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 , , , , and , . c) Calculate the Absolute Maximum Shear Stress , . The state of stress is repeated: = −3 = 2 = 1 = 0 = 0 = 0 The three Mohr’s circles that describe the state of stress are drawn: 2 MPa 3 MPa 1 MPa x y z YZ Plane Scale: 4 blocks = 1 MPa (1,0) (2,0)
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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