Problem Set 2-Chapter 4 - Static Body Stresses

 

SOLUTION (4.18) Known: A steel propeller shaft with a given diameter transmits a known power at a specified angular velocity. Find:  (a) Determine the nominal shear stress at the surface. (b) Determine the outside diameter re required quired to give the same outer surfac surfacee stress if a

hollow shaft of inside diamete diameterr 0.85 times the outside diameter is used. (c) Compare the weights of the solid and h hollow ollow shafts. Schematic and Given Data: Rotation

2.5 in.

do

3200 hp, n = 2000 rpm

 di

d

  Assumptions: 1. Bending and axial loads are negligible. 2. The bar is straight and round.

3. 4.

The material is homogeneous, and perfectly elastic within the stress range involved. The effect of stress raisers is negligible.

Analysis: 

1.

W  From Eq. (1.3), T = 5252 n

T =

2.

5252(3200)  = 8403 lb ft = 100, 838 lb in. 2000  

From Eq. (4.4),  ! = 16 T   " d

 ! =

3

16 (100, (100, 838)  = 32,868 32, 868 psi = 33 ksi 3   " (2.5)

4-24

! 

 

3.

4.

For a hollow shaft, do3 - (0.85)4 do3 = 2.53  0.478 do3 = 15.625 do = 3.20 in.

! 

2 20))2 hollow w = 3.20  - (0.85 ! 3. 20 Wt. ho Wt. hollow llow  = Area hollo  = 0.46  Area solid Wt. solid 2. 2.5 52

! 

Comment: It is more economical to use a hollow shaft when pure shear stress is involved.

SOLUTION (4.19) Known: The maximum shear stress is given for a hollow shaft of known geometry subjected to pure torsion. Find: Determine the torque that produces the given maximum shear stress. Schematic and Given Data:

d i = 20 mm

T

do  = 25 mm  !

max = 570 MPa

 

Assumptions:

1. 2. 3.

The shaft is straight. The material is homogeneous and perfectly elastic. There are no stress raisers.

Analysis:

1.

2.

From Eq. (4.3),  !  = 4

Tr  = J

16Tdo "

d4o - d 4i

 

4

6 (570)25 - 20 T=   = 10.325   " 10 N•mm = 10,3 10,325 25 N•m  16(25) !

4-25

! 

 

SOLUTION (4.20) Known: The same value of torque is applied to both a solid square shaft (b " b) and a solid round shaft of radius r. Find: (a) Determine the ratio b/r for both square aand nd round sshafts hafts so as to produce eequal qual

maximum shear values. (b) Compare the weight of the two sha shafts fts for the square aand nd round sshafts. hafts. (c) Compare the rati ratio o of sstrength-to-we trength-to-weight ight for the squa square re an and d round shafts. Schematic and Given Data:

T T

T

T

2r  b

  Assumptions:  1. The shafts are straight. 2. The material is homogeneous and perfectly elastic. 3. There are no stress raisers. 4. The shafts are made of the same material.

Analysis:  1. For equal stress, equate Eq. (4.4) with Eq. (4.5). 16 T  = 4. 8 T 3 b3   !(2r) 16 b3  = 4. 4. 8 8!r 3  

  b 3 = 7. 5398 5398 r ! 

b  = 1.96  r

4-26

 

2.

3.

Wt. square pe perr unit llength ength b2  = 2  = !r Wt. round pe perr unit length (Strength/wt.) square (Strength/wt.) round 

=

!

b2  = 1.22  b 2 1.96

T/Ws W   = r  = 1  = 0.82  T/Wr Ws 1.22

! 

! 

Comment: The round bar is more economica economicall (higher strength to weight ratio) than the square bar for the same shear stress.

4-27

 

SOLUTION (4.21) Known: The maximum shear stress produced in a shaft transmitting torque is given. Find: Determine the torque: (a) In a round shaft of 40 mm diameter. (b) In a square shaft, 40 mm on a side. Schematic and Given Data:

T

T

dia. = 40 mm

 !

40 mm

max = 400 MPa

  Assumptions: 1. The shafts are straight. 2. The material is homogeneous and perfectly elastic. 3. There are no stress raisers. Analysis: 1. From Eq. (4.4),  !  = T=

16T   3 "d

(400)(!)(40)3  = 5.03 "  106 N •mm 16

= 5030 N•m 

2.

! 

From Eq. (4.5),  !   =   4.8T   3 a

T=

400(40)3  = 5.33 ! 106 N •mm = 5330 N•m 4. 4.8 8  

4-28

! 

 

SOLUTION (4.22) Known: The geometries of a solid round shaft and a solid square shaft of the same size (circle diameter equal to side of square) are given. Find: Compare the torque transmitting strength, the weight, and the ratio of strength to weight of the two shafts. Schematic and Given Data:

T T

d d

  Assumptions:  1. The shafts are straight. 2. The material is homogeneous and perfectly elastic. 3. There are no stress raisers. 4. The shafts are made of the same material. Analysis:  1. For equal stress, equate Eq. (4.14) with Eq. (4.5). 16 Tround 3

!

d

 =

4. 8 Tsquare 3

d

 

T square = T round  16  = 1.061 1. 061 Tround 4. 4.8 8!  

2.

3.

2 Wt. squa Wt. square re per unit llengt ength h  = d  = 4  = 1.273 1. 273  ! 2 ! Wt.. round pe Wt perr unit length d 4

(Strength/wt.) square

= 1.061   = 0. 833  (Strength/wt.) round  1.273

! 

! 

! 

Comment: The round bar is more economica economicall (higher strength to weight ratio) than the square bar for the same shear stress.

4-29

 

SOLUTION (4.33) Known: A rectangula rectangularr beam has an initial curvature, r , equal to twice the section depth, h. Find: Compare its extreme-fiber bending stresses with those of an otherwise identical straight beam. Schematic and Given Data: b

h

r 

M

 

M

  Assumption: The material is homogeneous and perfectly elastic. Analysis: 

1.

For a straight beam, from Eq. (4.7) where Z  = - 6M2   ; bh

!i

2.

6

2

 (From Appendix B-1).

 = + 6M2 bh  

!o

For a curved beam, from Eq. (4.10) e=r-

A

bh

 = 2h -

dA /!

r o

= 2h -

d!/!  

b

h h = 2h ro ln ln 2.5h ri   1.5h

r i

= 2h -

3.

=

  bh

h  = .042385 h ln(5 3)

 

From Eq. (4.9), 4-43

 

 

!i

=

"ci

eAr i

=

M(0.5h M(0. 5h - . 042 042385 385h) h)   7.1978M  = 2 (.042385h)(bh)(1.5h) bh

 

-7. 20M But for direction of "M" shown, ! i  =   -7.   2 bh

!o

 =

M(0.5h + .042385h)

119M M  = - 5. 119 (.042385h)(bh)(2.5h) bh2  

But for direction shown of M: !o = 4.

+5.20M   bh2

From Eq. (4.11), with Z = bh2/6: 7. 1978 = 1. 20 and Ki = 7.1978 and Ko = 5.119  = 0.85 6 6  

which is consistent with Fig. 4.11. 5.

The inner and outer fiber bending stresses for the curved beam are 120% and 85% of the straight beam stresses.

4-44

! 

 

SOLUTION (4.34) Known: A known force is exerted on an S-hook. Find: Determine the location and magnitude of the maximum tensile stress. Schematic and Given Data: 200 lb

3 in.

1 in. dia. round rod

4 in.

200 lb

  Assumption: Material is homogeneous and perfectly elastic. Analysis: 1. At point A, the tensile stress due to bending is

 =

!

32M  K   t 3 "d

[From Eq. (4.11)]

The tensile stress due to tension is

 =

!

P  = 4P   2 A "d

[From Eq. (4.1)]

200 lb r = 3 in.

600 in. lb

A

200 lb

  4-45

 

Thus, the combined tensile stress is ! =

2. 3.

4.

32M  K  + 4P   t 2 3 "d "d

3  = 6 r 6,, Kt = 1. 1. 14. 14. From Fig. 4.11, for c  = 0.5 0. 5

 = (1.14) 32(600)  + 4(200)  = 6,967 + 255 3 2 "(1) "(1) = 7,222 psi   At point A, ! = 7.2 ksi. At point B, from Fig. 4.11, for !

r  = 4  = 8, 8, Kt = 1.10 c 0.5 0. 5

 = (1.10)

!

32(800) 4(200)    + 2 3 "(1) "(1)

= 8,964 + 255 = 9,219 psi. 200 lb

800 in. lb B

r = 4 in. 200 lb

  At point B, ! = 9.2 ksi. This point correspo corresponds nds to the loca location tion of the maxim maximum um tensile stress. !   Comment: The inner fiber is stressed more than the outer fiber because the stresses due to the direct tension and bending are of the same sign, and hence, add up to give a large resultant stress. 

4-46

 

SOLUTION (4.36) Known: The critical section of a crane hook is considered to be trapezoidal with dimensions as shown. Find: Determine the resultant stress (bending plus direct tension) at points P and Q. Schematic and Given Data:

40

80 120

 

A P

A  

60

Q

70,000 N

  Assumption: Material is homogeneous and perfectly elastic. Analysis: 1. 40

120

=

+ c

c

80

c (rec (recta tang ngle) le) = 60 60,,

40

c

40

c (tria (triang ngle) le) = 40 

4-49

 

Ac (rectangl (rectangl e) + Ac(triangle) A(trapezoid)    4800(60) + 2400(40) =   = 53 53.. 33 60(120)   c (trapezoid) =

2.

I of trapezoid about its ccentroidal entroidal axis = I of rec rectangle tangle + I of triangle, triangle, eeach ach about the centroidal axis of trapezoid. trapezoid. Using the parallel axis the theorem orem and equations from Appendix B-1: 3

2  40(120) I(trapezoid) =   + 4800( 4800(60 60 - 53 53.. 33) 12  

+ = 5.76

!  10

= 8.32

!

40(120)3  + 2400(40 - 53. 53. 33)2 36  

6

 + . 213

!

  106 mm4 

  106 + 1.92 !  106 + . 426 ! 106 

3.

53. 33  = 2.13. From Fig. 4.11, Ki = 1.52, Ko = 0.73. r (trapezoid) = 60 + 53. 53.33 c  

4.

The tensile stress due to tension is ! =

 P A

  [From Eq. 4.1)]

and tensile stress due to bending is ! =

5.

K Mc   I

[From Eq. (4.11)].

At P, the resultant stress is Mc P o  = A  - K   I  

!

70,000N

=

7200 mm

 - (0.73) 2

[70,, 000(6 [70 000(60 0 + 53. 53.33)N 33)N•mm](66.67 mm) 8.32 ! 106 mm4

= 9.72 MPa - 46.41 MPa = -36.69 MPa 6.

At Q, the resultant stress is

 =

!

P  + K   Mc   i I A

4-50

  ! 

 

= 9.72 MPa + (1.52) 

[70,000(113.33)N•mm](53.33 mm) 8.32 ! 106 mm4

 

= 9.72 MPa + 77.29 MPa = 87.01 MPa

! 

SOLUTION (4.37) Known: The critical section of a crane hook is considered to be circular with areas as shown. Find: Determine the resultant stress (bending plus direct tension) at points P and Q. Schematic and Given Data: A = 7,200 mm 2

A

A

60

Q

P

70,000 N

  Assumption:  The material is homogeneous and perfectly elastic. Analysis:

1. 2. 3.

c = d  = 2

I=

!d

4

64

4A /!  = 2

 =

!

96 4 64

A/!  =

 = 4.17

7200/!  = 47. 87 = 48 mm 

" 10

6

mm4

 

r  (circle) = 60 + 48  = 2. 2. 25 25.. From Fig. 4.11, Ki = 1.50, Ko = 0.75 c 48

4-51

 

4.

The tensile stress due to tension is ! =

 P A

  [From Eq. 4.1)]

and tensile stress due to bending is ! =

5.

K Mc   I

[From Eq. (4.11)].

At P, the resultant stress is

 =

!

P  - K   Mc   o A I

[70,, 000(60 + 48)N•mm](48 mm)   = 70,000N  - (0. 75) [70 7200 mm2 4.17 ! 106 mm 4 ! 

= 9.72 MPa - 65.27 MPa = -55.55 MPa 6.

At Q, the resultant stress is

 =

!

P  + K   Mc i A

I

 

= 9.72 MPa + (1.50) 

[70,000(108)N•mm](48 mm) 4.17 ! 106 mm 4

= 9.72 MPa + 130.53 MPa = 140.25 MPa

  !

  Comment: By comparing the resultant stresses in Solutions 4.21 and 4.22, it is evident that for the same area of cross section the trapezoidal secti section on is stronger and hence, more economical economical than a circular cross section section crane hook. This is the reason, for the use of trapezoidal shaped cross sections in crane hooks for practical p ractical applications.

4-52

 

SOLUTION (4.44) Known: An I-beam with given dimensions is simply supported at each end and subjected to a know load at the center. Find: Compute the maximum transverse transverse shear stres stress. s. Compare the aanswer nswer with the approximation obtained by dividing the shear load by the area of the web, with the web considered to extend for the full 8-in. depth. Schematic and Given Data: 1000 lb

3

1

1

in.

2

2

3 8

3 500 lb

500 lb

1 2

in.

8 in.

1

in.

in.

2

in.

  Assumption: The beam material is homogeneous and perfectly elastic. Analysis:  1.  $max exists at the neutral bending axis,  !

 = V   Ib

yd ydA A

2.

[Eq Eq.. (4. 12)]   3.5

3.5

8

=

8

3.125

-

7 2

1

 

4-60

 

From the above figure, 3

3

4   3.5(8)  3.125(7) I = I 1 - I 2 =     = 60 60.. 01 in in..   12 12

4

3.5

3.

 =

 !max

500   (60.01) (0.375)

y(0.375dy) + 3.5

0 2

2

2

= 22.22 0.375 3. 5  + 3.5 4  - 3. 5 2 2 2

4.

y(3 y(3.. 5dy)  

! 

 = 197 psi

To check,     500  = 167 psi ( 3 )(8) 8

 !max "

! 

Comment: The rough check is 15% low in this case.

SOLUTION (4.45) Known: A box section beam, where the top plate of the section is cemented in place, is loaded with a specified force. Find: Determine the shear stress acting on the cemented joint. Schematic and Given Data: 12 kN Cement

5

5

5

L

L

2

2

50 mm

5

40 mm

 

4-61

 

5

5

50 =

5

40

50 mm -

1

2

5

30

40 40 mm

 

Assumption: The beam material is homogeneous and perfectly elastic. Analysis:  y = 25

1. 2.

From Eq. (4.12),  ! = V Ib

ydA yd A  y = 20

From the above figure, I = I1 - I2  3

3

=  (40)(50)  -  (30)(40)  = 25 256, 6,667 667 mm 4 12 12   25

25

3.

 !

 =

6000   (256,667)(10)

y(40dy) = 20

y2

6000(40)   2,566,670 2

  20

! 

= 10.5 MPa

SOLUTION (4.46) Known: A shaft between self-aligning bearings A and B is loaded through belt forces applied to a central sheave. Find:  (a) Determine and make a sketc sketch h showing the stre stresses sses ac acting ting on the top and side elements, T and S. (b) Represent the state statess of stres stresss at T and S w with ith three-dime three-dimensional nsional Mo Mohr hr circles circles.. (c) At loca location tion S, sshow how the orientation aand nd stresse stressess acting on a principal element, and on a maximum shear eleme element. nt.

4-62

 

Schematic and Given Data:

100 mm 2000 N

100 mm

B Free end of  shaft

T S

20 mm dia. shaft A

120 mm dia. sheave 400 N Connected to flexible coupling and clutch

  T A

B

1200 N

2400 N 1200 N

1200 N V

-1200 N

M

T

120,000 N•mm

96,000 N•mm

  Assumptions:  1. The weights of the shaft and sheave are negligible.

4-63

 

2. 3. 4.

The shaft is straight. The effect of stress concentration concentrationss is negligible. The shaft material is homogeneou homogeneouss and perfectly elastic.

Analysis:  1. For torsion, Eq. (4.4),  !

2.

16(1600)(60)  = 16T  =  = 61.12 MPa 3 3 "d "(20)  

For bending, Eq. (4.8), 32M  = 32(1200)(100)  = 152.79 MPa ! = 3 3 "(20) "d

3.

 

For transverse shear, Eq. (4.13),  !

4(1200)  = 5.09 MPa  = 4   V  = 3 A 3(")(10)2  

y

4.

T  !

y x

S

= 66.21 MPa

 !

= 61.12 MPa  !

5.

 !

x (0,66)

152.79 MPa

x

max

 !

x (153,61) 38.66

0

0

"

"1

"

51.4˚

T

S

"

1

y (0,-61)

y (0,-66)

6.

"

25.7

2 = -21

o

y

S

"

x

153

o

S "

1

61.1

= +174

S  !

Original Element

19.3

= 77

= 98

o

Max Shear

Principal

4-64

 

 

SOLUTION (4.49) Known: A static vertical load is applied to the handle of a hand crank. Find: (a) Copy the drawing aand nd ma mark rk on it the location at highest bending str stress. ess. Make a three-dimensionall Mohr-circle representation of the stresses at this point. three-dimensiona p oint. (b) Mark on tthe he drawi drawing ng the llocation ocation at highest combined ttorsional orsional an and d transverse

shear stress. Make a three-dimen three-dimensional sional Mohr-circle repres representation entation of the stresse stressess at this point. Schematic and Given Data: 200 mm a

25-mm-dia. round rod 250 mm

bent into crank

100 mm 1000 N

a

b

  Assumptions:  1. The weight of the hand crank is negligible. 2. The effect of the stress concentration is negligible. 3. The crank material is homogeneous and perfectly elastic. Analysis: 1. For bending, Eq. (4.8) !

2.

32(300 mm)(1000 N)  = 195.6 MPa  =  = 32M 3 3 "(25 mm) "d  

For torsion, Eq. (4.4)  !

16(250 mm)(1000 N)  = 81.5 MPa  =  = 16T 3 3 "(25 mm) "d  

4-70

 

3.

For transverse shear, Eq. (4.13) 4(1000 N)  = 2.7 2. 7 MPa  = 4V  = 3A 3"(12.5 mm)2  

 !tr

4.

From the Mohr circle for "a",  !max = 128 " Pa, #max =

225  " Pa.  

From the Mohr circle for "b",  !max = 84.2 " Pa, #max =

5.

84.2 " Pa.  

  "

y

a

x  !tor

a b

y b

x  +  !tran.shr.

 !tor

  6. y (0, 81.5)

Highest  = 128 shear stress

+ "

0  = -30

+ "

 "

40°

y (0, 84.2)

Highest ! = 225 normal stress +!

+!

!

x (195.6, -81.5)

Mohr circle for "a"

x (0, -84.2) Mohr circle for "b"

 

4-71

 

SOLUTION (4.51) Known: An electric motor is loaded by a belt drive. Find: Copy the drawing and show on both views the location or locations on the shaft of the highest stress. Make a com complete plete Mohr-circle repre representation sentation of the stress aatt this location. Schematic and Given Data: 1 in. dia. shaft 3000 lb belt tension

Motor

6 in. dia.

1000 lb belt tension 1 in.

  Assumptions:  1. The weight of the structure is negligible. 2. The effect of stress concentration is negligible. 3. The shaft material is homogeneou homogeneouss and perfectly elastic. Analysis: 1. For torsion, Eq. (4.4)  !

2.

16(2000 lb)(3 in. 16(2000 in.))  = 31 ksi  =  = 16T 3 3 "(1 in.) "d  

For bending, Eq. (4.8) !

3.

32(4000 lb)(1 in. 32(4000 in.))  = 41 ksi  =  = 32M 3 3 "(1 in.) "d  

The Mohr circle representati representation on is given above.

4-74

 

4.

  Answer to 1 3000 lb tension of top belt

1000 lb

 

5. + "

 "

max = 37

y (0, 31)

31 ksi y !1

0 56.5°

 = -17

!2

 = 58 +!

x

41 ksi

x (41, -31) Mohr circle representation

Critical element

SOLUTION (4.52) Known: An electric motor is loaded by a belt drive. Find: Copy the drawing and show on both views the location or locations of the highest stress on the shaft. Make a comple complete te Mohr-circle repres representation entation of the stress at this location. Schematic and Given Data: 1 in. dia. shaft 3000 lb belt tension

Motor

5 in. dia.

1000 lb belt tension

1 in.

  4-75

 

Assumptions:  1. The weight of the structure is negligible. 2. The effect of stress concentration is negligible. 3. The shaft material is homogeneou homogeneouss and perfectly elastic. Analysis: 1. For torsion, Eq. (4.4)  !

2.

16(2000 lb)(2.5 in.)  = 25 ksi  =  = 16T 3 3 "(1 in.) "d  

For bending, Eq. (4.8) 32(4000 32(4 000 lb)(1 in. in.))  = 41 ksi  =  = 32M 3 3 "(1 in.) "d   The Mohr circle representati representation on is given below.

!

3. 4.

Answer to 1 3000 lb of tension top belt

1000 lb

  5. + "

 "max

= 32

y (0, 25)

25 ksi y 0

 = -12

!2

 = 53 +!

!1

50.7°

x

41 ksi

x (41,-25)

Mohr circle representation

Critical element

 

4-76

 

SOLUTION (4.53) Known: A solid round shaft with a known diameter is supported by self-aligning bearingss at A aand bearing nd B. Two chain sprockets that aare re transmitting a load are attached to the shaft. Find: Identify the specific shaft location subjected to the most severe state of stress, and make a Mohr-circle representation of this stress state. Schematic and Given Data: B 1 in. dia. shaft

2 in.

1000 lb A 3 in. 4 in. 500 lb

4 in.

3 in.

  Assumptions: 1. The loads are static. 2. Stress concentrati concentrations ons can be ignored. 3. The shaft is straight.

4.

The shaft material is homogeneou homogeneouss and perfectly elastic.

4-77

 

Analysis:  1. 3 in.

4 in.

3 in.

A

B 850 lb

650 lb 500 lb

1000 lb

650 lb V

150 lb

- 850 lb Mmax = 850 lb(3 in.) = 2550 in.lb M

  !MA = 0

 

1000(7) + 500(3) = F B(10)  FB = 8500/10 = 850 lb  !F v  = 0  

FA = 500 + 1000 - 850

= 650 lb 2. Bottom of shaft; just to left of gear

B

A

  4-78

 

The location subjected to the most severe state of stress is at the bottom of the shaft, just to the left of the smaller gear. 3.

For bending, Eq. (4.8) !

32(2550)  = 26 ksi  =  = 32M 3 3 " 1 "d  

For torsion, Eq. (4.4)  !

16(2000)  = 10.2 ksi  =  = 16T 3 3 " 1 "d

4. + "

 

 = 17

 "max

(26,10.2) -3.5

0

38°

30

26 ksi +! 10.2 ksi

(0,-10.2)

Mohr circle representation

Critical element

 

SOLUTION (4.54) Known: A solid round shaft with a known diameter is supported by self-aligning bearingss at A aand bearing nd B. Two chain sprockets that aare re transmitting a load are attached to the shaft. Find: Identify the specific shaft location subjecte subjected d to the most severe state of stress, and make a Mohr-circle representation of this stress state.

4-79

 

Schematic and Given Data: B 1 in. dia. shaft

2 in.

1000 lb A 3 in. 4 in. 500 lb

3 in.

3 in.

  Assumptions: 1. The loads are static. 2. Stress concentrati concentrations ons can be ignored. 3. The shaft is straight. 4. The shaft material is homogeneou homogeneouss and perfectly elastic.

4-80

 

Analysis:  1. 3 in.

3 in.

3 in.

A

B 750 lb

750 lb 500 lb

1000 lb

750 lb V

250 lb

- 750 lb Mmax = 750 lb(3 in.) = 2250 in.lb M

  !MA = 0

 

1000(6) + 500(3) = FB(10)  FB = 7500/10 = 750 lb  !F v  = 0  

FA = 500 + 1000 - 750

2.

= 750 lb

Bottom of shaft; just to left of gear

B

A

  4-81

 

The location subjected to the most severe state of stress is at the bottom of the shaft, just to the left of the smaller gear. 3.

For bending, Eq. (4.8) !

32(2250)  = 22. 92 = 23 ksi  =  = 32M 3 3 " 1 "d  

For torsion, Eq. (4.4)  !

4.

16(2000)  = 10.2 ksi  =  = 16T 3 3 " 1 "d   + "

 "

max  = 15.4

(23,10.2) -3.87

0

23 ksi

41.6° 26.9 +!

10.2 ksi

(0,-10.2)

Mohr circle representation

Critical element

 

SOLUTION (4.55) Known: A solid round shaft with a known diameter is supported by self-aligning bearingss at A aand bearing nd B. Two chain sprockets that are transmitting a load are attached to the shaft. Find: Identify the specific shaft location subjected to the most severe state of stress, and make a Mohr-circle representation of this stress state.

4-82

 

Schematic and Given Data: B F 100 mm dia.

50 mm dia.

A

30 mm 50 mm

  dia. 4000 N 100 mm

50 mm

  Assumptions:

1. 2. 3. 4.

The loads are static. Stress concentrati concentration on can be ignored. The shaft is straight. The shaft material is homogeneou homogeneouss and perfectly elastic.

Analysis: 1. Fg = 2000 N

  Top  view

1500 N D

1000 N

B Front  view 500 N

C

A 4000 N 3000 N

 

4-83

 

2. 2000 N 1500 N 500 N A

C

B

A

D

C D

B

1000 N 4000 N 3000 N Front view V

3000 N

Top view V

150 N•m

1500 N

-500 N

-1000 N

M

M

-75 N•m T

100 N•m

100 N•m

T

  3.

The most severe state of stress is at B M=

 2

  [(3000)(50)]

 2

+ [ (500)(50)] = 152,069 N•mm 

T = 4000 (25) = 100,000 N•mm 4.

For bending, Eq. (4.8) !

 !

32(152,069)  = 57.4 MPa  =  = 32M 3 3 "(30) "d  

16(100,000)  = 18.9 MPa  =  = 16T 3 3 "(30) "d  

4-84

 

5. + "

 = 34.3 MPa

 "max

(0,18.9) !1

0 !

2 = -5.7

 = 63 MPa

18.9

+!

33°

57.4 (57.4,-18.9)

  Mohr circle representation of point B SOLUTION (4.56) Known: A small pressurized cylinder is attached at one end and loaded with a pipe wrench at the other. The stresses due to the internal pressure and the pipe wrench are known. Find: (a) Draw a Mohr-circle representatio representation n of the sta state te of stress at point A. (b) Determine the m magnitude agnitude of tthe he maximum shear stress at A. (c) Sketch the orientation of a princ principal ipal element, and show all stre stresses sses ac acting ting on it. Schematic and Given Data:

400 MPa

300 MPa

A 200 MPa

  4-85

 

Assumption: The "positive-clockwise" rule is used. Analysis: 1. + !

 = 278 MPa

 !max

 = 144 MPa

"2

x 0

"3

300

103°

 = 556 MPa +"

"1

400 y

  2. 3.

The maximum shear stress at A is  ! max = 278   MPa 

!1 =

556 MPa !2 =

!3 =

! 

144 MPa

0

A !2

!1

51.5°

  SOLUTION (4.57) Known: A small pressurized cylinder is attached at one end and loaded with a pipe wrench at the other. The stresses due to the internal pressure and the pipe wrench are known. Find: (a) Draw a Mohr-circle representatio representation n of the sta state te of stress at point A. (b) Determine the m magnitude agnitude of tthe he maximum shear stress at A. (c) Sketch the orientation of a princ principal ipal element, and show all stre stresses sses ac acting ting on it.

4-86

 

Schematic and Given Data:

100 MPa

135 MPa

A 100 MPa

  Assumption: The "positive-clockwise" rule is used. Analysis: 1. + !

 = 110 MPa

 !max "

2

x 80°

"

100

3

0

135

 = 219 MPa

"1

+"

y

  2.

The maximum shear stress at A is  !max

 =

110 MPa 

4-87

 

! 

 

3. !1 =

219 MPa !2 =

!3 =

16 MPa

0

A !2

!1

40°

  Comment:  At the outer surface of an internal internally ly pressurized cylinder, cylinder, the tangential stress is analytically twice the axial stress. That is, the aaxial xial stress for a thick walled pr2 2p r 2 cylinder is !a = 2 i i 2  and the tangential stress is !t = 2 i i 2 . Therefore, !t  = 2. ro  - ri ro  - ri !a

We axial couldstress speculate problem that a stress concentration existed which increased increased the fromin 50this MPa to 60 MPa.

4-88

 

SOLUTION (4.77) Known: A stepped shaft with known dimensions is supported by bearings and carries a known load. Find: Determine the maximum stress at the shaft fillet.

Schematic and Given Data:

d = 40 mm

1000 N r = 5 mm

A

B

70 mm 500 mm R

250 mm

A

R

B

 

4-111

 

B A

 500 mm 250 mm

667 N

333 N 1000 N

70 mm V 333 N

-667 N C

M

 

168 N•m G

 47 N• m D

E

F

Assumptions: 1. The shaft remains straight . 2. The material is homogeneous and perfectly elastic.

 

Analysis: 1. * MB = 0 : Hence RA = 333 N * FY = 0 : RA + RB = 1000, Therefore RB = 667 N 2. From similar triangle, 'CDF and 'GEF, GE = 47 N•m. The stress due to bending at the critical shaft ffillet illet is equal tto o 32M  = 32(47)  = 7.5 MPa. !nom = 3 3 "(0.04) "d   3. r/d for the critical shaft fillet = 5/40 = 0.125 D/d = 80/40 = 2

From Fig. (4.35a), Kt = 1.65 Therefore, !max = !nom Kt = 7.5(1.65) = 12.4 MPa 4-112

 

SOLUTION (4.88) Known: Three notched tensile bars (see Fig. 4.39) have stress-conce stress-concentrations ntrations of 1, 1.5 and 2.5 respectively. respectively. Each is ma made de of ductile stee steell and have Sy = 100 ksi, a rectangular cross-section with a minimum area of 1 in. 2, and is initially free of residual stress. Find: Draw the shape of the stress-distrib stress-distribution ution curve for each case when

(a) A tensile load of 50,000 lb is applied. (b) The load is increase increased d to 100,000 lb. (c) The load is removed. Schematic and Given Data:

P

P

2

A = 1 in. Sy = 100 ksi

  Analysis: 

K=1

K = 1.5

100

K = 2.5

50

75

(a)

50 100

50 100

100

(b)

-50

 

-100

(c)

  4-127

 

SOLUTION (4.89) Known: Two rectangular steel beams beams having a known tensile yield strength are loaded in bending and Z = I/c is known. known. The dimensions and a stress stress concentration factor are given for both beams. Find: (a) For each beam, determine w what hat mom moment, ent, M, ccauses auses (1) initial yielding, and (2)

complete yielding. (b) Beam A is loaded to cause yielding to a depth of 1/4 in. Determin Determinee and plot the distribution of residual stresses which remain after the load lo ad is removed. Schematic and Given Data: 0.5 in. 0.5 in. 1 in.

A

1 in.

1.5 in.

B

Z = (1/12) in.3

3 Z = (1/12) in.

Syt = 80 ksi

S yt = 80 ksi Kt = 3

0.5 in.

 

Assumptions: 1. The idealized stress-strain curve is appropriate. 2. The beam is homogeneous. 3. There are no residual stresses initially. Analysis:  1. For initial yielding, using Eqs. (4.7) and (4.21),

Beam A:

 = S yt = M   Z

!

M = S yt Z = 80,000 1  = 6667 in•lb  12

! 

Beam B: ! = S yt = M  K t  Z 1 Syt Z 80,000 12 M=  =  = 2222 in•lb   Kt 3

4-128

!

 

 

2.

For complete yielding, 0 F

h/2

F

  Thus, for both beams, 1 1 1 (80, 0,000 000)( )( )( ) ( ) = 10,000 in.lb  M = F( h ) = (Syt)( h )(b) ( h ) = (8 2 2 2 2 2 2  

3.

! 

For loadi loading ng beam A to yield the beam to a depth of 1/4 in. and the then n rele releasing asing the load we have: 0

80 110

-30 ksi 0

(1/4) in.

+25 ksi -25 ksi

(1/4) in. "Moment on" curve  and Elastic curve

+30 ksi Residual stresses

  4.

The moment to yield the beam to a depth of 1/4 iin. n. is given by M = (stress)(area)(mom. arm) + (avg. ( avg. stress)(area)(mom. arm) 80,000 1 1 1 = 80,000 1 1 3  +   = 9167 in•lb 4 2 4 2 4 2 3

5.

The elastic stress for M = 9167 in•lb is calculated as !elastic =

6.

M  = 9167  = 110 ksi Z 1 12  

The residual stresses are shown o on n the right side in the aabove bove figure.

Comment:  An overload caus causing ing yielding pr produced oduced residual sstresses tresses that are are favorable to future loads in the same direction and unfavorable to future loads in the opposite direction.

4-129

 

SOLUTION (4.90) Known: Two rectangular steel beams beams having a known tensile yield strength are loaded in bending and Z = I/c is known. The dimensions and a stress concentration concentration factor are given for both beams. Find: (a) For each beam, determine what m moment, oment, M, causes (1) initial yielding an and d (2) complete yielding. (b) Beam A is loaded to cause yielding to a depth of 6.35 mm mm.. Determine and plot the distribution of residual stresses that remain after the load is removed. Schematic and Given Data: 12.5 mm 12.5 mm 25 mm

A

25 mm

37.5 mm

B

Z = 1302 mm3

Z = 1302 mm3

Syt = 550 MPa

S yt = 550 MPa K t = 2.5

12.5 mm

 

Assumptions: 1. The idealized stress-strain curve is appropriate. 2. The beam is homogenous. 3. There are no residual stresses initially. Analysis: 1. For initial yielding, using Eqs. (4.7) and (4.21),

Beam A:

 = S yt = M   Z

!

  M = Syt Z = (550 MPa)(1302 mm3) = 716.1 N·m

! 

Beam B: ! = S yt = M  K t  Z  S ytZ  550 MPa 1302 mm 3 = M=   = 286.4 286.44 4 N·m  Kt 2.5

4-130

!

 

 

2.

For complete yielding, 0 F

h/2

F

Thus, for both beams, M = F( h ) = (Syt)( h )(b) ( h ) = 2 2 2  

3.

  550 550 MPa MPa

  25 2

  12.5

  25 2

  = 1074.22

N·m  

! 

For loa loading ding be beam am A to yield the beam to a depth of 6.35 m mm m and and the then n rele releasing asing the load we have: 0 6.35 mm

550

-208.5 MPa 0 758.5

+176.3 MPa

-176.3 MPa

6.35 mm

+208.5 MPa "Moment on" curve

Residual stresses

 and Elastic curve

4.

 

The moment to yield the beam to a depth of 6.35 m mm m is given by M = (stress)(area)(mom (stress)(area)(mom.. arm) + (avg. stress)(area stress)(area)(mom. )(mom. arm) = 550 550 MPa6.35 m mm m 12.5 12.5 mm 18. 18.65 65 mm   +

  550 MPa   6. 6.15 15 mm 12. 12.5 5 mm 8. 8.20 20 mm 2

 

M = 987.54 N·m 5.

The elastic stress for M = 987.54 N·m is calculate calculated d as ! elastic =

6.

  M  987.54 N·m  =   = 758.5 MPa 3 Z 1302 mm  

The residual stresses are shown o on n the right side in the aabove bove figure.

Comment:  An overload caus causing ing yielding pr produced oduced residual stresses that aare re favorable to future loads in the same direction and unfavorable to future loads in the

opposite direction. 4-131

 

SOLUTION (4.91) Known: A 12 in. length of aluminum tubing with a cross-sectional area of 1.5 in. 2  expands 0.008 in. from a stress-free condition at 60 °F when the tube is heated to a uniform 260°F. Find: Determine the end loads on the aluminum tubing loads and the resultant compressive stresses. Schematic and Given Data:

12.000 in.

12.008 in. T = 260 oF

T = 60 oF

P = 0 lb

P = 0 lb

P=?

P=?

  Assumptions: 1. The tube material is homogenous and isotropic. 2. The material stresses remain within the elastic range. 3. No local or column bending occurs. Analysis: 1. For the unrestrained tube + 

= ,'&   =

'L

12!10

= L+ = 12 in. in.

-3 in.  / F (200 F ) = 2. 2.4x 4x10 10 in. 

-6

"

2.4!10

  "

-3

= 0.0288 in. 

4-132

 

2.

Since the me measured asured expansion was only 0.008 in., the constraints must aapply pply forces sufficient to to produce a defle deflection ction of 0.0208 in. From the rela relationship tionship !

  =

  PL AE 

which is from elementary elastic theory, where (  = 0.0208 in., L = 12.000 in., 6

A = 1.5 in.2, and E = 10.4 - 10  ksi.  ksi.   With substitution we have  0.0208 in. =

P 12.000 in. 2 (1.5 1.5 in. in. )(10.4

!

6

)

10 psi

 

yielding P = 27,040 lb 3.

The resultant stress is, ! =   P  =  27,040 lb  = 18,027 psi 2 A 1.5 in.  

  !

Comment: Since these answers are based on elastic relationships, they are valid only if the material has a yield strength of at least 18.03 ksi at 260 °F.

SOLUTION (4.92) Known: A 250 mm length of steel tubing with a cross-sectional area of 625 mm 2  expands longitudinally 0.20 mm from a stress-free condition at 26 °C when the tube is heated to a uniform 249°C. Find: Determine the end loads on the steel tubing and the resultant internal stresses. Schematic and Given Data:

250 mm

250.2 mm

T = 26 oC

T = 249 oC

P = 0 kN

P = 0 kN

P=?

P=?

  4-133

 

Assumptions: 1. The tube material is homogenous and isotropic. 2. The material stresses remain within the elastic range. Analysis: 1. For the unrestrained tube

2.

-6

+ =

,'& =   12!10

'L

= L+ = 250 mm

-3

(249249-26 26) = 2.68!10  

2.68 2.68!10

-3

= 0.669 mm 

Since the me measured asured eexpansion xpansion was only 0.20 mm, the constraints must aapply pply forces sufficient to to produce a defle deflection ction of 0.469 mm. From the rela relationship tionship !

  =

  PL AE  

which is from elementary elastic theory, where ( = 0.469 mm, L = 250 mm, and A = 625 mm2 = 625 mm E = 207!10

9

2 !

  1m

!

  1m

1000 mm 1000 mm

Pa = 207 207!10

9

N m

 = 207!10 2

N

3

mm

  = 0.000 0.000625 625 m

 

 

2

P 250 mm

Therefore, 0.469 mm =

625 mm

2

207!10

3

 

N mm

2

and P = 242,707 N 3.

The resultant stress is, ! =

 242,707 N  = 388 MPa 2 625 mm  

 

!

Comment: Since these answers are based on elastic relationships, they are valid only if

the material has a yield strength of at least 388 MPa at 249°C.

4-134