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Solutions to Chapter 8 Problems
SOLUTIONS TO CHAPTER 8 PROBLEMS S.8.2
From Eq. (7.8) Young’s modulus E is equal to the slope of the stress–strain curve. Then, since stress ¼ load/area and strain ¼ extension/original length, E ¼ slope of the load-extension curve multiplied by (original length/area of cross section). From the results given the slope of the load-extension curve ’402.6 kN/mm. Then E’
402:6 103 250 ’ 205 000 N=mm2 π 252 4
From Eq. (11.4) the modulus of rigidity is given by G¼
TL θJ
Therefore the slope of the torque-angle of twist (in radians) graph multiplied by (L/J ) is equal to G. From the results given the slope of the torque-angle of twist graph is ’12.38 kNm/rad. Therefore G’
12:38 106 250 ’ 80 700 N=mm2 π 254 32
Having obtained E and G the value of Poisson’s ratio may be found from Eq. (7.21), i.e.
E 1 ’ 0:27 v¼ 2G Finally, the bulk modulus K may be found using either of Eqs (7.22) or (7.23). From Eq. (7.22) K’ S.8.3
E ’ 148 500 N=mm2 : 3ð1 2vÞ
Suppose that the actual area of cross section of the material is A and that the original area of cross section is Ao. Then, since the volume of the material does not change during plastic deformation AL ¼ Ao Lo where L and Lo are the actual and original lengths of the material respectively. The strain in the material is given by
e114
Solutions to Chapter 8 Problems ε¼
L Lo A o ¼ 1 Lo A
e115 (i)
from the above. Suppose that the material is subjected to an applied load P. The actual stress is then given by σ ¼ P/A while the nominal stress is given by σ nom ¼ P/Ao. Therefore, substituting in Eq. (i) for A/Ao ε¼
σ 1 σ nom
Then σ nom ð1 þ εÞ ¼ σ ¼ C εn or σ nom ¼
C εn 1þε
(ii)
Differentiating Eq. (ii) with respect to ε and equating to zero gives dσ nom nC ð1 þ εÞεn1 C εn ¼ ¼0 dε ð1 þ εÞ2 i.e. nð1 þ εÞεn1 εn ¼ 0 Rearranging gives ε¼ S.8.4
n : ð1 nÞ
Substituting in Eq. (8.1) from Table P.8.4 104 105 106 107 þ 6þ ¼ 0:39 < 1 7þ 4 5 10 10 24 10 12 107 Therefore fatigue failure is not probable.
S.8.5
For the Goodman prediction Eq. (8.3) applies in which Sa,0 ¼ 460 N/mm2 and Sm ¼ Sa/2. Then Sa ¼ 460½1 ðSa =2 870Þ which gives Sa ¼ 363 N=mm2
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S.8.6
For the Gerber prediction m ¼ 2 in Eq. (8.4). Then Sa ¼ 460½1 ðSa =2 870Þ2 which simplifies to Sa2 þ 6582Sa 3:03 106 ¼ 0 Solving Sa ¼ 432 N=mm2
S.8.7
Suppose that the component fails after P sequences of the three stages. Then, from Eq. (8.5) P½ð200=104 Þ þ ð200=105 Þ þ ð600=2 105 Þ ¼ 1 which gives P ¼ 40 The total number of cycles is 1000 so that at the rate of 100 cycles per day the total number of days to fracture is 40 1000/100 ¼ 400 days.
S.8.8
From Eq. (8.10) S ¼ K0 =ðπaÞ1=2 so that S ¼ 3320=ðπ 2:0Þ1=2 from which S ¼ 1324 N=mm2
S.8.9
From Eq. (8.7) K ¼ SðπaÞ1=2 1:12 so that a¼
18002 π 1802 1:122
which gives a ¼ 25:4 mm
Solutions to Chapter 8 Problems
e117
From the above data and Eq. (8.14) it can be seen that C ¼ 30 1015 and n ¼ 4. Substituting these values in Eq. (8.17) Nf ¼
1 30 10
15
ð180 π1=2 Þ4
½ð1=0:4Þ ð1=25:4Þ
which gives Nf ¼ 7916 cycles: S.8.10
For a plate of finite width α ¼ ½secðπa=wÞ1=2 so that α ¼ ½secð3π=50Þ1=2 ¼ 1:01 From Eq. (8.9) K0 ¼ 3500=1:01 ¼ 3465:3 Then, from Eq. (8.11) 3465:3 ¼ ð3M 3=4 1:23 Þðπ 3Þ1=2 which gives M ¼ 866:9 Nmm=unit thickness The maximum bending moment the plate can withstand is therefore Mmax ¼ 866:9 5 ¼ 4334:5 Nmm Say Mmax ¼ 4300 Nmm
From Eq. (7.8) Young’s modulus E is equal to the slope of the stress–strain curve. Then, since stress ¼ load/area and strain ¼ extension/original length, E ¼ slope of the load-extension curve multiplied by (original length/area of cross section). From the results given the slope of the load-extension curve ’402.6 kN/mm. Then E’
402:6 103 250 ’ 205 000 N=mm2 π 252 4
From Eq. (11.4) the modulus of rigidity is given by G¼
TL θJ
Therefore the slope of the torque-angle of twist (in radians) graph multiplied by (L/J ) is equal to G. From the results given the slope of the torque-angle of twist graph is ’12.38 kNm/rad. Therefore G’
12:38 106 250 ’ 80 700 N=mm2 π 254 32
Having obtained E and G the value of Poisson’s ratio may be found from Eq. (7.21), i.e.
E 1 ’ 0:27 v¼ 2G Finally, the bulk modulus K may be found using either of Eqs (7.22) or (7.23). From Eq. (7.22) K’ S.8.3
E ’ 148 500 N=mm2 : 3ð1 2vÞ
Suppose that the actual area of cross section of the material is A and that the original area of cross section is Ao. Then, since the volume of the material does not change during plastic deformation AL ¼ Ao Lo where L and Lo are the actual and original lengths of the material respectively. The strain in the material is given by
e114
Solutions to Chapter 8 Problems ε¼
L Lo A o ¼ 1 Lo A
e115 (i)
from the above. Suppose that the material is subjected to an applied load P. The actual stress is then given by σ ¼ P/A while the nominal stress is given by σ nom ¼ P/Ao. Therefore, substituting in Eq. (i) for A/Ao ε¼
σ 1 σ nom
Then σ nom ð1 þ εÞ ¼ σ ¼ C εn or σ nom ¼
C εn 1þε
(ii)
Differentiating Eq. (ii) with respect to ε and equating to zero gives dσ nom nC ð1 þ εÞεn1 C εn ¼ ¼0 dε ð1 þ εÞ2 i.e. nð1 þ εÞεn1 εn ¼ 0 Rearranging gives ε¼ S.8.4
n : ð1 nÞ
Substituting in Eq. (8.1) from Table P.8.4 104 105 106 107 þ 6þ ¼ 0:39 < 1 7þ 4 5 10 10 24 10 12 107 Therefore fatigue failure is not probable.
S.8.5
For the Goodman prediction Eq. (8.3) applies in which Sa,0 ¼ 460 N/mm2 and Sm ¼ Sa/2. Then Sa ¼ 460½1 ðSa =2 870Þ which gives Sa ¼ 363 N=mm2
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S.8.6
For the Gerber prediction m ¼ 2 in Eq. (8.4). Then Sa ¼ 460½1 ðSa =2 870Þ2 which simplifies to Sa2 þ 6582Sa 3:03 106 ¼ 0 Solving Sa ¼ 432 N=mm2
S.8.7
Suppose that the component fails after P sequences of the three stages. Then, from Eq. (8.5) P½ð200=104 Þ þ ð200=105 Þ þ ð600=2 105 Þ ¼ 1 which gives P ¼ 40 The total number of cycles is 1000 so that at the rate of 100 cycles per day the total number of days to fracture is 40 1000/100 ¼ 400 days.
S.8.8
From Eq. (8.10) S ¼ K0 =ðπaÞ1=2 so that S ¼ 3320=ðπ 2:0Þ1=2 from which S ¼ 1324 N=mm2
S.8.9
From Eq. (8.7) K ¼ SðπaÞ1=2 1:12 so that a¼
18002 π 1802 1:122
which gives a ¼ 25:4 mm
Solutions to Chapter 8 Problems
e117
From the above data and Eq. (8.14) it can be seen that C ¼ 30 1015 and n ¼ 4. Substituting these values in Eq. (8.17) Nf ¼
1 30 10
15
ð180 π1=2 Þ4
½ð1=0:4Þ ð1=25:4Þ
which gives Nf ¼ 7916 cycles: S.8.10
For a plate of finite width α ¼ ½secðπa=wÞ1=2 so that α ¼ ½secð3π=50Þ1=2 ¼ 1:01 From Eq. (8.9) K0 ¼ 3500=1:01 ¼ 3465:3 Then, from Eq. (8.11) 3465:3 ¼ ð3M 3=4 1:23 Þðπ 3Þ1=2 which gives M ¼ 866:9 Nmm=unit thickness The maximum bending moment the plate can withstand is therefore Mmax ¼ 866:9 5 ¼ 4334:5 Nmm Say Mmax ¼ 4300 Nmm