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Question 1 (a) The truss of Figure 1 is subject to point loads at E and F and two members, 3 and 9, arriving on site 0.02 m t

(b) If the answer to Part (a) is an upwards deflection at D of 0.021 m, analyse the indeterminate truss of Figure 2 to find t

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Answer #1

SOKN HA VA { MA =0 5014) + 50(12 )- vel16)=0 Nc = SOKN VA +Vc -100 : 0 P-analysPsi: Fireling reastion fories magnitude in EacAt jont E Free bodey diagram,at, ĴointE 50 -7Besfno, BE Tane = AE AESPNO - 533 O = Tan --83-35 Efyi=0 =) Ef 4(-83-35)Sin(5Substitute BF, in Ef, then BC = 66:69 KN (tve) So tensfon Nou Consider foint F 8=53 3 SO KN GF EF CEsine BEIOSB, BE BÉsine EFEFy =0 =) 50+ CFSin(3686)+ Gcsin C36·86) -0 50+ (30. 56) sin36i86)+ Gc (sin (36-86)) 0 - 387193 KN GC (- ve) so comprersion ENow By P-analyis : unit load at nt & neglect AssLume loads ail thi External Except reaetion. Now, same precedete folloios asEfiy=0, sin - 0.5+AE (36.86) = 0 At = 0S 0.8335KN +ve Sin(3686) So tensi AELOSO + AB= 0 (o.8.235 (es(36-86) = - AB AB - -0.6{Fy=0 BESINO +BFsino · =d BF = - BE BF = -0-208 KN compresion(-ves AB+BC +(BE +BF)CoSO o - AB BC=0.666KN iotve, tension Joint{fy=o, CFSINE i CGsino+lis=0 0•208sin( 6.86) +les 7CG Sin(36.86) -D compression CG: -2:708KN - 0:666 + CO + 0:208cos(3686)-20GD 83.56 696id¢7 1.667 -94.44 -0:8335 314186 O8335 32.77 109.255 EKPL = 8455•205 %3D Deflecetion at D = Ed4 55.205 = 6455.20

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