Question

*ALL 19.90 FORCES IN KN 19.59 18.65 UN 15.86 17.47 16.42 15.00 A+ 3m A 54.65 B 31.02 356.2 347.8 40.03 38.84 101.1 283.1 33.8

How do I find internal force for Joint O & P?

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Answer #1

note

methods of joint is used

we have to consider joint A ,B, C in addition to joint O and P

, 19.90, 19.59 18.65 5 .86 11-47 16.42 is Zo> 356.2 37.02. lol. 347.8 > 28 3.1 * KN. 38.84 * - 33.81 > 34.61 13:05 4. 3 14.3

{H=0 Ax=0 since the support I have vertical component only (Reaction) The effect of load on support need not be considered. T

/ 37.02 34.61 I + 520.81 802.96 = 555.42 =839.81 KN. ( 1395.4 - 555.42) 1 Method of joints. considen Joint A. EV= 0 FAP SIN 5

consider Joint B TEBO FAB=FBc=569.79 KN FBO= 356.2 KN - FAB Į FB (? 135602 P considen Joint Fap= 1984.58 () ©= 90 - 54.64 ) =

Ev=o Fap Cos 35.36+ FPC Costs = 19.9+ Foo FPC Cos45= 19.9+356.2 - 984.58X (os 35.36 solving FPc =-603-66 KN hence is tensile.

consided Joint 603-66 & fco 450 d F CD FBC V 347.8 FCP= 663-66 KN. FBC=569.79 KN {H=0 Fco=FB ct 603-66X (0545 -99664 KN Ev=0

Join o V 19.59 14293 FPo=142-9 (c) 1905 Ft Fco= 79.05 (c) Ev=o Foo cos 45 +19.59 = 79.05 . For = 84.04 (T) &H=o 142.93 - Font

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