# An clement in plane stress is subjected to stresses a,. cry, and (see figure). Using Mohr’s circle, determine the stresses acting on an element oriented at an angle 0 from the x axis. Show these stresses on a sketch of an clement oriented at the angle 0. Now. The angle 0 is positive when count ere clockwise and negative when clockwise.

### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347

### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 7, Problem 7.4.12P
Textbook Problem
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## An clement in plane stress is subjected to stresses a,. cry, and (see figure). Using Mohr’s circle, determine the stresses acting on an element oriented at an angle 0 from the x axis. Show these stresses on a sketch of an clement oriented at the angle 0. Now. The angle 0 is positive when count ere clockwise and negative when clockwise.

To determine

The values of stress on the element oriented at angle 40°from x-axis using Mohr’s circle.

### Explanation of Solution

Given information:

The normal stress along x-axis is 27MPa, normal stress along y-axis is 14MPa,and shear stress along xy-plane is 6MPa, orientation of element is 40°.

The following are the steps TO-Draw the Mohr’s circle

Draw the x-axis as normal stress and y-axis as shear stress and indicate the origin as point O (0,0).

1. Write the expression to locate the center point C of the circle from the origin O(ie., OC).
2. OC=σavg=σx+σy2        …… (I)

Here, the normal stress along x-axis is σx, normal stress along y-axis is σy, and average stress is σavg.

3. Mark the center point as C: (σavg,0).
4. Write the expression for the radius of Mohr’s circle
5. R=(σxσavg)2+τxy2        …… (II)

Here, the radius of the Mohr’s circle is Rand shear stress along xy-plane is τxy.

6. Write the expression for principle angle.

θp=12tan1(τxyσxσavg)        …… (III)

Here, principle angle is θp.

Write the expression for the stress acting on element along x-direction.

σx=σavg+Rcos(2θ2θp)        …… (IV)

Here, the stress acting on element along x-direction is σxand orientation of the plane is θ.

Write the expression for the stress acting on element along y-direction.

σy=σavgRcos(2θ2θp)        …… (V)

Here, the stress acting on element along y-direction is σy.

The shear stress acting on element along x-y plane.

τxy=Rsin(2θ2θp)......VI)

Here, shear stress acting on element along x-y plane is τxy.

Calculation:

Substitute 27MPafor σxand 14MPafor σyin Equation (I).

σavg=27MPa+14MPa2=27MPa2=20.50MPa

Substitute 27MPafor σavgand 20.50MPafor σyand 6MPafor τxyin Equation (II).

R=(27MPa20.5MPa)2+(6MPa)2=78

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