Question

Problem 6.13 3 of 3 v Part A Determine the force in member GB of the truss and state if the member is in tension or compression. Set P- 0, P2 21 kN (Figure 1) Express your answer to three significant figures and include the appropriate units. Assume positive scalars for members in tension and negative scalars for members in compression. HA FGB2.5 kN Submit Previous Answers Request Answer XIncorrect: Try Again; 5 attempts remaining ? Part B Determine the force in member AF of the truss and state if the member is in tension or compression. Express your answer to three significant figures and include the appropriate units. Assume positive scalars for members in tension and negative scalars for members in compression. FAF= Value Units Submit Request Answer Figure 1 of 1 Part C Determine the force in member AB of the truss and state if the member is in tension or compression. Express your answer to three significant figures and include the appropriate units. Assume positive scalars for members in tension and negative calars for members in compression. FAB- Value Units Submit Request Answer

Answer 1

5 od D E am from triangle EDC tano = 4 Gm o=tant (4) 8 = 63.4349 free body diagrams of joints FED Efy = 0 Take itve; lave Fco sino =0 FOD FED Sino FD = OKN I Fx=0 Take the; ke -FCD COSO-FED=0 -(0) COSO-FED=0 -0-FED=0 FED=OKN 0 free body diagrams of Joint E . 1 FEC Ety = 0 Take atve; love EF=0 Take+Ve; ve - FEF - FED 0 - FEF E FEC -21 =0 FED FEF -0 =0 FEC = 21KN FeF = OKN B=21KN from triangle FCE tano = (4) o= tant(e) 0= 45° um EK F kam free body diagrams of bint c tano o = tant ) O=26.560 IB Gm VE 2m

FBC 45 26.54 NEED K FF FCE=P2 Take Apply farce equilibriums along y-direction. Egy co ↑ tve; v-ve - FF cos45° - FCE - FD Cos26.56° = 0 - Fer cos45°-21 (0) C0526.56=0 .29.6989 kn cos45° 29.6989 KN) (C) Apply force equilibriums alont re-direction. 21 > FCF - FCF FCF I Fe =0 --ke Take tve; --Foc - FeF Sin45'+ Fog sin26-56=0 - FBC - (-29.6994) Sin45+ Co) Sin 26.56=0 FBC 21kN =0 free body diagsans of joint F AFBF E Fg = 7 FCF Take atve; Sin 45º=0 FEF FBF - Pit FCF t-ke 45 FAF FBF - O+(-29.6984) Sing5-0 21kN GJ FBF P=0

take Apply fasce equilibrium along x-direction E Fx = 0 Take tve; - PF + FF C0545'+ FEF = 0 -FoF + (-29.6984)C0545 +0=0 FAF - 21KN from triande ABF tamor () O= tant (24) 0 = 26.569 Afzmy diagsans of >Гос PAFe = 21kN (c). am A free body joint B B FGB Fy=0 thre; v-ve --FBF - FAR C0526-56=0 -21 - FAB COS26.56=0 FAB WEBF 21 FAB Cosg56 FAB=-23.4776KN FAB = 23.6776KNIC) Apply force equilibrium along xe-direction EF=0 te;t - ve FBc - FB - Fag sin 26.56'=0 21 – EB-623.4776) Sin26.56°) = 6 GB = 31.9976KN) G)