Question

120 N 120 N $1 54 A T 2,5 m 53 S2 S5 B 6 m 6 m a) S1=220.50 N(C) S2=229.69 N(B) S3=110 N (B) S4=220.50 N(C) S5=229.69 N (B) b-) 51=287.69 N(C) 52=311.69 N (B) S3=120 N (B) S4=287.69 N(C) S5=311.69 N (B) S 1=239.74 N(C) S2=229.69 N(B) S3=100 N (B) S4=239.74 N(C) S5=229.69 N (B) d-) 51=239.74 N (C) S2=311.69 N (B) S3=120 N (B) S4=239.74 N(C) S5=100 N (B) e-) S1=287.69 N(C) S2=240.50 N (B) S3=130 N (B) S4=287.69 N() S5=240.50 N (B)

What is the stick strength of the bars in the fully connected and isostatic cage system in the figure? (drawing effect = (Ç), pressing effect = (B) )

Answer 1

Solution : Support reactica i PATRB = 2youn {moment about B=0 :) -120x12-12086+ Roxb.co =) | RB = 360 un =) IRA=-120 KN 2 take I wurd Joint (2) 90 = temn=645 79= 22.620 :51 lo Se = {fy zo =) Se sim 22.62 = 1round 311.99 KN (6) - Sq = =) & S1 = 2888ve) S2 = 312un (B) Są COLD - 311.99 Cos 22.62 Si = 288 kN (c) at joint 4 54 at is similar to joint (7) & hence 55 t 1201 S4 = 288 KN (C) 55= 3 12 kN (B)

S3 = 12000 force in member 120kN (B) (as two members are collinear & force at joint i acting in the direction of member? so the correct option is option B