ch8(16-55) - NQ{NEERING MECHANICS — STATICS. 2nd. Ed. W....

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: NQ{NEERING MECHANICS — STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-16* )etermine the internal resisting "fforces 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: 450 cos 60° — P = 0 225 N = 225 N +— v + 450 sin 60° - 250 = 0 ‘ -139.71 N 3 139.7 N ¢ + C IN” a M + 250i0.300) ~ 450 cos 60° (0.240: — 450 sin 50° (0.300) = M = —95.91 N-m'é 95.9 N'm C ENGINEERING MECHANICS - STATICS, 2nd. Ed. iDetermine the internal resisting forces and moment transmitted by section aa in bar ABC of the three-bar frame shown in Fig. P8~20. SOLUTION d Link BB 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 B = 6.00 kN = 6.00 kN T From a free—body diagram for the part of member ABC to the right of section as: _ -1 120 _ 0 ¢ - tan 566 - 30.95 + 2/ 23 = p - 6 sin 30.95° + 3 sin 30.96° = 0 1.5433 kN a 1.543 kN 2/ — V — 6 cos 3G.96° + 3 cos 30.950 = 0 2.573 kN e 2.57 kN ‘3 + C EM: = M + 3(0.100) — 3(0.300) = o M = 0.300 kN'm = 0.300 kN-m C Ans. W. F. RILEY AND L. D. STURGES ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8-35* A beam is loaded and supported as shown in Fig. P8-35. 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 < 16 ft. SOLUTION From a free-body diagram for the complete beam: : C 2MB = —A(16) + 900(16)(8) — 3200(6) = 0 A 3 6000 lb = 6000 lb T For the interval 0 S x S 16 ft: 7 = 6000 m 900x lb 6000x ~ 900!x)(x/2) = —450x2 + 6000x ft-lb yvbeam 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. ,,,£, W, . I“ H _ [in ;|I 1m ,.Hm $.ma .15! H u m m m m SOLUTION Siimtfiefé‘iiffi’ii‘: 3:23?” 1!!!!!!!! +C2ME;13—A(8) +5(6)(5) :0 4"" A = 21 kW = 21 kN T For the interval 0 S x S 6: V : 21 - 5X = *5X + 21 kN M = —18 + 21x - 5(x)(x/Z} = -2.5x2 + 21x — 18 kN-m ENGINEERING MECHANICS - STATICS. 2nd. 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) + 30(4H6) + 40(2) = 0 interval 0 < x < 4 m: ' = 100 - 30x = 100x - 30£xl(x/2) = y”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(6) = 0 the interval 4 ft < x < 12 ft: —30x + 100 kN Ed. W. F. RILEY AND L: D. STURGES 100 kN = 100 kN T -15x2 + 100x kN-m 1875 lb = 1875 lb T ' = 1875 — 750(x - 4) = -750x + 4875 lb 1875K - 750(x — 4)(x w 4)/2 -375x2 + 4875x — 6000 ft-lb 7E3‘1 ENG " RING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES rxiraw complete shear and moment diagrams for the beam shown - in Fig. P8-52. SOLUTION From a free—body diagram for the complete beam: a + C 2143 = -A(11) + 10(7) + 30(3) — 20(3) A a 9.091 kN = 9.091 L201 T + c EMA = 13111) —10(4)- 30(8) — 20(14) B = 50.909 kN = 50.909 kN T Load, shear, and moment diagrams for the beam are shown below: )OKN BOKN 4m. 3m. 3m. 745 IEI INEéfixNG MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Eifaw 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(3) — 300(10)(183 = o B = 7130 lb + q 2MB = —A(23) + 2500(15) + 800(10)(5) = o A = 3370 lb Load, shear, and moment diagrams for the beam are shown below: 25“ ‘5 . 800351-51: ...
View Full Document

This homework help was uploaded on 03/25/2008 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.

Page1 / 6

ch8(16-55) - NQ{NEERING MECHANICS — STATICS. 2nd. Ed. W....

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online