Support Reactions: a + M C = 0; 800(10) F DE(4) F DE(2) = 0. F DE = 2000 lb. + c F y = 0; (2000) - C y = 0 C y = 400 lb
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1 06 Solutions 46060_Part1 5/27/10 3:51 P Page The overhanging beam has been fabricated with a projected arm D on it. Draw the shear and moment diagrams for the beam C if it supports a load of 800 lb. Hint: The loading in the supporting strut DE must be replaced by equivalent loads at point on the axis of the beam. 800 lb D E 5 ft Support Reactions: 2 ft C a + C = 0; 800(10) F DE(4) F DE(2) = 0 6 ft 4 ft F DE = 2000 lb + c F y = 0; (2000) - C y = 0 C y = 400 lb : + F x = 0; -C x (2000) = 0 C x = 1600 lb Shear and oment Diagram: *6 12. reinforced concrete pier is used to support the stringers for a bridge deck. Draw the shear and moment diagrams for the pier when it is subjected to the stringer loads shown. ssume the columns at and exert only vertical reactions on the pier. 60 kn 35 kn 35 kn 35 kn 60 kn 1 m 1 m 1.5 m 1.5 m 1 m 1 m 334
2 06 Solutions 46060_Part1 5/27/10 3:51 P Page The shaft is supported by a smooth thrust bearing at and smooth journal bearing at D. f the shaft has the cross section shown, determine the absolute maximum bending stress in the shaft. C D 40 mm 25 mm 0.75 m 1.5 m 0.75 m Shear and oment Diagrams: s shown in Fig. a. 3 kn 3 kn aximum oment: Due to symmetry, the maximum moment occurs in region C of the shaft. Referring to the free-body diagram of the segment shown in Fig. b. Section Properties: The moment of inertia of the cross section about the neutral axis is = p = m 4 bsolute aximum ending Stress: s allow = maxc = (0.04) = 52.8 Pa ns. *6 76. Determine the moment that must be applied to the beam in order to create a maximum stress of 80 Pa.lso sketch the stress distribution acting over the cross section. 20 mm 300 mm The moment of inertia of the cross-section about the neutral axis is Thus, = 1 12 (0.3)(0.33 ) (0.21)(0.263 ) = (10-3 ) m 4 s max = c ; 80(106 ) = (0.15) (10-3 ) 260 mm 20 mm = (10 3 ) N # m = 196 kn # m ns. The bending stress distribution over the cross-section is shown in Fig. a. 373
3 06 Solutions 46060_Part2 5/26/10 1:17 P Page 395 * f w = 10 kn>m, determine the maximum bending stress in the beam. Sketch the stress distribution acting over the cross section. Support Reactions. The FD of the beam is shown in Fig. a The shear and moment diagrams are shown in Figs. b and c,respectively.s indicated on the moment diagram, max = 1.25 kn # m. The moment of inertia of the cross-section is 0.5 m 1 m 0.5 m 75 mm w 150 mm = 1 12 (0.075)0.153 = m 4 Here, c = m. Thus s max = max c = (0.075) = Pa = 4.44 Pa ns. The bending stress distribution over the cross section is shown in Fig. d 395
4 06 Solutions 46060_Part2 5/26/10 1:17 P Page f the beam is subjected to an internal moment of = 45 kn # m,determine the maximum bending stress developed in the -36 steel section and the 2014-T6 aluminum alloy section. 50 mm 150 mm Section Properties: The cross section will be transformed into that of steel as shown in Fig. a. Here, n = E al = = Thus, b st = nb al = (0.015) = m. The E st location of the transformed section is y = y 0.075(0.15)( ) + 0.2cp d = 0.15( ) + p = m The moment of inertia of the transformed section about the neutral axis is = + d 2 = 1 12 ( ) (0.15)( ) 2 4 p p ( ) 2 = m 4 aximum ending Stress: For the steel, (s max ) st = c st = ( ) = 154 Pa ns. For the aluminum alloy, (s max ) al = n c al = C (0.1882) S = 171 Pa ns. 424
5 06 Solutions 46060_Part2 5/26/10 1:17 P Page 427 * The low strength concrete floor slab is integrated with a wide-flange -36 steel beam using shear studs (not shown) to form the composite beam. f the allowable bending stress for the concrete is (s allow ) con = 10 Pa, and allowable bending stress for steel is (s allow ) st = 165 Pa, determine the maximum allowable internal moment that can be applied to the beam. 1 m 100 mm 400 mm Section Properties: The beam cross section will be transformed into that of steel. Here, n = E con = Thus, E st 200 = b st = nb con = (1) = m.the location of the transformed section is 200 mm y = y = (0.015)(0.2) + 0.2(0.37)(0.015) (0.015)(0.2) (0.1)(0.1105) 0.015(0.2) (0.015) (0.2) + 0.1(0.1105) = m The moment of inertia of the transformed section about the neutral axis is = + d 2 = 1 12 (0.2) ending Stress: ssuming failure of steel, (s allow ) st = c st = m 4 ssuming failure of concrete, + 0.2(0.015)( ) 2 12 (0.015) (0.37)( ) 2 12 (0.2) (0.015)( ) 2 12 (0.1105) (0.1)( ) 2 ; = (0.3222) = N # m = 332 kn # m (s allow ) con = n c con ; ( ) = C S = N # m = 330 kn # m (controls) ns. 427
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