b A beam is made of square tubing, with 2 in on a side, and 0.15 in wall thickness. It is being bent about the a-a axis

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B A Beam Is Made Of Square Tubing With 2 In On A Side And 0 15 In Wall Thickness It Is Being Bent About The A A Axis 1 (1.83 MiB) Viewed 29 times

b A beam is made of square tubing, with 2 in on a side, and 0.15 in wall thickness. It is being bent about the a-a axis shown. Find the value of k, using Equation 6-19. (Hint: this shape is not found in Table 6-3, but you can use the same A0.95 idea described in the text. Use this area, and Eq. 6-21.) to- 12 A. 0.84 B. 0.97 C. 1.2 D. 0.76 E. None of the above |1 This Table 6-3 A0.950 Areas of Common Nonrotating Structural Shapes Loaded in Bending 1 H |1 k₂ =< Opet s A zbriammanos y ovu ballon-jod on A -0.107 (d/0.3)~ = 0.879d-0.107 0.91d-0.157 (d/7.62)-0.107 = 1.24d-0.107 1.51d-0.157 min For d less than 0.3 inches (7.62 mm), the data is quite scattered. Unless more specific data is available to warrant a higher value, a value of k = 1 is recommended. For axial loading there is no size effect, so kb = 1 A0.950 = 0.01046d² de = 0.370d A0.950 = 0.05hb -8797 but see ke. Equation (6-19) applies to round rotating bars in bending and torsion, in which the highly stressed volume is around the outer circumference. If a round bar is not rotating, the highly stressed volume is the same for torsion (all the way around the circumference), but is much less for bending (e.g., just on opposite sides of the cross section). Kuguel introduced a critical volume theory in which the volume of material experiencing a stress above 95 percent of the maximum stress is considered to be critical.10 The method employs an equivalent diameter de obtained by equating the volume of material stressed at and above 95 percent of the maximum stress to the same volume in the rotating-beam specimen. When these two volumes are equated, the lengths cancel, and so we need only consider the areas. For a rotating round sec- tion, the 95 percent stress area is the area in a ring having an outside diameter d and an inside diameter of 0.95d. So, designating the 95 percent stress area A0.950, we have A0.95a = [d² - (0.95d)²] = 0.0766d² de = 0.808 √hb A0.950 = (0.10atf 0.05ba [rous room A0.950 = 0.3 < d < 2 in 2 < d ≤ 10 in viso 7.62 ≤ d ≤ 51 mm 51 < d < 254 mm tf > 0.025a (0.05ab 0.052xa + 0.1t, (b - x) d axis 1-1011 axis 2-2 (6-19) axis 1-1 axis 2-2 (6-20) (6-21)