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EMA 3702
Mechanics & Materials Science
(Mechanics of Materials)
Chapter 7 Transformation of
Stress & Mohr’s Circle
Homework Hints/Answers
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.1
For the given 2D plane stress state, determine
(a) the principle stresses in that plane, (b) the orientation of principle
axes in that plane, (c) the maximum in-plane shearing stress, (d) the
orientation of the plane for maximum in-plane shearing stress, and (e)
the corresponding normal stress for those planes where maximum in-
plane shearing stress occur
9 MPa
3 MPa
4 MPa
x
y
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.1
(a) The coordinates for the two related points
X (?, ?) and Y (?, ?)
Therefore, for the corresponding Mohr’s circle:
Center at C (?, ?)
Radius R = ? MPa
Principle stress σA = 11 MPa
Principle stress σB = 1 MPa
(b) From X to A, tan2θp = ?/?,
θp = 0.5 ∙ atan (?/?) = ?? rad = 26.6o
From X, rotate CW by 26.6o,
will be the principle plane with max
normal stress (A point on Mohr’s circle)
From Y, rotate CW by 26.6o
will be the principle plane with min
normal stress (B point on Mohr’s circle)
x’
x’y’
X (?, ?)
Y (?, ?)
C (?, ?)
R = ?
AB
D
O
9 MPa
3 MPa
4 MPa
y
1 MPa
11 MPa
p
2p
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.1
(c) The max shearing stress
τmax = R = 5 MPa
(d) Orientation for the max shearing would
occur at 45o CCW from A or B or
45-26.6 = 18.4o CCW from X (reaching D point)
or Y (reaching E point)
(e) The corresponding normal stress for the
planes where τ = τmax is ave = 6 MPa
x’
x’y’
X (?, ?)
Y (?, ?)
C (?, ?)
R = ?
AB
D
O
9 MPa
3 MPa
4 MPa
6 MPa
6 MPas
2s
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.2
Given original 2D plane stress state as illustrated, determine the
normal and shearing stress after the element shown is rotated (a)
26.6o clockwise (CW) (b) 30o counter clockwise (CCW)
9 MPa
3 MPa
4 MPa
x
y
Solution using equations on slides 6-8
(a) Rotate 26.6o CW, therefore
2 = -2 26.6o = -53.2o
Note: to use the equations on slides 6/7, sign
convention for shearing stress is opposite as in
Mohr’s circle, i.e.,
τxy = -4 MPa !!
𝜎𝑥′ =𝜎𝑥 + 𝜎𝑦
2+𝜎𝑥 − 𝜎𝑦
2𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦𝑠𝑖𝑛2𝜃
𝜎𝑥′ = 11.0 MPa
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.2
9 MPa
3 MPa
4 MPa
x
y(a) Rotate 26.6o CW, therefore
2 = -2 26.6o = -53.2o
𝜏𝑥′𝑦′ = −𝜎𝑥 − 𝜎𝑦
2𝑠𝑖𝑛2𝜃 + 𝜏𝑥𝑦𝑐𝑜𝑠2𝜃
𝜏𝑥′𝑦′ = 0.0 MPa
𝜎𝑦′ =𝜎𝑥 + 𝜎𝑦
2−𝜎𝑥 − 𝜎𝑦
2𝑐𝑜𝑠2𝜃 − 𝜏𝑥𝑦𝑠𝑖𝑛2𝜃
𝜎𝑦′ = 1.0 MPa
= 26.2o
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.2
(b) Rotate 30.0o CCW, therefore
2 = 2 30.0o = 60.0o
Again, note to use the equations on
slides 6-8, sign convention for shearing
stress is opposite as in Mohr’s circle:,
τxy = -4 MPa
9 MPa
3 MPa
4 MPa
x
y
𝜎𝑥′ = 4.04 MPa
𝜏𝑥′𝑦′ = −4.60 MPa
𝜎𝑦′ = 7.96 MPa
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.2
Solution using Mohr’s circle:
For (a), CW 26.6o, meaning CW ??.?o
on Mohr’s circle, reaching A and B.
The resulting normal stress x’ = 11.0 MPa, y’ = 1.0 MPa, τx’y’ = 0 MPa
For (b), CCW 30.0o, meaning CCW 60.0o on Mohr’s circle, reaching X” and Y”
The resulting normal stress x’ = ave - 5cos???.?o = 4.03 MPa,
y’ = ave + R cos113.2o = 7.97 MPa, τx’y’ = 5sin113.2o = 4.60 MPa
9 MPa
3 MPa
4 MPa
y
x’
x’y’
X (?, ?)
Y (?, ?)
C (?, ?)
R = ?
ABO
2 = ??.?o
X’’
Y ”
2” = 230o +2 = ???.?o
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.3
For the state of stress shown, determining the maximum shearing
stress when (a) y = 1 MPa and x = 7 MPa and (b) when y = 3 and x =
9 MPa (Hint: may need to consider both in-plane and out-of-plane
shearing stresses)
x
y
4 MPa
x
y
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
(a) When y = 1 MPa and x = 7 MPa
the coordinates for the two points on Mohr’s
circle are: X (?, ?), Y (?, ?)
Therefore, the center for the Mohr’s circle
is C (?, ?), and the radius
Because max in-plane normal stress at
A and B are ? sign, the max
in-plane shearing stress will be the 3D
max shearing stress, and
τmax = R = 5 MPa
Homework 7.3
x = 7 MPa
y =1 MPa
4 MPa
x’
x’y’
X (?, ?)
Y (?, ?)
C (?, ?)
R = ?
AB O
𝑅 = ?2+?2=?𝑀𝑃𝑎
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.3
(b) When y = 3 MPa and x = 9 MPa
the coordinates for the two points on Mohr’s
circle are: X (?, ?), Y (?, ?)
Therefore, the center for the Mohr’s circle
is C (?, ?), and the radius
Because max in-plane normal stress at
A and B are ?? sign, the max
in-plane shearing stress will NOT be the
3D max shearing stress.
Out of plane (3D) max shearing stress
τmax = R’ = 0.5 (?+?) = 5.5 Mpa
at D’
x = 9 MPa
y =3 MPa
4 MPa
x’
x’y’X (?, ?)
Y (?, ?)
C (?, ?)
R = ?
ABO
𝑅 = ?2+?2=?𝑀𝑃𝑎D’
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.4
A spherical container made of steel has 20 ft outer diameter and wall
thickness of 1/2 inch. Knowing the internal pressure is 50 psi, estimate
the maximum normal stress and the maximum shearing stress in the
container.
For spherical container, the in plane normal stress
The corresponding shearing stress are both zero,
therefore, the two points on Mohr’s circle are
X(?,?) and Y(?,?)
Max normal stress 𝜎𝑚𝑎𝑥 = 𝜎1 = 6000𝑝𝑠𝑖
Max shearing stress 𝛕𝑚𝑎𝑥 = 𝟎. 𝟓𝜎1 = 3000𝑝𝑠𝑖
𝜎1 = 𝜎2 =𝑝𝑟
2𝑡
𝜎1 = 𝜎2 = ⋯ = 6000𝑝𝑠𝑖
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.5
A 4 m diameter 10 m tall cylinder shaped storage tank has wall
thickness of 10 mm. Please calculate the maximum normal stress and
maximum shearing stress in the cylinder side wall when the tank is
filled all the way up with water.
When the tank is filled up with water, the max hydraulic pressure is occurring at the
bottom of the tank with
p = ρgh
p = 1000 kg/m3 9.8 m/sec2 10 m = 9.8104 (kg m/sec2)/m2
p = 9.8104 N/m2 = 9.8104 Pa
For cylinder shaped vessel,
Hoop stress
Longitudinal stress
10 m4 m
𝜎1 =𝑝𝑟
𝑡= 19.6𝑀𝑃𝑎
𝜎1 =𝑝𝑟
2𝑡= 9.8𝑀𝑃𝑎
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle
Homework 7.5
The two points on the Mohr’s circle
are X(?, ?) and Y(?, ?)
Maximum normal stress
m = 1 = 19.6 MPa
Maximum shearing stress
m = 0.51 = 9.8 MPa