Chapter 19, Problem 19.1P

To determine

The allowable tensile loadfor the single shear lap joint.

Answer to Problem 19.1P

  $P_g=129.6ksi$

Explanation of Solution

Given:

The lap joint is single shear.

So, $n=1$

The diameter of the bolt is $d=\frac{7}{8}\text{'}\text{'}$

ASTM $A36$ steel plates- $12\text{'}\text{'}\times \frac{1}{2}\text{'}\text{'}$

Calculation:

Now, calculate the area of the bolt

That is,

  $\begin{array}{l}{A}_B=\frac{\pi }{4}{d}^2\\ A_B=\frac{\pi }{4}{\left(\frac{7}{8}\right)}^2\\ A_B=0.60i{n}^2\end{array}$

The allowable load for the connection, based on bolt shear

  $P_s=n.A_B{S}_{s\left(allow\right)}N$

Where, N is the number of bolts.

And, from the table of allowable stresses of steel on steel fastener (A325-X)

  $S_{s\left(allow\right)}=30ksi$

  $\begin{array}{l}{P}_s=n.A_B{S}_{s\left(allow\right)}N\\ P_s=1\times 0.6\times 30\times 9\\ P_s=162kips.\end{array}$

Now, calculate the allowable load for the connection based on bearing on connected material

  $P_b=d.t.S_{p\left(allow\right)}N$

Where, $S_{s\left(allow\right)}=87ksi$ from the ASTM A36.

That is,

  $\begin{array}{l}{P}_b=d.t.S_{s\left(allow\right)}N\\ P_b=0.875\times 0.5\times 87\times 9\\ P_b=342.56kips\end{array}$

Now, calculating allowable tensile load based on the gross area.

  $P_g=A_g\times S_{t(all)}$

Where, $A_g$ is the gross area of the plate.

  $A_g=12\times 0.5=6i{n}^2$

And, fom the ASTM 36 $S_{t(allow)}=21.6ksi$ .

  $\begin{array}{l}{P}_g=6\times 21.6\\ P_g=129.6ksi\end{array}$

Now, calculating the allowable tensile load based on the net area.

  $P_n=A_n\times S_{t(all)}$

Where, $A_n$ is the net area.

  $A_n=bt-N_F{d}_{H}t$

  $\begin{array}{l}{d}_H=\frac{1}{8}+d\\ d_H=\frac{1}{8}+\frac{7}{8}\\ d_H=1in\end{array}$

So,

  $\begin{array}{l}{A}_n=12\times 0.5-3\times 1\times 0.5\\ A_n=4.5i{n}^2\end{array}$

Therefore, load will be

  $\begin{array}{l}{P}_n=4.5\times 29\\ P_n=130.5ksi\end{array}$

Conclusion:

By comparing all, the tensile load for the single shear joint is $P_g=129.6ksi$