Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 7, Problem 7.2.1P

An clement m plane stress from the frame of a racing car is oriented at a known angle 8 (sec figure). On this inclined clement, the normal and shear stresses have the magnitudes and directions shown in the figure.

Determine the normal and shear stresses acting on an clement whose sides are parallel to the xy axes, that is, determine crv, tr(_, and t. Show the results on a sketch of an clement oriented at B = 10

  Chapter 7, Problem 7.2.1P, An clement m plane stress from the frame of a racing car is oriented at a known angle 8 (sec

Expert Solution & Answer
Check Mark
To determine

The stress acting on the element oriented at an angle.

Answer to Problem 7.2.1P

The normal stress is in x direction is 4221.42psi.

The normal stresses is in y direction is 4403.89psi.

The shear stresses is in xy direction is 3410.74psi.

Explanation of Solution

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

σx=(σx+σy2)+(σxσy2)cos2θ+τxysin2θ        …… (I)

Here, stress acting along the x direction is σx, stress is acting along y direction σy,shear stress acting in xy is τxy, angle oriented on an element is θ.

Write the expression for normal stress acting on the yaxis inclined to the original axis.

σy=(σx+σy2)(σxσy2)cos2θτxysin2θ        …… (II)

Write the expression for shear stress acting along the x1y1plane inclined to the original axis.

τxy=(σxσy2)sin2θ+τxycos2θ        …… (III)

Calculation:

The following figure shows the stress acting on an element at θ=30°

    Mechanics of Materials (MindTap Course List), Chapter 7, Problem 7.2.1P , additional homework tip  1

    Figure-(1)

Substitute, 7750psifor σx, 1175psifor σy, and 940psifor τxy, 55°for θin the Equation (I)

σx=(7750psi+1175psi2)+(7750psi11752)cos(2×55°)+(940psi)sin(2×55°)=(8925psi2)+(6575psi2)×(0.342020143)+(940psi×0.939692621)=90psi+3287.5psi×(0.342020143)+(940psi×0.939692621)=4221.42psi

The normal stress along x direction is 4221.42psi.

Substitute, 7750psifor σx, 1175psifor σy, and 940psifor τxy, 55°for θin the Equation (II)

σy=(7750psi+1175psi2)(7750psi1175psi2)cos(2×55°)(940psi)sin(2×55°)=(8925psi2)(6575psi2)×(0.342020143)(940psi×0.939692621)=4462.5psi+1124.39psi883.31psi=4403.89psi

The normal stress is along y direction is 4403.89psi.

Substitute, 7750psifor σx, 1175psifor σy, and 940psifor τxy, 55°for θin the Equation (III)

τxy=(7750psi1175psi2)sin(2×55°)(940psi)cos(2×55°)=(6575psi2)×0.939692621(940psi)×(0.342020143)=3089.23321.49=3410.74psi

The normal stress is along xy direction is 3410.74psi.

Conclusion:

The following figure represents the stress acting on an element oriented at an angle 55°.

    Mechanics of Materials (MindTap Course List), Chapter 7, Problem 7.2.1P , additional homework tip  2

    Figure (2)

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Chapter 7 Solutions

Mechanics of Materials (MindTap Course List)

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