Chapter 7.20, Problem 1QP

To determine

To derive:

The Equation 6.4a and 6.4b using mechanics of materials principles.

Answer to Problem 1QP

The Equation 6.4a and 6.4b using mechanics of materials principles is shown in conclusion.

Explanation of Solution

Write the expression for engineering stress.

$\sigma =\frac{P}{A}$ (I)

Here, the tensile force is P and cross-sectional area is A.

Write the expression for shear stress.

$\tau =\frac{V}{A}$ (II)

Here, the normal force is P and shear forces is V.

Conclusion:

Show the diagram of a block element of material of cross-sectional area A that is subjected to a tensile force P, the orientations of the applied stress, the normal stress to this plane, as well as the shear stress taken parallel to this inclined plane.

For static equilibrium in the x-direction.

${\displaystyle \sum F_{x^{\prime }}}=0$

Which means that

$\begin{array}{l}{P}^{\prime }-P\cos \theta =0\\ P^{\prime }=P\cos \theta \end{array}$

Write the expression for the stress ${\sigma }^{\prime }$ in terms of $P^{\prime }$ and $A^{\prime }$ using the above expression and the relationship between $A$ and $A^{\prime }$.

$\begin{array}{c}{\sigma }^{\prime }=\frac{P^{\prime }}{A^{\prime }}\\ =\frac{P\cos \theta }{\frac{A}{\cos \theta }}\\ =\frac{P}{A}{\cos }^2\theta \\ =\sigma {\cos }^2\theta \end{array}$

Write the expression for ${\tau }^{\prime }$.

$\begin{array}{c}{\tau }^{\prime }=\frac{V^{\prime }}{A^{\prime }}\\ =\frac{P\sin \theta }{\frac{A}{\cos \theta }}\\ =\frac{P}{A}\sin \theta \cos \theta \\ =\sigma \sin \theta \cos \theta \end{array}$

Thus, the ${\tau }^{\prime }$ value is $\underset{\_}{\sigma \sin \theta \cos \theta }$.