Chapter 6.2, Problem 5P

The American Standard rolled-steel beam shown has been reinforced by attaching to it two 16 × 200-mm plates, using 18-mm-diameter bolts spaced longitudinally every 120 mm. Knowing that the average allowable shearing stress in the bolts is 90 MPa, determine the largest permissible vertical shearing force.

To determine

The largest permissible vertical shearing force.

Answer to Problem 5P

The largest permissible vertical shearing force is $\underset{\_}{193.4\text{kN}}$.

Explanation of Solution

Given information:

The diameter of bolt is $18\text{mm}$.

The longitudinal spacing is $120\text{mm}$.

The average shearing stress in the bolts is $90\text{MPa}$.

Calculation:

Provide the section properties of the rolled steel beam $\text{S}310\times 52$ as shown below.

The area of the section is $A=6,650{\text{mm}}^2$.

The moment of inertia of the section $I=95.3\times {10}^6{\text{mm}}^4$.

The Overall depth of the member $d=305\text{mm}$.

Calculate the moment of inertia as shown below.

$I=\frac{b{h}^3}{12}+A{\left(y-\overline{y}\right)}^2$

Here, b is the breadth of the beam, h is the height of the beam, A is the area of the beam, and $\left(y-\overline{y}\right)$ is the centroid of the beam from the neutral axis.

For the top plate.

$\begin{array}{c}{I}_1=\left[\frac{1}{12}\times 200\times {16}^3\right]+\left(200\times 16\right)\times {\left(\frac{305}{2}+\frac{16}{2}\right)}^2\\ =68,266.67+82,432,800\\ =82.5\times {10}^6{\text{mm}}^4\end{array}$

Calculate the area $\left(A_1\right)$ of the top plate as shown below.

$\begin{array}{c}{A}_1=200\times 16\\ =3,200{\text{mm}}^2\end{array}$

Calculate the location of the centroid $\left({\overline{y}}_1\right)$ above the neutral axis as shown below.

$\begin{array}{c}\overline{y}=\frac{305}{2}+\frac{16}{2}\\ =160.5\text{mm}\end{array}$

Similarly calculate the moment of inertia for the bottom plate and rolled steel beam $\text{S}310\times 52$ as shown in table 1.

Part	Area, $A\left({\text{mm}}^2\right)$	$\left(y-\overline{y}\right)$$\left(\text{mm}\right)$	$A{\left(y-\overline{y}\right)}^2\left({\text{mm}}^4\right)$	Moment of Inertia, $I\left({\text{mm}}^4\right)$
Top plate	$3,200$	$160.5$	$82.4328\times {10}^6$	$82.5\times {10}^6$
$\text{S}310\times 52$	$6,650$			$95.3\times {10}^6$
Bottom plate	$3,200$	$160.5$	$82.4328\times {10}^6$	$82.5\times {10}^6$

Calculate the moment of inertia for the whole section as shown below.

$\begin{array}{c}I=82.5\times {10}^6+95.3\times {10}^6+82.5\times {10}^6\\ =260.3\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4\\ =2.603\times {10}^{-4}{\text{m}}^4\end{array}$

Calculate the first moment of area $\left(Q_{\text{plate}}\right)$ of the plate as shown below.

$Q_{\text{plate}}=A_{\text{plate}}{\overline{y}}_{\text{plate}}$

Substitute $3,200{\text{mm}}^2$ for $A_{\text{plate}}$ and $160.5\text{mm}$ for ${\overline{y}}_{\text{plate}}$.

$\begin{array}{c}{Q}_{\text{plate}}=3,200\times 160.5\\ =513.6\times {10}^3{\text{mm}}^3\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^3\\ =5.136\times {10}^{-4}{\text{m}}^3\end{array}$

Calculate the area of bolt as shown below.

$A_{\text{bolt}}=\frac{\pi d_{\text{bolt}}^2}{4}$

Here, $d_{\text{bolt}}$ is the diameter of the bolt.

Substitute $18\text{mm}$ for $d_{\text{bolt}}$.

$\begin{array}{c}{A}_{\text{bolt}}=\frac{\pi \times {18}^2}{4}\\ =254.47{\text{mm}}^2\end{array}$

Calculate the force acting on the bolt $\left(F_{\text{bolt}}\right)$ as shown below.

$F_{\text{bolt}}={\tau }_{\text{bolt}}{A}_{\text{bolt}}$

Here, ${\tau }_{\text{bolt}}$ is the shear stress in the bolt.

Substitute $90\text{MPa}$ for ${\tau }_{\text{bolt}}$ and $254.47{\text{mm}}^2$ for $A_{\text{bolt}}$.

$\begin{array}{c}{F}_{\text{bolt}}=\left(90\text{MPa}\times \frac{{10}^6\text{N}/{\text{m}}^2}{1\text{MPa}}\right)\times 254.47{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =22.90\times {10}^3\text{N}\end{array}$

Calculate the horizontal shear per unit length as shown below.

$qs=2F_{\text{bolt}}$

Substitute $22.90\times {10}^3\text{N}$ for $F_{\text{bolt}}$ and $120\text{mm}$ for s.

$\begin{array}{l}q\times 120=2\times 22.90\times {10}^3\\ q=381.67\text{N}/\text{mm}\times \left(\frac{1,000\text{mm}}{1\text{m}}\right)\\ q=381.67\times {10}^3\text{N}/\text{m}\end{array}$

Calculate the vertical shearing force (V) as shown below.

$q=\frac{VQ}{I}$

Substitute $5.136\times {10}^{-4}{\text{m}}^3$ for Q, $2.603\times {10}^{-4}{\text{m}}^4$ for I, and $381.67\times {10}^3\text{N}/\text{m}$ for q.

$\begin{array}{l}381.67\times {10}^3=\frac{V\times 5.136\times {10}^{-4}}{2.603\times {10}^{-4}}\\ V=193.44\times {10}^3\text{N}\times \frac{1\text{kN}}{1,000\text{N}}\\ V=193.44\text{kN}\end{array}$

Therefore, the largest permissible vertical shearing force is $\underset{\_}{193.4\text{kN}}$.