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5-1. The cable segments support the loading shown.
Determine the distace ye from the force at C to point
D. Set P = 20 KN.
1m
Tip
GBMy = 0; Typ 605 38.69°(6.5) ~ Ty in 33.69°(15) + 30(6.5) + 202.5) = 0
Tyg = 3926 N
SPER <0; -20- 30+ 39.26 c0s 33.69% +B, =0
d,
7.33 KN
SBA = 0; D,~3926 sin 33.69" =
D,=21.78 kN
Joint D
EA = 0; —Tacc0s 6+ 21.7
+t Ee, 17.33~ Tae sin 0= 0
Solving,
= 3852
Toc = 27.84 N
ye= 25 tan 3852° = 1.99 m ns The
144 5-2. The cable segments support the loading
shown, Determine the magnitude of the vertical force
P so that y
Tae cos 39.81%(S) + Toc sin 39.81%) ~ 30(1) P15) = 0
$:18Tac~5P = 30 a
~ Peos 205° =0 °
Solving Eqs. (1) and (2) yields
P=420KN Ans
Tac= 41.68N
145 5-3, The cable supports the three loads shown.
Determine the sags yp and yp of points B and D.
Take P, = 4 KN, P; = 2.5 KN.
yc + Ty - 25
Vg + ae
Ls 2
ass” oe?
WiamyyP iis” Vikd=yP ee
14 = yp La- yy
Teo + Tye -4=0
Yasmr ets dane ee
rip + 49-1 5 @
YaA=yy +4
@)
12 oe 1s Ico «0
Voasyr sie Via + 225
ht we __y, - —14= __p, - 2520
Oss wp eee Vla=yP + 225
=1.08 + 2790,
ttl ps
va. mre 2s? @
Combining Ex. (1) and 2)
28pp + 2p = 4.26
‘Combining Eqs. (3) and (4)
4.59 + 27.6 = B34
yn = 0.867 Ama
yp = 0.704 m Ans
146
