Question 8.6: Compute the distribution of shearing stress with position in...
Compute the distribution of shearing stress with position in the cross section for a beam with the rectangular shape shown in Figure 8–7. The actual dimensions are 50 mm by 250 mm. Plot the results. The shearing force, V, on the section of interest is 5400 N.
Objective Compute the shearing stress at several axes and plot τ versus position.
Given Cross-section shape and dimensions in Figure 8–7. V = 5400 N.
Analysis Use the Guidelines for Computing Shearing Stresses in Beams. Because the shape is symmetrical with respect to the centroidal axis, we choose to compute the shearing stresses in the upper part at the axes a–a, b–b, c–c, and d–d, as shown in Figure 8–11. Then, the values
of stresses in the lower part at sections b′–b′, c′–c′, and d′–d′ will be the same as the corresponding points computed earlier.
Results Step 1. V = 5400 N (given).
Step 2. For the rectangular shape, the centroid is at the midheight, as shown in Figure 8–7, coincident with axis a–a. \bar{Y} =125 mm.
Step 3. I = bh^{3} /12 = (50 mm)(250 mm)³/12 = 6.510 × 10^{7} mm^{4} .
Step 4. Thickness = t = 50 mm at all axes.
Step 5. We will compute Q = A_{p} y for each axis using the method shown earlier in this chapter. Recall that the value of Q for this section at the centroidal axis was computed in Example Problem 8–1. where we found Q = 390 625 mm³. A similar calculation is summarized in the table following step 6, using data from Figure 8–11.
Step 6. Using Equation (8–1), the calculation for shearing stress at the neutral axis a–a is shown here:
\tau = \frac{VQ}{It} = \frac{(5400 N)(390 625 mm^{3})}{(6.51 \times 10^{7} mm^{4})(50 mm)} = 0.648 N/mm² = 648 kPa
The calculation would be the same at the other axes with only the value of Q changing. See the following table.
| Axis | V | I | t | A_{p} | y | q=A_{p}y | \tau = VQ/It (kPa) |
| a-a | 5400 | 6.51 \times 10^{7} | 50 | 6250 | 63 | 390 625 | 648 |
| b-b | 5400 | 6.51 \times 10^{7} | 50 | 5000 | 75 | 375 000 | 622 |
| c-c | 5400 | 6.51 \times 10^{7} | 50 | 2500 | 100 | 250 000 | 415 |
| d-d | 5400 | 6.51 \times 10^{7} | 50 | 0 | 125 | 0 | 0 |
The results of shearing stress versus position are shown in Figure 8–12 alongside the rectangular section itself.
Comment Note that the maximum shearing stress does occur at the neutral axis as predicted. The variation of shearing stress with position is parabolic, ending with zero stress at the top and bottom surfaces.
