How to get the P = 1990.8N? Let T be the Tension in The Cable.TiosoTsine+2EMA =0T'Tsino (6) – P(3:5) = 0=>T= l:05 PAStress criteriatano = Ź41.05PTT (4)'4=> e =33.69°BcA= 0•334 PL=14462Limit,Ooc = 190 Mpalox) N7.21 m2.0:334P190P = 568.19 NewtonsE lon gation crifexia3A = TLAEl-o5P (7.21810)= 0.003 P(4) x42200 x10Limi t : A = 6mm =)O. 003 P = 6P=1990.8 N-%3D

Answered: Image /qna-images/answer/ff6f75ac-9480-4cce-a7ff-9ebf82274873.jpg

Let T be the Tension in The Cable. Toso Tsrne +2EMA =0 T' T'sine (6) – P(3:5) = 0 => T= 1.05 P LA Stress criteri a tan 0 = 4 1.05P => e = 33.69° %3D ニ T (4) 4 2 = 0:334 P L= [4462 Limit, Oec = 190 MPalox) N 7.21 m 0:334 P 190 = P = 568.19 Newtons Elangation critaria A = TL AE losp (7.21418) = 0.003 P 立 4 200 X10 Limit : A= 6mm ) O. 003 P = 6 P=1990.8 N-

Let T be the Tension in The Cable. Toso Tsrne +2EMA =0 T' T'sine (6) – P(3:5) = 0 => T= 1.05 P LA Stress criteri a tan 0 = 4 1.05P => e = 33.69° %3D ニ T (4) 4 2 = 0:334 P L= [4462 Limit, Oec = 190 MPalox) N 7.21 m 0:334 P 190 = P = 568.19 Newtons Elangation critaria A = TL AE losp (7.21418) = 0.003 P 立 4 200 X10 Limit : A= 6mm ) O. 003 P = 6 P=1990.8 N-

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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