MEEN 221- Chapter8

MEEN 221- Chapter8 - ENGINEERING MECHANICS - STATICS. 2nd....

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Unformatted text preview: ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-1* A steel bar with a rectangular cross section is used to transmit four axial loads as shown in Fig. P8—1. (a) Determine the axial forces transmitted by cross sections in intervals AB, BC, and CD of the bar. (b) Draw an axial force diagram for the bar. SOLUTION For overall equilibrium of the bar: +-—+ EFX = -35 + P - 20 + 18 = 0 P = 37 kip = 37 kip —% A ioad diagram for the bar, free-body diagrams for parts of the bar to the left of sections in intervals AB, BC, and CD of the bar, and an axial force diagram for the bar are shown below. 35 37 2° )8 H ao-v <—-°c u Lea—>5 1-33 3: 37 'Illl'l'l'll'. iiicsf From the free-body diagrams: +-—+ EFX = FAB — 35 = 0 FAB = 35 kip = 35 kip (T) Ans. +-—+ EFX = FBC - 35 + 37 ='0 FBC = -2 kip = 2 kip (C) Ans. +-—+ ZFX = FCD - 35 + 37 - 20 = 0 FCD = 18 kip = 18 kip (T) Ans. ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 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 5 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. 5°C! From the free-body diagrams: +C2Mx TBC — 500:0- +(j53Mx '_ +100=0 +C2Mx 100+150=0 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-10 For the steel shaft shown in Fig. P8-10, (a) Determine the maximum torque transmitted by any transverse cross section of the shaft. (b) Draw a torque diagram for the shaft. I5 kN-l'n 6 kN-m SOLUTION For overall equilibrium of the shaft: +CXM‘=T-8-15-6+14=0 T=15kN-m=15kN-m 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. 8 IS I. E-_[-(GH B . C The 'IlIII/II/l/l/l/l/lfl uuuvzzngnn From the free-body diagrams: + C XMX = TAB - 8 kN'm = 8 kN-m —C— + C XMX TBC - -7 kN~m = 7 kN-m -Q— + C 2Mx TCD 8 kN'm = 8 kN-m —C— + C >2Mx TDE - i4 kN-m = 14 kN-m —C— Ans. 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. (b) Draw an axial force diagram for the bar. SOLUTION For overall equilibrium of the bar: + T 2Fy = R - 2(40) + 2(50) .U - 2(10) - 10 = 0 fig.P842 R = 10 RN = 10 RN T 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: + T 2F = -FAB - l0 - 2(10) = 0 FAB = —30 kN = 30 kN (C) + T 2F -FBC - 10 - 2(10) 2(50) = 0 FBC = 70 kN = 70 kN (T) + T [F -FCD + 10 + 2(10) - 2(50) + 2(40) = CD = -10 kN = 10 kN (C) BC = 70 kN = 70 kN (T) Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Five BOO-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. Fig. P844 SOLUTION A load diagram for the shaft, free-body diagrams for parts of the shaft toathe left of sections in intervals AB, BC, CD, and DE of the shaft, and a torque diagram for the Shaft are shown below. (Ocali-nL '7ZC>)i-nt From the free-body diagrams: + C 251x = TAB — 600 = o - = 600 um —C- + C 2Mx TBC - 600 + 480 = o = 120 N'm 120 N-m -C- TCD - 600 + 480 - 720 = o 840 Nm 840 N-m -C— - 600 + 480 — 720 + 720 = o - 120 N-m = 120 N-m -C— = 840 Mm = 840 N-m -C— Ans. 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-l7. SOLUTION From a free-body diagram for the part of the curved bar to the right of section aa: II D -P - 750 sin 30° + /’ 2F , X ,-375 lb = 375 lb 1’ v — 750 cos 30° = o 649.5 lb 3 650 lb “\ Ans. M - 750 cos 30° (30 cos 30°) - 750 sin 30° (30 - 30 sin 30°) = 0 22,500 in.-lb = 22.5 in.-kip C 707 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-20* Determine the internal resisting forces and moment transmitted by section aa in bar ABC of the three-bar frame P8—20. shown in Fig. IOOmm IOOmm IOOmm 100mm Fig. P8-20 SOLUTION Link BD is a two-force member: therefore, the line of action of force 3 is known and the free—body diagram for member ABC can be drawn as shown. + C EMA = 3(200) - 3(400) = 0 )oo B = 6.00 kN = 6.00 kN T From a free—body diagram for the part of member ABC to the right of section aa: 30.96° + v/ 2F = p - 6 sin 30.960 + 3 sin 30.96° = 0 1.5433 kN a 1.543 kN v/ v - 6 cos 30.96° + 3 cos 30.960 = o v = 2.573 kN z 2.57 kN ‘3 Ans. + C 2M2 = M + 6(0.100) - 3(0.300) = 0 M = 0.300 kN'm = 0.300 kN'm C ENGINEERING MECHANICS - STATICS, 2nd. 8-30 A three-bar frame is loaded and supported as shown in Fig. P8-30. Determine the internal resisting forces and moment transmitted by (a) Section aa in bar BD. (b) Section bb in bar ABC. SOLUTION .a From a free-body diagram for the complete frame: E(960) - 900(1140) = 0 +CEMA= E 1068.8 N = 1068.8 N T + —% SE A = 0 A = 0 X X X + T SE = A y y y = -168.8 N = 168.8 N ¢ + 1068.2 - 900 = 0 From a free—body diagram for bar BD: + C EMD = -By(600) - 900(360) = 0 B = -540 N = 540 N ¢ Y From a free-body diagram for bar ABC: + C EM = -B (500) — 540(300) C x + 168.8(480) = 0 . = -162.0 N = 162.0 N -+ X Ed. W. F. 722 RILEY AND L. D. STURGES ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 8—30 (Continued) (a) From a free-body diagram for the part of bar BD to the left of section aa: + -4 BF“ = P - 162.0 = 0 P = 162.0 N = 162.0 N -9 V - 540 = 0 = 540 N = 540 N T ’ + C ZM’ = M + 540(0.300) = o M = -162.0 N-m = 162.0 N'm D (b) From a free-body diagram for the part of bar ABC below section bb: - -1M_ 0 ¢ - tan 300 - 59.04 0 + /’ 2F“ = p + 540 sin 59.04 + 162.0 cos 59.04° - 168.8 sin 59.04° = o = ~401.6 N = 402 N v/ v + 540 cos 59.o4° - 162.0 sin 59.04° — 168.8 cos 59.04° = o v = -52.0 N = 52.0 N \N C + 2M: = M + 162.0(0.250) — 540(o.150) + 163.3(o.330) = o = —15.20 N-m = 15.20 N-m D Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-32* A three-bar frame is loaded and supported as shown in Fig. P8-32. Determine the internal resisting forces and moment transmitted by (a) Section an in bar CGE. (b) Section bb in bar DEF. SOLUTION From a free-body diagram of the pulley: '+——92F X From free-body diagrams of members DEF and CGE: D quanta. E Anon-n. F Ey(400) - 500(800) = 0 500(500) - Ex(600) - 1000(400) = 0 726 ENGINEERING MECHANICS - STATICS. 2nd. Ed. H. F. RILEY AND L. D. STURGES 8—32 (Continued) (a) From a free-body diagram for the part of bar CGE below section 38: aw- _ 0 200 — 06.31 ¢ = tan = p - 500 cos 56.31° 250 cos 56.31° 1000 sin 56.31° = 0 1248 N = 1248 N ‘\ Ans. — v - 500 sin 56.31° - 250 sin 56.31° + 1000 cos 56.31°= 0 69.3 N = 69.3 N v/ M + 500(0.200) + 250(0.300) - 1000(0.200) = 25.0 N-m = 25.0 N-m C (b) From a free-body diagram of the part of member DEF to the right of section bb: ZoomE 400mm. F + 6— 2F = P + 250 + 500 = 0 —750 N = 750 N -* V — 1000 + 500 = 0 500 N = 500 N ¢ M + 1000(0.200) 500(0.600) = 0 100 N'm = 100 N-m C ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-37 A beam is loaded and supported as shown in Fig. P8-37. Using the coordinate axes shown, write equations for the shear V and bending moment M for any section 'w»" of the beam in the interval L3“*F————'°" 0 < x < 10 ft. Fm P&37 SOLUTION 5 75035 4Soogt.n, I 300 lLHt " From a free-body diagram for the complete beam: + C 2MB = 4500 - A(10) + 300(10)(5) - 750(2) = 0 ,a A = 1800 lb = 1800 lb T the interval 0 S x S 10 ft: = 1800 - 300x = -300x + 1800 lb -4500 + 1800x - 300(x)(x/2) = -150x2 + 1800x - 4500 ft'lb A beam is loaded and supported as shown in Fig. P8-38. 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 < 3 m. SOLUTION From a free—body diagram for the complete beam: + T EFy = VO - 2(3) - 3 = 0 o = 9 kN = 9 kN T + c 2M0 = —M0 - 2(3)(1.5) — 3(4) Mo = —21 kN'm = 21 kN'm C For the interval 0 S x S‘3 m: V = 9 — 2x = -2x + 9 kN —21 + 9x - 2(x)(x/2) = ‘x2 + 9x — 21 kN-m 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: +-g 2MB = -Ay(8) + 30(4)(6) + 40(2) = 0 ; 100 kN = 100 kN T For the interval 0 < x < 4 m: V = 100 - 30x = -30x + 100 kN M 100x — 30(x)(x/2) = -15x2 + 100x kN'm A beam is loaded and supported as shown in Fig P8-41. Using the coordinate axes shown. write equations for the shear V and bending moment M for any section of the beam in the interval 4 ft < x < 12 ft. SOLUTION From a free-body diagram for the complete beam: + C 2MB = -A(16) + 750(8)(8) - 3000(5) = 0 the interval 4 ft < x < 12 ft: ’ = 1875 - 750(x - 4) = -750x + 4875 lb 1875x — 750(x — 4)(x — 4)/2 = -375x2 + 4875x - 6000 ft-lb 7H3‘7 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8—51* Draw complete shear and moment diagrams for the beam shown in Fig. P8-51. SOLUTION From a free-body diagram' for the complete beam: + C 2MB = -A(13.5) + 5000(9.5) + 3000(4.5) = o A = 6185 lb = 6185 lb T + C XMA = B(13.5) - 5000(4) - 8000(9) = 0 B = 6815 lb = 6815 lb T Load, shear, and moment diagrams for the beam are shown below: 5000 IL 8090 lb 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 = 3(23) - 2500(8) - 800(10)(18) = o B = 7130 lb + C EM = -A(23) + 2500(15) + 800(10)(5) = 0 B A = 3370 lb Load, shear, and moment diagrams for the beam are shown below: 25“ ’5 800 mm: 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: ,4 + C EN = 3(8) - 30(4)(2> - 40(6) = o A B = 60 kN + C 2MB = -A(8) + 30(4)(6) + 40(2) = 0 A = 100 kN Load, shear, and moment diagrams for the beam are shown below: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-67 Draw complete shear lOOOlb 2501b,“ 375Ib/n and moment diagrams for the beam shown in Fig. P8-67. I ' 6n——*—-6n——*—4n Fig. P8-67 SOLUTION ‘4§t L§t H Zooogtqb From a free-body diagram for the complete beam: + C EMA = 1000(4) - 2000 - 250(12)(10) + B(20) - %(375)(4)[20 + a] : 0 B = 2250 lb H O + ChZM = 1000(24) - A(20) - 2000 + 250(12)(10) - %(375)(4)[] B A = 2500 lb Load, shear, and moment diagrams for the beam are shown below: )Ooo“, 395$) 250 )L/fi; a ...
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MEEN 221- Chapter8 - ENGINEERING MECHANICS - STATICS. 2nd....

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