ME 323: Mechanics of Materials
Spring 2020
Homework Set 12
Due: Wednesday, Apr. 29
Problem 12.1 (10 points)
A section is subjected to an axial load 𝑃, a torque 𝑇 and a moment 𝑀. It has a diameter 2D. The
material is ductile and has a yield stress 𝜎% = 150 𝑀𝑃𝑎. Determine the factors of safety using
the maximum shear stress theory and the maximum distortion energy theory.
Use: 𝐷 = 10 𝑚𝑚, 𝑃 = 100 𝑁, 𝑇 = 80 𝑁 ∙ 𝑚, 𝑀 = 50 𝑁 ∙ 𝑚
Solution:
Point A has the absolute maximum normal stress.
𝜎2 =
𝑃 𝑀𝑦
𝑃
𝑀∗𝐷
100
50 ∗ 0.01
+
=
− 𝜋
=−
− 𝜋
= −63.98 𝑀𝑃𝑎
9
9
𝐴
𝐼7
𝜋𝐷
𝜋 ∗ 0.01
𝐷=
0.01=
4
4
𝜏27 =
𝑇𝐷 𝑇 ∗ 𝐷
80 ∗ 0.01
= 𝜋
=− 𝜋
= −50.93 𝑀𝑃𝑎
𝐼C
𝐷=
0.01=
2
2
𝜎EFG =
𝑅=
𝜎2
2
9
+
9
𝜏2I
=
𝜎2
= −31.99 𝑀𝑃𝑎
2
63.98
−
2
9
+ 50.939 = 60.14 𝑀𝑃𝑎
Principal stresses:
𝜎J = 𝜎EFG + 𝑅 = −31.99 + 60.14 = 28.15 𝑀𝑃𝑎
𝜎9 = 0
𝜎K = 𝜎EFG − 𝑅 = −31.99 − 60.14 = −92.13 𝑀𝑃𝑎
Absolute maximum shear stress:
𝜏LE2,EMN =
𝜎J − 𝜎K
= 60.14 𝑀𝑃𝑎
2
Von mises stress:
𝜎O =
1
2
𝜎J − 𝜎9
9
+ 𝜎9 − 𝜎K
For maximum shear stress theory:
Safety factor:
𝑆𝐹 =
𝜎%
𝜏LE2,EMN ∗ 2
9
=
+ 𝜎K − 𝜎J
= 108.97 𝑀𝑃𝑎
150
= 1.25
120.28
For maximum distortion energy theory:
Safety factor:
𝑆𝐹 =
9
𝜎%
150
=
= 1.38
𝜎O 108.97
Problem 12.2 (10 points)
For the given stress element,
(1) Calculate the three principal stresses, the absolute maximum shear stress 𝜏LE2,EMN and
the von-Mises stress.
(2) If the material is ductile and the yield stress is 75 𝑀𝑃𝑎, determine the factor of safety
using the maximum shear stress theory and the maximum distortion energy theory.
(3) If the material is brittle, the ultimate tensile stress is 100 𝑀𝑃𝑎 and the ultimate
compression stress is 120 𝑀𝑃𝑎 . Determine the factor of safety using the maximum
normal stress theory and Mohr’s failure criterion.
Solution:
𝜎2 = −10 𝑀𝑃𝑎, 𝜎I = 30 𝑀𝑃𝑎, 𝜏2I = 15 𝑀𝑃𝑎
𝜎EFG =
𝑅=
𝜎2 − 𝜎I
2
9
𝜎2 + 𝜎I
= 10 𝑀𝑃𝑎
2
9 =
+ 𝜏2I
209 + 159 = 25 𝑀𝑃𝑎
(1) Principal stresses:
𝜎J = 𝜎EFG + 𝑅 = 35 𝑀𝑃𝑎
𝜎9 = 0
𝜎K = 𝜎EFG − 𝑅 = −15 𝑀𝑃𝑎
Absolute maximum shear stress:
𝜏LE2,EMN =
𝜎J − 𝜎K
= 25 𝑀𝑃𝑎
2
Von mises stress:
𝜎O =
1
2
𝜎J − 𝜎9
9
+ 𝜎9 − 𝜎K
9
+ 𝜎K − 𝜎J
(2) Ductile material:
For maximum-shear-stress theory:
Safety factor:
𝑆𝐹 =
𝜎%
𝜏LE2,EMN ∗ 2
For maximum distortion energy theory:
Safety factor:
𝑆𝐹 =
(3) Brittle material:
=
75
= 1.5
50
𝜎%
75
=
= 1.69
𝜎O 44.44
9
= 44.44 𝑀𝑃𝑎
For maximum normal stress theory:
In tension:
𝜎ST = 𝜎J = 35 𝑀𝑃𝑎
Safety factor:
𝑆𝐹T =
In compression:
100
= 2.86
35
𝜎SU = 𝜎K = −15 𝑀𝑃𝑎
Safety factor:
𝑆𝐹U =
120
=8
15
So, 𝑆𝐹 = 𝑆𝐹T = 2.86
For Mohr’s failure criterion:
𝜎J
𝜎K
35
15
−
=
+
= 0.475
100 120 100 120
Safety factor:
𝑆𝐹 =
1
= 2.11
0.475
Problem 12.3 (10 points)
A pipe has outer diameter 4𝑑 and inner diameter 2𝑑 . The material has a yield stress 𝜎% =
200 𝑀𝑃𝑎. If the factor of safety at point A is 2, determine the minimum value of 𝑑 using the
maximum shear stress theory and the maximum distortion energy theory.
Solution:
𝐼C =
𝜋
[ 2𝑑
2
=
− 𝑑= ] =
15 =
𝜋𝑑
2
𝐼7 =
𝜋
[ 2𝑑
4
=
− 𝑑= ] =
15 =
𝜋𝑑
4
Equilibrium:
Y 𝐹I
Y𝑇
(1)
= 0 : 𝑉 − 900 + 900 = 0
= 0 : − 0.2 ∗ 900 − 0.2 ∗ 900 + 𝑇 = 0
Y 𝑀7
= 0 : 0.25 ∗ 900 − 0.15 ∗ 900 + 𝑀7 = 0
solving for equations (1), (2) and (3):
𝑉=0
𝑇 = 360 𝑁 ∙ 𝑚
𝑀7 = −90 𝑁 ∙ 𝑚
stress state at element A:
(2)
(3)
𝜏27 =
𝑇 ∗ 2𝑑
720𝑑
96
=
= K
15 =
𝐼C
πd
𝜋𝑑
2
𝜎2 = −
𝑀 ∗ 2𝑑
180𝑑
48
=
=
15 = 𝜋𝑑 K
𝐼7
𝜋𝑑
4
𝜎I = 𝜎7 = 0
𝜏I7 = 𝜏2I = 0
Principal stresses:
𝜎J,K =
𝜎2 + 𝜎7
±
2
𝜎2 − 𝜎7
2
9
9 =
+ 𝜏27
24 24 17
±
𝜋𝑑 K
𝜋𝑑 K
𝜎9 = 0
Absolute maximum shear stress:
𝜏LE2,EMN =
𝜎J − 𝜎K
24 17
=
2
𝜋𝑑 K
For maximum-shear-stress theory:
𝐹𝑆 =
𝜏LE2,EMN =
𝜎%
=2
2 ∗ 𝜏LE2,EMN
200 ∗ 10_
= 50 𝑀𝑃𝑎
2∗2
𝑑 = 8.57 𝑚𝑚
Absolute maximum shear stress:
𝜎O =
1
2
𝜎J − 𝜎9
9
+ 𝜎9 − 𝜎K
9
+ 𝜎K − 𝜎J
9
=
48 13
𝜋𝑑 K
𝐹𝑆 =
𝜎%
=2
𝜎O
200 ∗ 10_
𝜎O =
= 100 𝑀𝑃𝑎
2
𝑑 = 8.20 𝑚𝑚
Problem 12.4 (10 points)
A horizontal rigid bar DCB is supported by a pin-fixed column AC and is subjected to a uniformly
distributed load 𝑞. The column AC has a Young’s modulus 𝐸 and a rectangular cross section as is
shown. Consider supports A and C to act as pinned-pinned when buckling in x-y plane and fixedfixed when buckling in y-z plane. Determine the maximum distributed load 𝑞 can be applied
without buckling.
Use: 𝐿 = 2 𝑚, 𝑏 = 40 𝑚𝑚, ℎ = 60 𝑚𝑚, 𝐸 = 150 𝐺𝑃𝑎
Solution:
FBD:
f𝑀
(1)
= 0 : 𝑞 ∗ 2𝐿 ∗ 2𝐿 − 𝐹U ∗ 2𝐿 = 0
solving for equation (1):
𝐹U = 2𝑞𝐿
For buckling in x-y plane,
𝐼7 =
1
ℎ𝑏 K
12
1
9
k
K
𝜋 9 𝐸𝐼i 𝜋 150 ∗ 10 ∗ 12 0.06 ∗ 0.04
𝑃gh =
=
= 29.61 𝑘𝑁
𝐾𝐿 9
1 ∗ 2𝐿 9
𝐹U = 2𝑞𝐿 = 𝑃gh
𝑞gh =
𝑃gh
= 7.4 𝑘𝑁/𝑚
2𝐿
For buckling in y-z plane,
𝐼2 =
1
𝑏ℎK
12
1
9
k
K
𝜋 9 𝐸𝐼i 𝜋 150 ∗ 10 ∗ 12 0.04 ∗ 0.06
𝑃gh =
=
= 266.48 𝑘𝑁
𝐾𝐿 9
0.5 ∗ 2𝐿 9
𝐹U = 2𝑞𝐿 = 𝑃gh
𝑞gh =
𝑃gh
= 66.62 𝑘𝑁/𝑚
2𝐿
So, the maximum distributed load 𝑞 = 7.4 𝑘𝑁/𝑚