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Unformatted text preview: NQ{NEERING MECHANICS — STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 816* )etermine the internal resisting "fforces and moment transmitted by
section aa in the bracket shown SOLUTION ,.' From a freebody 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 Nm'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 threebar 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
freebody 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 kNm C Ans. W. F. RILEY AND L. D. STURGES ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 835* A beam is loaded and
supported as shown in
Fig. P835. 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 freebody 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 ftlb yvbeam is loaded and . supported as shown in
Fig. P836. 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 kNm ENGINEERING MECHANICS  STATICS. 2nd. 8—40 A beam is loaded and supported as shown in
Fig P840. 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 freebody 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 freebody 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 kNm 1875 lb = 1875 lb T ' = 1875 — 750(x  4) = 750x + 4875 lb 1875K  750(x — 4)(x w 4)/2 375x2 + 4875x — 6000 ftlb
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. P852. 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 freebody 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 . 80035151: ...
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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.
 Spring '08
 McVay

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