### 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 7.2.22P
Textbook Problem
1 views

## Solve the preceding problem for the element shown in the figure.

To determine

The normal stresses acting on the element.

The shear stresses acting on the element.

### Explanation of Solution

Given information:

The normal stress acting on the element in x −direction is 18MPa , the normal stress acting on the element in y -direction is 55MPa , the shear stress acting on the element is 27MPa , the angle subtended from x-axis is 50° .

The figure below shows the stresses acting on the element.

Figure-(1)

Write the expression for normal stress acting along x-direction.

σx=σx+σy2+σxσy2cos(2θ)+τxysin(2θ)    ...... (I)

Here, normal stress acting on the element is σx , normal stress acting on the element is σy , shear stress acting on the element is τxy and the angle subtended from x-axis is θ .

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

σy=σx+σy2σxσy2cos(2θ)τxysin(2θ)  ...... (II)

Here, normal stress along y direction is σy .

Write the expression for shear stress.

τxy=σxσy2sin(2θ)+τxycos(2θ)  ...... (III)

Here, shear stress is τxy .

Calculation:

Substitute, 18MPa for σx , 55MPa for σy , 27MPa for τxy and 50° for θ in Equation (I).

σx=[ 18MPa+55MPa2+[ ( 18MPa55MPa ) 2×cos( 100°)]+( ( 27MPa )×sin( 100° ))]=18.5MPa+(36.5MPa×(0.1737))+((27MPa)×sin(0

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