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EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 7 Transformation of Stress & Mohr’s Circle Homework Hints/Answers

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Page 1: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

EMA 3702

Mechanics & Materials Science

(Mechanics of Materials)

Chapter 7 Transformation of

Stress & Mohr’s Circle

Homework Hints/Answers

Page 2: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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

Page 3: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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

Page 4: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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

Page 5: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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

Page 6: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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

Page 7: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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

Page 8: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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

Page 9: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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

Page 10: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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

Page 11: EMA 3702 Mechanics & Materials Science (Mechanics of ... · EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 7 Stress Transformation & Mohr’s Circle Homework 7.2 Solution

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’

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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

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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

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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