## Q. 7.6

The beam shown is subjected to a distributed load. For the cross section at x = 0.6 m, determine the average shear stress (a) at the neutral axis and (b) at z = 0.02 m.

Find: Shear stress at two locations along height of cross section at x = 0.6 m.
Assume: Hooke’s law applies. ## Verified Solution

First we need to consult our FBD and find the reactions at the supports.

\begin{aligned} \sum F_y=0 & \rightarrow R_A+R_B=91 \mathrm{~kN}, \\ \sum M_B & =0=R_A(0.8 \mathrm{~m})+91 \mathrm{~kN}(0.133 \mathrm{~m}), \\ & R_A=-15.1 \mathrm{~kN} \text { (downward) } ,\\ & R_B=106.1 \mathrm{~kN} \text { (upward) }. \end{aligned}

We are interested in the cross section at x = 0.6 m. We know that the average shear stress depends on the internal shear force in the beam at the point of interest, so we need to calculate the shear V(x = 0.6 m). To do this, we will make an imaginary cut at x = 0.6 m:

In this 0.6-m-long span, the distributed load has a maximum intensity of $\left(130 \frac{\mathrm{kN}}{\mathrm{m}}\right) \frac{0.6 \mathrm{~m}}{1.4 \mathrm{~m}}=55.7 \frac{\mathrm{kN}}{\mathrm{m}}$, so the equivalent concentrated load acting on the 0.6-m-long segment is the area under this load: $\frac{1}{2}\left(55.7 \frac{\mathrm{kN}}{\mathrm{m}}\right)(0.6 \mathrm{~m})=16.7 \mathrm{~kN}$.
Equilibrium of our 0.6 m segment:

$\begin{gathered} \sum F_y=0=V-R_A-16.7 \mathrm{~kN} ,\\ \rightarrow \quad V=16.7 \mathrm{~kN}+15.1 \mathrm{~kN}=31.8 \mathrm{~kN}. \end{gathered}$

The second moment of area for the cross section is

$I=\frac{1}{12} b h^3=\frac{1}{12}(0.04)(0.06)^3=7.2 \times 10^{-7} \mathrm{~m}^4$.

At the height of the centroid or neutral axis, $Q^{\prime}$ is the first moment of area of the area above (or below) the centroid:

$Q^{\prime}=z^{\prime} A^{\prime}=0.015(0.03 \times 0.04)=1.8 \times 10^{-5} \mathrm{~m}^3$.

Then the shear stress at the centroid is

$\sigma_{x z}=\frac{V Q^{\prime}}{I b}=\frac{(31.8 \mathrm{~kN})\left(1.8 \times 10^{-5} \mathrm{~m}^3\right)}{\left(7.2 \times 10^{-7} \mathrm{~m}^4\right)(0.04 \mathrm{~m})}=19.9 \mathrm{~MPa}$.

At z = 0.02 m above the neutral axis

$Q^{\prime}=z^{\prime} A^{\prime}=0.02(0.02 \times 0.04)=1.0 \times 10^{-5} \mathrm{~m}^3$.

This is the only change from the calculation at z = 0, and the shear stress at this point is 11.0 MPa.  