   # Solve Problem 7.7-9 by using Mohr’s circle for plane strain. ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 7, Problem 7.7.27P
Textbook Problem
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## Solve Problem 7.7-9 by using Mohr’s circle for plane strain.

To determine

The principle strains on the element oriented at 35 ° .

### Explanation of Solution

Given information:

The normal strain along x-axis is 280 × 10 6 , normal strain along y-axis is 420 × 10 6 ,and shear strain along xy-plane is 150 × 10 6 , orientation of element is 35 ° .

Explanation:

The following are the steps to draw the Mohr’s circle.

1. Draw the x-axis as normal strain and y-axis as shear strain and indicate the origin as point O ( 0 , 0 ) .
2. Write the expression to locate the center point C of the circle from the origin O(ie., OC).
3. OC = ε a v g = ε x + ε y 2 .....(I)

Here, the normal strain along x-axis is ε x , normal strain along y-axis is ε y , and average strain is ε a v g .

4. Mark the center point C: ( ε a v g , 0 ) .
5. Write the expression for the radius of Mohr’s circle.

R = ( ε x ε a v g ) 2 + ( γ x y 2 ) 2 .....(II)

Here, the radius of the Mohr’s circle is R and shear strain along xy-plane is γ x y .

Write the expression for principle angle.

α = tan 1 ( γ x y 2 ε x ε a v g ) .....(III)

Here, principle plane angle is α .

Write the expression for angle of oriented strain plane.

β = 180 ° - 2 θ + α .....(IV)

Here, angle of oriented strain plane is β , and θ is the orientation of element.

Write the expression for the strain at point G .

ε x = ε a v g + R cos β .....(V)

Here, the strain at point G is ε x .

Write the expression for the strain at point H .

ε y = ε a v g R cos β ...... (VI)

Here, the strain at point H is ε y .

Write the expression for the shear strain at point H .

( γ x y ) 1 2 = R sin β ...... (VII)

Here, the shear strain at point H is ( γ x y ) 1 .

Write the expression for the shear strain at point G

( γ x y ) 2 2 = R sin β ...... (VIII)

Here, the shear strain at point G is ( γ x y ) 2 .

Calculation:

Substitute 280 × 10 6 for ε x , and 420 × 10 6 for ε y in Equation (I).

ε a v g = 280 × 10 6 + 420 × 10 6 2 = 700 × 10 6 2 = 350 × 10 6

Substitute 280 × 10 6 for ε x , 350 × 10 6 for ε a v g and 150 × 10 6 for γ x y in Equation (II).

R = ( 280 × 10 6 350 × 10 6 ) 2 + ( 150 × 10 6 2 ) 2 = ( 70 × 10 6 ) 2 + ( 75 × 10 6 ) 2 = 102

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