Question 8.7: For the triangular beam cross section shown in Figure 8–13, ...
For the triangular beam cross section shown in Figure 8–13, compute the shearing stress that occurs at the axes a through g, each 50 mm apart. Plot the variation of stress with posi-tion on the section. The shearing force is 50 kN.
Objective Compute the shearing stress at seven axes and plot τ versus position.
Given Cross-section shape and dimensions in Figure 8–13. V = 50 kN.
Analysis Use the Guidelines for Computing Shearing Stresses in Beams.
Results In the general shear formula, the values of V and I will be the same for all computations. V is given to be 50 kN and
I \frac{bh^{3}}{36} = \frac{(300)(300)³}{36} = 225 \times 10^{6} mm^{4}
Table 8–1 shows the remaining computations. Obviously, the value for Q for axes a–a and g–g is zero because the area outside each axis is zero. Note that because of the unique shape of the given triangle, the thickness t at any axis is equal to the height of the triangle above the axis.
| TABLE 8–1 Calculations for shearing stresses. | |||||
| Axis | A_{p} (mm^{2}) | \bar{y} (mm) | Q = A_{p} \bar(y) (mm^{3}) | t (mm) | τ(Mpa) |
| a-a | 0 | 100 | 0 | 300 | 0 |
| b-b | 13 | 75.8 | 1.042 \times 10^{6} | 250 | 0.92 |
| c-c | 20 | 66.7 | 1.333 \times 10^{6} | 200 | 1.48 |
| d-d | 11 | 100 | 1.125 \times 10^{6} | 150 | 1.67 |
| e-e | 5000 | 133.3 | 0.667 \times 10^{6} | 100 | 148 |
| f-f | 1250 | 1667 | 0.208 \times 10^{6} | 50 | 0.92 |
| g-g | 0 | 200 | 0 | 0 | 0 |
Figure 8–14 shows a plot of these stresses. The maximum shearing stress occurs at half the height of the section, and the stress at the centroid (at h/3) is lower. This illustrates the general statement made earlier that for sections whose minimum thickness does not occur at the centroidal axis, the maximum shearing stress may occur at some axis other than the centroidal axis.
Comment One further note can be made about the computations shown for the triangular section. For the axis b–b, the partial area A_{p} was taken as that area below b–b. The resulting section is the trapezoid between b–b and the bottom of the beam. For all other axes, the partial area A_{p} was taken as the triangular area above the axis. The area below the axis could have been used, but the computations would have been more difficult. When computing Q, it does not matter whether the area above or below the axis of interest is used for computing A_{p} and \bar{y} .
