Question

1 m 1 m Q5. Determine the member forces in each member of the the truss and state if they are in tension or compression 1 m 0.5 m 0.5 m 10 KN B 1 m

Answer 1

0.5) Im im > IF Im (1) e = tant (os) 63.43' D. li O: G' OE 0.5m ay (P) HE orom 1 ors IVE (2) 2 = 1917 (155) B X(+) (+) 10 kn Im La = 33.699 RA Fво of bruss Step 2: Calculation of support reaction 0) i H E Efx = 0 -RA HE ERA Ety =0 -/ot VE 0 (ii) Efy .. =0 n's VE = 10 KN sign convention (10) Em E G 2 :: 10x2 - RAX 1.5 =0 (+) (-) : RA = 13:33 KN putting in equation o, .. HE HE = RA - 13.33 EN

③ force in member AB and AC Y(+) LE -X(+) Fre comider foo of Joint A, (1) E Fx : FACSINX-RA FACE 13.33 sin 33.69 FAC Re=13:33 Er FAC 24.031 KN (c) (13) {fy =0 FAB - Fac cosa zo FAB = 24.03) (0533.69 FAR 20 (T) EN force in member DE and BE POE ei E consider FBO o of Joint E, (0) Ex=0 : -Foe sino -Fre since the=o fpe sin 63.43t FBE sin 63.43 = 13:33 FOE +FBE 13.33 Sin 63.43" fpe + FRE = 14.goy HE = 13.33 En е :- FBE VE = IO EN ... (2) Efy 70 . FDE COSO - FBE COTO + VE Foe - Fee = -VE COSO

:: Foe -Fee = -lo cos 63.43' FOE -FBE -22.357 KN 2 on solving equation o 0 and ①, Foe = -3.73 KN indicates our assumed direction Negative sign wrong ...FE 3.73 IN C () and FBE 18.63 KN (T) 3 ho the members on (3) force in member CF and of in When any joint fress is made up of only 2 members and external load acts the Joint then both identified as zero force member, are F F CF For member EO n zero o force

force in member CD and BD consider FBO of Joint D 19 (0) Erx = 0 D .. Fio sino -fesine =0 ->x FDE oo Fcp = FRE FLD FBD : - FD 3.73 KN (C) =0 (i) Efy :: -FBpt fcp CODE 7 FDE COSO 70 FBB = (Fco & B Foe) coso Fepa (3.73 + 3.73) cos 63-43 3.34 kN (T) For 5 force in member BC consider Fon of Joint B, FBO foc ( i) Efx=0 FRE :-FBc sino + FBE sino =o B FAB FBE foc FBC 18.63 EN CTU Note: T - Tension cu compression