Given - Length of Cable PQ -25mHorizental distonce between Pa=20 mN Jom - Jo mTeTa 853-1353-1336.812.5m36-870150KNZet tension in cable PR and a R be Tp and l8mespectivilyNow in A P&R,PsE RQ - 12'5 m'ニPR =L PRS = ESR = sint( ) =53.13°12.5VSing lami's, theomem ,-TpSin (90+36.87)150Si'n(53.13+53.13Sin(go+36'87)150 xsin(90+36-87)Sin (53-13 + 5313).Tp- 124.98 KNNow we gst, the tension m the cable due to theapplied load is > 125 KNいoitan of 212:5m

This can be done as shown below

Answered: lom * Jom Te Ta & 53-13 53-13 12.5m…

Engineering

Civil Engineering

lom * Jom Te Ta & 53-13 53-13 12.5m 36-87° 36.87° R 150 KN 12-5m

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Given - Length of Cable PQ -25m
Horizental distonce between Pa=20 m
N Jom - Jo m
Te
Ta 8
53-13
53-13
36.8
12.5m
36-870
150KN
Zet tension in cable PR and a R be Tp and l8
mespectivily
Now in A P&R,
Ps
E RQ - 12'5 m'
ニ
PR =
L PRS = ESR = sint( ) =53.13°
12.5
VSing lami's, theomem ,-
Tp
Sin (90+36.87)
150
Si'n(53.13+53.13
Sin(go+36'87)
150 xsin(90+36-87)
Sin (53-13 + 5313).
Tp
- 124.98 KN
Now we gst, the tension m the cable due to the
applied load is > 125 KN
い
oitan of 2
12:5m

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