Question 7.4: Refer back to Figure 5–33, which shows a beam carrying a uni...
Refer back to Figure 5–33, which shows a beam carrying a uniformly distributed load over part of its length. Note that the maximum bending moment is 2139 N · m at point D, 1.308 m to the left of support C. It has been proposed that the beam be made from a standard metric rectangular tube, 20×30×3 with the 30 mm dimension vertical. Compute the maximum stress due to bending in the beam.
Objective Compute the maximum stress due to bending.
Given Maximum bending moment = 2139 N · m
Beam is a metric rectangular tube, 20×30×3 with the 30 mm dimension vertical.
Analysis Use Equation (7–1) and determine the section properties for the tube from Appendix A–8(c).
Results I_{x} = 2.887 \times 10^{4} mm^{4}.
c = 30 mm/2 = 15 mm
\sigma_{max} = \frac{Mc}{I} = 2139 N·m \times \frac{1000 mm}{m} \frac{15 mm}{2.887 \times 10^{4} mm^{4}} = 1.111 × 10³ N/mm² = 1111 MPa
Comment This maximum stress would occur as a tensile stress on the bottom surface of the beam and as a compressive stress on the top surface at the position D in Figure 5–33. Note that the bending moment is positive, above the axis, causing the beam to bend in the concave upward shape. Note also the advantage of orienting the tube with the 30 mm dimension vertical. If the tube were laid with the 20 mm side vertical, bending would occur with respect to the y-axis of the shape shown in Appendix A–8(c). Then,
I_{y} = 1.451 \times 10^{4} and c=20 mm/2 = 10 mm
\sigma_{max} \frac{Mc}{I} = 2139 N·m \times \frac{1000 mm}{m} \frac{10 mm}{1.451 \times 10^{4} mm^{4}} = 1.474 × 10³ N/mm² = 1474 MPa
This stress is about 32% higher than for the preferred orientation.