Question 12.6: Nails, each having a total shear strength of 40 lb, are used......
Nails, each having a total shear strength of 40 \ \mathrm{lb}, are used in a beam that can be constructed either as in Case \text{I} or as in Case \text{II} , Fig. 12-19. If the nails are spaced at 9 in., determine the largest vertical shear that can be supported in each case so that the fasteners will not fail.
Since the cross section is the same in both cases, the moment of inertia about the neutral axis is calculated using one large rectangle and two smaller side rectangles.
I=\frac{1}{12}(3 \text { in. })(5 \text { in. })^{3}-2\left[\frac{1}{12}(1 \text { in. })(4 \text { in. })^{3}\right]=20.58 \mathrm{in}^{4}
Case \text{I}. For this design a single row of nails holds the top or bottom flange onto the web. For one of these flanges,
Q=\bar{y}^{\prime} A^{\prime}=[2.25 \text { in. }](3 \text { in. }(0.5 \text { in. }))=3.375 \mathrm{in}^{3}
so that
\begin{aligned} q=\frac{V Q}{I} ; \qquad \frac{40 \mathrm{lb}}{9 \ \mathrm{in} .} & =\frac{V\left(3.375 \ \mathrm{in}^{3}\right)}{20.58 \ \mathrm{in}^{4}} \\ V & =27.1 \mathrm{lb} \end{aligned}
Case \text{II}. Here a single row of nails holds one of the side boards onto the web. Thus,
\begin{gathered} Q=\bar{y}^{\prime} A^{\prime}=[2.25 \text { in. }](1 \ \mathrm{in} .(0.5 \ \mathrm{in} .))=1.125 \ \mathrm{in}^{3} \\ q=\frac{F}{s}=\frac{V Q}{I} ; \quad \frac{40 \ \mathrm{lb}}{9 \ \mathrm{in} .}=\frac{V\left(1.125 \ \mathrm{in}^{3}\right)}{20.58 \ \mathrm{in}^{4}} \\ V=81.3 \ \mathrm{lb} \end{gathered}
Or, we can also say two rows of nails hold two side boards onto the web, so that
\begin{aligned} q=\frac{F}{s}=\frac{V Q}{I} ; \qquad \frac{2(40 \mathrm{lb})}{9 \ \mathrm{in.}} & =\frac{V\left[2\left(1.125 \mathrm{in}^{3}\right)\right]}{20.58 \mathrm{in}^{4}} \\ V & =81.3 \mathrm{lb} \end{aligned}