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ME323: Mechanics of Materials Homework Set 11
Spring 2018 Due: Wednesday, April 18
Problem 11.1 – 10 points
For following state of stress,
a) Draw the three Mohr’s circles that describe the state of stress.
b) Calculate the three Maximum In-Plane Shear Stresses 𝜏𝑚𝑎𝑥,𝑦𝑧, 𝜏𝑚𝑎𝑥,𝑥𝑧, and 𝜏𝑚𝑎𝑥,𝑥𝑦.
c) Calculate the Absolute Maximum Shear Stress 𝜏𝑚𝑎𝑥,𝑎𝑏𝑠.
The state of stress is repeated:
𝜎𝑥 = −3 𝑀𝑃𝑎
𝜎𝑦 = 2 𝑀𝑃𝑎
𝜎𝑧 = 1 𝑀𝑃𝑎
𝜏𝑦𝑧 = 0 𝑀𝑃𝑎
𝜏𝑥𝑧 = 0 𝑀𝑃𝑎
𝜏𝑥𝑦 = 0 𝑀𝑃𝑎
The three Mohr’s circles that describe the state of stress are drawn:
2 MPa
3 MPa
1 MPa
x
y
z
𝜏
𝜎
YZ Plane
Scale: 4 blocks = 1 MPa
(1,0) (2,0)
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
These may also be combined:
𝜏
𝜎
XZ Plane
Scale: 4 blocks = 1 MPa
(1,0) (2,0)
𝜏
𝜎
XY Plane
Scale: 4 blocks = 1 MPa
(-3,0) (2,0)
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
The maximum in-plane shear stresses are determined:
𝜏𝑚𝑎𝑥,𝑦𝑧 =|𝜎𝑦 − 𝜎𝑧|
2=
|2 𝑀𝑃𝑎 − 1 𝑀𝑃𝑎|
2= 0.5 𝑀𝑃𝑎 = 𝜏𝑚𝑎𝑥,𝑦𝑧
𝜏𝑚𝑎𝑥,𝑥𝑧 =|𝜎𝑥 − 𝜎𝑧|
2=
|−3 𝑀𝑃𝑎 − 1 𝑀𝑃𝑎|
2= 1 𝑀𝑃𝑎 = 𝜏𝑚𝑎𝑥,𝑦𝑧
𝜏𝑚𝑎𝑥,𝑥𝑦 =|𝜎𝑥 − 𝜎𝑧|
2=
|−3 𝑀𝑃𝑎 − 2 𝑀𝑃𝑎|
2= 2.5 𝑀𝑃𝑎 = 𝜏𝑚𝑎𝑥,𝑦𝑧
The absolute maximum shear stress 𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 is the largest of the three in-plane values 𝜏𝑚𝑎𝑥,𝑦𝑧, 𝜏𝑚𝑎𝑥,𝑥𝑧,
and 𝜏𝑚𝑎𝑥,𝑥𝑦 :
𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝜏𝑚𝑎𝑥,𝑥𝑦 = 2.5 𝑀𝑃𝑎 = 𝜏𝑚𝑎𝑥,𝑎𝑏𝑠
This is verified using the principal stresses. By inspection, the Principal Stresses, in the order 𝜎1 > 𝜎2 >
𝜎3, are:
𝜎1 = 𝜎𝑦 = 2 𝑀𝑃𝑎
𝜎2 = 𝜎𝑧 = 1 𝑀𝑃𝑎
𝜎3 = 𝜎𝑥 = −3 𝑀𝑃𝑎
When the principal stresses are in the order given, the absolute maximum shear stress is given by
𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 =𝜎1 − 𝜎3
2=
(2 𝑀𝑃𝑎) − (−3 𝑀𝑃𝑎)
2= 2.5 𝑀𝑃𝑎
which matches the value calculated above.
The absolute maximum shear stress may also be verified graphically; using the scale shown in the
figures, the radius of the largest Mohr’s circle measures 10 cells or 2.5 MPa, so that again,
𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 2.5 𝑀𝑃𝑎
𝜏
𝜎
3D Mohr’s Circle
Scale: 4 blocks = 1 MPa
𝜎1 𝜎2 𝜎3
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
Problem 11.2 – 10 points
For each of the states of stress shown below,
a) State the principal stresses 𝜎1, 𝜎2, 𝜎3 in the order 𝜎1 > 𝜎2 > 𝜎3.
b) Draw and label the three Mohr’s circles that describe the states of stress.
c) Calculate the three Maximum In-Plane Shear Stresses 𝜏𝑚𝑎𝑥,𝑦𝑧, 𝜏𝑚𝑎𝑥,𝑥𝑧, 𝜏𝑚𝑎𝑥,𝑥𝑦.
d) Calculate the Absolute Maximum Shear Stress 𝜏𝑚𝑎𝑥,𝑎𝑏𝑠.
2 MPa
5 MPa
4 MPa
1 MPa
2 MPa
1 MPa
(iii) (iv)
(i)
(v)
(ii)
x
y
z
3 MPa
4 MPa
3 MPa
1 MPa
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
𝜏
𝜎
(i)
Scale: 4 blocks = 1 MPa
𝜎1 𝜎2, 𝜎3
𝜎1 = 𝜎𝑦 = 1 𝑀𝑃𝑎
𝜎2 = 0 𝑀𝑃𝑎 𝜎3 = 0 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑦𝑧 =|𝜎𝑦−𝜎𝑧|
2= 0.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑧 =|𝜎𝑥−𝜎𝑧|
2= 0.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑦 =|𝜎𝑥−𝜎𝑦|
2= 1 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝜏𝑚𝑎𝑥,𝑥𝑦 = 1 𝑀𝑃𝑎
𝜏
𝜎
(ii)
Scale: 4 blocks = 1 MPa
𝜎1 𝜎2 𝜎3
𝜎1 = 𝜎𝑦 = 2 𝑀𝑃𝑎
𝜎2 = 𝜎𝑧 = 1 𝑀𝑃𝑎 𝜎3 = 𝜎𝑥 = 0 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑦𝑧 =|𝜎𝑦−𝜎𝑧|
2= 0.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑧 =|𝜎𝑥−𝜎𝑧|
2= 0.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑦 =|𝜎𝑥−𝜎𝑦|
2= 1 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝜏𝑚𝑎𝑥,𝑥𝑦 = 1 𝑀𝑃𝑎
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
𝜏
𝜎
(iii)
Scale: 4 blocks = 1 MPa
𝜎1 𝜎2 𝜎3
𝜎1 = 𝜎𝑧 = 1 𝑀𝑃𝑎 𝜎2 = 𝜎𝑥 = 0 𝑀𝑃𝑎 𝜎3 = 𝜎𝑦 = −3 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑦𝑧 =|𝜎𝑦−𝜎𝑧|
2= 2 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑧 =|𝜎𝑥−𝜎𝑧|
2= 0.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑦 =|𝜎𝑥−𝜎𝑦|
2= 1.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝜏𝑚𝑎𝑥,𝑦𝑧 = 2 𝑀𝑃𝑎
𝜏
𝜎
(iv)
Scale: 4 blocks = 1 MPa
𝜎1 𝜎2 𝜎3
𝜎1 = 𝜎𝑥 = 0 𝑀𝑃𝑎 𝜎2 = 𝜎𝑦 = −3 𝑀𝑃𝑎
𝜎3 = 𝜎𝑧 = −4 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑦𝑧 =|𝜎𝑦−𝜎𝑧|
2= 0.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑧 =|𝜎𝑥−𝜎𝑧|
2= 2 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑦 =|𝜎𝑥−𝜎𝑦|
2= 1.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝜏𝑚𝑎𝑥,𝑥𝑧 = 2 𝑀𝑃𝑎
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
𝜏
𝜎
(v)
Scale: 4 blocks = 1 MPa
𝜎1 𝜎2 𝜎3
𝜎1 = 𝜎𝑥 = 5 𝑀𝑃𝑎 𝜎2 = 𝜎𝑦 = 2 𝑀𝑃𝑎
𝜎3 = 𝜎𝑧 = −4 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑦𝑧 =|𝜎𝑦−𝜎𝑧|
2= 3 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑧 =|𝜎𝑥−𝜎𝑧|
2= 4.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑥𝑦 =|𝜎𝑥−𝜎𝑦|
2= 1.5 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝜏𝑚𝑎𝑥,𝑥𝑧 = 4.5 𝑀𝑃𝑎
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
Problem 11.3 – 10 points
A sign of weight 𝑊𝑆 = 400 𝑁 is suspended from a horizontal arm of weight 𝑊𝐴 = 3200 𝑁, which is
cantilevered from a tubular steel column with a weight 𝑊𝐶 which is to be calculated from its unit mass
of 450𝑘𝑔
𝑚. The inner and outer diameter of the steel column are 𝑑𝑖 = 43 𝑐𝑚 and 𝑑𝑜 = 50 𝑐𝑚,
respectively. The force of the wind blowing on the sign is directed into the page and is estimated as
𝐹𝑊 ≈ 800 𝑁.
a) For a height of 𝐻 = 3.5𝑚, calculate the state of stress [𝜎𝑥, 𝜎𝑦, 𝜎𝑧, 𝜏𝑦𝑧, 𝜏𝑥𝑧, 𝜏𝑥𝑦]𝑇
at point A on
the cross-section shown.
b) For a height of 𝐻 = 3.5𝑚, calculate the state of stress [𝜎𝑥, 𝜎𝑦, 𝜎𝑧, 𝜏𝑦𝑧, 𝜏𝑥𝑧, 𝜏𝑥𝑦]𝑇
at point B on
the cross-section shown.
b) If the Absolute Maximum Shear Stress allowable at points A and B is given as 𝜏𝑎𝑙𝑙𝑜𝑤 =
250 𝑀𝑃𝑎, determine the maximum allowable height 𝐻𝑎𝑙𝑙𝑜𝑤.
H
Stee
l Co
lum
n
Purdue SIAM
Computational Science
and Engineering
Student Conference
14 April 2018
Arm
FW
WS
2m
WA
WC
x
y
z
1m
z
x
A
B
Cross-Section at 𝑦 = 0
1m
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
Summing forces and moments,
Σ𝐹𝑥 = 0 → 𝑅𝑥 = 0 𝑁
Σ𝐹𝑦 = 0 → 𝑅𝑦 = −400𝑁 − 3200𝑁 − (450𝑘𝑔
𝑚) (9.81
𝑚
𝑠2) (𝐻) = −8829𝐻
2
𝑁
𝑚 − 3600𝑁 ≈ −19051 𝑁
Σ𝐹𝑧 = 0 → 𝑅𝑧 = −800 𝑁
Σ𝑀𝑥 = 0 → 𝑀𝑥 = −(800 𝑁)(𝐻 − 1 𝑚) = 800 𝑁 ∙ 𝑚 − 800𝐻 𝑁 = −2000 𝑁 ∙ 𝑚
Σ𝑀𝑦 = 0 → 𝑀𝑦 = −(800 𝑁)(3 𝑚) = −2400 𝑁 ∙ 𝑚
Σ𝑀𝑧 = 0 → 𝑀𝑧 = −(3200 𝑁)(2 𝑚) − (400 𝑁)(3 𝑚) = −7600 𝑁 ∙ 𝑚
Geometric properties are calculated,
𝐴 =𝜋
4(𝑑𝑜
2 − 𝑑𝑖2) ≈ 0.0511 𝑚2
𝐼𝑥 = 𝐼𝑧 =𝜋
64(𝑑𝑜
4 − 𝑑𝑖4) ≈ 0.00138 𝑚4
𝐼𝑦 = 𝐼𝑝 =𝜋
32(𝑑𝑜
4 − 𝑑𝑖4) ≈ 0.00278 𝑚4
�̅�′ =
(4𝑑𝑜2
3𝜋)(
𝜋(𝑑𝑜2 )
2
2)−(
4𝑑𝑖2
3𝜋)(
𝜋(𝑑𝑖2 )
2
2)
(𝜋(
𝑑𝑜2
)2
2)−(
𝜋(𝑑𝑖2
)2
2)
≈ 0.1483 𝑚
[𝑑′ is defined here as the distance of the centroid of half of the cross-section of the column from the
center of the column.]
H
800 𝑁
400 𝑁
2m
3200 𝑁
(450𝑘𝑔
𝑚) (𝑔𝐻)
x
y
1m
1m
𝑅𝑥
𝑅𝑦
𝑅𝑧
𝑀𝑥
𝑀𝑦
𝑀𝑧
z
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
The state of stress at ‘A’ is calculated,
[ 𝜎𝑥
𝜎𝑦
𝜎𝑧
𝜏𝑦𝑧
𝜏𝑥𝑧
𝜏𝑥𝑦]
=
[
0𝑅𝑦
𝐴+
𝑀𝑥(𝑑𝑜 2⁄ )
𝐼𝑥
000
−𝑀𝑦(
𝑑𝑜2
)
𝐼𝑝+
𝑅𝑥(𝐴
2)�̅�′
𝐼𝑧(𝑑𝑜−𝑑𝑖)]
≈
[
0−732.37
000
215.86 ]
𝑘𝑃𝑎 ≈
[ 𝜎𝑥
𝜎𝑦
𝜎𝑧
𝜏𝑦𝑧
𝜏𝑥𝑧
𝜏𝑥𝑦]
The state of stress at ‘B’ is calculated,
[ 𝜎𝑥
𝜎𝑦
𝜎𝑧
𝜏𝑦𝑧
𝜏𝑥𝑧
𝜏𝑥𝑦]
=
[
0𝑅𝑦
𝐴−
𝑀𝑧(𝑑𝑜 2⁄ )
𝐼𝑧
0𝑀𝑦(
𝑑𝑜2
)
𝐼𝑝+
𝑅𝑧(𝐴
2)�̅�′
𝐼𝑥(𝑑𝑜−𝑑𝑖)
00 ]
≈
[
0994.54
0−247.04
00 ]
𝑘𝑃𝑎 ≈
[ 𝜎𝑥
𝜎𝑦
𝜎𝑧
𝜏𝑦𝑧
𝜏𝑥𝑧
𝜏𝑥𝑦]
The absolute maximum shear stress is calculated at points A and B as a function of the height 𝐻 of the
column,
250 𝑀𝑃𝑎 = 𝜏𝑎𝑏𝑠,𝑚𝑎𝑥(𝐴)
= √(𝜎𝑦
2)2+ (𝜏𝑥𝑦)
2=
√(
−8829𝐻𝐴
2𝑁𝑚
−3600𝑁
𝐴+
(800 𝑁∙𝑚−800𝐻𝑁𝑚)(𝑑𝑜 2⁄ )
𝐼𝑥
2)
2
+ (−𝑀𝑦(
𝑑𝑜2
)
𝐼𝑝+
𝑅𝑥(𝐴
2)�̅�′
𝐼𝑧(𝑑𝑜−𝑑𝑖))
2
→ 𝐻𝐴 ≈ 1086 𝑚
250 𝑀𝑃𝑎 = 𝜏𝑎𝑏𝑠,𝑚𝑎𝑥(𝐵)
= √(𝜎𝑦
2)2+ (𝜏𝑦𝑧)
2= √(
(−8829𝐻𝐴
2
𝑁
𝑚 −3600𝑁)
2𝐴−
𝑀𝑧(𝑑𝑜 2⁄ )
2𝐼𝑧)
2
+ (𝑀𝑦(
𝑑𝑜2
)
𝐼𝑝+
𝑅𝑧(𝐴
2)�̅�′
𝐼𝑥(𝑑𝑜−𝑑𝑖))
2
→ 𝐻𝐵 ≈ 2911 𝑚
The maximum allowable height is the minimum of 𝐻𝐴 and 𝐻𝐵, so
𝐻𝑎𝑙𝑙𝑜𝑤 ≈ 1086 𝑚
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
Problem 11.4 – 10 points
A rubber-capped solid circular shaft of length 𝐿 = 90𝑐𝑚 and diameter 𝑑 = 3𝑐𝑚 is wedged in a door-
frame and used for pull-ups, leading to the loading conditions shown.
a) Determine the magnitude of the maximum principal stresses and indicate the location(s) at
which they occur.
b) Determine the magnitude of the absolute maximum shear stress and indicate the location(s)
at which it occurs.
A Free-Body Diagram is drawn,
Summing the forces and moments,
Σ𝐹𝑥 = 0 → 𝐹1 + 𝐹2 = 0
Σ𝐹𝑦 = 0 → 𝑉1 − (40 ∗ 9.81)𝑁 − (40 ∗ 9.81)𝑁 − 𝑉2
Σ𝑇𝑥 = 0 → −𝑇1 + 120 𝑁 ∙ 𝑚 + 𝑇2 = 0
Σ𝑀𝑧 = 0 → −𝑀1 −1
4(𝐿)(392.4 𝑁) −
3
4(𝐿)(392.4 𝑁) + 𝑀2
The problem is indeterminate. However, the symmetry of the problem is noted, meaning that
𝐹2 = −𝐹1
𝑉2 = −𝑉1
𝑇2 = −𝑇1
𝑀2 = −𝑀1
𝐿 4⁄ 𝐿 4⁄
40 kg 40 kg
120 N ∙ m
d
𝐿 4⁄ 𝐿 4⁄
22.5 𝑐𝑚 22.5 𝑐𝑚
(40kg ∗ 9.81m
s2)
120 N ∙ m
d
22.5 𝑐𝑚 22.5 𝑐𝑚
T1 T2 𝐹1 𝐹2
𝑉1 𝑉2
(40kg ∗ 9.81m
s2)
x
y
z
O
𝑀1 𝑀2
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
Plugging these into the equilibrium equations gives
𝐹1 = −𝐹2 = 0 𝑁
𝑉1 = −𝑉2 = 392.4 𝑁
𝑇1 = −𝑇2 = 60 𝑁 ∙ 𝑚
𝑀1 = −𝑀2 = 176.58 𝑁 ∙ 𝑚
A torque diagram is drawn
A shear force diagram is drawn
A bending moment diagram is drawn
The maximum values of the torque, shear stress, and bending moment align at 𝑥 =1
4𝐿 and 𝑥 =
3
4𝐿, so
the maximum principal stresses are expected at these two cross-sections.
Only the stresses at 𝑥 =1
4𝐿 are calculated because the loading is symmetrical.
𝐿 4⁄ 𝐿 4⁄ 𝐿 4⁄ 𝐿 4⁄
60 𝑁 ∙ 𝑚
−60 𝑁 ∙ 𝑚
𝐿 4⁄ 𝐿 4⁄ 𝐿 4⁄ 𝐿 4⁄
−392.4 𝑁
392.4 𝑁
𝐿 4⁄ 𝐿 4⁄ 𝐿 4⁄ 𝐿 4⁄
x (m)
−568.98
−176.58 −176.58
M (N∙m)
ME323: Mechanics of Materials Homework Set 11 - continued
Spring 2018 Due: Wednesday, April 18
The maximum stresses due to the torque, shear stress, and bending moment are given by
𝜏𝑚𝑎𝑥,𝑇 = |𝑇𝑚𝑎𝑥𝑟
𝐼𝑝| =
(60 𝑁 ∙ 𝑚)(0.015 𝑚)
(𝜋(0.015 𝑚)4
2)
≈ 11.32 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥,𝑉 = |4
3
𝑉𝑚𝑎𝑥
𝐴| =
(4)(392.4 𝑁)
(3)(𝜋)(0.015𝑚)2≈ 0.01 𝑀𝑃𝑎
𝜎𝑚𝑎𝑥,𝐵 = |𝑀𝑚𝑎𝑥𝑟
𝐼𝑧| =
(568.98 𝑁 ∙ 𝑚)(0.015 𝑚)
(𝜋(0.015 𝑚)4
4)
≈ 214.65 𝑀𝑃𝑎
The shear stress is neglected because it is several orders of magnitude smaller than the remaining
stresses, leaving
𝜏𝑚𝑎𝑥,𝑇 ≈ 11.32 𝑀𝑃𝑎
𝜎𝑚𝑎𝑥,𝐵 ≈ 214.65 𝑀𝑃𝑎
The maximum principal stresses for this stress state are given by
𝜎1 = (𝜎𝑚𝑎𝑥,𝐵
2) + 𝑠𝑞𝑟𝑡 ((
𝜎𝑚𝑎𝑥,𝐵
2)2
+ (𝜏𝑚𝑎𝑥,𝑇)2) ≈ 322 𝑀𝑃𝑎
𝜎2 = 𝜎3 = 0
The absolute maximum shear stress for this stress state is given by
𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝑠𝑞𝑟𝑡 ((𝜎𝑚𝑎𝑥,𝐵
2)2
+ (𝜏𝑚𝑎𝑥,𝑇)2) ≈ 215 𝑀𝑃𝑎
These stresses occur at all points around the perimeter of the bar, at 𝑥 =1
4𝐿 and at 𝑥 =
3
4𝐿.
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