32



(3.3482)

 4.9580 in 4 di  1.492 in.  (b)

  

Tc0 J (25  103 )(1.25) 2.5112

 9.33  103 psi

  9.33 ksi 

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

A 2c

U c L

T

7 T 2L 48 GJ min

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

B

SOLUTION r J

cx L



r4 

 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

 2

c4

T 2 dx   c4 4  x  2G  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

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PROBLEM 11.36

y

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

20 MPa

75 MPa

σz

100 MPa

z x

SOLUTION 1 2  60 MPa

 ave  (100  20) x  y 2

100  20 2  40 MPa 

 xy  75 MPa 2

x  y  2 R     xy 2    402  752  85 MPa  a   ave  R  145 MPa

 b   ave  R  25 MPa

c   z ( a   b )2  ( b   c ) 2  ( c   a ) 2  2 Y2 (145  25) 2  (25   z ) 2  ( z  145) 2  (2)(160) 2 28,900  (625  50 z   z2 )  ( z2  290 z  21,025)  51,200 2 z2  240 z  650  0

240  2402  (4)(2)(650)  60  62.65 (2)(2)  z  122.65 MPa, 2.65 MPa

z 

2.65 MPa <  z < 122.65 MPa 

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PROBLEM 11.37

y

The state of stress shown occurs in a machine component made of a brass for which  Y  160 MPa. Using the maximum-distortion-energy criterion, determine whether yield occurs when (a)  z  45 MPa, (b)  z  45 MPa.

20 MPa

75 MPa

σz

100 MPa

z x

SOLUTION 1 2  60 MPa

 ave  (100  20) x  y 2

100  20 2  40 MPa 

 xy  75 MPa 2

x  y  2 R     xy 2    402  752  85 MPa

 a   ave  R  145 MPa  b   ave  R  25 MPa c   z ?

( a   b ) 2  ( b   c )2  ( c   a )2  2 Y2

(a)

 c   z  45 MPa ?

(145  25) 2  (25  45) 2  (45  145) 2  2(160)2  51,200 28,900  4900  10,000  43,800  51,200

(b)

(No yield.) 

 c   z  45 MPa ?

(145  25) 2  (25  45)2  (45  145)2  51,200 28,900  400  36,100  65,400  51,200

(Yield occurs.) 

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PROBLEM 11.38

y

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

σy

8 ksi z

14 ksi

x

SOLUTION 1 2  4 ksi x z 8  0  2 2  4 ksi  xz  14 ksi

 ave  (0  8)

2

 z  2 R  x    xz 2    42  142  14.56 ksi  a   ave  R  18.56 ksi  b   ave  R  10.56 ksi c   y    ( a   b ) 2  ( b   c )2  ( c   a )2  2  Y   F .S .   65  (18.56  10.56)2  (10.56   y ) 2  ( y  18.56)2  2    2.2 

2

2

847.97  (111.51  21.12 y   y2 )  ( y2  37.12 y  344.47)  1745.87 2 y2  16 y  441.92  0 16  162  (4)(2)(441.92) (2)(2)  4  15.39

y 

 y  19.39 ksi, 11.39 ksi 11.39 ksi   y  19.39 ksi 

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PROBLEM 11.39

y

The state of stress shown occurs in a machine component made of a grade of steel for which  Y  65 ksi. Using the maximum-distortionenergy criterion, determine the factor of safety associated with the yield strength when (a)  y  16 ksi, (b)  y  16 ksi.

σy

8 ksi z

x

14 ksi

SOLUTION 1 2  4 ksi x z 8  0  2 2  4 ksi  xz  14 ksi

 ave  (0  8)

2

 z  2   xz R  x  2    42  142  14.56 ksi  a   ave  R  18.56  b   ave  R  10.56 c   y

   ( a   b ) 2  ( b   c )2  ( c   a )2  2  Y   F .S .  (a)

2

 c   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 

 c   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 

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PROBLEM 11.40

b

M0

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

d

B

A L

SOLUTION Reactions:

M0 M , RB  0  L L

RA 

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 



 xy 

3V  y2  1  2  2 A c 

0



For shear,

u

2  xy

2G







2

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

9V 2 8GA2

U 2  u dv 

L

  0

9M 02 2 2

8Gb d L



1 d 2

c

 y2 y4  1  2 2  4  c c   2 c L c  9M 0 b y2 y4    ub dy dx  1 2   dy dx c c 2 c 4  8Gb 2 d 2 L2 0 c 

 

c

L 0

9 M 02 2 y3 1 y5    dx  y   3 c 2 5 c 4  8Gb d 2 L2  c



L 0

4 2   2c  3 c  5 c  dx  

9M 02

6 M 02 c 3 M 02  16    c L   5 Gb d 2 L 5 Gb dL 8Gb d 2 L2  15 

Total: with

U  U1  U 2  I U

1 bd3 12 2 M 02 L Eb d 3



M 02 L 3 M 02  6 EI 5 Gb dL

3 M 02 5 Gb dL

U

2 M 02 L  3Ed 2  1  3  Eb d  10GL2 

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PROBLEM 11.41* B R2

R1

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.

Q

A A

B A

L Q

(a)

(b)

SOLUTION Fx  0:  (2 rL)  Q  0

 2

Q2 2G 8 2 r 2 L2 G Q2 U  u dV  2 2 8 GL u







Q 2 rL

Q2 4 GL2

L

  0

R2 R1



dV Q2  r 2 8 2 GL2

dr Q2 dx  r 4 GL2

L

R2

0

R1

 

2 r dr dx r2

  lnr  dx L

0

R2 R1

U

R Q2 ln 2  4 GL R1

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PROBLEM 11.42

V0

A

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.

B D 1.2 m

SOLUTION 1 2 1 mv0  (5)(6) 2  90 J 2 2 2 P L ( A max ) 2 L Um  m  2 EA 2 EA 2 EU m (2)(200  109 )(90)   480  106 m 2 A 2  max L (250  106 )2 (1.2) Um 

 4

d2  A

d

4A





(4)(480  106 )



 24.7  103 m

d  24.7 mm 

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B

A v0 E C

D 3.5 ft

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  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   Y A. Corresponding strain energy: U AB 

2 FAB LAB  2 A2 L  Y2 AL  Y  2 EAAB 2EA 2E

U CD 

2 FCD LCD  2 AL  Y 2 EACD 2E

Total:

U m  U AB  U CD  Um 

Solving for v02 ,

v02 

 Y  50  103 psi

 4

d2 

 7

2

2    0.60132 in , 48

L  3.5 ft  42 in. v0 

1 2 1W 2 mv0  v0 2 2 g

2 g Y2 AL EW

g  32.17 ft/s 2  386 in./s 2 A

E

2 gU m 2 g Y2 AL  W EW

v0  Data:

 Y2 AL

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 

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B

A v0 E C

D 3.5 ft

PROBLEM 11.44 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  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   Y A. Corresponding strain energy: U AB 

2 FAB LAB  Y2 A2 L  Y2 AL   2 EAAB 2 EA 2E

U CD  U AB 

 Y2 AL 2E

U m  U AB  U CD 

 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 g Y2 AL v02 ( F .S .) E

g  32.17 ft/ sec2  386 in./ sec 2 ,

A

 4

d2 

 7

 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 

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PROBLEM 11.45

A 2m

40-mm diameter

B 1.5 m D

The 35-kg collar D is released from rest in the position shown and is stopped by a plate attached at end C of the vertical rod ABC. Knowing that E  200 GPa for both portions of the rod, determine the distance h for which the maximum stress in the rod is 250 MPa.

30-mm diameter

m h C

SOLUTION Portion BC has smaller cross section, thus Pm   all ABC  (250 MPa)

 4

(0.030 m) 2  176.7 kN

Maximum strain energy (for Pm  176.7 kN ) Um

   Pm2 Li Pm2 Li (176.7 kN) 2  2m 1.5 m         2 Ai E 2 E Ai 2(200 GPa)  (0.040 m)2 (0.030 m)2  4 4 

U m  78.06  103[1591.6  2122.1]  289.9 J

Max. deflection:

1 Pm  m  U m 2

1 (176.7 kN) m  289.9 J 2  m  3.28  103 m  3.28 mm

Work of weight  U m W (h   m )  U m (mg )(h   m )  U m (35 kg)(9.81 m/s 2 )(h   m )  289.9 J

h   m  0.8443 m  844.3 mm h  844.3   m  844.3  3.28

h  841 mm 

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PROBLEM 11.46

A 2m

40-mm diameter

B 1.5 m D

30-mm diameter

The 15-kg collar D is released from rest in the position shown and is stopped by a plate attached at end C of the vertical rod ABC. Knowing that E  200 GPa for both portions of the rod, determine (a) the maximum deflection of end C, (b) the equivalent static load, (c) the maximum stress that occurs in the rod.

m h C

SOLUTION

E  200 GPa

(a)

m

   Pm Li Pm Li Pm 2m 1.5 m         Ai E E Ai 200 GPa   (0.04 m)2  (0.03 m) 2  4 4 

 m  18.57  109 Pm Pm  53.85  106 m Strain energy:

Um 

1 1 2 Pm m  (53.85  106 ) m 2 2

Weight falls distance of h   m Work of weight  strain energy W (h   m ) 



1 2 53.85  106  m 2



(1)

W  mg  (35 kg)(9.81 m/s)  343.4 N h  0.5 m

For Eq. (1):

343.4(0.5   m )  26.93  106 2m  m  2.531  103 m  2.53 mm 

Solve quadratic: (b)

Pm  53.85  106 m  (53.85  106 )(2.531  103 m)

(c)

m 

Pm 136.3 kN   192.82 MPa  ABC (0.030 m) 2 4

Pm  136.3 kN 

 m  192.8 MPa 

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A

C

PROBLEM 11.47

E

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  all  180 MPa and E  200 GPa, determine the largest allowable distance h.

2.5 m

G h B

D

F

SOLUTION Let  m be the maximum elongation.

m 

 AB L E



 CD L E



 EF L E

 AB   CD   EF  180 MPa  180  106 Pa E  200  109 Pa

L  2.5 m

For each rod, ACD 

Rod CD:

U CD 

 4

m 

(180  106 )(2.5)  0.00225 m 200  109

U 

2 Fm2 L ( EA m /L) 2 L EA m   2 EA 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 

Rods AB and EF:

U AB  U EF 

 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

Total strain energy: Work of falling collar:

U m  mg (h   m )  (48)(9.81)(h   m ) Equating,

(48)(9.81)(h   m )  134.97

h   m  0.28662 m

h  0.28662  0.00225  0.285 m

h  285 mm 

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PROBLEM 11.48

A

v0 C 7.5 ft

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.

B

W5  16

SOLUTION I x  21.4 in 4 ,

W5  16:

S x  8.55 in 3

 m  18 ksi

Maximum stress:

Maximum bending moment: M m   m S x  (18 ksi)(8.55 in 3 )  153.9 kip  in.

Pm L  M m

Equivalent force: Pm 

M m 153.9 kip  in.   1.71 kips  1710 lb L 90 in.

From Appendix D, ym  Um 

Kinetic energy:

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

T 

1W 2 v0 2 g

T 

25 v02  0.3882v02 ft  lb (2)(32.2)

T  Um 0.3882v02  47.706 v0  11.09 ft/s 

Maximum speed.

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PROBLEM 11.49

A

v0 C

Solve Prob. 11.48, assuming that the post AB has rotated 90 about its longitudinal axis. PROBLEM 11.48 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.

7.5 ft

B

W5  16

SOLUTION I y  7.51 in 4 ,

W5  16:

S y  3.00 in 3

 m  18 ksi

Maximum stress:

Maximum bending moment: M m   m S y  (18 ksi)(3.00 in 3 )  54.0 kip  in. Pm L  M m

Equivalent force: Pm 

Mm 54.0 kip  in.   0.600 kips  600 lb L 90 in.

From Appendix D, ym  Um 

Kinetic energy:

Equating,

Pm L3 600)(90)3   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

T 

1W 2 v0 2 g

T 

25 v02  0.3882v02 ft  lb (2)(32.2)

T  Um 0.3882v02  16.736 v0  6.57 ft/s 

Maximum speed.

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PROBLEM 11.50

0.9 m

An aluminum tube having the cross section shown is struck squarely in its midsection by a 6-kg block moving horizontally with a speed of 2 m/s. Using E  70 GPa, determine (a) the equivalent static load, (b) the maximum stress in the beam, (c) the maximum deflection at the midpoint C of the beam.

B

0.9 m

t = 10 mm 80 mm

C 100 mm

100 mm

v0

A

SOLUTION T 

Kinetic energy:

1 2 1 mv0  (6 kg)(2 m/s) 2  12 J 2 2

Moment of inertia. Aluminium E  70 GPa

I aa 

1 [80  1003  60  803 ] 12

 4.1067  106 mm 4  4.1067  106 m 4

From Appendix D: ym  Um 

(a)

Um  T :

Pm L3 48EI 1 P 2 L3 Pm ym  m 2 96 EI

Pm2 (1.8 m)3  12 J 96(70 GPa)(4.1067  106 m 4 )

Pm  7535.5 N

Pm  7.54 kN  (b)

1 1 Pm L  (7535.5 N)(1.8 m)  3391 N  m 4 4 M c (3391 N  m)(0.050 m)  m   41.28 MPa I 4.1067  106 m 4

Mm 

m

 m  41.3 MPa  (c)

ym 

3

3

Pm L (7535.5 N)(1.8 m)   3.184  103 m 48EI 48(70 GPa)(4.1067  106 m 4 )

ym  3.18 mm 

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PROBLEM 11.51 0.9 m B

0.9 m

t = 10 mm 80 mm

C 100 mm

100 mm

v0

A

Solve Prob. 11.50, assuming that the tube has been replaced by a solid aluminum bar with the same outside dimensions as the tube. PROBLEM 11.50 An aluminum tube having the cross section shown is struck squarely in its midsection by a 6-kg block moving horizontally with a speed of 2 m/s. Using E  70 GPa, determine (a) the equivalent static load, (b) the maximum stress in the beam, (c) the maximum deflection at the midpoint C of the beam.

SOLUTION For solid aluminum bar, Ia 

1 80  1003  6.667  106 m 4 12

Follow solution of Prob. 10.50: T  12 J ym 

Pm L2 1 P 2 L3 ; U m  Pm ym  m 48EI 2 96EI

Um  T :

(a)

Pm2 (1.8 m)3  12 J 96(70 GPa)(6.667  106 m 4 )

Pm  9601 N Pm  9.60 kN 

(b)

1 1 Pm L  (9601 N)(1.8 m)  4320.5 N  m 4 4 M c (4320.5 N  m)(0.05 m)  m   32.40 MPa I 6.667  106 m 4

Mm 

m

 m  32.4 MPa  (c)

ym 

Pm L3 (9601 N)(1.8 m)3   2.500  103 m 48EI 48(70 GPa)(6.667  106 m 4 )

ym  2.50 mm 

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D

PROBLEM 11.52 2 kg

40 mm A

B 0.6 m

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

 d

4

4

  16      3.2170  103 mm 4   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  3 (0.6) 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)

m 

|M m |  83.8 N  m 

|M m | c (83.8)(8  103 )   208  106 Pa 9 I 3.2170  10

 m  208 MPa 

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D

m

40 mm h B

A

E

60 mm

0.4 m

PROBLEM 11.53 The 10-kg block D is dropped from a height h  450 mm onto the aluminum beam AB. Knowing that E  70 GPa, determine (a) the maximum deflection of point E, (b) the maximum stress in the beam.

1.2 m

SOLUTION

1 (40)(60)3  720  103 mm 4  720  109 m 4 12 1 S  (40)(60) 2  24  103 mm3  24  106 m3 6 I 

Appendix D:

(a)

ym  0.5644  106 Pm

Um 

or

ym 

Pma 2b 2 3EIL

ym 

Pm (0.4 m)2 (0.8 m) 2 3(70 GPa)(720  109 )(1.2 m)

Pm  1.772  106 ym

1 1 Pm ym  (1.772  106 ym ) ym  885.9  103 ym2 2 2

Work of weight  W (h  ym )  mg (h  ym ) (10 kg)(9.81 m/s 2 )(0.45 m  ym )  885.9  103 ym2 885.9  103 ym2  98.1ym  44.145  0 ym  7.114  103 m

Solve quadratic: (b)

ym  7.11 mm 

Pm  1.772  106 ym  1.772  106 (7.114  103 m)

Pm  12.61 kN M m  M E  Pm

ab (0.4 m)(0.8 m)  (12.61 kN) L 1.2 m

M m  3.363 kN  m

m  Mm

c M 3.363 kN  m  m  I S 24  106 m3

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D

PROBLEM 11.54

4 lb

1.5 in.

B

C

A

2 ft

The 4-lb block D is dropped from the position shown onto the end of a 5 -in.-diameter rod. Knowing that E  29  106 psi, determine (a) the 8 maximum deflection at point A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the rod.

2 ft

SOLUTION Use Appendix D.

y2 

Pm L3 3EI

y1  L B 

Pm L3 3EI

2Pm L3 3EI

ym 

ym  y1  y2 

Um 

1 1  3EI Pm ym   3 2 2  2L

2 Pm L3 3EI

3EI 2  ym  ym  4 L3 

W ( h  ym )  U m W ( h  ym ) 

3EI 2 ym 4 L3

(1)

Substitute given data: (4 lb)(1.5 in.  ym ) 

(a)

6  4 ym  11.784 ym2

Solve quadratic: (b)

3(29  106 psi)(7.49  103 in 4 ) 2 ym 4(24 in.)3

Pm 

ym  0.9032 in.

ym  0.903 in. 

3EI 3(29  106 psi)(7.49  103 m 4 ) y  (0.9032 in.)  21.29 lb m 2L3 2(24 in.)3

M m  M E  PmL  (21.29 lb)(24 in.)  510.96 lb  in.

M m  511 lb  in.  (c)

m 

M mc M 510.96 lb  in.  m  I S 23.97  103 in 3

 m  21.3 ksi 

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PROBLEM 11.55

A

2.65 in.

20 in.

B C 9.5 ft

2.5 ft

16 in.

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)  y ym  71.598 ym m (114)2 (114  30) L2 (a  L) 1 U m  Pm ym  35.799 ym2 2 Pm 

 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)  97,535 lb  in. (b)

m 

Mm c I



(97,535)(1.325)  5210 psi 24.813

 m  5.21 ksi 

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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

W h D A

 2h  ym  yst 1  1    yst  

B ym D'

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. D

2 kg

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.

40 mm A

B 0.6 m

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 

where

yst 

W k  2h  ym  yst 1  1     yst  

2 yst  4 yst2  8 yst h 2

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PROBLEM 11.56 (Continued)

For Problem 11.52,

W  mg  (2)(9.81)  19.62 N 9

E  200  10 Pa

I

  16 

4

 3.217  103 mm 4  3.217  109 m 4   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 

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PROBLEM 11.57 W

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.

h D A

B ym D'

D

2 kg

40 mm A

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.

B 0.6 m

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 kym  Wh (2) approximate 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 )

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PROBLEM 11.57 (Continued)

2W ym2  k h

But



from Eq. (2).

ym  ym ym2 y   m  ym h ( ym  ym ) 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

 d

4

4

  10      3.217  103 mm 4   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

relative error  0.152  ym  0.166  2h



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PROBLEM 11.58

P D

A

B

a

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

b L

SOLUTION Pb Pa , RB  L L

Reactions:

RA 

Over AD:

M  RA x 

Pbx L

M2 P 2b2 dx  0 2 EI 2 EIL2 P 2b 2 a3  6 EIL2

U AD 

Over DB:



a

M  RB v 

Total:



b

U  U AB  U BC  1 P D  U 2

0

x 2dx

Pav L

M2 P2a2 dv  0 2 EI 2 EIL2 P 2 a 2 b3  6 EIL2

U DB 



a

D 



b 0

v 2dv

P 2 a 2b 2 (a  b) P 2a 2 b 2  6 EIL 6 EIL2

2U P

D 

Pa 2 b 2   3EI

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PROBLEM 11.59

P

A

D B L

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

a

SOLUTION

See solution of Prob. 11.28 for beam of length L with overhang of length a, load at end of overhang. U 

P 2a 2 (a  L ) 6 EI

1 1 P 2a 2 PyD  U : PyD  (a  L) 2 2 6 EI

yD 

Pa 2 (a  L)   3EI

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PROBLEM 11.60

M0 A

B D a

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

b L

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 0 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 M 02b



b



b

0

v 2 dv

6 EIL2

U  U AD  U DB 

Total:



RB 

D 

2U M0

M 02 (a3  b3 ) 6 EIL2

D 

M 0 ( a 3  b3 ) 3EIL2



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M0

B A

PROBLEM 11.61

D

L

a

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

SOLUTION

Reactions: Over portion AB:

M0  L M x M  0 L

RA 

L

U AB   0 

Over portion BD:

RB 

M0  L

M2 M 02 L 2 dx   x dx 2 EI 2 EIL2 0

M 02 L 6EI

M  M 0 U BD 

M 02a 2 EI

U  U AB  U BD

Total:



M 02 ( L  3a) 6 EI

1 M 00  U 2 2U D  M0

D 

M 0 ( L  3a) 3EI



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2EI

PROBLEM 11.62

C

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

EI B

A

P

L/2

L/2

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 P C  0 2

C 

3 P 2 L3 32 EI 2U P

C 

3PL3   16 EI

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PROBLEM 11.63

P EI

EI

C

B

A

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

2EI a

a

a

a

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 8 EI



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 3 3  (2 a )  a   16 EI 24 EI 48 EI  By symmetry, Total:

U CB  U AB 

3 P 2 a3 16 EI

U  U AB  U BC  1 P C  U 2

C 

3 P 2 a3 8 EI 2U P

C 

3Pa3   4 EI

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B A

M0

Using the method of work and energy, determine the slope at point B caused by the couple M0.

EI 2EI

C

L/2

PROBLEM 11.64

L/2

SOLUTION M B  0:  RA L  M 0  0 M  RA X   Over portion AC:

M0 L

M0 X L U AC  U AC 

Over portion CB:

RA  

U CB 

U CB 



L/2

0

M2 dx 2(2 EI )

M 02 4 EIL2



L/2

0

x 2 dx 

1 M 02 96 EI

2

M dx L / 2 2 EI



L

M 02 2

2 EIL



L

L/2

x 2 dx

3 M 02  3  L    L     6 EIL2   2  

 Total:

5 M 02 L 32 EI 2U B  M0

7 M 02 L 48 EI

U  U AC  U CB 

1 M 0 B  U 2

B 

5M 0 L 16 EI



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PROBLEM 11.65

M0 B A

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

EI 2EI

D

L/2

L/2

SOLUTION RA 

Reactions:

M0 , L

RB 

M0  L

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

Strain energy:

U  U AD  U BD 

U 



L/2

0



M0 L

x



0

2 M AD dx  2(2 EI )

2

2(2 EI )

M 02 L 



L/2

dx 



L/2

0



M0 L

v



2 EI



L/2

0

M 0x L M v M  0 L

M 

2 M BD dv 2 EI

2

dv 

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    32 EI EI  96 48  96 EI

1 M 0 D  U : 2

M 02 L

1 1 M 02 L M 0 D  2 32 EI

D 

1 M 0L 16 EI



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PROBLEM 11.66 450 N

L 500 mm C

A

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.

B

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 a3  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



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 P A  U 2

A 

2U (2)(13.4617)   59.8  103 m P 450

 A  59.8 mm  

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PROBLEM 11.67 B

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.

T 60 in. 2 in.

A

D E

T L C 1.5 in.

SOLUTION T is the same for each shaft.

C  1.25 A

U AB 

1 T A 2

U CD 

and

1 T C 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:

LCE  L,



c 2

4 o



 ci4 

 2

c4 

 2

(2) 4  25.133 in 4

co  2 in.,

 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 L  32.4 in. 

0.018416 L  (0.25)(2.3873)

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PROBLEM 11.68 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. (Neglect the strain energy due to the bending of the shafts.)

C 3 in. F

B

4 in.

E

T

8 in. A D

6 in. 5 in.

SOLUTION Work-energy equation: 1 TA A  U 2 2U A  TA

Portion AB of shaft ABC: TAB  TA  750 lb  in. LAB  5  6  11 in. J AB  U AB 

 d

2 TAB LAB (750) 2 (11)   8.892 in.  lb 2GJ AB (2)(11.2  106 )(31.063  103 )

U BC  0

Gear B:

FBE 

Portion DE of shaft DEF:

4

  0.75     31.063  103 in 4 2  2  2  2 

Portion BC of shaft ABC:

Gear E:

4

TB TAB 750    250 lb rB rB 3

TE  rE FBE  (4)(250)  1000 lb  in. U DE  0

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PROBLEM 11.68 (Continued)

Portion EF of shaft DEF:

TEF  TE  1000 lb  in. LEF  8 in. U EF 

Total:

J EF 

 d

4

 31.063  103 in 4 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

A 

2U (2)(20.389)   54.4  103 rad TA 750

 A  3.12 

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PROBLEM 11.69 70 mm

200 mm

TB

B D A

C 300 mm

The 20-mm-diameter steel rod CD is welded to the 20-mmdiameter steel shaft AB as shown. End C of rod CD is touching the rigid surface shown when a couple TB is applied to a disk attached to shaft AB. Knowing that the bearings are self aligning and exert no couples on the shaft, determine the angle of rotation of the disk when TB  400 N  m. Use E  200 GPa and G  77.2 GPa. (Consider the strain energy due to both bending and twisting in shaft AB and to bending in arm CD.)

SOLUTION M AB  0: rCD FC  TB

FC 

TB rCD

400  1333.3 N, 300  103 FD  1333.3 N FC 

Bending of rod CD: I  

 d 

4

  42

  20 

4

  4 2 

 7.854  103 mm 4  7.854  109 m 4 M  FC x U 



LCD

0

( FC x)2 F 2 L3  C CD 2 EI 6 EI

(1333.3) 2 (300  103 )3  5.093 J (6)(200  109 )(7.854  109 )

Bending of shaft ADB: M B  0:  FA LAB  FDb  0

FA 

FDb L AB

M A  0:  FA LAB  FDb  0

FA 

FD a LAB

1 U  2 EI

2 2  a  FDb   FD2a 2b 2 b  FD a  0   dx   0   dx   L 6 EILAB  LAB    AB  

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PROBLEM 11.69 (Continued)

I 

 d

4

9 4 3 4    7.854  10 mm  7.854  10 m 42

LAB  (270  103 ) m

U  Torsion:

Only portion DB carries torque. U 

Total:

(1333.3)2 (70  103 )2 (200  103 )2  0.137 J (6)(200  109 )(7.854  109 )(270  109 ) J  2 J  15.708  109 m 4

TB2 LDB (400)2 (200  103 )   13.194 J 2GJ (2)(77.2  109 )(15.708  109 )

U  5.093  0.137  13.194  18.424 J

1 TB B  U 2 2U (2)(18.424) B   TB 400  92.1  103 rad

 B  5.28 

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PROBLEM 11.70 T'

ds

The thin-walled hollow cylindrical member AB has a noncircular cross section of nonuniform thickness. Using the expression given in Eq. (3.50) of Sec. 3.10 and the expression for the strain-energy density given in Eq. (11.17), show that the angle of twist of member AB is

t

A B x

L

T



TL 4A 2 G

ds t

where ds is the length of a small element of the wall cross section and A is the area enclosed by the center line of the wall cross section.

SOLUTION From Eq. (3.53),



T 2t A

Strain energy density:

u U

 Work of torque:

2 2G





L

0 L 0



T2 8Gt 2 Ꮽ2

ut ds dx T2 8GᏭ2

ds T 2L dx  t 8GᏭ 2

T 2L 1  T  2 8GᏭ 2

ds t

ds t



TL 4GᏭ 2

ds  t

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P

A

3 4

PROBLEM 11.71

B

l C

D

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

l

SOLUTION Members AB and BD are Zero force members. Joint A:

4 5 FAD  P  0 FAD   P 5 4 3 3 Fy  0:  FAC  FAD  0 FAC  P 5 4 Fx  0:

Member

F

L

F2 L

AB

0

0

BD

0

AD

5  P 4

l 3 l 4 5 l 4

CD

P

AC

3 P 4

l 3 l 4



Joint D:

Fx  0:

0 125 2 Pl 64 P2 l 27 2 Pl 64 27 2 Pl 8

4 5  P  FCD  0 5 4 FCD  P

U  



2

F L 1  F 2 L 2 EA 2 EA

1 27 P 2l Work of P  P   U 2 16 EA 2U 27 Pl  P 8 EA

  3.38

Pl   EA

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A

B

3 4

C

PROBLEM 11.72

P

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

l

D l

SOLUTION Members AB, AC and CD are zero force members. 4 5 FBC  0 FBC  P 5 4 3 3 Fy  0:  FBD  FBC  0 FBD   P 5 4 Fx  0: P 

Joint B:

U 

F 2L 1  F 2 L 2 EA 2 EA 

19 P 2l 16 EA

1 P  U 2 2U 19 Pl   P 8 EA

Work of P 

Member

F

L

F2 L

AB

0

0

AC

0

CD

0 5 P 4 3  P 4

l 3 l 4 l 5 l 4 3 l 4

BC BD ∑

0 0 125 2 Pl 64 27 2 Pl 64 19 2 Pl 8   2.38

Pl   EA

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PROBLEM 11.73

20 kips B D

A

C 6 ft

2.5 ft

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 B caused by the application of the 20-kip load.

6 ft

SOLUTION RA  RB  10 kips  LAC  LCD  6 ft  72 in. LBC  2.5 ft  30 in. LAB  LBC 

62  2.52  6.5 ft  78 in.

Equilibrium of joint A. Fy  0:

Fx  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 

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

52,728

AC

24

72

41,472

BC

0

30

0

CD

24

72

41,472

BD

26

78

52,728



188,400 U 

188,400  0.64966 kip  in. (2)(29  103 )(5)

Vertical deflection of point B. 1 P B  U 2 2U (2)(0.64966) B   P 20

 B  0.0650 in.  

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PROBLEM 11.74

60 kN

D

B

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.

0.5 m A

C 1.2 m

1.2 m

SOLUTION M A  0: 2.4 RD  (0.5)(60)  0

Entire truss:

Fy  0: 12.5 

Joint D:

0.5 FCD  0 1.3

Fx  0: 60  FBD 

Fy  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

Fx  0: 30 

Joint B:

1.2 (32.5)  0 1.3 F 2L 1 F 2L U    2 EA 2 E A

Fx  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



E  200  109 Pa U

12.3633  1012  30.908 J (2)(200  109 )

Work-energy: 1 P  U 2



2U (2)(30.908)   1.030  103 m 3 P 60  10

  1.030 mm  

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6 ft A

PROBLEM 11.75

6 ft B

C

2.5 ft

Each member of the truss shown is made of steel and has a cross-sectional area of 5 in2. Using E  29  106 psi, determine the vertical deflection of point C caused by the 15-kip load.

15 kips E

D

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:

Fy  0: 

2.5 FCD  15  0 6.5

Fx  0: 

6 FCD  FBC  0 6.5

Fx  0:

Joint D:

FCD  39 kips FBC  36 kips  Fx  0:

Joint B:

 FAB  FBC  0

6 ( FAD  39)  0 6.5  39 kips

72  FAD

Um  

Strain energy: Member

FAB  36 kips

F 2L 1  F 2 L 2 EA 2 EA

F (kips)

L (in.)

F 2 L (kip 2  in.)

AB

36

72

93,312

BC

36

72

93,312

CD

39

78

118,638

DE

72

72

373,248

BD

0

30

0

AE

0

30

0

AD

39

78

118,638



Data:

3

797,148

E  29  10 ksi A  5 in Um 

2

797,148  2.7488 kip  in. (2)(29  103 )(5)

1 Pm  m  U m 2

m 

2U m (2)(2.7488)  Pm 15

 m  0.366 in.  

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480 mm

PROBLEM 11.76

480 mm A

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.

360 mm C

B 360 mm D 12 kN

SOLUTION FAB  FBD  FDC  FCA

Owing to symmetry,

U AB  U BD  U DC  U CA

U  4U BD  U BC  4

2 2 LBC FBD LBD FBC  2 EABD 2 EABC

Let P be the load at D. 1 P D  U 2

D 

U P

4

FBD 2 LBD FBC 2 LBC  EABD P EABC P

Joint B:

 Fy  0:

Joint D:

Fx  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       D  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 

D 

12  103 200  109

 4



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

 D  1.111 mm  

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PROBLEM 11.77

P M0

B

A

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.

L

SOLUTION From Appendix D, Case 1, yAP 

PL3 3EI

AP  

yAM 

M 0 L2 2 EI

AM 

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 0 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 0 AP  M 0 AM 2 2

U

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



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P M0 B

L/2

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.

C

A

PROBLEM 11.78

L/2

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 

CP 

yBM

 BM

First P, then M0. U  A1  A2  A3 

1 1 PyBP  PyBM  M 0CM 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 0CP  M 0CM 2 2

U

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









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P

P A

B

L/2

C L/2

PROBLEM 11.79 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. 11.2A 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,

 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

 CB   BB   B 

 CC 

1 PC L3 3 EI

 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  BB  PB BC  PC 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

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PROBLEM 11.79 (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  1 2 L  1 2 L   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

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PROBLEM 11.80 M0

M0 C

A

B L/2

L/2

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.11.2A 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,

 AA  U1 

M AL EI 1 M 2 AL M B AA  2 2 EI

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

 BB 

M B ( L/2) M L  B EI 2 EI

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

 AB   BB  U2  

M BL 2 EI

1 M B BB  M A AB 2 M B2 L M AM B  4EI 2 EI

Total strain energy: U  U1  U 2 

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

M A  M B  M 0.

Set

U 

5M 0 2 L  4EI

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PROBLEM 11.80 (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  2 M 0



U BC 

(2M 0 ) 2 M 2L  dx  0 L/ 2 2 EI EI



L

Total strain energy:

U  U AB  U BC 



U 



5M 0 2 L  4EI

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P D

A

PROBLEM 11.81

P E

L 4

L 2

B L 4

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. 11.2A 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

 EE  DE Likewise,

 DD 

2

3 PD L3 256 EI

and  ED 

7 PD L3 768 EI

Let PD be applied first. U  A1  A2  A3 1 1 PD DD  PD DE  PE  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  48 EI

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PROBLEM 11.81 (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

P2 1  L  1 P 2 L3 x dx      2 EI 3  4  384 EI 2

U DE 

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

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PROBLEM 11.82 M0

M0 A

B L

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. 11.2A and show that it is equal to the work obtained in part a.

SOLUTION

(a)

Label the couples M A and M B . Using Appendix D, Case 7.

 AA 

M AL 3EI

 BA 

M AL 6EI

 BB 

M BL 3EI

 AB 

M BL 6 EI

Apply MA first, then M B . U  A1  A2  A3 U 

1 1 M A AA  M A AB  M B BB 2 2



1 M A2 L 1 M AM B L 1 M B2 L   EI 6 EI 6 6 EI

With

M A  M B  M0 U 

(b)

1 M 02 L 2 EI

Bending moment:

U 

M 02 L  2EI

U 

M 02 L  2EI

M  M0 L

U  0

M2 M 2L dx  0 2 EI 2 EI

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PROBLEM 11.83

w

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

A

B

D L/2

L/2

SOLUTION Add force Q at point D. M2 dx 0 2 EI U 1  D   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.

D  



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  2 x  dx L/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

 D  0.0443

wL4   EI

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PROBLEM 11.84

P

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

A

B

D L/2

L/2

SOLUTION Add force Q at point D. M2 dx 0 2 EI L M M U 1  D  dx  0 EI  Q Q EI 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   M   Px  Q  x   ,   x   2  Q 2  

Set Q  0.

D 

1 EI



P EI





P EI

 1 3 1  L 3  L  1 2 L 1  L 2   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  2 x  dx 

L/2 

3  1 1 1 1  PL       3 24 4 16  EI

D 

5 PL3   48EI

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PROBLEM 11.85

w D

A

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

B E

L/2

L/2

L/2

SOLUTION Add force Q at point D. Reactions:

1 RA   Q, 2

RB  wL 

1 Q 2

Over portion AD,

with Q  0

Over portion DE,

M 0

M  RBv 

U Q

D 

U  U AD  U DE  U EB

U AD 0 Q

1  L w v   2  2

2

2

 wLv 

1  L 1 w  v    Qv 2  2 2

M 1  v Q 2

U DE 

1 L2 2  M dv 2 EI 0

Set Q  0

1 L2 M U DE   M Q dv EI 0 Q

Over portion EB,



2 1 L2  1  L   1  wLv w v          v  dv EI 0  2  2    2 



w L2  1 3 1 2  2 2    Lv  2  v  Lv  4 L v   dv 2EI 0   



w  L  2EI  



1 1 1 1 1  wL4 1 wL4        2  24 128 48 64  EI 768 EI

1 M   wu 2 2

D 

3 4 3 2 1 L  11 L 1 L 1 2 1  L      L   L          3 2  2  4  2  3 2  4 2  2    

M 0 Q

U EB 0 Q

U AD U DE U EB 1 wL4 1 wL4   0 0  Q Q Q 768 EI 768 EI



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PROBLEM 11.86

w

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

A

B

D L/2

L/2

SOLUTION Add couple M 0 at point D. L

M2 dx 2 EI

U



D 

U   M0

0



L 0

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.

D 



1  1 2   wx  (0)dx  EI L/2  2  L



w 2 EI



1  1  wL3 1 6  8  EI



L L/2

x 2 dx 

w 2 EI



L 0

M

M dx  M0

M 0  M0

Over portion AD:

1 EI

M M 1 dx  EI  M 0 EI



M  1  M0 L

 1 2   2 wx  (1)dx 

L/2 

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

D 

7 wL3 48EI



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PROBLEM 11.87

P

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

A

B

D L/2

L/2

SOLUTION Add couple M 0 at point D. U

D 



L 0

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.

D 

1 EI



P EI



L L/2 L L/2

( Px)(0)dx  x dx 

P 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



L L/2

M  1  M0 ( Px)(1)dx

2 1 1L   L2     2  2    2

2  1 1  PL     2 8  EI

D 

3PL2 8 EI



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PROBLEM 11.88

w D

A

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

B E

L/2

L/2

L/2

SOLUTION Add couple M0 at point D. Reactions:

RA 

M0 L

RE  wL 

M0 L

U  U AD  U DE  U EB , M0 x  0 with M 0  0 L

Over portion AD,

M 

Over portion DE,

M  RBv 

D 

U M 0

U AD 0 M 0

2

2

1  L 1  L M w  v    wLv  w  v    0 v L 2  2 2  2

M 1   v, M 0 L

U DE 

1 L2 2  M dv 2EI 0

Set M 0  0

2 1 L2 1 L2  1  L   1  U DE M M dv wLv w v             v  dv EI 0 EI 0  2  2    L  M 0 M 0



w L2  1 1   Lv 2   v3  Lv 2  L2v   dv  0  EIL  2 4 



3 4 3 2 w  1 L 11 L 1 L 1 1 L    L         L     L2     EIL  3 2  2  4  2  3 2  4 2  2     

1 1 1  wL3 1 wL3  1        384 EI  24 128 48 64  EI Over portion EB,

1 M   wu 2 2

D 

M 0 M 0

U EB 0 M 0

U AD U DE U EB 1 wL3 1 wL3   0 0 M 0 M 0 M 0 384 EI 384 EI



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PROBLEM 11.89

P

A

B

For the prismatic beam shown, determine the slope at point A.

D a

b L

SOLUTION Add couple M A at point A. Reactions:

RA 

Pb M A Pa M A  , RB   L L L L

M2 1 a 2 1 b 2 dx   M dx  2EI  0 M dv 2EI 2 EI 0 U 1 a M 1 b M A  M dx  M dv    0 0 EI EI M A M A M A L

U  0

Over portion AD (0  x  a),

x  Pbx  , M  M A  RA x  M A  1    L L 

Over portion DB (0  v  b),

M  RBv 

Set M A  0

A  

Pav M Av  , L L

M x 1 M A L

M v  L M A

1 a  Pbx  x 1 b Pav v dv    1   dx  EI 0  L  L EI 0 L L P 1 1 1  bLa 2  ba3  ab3  2 3 3 EIL  2 

A 

Pab (3La  2a 2  2b 2 ) 2 6EIL



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M0

PROBLEM 11.90

C A

B

L/2

For the prismatic beam shown, determine the slope at point B.

L/2

SOLUTION Add couple M B at point B as shown. Reactions. Strain energy. Slope at point B.

RA 

1 (M 0  M B )  L

M2 dx 2EI U B  M B L

U  0

M  RA x  M 0  (M 0  M B ) M X  M B L

With M B  0

x  M0 L

 x M  M 0   1 L 

U L M M  0 dx M A EI M B M0 L x  x  1 dx  0  EI L L M0 L   ( x  L) x dx EIL2 0 M0 L 2   ( x  Lx) dx EIL2 0 

L

M  x3 Lx 2  M L  0     0 EI  3 2  6 EI 0

B  

M 0L 6 EI

B 

M 0L 6 EI



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1.5 kips

PROBLEM 11.91

1.5 kips

A B

C

S8 3 13

For the beam and loading shown, determine the deflection of point B. Use E  29  106 psi.

5 ft

5 ft

SOLUTION Add force Q at point B. Units: forces in kips,

lengths in ft

E  29  103 ksi I  39.6 in 4 EI  (29  103 )(39.6)  1.148  106 kip  in 2  7975 kip  ft 2 2 M2 5 M dx   0 dv 2 EI 2EI U M  1  5 M 5  B  dx   0 M dv  0 M Q Q Q  EI  5

U  0

M M 5  0 0 M dx  0 Q Q

Over AB:

M  1.5 x,

Over BC:

M  1.5(v  5)  1.5v  Qv  3v  7.5  Qv;

M

M  v Q

5 5   3   2 2  0 Q dv   0 (3v  7.5v) dv  (3)  3  (5)  (7.5)  2  (5)  218.75    

1

B 

1

1 218.75  27.43  103 ft 0  143.75  EI 7975

 B  0.329 in.  

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1.5 kips

PROBLEM 11.92

1.5 kips

A B 5 ft

C

S8 3 13

For the beam and loading shown, determine the deflection of point A. Use E  29  106 psi.

5 ft

SOLUTION Add force Q at point A. Units:

forces in kips,

length in ft

E  29  103 ksi, I  39.6 in 4 EI  (29  103 )(39.6)  1.148  106 kip  in 2  7975 kip  ft 2

M2 dx 2 EI U 1 10 M  A   M Q dx Q EI 0 10

U  0

Over portion AB:

M  x Q

0  x  5, M  1.5 x  Qx

M

5 5 5 2   3  0 M Q dx   0 (1.5x)( x) dx  1.5 0 x dx  (1.5)  3  (5)  62.5  

Over portion BC:

1

5  x  10 M  1.5 x  1.5( x  5)  Qx M  3x  7.5  Qx

M  x Q

M

10 10   3   2 2 3 2  0 M Q dx   5 (3x  7.5x) dx  (3)  3  (10  5 )  (7.5)  2  (10  5 )    

1

 593.75 1 656.29 A   82.29  103 ft 62.5  593.75  7975 EI

1

 A  0.987 in.  

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PROBLEM 11.93

18 kN/m

8 kN A

B 1m

C

W250  22.3

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

1.5 m 2.5 m

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



B 

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  8(v  1)  (18)v 2  (v)dv   (9v  8v 2  8v)dv 0   0  0 2 EI  EI   1 9 8 8 29.391 29.391 4 3 2    5.12  103 m  (1.5)  (1.5)  (1.5)   3 2 5740 EI  4 EI 







 B  5.12 mm  

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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1890

5 kN/m A B

40 mm

PROBLEM 11.94

80 mm

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

C 4 kN

0.6 m

0.9 m

SOLUTION M2 dx  0 2 EI U B  P U



Portion AB:



a



a 0

a

M M dx  EI  P

M2 dx 2 EI



L a

M M dx EI  P

(0  x  a )

M 0 P

1 M   wx 2 2

 Portion BC:



L

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  2 EI EI  P 

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

w  L4 aL3 a 4 a 4  Pb3      2 EI  4 3 4 3  3EI w  L4 aL3 a 4  Pb3 B  0      2 EI  4 3 12  3EI 

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PROBLEM 11.94 (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 I

1 (40)(80)3  1.70667  106 mm 4 12  1.70667  106 m 4

EI  (200  109 )(1.70667  106 )  341,333 N  m 2

B  0 

 (1.5) 4 (0.6)(1.5)3 (0.6)3  (4  103 )(0.9)3 5  103     (2)(341,333)  4 3 12  (3)(341,333)

 7.25  103 m

 B  7.25 mm  

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.

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PROBLEM 11.95

160 kN W310  74

C

A

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

B

2.4 m

2.4 m 4.8 m

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  32,600 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:

2 M2 2.4 M dx   0 dv 2EI 2EI

M  M A  RA x  M A  80 x  U AB 1  M A EI



2.4 0

MA x 4.8

A 

U UAB UBC   M A M A M A

M x    1   M A  4.8 

1 x   dx  (80 x) 1   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  RBv  80v 

Set M A  0.

MA 4.8

MA v, 4.8

U BC 1  EI M A



2.4 0

M 1  v M A 4.8

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

(16.6667)(2.4)3 76.8  3EI EI 1 230.4 A  {153.6  76.8}  EI 32,600



2.4 0

v 2 dv



A  7. 07  103 rad



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90 kN

A

PROBLEM 11.96

90 kN

D

E

B S250  37.8

0.6 m

2m

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

0.6 m

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 )  10,240 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  0.6Q  (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)

D 

1 U  Q EI



3.2 0

M

M dx Q

Over portion AD: (0  x  0.6 m) M  (16.875  0.8125Q) x M  0.8125 x; Q M  90 x



0.6 0

M M  Q



0.6 0

Set Q  90 kN.

x3 (90 x )(0.8125 x )  73.125 3

0.6

 5.265 kN  m3 0

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PROBLEM 11.96 (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  0.1875 x  0.6; Set Q  90 kN. 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  0.1875Q)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)(0.1875v)dv 2.6 0 0 Q Q





v3  16.875 3



0.6

 1.215 kN  m3 0

Deflection at point D: (calculated)

D 

5.265  32.4  1.215 38.88   3.797  103 m EI 10,240

 D  3.80 mm  

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PROBLEM 11.97

8 kips

3 ft C

A

D

For the beam and loading shown, determine the slope at end A. Use E  29  106 psi.

B S8  18.4

6 ft

3 ft

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  11,580 kip  ft 2

Add couple M A at end A. Reactions:

RA  4 

MA M , RB  12  A 6 6

A 

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

A 

MA x 6

Set M A  0.

x 1  (4 x) 1   dx  0 EI  6 6

 Over DB: (0  u  3)

 UAD  UDB U    MA  MA  MA

M 0  MA

24 24 0 0 EI 11,580



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

24 EI

 U DB 0  MA  A  2.07  103 rad



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PROBLEM 11.98

8 kips

3 ft C

A

D

For the beam and loading shown, determine the deflection at point C. Use E  29  106 psi.

B S8  18.4

6 ft

3 ft

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  11,580 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. 2 Q

Over CD: (0  v  3)



 U  UAC  UCD  UDB    Q Q Q Q

1  M 1   x Set Q  0. M   4  Q x 2  2 Q   UAC 1 3 2 3 2 (2)(3)3 18 1    (4 x)  x  dx  x dx  Q 3EI EI 0 EI 0 EI 2 

Over AC: (0  x  3)

 UCD 1  EI Q

C 

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

 Over DB: (0  u  3)

M  8u

C 



3 0

(12v  2v 2 )dv 

1 EI

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

36 EI

M 0 Q

 U DB 0 Q

18 36 54  0  4.663  103 ft EI EI 11,580

 C  0.0560 in.  

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PROBLEM 11.99

B 1 2

l

1 2

l

A

P

C D

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

A l

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 E Ai Q Q

Vertical:

yC 

U 1 F L F   i i i P E Ai P

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PROBLEM 11.99 (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 4 EA



5 P 5 A 16 0 xC  0  yC  2.80

P  EA

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2A

B A

A

1 2

C l

PROBLEM 11.100

D l

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

l P

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

5 l 2

Equilibrium of joint C. 2 2  Fx  0:  FBC  FCD  Q  0 5 5 1 1  Fy  0: FBC  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

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PROBLEM 11.100 (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 A 8

5 Pl 5 A 16

5 5 P Q 2 4 1 P  Q 2

5 l 2

A

5 2

5 Pl 5 A 8

5 Pl 5 A 16

2l

2A

1

5 4 1  2



5

  4 xC 

1 Fi Li Fi Pl   E Ai Q 2 EA

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

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PROBLEM 11.101

B 4 in2 2.5 ft

3 in2

C 48 kips

2.5 ft

80 kips

6 in2

Each member of the truss shown is made of steel and has the cross-sectional area shown. Using E  29  106 psi, determine the deflection indicated. Vertical deflection of joint C.

D 6 ft

SOLUTION

Joint C:

Fx  0: 

12 12 FBC  13 13

FCD  Q  0

13 Q 12 5 5 Fy  0: FBC  FCD  P  0 13 13

FBC  FCD  

FBC  FCD 

13 P 5

(1)

(2)

Solving (1) and (2) simultaneously, 13 13 P Q 10 24 13 13 FCD   P  Q 10 24 5 Fy  0: FCD  FBD  0 13 5 1 5 FBD   FCD  P  Q 13 2 24 FBC 

Joint D:

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PROBLEM 11.101 (Continued) LBC  78 in.

Lengths of members:

LCD  78 in. LBC  60 in. F 2L 2 EA U FL F 1 FL F P     P EA P E A P U 

Member BC CD BD

F 13 13 P Q 10 24 13 13  P Q 10 24 1 5 P Q 2 24

L (in.) 78 78 60

F P 13 10 13  10 1 2

A (in 2 )

FL F A P

4

32.955P  13.73125Q

6

21.97 P  9.15417Q

3

5.00 P  2.08333Q 59.975P  2.49375Q



Further data:

E  29  106 psi  29,000 ksi P  80 kips Q  48 kips

P 

(59.975)(80)  (2.49375)(48)  0.1613 in.  29,000



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PROBLEM 11.102

B 4 in2 2.5 ft

3 in2

C 48 kips

2.5 ft

80 kips

6 in2

Each member of the truss shown is made of steel and has the cross-sectional area shown. Using E  29  106 psi, determine the deflection indicated. Horizontal deflection of joint C.

D 6 ft

SOLUTION

Joint C: Fx  0: 

12 12 FBC  13 13

FCD  Q  0 FBC  FCD  

Fy  0:

5 5 FBC  FCD  P  0 13 13 13 FBC  FCD  P 5

13 Q 12

(1)

(2)

Solving (1) and (2) simultaneously, 13 13 P Q 10 24 13 13 Q  P 10 24

FBC  FCD Joint D:

5 FCD  FB  0 13 5 1 5 Q   FCD  P  13 2 24

Fy  0: FBD

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PROBLEM 11.102 (Continued) LBC  78 in.

Lengths of members:

LCD  78 in. LBC  60 in. F 2L 2 EA U FL F 1 LF F P     Q EA P E A Q U 

F Q

Member

F

L (in.)

BC

13 13 P Q 10 24

78



78



CD



13 13 P Q 10 24

1 5 P Q 2 24

BD

60

A (in 2 )

FL F A Q

13 24

4

13.73125P  5.72135Q

13 24

6

9.15467 P  3.81424Q

5 24

3

2.08333P  0.86806Q 2.49325P  10.40365Q

 Further data:

E  29  106 psi  29,000 ksi P  80 kips Q  48 kips

Q 

(2.49325)(80)  (10.40365)(48)  0.01034 in.  29,000



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PROBLEM 11.103

1.6 m A 1.2 m

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.

B 1.2 m C

D 4.8 kN 2.5 m

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

B 

Joint C:

U  F 2L 1 F    F L  Q  Q 2 EA EA Q

Fy  0:

4 FCB  4.8  0 FCB  6.0 kN 5

Joint B:

3 FCB  FCD  0 FCD  3.6 kN 5 4 4 Fx  0: FAB  FBD  3.6  0 5 5 3 3 Fy  0: FAB  FBD  4.8  Q  0 5 5

Solving simultaneously,

FAB  6.25  0.8333Q

Fx  0:

kN

FBD  1.75  0.8333Q kN Joint D:

Fy  0:

3 FBD  FAD  0 5

3 FAD   FBD  1.05  0.5 Q 5

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PROBLEM 11.103 (Continued)

Member

F (103 N)

 F/ Q

L (m)

with Q  0 F ( F/ Q) L (103 N  m)

AB

6.25  0.8333Q

0.8333

2.0

10.4167

AD

1.05  0.5Q

0.5

2.4

1.26

2.0

2.9167

BD

0.8333

1.75  0.8333Q

BC

6.0

0

1.5

0

CD

3.6

0

2.5

0



14.593 1 F ( F / Q) L EA 14.593  103   (200  109 )(500  106 )

B  

 145.9  106 m   B  0.1459 mm  



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PROBLEM 11.104

1.6 m A

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.

1.2 m B 1.2 m C

D 4.8 kN 2.5 m

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

B 

U  F 2L 1 F    F L  Q  Q 2 EA EA Q

Joint C:

Fy  0:

4 FCB  4.8  0 5

Joint B:

Fx  0:

3 FCB  FCD  0 5

Fx  0:

4 4 FAB  FBD  3.6  Q  0 5 5

Fy  0:

3 3 FAB  FBD  4.8  0 5 5

Solving simultaneously,

FCB  6.0 kN FCD  3.6 kN

FAB  6.25  0.625Q kN FBD  1.75  0.625Q kN

Joint D:

Fy  0:

3 FBD  FAD  0 5

3 FAD   FBD  1.05  0.375Q 5

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PROBLEM 11.104 (Continued)

Member

F (103 N)

 F/ Q

L (m)

F ( F/ Q) L (103 N  m)

AB

6.25  0.625Q

0.625

2.0

7.8125

AD

1.05  0.375Q

0.375

2.4

0.9450

BD

1.75  0.625Q

0.625

2.0

2.1875

BC

6.0

0

1.5

0

CD

3.6

0

2.5

0



4.680 1 F ( F / Q) L EA 4.680  103   (200  109 )(500  106 )

B  

 46.8  106 m

 B  0.0468 mm  



PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.

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PROBLEM 11.105

P

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.

A L

60 B C

L

SOLUTION Add horizontal force Q at point A. 1 3 Over AB: M  Pv  Qv 2 2 3 M 1 M  v  v 2 P 2 Q UAB 



M2 dx 2 EI

L 0

Set Q  0.

 UAB 1  P EI  UAB 1  Q EI Over BC:

 

M 1 dv  0 P EI L M 1 M dv  0 Q EI L

M

 

1  1  1 PL3 Pv  v  dv   0 2 12 EI  2  L 1 3 PL3  3 Pv dv    2 0 2 12 EI  L

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  EI P  UBC 1  EI Q (a)

(b)

 

L 0

L 0

1 M M dx  EI P 1 M M dx  EI Q

 

L 0

L 0

2

L P  L  P  x   dx  x   2 3EI  2 

3

L

 0

1 PL3 12 EI

2 3P  L  3  L   P x   L dx   x  2   2  4 EI  2  

L

0 0

Vertical deflection of point A.

P 

 UAB  UBC  P P

Q 

 UAB  UBC 3 PL3   Q Q 12 EI

P 

PL3   6 EI

 Q  0.1443

PL3   EI

Horizontal deflection of point A.

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.

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PROBLEM 11.106 A

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

SOLUTION Use polar coordinate  . Calculate the bending moment M ( ) using free body BJ. M J  0: Px  M  0

M  Px  PR sin 

Strain energy:

U

U



0

 0

M2 ds 2EI

( PR sin  ) 2 ( Rd ) 2 EI



P 2 R3 2EI





P 2 R3 2EI





P2 R2 2EI

1    2

 By Castigliano’s theorem,



L



 0

sin 2  d

 1  cos

2

0



 0

2

d

1 sin 2 4

 0

   

P R 2

4EI

U P



 PR3 2EI

 

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PROBLEM 11.107

P

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.

B

R A

SOLUTION Add horizontal force Q at point B. Use polar coordinate  . U



 /2 0

M2 Rd 2EI

Bending moment.

M J  0: M  Pa  Qb  0 M  Pa  Qb  PR sin   QR (1  cos  ) M M  R sin   R (1  cos  ) P Q Set Q  0. (a)

(b)

U 1   Q EI





PR3 EI

(sin   sin  cos  )d 



PR3   1  1   cos  cos 0  sin 2  sin 2 0   EI  2 2 2 2 



PR3  1  0 1  0  EI  2 

Q 



 /2 0

 /2 0

M

M 1 Rd  EI Q



 /2 0

PR sin  R (1  cos  ) Rd  /2

PR3 1 ( cos   sin 2  ) EI 2 0

Q 

U 1   P EI





PR3 EI

sin 2  d 



PR3  1 1 PR3  1  1 1 1      0  sin    sin 0    sin 2     EI  2 2 EI  2 2 2 2 2 0 



PR3     0  0  0  EI  4 

P 



 /2 0

 /2 0

M

M 1 Rd  P EI PR3 EI



 /2 1 0

2



 /2 0

PR3  2EI

PR sin  R sin  Rd

(1  cos 2 )d

 /2



P 

 PR3 4 EI

 

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PROBLEM 11.108

l P B

C

l

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.

A

SOLUTION Add horizontal force Q and couple M C at C. M A  0: RC l  M C  ( P  Q)l  0 RC  P  Q 

MC l

Fx  0: P  Q  RAx  0

RAx  P  Q 

M  y, Q

M  RAx y  ( P  Q) y,

Member AB:

U AB 



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



0

( Py )( y )dy 

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 2EI l

Set Q  0 and M C  0.

 UBC 1  EI Q 1 U   M A EI

 

1 l 1 Pl 3 M dx  ( Px) x dx  0 EI 0 3 EI Q l 1 l x M  ( Px) 1   dx M dx  0 EI 0 l  MA  l



M



 

P EI

P EI



l

x2   x   dx 0 l  

2  1 2 1 2  1 Pl   l l 2 3  6 EI 

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PROBLEM 11.108 (Continued)

(a)

Deflection at C.

C 

 UAB  UBC  Q Q

(b)

Slope at C.

C 

 UAB  UBC   MA  MC

C  C 

2Pl 3  3EI

Pl 2 6EI



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PROBLEM 11.109

P B

C

Three rods, each of the same flexural rigidity EI, are welded to form the frame ABCD. For the loading shown, determine the deflection of point D. L

A

D

L

SOLUTION Add dummy force Q at point D as shown. M A  0: DL  PL  0

Statics

D  P

Fx  0:  Ax  P  Q  0

Ax  ( P  Q) 

Fy  0: Ay  D  0

Ay  P 

U  U AB  U BC  U CD By Castigliano’s theorem,

D 

D 

U Q

U AB U BC U CD   Q Q Q

Member AB:

M  ( P  Q) y

Set

Q0

M  y Q

M  Py L

U AB   0

M 2dy 2EI

U AB p L 2 PL3 L M M  0 dy  y dy   3EI Q EI Q EI 0 Member BC: M  Px  QL

M  L Q

Q  0 M  Px

Set

L

U BC   0

M 2dx 2EI

U BC PL L PL3 L M M  0 dx  x dx   2EI Q EI Q EI 0

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PROBLEM 11.109 (Continued) M  y Q

Member CD:

M  Qy

Set

Q0 M 0 L

U CD   0

M 2dy 2 EI

U CD L M M dy  0  0 EI Q Q

D 

PL3 PL3 5PL3  0   3EI 2EI 6EI

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PROBLEM 11.110

P B

C

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.

L

A

D

L

SOLUTION Add couple M 0 at point D. Statics:

M A  0: M 0  DL  PL  0 DP

M0  L

Fx  0: Ax  P  0 Fy  0: Ay  D  0 Strain energy:

Ax  P  Ay  P 

M0  L

U  UAB  UBC  UCD

U  M0  U AB  U BC  U CD D     M0  M0  M0

By Castigliano’s theorem,  D 

Member AB:

M  0 UAB   M0 L M M  UAB  dy  0 0 EI  M  M0 0 M  Py



L

0

M2 dy 2EI



Member BC:

M  M 0  Dx  M 0  Px 

Set M 0  0

M  Px UBC 

 UBC   M0

 

L

0 L

0

M0x L

M x 1 L M 0 M2 dx 2 EI M M P dx  EI  M 0 EI



L

0

x PL2  x 1   dy  L 6 EI 

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PROBLEM 11.110 (Continued)

Member CD:

M  M0

Set M 0  0

M 0 U



L

0

D  0 

M 1  M0 M2 dy 2EI 2

PL 0 6 EI

U   M0



L

0

M M dx  0 EI  M 0

D 

PL2 6 EI



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PROBLEM 11.111

P C B

A L/2

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

L/2

SOLUTION Remove support B and add reaction RB as a load. U  U AC  U CB  yB 

Over AC:

 U  U AB   RB  RB

M2 du  0 2 EI  U CB  0  RB



L/2

0

M2 dv 2EI

L M  L  M  RB  u    Pu ,  u   2 2  RB  

 U AB 1   RB EI

L/2 



0

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



RB EI



L/2 

2

L  u  2  dv  0   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  EI  RB



L/2

0

( RB v)v dv 

RB 3EI

3

1 RB L3 L  2 24 EI  

1  RB L3 5 PL3  7   0 yB    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 

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M0 B

A

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

L

SOLUTION Remove support B and add reaction RB as a load. M2 dv 2EI U 1 L M yB    M R dv  0 RB EI 0 B L

U  0

M  RBv  M 0

M v RB

1 L  ( RBvM 0 ) v dv EI 0 R L M L  B  0 v 2dv  0  0 v dv EI EI

yB 

RB L3 M 0 L2 3 M0 RB   0  3EI 2 EI 2 L 3 1 M A  RB  M 0  M 0  M 0  M 0 2 2 

 

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PROBLEM 11.113

M0 A

D

B

a

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

b L

SOLUTION Remove support A and add reaction RA as a load. M2 dx 0 2EI 1 U A    RA EI U

Portion AD:

(0  x  a )



L

(a  x  L)

 U DB 1   RA EI A 



0

L a



M

M dx  0  RA

M x  RA

M  RA x

 U AD 1   RA EI Portion DB:



L

a

0

( RA x)( x) dx 

RA a 3 3EI

M x  RA

M  RA x  M 0 ( 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 0 M B  RA L  M 0 Bending moment diagram drawn to scale for a 

M D  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 0b( L  a ) x/2 L3  M 0  L  a 0 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.

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PROBLEM 11.114

w C A

B L/2

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

L/2

SOLUTION Remove support A and add reaction RA as a load. L M2 M2 dx   02 dv 2 EI 2EI U 1 L2 M 1 L2 M  M dx  A    M R dv  0 0 RA EI RA EI 0 A L

U   02

Portion AC:

L  O  x   2 

M  x RA

M  RA x

U AC 1 L2 RA L3  ( R x )( x ) dx   A RA EI 0 24 EI Portion CB:

L  0  v   2  L 1  M  RA  v    wv 2 2 2 

M L   v   RA  2

 U CB 1 L  L 1 L  R v    wv 2   v   dv  0  A RA EI   2 2 2  1  EI 

2  L L L 1 L 2   3 2  RA  0  v   dv  w 02  v  v  dv  2 2 2     

3 RA  1 3 1  L   w  L     EI  3 3  2   2 EI

 1  L  4 L 1  L 3         2 3  2    4  2 

1  RA L3 7 wL4 1     384 EI  3 24  EI

A 

U AC U CB 1 RA L3 7 wL4    0 3 EI 384 EI RA RA RA 

7 wL   128

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PROBLEM 11.114 (Continued)

Bending moments.

Over AC:

MC Over CB:

7 wLx 128 7  wL2  0.02734wL2 256

M 

M 



7 L 1  wL  v    wv 2 128 2 2  2

MB 

7 1 L 9 wL2  w     wL2 128 2 2 128 M B  0.07031wL2 

dM 7  wL  wvm  0 dv 128 Mm 

or

vm 

7 L 128

7 L 1  7   7 wL  L    w L 128 128 2  2  128  

2



945 wL2  0.02884wL2 32,768



M 

7 wLx  w x  L /2 2 /2 128



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PROBLEM 11.115

P D

A L 3

B

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

2L 3

SOLUTION Remove support A and add reaction RA as a load. M2 dx 0 2 EI U 1 A    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 L   Portion DB:   x  L  M  RA x  P  x   3 3    M x  RA  U DB 1 L 1 M M dx    RA EI L/3  RA EI 

 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 





R  A 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  EI  3 81 6 54  EI

A 

 U AD  U DB  1 1 1  RA L3 14 PL3      81 EI  RA  RA  81 3 81  EI

1 RA L3 14 PL3  3 EI 81 EI 0 

RA 

14 P  27

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PROBLEM 11.115 (Continued)

Bending moments:

 L  14 M D  RA    PL  3  81

M D  0.1728PL 

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

M B  0.1481PL  M

By singularity functions,

14 Px  P x  L/31  27

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PROBLEM 11.116

w B

A

L/2

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

C L

SOLUTION Remove support A and add reaction RA as a load. L 1 M B  0  RA  wL2  RC L  0 2 2 1 1 RC  RA  wL 2 2 2 L M2 L M U  U AB  U BC   02 dx   0 dv 2 EI 2 EI U U AB U BC   0 A  RA RA RA Portion AB:

M  RA x,

M  x QA 3

Portion BC:

U AB 1 L2 M 1 L2 RA  L  1 RA L3  M dx  ( R x )( x ) dx     A   RA EI 0 RA EI 0 3EI  2  24 EI 1 2 1 1 1 M  RC v  wv  RAv  wLv  wLv 2 2 2 2 2 M 1  v RA 2 U BC 1 L 1 1  1   RAv  w( Lv  v 2 )   v  dv  RA EI 0  2 2  2  1 L 2 2 3  RAv  w( Lv  v )  dv 4EI 0   L4 L4   RA L3 1  L3 wL4   w     RA    4EI  3 4   12 EI 48EI  3 U AB U BC  1 1  R L3 wL4     A  0 A  RA RA 48EI  24 12  EI 1 RA   wL 6 1 1  1 RC    wL   wL 2 6  2 

1 RA   wL   6 5 RC  wL   12

Fy  0: RA  RB  RC  wL  0

1 5  wL  RB  wL  wL  0 6 12

RB 

3 wL   4

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PROBLEM 11.117 D

C



E



l

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

B P

SOLUTION Detach member BC at support C. Add reaction RC as a load. U 

F 2L 2 EA

yC 

U FL  F  0 EA  RC  RC

Joint C:

FBC  RC

Joint B:

Fx  0: FBE sin   FBD sin   0

FBE  FBD

Fy  0: FBD cos   FBE cos   RB  P FBD  FBE 

P  RB 2cos  L

( FL/EA) ( F/ RB )

l/ cos 

( RB  P)l/4 EA cos3

( P  RB )/2cos 

 F/ RB 1/2 cos  1/2 cos 

l/ cos 

( RB  P)l/4 EA cos3

RB

1

l

RB l/EA

Member

F

BD

( P  RB )/2cos 

BE BC

yB   Pl/2 EA cos3   RB l/2 EA cos3   RB l/EA  0 RB 

P 1  2cos3

FBC  RB

FBC 

P  1  2 cos3

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PROBLEM 11.118

C E

R

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

f

D B P

SOLUTION LBD  LCB  LBE  R A  Constant E  Constant

Fx  0: FBE cos   FBD  0

Joint E:

FBE  FBD / cos  Fy  0: FBC  ( FBD / cos  )sin   P  0; FBC  P  FBD tan 

xB   We have

Fi

Fi Li Fi L Fi  FL 0 EAi FBD AE FBD

Fi 0 FBD

Fi

Fi /FBD

Fi Fi /FBD

BD

FBD

1

FBD

BC

P  FBD tan 

 tan 

BE

FBD / cos 

1 cos 

 P tan   FBE tan 2  FBD / cos 2 

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PROBLEM 11.118 (Continued)

Fi

 1  Fi  FBD 1  tan 2     P tan   0 FBD cos 2    FBD  P

tan  1 1  tan 2   cos 2 

sin  cos 

 1

sin  1  2 cos  cos 2  2



sin  cos  cos   sin 2   1 2

P P sin  cos   sin 2 2 4  P  FBD tan 

FBD  FBC

1  sin   P   sin  cos   2  cos  1    P 1  sin 2   2   2  11  7 FBC  P 1      P   2  2   8 

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PROBLEM 11.119

B

D

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

308 l A

C l P

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

U  F 2L 1 F    F L0  FB  FB EA EA  FB

B 

Fy  0:

Joint C:

FCD 

2 3

3 FCD  FBC  P  0 2 P

2 3

Fx  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



1 3

B    FB  



l

1 3



4 3



8 3 8 3

8  Pl  4 8  FB l   0   3  EA  3 3  EA P

8 3 84 3

FBC  FB 

P  0.652 P FBC  0.652 P 

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PROBLEM 11.120 C 3 4

D

l

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

E B l P

SOLUTION Detach member BC from support C. Add reaction FC as a load. F 2L 1  F 2 L 2 EA 2 EA 1 U F L C   F  FC EA  FC U 

Joint B:

Fy  0: FC  P  Fx  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



C 

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

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A

3 4

P

PROBLEM 11.121

B

Knowing that the eight members of the indeterminate truss shown have the same uniform cross-sectional area, determine the force in member AB.

C

l

D

E l

SOLUTION Cut member AB at end A and replace member force FAB by load FA  acting on member AB at end A.

A  Joint B:

U  F 2L 1 F L0    F  FA  FA 2 EA EA  FA

Fx  0: FA 

4 FBD  0 5

5 FBD   FA 4

3 Fy  0: P  FBE  FBD  0 5 Joint E:

Fy  0: FBE 

Fx  0:  Joint D:

4 FAE  FDE  0 5

Fy  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

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PROBLEM 11.121 (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

 1  9 27   Pl  FAl   0  4 EA  2  2 FA  P 3

A 



FAB  FA 

FAB 

2 P  3

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A

3 4

PROBLEM 11.122

B

Knowing that the eight members of the indeterminate truss shown have the same uniform cross-sectional area, determine the force in member AB.

C

l

D

E l P

SOLUTION Cut member AB at end A and replace member force FAB by load FA  acting on member AB at end A.

A 

U  F 2L 1 F L0    F  FA  FA 2EA EA  FA

Joint B:

5 FBD   FA 4

Joint E:

Fy  0: FBE  P 

FBE 

3 FA 4

3 FAE 5

5 5 P  FBE 3 3 5 5  P  FA 3 4

FAE 

Fx  0: 

4 FAE  FDE  0 5

4 4 FDE   FAE   P  FA 5 3 Joint D:

Fy  0: FAD 

3 FDB  0 5

3 3 FAD   FDB   FA 5 4

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PROBLEM 11.122 (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





63 27 Pl  FAl 16 4

1  63 27   Pl  FAl   0  4 EA  16  7 FA  P 12

A 



FAB  FA 

FAB 

7 P  12

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PROBLEM 11.123

1.6 m 1.2 m

C B

A P

14-mm diameter

Rod AB is made of a steel for which the yield strength is  Y  450 MPa and E  200 GPa; rod BC is made of an aluminum alloy for which  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.

10-mm diameter

SOLUTION AAB 

ABC 

 4

 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   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 

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PROBLEM 11.124

B 3 in2

Each member of the truss shown is made of steel and has the cross-sectional area shown. Using E  29  106 psi, determine the strain energy of the truss for the loading shown.

4 ft D

C

20 kips

2

4 in

24 kips 7.5 ft

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  29,000 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 2EABC 2 EACD

(51)2 (102) (25) 2 (90)  (2)(29,000)(3) (2)(29,000)(4)

 1.5247  0.2425 U  1.767 in.  kip 

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PROBLEM 11.125

A

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.

5000 ft

B

SOLUTION

  (2) (2 )  4 rad L  5000 ft  60  103 in. co  J

do  4 in. 2



c

2 TL  GJ

4 o

ci  co  t  3.5 in.



 ci4  166.406 in 4 T

GJ  L 2

U

T 2 L  GJ   L GJ  2   2GJ  L  2GJ 2L

U

(11.2  106 )(166.406)(4 ) 2 (2)(60  103 )

U  2.45  106 in.  lb 

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A 4m B 2.5 m D

PROBLEM 11.126 Bronze E ⫽ 105 GPa 12-mm diameter Aluminum E ⫽ 70 GPa 9-mm diameter 0.6 m

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.

C

SOLUTION

 m  125  106 Pa  2

Portion BC:

ABC 

4

(9)  63.617 mm 2  63.617  106 m 2

Pm   m ABC  7952 N Corresponding strain energy: U BC 



(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  m : 1 Pm  m  U m 2 2U m (2)(28.324) m    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 

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1.5 m A

B

C

D

0.8 m

PROBLEM 11.127 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 point D caused by the 16-kN load.

E

16 kN

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)

By symmetry, FDE  30 kN FBCD  34 kN

(compression) (tension)

Equilibrium of joint B.

Fy  0:  

8 FACE  FBE  0 17

8 (34 kN)  FBE  0 FBE  16 kN 17

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PROBLEM 11.127 (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 

Strain energy:

Fi 2 Li 1  Fi 2 Li 2 EA 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 ( P  16 kN  16  103 N) P  U 2 1   5.34  103 m (16  103 )  42.72 2

Deflection of point D.

  5.34 mm  

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PROBLEM 11.128 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

 st 

W . k

ym  2 st 

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PROBLEM 11.129

C 50 mm

60 mm

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.

40 mm

A

0.40 m

B

100 mm

D T

0.60 m

SOLUTION Shaft CD:

T  820 N  m J 

 2

c4 

 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 (820 N  m)  1366.67 N  m TCD  60 mm rC

T  1366.67 N  m J 

 2

c4 

 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

2

T L (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 T  U 2

 

2U (2)(18.283)   0.04459 rad TA 820

  2.55 

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PROBLEM 11.130

A

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.

B

l ⫽ 200 mm

l ⫽ 200 mm

C

P ⫽ 150 N

SOLUTION J



c4 

2

  12 

4

 2.0358  103 mm 4 2  2 

 2.0358  109 m 4 I

Portion AB:

bending

1 J  1.0179  109 m 4 2

M   Px UAB , b 



LAB

0

M2 P2 dx  2EI 2EI



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 UBC  

Total: Work-energy:



LBC

0

M2 P2 dx  2EI 3EI



LBC

0

x 2dx 

P 2 L3BC 6 EI

(150) 2 (200  103 )3  0.14736 J (6)(200  109 )(1.0179  109 )

U  UAB, b  UAB, t  UBC  0.86737 J 1 P  U 2



2U (2)(0.86737)  P 150  11.57  103 m

  11.57 mm  

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P D

PROBLEM 11.131

B

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

E

A

L/2

P

L/2

L/2

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).

D 

U U AD U DE U EB    M 0 M 0 M 0 M 0 L  0  x   2 

Portion AD:

M2 dx 0 2 EI M M x M 0x  L M 0 L

UAD 



L/2

Set M 0  0 so that M  0. U AD 1  M 0 EI Portion DE:

L    x  L 2 

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 L 2     

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PROBLEM 11.131 (Continued)

M x   1; L M 0 U DE 1  M 0 EI



L

L/2

L  Set M 0  0 so that M   P  x   . 2  M

P M dx  M 0 EI



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



PL2 EI

L  x   x  2 1  L  dx  

L/2 

L

   L /2   L

3  3 3 1 1 1 1  PL          4 16 3 24 2 4  48 EI

L  0  v   2 

Portion EB: M   Pv

Slope at point D.

U EB 

0

U EB 1  M 0 EI

M 0 M 0

D  0 



L /2



M2 dx 2EI L/2

0

M

M dv  0 M 0

PL2 0 48EI

D 

PL2 48EI



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PROBLEM 11.132 L

A

a

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 a 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

B

C 

PL3  Ea 2  1  1.5 2  . 3EI  GL 

C P

SOLUTION Torsion:

T  Pa Ut 

Bending:

M  Pv Ub  

Total:

T 2 L P2 a2 L  2GJ 2GJ



L 0

M 2 dv  2 EI



L 0

P 2 v 2 dv 2 EI

2 3

P L 6 EI

P 2 a 2 L P 2 L3 1   P C 2GJ 6 EI 2 2 3 Pa L PL PL3  3EIa 2 C    1  GJ 3EI 3EI  GJL2 U

  

C 

Since J  2I ,

PL3  Ea 2 1  1.5 2 3EI  GL

   

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B

PROBLEM 11.133

C

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.

l A

D

P

l

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  P y

U AB 



M 1  MD

M2 dy 0 2EI 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 0



( Py ) y dy  l 0

M  M A  RAl  MD  Pl UBC 



Pl 3 3EI

( Py )(1) dy 

Pl 2 2EI

M  l, P

M 1  MD

M2 dx 0 2 EI l

Set M D  0.

 UBC 1  EI P  UBC 1   MD EI

 

l 0 l 0

M

1 M dx  EI P

M

1 M dx  EI  MD



l 0

( Pl )(l ) dx 



l 0

Pl 3 EI

( Pl )(1) dx 

Pl 2 EI

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PROBLEM 11.133 (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  EI P  U CD 1   M D EI (a)

(b)

 

l

0 l

0

M

1 M dy  EI P

M

1 M dy  EI  MD



l

0

( Py )( y ) dy 



l

0

Pl 3 3EI

( Py )(1) dy 

Pl 2 2 EI

Horizontal deflection of point D.

P 

 UAB  UBC  UCD  1 1  Pl 3     1  3  EI P P P 3

D 

 UAB  UBC  UCD  1 1  Pl 2     1  2  EI  MD  MD  MD  2

P 

5Pl 3   3EI

Slope at point D.

D 

2 Pl 2 EI



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PROBLEM 11.134

D

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  106psi for the rod BD and the bar, determine the deflection of point C.

0.2-in. diameter 25 in. P C

A

B 10 in.

30 in.

SOLUTION Assume member BD is a two-force member. M A  0: 10 FBD  (40)(50)  0 ABD 

Member ABC: Portion AB:



FBD  200 lb

(0.2)2  31.416  103 in 2

4 2 FBD LBD (200) 2 (25)  U BD  2 EA (2)(29  106 )(31.416  103 )  0.5488 in.  lb 1 I  (0.75)(0.75)3  26.367  103 in 4 12 x M  1500  150 x 10 10 M 2 1502 10 2 UAB  dx  x dx 0 2 EI 2 EI 0





2

(150) (103 ) (2)(29  106 )(26.367  103 )(3)  4.904 in.  lb 

Portion BC:

M  50v UBC  

Total:



30 0

M2 502 dv  2 EI 2 EI



30 0

v 2dv

(50)2 (30)3  14.713 in.  lb (2)(29  106 )(26.367  103 )(3)

U  U BD  U AB  U BD  20.166 in.  lb 1 P C  U 2

C 

2U (2)(20.166)  P 50

 C  0.807 in.  

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PROBLEM 11.C1 Element n

Element i

Element 1

P

A rod consisting of n elements, each of which is homogeneous and of uniform cross section, is subjected to a load P applied at its free end. The length of element i is denoted by Li and its diameter by di . (a) Denoting by E the modulus of elasticity of the material used in the rod, write a computer program that can be used to determine the strain energy acquired by the rod and the deformation measured at its free end. (b) Use this program to determine the strain energy and deformation for the rods of Probs. 11.9 and 11.10.

SOLUTION Enter: P and E For each element Enter Ai and Di Compute: Normal stress:

i 

P Ai

Strain energy:

Ui 

P 2 Li 2 Ai E

Strain energy density:

u

 i2 2E

Total strain energy. Update through n elements. U  U  Ui

Total deformation. 1 2U P  U :   2 P Program Outputs Problem 11.9

Axial load  8.000 kips

Modulus of elasticity  29  106 psi

Element

Length in.

L in.

Stress ksi

Strain Energy in.  lb

Strain Energy Density lb  in./in3

1

24.000

0.022

26.08

86.32

11.72

2

36.000

0.022

18.11

89.92

5.65

Total strain energy  176.24 in.  lb Total deformation  0.0441 in. PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1951

PROBLEM 11.C1 (Continued) Program Outputs (Continued) Problem 11.10 Axial load  25.000 kN

Modulus of elasticity  200 GPa

Element

Length m

L mm

Stress MPa

Strain Energy J

Strain Energy Density kJ/m3

1

0.80

0.497

124.34

6.22

38.65

2

1.20

0.477

79.58

5.97

15.83

Total strain energy  12.1853 J Total deformation  0.9748 mm

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PROBLEM 11.C2 D

F C

1500 lb h

3 4

E

⫻ 6 in.

B

A

W8 ⫻ 18 a

a 60 in.

60 in.

Two 0.75  6-in. cover plates are welded to a W8  18 rolled-steel beam as shown. The 1500-lb block is to be dropped from a height h  2 in. onto the beam. (a) Write a computer program to calculate the maximum normal stress on transverse sections just to the left of D and at the center of the beam for values of a from 0 to 60 in. using 5-in. increments. (b) From the values considered in part a, select the distance a for which the maximum normal stress is as small as possible. Use E  29  106 psi.

SOLUTION Compute and enter moments of inertia and section moduli. For AD and EB: W8  18: I1  61.9 in 4

S1  15.2 in 3



For DCE: W8  18 plus cover plates:

I 2  61.9  2(6  0.75)(4.445) 2  239.72 in 4



I2 (4.07  0.75) 239.72  4.82  49.7 in 3



S2 

------------------------------------------------------------------------------------------------------------------------------------ym  Pm where   influence coefficient. See next page for determination of  . Pm  equivalent static load U2 

1 1 ym2 Pm ym  2 2 

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PROBLEM 11.C2 (Continued) Work done by w is w(h  ym ) 1 ym2  wh  wym 2  A ym2  2 w ym  2 wh 

or

Position 1

Position 2

A for ym Program solution of 

Enter L  120 in., h  2 in., W  1500 lb, E  29  106 psi For a  0 to 60 in., Step 5 in.: A for ym , Pm  ym / , yst  w Solve 

 D  1 

1 1 Pm a/S1 ;  c   2  Pm L/S 2 2 4

Print: a, yst , ym , Pm ,  1 ,  2 , and ( 1   2 ) Repeat with smaller intervals to find a for ( 1   2 )  0 This is the distance a for  max as small as possible. Determination of :  is deflection at c for a unit load at C.

 2a 

L

  t A /C  A1    A2    3  3  1 a  1 1  a  2a  1 L L  L           2 EI1  I1 I 2  2  3  2 4 EI 2 2  3  1 1  1 3      a 3  L  6E 8 I 2   I1 I 2 



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PROBLEM 11.C2 (Continued)

Program Output Beam  W8  18 with two 6 by 0.75-in. cover plates h  2 in. W  1500 lb

L  120 in.

a in.

ystat in.

ymax in.

Pmax lb

ksi

1

ksi

2

1   2

0.00

0.00777

0.1842

35,572

0.00

21.46

–21.46

5.00

0.00778

0.1844

35,544

5.85

21.44

–15.59

10.00

0.00787

0.1855

35,348

11.63

21.32

–9.69

15.00

0.00812

0.1885

34,834

17.19

21.01

–3.82

20.00

0.00859

0.1942

33,896

22.30

20.45

1.85

25.00

0.00938

0.2033

32,509

26.73

19.61

7.13

30.00

0.01056

0.2163

30,736

30.33

18.54

11.79

35.00

0.01220

0.2334

28,706

33.05

17.32

15.73

40.00

0.01438

0.2546

26,563

34.95

16.02

18.93

45.00

0.01718

0.2799

24,436

36.17

14.74

21.43

50.00

0.02068

0.3090

22,415

36.87

13.52

23.35

55.00

0.02496

0.3419

20,550

37.18

12.40

24.78

60.00

0.03008

0.3783

18,862

37.23

11.38

25.85

ksi

  

-----------------------------------------------------------------------------------------------------------------Use smaller increments to seek the smallest maximum normal stress. 18.33

0.00840

0.1919

34,259

20.657

20.665

–0.01

18.34

0.00840

0.1920

34,257

20.667

20.664

0.00

18.35

0.00841

0.1920

34,255

20.677

20.663

0.01

---------------------------------------------------------------------------------------------------------------Max. stress small as possible for a  18.34 in. Smallest max. stress

 20.67 ksi

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PROBLEM 11.C3 24 mm D h 24 mm

A B L

The 16-kg block D is dropped from a height h onto the free end of the steel bar AB. For the steel used  all  120 MPa and E  200 GPa. (a) Write a computer program to calculate the maximum allowable height h for values of the length L from 100 mm to 1.2 m, using 100-mm increments. (b) From the values considered in part a, select the length corresponding to the largest allowable height.

SOLUTION Enter

 all  120 MPa, E  200 GPa, d  0.024 m m  16 kg, g  9.81 m/s 2 I I I  d 4 /12 S    d 3 /b c d/2

For

L  100 m to 1200 m, Step 100 mm L  L /1000 yst  mgL3 /3EI M max   all S Pmax  M max /L ymax  Pmax L3 /3EI

From Problem 11.69, Page 705, 2  y y   2h  Solve for max   h   1  1 st ym  yst 1  1    yst   yst  2    

Print: L, yst , ymax , Pmax , M max , h Return

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PROBLEM 11.C3 (Continued)

Program Output

Problem 11.C3 m  16.0 kg d  24 mm   120 MPa L mm

ystat mm

G  200 GPa

ymax mm

Pmax N

Mmax Nm

h mm

100

0.00946

0.167

2764.8

276.48

1.301

200

0.07569

0.667

1382.4

276.48

2.269

300

0.25547

1.500

921.6

276.48

2.904

400

0.60556

2.667

691.2

276.48

3.205

500

1.18273

4.167

553.0

276.48

3.173

600

2.04375

6.000

460.8

276.48

2.807

700

3.24540

8.167

395.0

276.48

2.109

800

4.84445

10.667

345.6

276.48

1.076

900

6.89766

13.500

307.2

276.48

–0.289

1000

9.46181

16.667

276.5

276.48

–1.988

1100

12.59367

20.167

251.3

276.48

–4.020

1200

16.35000

24.000

230.4

276.48

–6.385

-------------------------------------------------------------------------------------------------------------------Use smaller increments to seek the largest height h. 435

0.77883

3.154

635.6

276.48

3.2316

440

0.80599

3.227

628.4

276.48

3.2320

445

0.83378

3.300

621.3

276.48

3.2317

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D

PROBLEM 11.C4

m h

A

B E

W150 ⫻ 13.5

a 1.8 m

The block D of mass m  8 kg is dropped from a height h  750 mm onto the rolled-steel beam AB. Knowing that E  200 GPa, write a computer program to calculate the maximum deflection of point E and the maximum normal stress in the beam for values of a from 100 to 900 mm, using 100-mm increments.

SOLUTION Enter

L  1.8 m, E  200 GPa, h  0.75 m m  8 kg, g  9.81 m/s 2 I  6.87  106 m 4 S  91.6  106 m 4

For

a  100 mm to 900 mm, Step 100 mm a  a /1000 b La

See Prob. 11.71, page 705 

yst  mga 2b 2 /3EIL

Influence coefficient for  E   for unit load at E 

  a 2b 2 /3EIL

See Prob. 11.69, page 705 

 2h  ym  yst 1  1   yst   Pmax  ym /

M max  Pmax ab /L

 max  M max /S Print:

a, yst , ym , Pmax ,  max

Return Problem 11.C4 Beam:

W150  13.5 I  6.87  106 m 4 S  91.6  106 m3 L  1.8 m h  750 mm m  8 kg g  9.81 m/s 2

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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1958

PROBLEM 11.C4 (Continued)

a mm

ystat mm

ymax mm

Pmax N

 max

100

0.0003

0.6775

173.93

179.33

200

0.0011

1.2757

92.43

179.40

300

0.0021

1.7946

65.75

179.46

400

0.0033

2.2339

52.85

179.51

500

0.0045

2.5936

45.55

179.55

600

0.0055

2.8734

41.13

179.59

700

0.0063

3.0734

38.46

179.61

800

0.0068

3.1934

37.02

179.63

900

0.0069

3.2334

36.56

179.63

MPa

Note the small variation in max. This is due to the energy acquired by the mass as it falls through ymax. See Prob. 11.147, page 731, for a case where energy delivered is constant and max is also constant.

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PROBLEM 11.C5 10-mm diameter B

A

6-mm diameter

a

C P 6m

The steel rods AB and BC are made of a steel for which  Y  300 MPa and E  200 GPa. (a) Write a computer program to calculate for values of a from 0 to 6 m, using 1-m increments, the maximum strain energy that can be acquired by the assembly without causing any permanent deformation. (b) For each value of a considered, calculate the diameter of a uniform rod of length 6 m and of the same mass as the original assembly, and the maximum strain energy that could be acquired by this uniform rod without causing permanent deformation.

SOLUTION

 Y  300 MPa, E  200 GPa, L  6 m   2

Enter:

Area AB 

4

(0.010 m) , Area BC 

Pm   Y Area BC

4

(0.006 m) 2

For a  0 to 6 m, Step 1 m U

Pm2  a La     2 E  Area AB Area BC 

For uniform rod of same volume, vol  a (Area AB )  ( L  a )(Area BC ) 4vol L

d Area new 

 4

d2

Pnew   Y (Area new ) U new 

Print:

2 Pnew L 2 E (Area new )

a, U , vol, d , Pnew , U new

Return

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PROBLEM 11.C5 (Continued)

Program Output

Problem 11.C5

 Y  300 MPa,

Pm  8482 N, L  6 m,

E  200 GPa

a m

U J

Vol m3

d mm

New P N

New U J

0.00

38.17

169.65

6.00

8482.30

38.17

1.00

34.10

219.91

6.83

10,995.58

49.48

2.00

30.03

270.18

7.57

13,508.85

60.79

3.00

25.96

320.44

8.25

16,022.12

72.10

4.00

21.88

370.71

8.87

18,535.40

83.41

5.00

17.81

420.97

9.45

21,048.67

94.72

6.00

13.74

471.24

10.00

23,561.95

106.03

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PROBLEM 11.C6

2.65 in.

20 in.

B

A

A 160-lb diver jumps from a height of 20 in. onto end C of a diving board having the uniform cross section shown. Write a computer program to calculate for values of a from 10 to 50 in., using 10-in. increments, (a) the maximum deflection of point C, (b) the maximum bending moment in the board, (c) the equivalent static load. Assume that the diver’s legs remain rigid and use E  1.8  106 psi.

C a

16 in. 12 ft

SOLUTION L  12 ft, h  20 in., W  160 lb

Enter:

E  1.8  106 psi I  (16 in.)(2.65 in.)3 /12 S  (16 in.)(2.65 in.)2 /6

ym  Pm where   influence coefficient. See below for determination of  where Pm  equivalent static load. 1 1 ym2 Pm ym  2 2  work  w(h  ym ) U2 

work  U 2 w(h  ym ) 

1 ym2 A 2 

Position 1

Position 2

Program solution of a for ym . Enter  For a  10 in. to 50 in., Step 10 in. Solve A for ym ,

Pm  ym / M max  M B  Pm ( L  a)

  M max /S Print a, ym , Pm , M m , 

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PROBLEM 11.C6 (Continued)

Program Output



a in.

ym in.

Pm lb

Max. M kip  in.

psi

10

14.622

757.7

101.532

5422

20

13.262

802.6

99.519

5314

30

11.950

855.6

97.536

5208

40

10.683

919.1

95.583

5104

50

9.462

996.4

93.661

5001

Determination of influence of coefficient  :

M-Diagram U

 2



1 M2 (1 lb)   dx 2 2EI



1 2EI

  



0

2

La 2  a  x dx   

a



La 0

1  ( L  a ) 2 a 3 ( L  a )3     3 3  EI  a 2 1   ( L  a) 2 a  ( L  a )3   3EI

v 2 dv





PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1963