Chapter 6.5, Problem 53P

To determine

To provide: The expression of maximum shearing stress as ${\tau }_{\max }=k\left(\frac{V}{A}\right)$.

The constant k for each orientation.

Answer to Problem 53P

The constant k for orientation (a) is $\underset{\_}{2.08}$.

The constant k for orientation (b) is $\underset{\_}{2.10}$.

Explanation of Solution

Given information:

The beam is a hollow square of side a and thickness t.

The beam is subjected to a vertical shear of V.

Calculation:

Orientation (a)

Sketch the cross section of an extruded beam as shown in Figure 1.

Refer to Figure 1.

The height of the section above neutral axis is $h=\frac{\sqrt{3}}{2}a$.

Area of the cross member is $A_1=A_2$$=at$.

Calculate the area of the member A as shown below.

$A=6at$

Calculate the moment of inertia of the beam I as shown below.

$\begin{array}{c}I=2at{h}^2+4\times \frac{1}{3}at{h}^2\\ =\frac{10at{h}^2}{3}\end{array}$

Substitute $\frac{\sqrt{3}}{2}a$ for h.

$\begin{array}{c}I=\frac{10at}{3}\times {\left(\frac{\sqrt{3}}{2}a\right)}^2\\ =\frac{10at}{3}\times \frac{3}{4}{a}^2\\ =\frac{5}{2}{a}^{3}t\end{array}$

Calculate the first moment of area as shown below.

$Q={\displaystyle \sum A\overline{y}}$

Here, A is the area of the section and $\overline{y}$ is the centroid of the section.

Calculate the first moment of area along the neutral axis as shown below.

$\begin{array}{c}{Q}_{\max }=at\times h+\left(2\times at\times \frac{h}{2}\right)\\ =ath+ath\\ =2ath\end{array}$

Substitute $\frac{\sqrt{3}}{2}a$ for h.

$\begin{array}{c}{Q}_{\max }=2at\times \frac{\sqrt{3}}{2}a\\ =\sqrt{3}{a}^{2}t\end{array}$

Calculate the shear stress $\tau$ as shown below.

$\tau =\frac{VQ}{It}$

Here V is the vertical shear.

Substitute $\sqrt{3}{a}^{2}t$ for Q, $2t$ for t, and $\frac{5}{2}{a}^{3}t$ for I in Equation (2).

$\begin{array}{c}\tau =\frac{V\times \sqrt{3}{a}^{2}t}{\frac{5}{2}{a}^{3}t\times 2t}\\ =\frac{\sqrt{3}V}{5at}\end{array}$

Substitute $\frac{1}{6}A$ for $at$.

$\begin{array}{c}\tau =\frac{\sqrt{3}V}{5\frac{1}{6}A}\\ =\frac{6\sqrt{3}}{5}\frac{V}{A}\end{array}$

Hence, the expression for maximum shearing stress is $\frac{6\sqrt{3}}{5}\frac{V}{A}$.

Calculate the constant as shown below.

$\tau =2.08\frac{V}{A}$

Therefore, the constant k is $\underset{\_}{2.08}$.

(b)

Sketch the cross section of an extruded beam as shown in Figure 2.

Refer to Figure 2.

The height of the section above neutral axis is $h=\frac{a}{2}$.

Area of the cross member is $A_1=A_2$$=at$.

Calculate the area of the member A as shown below.

$A=6at$

Calculate the moment of inertia I of the beam as shown below.

$\begin{array}{c}I=4\times \left(\frac{1}{12}at{h}^2+at{\left(\frac{a}{2}+\frac{h}{2}\right)}^2\right)+4\times \frac{1}{3}t\times {\left(\frac{a}{2}\right)}^3\\ =4\times \left[\frac{1}{12}at{h}^2+at{\left(\frac{a}{2}+\frac{h}{2}\right)}^2+\frac{1}{24}{a}^{3}t\right]\end{array}$

Substitute $\frac{a}{2}$ for h.

$\begin{array}{c}I=4\left(\frac{1}{12}at{\left(\frac{a}{2}\right)}^2+at{\left(\frac{a}{2}+\frac{\frac{a}{2}}{2}\right)}^2+\frac{1}{24}{a}^{3}t\right)\\ =4\times \left(\frac{1}{48}{a}^{3}t+at{\left(\frac{3}{4}a\right)}^2+\frac{1}{24}{a}^{3}t\right)\\ =4\times \left(\frac{3}{48}{a}^{3}t+\frac{9}{16}{a}^{3}t\right)\\ =4\times \left(\frac{3+27}{48}\right)a^{3}t\end{array}$

$\begin{array}{c}=\frac{30}{12}{a}^{3}t\\ =\frac{5}{2}{a}^{3}t\end{array}$

Calculate the first moment of area as shown below.

$Q={\displaystyle \sum A\overline{y}}$

Here, A is the area of the section and $\overline{y}$ is the centroid of the section.

Calculate the first moment of area along the neutral axis as shown below.

$\begin{array}{c}{Q}_{\max }=2\times at\times \left(\frac{a}{2}+\frac{h}{2}\right)+2\times \frac{1}{2}at\times \frac{a}{4}\\ =2at\left(\frac{a}{2}+\frac{h}{2}\right)+\frac{a^{2}t}{4}\end{array}$

Substitute $\frac{a}{2}$ for h.

$\begin{array}{c}{Q}_{\max }=2at\left(\frac{a}{2}+\frac{\frac{a}{2}}{2}\right)+\frac{a^{2}t}{4}\\ =2at\times \frac{3a}{4}+\frac{a^{2}t}{4}\\ =\frac{3a^{2}t}{2}+\frac{a^{2}t}{4}\\ =\frac{7}{4}{a}^{2}t\end{array}$

Calculate the shear stress as shown below.

$\tau =\frac{VQ}{It}$

Here V is the vertical shear.

Substitute $\frac{7}{4}{a}^{2}t$ for Q, $2t$ for t, and $\frac{5}{2}{a}^{3}t$ for I in Equation (2).

$\begin{array}{c}\tau =\frac{V\times \frac{7}{4}{a}^{2}t}{\frac{5}{2}{a}^{3}t\times 2t}\\ =\frac{7V}{4\times 5at}\\ =\frac{7}{20}\frac{V}{at}\end{array}$

Substitute $\frac{1}{6}A$ for $at$.

$\begin{array}{c}\tau =\frac{7}{20}\frac{V}{\frac{1}{6}A}\\ =\frac{21}{10}\frac{V}{A}\end{array}$

Hence, the expression for maximum shearing stress is $\underset{\_}{\frac{21}{10}\frac{V}{A}}$.

Calculate the constant as shown below.

$\tau =2.1\frac{V}{A}$

Therefore, the constant k is $\underset{\_}{2.1}$.