HW Solutions Chapter 4-5

Chapter 4 (MECH of MAT)

4-4 The copper shaft id subjected to the axial loads shown. Determine the displacement of end A with respect to end D if the diameters of each segment are

mm, mm, and mm AB BC CD d 20 d 25 d 12 . Take E_cu  126GPa. kN kN kN kN kN kN kN x AB AB x BC AB x CD CD F F F F F F F F F                              







0 40 0 40 0 40 50 0 10 0 30 0 30

















































/ / / N m N . m N . m Pa . m Pa . m Pa . m . m i i AB AB BC BC CD CD D A i i i AB AB BC BC CD CD D A D A F L F L F L F L E A E A E A E A                          _{ }  _{ }  _{ }        



3 3 3 2 2 2 9 9 9 3 40 10 2 10 10 3 75 30 10 2 5 126 10 0 02 126 10 0 025 126 10 0 012 4 4 4 3 8484 10 / . mm Ans. D A



  3 85

4-13 A spring-supported pipe hanger consists of two springs which are originally unstretched and have a stiffness of k = 60 kN/m, three 304 stainless steel rods, AB and CD which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe and the fluid it carries have a total weight of 4 kN, determine the

displacement of the pipe when it is attached to the support.

 

 





 

kN kN kN kN kN y EF EF y AG CH AG CH y S AG S AG F F F T F F F F F T symmetry F F F F F C                              







0 4 0 4 0 4 0 2 0 0 2

The displacement of pipe is from 3 parts 1. The elongation of rod EF

2. The contraction of spring (Consider force F that produced the contraction in one spring) _S

3. The elongation of rod AG or CH (Consider force F_AG that produced the elongation in one rod)

































N . m _N N . m N/m Pa . m Pa . m . pipe EF S AG S AG AG EF EF pipe EF EF spring AG AG pipe pipe FL EA F F L F L E A k E A                 _  _       _           _{ }  _{ }       3 ₃ 3 3 2 2 9 9 4 10 0 75 _{2 10} 2 10 0 75 60 10 193 10 0 012 193 10 0 005 4 4 33 867 10 m3 . mm Ans. pipe



  33 9

stainless steel  E GPa

4-21 The rigid beam is supported at its ends by two A-36 steel tie rods. The rods have diameters

mm AB

d  12 and . d_CD  7 5mm. If the allowable stress for the steel is

MPa allow

  115 , determine the intensity of the distributed load w and its length x on the beam so that the beam remains in the horizontal position when it is loaded.



  





 

. m N where : N/m and : m . N w . A CD CD y AB CD AB CD AB x M F wx wx F w x F F F wx F F wx wx F wx       _{ }   _{ }             





2 2 0 2 4 0 2 1 4 8 0 0 4 8



here : N/m and : m w x



 

2 Given;

1. The beam remains in the horizontal position





and or

Member and will reach allowable stress at the same time AB CD CD CD AB AB AB CD AB CD AB AB CD CD CD AB AB CD AB CD FL AE F L F L E E L L A E A E F F A A AB CD         _  _       

2. _allow  115MPa





MPa MPa . m . kN AB allow AB AB AB AB F A F A F F        _  _      2 115 115 0 012 4 13 006





MPa MPa . m . kN CD allow CD CD CD CD F A F A F F        _  _      2 115 115 0 0075 4 5 0805 From (1); . . N N N/m . wx w x   3  2   3 2 24 386 10 5 0805 10 4 8 From (2);





. . . N N . . . N . N : m . m . N/m . N/m . x x x x where x x x w        _{ } _{ }               3 3 2 3 2 2 3 3 3 3 3 2 24 386 10 24 386 10 13 006 10 4 8 24 386 10 13 006 10 5 0804 10 1 3483 24 386 10 13 414 10 1 3483 . kN/m . m Ans. w x    13 4 1 35

4-36 The A-36 steel pipe has an outer radius of 20 mm and an inner radius of 15 mm. If it is fits snugly between the fixed walls before it is loaded, determine the reaction at the walls when it is subjected to the load shown.

Compatibility;









  



 

where and . m . m C A B A C B BC BC AB AB AB BC AB BC AB AB BC BC A B A B FL EA F L F L E E A A E A E A F F F F        _  _             0 0 0 0 3 0 7 0 7 2 3 Insert (2) into (1); kN . kN kN . kN . kN B B B A F F F F       7 16 3 4 8 16 4 8 11 2 . kN . kN Ans. A A F F      11 2 4 8 x AB A AB A x C BC BC C F F F F F F F F F F                   





0 0 0 0

 

Equilibrium; kN kN x A B A B F F F F F          



0 16 0 16 1

4-45 The distributed loading is supported by the three suspender bars. AB and EF are made from aluminum and CD is made from steel. If each bar has a cross-sectional area of 450 mm2, determine the maximum intensity w of the distributed loading so that an allowable stress of



_{allow st}



 180 MPa in the steel and



_{allow al}



 94 MPa in the aluminum is not exceeded. E_st 200 GPa, E_al 70 GPa.

Compatibility; the system is symmetry, so same displacement at A, C, and E





 

where and GPa GPa al st al al st st al st al st al al st st al st st al FL EA F L F L L L A A E A E A F F F F   _  _        70 200 20 2 7 Insert (2) into (1); . . al al al st al F F w F w w F F w w w         20 2 3 7 21 0 61765 34 20 20 21 30 1 7647 7 7 34 17

Because F_st 3F_al and A_st  A_al, the stress in steel is 3 times of in aluminum while _stallow  2_alallow, therefore the system is controlled by steel.

Case 1; Assume steel failed

MPa Pa m . N/m allow st st st st F A w w           1 6 6 2 3 1 180 30 17 _{180 10} 450 10 45 9 10

Case 2; Assume aluminum failed

MPa Pa m . N/m allow al al al al F A w w           2 6 6 2 3 2 94 21 34 _{94 10} 450 10 68 486 10

Because w₁w₂; so the system is controlled by steel member and the maximum intensity w is 45.9 kN/m Ans.









 

Equilibrium; . m . m C E A E A al y al st al st M F F F F F F F F w F F w                   





0 1 5 1 5 0 0 2 3 0 2 3 1

4-112 The rigid link is supported by a pin at A and two A-36 steel wires, each having an unstretched length of 300 mm and cross-sectional area of 7.8 mm2. Determine the force developed in the wires when the link supports the vertical load of 1.75 kN.

Equilibrium;















 

. kN mm mm mm . . kN A B C B C M F F F F          



0 1 75 150 100 225 0 2 25 2 625 1 Compatibility; From similar triangle





 

mm mm . . where and . C B C B C C B B C B C C B B C C B B C B FL AE F L F L L L A E A E A E A E F F   _      _  _        225 100 2 25 2 25 2 25 2 Insert (2) into (1);









. . . kN . kN . . . kN . kN B B B C B F F F F F       2 25 2 25 2 625 0 43299 2 25 2 25 0 43299 0 97423 . kN . kN Ans. B C F F    0 433 0 974

4-84 The rigid block has a weight of 400 kN and is to be supported by posts A and B, which are made of A-36 steel, and the post C, which is made of C83400 red brass. If all the posts have the same original length before there are loaded, determine the average normal stress developed in each post when post C is heated so that its temperature is increased by 10 oC. Each post has a cross-sectional area of 5000 mm2.

Compatibility; The system is symmetry, the displacement of all bars are same









 



 

o



o



where and / C C m Pa m Pa . N . A B st st B B br B st B st B st st B brass st B B st B st FL AE F L F L TL L L A A A E A E F F F F F F           _  _                       6 6 2 9 6 2 9 18 10 10 5000 10 200 10 5000 10 101 10 0 505 90900 90900 0 505 2

 

Insert (2) into (1);









N . N . N N . N . . N . N st st st B st F F F F F              3 3 3 3 2 90900 0 505 400 10 123 39 10 90900 0 505 90900 0 505 123 39 10 153 21 10

Determine the stress in each post;

. N . MPa mm . N . MPa mm st A C B B F A F A F A      _               3 2 3 2 123 39 10 24 678 5000 153 21 10 30 642 5000

. MPa and . MPa Ans.

A C B      24 7 30 6

 

 

 

Equilibrium; m m kN kN G C A A C st y st B st B M F F F F F F F F F F                   





0 1 1 0 0 2 400 0 2 400 1

4-114 The 2014-T6 aluminum rod has a diameter of 12 mm and is lightly attached to the rigid supports at A and B when T1 =25oC. If

the temperature becomes T2 = -20oC, and an

axial force of P = 80 N is applied to the rigid collar as shown, determine the reactions at A and B. Compatibility;









where and . m . P B A C A B C

Force Temp Force Temp C A C A B C B C AC AC BC BC AC AC BC BC AC BC al AC BC AC AC BC BC A FL EA F L F L TL TL E E E A A E A E A F              _  _                                     9 0 0 0 0 0 125 73 1 10











































 

o o o o o o / C C C . m a . m . m / C C C . m . Pa . m . N . N B B A B A F F F F F                      _{ }       _             _     _{ }         6 2 6 2 9 23 10 20 25 0 125 0 012 4 0 2 23 10 20 25 0 2 0 73 1 10 0 012 4 0 625 13905 0 625 13905 2 Insert (2) into (1);









. N N . N . N . . N N . N A A A B A F F F F F           0 625 13905 80 8606 15 0 625 13905 0 625 8606 15 13905 8525 2 . kN and . kN Ans. A B F F  8 61  8 53 

Chapter 5 (MECH of MAT)

x AC A AC A x BC B BC B F F F F F F F F F F                    





0 0 0 0

 

Equilibrium; N N x A B A B F F F F F           



0 80 0 80 1 o

5-12 The solid shaft is fixed to the support at C and subjected to the torsional loading shown. Determine the shear stress at points A and B and sketch the shear stress on volume elements located at these points.





































. m N.m . m . Pa . m . m . m N.m . m . Pa . m . m A B T J T T    _{ }  _{ }  _              2 6 4 4 1 6 4 4 0 035 500 0 035 7 4241 10 0 035 0 035 2 2 0 02 800 0 02 6 7878 10 0 035 0 035 2 2 . MPa . MPa Ans. A B      4 45 6 79 N.m 800 N.m 300 C T N.m 800 N.m 800 N.m 300 T₁ T2 N.m N.m x M T T        



1 1 0 800 0 800 x N.m N.m N.m x M T T         



2 2 0 800 300 0 500

5-13 A steel tube having an outer diameter of 62.5 mm is used to transmit 3 kW when turning at 27 rev/min. Determine the inner diameter d of the tube to the nearest multiples of 5 mm if the allowable shear stress is _allow  70MPa.







rev/min



rad/rev W



min/ sec



N.m

P T P T         3 103 10000 27 2 1 60 3









max . m Pa . m . m allow allow Tc J Tc J d d                   _{  }  _      6 4 ₄ 3 10000 0 0625 3 2 70 10 0 0625 32 56 8345 10 mm Ans. d   60

5-56 The motor delivers 32 kW to the 304 stainless steel solid shaft while it rotates at 20 Hz. The shaft has a diameter of 37.5 m and is supported on smooth bearings at A and B, which allow free rotation of the shaft. The gears C and D fixed to the shaft remove 20 kW and 12 kW, respectively. Determine the absolute maximum stress in the shaft and the angle of twist of gear C with respect to gear D.

















W

N.m

Hz

W

N.m

Hz

W

N.m

Hz

C D

P

T

P

T

P

T

P

T















































3 3 3

32 10

800

2

20

20 10

500

2

20

12 10

300

2

20

Determine the internal load in each section; N.m x AC AC M T T T T



        



0 0 800 N.m N.m x CD C CD C M T T T T T T







            



0 0 800 500 300 x BD BD M T T        



0 0 0

Determine the shear stress in each section;









. m N.m . Pa . m . m N.m . Pa . m AC AC CD CD BD BD Tc J T c J T c J T c J    _   _   _                                  6 4 6 4 800 0 0375 2 24 593 10 0 0375 32 300 0 0375 2 9 2225 10 0 0375 32 0 Angle of twist;













o N.m . m . rad . Pa . m CD CD CD CD CD TL GJ T L G J    _  _                 ₉ 4 300 0 2 0 0013116 0 075152 75 10 0 0375 32 max o . MPa _CD . CCW Ans.      24 6 0 0752

5-66 The device serves as a compact torsion spring. It is made of A-36 steel and consists of a solid inner shaft CB which is surrounded by and attached to a tube AB using a rigid ring at

B. The ring at A can also be assumed rigid and

is fixed from rotating. If the allowable shear stress for the material is _allow  84 MPa and the angle of twist at C is limited to o

allow   3 , determine the maximum torque T that can be applied at the end C.

Case 1; Consider _allow 84 MPa in tube AB











 









 



max . m MPa . m . m . m Pa . m . m . N.m . N.m AB allow tube case Tc J T T T T          _       _       4 4 6 4 4 1 0 025 84 0 025 0 01875 2 0 025 84 10 0 025 0 01875 2 1409 34 1409 34

Case 1; Consider _allow 84 MPa in shaft BC





















max . m MPa . m . m Pa . m . N.m . N.m BC allow shaft case Tc J T T T T              4 6 4 2 0 0125 84 0 0125 2 0 0125 84 10 0 0125 2 257 71 257 71

Case 3; Consider _allow  3 o











 















o o and . m . m radian Pa . m . m Pa . m . N.m c allow c B A C B shaft shaft tube tube allow tube tube shaft shaft

TL GJ T L T L G J G J T T T              _    _         _  _    4 4 4 9 9 0 3 0 6 3 180 75 10 0 025 0 01875 75 10 0 0125 2 2 240 02 Tcase3 240 02. N.m

Because ,T_case3T_case2 T_case1, therefore the system is controlled by _allow and T_max  240 N.m Ans.

x tube tube x shaft shaft M T T T T M T T T T                  





0 0 0 0

5-77 The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the couple, determine the

maximum shear stress in regions AC and CB.

Equilibrium;

 

N.m N.m x A B A B x AC A AC A x BC B BC B M T T T T M T T T T M T T T T                              







0 200 0 200 1 0 0 0 0 Compatibility;

















 

where . m . m B A C A B C AC AC BC BC AC AC BC BC AC AC BC BC A B A B TL GJ T L T L G J G J G J G J T T T T               0 0 0 0 4 0 6 0 3 2 2 Insert (2) into (1);





N.m N.m N.m N.m B B B A B T T T T T       3 200 2 80 3 3 80 120 2 2

Shear stress in each section;

























N.m . m . Pa . m N.m . m . Pa . m AC AC BC BC Tc J T c J T c J   _  _          6 4 6 4 120 0 04 2 9 5493 10 0 04 32 80 0 04 2 6 3662 10 0 04 32 . MPa . MPa Ans. AC BC      9 55 6 37

5-79 The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core. If it is fixed to a rigid support at A, and a torque of T = 50 N.m is applied to it at C, determine the angle of twist that occurs at C and compute the maximum shear stress and maximum shear strain in the brass and steel. Take G_st 80 GPa, G_br 40 GPa.

Consider section BC;

 







 

 





 



 

N.m N.m ; where GPa . m . m GPa . m x br st br st st br st st br br st br st st br br st br st br M T T T T Compatibility T L T L L L G J G J T T T T                     _    



4 4 4 0 50 0 50 1 80 0 02 0 01 40 0 01 2 2 30 2 Insert (2) into (1); N.m N.m N.m N.m br br br st br T T T T T        _{ }   30 50 50 31 50 1500 30 30 31 31

Determine the deformation of section BC;

 









o N.m m . rad . CW Pa . m br br BC br br T L G J  _            ₉ 4 50 1 31 0 0025670 0 14708 40 10 0 01 2 Consider section AB;















o N.m N.m N.m . m . rad . CW Pa . m x AB AB AB AB B A AB AB M T T T L G J  _              



4 9 0 50 0 50 50 1 5 0 0037302 0 21372 80 10 0 02 2

Determine the rotation at end C; o . rad . rad . rad . C C B B A C           0 0025670 0 0037302 0 0062972 0 36080

Determine the stress and strain in steel;

















 



N.m . m . Pa . m N.m . m . Pa . m . m . Pa . rad Pa AB AB st AB BC st st st BC BC BC st st Tc J T c J T c J G   _  _     _                     _          6 4 6 4 4 6 6 9 50 0 02 3 9789 10 0 02 2 1500 0 02 31 4 1072 10 0 02 0 01 2 4 1072 10 51 340 10 80 10

Determine the stress and strain in brass;









N.m . m . Pa . m . Pa . rad Pa BC br br br BC BC BC br br Tc J T c J G   _     _                      6 4 6 6 9 50 0 01 31 1 0268 10 0 01 2 1 0268 10 25 670 10 40 10 o . CW Ans. C    0 361 max max max max . MPa in section . rad in section . MPa . rad Ans. st st br st BC BC              6 6 4 11 51 3 10 1 03 25 7 10