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 cr_v, tr₍_, and t. Show the results on a sketch of an clement oriented at B = 10

  

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 $4221.42\text{psi}$.

The normal stresses is in y direction is $4403.89\text{psi}$.

The shear stresses is in xy direction is $-3410.74\text{psi}$.

Explanation of Solution

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

${\sigma }_{x^{\prime }}=\left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)+\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)\cos 2\theta +{\tau }_{xy}\sin 2\theta$        …… (I)

Here, stress acting along the x direction is ${\sigma }_x$, stress is acting along y direction ${\sigma }_y$,shear stress acting in xy is ${\tau }_{xy}$, angle oriented on an element is $\theta$.

Write the expression for normal stress acting on the $\text{y}$axis inclined to the original axis.

${\sigma }_{y^{\prime }}=\left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)-\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)\cos 2\theta -{\tau }_{xy}\sin 2\theta$        …… (II)

Write the expression for shear stress acting along the $\text{x1y1}$plane inclined to the original axis.

${\tau }_{x^{\prime }{y}^{\prime }}=-\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)\sin 2\theta +{\tau }_{xy}\cos 2\theta$        …… (III)

Calculation:

The following figure shows the stress acting on an element at $\theta \text{=30}°$

    

    Figure-(1)

Substitute, $7750\text{psi}$for ${\sigma }_x$, $1175\text{psi}$for ${\sigma }_y$, and $940\text{psi}$for ${\tau }_{xy}$, $55°$for $\theta$in the Equation (I)

$\begin{array}{c}{\sigma }_{x^{\prime }}=\left(\frac{7750\text{psi}+1175\text{psi}}{2}\right)+\left(\frac{7750\text{psi}-1175}{2}\right)\cos \left(2\times 55°\right)\\ +\left(940\text{psi}\right)\sin \left(2\times 55°\right)\\ =\left(\frac{8925\text{psi}}{2}\right)+\left(\frac{6575\text{psi}}{2}\right)\times \left(-0.342020143\right)+\left(940\text{psi}\times 0.939692621\right)\\ =90\text{psi}+3287.5\text{psi}\times \left(-0.342020143\right)+\left(940\text{psi}\times 0.939692621\right)\\ =4221.42\text{psi}\end{array}$

The normal stress along x direction is $4221.42\text{psi}$.

Substitute, $7750\text{psi}$for ${\sigma }_x$, $1175\text{psi}$for ${\sigma }_y$, and $940\text{psi}$for ${\tau }_{xy}$, $55°$for $\theta$in the Equation (II)

$\begin{array}{c}{\sigma }_{y^{\prime }}=\left(\frac{7750\text{psi}+1175\text{psi}}{2}\right)-\left(\frac{7750\text{psi}-1175\text{psi}}{2}\right)\cos \left(2\times 55°\right)\\ -\left(940\text{psi}\right)\sin \left(2\times 55°\right)\\ =\left(\frac{8925\text{psi}}{2}\right)-\left(\frac{6575\text{psi}}{2}\right)\times \left(-0.342020143\right)-\left(940\text{psi}\times 0.939692621\right)\\ =4462.5\text{psi}+1124.39\text{psi}-883.31\text{psi}\\ =4403.89\text{psi}\end{array}$

The normal stress is along y direction is $4403.89\text{psi}$.

Substitute, $7750\text{psi}$for ${\sigma }_x$, $1175\text{psi}$for ${\sigma }_y$, and $940\text{psi}$for ${\tau }_{xy}$, $55°$for $\theta$in the Equation (III)

$\begin{array}{c}{\tau }_{x^{\prime }{y}^{\prime }}=-\left(\frac{7750\text{psi}-1175\text{psi}}{2}\right)\sin \left(2\times 55°\right)-\left(940\text{psi}\right)\cos \left(2\times 55°\right)\\ =-\left(\frac{6575\text{psi}}{2}\right)\times 0.939692621-\left(940\text{psi}\right)\times \left(-0.342020143\right)\\ =-3089.23-321.49\\ =-3410.74\text{psi}\end{array}$

The normal stress is along xy direction is $-3410.74\text{psi}$.

Conclusion:

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

    

    Figure (2)