Question

Determine the statical determinacy of the truss below and find the forces in all members of the truss due to the loading shown below. State if the member is in tension or compression.

3m EG. 15KN 15KN 4 panels at 4 m = 16 m

Answer 1

The detailed solution is given below.
Number of members(m) = 13 Number of joints (g) = 8 hos 21-3 = 2x8-3 3 13 am So, (m=2; 3) The truss is stable and statically determinate. um । um 15KN 1SKN. *EFxzo → HA=0 * Efy=0 VA-15-15 + RH=0 VA+RH = 30) * EMA=O » Isx +1Sx8 – RCs)= o RH = 11.25kN] so, VA = 30-11.25 Na=18.75kN

using Method of Joints Initially we assume all the members are, an tension. If the value obtained as a negative valued, then that he member is in compression. Joint a . FAB O= tant (2) Fac 18.75kN = 36.870 * Eff=0 → FABsino +18.75=0 + FAB =-31.25 kN * Efs=0 → FAB Coso + FAC=0 → FAC = 25kN] Joint a Sfx=0 * FAC - F E=0 → ECE = FAC = 25KN) to ISKN a sfyzo FCB=15KN)

Toint H FAF 0-36.890 RH FAG T 11:25kn *£Fy=0 => Fue Sino+11.25 = 0 => FHF = -18.75kNT A E Fx = 0 → FHF Coso + FAG=0 → FHG = 15kN] Toint a nd * Efy=0 (FGE=0 was e * EFx=0 - FGE - PGH =0 (FGE=FGH = 15kn| Joint D * <Fy -0 FGE i Foe v * £ Fx = 0 FOB = For Jant E 2=90-0 a = 90-36.87 [a=5301307 RED [FFG=0 FET FEG FF 18.75

Efyzo FFHCosa + FFE Cosa so * (FFE = - FHF = 18.75 kN] * EFxzo > FAF Sind- FFDFFE Sind=0 > FED = -18.75 sud - 18.75sma → FED = - 2x18.754 Sin 53,13 FRD = -30kn Fo= FBD = -30 kN] Toint E FED=0 o=0 "FEF=18.75kn EBITED * Efyzo - 25kN FEG=15kN - 25kN 15kn "FEF Cosa A FER Cosa_15=0 | 18.75 Cos 53013 –15 + FED Cos 5301350 I FER = 6.25kN] FAR= 31.25N, FRE=6.25kN, T 1 FEGEISKN, T Fac=25kN, T Fe=0 FGH = 15KN,T FBD = For 30kn, a SFEE=25KN,T FFG=0 FBC = 15kN, T FDE = 30kN, C IFFH=18.75kn, JFEF 318.75kN, TI De

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