## Chapter 7

## Q. 7.5

An I-beam is made by gluing five wood planks together, as shown. At a given axial position, the beam is subjected to a shear force V = 6000 lb. (a) What is the average shear stress at the neutral axis z = 0? (b) What are the magnitudes of the average shear stresses acting on each glued joint?

Given: Cross section, local loading.

Find: Average shear stresses.

Assume: Hooke’s law applies.

## Step-by-Step

## Verified Solution

We obtained a formula for shear stress at a given height, σ_{xz} = V Q^{\prime} /I b, where Q^{\prime} and b depend on the z position in question. I in this relationship is always the second moment of area of the entire cross section about the z-axis. We have been given V. So, we must calculate I once and then calculate the appropriate values of Q^{\prime} and b for both parts of this problem.

By inspection of the cross section’s symmetry, we see that the centroid is at the geometric center of the I-beam. For the central vertical segment, therefore d, the distance between the centroid of the segment and the centroid of the entire cross section, is zero. The four remaining segments will each have the same second moment of area about their own horizontal bisectors, and the same areas and distances d. Thus, we can write

\begin{aligned} & I=I_{\text {vertical }}+4 I_{\text {smaller }}=\left(\frac{1}{12} b h^3\right)_{\text {vertical }}+4\left[\frac{1}{12} b h^3+A d^2\right]_{\text {smaller }}{ }^{\prime} \\ & I=\frac{1}{12}(2 \text { in })(8 \text { in })^3+4\left[\frac{1}{12}(4 \text { in })(2 \text { in })^3+(2 \text { in } \times 4 \text { in })(3 \text { in })^2\right]=384 \text { in }^4 \end{aligned}We can calculate Q^{\prime} at the neutral axis by finding the centroid and area of the shaded area on the left, or by summing the contributions due to the individual planks, as shown at right. The values of z for planks 2 and 3 are the same as their d values used in the I calculation.

\begin{aligned} Q^{\prime} & =\int_{A^{\prime}} z \mathrm{~d} A=\sum z^{\prime} A^{\prime} \\ & =(2)(2 \cdot 4)+3(4 \cdot 2)+3(4 \cdot 2)=64 \mathrm{in}^3 . \end{aligned}So, the average shear stress at the neutral axis is \sigma_{x z}=V Q^{\prime} / I b=(6000 \mathrm{lb})\left(64 \mathrm{in}^3\right) / \left(384 \text { in }^4\right)(2 \text { in })=500 \text { psi. }

As an exercise, verify that each glued joint is subjected to the same average shear stress. We will determine only the average shear stress acting on the lower-right glued joint by using the area A and length of contact b as shown below. The value of Q^{\prime} is (3)(4 · 2) = 24 in³, and the average shear stress is V Q^{\prime} /I b = 188 psi.