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hw14_sol

# hw14_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES ' 8-12* A bar is loaded and supported as shown in Fig. P8—12. (a) Determine the maximum axial load transmitted by any transverse cross section of the bar. Draw an axial force diagram for the bar. SOLUTION For overall equilibrium of the bar: + T ZFY : R - 2(40) + 2(50) - 2(10) — 10 = 0 R = 10 kN = 10 kN T Fig. P8-12 A load diagram for the bar. free—body diagrams for parts of the bar above sections in intervals AB, BC, and CD of the bar. and an axial force diagram for the bar are shown below. From the free-body diagrams: = --FAB — 10 - 2(10) = 0 FAB = -30 kN i 30 kN (C) —FBC - 10 - 2(10) 2(50) = 0 FBC = 70 kN = :9 kN (Tl -FCD + 10 + 2(10) 2(50) + 2(40) = CD = -10 kN = 10 kN (C) _ _ “*2 BC _ 70 kN - 70 RN (T) _ Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Five GOO-mm diameter pulleys are keyed to a steel shaft as shown in Fig. P8—14. The pulleys carry belts that are used to drive machinery in a factory. Belt tensions for normal operating conditions are indicated on the figure. Determine the maximum torque transmitted by any transverse cross section of the shaft. 1‘ SOLUTION ll A load diagram for the shaft, free-body diagrams for parts of the shaft to the left of sections in intervals AB, BC, CD, and DE of the shaft, and a torque diagram for the shaft are shown below. LooN-m. 720 N-m ll 1 - From the free-body diagrams: l + C ZMX = TAB - 600 = 0 L + C zMx TBC - 600 + 480 = 0 = 120 N-m 120 N'm -C— l ‘ + C XMX TCD — 600 + 480 — 720 = 0 = 840 N-m 840 N;m -C— - 600 + 480 - 720 + 720 = 0 ' 120 N'm = 120 N'm -C- DE TCD = 840 N-m = 340 N'm -C— I Ans. ESGINEERING 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 in Fig. PH-15. SOLUTION From a free—body diagram for the part of the bracket to the right of section aa: 2F = 300 - P = 0 X P = 300 lb = 300 lb +- = V + 500 = 0 -500 lb 2 500 lb ¢ M + 500{12) - 300(8) = 0 M —3600 in-lb = 3500 in.°lb C ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-17 Determine the internal resisting forces and moment transmitted by section aa in the curved bar shown in Fig. P8-17. SOLUTION From a free-body diagram for the part of the curved bar to the right of section aa: + x7 ZFX, —P - 750 sin 30° = 0 P -375 lb = 375 1b /” v — 750 cos 30° 0 649.5 lb a 650 lb “\ Ans. M — 750 cos 30° (30 cos 30°) - 750 sin 30° (30 - 30 sin 30°) 22,500 in.-lb = 22.5 in.'kip 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 From a free-body diagram for the complete beam: I! Q + C 2MB = 18 - A{8} + 5(6){5} A = 21 kN = 21 kN T - S 6: If\ y interval 0 V a 21 - 5x = —5x + 21 RN —18 + 21x — 5{x1{x/2i z ~2.5x2 + 21x — 18 kN'm '73) ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-44 A beam is loaded and supported as shown in Fig P8-44. 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 = 9 - AllO} + 18(6){9} + 36(41 = 0 112.5 kN = 112.5 kN T For the interval 0 < x < 4 m: V = 112.5 - 18(x + 2} = -18x + ?6.5 kN M -9 + 112.5x - 18(x + lex + 2)/2 —9x2 + ?6.5x — 45 kN-m ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-56* Draw complete shear and moment diagrams for the beam shown in Fig. P8-56. SOLUTION From a free-body diagram for the complete beam: + C EMA = 8(8) — 30(4)(2) — 40(6) = 0 B = 60 kN + c 2MB = -A(8) + 30(4)(6) + 40(2) = A = 100 kN Load, shear, and moment diagrams for the beam are shown below: 40 RM 30 Km»... ...
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