Question 12.2: Determine the distribution of the shear stress over the cros......
Determine the distribution of the shear stress over the cross section of the beam shown in Fig. 12-11a.
The distribution can be determined by finding the shear stress at an arbitrary height y from the neutral axis, Fig. 12-11b, and then plotting this function. Here, the dark colored area A^{\prime} will be used for Q.{ }^{*} Hence
Q=\bar{y}^{\prime} A^{\prime}=\left[y+\frac{1}{2}\left(\frac{h}{2}-y\right)\right]\left(\frac{h}{2}-y\right) b=\frac{1}{2}\left(\frac{h^{2}}{4}-y^{2}\right) b
Applying the shear formula, we have
\tau=\frac{V Q}{I t}=\frac{V\left(\frac{1}{2}\right)\left[\left(h^{2} / 4\right)-y^{2}\right] b}{\left(\frac{1}{12} b h^{3}\right) b}=\frac{6 V}{b h^{3}}\left(\frac{h^{2}}{4}-y^{2}\right) (1)
This result indicates that the shear-stress distribution over the cross section is parabolic. As shown in Fig. 12-11 c, the intensity varies from zero at the top and bottom, y= \pm h / 2, to a maximum value at the neutral axis, y=0. Specifically, since the area of the cross section is A=b h, then at y=0 Eq. 1 becomes
\tau_{\max }=1.5 \frac{V}{A} (2)
\text { Rectangular cross section }
This same value for \tau_{\max } can be obtained directly from the shear formula, \tau=V Q / It , by realizing that \tau_{\max } occurs where Q is largest, since V, I, and t are constant. By inspection, Q will be a maximum when the entire area above (or below) the neutral axis is considered; that is, A^{\prime}=b h / 2 and \bar{y}^{\prime}=h / 4, Fig. 12-11 d. Thus,
\tau_{\max }=\frac{V Q}{I t}=\frac{V(h / 4)(b h / 2)}{\left[\frac{1}{12} b h^{3}\right] b}=1.5 \frac{V}{A}
By comparison, \tau_{\max } is 50 \% greater than the average shear stress determined from Eq. 7-4; that is, \tau_{\text {avg }}=V / A.
It is important to realize that \tau_{\max } also acts in the longitudinal direction of the beam, Fig. 12-11e. It is this stress that can cause a timber beam to fail at its supports, as shown Fig. 12-11f. Here horizontal splitting of the wood starts to occur through the neutral axis at the beam’s ends, since there the vertical reactions subject the beam to large shear stress, and wood has a low resistance to shear along its grains, which are oriented in the longitudinal direction.
It is instructive to show that when the shear-stress distribution, Eq. 1, is integrated over the cross section it produces the resultant shear V. To do this, a differential strip of area d A=b d y is chosen, Fig. 12-11c, and since \tau acts uniformly over this strip, we have
\begin{aligned} \int_{A} \tau d A & =\int_{-h / 2}^{h / 2} \frac{6 V}{b h^{3}}\left(\frac{h^{2}}{4}-y^{2}\right) b d y \\ & =\frac{6 V}{h^{3}}\left[\frac{h^{2}}{4} y-\frac{1}{3} y^{3}\right]_{-h / 2}^{h / 2} \\ & =\frac{6 V}{h^{3}}\left[\frac{h^{2}}{4}\left(\frac{h}{2}+\frac{h}{2}\right)-\frac{1}{3}\left(\frac{h^{3}}{8}+\frac{h^{3}}{8}\right)\right]=V \end{aligned}
*The area below y can also be used \left[A^{\prime}=b(h / 2+y)\right], but doing so involves a bit more algebraic manipulation.
