Question 12.3: A steel wide-flange beam has the dimensions shown in Fig. 12......

Since the flange and web are rectangular elements, then like the previous example, the shear-stress distribution will be parabolic and in this case it will vary in the manner shown in Fig. 12-12b. Due to symmetry, only the shear stresses at points B^{\prime}, B, and C have to be determined. To show how these values are obtained, we must first determine the moment of inertia of the cross-sectional area about the neutral axis. Working in meters, we have

For point B^{\prime}, t_{B}{ }^{\prime}=0.300 \mathrm{~m}, and A^{\prime} is the dark shaded area shown in Fig. 12-12c. Thus,

Q_{B^{\prime}}=\bar{y}^{\prime} A^{\prime}=[0.110 \mathrm{~m}](0.300 \mathrm{~m})(0.02 \mathrm{~m})=0.660\left(10^{-3}\right) \mathrm{m}^{3}

so that

\tau_{B^{\prime}}=\frac{V Q_{B^{\prime}}}{I t_{B^{\prime}}}=\frac{80\left(10^{3}\right) \mathrm{N}\left(0.660\left(10^{-3}\right) \mathrm{m}^{3}\right)}{155.6\left(10^{-6}\right) \mathrm{m}^{4}(0.300 \mathrm{~m})}=1.13  \mathrm{MPa}

For point B, t_{B}=0.015 \mathrm{~m} and Q_{B}=Q_{B^{\prime}}, Fig. 12-12c. Hence

\tau_{B}=\frac{V Q_{B}}{I t_{B}}=\frac{80\left(10^{3}\right) \mathrm{N}\left(0.660\left(10^{-3}\right) \mathrm{m}^{3}\right)}{155.6\left(10^{-6}\right) \mathrm{m}^{4}(0.015 \mathrm{~m})}=22.6  \mathrm{MPa}

Note from our discussion of the limitations on the use of the shear formula that the calculated values for both \tau_{B^{\prime}} and \tau_{B} are actually very misleading. Why?

For point C, t_{C}=0.015 \mathrm{~m} and A^{\prime} is the dark shaded area shown in Fig. 12-12d. Considering this area to be composed of two rectangles, we have

\begin{aligned} Q_{C}=\Sigma \bar{y}^{\prime} A^{\prime}=[0.110 \mathrm{~m}](0.300 \mathrm{~m})(0.02 \mathrm{~m}) \\ \quad+[0.05 \mathrm{~m}](0.015 \mathrm{~m})(0.100 \mathrm{~m}) \\ =0.735\left(10^{-3}\right) \mathrm{m}^{3} \end{aligned}

Thus,

\tau_{C}=\tau_{\max }=\frac{V Q_{C}}{I t_{C}}=\frac{80\left(10^{3}\right) \mathrm{N}\left[0.735\left(10^{-3}\right) \mathrm{m}^{3}\right]}{155.6\left(10^{-6}\right) \mathrm{m}^{4}(0.015 \mathrm{~m})}=25.2  \mathrm{MPa}

NOTE: From Fig. 12-12b, the largest shear stress occurs in the web and is almost uniform throughout its depth, varying from 22.6 \ \mathrm{MPa} to 25.2 \ \mathrm{MPa}. It is for this reason that for design, some codes permit the use of calculating the average shear stress on the cross section of the web, rather than using the shear formula; that is,

\tau_{\text {avg }}=\frac{V}{A_{w}}=\frac{80\left(10^{3}\right) \mathrm{N}}{(0.015 \mathrm{~m})(0.2 \mathrm{~m})}=26.7  \mathrm{MPa}

This will be discussed further in Chapter 15 .