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# hw 11 - 8-6 The motor shown in Fig P8-6 supplies a torque...

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Unformatted text preview: 8-6 The motor shown in Fig. P8-6 supplies a torque of 500 N'm to shaft BCDE. The torques removed at gears C, D, and E are 100 N°m, 150 N'm, and 250 N-m. respectively. ’ (a) Determine the torques ' transmitted by transverSe cross sections in intervals BC. CD, and DE of the shaft. (b) Draw a torque diagram for the shaft. SOLUTION A load diagram for the shaft, free-body diagrams for parts of the shaft to the left of sections in intervals BC, CD, and DE of the shaft, and a torque diagram for the shaft are shown below. From the free—body diagrams: +C2Mx=TBC~V =0 +C2Mx Ten: 100:0 +C>ZMx TDE- 100+150=0 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-15* Determine the internal resisting forces and moment transmitted by section aa in the bracket shown SOLUTION From a free-body diagram for the part of the bracket to the right of section aa: + —9 ZFX = 300 - P = 0 P = 300 lb = 300 lb +- + T 2Fy = V + 500 = 0 V = -500 lb = 500 lb I + C.ZMZ = M + 500(12) - 300(8) = 0 M = -3600 in-lb = 3600 in.-lb C 8—36* A beam is loaded and supported as shown in Fig. P8-36. Using the coordinate axes shown, write equations for the shear V and bending moment M for any section of the beam in the interval 0 < x < 6 m. SOLUTION 523%:323332‘5: 3:23?” ' !!!!!!!!!!!!!!!! + C 2M3 = 18 - A(8) + 5(6)(5) = o = 21 kN = 21 kN T interval 0 S x S 6: = 21 - 5x = -5x + 21 RN -13 + 21x - 5(x)(x/2) = -2.5x2 + 21x — 18 kN-m '75?) ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-40 A beam is loaded and supported as shown in Fig P8-40. Using the coordinate axes shown, write equations for the shear V and bending moment M for any section of the beam in the interval 0 < x < 4 m. SOLUTION From a free-body diagram for the complete beam: + C 2MB : -Ay(8) + + 40(2) = 0 100 kN = 100 kN T interval 0 < x < 4 m: ' = 100 - 30x = ~30x + 100 kN 100x - 30(x)(x/2) = -15x2 + 100x kN-m ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8—55* Draw complete shear and moment diagrams for the beam shown in Fig. P8-55. SOLUTION From a free—body diagram for the complete beam: + C EMA = B(23) - 2500(8) - 800(10)(18) = 0 B = 7130 lb + G 2:48 -A(23) + 2500(15) + 300(10H5) = o A = 3370 lb Load, shear, and moment diagrams for the beam are shown below: 2500 N, 800 “out ...
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