ME323: Mechanics of Materials Homework Set 11 10 … · ME323: Mechanics of Materials Homework Set...

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


These may also be combined:

𝜏

𝜎

XZ Plane


(1,0) (2,0)

𝜏

𝜎

XY Plane


(-3,0) (2,0)



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


𝜎1 𝜎2 𝜎3




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



𝜏

𝜎

(iii)


𝜎1 𝜎2 𝜎3

𝜎1 = 𝜎𝑧 = 1 𝑀𝑃𝑎 𝜎2 = 𝜎𝑥 = 0 𝑀𝑃𝑎 𝜎3 = 𝜎𝑦 = −3 𝑀𝑃𝑎


2= 2 𝑀𝑃𝑎


2= 0.5 𝑀𝑃𝑎


2= 1.5 𝑀𝑃𝑎

𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝜏𝑚𝑎𝑥,𝑦𝑧 = 2 𝑀𝑃𝑎

𝜏

𝜎

(iv)


𝜎1 𝜎2 𝜎3

𝜎1 = 𝜎𝑥 = 0 𝑀𝑃𝑎 𝜎2 = 𝜎𝑦 = −3 𝑀𝑃𝑎

𝜎3 = 𝜎𝑧 = −4 𝑀𝑃𝑎


2= 0.5 𝑀𝑃𝑎


2= 2 𝑀𝑃𝑎


2= 1.5 𝑀𝑃𝑎

𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝜏𝑚𝑎𝑥,𝑥𝑧 = 2 𝑀𝑃𝑎



𝜏

𝜎

(v)


𝜎1 𝜎2 𝜎3

𝜎1 = 𝜎𝑥 = 5 𝑀𝑃𝑎 𝜎2 = 𝜎𝑦 = 2 𝑀𝑃𝑎

𝜎3 = 𝜎𝑧 = −4 𝑀𝑃𝑎


2= 3 𝑀𝑃𝑎


2= 4.5 𝑀𝑃𝑎


2= 1.5 𝑀𝑃𝑎

𝜏𝑚𝑎𝑥,𝑎𝑏𝑠 = 𝜏𝑚𝑎𝑥,𝑥𝑧 = 4.5 𝑀𝑃𝑎




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



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



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




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



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)



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.

ME323: Mechanics of Materials Homework Set 11 10 … · ME323: Mechanics of Materials Homework Set...

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