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 d AB  20 mm, d BC  25 mm, and dCD  12 mm . Take Ecu  126 GPa .

  Fx  0  FAB  40 kN  0   Fx  0  FBC  40 kN  50 kN  0



FAB  40 kN



FAB  10 kN

  Fx  0 



FCD  30 kN

 D/ A   i

 FCD  30 kN  0

Fi Li FAB LAB FBC LBC FCD LCD    Ei Ai E AB AAB EBC ABC ECD ACD

 40 10 N   2 m   126 10 Pa   4  0.02 m  3

 D/ A

9

10 10 N   3.75 m      126  10 Pa    0.025 m   4 3

2

9

 D / A  3.8484  103 m

 D / A  3.85 mm

Ans.

 30 10 N   2.5 m      126  10 Pa    0.012 m   4 3

2

9

2

  

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.

304 stainless steel  E  193 GPa

  Fy  0  FEF  4 kN  0  FEF  4 kN T    Fy  0  FAG  FCH  4 kN  0  FAG  FCH  2 kN T   symmetry    Fy  0  FS  FAG  0

 FS  FAG  2 kN C 

The displacement of pipe is from 3 parts 1. The elongation of rod EF 2. The contraction of spring (Consider force FS that produced the contraction in one spring) 3. The elongation of rod AG or CH (Consider force FAG that produced the elongation in one rod)

 pipe   EF   S   AG  pipe 

F FEF LEF  S EEF AEF kspring

FL     EA  F L  AG AG E AG AAG

 4 10 N   0.75 m   193 10 Pa   4  0.012 m  3

 pipe

9

2  103 N   0.75 m   2  103 N   2 2 60  103 N/m  193  109 Pa    0.005 m      4 

 pipe  33.867  103 m

 pipe  33.9 mm

Ans.

4-21 The rigid beam is supported at its ends by two A-36 steel tie rods. The rods have diameters d AB  12 mm and dCD  7.5 mm . If the allowable stress for the steel is  allow  115 MPa , 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.

x   M A  0  FCD  2.4 m    wx     0 2 2 wx N  where w : N/m and x : m  FCD  4.8   Fy  0  FAB  FCD  wx  0

1

FAB  FCD  wx FAB  wx 

wx 2 N 4.8

 where w : N/m and x : m 

Given; 1. The beam remains in the horizontal position FL    AB   CD   AE  FAB LAB FCD LCD   EAB  ECD and LAB  LCD  AAB E AB ACD ECD FAB FCD  AAB ACD

or

 AB   CD

Member AB and CD will reach allowable stress at the same time

 2

2.  allow  115 MPa F    A 

 AB   allow FAB  115 MPa AAB



FAB

 0.012 m 

F    A 

 CD   allow FCD  115 MPa ACD

 115 MPa 2

4 FAB  13.006 kN



FCD

 115 MPa

 0.0075 m  4 FCD  5.0805 kN 2

From (1); 5.0805  103 N 

wx 2 N 4.8



w

24.386  103 N/m x2

From (2);  24.386  103   24.386  103  x 2 13.006  10 N   N x   x2 x2     4 .8 24.386  103  5.0804  103 N 13.006  103 N   where x : m  x x  1.3483 m 3

w

24.386  103 N/m  13.414  103 N/m 1.34832

 w  13.4 kN/m x  1.35 m

Ans.

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.

Equilibrium;   Fx  0  FA  FB  16 kN  0 FA  FB  16 kN   Fx  0  FAB  FA  0

 FAB   FA

  Fx  0  FC  FBC  0

 FBC  FC

Compatibility;

C A  0

FL     EA 

 B A  C B  0 FAB LAB FBC LBC  0 E AB AAB EBC ABC

 where E AB  EBC

and AAB  ABC 

  FA  0.3 m    FB  0.7 m   0 7 FB  2 3 Insert (2) into (1); 7 FB  FB  16 kN 3 FB  4.8 kN FA 

FA  16 kN  4.8 kN  11.2 kN

 FA  11.2 kN FA  4.8 kN

 

1

Ans.

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. Est  200 GPa, Eal  70 GPa . Equilibrium;   M C  0  FE 1.5 m   FA 1.5 m   0 FE  FA  Fal   Fy  0  2 Fal  Fst  3 w  0 2 Fal  Fst  3 w

Compatibility; the system is symmetry, so same displacement at A, C, and E FL    al   st   EA  Fal Lal Fst Lst   where Lal  Lst and Aal  Ast  Eal Aal Est Ast

1

Insert (2) into (1); 20 2 Fal  Fal  3 w 7 21 Fal  w  0.61765 w 34 20 20 21 30 Fst  Fal  w w  1.7647 w 7 7 34 17

Fal Fst  70 GPa 200 GPa 20 Fst  Fal  2 7 Because Fst  3 Fal and Ast  Aal , 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 Case 2; Assume aluminum failed allow  al   alallow  st   st

Fst  180 MPa Ast

Fal  94 MPa Aal

30 w1 17  180  10 6 Pa 450  10 6 m2 w1  45.9  10 3 N/m

21 w2 34  94  10 6 Pa 2 6 450  10 m w2  68.486  10 3 N/m

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

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;

  M A  0 

Compatibility; From similar triangle

C

1.75 kN 150 mm   FB 100 mm   FC  225 mm   0 FB  2.25 FC  2.625 kN 1 

B

FL     AE 

225 mm 100 mm  C  2.25 B FC LC FL  2.25 B B AC EC AB EB FC  2.25 FB Insert (2) into (1);

 where LC  LB

and AC EC  AB EB 

 2

FB  2.25  2.25 FB   2.625 kN FB  0.43299 kN

FC  2.25 FB  2.25  0.43299 kN   0.97423 kN  FB  0.433 kN FC  0.974 kN

Ans.

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.

Equilibrium;   M G  0  FC 1 m   FA 1 m   0 FA  FC  Fst   Fy  0  2 Fst  FB  400 kN  0 2 Fst  FB  400 kN

1

Compatibility; The system is symmetry, the displacement of all bars are same FL    A  B   AE  F L FL  st st   B B   br TLB  where Lst  LB and Ast  AB  Ast Est AB Ebrass 

 5000 10

6

Fst FB   18  106 / o C 10 o C  6 2 9 2 9 m  200  10 Pa   5000 10 m 10110 Pa 

FB  0.505 Fst  90900 N FB  90900  0.505 Fst

 2

Insert (2) into (1); 2 Fst   90900 N  0.505 Fst   400  103 N Fst  123.39  103 N

FB  90900 N  0.505 Fst  90900 N  0.505 123.39  103 N   153.21 103 N Determine the stress in each post; F    A  Fst 123.39  103 N   24.678 MPa  A  C  A 5000 mm 2 F 153.21 103 N B  B   30.642 MPa A 5000 mm 2  A   C  24.7 MPa and  B  30.6 MPa

Ans.

For 2014-T6 aluminum,  al  23  106 / o C, Eal  73.1 GPa 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.

Equilibrium;   Fx  0 

 FA  FB  80 N  0 FA  FB  80 N

  Fx  0  FAC  FA  0   Fx  0 

Compatibility;

B A  0

1  FAC  FA

 FBC  FB  0

 FBC   FB

FL     EA 

C A   B C  0 Temp Temp  CForce   Force  A   C A    B C   B C   0  FAC LAC  F L    AC TLAC    BC BC   BC TLBC   0   E AC AAC   EBC ABC 

 where E AC  EBC  Eal

and AAC  ABC 

    FA  0.125 m     23  106 / o C  20 o C  25 o C   0.125 m    73.1 109 Pa    0.012 m 2     4            FB  0.2 m     23  10 6 / o C  20 o C  25 o C   0.2 m    0  73.1 109 Pa    0.012 m 2     4      FB  0.625 FA  13905 N

FB  0.625 FA  13905 N  2 Insert (2) into (1); FA   0.625 FA  13905 N   80 N FA  8606.15 N

FB  0.625 FA  13905 N  0.625  8606.15 N   13905 N  8525.2 N  FA  8.61 kN  and FB  8.53 kN  Chapter 5 (MECH of MAT)

Ans.

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.

x

800 N.m

TC 300 N.m

800 N.m

 M x  0 T1  800 N.m  0 T1  800 N.m

T1

800 N.m

T2 300 N.m

Mx  0 T2  800 N.m  300 N.m  0 T2  500 N.m

T     J  T  0.035 m   500 N.m  0.035 m  A  2   7.4241 106 Pa   4 4  0.035 m   0.035 m  2 2 T  0.02 m   800 N.m  0.02 m   6.7878  106 Pa B  1    4 4  0.035 m   0.035 m  2 2

 A  4.45 MPa

 B  6.79 MPa

Ans.

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  70 MPa.

 P  T  3  103 W 10000  T  N.m   27 rev/min  2 rad/rev 1 min/ 60 sec  3 P

 max   allow

  Tc J 

Tc   allow J  10000  0.0625 m     2  3    70  106 Pa   4  0.0625 m   d 4  32  d  56.8345  103 m  d  60 mm

Ans.

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.

 P  T  32  103 W 800 N.m T    2  20 Hz   P

20  103 W 500 N.m TC     2  20 Hz   P

12  103 W 300 N.m TD     2  20 Hz   P

Determine the internal load in each section;

 M x  0 TAC  T  0 TAC  T 

800



N.m

 M x  0 TCD  TC  T  0 800 500 300 TCD  T  TC  N.m  N.m 

 M x  0







 TBD  0 TBD  0

Determine the shear stress in each section; Tc     J 

 AC

 CD  BD

 800  0.0375 m  N.m    T c  2    24.593  106 Pa  AC    4 J  0.0375 m  32  300   0.0375 m  N.m     T c 2     9.2225  106 Pa  CD    4 J  0.0375 m  32 T c  BD  0 J

Angle of twist; TL     GJ 

CD 

 300  N.m   0.2 m     

TCD LCD   0.0013116 rad  0.075152o  4 GCD J CD  0.0375 m   75 109 Pa  32

 max  24.6 MPa

CD  0.0752o

CCW

Ans.

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

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

 4 4  0.025 m    0.01875 m   2

 4 4  0.025 m    0.01875 m  

2 T  1409.34 N.m

 Tshaft  T

Tshaft  0.0125 m 

 84 MPa



 0.0125 m  2 T  0.0125 m 

 84  106 Pa







  M x  0  Tshaft  T  0

max  BC   allow   Tc J 



T  0.025 m 

 Ttube  T

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

max  AB   allow   Tc J 

Ttube  0.025 m 

  M x  0  Ttube  T  0

Tcase1  1409.34 N.m

 84 MPa

4

 0.0125 m  2 T  257.71 N.m

 84  106 Pa

4



Tcase 2  257.71 N.m

Case 3; Consider allow  3o

c  allow Ttube Ltube Gtube J tube

c  B A  C B  T L  shaft shaft  allow Gshaft J shaft

and   TL GJ 

T  0.3 m 

 75  10

9

Pa 

 4 4  0.025 m    0.01875 m  

2 T  240.02 N.m



Tcase3



  240.02 N.m

T  0.6 m 

 75 10

9

Pa 



2

 0.0125 m 

 4

3o   radian 180o

Because Tcase 3  Tcase 2 , Tcase1 , therefore the system is controlled by allow and Tmax  240 N.m

Ans.

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;

  M x  0  200 N.m  TA  TB  0 TA  TB  200 N.m

1

  M x  0  TAC  TA  0  TAC  TA   M x  0   TBC  TB  0  TBC  TB Compatibility;

B A  0

  TL GJ 

C A  B C  0 TAC LAC TBC LBC  0 GAC J AC GBC J BC

 where GAC J AC  GBC J BC 

TA  0.4 m   TB  0.6 m   0 3 TA  TB 2

 2

Insert (2) into (1);

3 TB  TB  200 N.m 2 TB  80 N.m 3 3 TA  TB   80 N.m   120 N.m 2 2 Shear stress in each section;   Tc J  TAC c 120 N.m  0.04 m 2    9.5493  106 Pa  4 J  0.04 m  32 T c  80 N.m  0.04 m 2   BC   6.3662  106 Pa  4 J  0.04 m  32

 AC 

 BC

 AC  9.55 MPa

 BC  6.37 MPa Ans.

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 Gst  80 GPa, Gbr  40 GPa .

Consider section BC;   M x  0  50 N.m  Tbr  Tst  0  Tbr  Tst  50 N.m 1 Compatibility; st  br Tst Lst Tbr Lbr  Gst J st Gbr J br

 where Lst  Lbr  Tst

 80 GPa 

 4 4  0.02 m    0.01 m   2

Tst  30Tbr  2  Insert (2) into (1); Tbr  30Tbr  50 N.m

Tbr 





Tbr



 40 GPa   0.01 m 

4

2

50 N.m 31

 50  1500 Tst  30Tbr  30  N.m   N.m 31  31  Determine the deformation of section BC;  50  N.m  1 m   T L  31   0.0025670 rad  0.14708o CW BC  br br  Gbr J br  40 109 Pa  2  0.01 m 4 Consider section AB;   M x  0  50 N.m  TAB  0  TAB  50 N.m

B A 

 50 N.m 1.5 m  TAB LAB   0.0037302 rad  0.21372o CW  4 GAB J AB  80 109 Pa  2  0.02 m 

Determine the rotation at end C; C  C B  B A  0.0025670 rad  0.0037302 rad

C  0.0062972 rad  0.36080

C  0.361o CW

o

Determine the stress and strain in steel; Tc     J  T c  50 N.m  0.02 m   3.9789  106 Pa  stAB  AB   4 J AB  0.02 m  2  1500  N.m   0.02 m   T c 31   4.1072  106 Pa  stBC  stst    4 4 J BC  0.02 m    0.01 m    2

 stBC 

 stBC G



4.1072  106 Pa  51.340  106 rad 9 80  10 Pa

Determine the stress and strain in brass; Tc     J   50  N.m   0.01 m   T c 31   brBC  brbr    1.0268  106 Pa  4 J BC  0.01 m  2  BC 1.0268  106 Pa  brBC  br   25.670  106 rad G 40  109 Pa

 stmax  4.11 MPa in section BC

 stmax  51.3  10 6 rad in section BC  brmax  1.03 MPa  stmax  25.7  10 6 rad

Ans.

Ans.