Chapter 5 Ansel C.ugural - Solution Manual Mechanics of Materials PDF

Preview

Summary

This is a rare solution manual for Ansel C.ugural - Solution Manual Mechanics of Materials. Excellent for Mechanical engineering students....

Description

CHAPTER 5 PROBLEM (5.1) A hollow steel shaft of outer radius c is fixed at one end and subjected to a torque T at the other end , as shown in Fig. P5.1. If the shearing stress is not to exceed  all , what is the required inner radius b ? Given:

T = 2 kN m,

c = 30 mm ,

 all

= 80 MPa

SOLUTION

 all  c

Tc

 2

b

(c  b ) 4

4

or

2(2 10 )(0.03) [(0.03) 4  b 4 ] 3

80(106 )  or 80(106 )[0.81(106 )  b4 ]  38.197 Solving, b  0.024 m  24 mm

________________________________________________________________________ PROBLEM(5.2) A solid shaft of diameter d = 2 in. is to be replaced by a hollow circular tube of the same material, resisting the same maximum shear stress and the same torque T (Fig. P5.2). Determine the outer diameter D of the tube if its wall thickness is t = D/25.

SOLUTION t

Di  23 D 25

d=2in.

D

 max 

D 25

T ( d 2) T ( D 2)  4  d 32  ( D 4  Di4 ) 32

or 23 4 ) ] 25 (2) 3  D3 (0.2836) D  3.046 in. d 3  D 3 [1 (

5-1

________________________________________________________________________ PROBLEM(5.3) A solid shaft of diameter d and a hollow shaft of outer diameter D and thickness t = D/4 are to transmit the same torsional loading at the same maximum shear stress (Fig. P5.2). Compare the weights of these two shafts of equal length.

SOLUTION Di 

D 2

d D

 max 

T ( D 2) 16T 16T   3  1 d (D 4  Di4 )  D 3 (1 ) 32 16

or 15 d 3  D 3 ( ), 16

D  1.0217d

Ratio of weight: D2  (D 2)2 0.75(1.0217d )2   0.783 d2 d2 ________________________________________________________________________ PROBLEM (5.4) The circular shaft is subjected to the torques shown in Fig. P5.4. What is the largest shearing stress in the member and where does it occur?

SOLUTION Apply the method of sections between the change of load points: TDE  30 N  m TEF  50 N  m TCD  TBC  120 N  m TAB  80 N  m We have Jh  2 [(0.025)4  (0.015)4 ]  0.534 106 m4 Thus,  max  Tc J : 2T 2(80)   3.26 MPa 3  c  (0.025) 3 2(120) 120(0.025)   4.89 MPa,  5.62 MPa  CD  3 0.534  10 6 (0.025)

 AB 

 BC

5-2

 DE 

30(0.025)  1.40 MPa , 0.534  106

So,

 EF 

50(0.025)  2.34 MPa 0.534 106

 CD   max  5.62 MPa

________________________________________________________________________ PROBLEM (5.5) Four pulleys, attached to a solid stepped shaft, transmit the torques shown in Fig. P5.5. Calculate the maximum shear stress for each segment of the shaft.

SOLUTION Apply the method of sections between the change of load points: TCD  1 kN  m TBC  2.5 kN  m TAB  2.5 kN  m

Therefore, max  2T  c3 :

2(2.5 10 3)  58.9 MPa  (0.03) 3 2(2.5  103 )  101.9 MPa  BC   (0.025)3 2(1 103 )  79.6 MPa  CD   (0.02)3 ________________________________________________________________________ PROBLEM (5.6) Redo Prob. 5.5, with a hole of 20-mm diameter drilled axially through the shaft

 AB 

to form a tube.

SOLUTION We have, applying the method of sections: TCD  1 kN  m TBC  2.5 kN  m TAB  2.5 kN  m

Hence,

 max 

2Tc  ( c4  b4 )

gives,

2(2.5 10 )(0.03)  59.7 MPa  [(0.03)4  (0.01)4 ] 3 2(2.5 10 )(0.025)  104.5 MPa  BC   [(0.025)4  (0.01)4 ] 3 2(1 10 )(0.02)  84.9 MPa  CD   [(0.02) 4  (0.01) 4 ] 3

 AB 

5-3

________________________________________________________________________ PROBLEM(5.7) A stepped shaft ABC is fixed at the left end and carries the torques TB and TC at sections B and C as illustrated in Fig. P5.7. Calculate the maximum shearing stress in the shaft. Given:

d1 = 2 in.,

d2 = 1 ¾ in.,

TB = 30 kip  in.,

TC = 12 kip  in.

SOLUTION TBC  12 kip  in.

TAB  18 kip  in.

Thus,  max  2T  c gives 3

2(18 10 )  11.46 ksi  (1) 3 2(12  103 )  BC   11.4 ksi  (7 8) 3 ________________________________________________________________________ PROBLEM (5.8) A stepped shaft ABC with fixed end at A is subjected to the torques TB and TC 3

 AB 

at sections B and C (Fig. P5.7). Determine the maximum shearing stress in the shaft. Given: d1 = 60 mm, d2 = 50 mm, TC = 3 kN  m, TB = 2 kN  m

SOLUTION TBC  3 kN  m

TAB  1 kN  m

Then,  max  2T  c yields 3

 BC 

2(3 103 )  122.2 MPa  (0.025) 3

 AB 

2(1 103 )  23.6 MPa  (0.03) 3

________________________________________________________________________ PROBLEM (5.9) Determine the values of the torques TB and TC applied at sections B and C so that each segment of the shaft shown in Fig. P5.7 is stressed to a permissible shear strength of  all . Given:

d1 = 2 3/8 in.,

d2 = 2 in.,

 all

= 14 ksi

SOLUTION

 max



2T , c 3

T

 max c 3 2

Thus, 14(10 ) (1)  21.99 kip  in. 2 3 3 14(10 ) (19 16) TBA  TB  TC   36.83 kip  in. 2 3

3

TC  TBC 

5-4

or TB  21.99  36.83  58.82 kip  in .

________________________________________________________________________ PROBLEM (5.10) The torques TB and TC are applied at sections B and C of the stepped shaft shown in Fig. P5.7. Determine: (a) The maximum permissible value of the torque Tall , if the allowable tensile stress in part BC is

 all . (b) The allowable compressive stress in part BC. Given:

d1 = 100 mm,

d2 = 50 mm,

TB = 12 kN  m,  all = 140 MPa

SOLUTION The polar moment of inertias are  J AB  (100) 4  9.817(10 6 ) mm 4 32  J BC  (50) 4  0.614(106 ) mm 4 32 In tension and compression (Figs. 5.8 and 5.10): Tc  max   max  J TBC CBC T (0.025) ; 140(10 6 )  C J BC 0.614(106 ) TC  TBC  3.438 kN  m

(a) We have  all  or

(b) TAB  TB  TC  12  3.438  8.562 kN  m TAB cAB 8.562(103 )(0.05)  J AB 9.817(10 6 )  43.61 MPa

 all 

________________________________________________________________________ PROBLEM (5.11) A solid axle of diameter d is made of cast iron having an ultimate strength of in tension  u , ultimate strength in compression  u ’ , and ultimate strength in shear  u . Calculate the largest torque that may be applied to the axle. Given: d = 50 mm,  u = 170 MPa,  u ’ = 650 MPa,

u

= 240 MPa

SOLUTION Formula (5.14a): at   45o  max   170 MPa

at   135o  min    650 MPa

5-5

The critical criterion is thus Tc 170 MPa J gives 6 3 170(10 ) (0.025) Tmax   4.17 kN  m 2 ________________________________________________________________________ PROBLEM (5.12) A hollow shaft is made by rolling a plate of thickness t into a cylindrical shape and welding the edges along the helical seams oriented at an angle  to the axis of the member (Fig. P5.12). If the permissible tensile and shear stresses in the weld are  all and  all , respectively, what is the maximum torque that can be applied to the shaft?  = 60o,  all = 100 MPa, D = 120 mm, t = 5 mm,

Given:

 all

= 55 MPa

SOLUTION

  60  90  150 o  J  [(0.06)4  (0.055)4 ]  5.984  106 m4 2 From Eq. (5.14a) at   150o :  x '   sin 300o   0.866 Tc T (0.06) 100  106  0.866  0.866 6 J 5.984 10 or T  11.52 kN  m Similarly, Eq. (5.14b):  x ' y '   cos300 o  0.5 gives T (0.06) 55  106  0.5 , T  10.97 kN  m 5.986 10 6 Thus, Tall  10.97 kN  m ________________________________________________________________________ PROBLEM (5.13) Redo Prob. 5.12, assuming that the helical seam is oriented at an angle  = 40 o to the axis of the member (Fig. P5.12).

SOLUTION From solution of Prob. 5.12: 6 4 J  5.984 10  m

5-6

From Equations (5.14) at   40  90  130o ,  x '   sin 260o   0.985 Tc T (0.06) 100  (106 )  0.985  0.985 5.984 106 J or T  10.13 kN  m and  x ' y '   cos 260 o  0.174 55  106  0.174

T (0.06) , 5.984  106

T  31.52 kN  m

Thus, Tall  10.13 kN  m

________________________________________________________________________ PROBLEM (5.14) A solid steel bar having the shear modulus of elasticity G and diameter d is subjected to a torque T. Determine the maximum shear strain in the member. Given: d = 2 in., T = 20 kip in., G = 12 x 10 6 psi

SOLUTION

 max 

2T 2(20 103 )   12.73 ksi c3  (1)3

 max

12.73(103 )   1060  G 12(106 ) ________________________________________________________________________ PROBLEM (5.15) A tube of outer diameter D and inner diameter d experiences a torque at its

 max 

ends T, as shown in Fig. P5.15. At this loading, from measurement of a strain gage positioned at an angle  it is calculated that shear strain  =1155  . Determine the shear modulus of elasticity G of the material.

Given: D = 100 mm ,

d = 80 mm,

T = 10 kN  m,  = 60o

SOLUTION J



[(0.05)4  (0.04)4 ]  5.796  106 m4

2 Equations (5.14b) with   30o gives   max cos(60o )  0.5 max Thus, 0.5Tc  G    J (1155 10 6 )



0.5(10 103 )(0.05)  37.3 GPa 5.796 10 6 (1155 10 6)

5-7

________________________________________________________________________ PROBLEM (5.16) Rework Prob. 5.15 for the case in which a strain gage positioned at an angle  = 25° (Fig. P5.15).

SOLUTION From solution of Prob. 5.15: 6 4 J  5.796 10  m Equations (5.14b) with   65o :   max cos(130o )  0.643 max Therefore, G

0.643Tc (0.643)(10  103 )(0.05)     J (1155  106 ) (5.796 106 )(1155 10 6 )

 48 GPa

________________________________________________________________________ PROBLEM (5.17) An electric motor at A exerts a torque of TA on part AB of the shaft ABCD of the assembly shown in Fig. P5.17. Torques TB , TC , and TD transmitted through a gears B and C, and pulley D, respectively. Calculate the maximum shear stresses in the parts AB, BC, and CD of the shaft. Given:

dAB = 2 in.,

dBC = 1½ in.,

TC = 8 kip  in., Assumptions:

dCD = 1½ in.,

TA = 25 kip  in., TB = 12 kip in.

TD = 5 kip  in.

The entire shaft is solid. Friction in the bearings may be neglected.

SOLUTION For a circular shaft :  max  Tc J  2T  c3 . 1 d AB  1 in. 2 2(25  103 )   15.92 ksi  (1) 3

Shaft AB: TAB  25 kip  in. cAB 

 AB 

Shaft BC:

2TAB  c 3AB

1 dBC  7 8 in. 2 2(13 10 3 )   12.35 ksi  (7 8) 3

TBC  13 kip  in. cBC 

 BC 

2TBC 3  c BC

5-8

1 Shaft CD: TCD  5 kip  in. cCD  d CD  3 4 in. 2 3 2T 2(5 10 )  CD  CD   7.54 ksi 3  c CD  (3 4)3 ________________________________________________________________________ PROBLEM (5.18) An electric motor A exerts a torque TA on the assembly shown in Fig. P5.17. The shaft is made of steel having an allowable shear stress all . Determine the required diameters dAB , dBC , and dCD for each part of the shaft. Given:

TA = 3 kN  m, TB = 1.4 kN m, TC = 1.0 kN  m,

TD = 0. 6 kN  m,  all = 80 MPa

SOLUTION For a circular shaft  16T  d 3 or d  3 16T  . We have  all  80 MPa Shaft AB: TA  3 kN  m 16(3 10 )  0.0576 m  57.6 mm  (80 10 6 ) 3

d AB 

3

Shaft BC: TBC  1.6 kN  m 16(1.6 10 )  0.0467 m  46.7 mm  (80 10 6) 3

d BC 

3

Shaft CD: TCD  0.6 kN  m d CD 

3

16(0.6 10 3)  0.0337 m  33.7 mm  (80 106 )

________________________________________________________________________ PROBLEM (5.19) Solve Prob. 5.18 for the case in which shaft ABCD (shown in Fig. P5.17) is hollow with outer diameter d and inner diameter d/2 throughout its entire length.

SOLUTION T ( d 2) 16Td (16)  d  [15d 4 ] 4 4 [ d ( ) ] 32 2 6 3 T  14.726(10 ) d

 all  80 106  or



Shaft AB: TA  3 kN  m 3

d AB 

3

3(10 )  0.0588 m  58.8 mm 14.726(10 6)

5-9

(1)

Shaft BC: TBC  1.6 kN  m d BC 

3

1.6(103 )  0.0477 m  47.7 mm 14.726(10 6)

Shaft CD: TCD  0.6 kN  m d CD 

3

0.6(103 )  0.0344 m  34.4 mm 14.726(10 6)

________________________________________________________________________ PROBLEM (*5.20) The two intersecting shafts are connected by bevel (or conical) gears as shown in Fig. P5.20. Calculate the maximum permissible torque TA that can be applied to shafts A. Given:  = 71o , dA = 15 mm, dB = 20 mm Assumption: The gears are made of heat-treated steel for which the allowable shear stress is all = 120 MPa

*SOLUTION Shaft A :  all  120 MPa J A all  3  cA all cA 2  79.52 N  m

TA 

1 c A  d A  7.5 mm 2   (7.5 10 3) 3(120 10 6 ) 2

rA



Shaft B :  all  120 MPa

rB

1 c B  d B  10 mm 2

J B all  3   cB all  (10 10 3 ) 3 (120 10 6 ) cB 2 2  188.5 N  m

TB 

rA 1  TB   TB rB  tan    1  (188.5)  64.91 N  m  o   tan 71  Permissible value of TA is the smaller of the three: Statics :

TA 

(TA )all  64.91 N  m

5-10

________________________________________________________________________ PROBLEM (5.21) Consider the aluminum shaft AF having shear modulus of elasticity G loaded by the torques applied at the sections B , D , E , and F as illustrated in Fig. P5.21. Determine: (a) The angle of twist at C (b) The angle of twist at F . Given: a = 100 mm, G = 28 GPa

SOLUTION We have: Jh 



[(0.025) 4  (0.015)4 ]  534  109 m4

2 Applying method of sections, as needed: TEF  70 N  m TDE  30 N  m TAB  100 N  m TBC  TCD  150 N  m

(a)  C   (TL GJ ) gives: 0.1 C  (100  150) 9  4 (28 10 ) (0.025) 2 3  0.291 10 rad  0.017o (b)  F  C   CF result in 0.1 (150  30 70) 9 (28 10 )(534 10  ) rad  0.089o

 F  0.291 10 3   1.562  10  3

9

________________________________________________________________________ PROBLEM (5.22) Pulleys A, B, C , and D are attached to a solid steel shaft with shear modulus of elasticity G and transmit the torques as illustrated in Fig. P5.22. Determine: (a) The relative angle of twist in degrees between pulleys A and D. (b) The relative angle of twist in degrees between pulleys A and C. Given: L 1 = 0.5 m, L 2 = 1.5 m, G = 80 GPa, L 3 =1 m

SOLUTION Apply the method of sections: TCD 1.5 kN  m TAB  3 kN  m

TBC  3.5 kN  m

(a)  AD   (TL GJ ) gives

5-11

 AD 

103 (80 109 )



[

3(0.5) 3.5(1.5) 3(1) ]   4 4 (0.035) (0.03) (0.025) 4

2 1 ( 1.000  6.48 7.68)  0.105 rad  6o  40

(b)  AC 

1 ( 1.000  6.48)  0.044 rad  2.5o 40

________________________________________________________________________ PROBLEM (5.23) A brass rod AB is bonded to an aluminum rod BC as shown in Fig. 5.23. What is the angle of twist at C? The shear moduli of elasticity for brass and aluminum are Gb and Ga , respectively. Given: Tb = 2 Tc = 40 kip-in., d 1 = 2d 2 = 4 in., L 1 = 2L 2 = 16 in., Gb = 6 x 10 6 psi, Ga = 4 x 10 6 psi

SOLUTION We have TBC  20 kips  in. Then  C   (TL GJ ) yields 20(10 3)

16 1 2 ] (8  ) 3  4 100 3 (2) 2 2 3  23.34 10 rad  1.34o

C 

[

8

TAB  20 kips  in.

 (1)4 (4 10 )



6

________________________________________________________________________ PROBLEM (5.24) Determine the torques TA , TB, and TC of the aluminum shaft having shear modulus of elasticity G = 28 GPa in equiibrium, as shown in Fig. P5.24. The maximum shear stress in segment AB is 80 MPa, and the rotation is 0.02 rad clockwise as viewed at A with respect to C.

SOLUTION

 T 0 :

TB  TA  TC

We have TBA  TA So,

(1) TBC  TA  TB  TC  TC

2T 2TA ; 80(10 6 )  3 c  (0.05) 3 TA 15.71 kN  m or Therefore,  AD   (TL GJ ) :

 max   AB 

5-12

(15.71)2  TC (1.5) ]  (0.05) 4 (0.0375) 4 9  (80 10 ) 2 4 4 4 87.965(10 )  502.72(10 ) 75.85(10 )T C 103

0.02 

[

or TC  5.47 kN  m Equation (1) results in then TB 15.71  5.47  21.18 kN  m

________________________________________________________________________ PROBLEM (5.25) Redo Prob. 5.24, given that the relative angle of twist between A and C is zero.

SOLUTION From solution of Prob. 5.24: TBA  TA 15.72 kN  m

TBC  TC

Thus,    (TL GJ ) gives 3

0

10

[

(15.71)2

 (0.05) 4 (28 10 9 )



TC (1.5) (0.0375)

4

]

2 TC  6.63 kN  m 502.72  75.85TC , TB  TA  TC  15.71  6.63  22.34 kN  m Then,

________________________________________________________________________ PROBLEM (5.26) The solid aluminum rod AB is bonded to the solid bronze rod BC to form a stepped composite shaft that supports the torques T 1 and T 2 acting as shown in Fig. P5.26. Determine the permissible value of the torque T al l , if the angle of twist at the free end is not to exceed 3.5 o . Given: Aluminum rod AB: d 1 = 60 mm, L 1 = 0.4 m, G a = 28 GPa, T1 = T, ( a ) all = 205 MPa, (a )all = 110 MPa Bronze rod BC : d 2 = 75 mm, L 2 = 0.6 m, G b = 41 GPa, (b )all = 260 MPa, (b )all = 140 MPa

T 2 = 3T

SOLUTION TBC  4T We have TAB  T   JAB  (60) 4  1.272(106 ) mm4 JBC  (75)4  3.106(106 ) mm4 32 32 Therefore 6 6 ( ) J 110(10 )(1.272)(10 ) T  a all AB   4.664 kN  m 0.03 c AB

5-13

T

( b )all J BC 140(106 )(3.106)(10 6 )   11.596 kN  m c BC 0.0375

Hence T AB L1 TBC L2  G a J AB G bJ BC T (0.4) 4T (0.6)  (3.5)  6  6  9 9 28(10 )(1.272)(10 ) 41(10 )(3.106)(10 ) 180

 AC   AB   BC 

Solving, T  2.047 kN  m  4.664 kN  m Tall  2.047 kN  m ________________________________________________________________________ PROBLEM (5.27) Three pulleys are attached to a solid stepped shaft with shear modulus of elasticity G and transmit the torques as illustrated in Fig. P5.27. Determine: (a) The maximum shear stress  max in the shaft.

 BC The angle of twist  BD

(b) The angle of twist

between B and C .

(c)

between B and D .

Given: d 1 = 1¼ in., d 2 = 1 in., L 1 = 20 in., L 2 = 25 in., TD = 3.5 kip  in. TB = 4.5 kip  in., TC = 8 kip in.,

G = 4.1 x ...