Chapter 6, Problem 89RP

Three boards are nailed together to form the beam shown, which is subjected to a vertical shear. Knowing that the spacing between the nails is s = 75 mm and that the allowable shearing force in each nail is 400 N, determine the allowable shear when w = 120 mm.

Fig. p6.89

To determine

The allowable shear in the beam.

Answer to Problem 89RP

The allowable shear in the beam is $\underset{\_}{738\text{N}}$.

Explanation of Solution

Given information:

The spacing between the nails is $s=75\text{mm}$.

The allowable shearing in each nail is $400\text{N}$.

Calculation:

Sketch the cross section as shown in Figure 1.

Refer to Figure 1.

Calculate the moment of inertia (I) as shown below.

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

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

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

$\begin{array}{c}I=\left[2\times \left[\frac{120\times {60}^3}{12}+120\times 60\times {60}^2\right]+\left[\frac{200\times {60}^3}{12}\right]\right]\\ =56.16\times {10}^6+3.6\times {10}^6\\ =59.76\times {10}^6{\text{mm}}^4\end{array}$

Calculate the first moment of area (Q) as shown below.

$\begin{array}{c}Q=A\overline{y}\\ =120\times 60\times 60\\ =432,000{\text{mm}}^3\end{array}$

Calculate the horizontal shear per unit length (q) as shown below.

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

Here, V is the vertical shear.

Substitute $432,000{\text{mm}}^3$ for Q and $59.76\times {10}^6{\text{mm}}^4$ for I.

$\begin{array}{c}q=\frac{V\times 432,000}{59.76\times {10}^6}\\ =7.2289\times {10}^{-3}V\end{array}$

Calculate the allowable shear $\left(F_{\text{nail}}\right)$ as shown below.

$q=\frac{F_{\text{nail}}}{s}$

Substitute $7.2289\times {10}^{-3}V$ for V, $400\text{N}$ for $F_{\text{nail}}$, and $75\text{mm}$ for s.

$\begin{array}{l}7.2289\times {10}^{-3}V=\frac{400}{75}\\ 542.1675\times {10}^{-3}V=400\\ V=737.78\text{N}\\ V=738\text{N}\end{array}$

Therefore, the allowable shear is $\underset{\_}{738\text{N}}$.