Question 8.5: Compute the shearing stress at the axes a–a and b–b for a be...
Compute the shearing stress at the axes a–a and b–b for a beam with the T-shaped cross section shown in Figure 8–9. Axis a–a is at the very top of the vertical web, just below the flange. Axis b–b is at the very bottom of the flange. The shearing force, V, on the section of interest is 5400 N.
Objective Compute the shearing stress at the axes a–a and b–b.
Given Cross section shape and dimensions in Figure 8–9. V = 5400 N.
Analysis Use the Guidelines for Computing Shearing Stresses in Beams.
Results For the axis a–a:
Step 1. V = 5400 N (given).
Step 2. This particular T-shape was analyzed in Example Problem 8–3. Use \bar{Y} =171mm.
Step 3. We will use the methods of Chapter 6 to compute I. Let the web be part 1 and the flange be part 2. For each part, I = bh^{3} /12 and d = \bar{Y} – \bar{y} .
| Part | I | A | d | Ad^{2} | I + Ad^{2} |
| 1 | 2.53 \times 10^{7} | 7600 | 71 | 3.83 \times 10^{7} | 6.36 \times 10^{7} |
| 2 | 2.08 \times 10^{7} | 10 000 | 54 | 2.92 \times 10^{7} | 3.12 \times 10^{7} |
| Total I = 9.484 \times 10^{7} mm^{4} | |||||
Step 4. Thickness = t = 38 mm at axis a–a in the web.
Step 5. Normally we would compute Q = A_{p} \bar{y} using the method shown earlier in this chap-ter. But the value of Q for the section in Figure 8–9 was computed in Example Problem 8–3. Use Q = 540 000 mm³
Step 6. Using Equation (8–1),
\tau = \frac{VQ}{It} = \frac{(5400 N)(540 000 mm^{3})}{(9.484 \times 10^{7} mm^{4})(38 mm)} = 0.809 N/mm² = 0.809 MPa = 809 kPa
For the axis b–b: Some of the data will be the same as at a–a.
Step 1. V = 5400 N (given).
Step 2. Again, use \bar{Y} =171 mm
Step 3. I = 9.484 × 10^{7} mm^{4}
Step 4. Thickness = t = 200 mm at axis b–b in the flange.
Step 5. Again, use Q = 540 000 mm³. The value is the same as at axis a–a because both A_{p} and \bar{y} are the same.
Step 6. Using Equation (8–1),
\tau = \frac{VQ}{It} = \frac{(5400 N)(540 000 mm^{3})}{(9.484 \times 10^{7} mm^{4})(200 mm)} = 0.154 N/mm² = 0.154 MPa = 154 kPa
Comment Note the dramatic reduction in the value of the shearing stress when moving from the web to the flange. The larger value of t in the denominator lowered the shearing stress significantly.