Question

A truss is loaded as shown in the figure.

a.) compute the forces in member GD given that P=300kN and Q=50kN

b.) If Pcd=600 kNand Pgd= 100kN (both compression), find P

c.) If Pcd=600 kNand Pgd= 100kN (both compression), find Q

SIT. A: A truss is loaded as shown in the figure. -2 m -2 m- E H А 3 m B 3 m 3 m D 1. 2. Compute the forces in member GD given that P = 300 kN and Q = 50 kN. A. 412.5 kN B. 135.2 KN C. 202.81 kN D. 243.74 kN If PCD = 600 kN and Pcp = 100 kN (both compression), find P. A. 517 kN B. 37 kN C. 579 kN D. 572.3 kN If Pcp =600 kN and Pcp =100 kN (both compression), find Q. A. 579 KN B. 5.17 kN C. 572.3 kN D. 37 kN 3.

Answer 1

Р 2m 2 m E H А hee Q 1 I B o elle we 2 G ( J e 10 e = tan (232) Socialinio 56.3;º ne Kx DA 1K 02 Dj Kg moment about it! - EMK = 0. Taking =P Dy x4 - PX4 -Qx9-Q46-Qx3 20 >> Dy = (P+4.52 Efy = 0; P_P-4.5Q ky =634.5@ + Dy 2 : Ky ky = 2 => alut EFxz0; Ka =3Q = 525 KN. .. P=300 KN, Q=50 KN, Dy = (300+4.5X50) .. Ky = 634.5*50 = 6).225KN. = ISO KN. Ka = 3x so Joint A: EF=0; - FR = Q = SO KN. =) FAE =SO KN Q. A 'AE FAB +P=0 AB Egzo; REDMI NOTE PAR C=1–300 kN. MI DUAL CAMERA -P

Joint te Ex=0; . EH = 0 : Flv = 0 H Efy=0; HI HI Joint Eu E AE A o ਨਿਤ - Q Cose FEL EF=0, - (EL - Fee) cose + Ex - Pae=0 FEE CS56-31° +0-50=0 - FBE = 90.14 FEE {fy=0; : (Ft + Fee Sin O=0 => FI + Pere Equation; ③ +6, - 2 FI = 90.14 =) FEL = 45.07 KN From equation ; F =645.07 kv. Q 20.90 a =(-) 2ceso BE Joint B E F = 0; BE В FBF 30 2 FAG . For + FoE Caso =2 FBF - 45.07 608 56-31°= 50 :- FeF = 75 kN. = EFy=0; FAB - Foc + PoE Sinozo Foc -300 - Fac – 45.07 Sin 56:31°=0 Fee =(-) 337.SO KN7=6-(P+Qtous) Fes + Joint 1% E Fresoj HI I FEL IJ i FFI & FEI Case=0 FFI + 45.07 Cos 56-31°=0 Fx =6) 25KN. 7=GQ EFyso; FHI + FI Sino = Pes → Ex = 0 + 45.07 Sim 56.31° REDMI NOTE FRO= 37-50KN. a tant MI DUAL CAMERA

Joint F: FBF A For FI 2a Cso ↑ T ^ - ES FJ E) Cosa Foc E F = 0; FI-FR) CSE + FI - Fer=0 IF FI (Fs - FF) 68856-319-25-75=0 F -Fr= 180.28 = » FfJ + F F = 180-28 [From em fy=0; a (Ffs + Fer) Sim O=o. = 90.14 KN. COSO & FF =(-) Ffs Joint co .. Fee 90.14 KN. EFxzo; FcG + Ef Caso -Q=0 FCF > Fca – 90.14C85631–50-0 -- Fea 2a EFy=0; : Fec + Ef Sind - Ep = =) -337-5-90-14 Siu56-31° = FD :- FeD = 6-) 412.50 KN. =6) (P+ 3a tano) Efy =0; FED : FED + Fad Simo+ Dy =o. 3-412.5 + Fan Sin56-31° +525=0 •Fox 112.5 Sim 56.31° = 100 KN. >EG FED Joint Do A Fad = > ED=67 IDY Fan = (-) 135.20 KN. --The force in member GD, (Ford) = 6-) 135.20 kn/compressing (-P-30 tand) + Ed Süno +P+4.5Q =0 =-4.50 - 3Q tend) = Fap Sine FGD = [45 4-52 30 y 20oso = ) Sind

구 Adi E=o[ 4.5Q 7. + = PD = Sino 8 30 2 case Fa=6 [4 še] If Pep = 600k N, and Pap=100KN. : From equation , 4.50 за 2coso i 100 =- Sino +0=56-330] = ) 180 = 2 [5.41Q – 2.70 :.Q=6) 37 KN. (as) From equation 600=-P 3x(-37) tam 56-30° 2 -p+ 30 tane > 660 =-p-316-3 > P=(-600+83) P = 517 KN. Asi CEPE+ em Go at