   # The strains for an element of material in plane strain (see figure) are as follows: ε x = 480 ×10 -6 . ε y = 140 × l0 -6 , and γ x y = —350 x 10”. Determine the principals strains and maximum shear strains, and show these strains on sketches of properly oriented elements. ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 7, Problem 7.7.11P
Textbook Problem
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## The strains for an element of material in plane strain (see figure) are as follows: ε x = 480 ×10-6. ε y = 140 × l0-6, and γ x y = —350 x 10”. Determine the principals strains and maximum shear strains, and show these strains on sketches of properly oriented elements. To determine

The principal strains and maximum shear strain.

### Explanation of Solution

Given:

Normal strain along x-direction is 480×106, and normal strain along y direction is 140×106and the shear strain along x-y plane is 350×106.

Write the expression for the principal strains.

ε1,2=εx+εy2+(εxεy2)2+(γxy2)2        …… (I)

Here, the maximum principal strain is ε1, the minimum principal strain is ε2.

The normal strain along x direction is εx, the normal strain along the y direction is εyand shear strain along x-y plane is γxy.

Write the expression for maximum shear strain.

γmax=ε1ε2     …… (II)

Here, the maximum shear strain is γmax.

Write the expression for principal strain angle.

tan2θp1=γxyεxεy        …… (III)

Here, principal strain angle is θp1.

Write expression for orientation of first shear strain angle.

θs1=θp145°        …… (IV)

Here, first shear angle is θs1.

Write expression for orientation of second shear strain angle.

θs2=θs1+90°     …… (V)

Here, second shear angle is θs2.

Calculation:

Substitute 480×106for εx, 140×106for εyand 350×106for γxyin Equation (I).

ε1,2=480×106+140×1062±(480×106140×1062)2+(350×1062)2=310×106±2.89×108+3

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