Chapter 6.6, Problem 80P

(a)

To determine

The points where the shearing stress is maximum and the values of the stress.

Answer to Problem 80P

The shearing stress is maximum at the points of $\underset{\_}{y=\frac{2}{5}a\text{ and }z=\frac{2}{3}a}$.

The maximum shear stress along the vertical leg is $\underset{\_}{\frac{27}{20}\frac{P}{ta}}$.

The maximum shear stress along the horizontal leg is $\underset{\_}{-\frac{1}{4}\frac{P}{ta}}$.

The maximum shear stress at the corner of the leg  is $\underset{\_}{\frac{3}{4}\frac{P}{ta}}$.

Explanation of Solution

Calculation:

Refer to sample problem 6.6 in the text book.

Combined stress along the vertical leg ${\tau }_e=\frac{3P\left(a-y\right)\left(a+5y\right)}{4t{a}^3}$ (1)

Combined stress along the horizontal leg ${\tau }_f=\frac{3P\left(a-z\right)\left(a-3z\right)}{4t{a}^3}$ (2)

Modify Equation (1).

$\begin{array}{c}{\tau }_e=\frac{3P}{4t{a}^3}\left(a^2-ay+5ay-5y^2\right)\\ =\frac{3P}{4t{a}^3}\left(a^2+4ay-5y^2\right)\end{array}$

Calculate the point along the vertical leg differentiate both sides of the equation with respect to y as shown below.

$\begin{array}{c}\frac{d{\tau }_e}{dy}=\frac{d}{dy}\left(\frac{3P}{4t{a}^3}\left(a^2+4ay-5y^2\right)\right)\\ =\frac{3P}{4t{a}^3}\left(4a-10y\right)\end{array}$

Consider the condition $\frac{d{\tau }_e}{dy}=0$. Hence,

$\begin{array}{l}\frac{3P}{4t{a}^3}\left(4a-10y\right)=0\\ 4a-10y=0\\ 10y=4a\\ y=\frac{2}{5}a\end{array}$

Hence, the shearing stress is maximum at the points of $\underset{\_}{y=\frac{2}{5}a}$.

Calculate the maximum shear stress along the vertical leg as shown below.

Substitute $\frac{2}{5}a$ for y in Equation (1).

$\begin{array}{c}{\tau }_m=\frac{3P}{4t{a}^3}\left(a-\frac{2}{5}a\right)\left(a+5\times \frac{2}{5}a\right)\\ =\frac{3P}{4t{a}^3}\times \frac{3}{5}a\times 3a\\ =\frac{27}{20}\frac{P}{ta}\end{array}$

Hence, the maximum shear stress along the vertical leg is $\underset{\_}{\frac{27}{20}\frac{P}{ta}}$.

Modify Equation (2).

$\begin{array}{c}{\tau }_f=\frac{3P}{4t{a}^3}\left(a^2-az-3az+3z^2\right)\\ =\frac{3P}{4t{a}^3}\left(a^2-4az+3z^2\right)\end{array}$

Calculate the point along the horizontal leg differentiate both sides of the equation with respect to z as shown below.

$\begin{array}{c}\frac{d{\tau }_f}{dz}=\frac{d}{dz}\left(\frac{3P}{4t{a}^3}\left(a^2-4az+3z^2\right)\right)\\ =\frac{3P}{4t{a}^3}\left(-4a+6z\right)\end{array}$

Consider the condition $\frac{d{\tau }_f}{dy}=0$. Hence,

$\begin{array}{l}\frac{3P}{4t{a}^3}\left(-4a+6z\right)=0\\ -4a+6z=0\\ 6z=4a\\ z=\frac{2}{3}a\end{array}$

Hence, the shearing stress is maximum at the points of $\underset{\_}{z=\frac{2}{3}a}$.

Calculate the maximum shear stress along the horizontal leg as shown below.

Substitute $\frac{2}{3}a$ for z in Equation (2).

$\begin{array}{c}{\tau }_m=\frac{3P}{4t{a}^3}\left(a-\frac{2}{3}a\right)\left(a-3\times \frac{2}{3}a\right)\\ =\frac{3P}{4t{a}^3}\left(\frac{1}{3}a\right)\left(-a\right)\\ =-\frac{1}{4}\frac{P}{ta}\end{array}$

Hence, the maximum shear stress along the horizontal leg is $\underset{\_}{-\frac{1}{4}\frac{P}{ta}}$.

The corner points of the horizontal and vertical legs are $y=0\text{and }z=0$.

Calculate the maximum shear stress at the corner point as shown below.

Substitute 0 for y in Equation (1).

$\begin{array}{c}{\tau }_m=\frac{3P}{4t{a}^3}\left(a-0\right)\left(a+5\times 0\right)\\ =\frac{3P}{4t{a}^3}\times a^2\\ =\frac{3}{4}\frac{P}{ta}\end{array}$

Substitute 0 for z in Equation (2).

$\begin{array}{c}{\tau }_m=\frac{3P}{4t{a}^3}\left(a-0\right)\left(a-3\times 0\right)\\ =\frac{3P}{4t{a}^3}{a}^2\\ =\frac{3}{4}\frac{P}{ta}\end{array}$

Therefore, the maximum shear stress at the corner of the leg  is $\underset{\_}{\frac{3}{4}\frac{P}{ta}}$.

(b)

To determine

Show that the points are located on the neutral axis for the loading P.

Answer to Problem 80P

The points y and z are located on the neutral axis for the loading P.

Explanation of Solution

Calculation:

Sketch the cross section along the neutral axis as shown below.

Refer to Figure 1.

Calculate the moment of inertia as shown below.

Along $y^{\prime }$ axis:

$I_{y^{\prime }}=\frac{1}{3}t{a}^3$

Along $z^{\prime }$ axis:

$I_{z^{\prime }}=\frac{1}{12}t{a}^3$

Consider the angle $\theta =45°$.

Calculate angle $\varphi$ as shown below.

$\tan \varphi =\frac{I_{z^{\prime }}}{I_{y^{\prime }}}\tan \theta$

Substitute $\frac{1}{3}t{a}^3$ for $I_{y^{\prime }}$, $\frac{1}{12}t{a}^3$ for $I_{z^{\prime }}$, and $45°$ for $\theta$.

$\begin{array}{l}\tan \varphi =\frac{\frac{1}{12}t{a}^3}{\frac{1}{3}t{a}^3}\tan 45°\\ \tan \varphi =\frac{1}{4}\\ \varphi ={\tan }^{-1}\left(\frac{1}{4}\right)\\ \varphi =14.036°\end{array}$

Calculate the angle of the neutral axis from the horizontal as shown below.

$\begin{array}{c}\theta -\varphi =45°-14.036°\\ =30.964°\end{array}$

Calculate the location of the centroid as shown below.

Along y axis:

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum A\overline{y}}}{{\displaystyle \sum A}}\\ =\frac{at\times \frac{a}{2}}{2at}\\ =\frac{1}{4}a\end{array}$

Along z axis:

$\begin{array}{c}\overline{z}=\frac{{\displaystyle \sum A\overline{z}}}{{\displaystyle \sum A}}\\ =\frac{at\times \frac{a}{2}}{2at}\\ =\frac{1}{4}a\end{array}$

Calculate the neutral axis intersects for vertical leg as shown below.

$y=\overline{y}+\overline{z}\tan 36.964°$

Substitute $\frac{1}{4}a$ for $\overline{y}$ and $\frac{1}{4}a$ for $\overline{z}$.

$\begin{array}{c}y=\frac{1}{4}a+\frac{1}{4}a\times \tan 36.964°\\ =0.4a\end{array}$

Calculate the neutral axis intersects for horizontal leg as shown below.

$z=\overline{z}+\overline{y}\tan \left(45°+\varphi \right)$

Substitute $\frac{1}{4}a$ for $\overline{y}$, $\frac{1}{4}a$ for $\overline{z}$, and $14.036°$ for $\varphi$.

$\begin{array}{c}z=\frac{1}{4}a+\frac{1}{4}a\times \tan \left(45°+14.036°\right)\\ =\frac{1}{4}a+\frac{1}{4}a\times \tan \left(59.036°\right)\\ =0.67a\end{array}$

Therefore, the points y and z are located on the neutral axis for the loading P