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Unformatted text preview: Review 2 593. If the maximum intensity of the distributed load n
acting 011 the beam is w = 4 kN/m, determine the reactions at the pin A and roller B.
II M. 6 kl LXE=0: AK=0 Ans
(+IMA=0; ~12(1.5)—6(4)+B,(6)=0; 31:75:»; An, ”213:0; A, +7—1s=o; A, =11kN Ans
594. Determine the normal reaction at the rollerA and
horizontal and vertical components at pin B for
equilibrium of the member. [OkN Equation of Equilibrium : The annual auction N‘ an heobtained
directly by summing moms about point B. (+ m‘ =0; 10(o.a+ 1.2003 «mama
N, (1.2+ 1.200: 60") = o NA = 8.00 icN Ans ’0 “’ 1.25:0; 8,—6cos30°=0 B,=5.20kN Ans mu; :0; n,+s.oo6smso°ioo
a, =5.00kN Ans 597. The uniform ladder rests along the wall of a
building at A and on the roof at B. If the ladder has a
weight of 25 lb and the surfaces at A and B are assumed smooth, determine the angle 9 for equilibrium. x W's. (+234. = o; —R4(185ine)+ 25(9cose) = o $m=o; m—Rssinw =0 +TZF} =0; Racos40°—25=0 Solving,
R5 = 32.641b
R4 = 20.93 lb
m 9 = [ﬂ]
20.9808) 0 = 30.8” An: 599. A vertical force of 80 lb acts on the crankshaft.
Determine the horizontal equilibrium force P that must
be applied to the handle and the x, y, 2 components of
force at the smooth journal bearing A and the thrust
bearing 8.11m bearings are properly aligned and exert
only force reactions on the shaft. m,=o; P(8)80(10)=0 P=1001b Ans
zu=; H,(28)—80(14)=o 19,:wa Ans EM: == 0;  B,(28)  100(10) = 0 B, = 3§.7 lb Ans m4; 0; A,+(35.7)100=0 A, = 136 "J A.
25:0; By=0 AIIS
“=0: A¢+40—80=0 Al=401b Ans Negative sign indicates that B. acts in the opposite seme to that shewn on the FBD. 6—127. The structure is subjected to the loadings shown.
Member AB is supported by a ballandsocket at A and
a smooth collar at B. Member CD is supported by a pin at C.
Determine the x, y, z components of reaction at A and C. From FBD(a) 2M, =0; My, =0 £M=0; —MB,+800=0 MB,=800Nm 2M: =0; By(3)_3x(2)=0 (1)
2Fl =0; Az=0 Ans 214} :0; —A,+Bx =0 (2)
2F, =0; —A,+B, =0 (3)
From FBD(b) EM, = 0; B, (1.5)+800—250 cos 45°(5.5) =0 B, = 114.85 N From Eq.(1) 114.85(3)—B, (2) = 0 B, = 172.27 N
From Eq. (2) A, = 172 N Ans
From Eq.(3) A, = 115 N Ans 2F, =0; C, +250 cos 60°172.27 =0 C, =47.3 N Ans
2]“ =0; 250 cos 45°;— 114.85—Q, :0 C; =61.9N ' Ans
EF;=0; 250cos 60°—Q=0 Q=125N Am EM, = 0; MC,  l72.27(1.5)+ 250 cos 60°(5.5) = 0 MC,=—429N. m Ans 224:0; Mez=0 Ans Negative sign indicates that M; , acts in the. opposite sense to that shown
on FBD. *6—128. Determine the resultant forces at pins B and C on
member ABC of the fourmember frame. Prob. 6—128 5 FEE
4 A
Fbpfn " 3155;?(2) = 0 F}. ——> T (,‘f‘ZMF = O.
4 F1 F0
Hm = 0; —150(7)(3.5) + 3%(5) — FCD(7) = 0 . 15mm,
211 5’1?
A ' —————— “l
5,5 = 153111; = 1.53 kip Ans " 4T 3 ., Fm = 350 lb Ans 6130. Determine the force in each member of the truss
and state if the members are in tension or compression. 20 kN
“my =0; 20(1.5) — 10(4.5) + E,(6) = o
= 3M
+ ‘ _ A: —)T ¢ & T
425:, Are A 9040* low E
3 3
+T2F, =0; Ay —2o10+12.5=0 W5“ L5“ L5," 15m
A” = 17.5kN
Joint A:
4
+T21v; =0; 175?” =0
«9 :4» F”
FB = 21.88 = 21.9kN (C) Ans 0 49F”. r
3 .5»:
$21; = 0; F16  g(21.88) = 0 ’7
50 = 13.125 = 13.1 kN (T) Ans
Joint B:
$21; = ; «Em + 201.88) = o 1
.._
Em = 13.1 kN (C) Ans iii F‘c—“ x
4 2m n
+T2F, =0; 3(2188)  Fsa = 0 k F”
Fm = 17.51;»: (T) Ans
Joint G: . ”E
4
+T2F, =0; 17.5—20+ EFGC=0
Vac
Fac = 3.125: 3.12kN (T) Ans 3 (2“,: :[email protected] x
$23 = ; 39125) + E”, — 13.125: 0 “if.
Fm, = 11.2kN (T) Ans .Cont'd 6  130 .Cont'd Joint C:
+1321; =0;
+
all”; = 0,
Joint D:
3215 = 0;
+TEI'; =0;
Joint F:
+
~92}; = 0,
+T2F} =0; 4 4
3 CF — 3(3.125) = 0 Fa. = 3.12m (C) Ans 1%”??ch —ac
., s
3 .1125an3
3 3 F“
13.12 — 3(3125) — 3(3125) — FCD = 0
FUD = 9.375 = 9.38 kN (C) Ans
3 ‘9’
9375—3013”) = o I
1.315401 ®
a ‘5
13,). = 15.63 = 15.6kN (C) Ans '35
l: F175
4 F9:
3(15‘63) — 13,. = 0
For = 12.5 kN (T) Ans
3 f
38.125) — 11.25 + F)"; = O 3.125%” 1251‘»)
:_:_T® . ,_
_ 11.: Far
15” — 9.38 kN (T) Ans 5m 1, 10k» 4
12.5 — 10 — 5(3'125) = 0 Check! 6134. Determine the horizontaland vertical components
of force that pins A and B exen on the twomember frame.
Set 1’ = 500 N. MunberAC:
“EMA = 0; 600(0.75) — C, (1.5605 60°) + C, (1.5 sin 60") = 0 Mamba CB : (32% = o. c,(1) — 6,0) + 500(1) = o Solving.
c, = 402.6N
C, = 97.4N
MemberAC:
imp, = o; —A, +600sin60° —402.6 = o A. =117N Ans
+T2F, =0; A,—60000560°—97.4=0 A, =397N Ans l>z,=o, 402.5—soo+a,=o
B,=97.4N Ans +TZF; =0; —B,+9'7.4=0 B, = 97.4N A‘ns ...
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 Fall '08
 Troxel

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