Question 7.2: The tee section shown in Figure 7–12 is from a simply suppor...
The tee section shown in Figure 7–12 is from a simply supported beam that carries a bending moment of 11.3 kN · m due to a load on the top surface. It has been determined that I = 7.10 × 10^{6} mm^{4} . The centroid of the section is 81.25 mm up from the bottom of the beam. Compute the stress due to bending in the beam at the six axes a to f in Figure 7–13. Then, plot a graph of stress versus position in the cross section.
Objective Compute the bending stress at six axes a–f. Plot a graph of stress versus position in the cross section.
Given M = 11 300 N · m; tee shape of cross section shown in Figure 7–12; I = 7.10 × 10^{6} mm^{4} ; \bar{Y} = 81 25 .mm from bottom of beam.
Analysis Equation (7–1) will be used to compute \sigma_{max} , which occurs at the bottom of the beam (axis a), because that is the location of the outermost fiber of the beam, farthest from the centroidal axis. The stress, \sigma_{a} , is tension because that section is below the neutral axis. Then, the stress at other axes will be computed using Equation (7–2), giving results to four significant figures to demonstrate the principle. See Figure 7–13 for values of y.
Results Axis a: In Equation (7–1), use c = \bar{Y} =81 25 .mm
\sigma_{max} = \sigma_{a} = \frac{Mc}{I} = \frac{(11 300 N·m)(81.25 mm)(1000 mm/1 m)}{7.10 × 10^{6} mm^{4} }
\sigma_{a} = 129.3 MPa (tension)
Axis b: Note that this axis is located by being 25 mm above the base and also as y_{b} —the distance from the neutral axis down to axis b.
\sigma_{b} = \sigma_{max} \frac{y_{b}}{c} = \sigma_{a} \frac{y_{b}}{c}
y_{b} = 81.25 mm – 25.0 mm = 56.25 mm
\sigma_{b} = 129.3 MPa \times \left\lgroup \frac{56.25 mm}{81.25 mm}\right\rgroup = 89.5 MPa
Axis c:
y_{c} = 81.25 mm – 50.0 mm = 31.25 mm
\sigma_{c} = 129.3 MPa \times \left\lgroup \frac{31.25 mm}{81.25 mm}\right\rgroup = 49.7 MPa
Axis d: At the centroid (neutral axis), y_{d} = 0 and σ_{d} = 0.
Axis e: This is coincident with the bottom of the flange of the tee shape and in the part above the neutral axis where the stress is compressive.
y_{f} = 100.0 mm – 81.25 mm = 18.75 mm
\sigma_{f} = 129.3 MPa \times \left\lgroup \frac{18.75 mm}{81.25 mm}\right\rgroup = 29.8 MPa (compression)
Axis f:
y_{f} = 125.0 mm – 81.25 mm = 43.75 mm
\sigma_{f} = 129.3 MPa \times \left\lgroup \frac{43.75 mm}{81.25 mm}\right\rgroup = 69.6 MPa
The graph of these data is shown in Figure 7–14.
Comment Notice the linear variation of stress with distance from the neutral axis and that the stresses above the neutral axis are compressive while those below are tensile.
