# An element in biaxial stress is subjected to stresses a = -29 MPa and s y = 57 MPa. as shown in the figure. Using Mohr’s circle, determine the following. (a) The stresses acting on an element oriented at a slope of I on 2.5 (see figure). (b) The maximum shear stresses and associated norm al stresses. Show all results on sketches of properly oriented elements.

### 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 7.4.6P
Textbook Problem
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## An element in biaxial stress is subjected to stresses a = -29 MPa and sy = 57 MPa. as shown in the figure. Using Mohr’s circle, determine the following. (a) The stresses acting on an element oriented at a slope of I on 2.5 (see figure). (b) The maximum shear stresses and associated norm al stresses. Show all results on sketches of properly oriented elements.

(a)

To determine

The stresses acting one an element oriented at a slope of 1on 2.5.

### Explanation of Solution

The stress acting on element along x-direction is 29MPaand along y-direction is 57MPa.

Write the expression to calculate the distance of center C from origin O along the horizontal direction.

OC=σx+σy2..…. (I)

Here the stress along the x-direction is σxand stress along the y-direction is and σy.

• Mark point A and B on the x-axis. OA represents stress acting on element along x-axis and OB represents stress acting on element along y-axis. Here shear stress is zero

Write the expression for the radius of Mohr’s circle.

R=|σx|+|σy|2        …… (II)

Here, radius of Mohr’s circle is R.

Write the expression for orientation of element.

θ=tan1(yx)        ……. (III)

Here, slope of the oriented element is θ, and intercept at x-axis is yand intercept at y-axis is x.

Write the expression for stress along center point.

σC=σx+R        …… (IV)

Here, stress along center point is σC.

Write the expression for minimum principle stress.

σx1=σCRcos(2θ)        …… (V)

Here, minimum principle stress is σx1.

Write the expression for maximum principle stress.

σx2=σC+Rcos(2θ)        …… (VI)

Here, maximum principle stress is σx2.

Write the expression for shear stress on oriented plane.

τx1y1=Rsin(2θ)        …… (VII)

Here, principal shear stress is τx1y1

Calculation:

Substitute 29MPafor σxand 57Mpafor σyin Equation (I).

OC=29MPa+57MPa2=28MPa2=14MPa

Substitute 29MPafor σxand 57Mpafor σyin Equation (II).

R=|29MPa|+|57MPa|2=(86MPa2)=43MPa

Substitute 1unitfor yand 2.5unitsfor xin equation (III)

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