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# Resolução do Capítulo 6 - Análise de Estruturas

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PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: Joint FBDs: Joint B: N Joint C: N ( ) ( )( )0: 6.25 m 4 m 315 N 0 240 NB y yM C\u3a3 = \u2212 = =C 0: 315 N 0 75 Ny y y yF B C\u3a3 = \u2212 + = =B 0: 0x xF\u3a3 = =B 75 N 5 4 3 AB BCF F = = 125.0 N C ABF = W 100.0 N T BCF = W By inspection: 260 N CACF = W PROBLEM 6.2 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: Joint FBDs: Joint C: Joint A: ( ) ( )( )0: 14 ft 7.5 ft 5.6 kips 0 3 kipsA x xM C\u3a3 = \u2212 = =C 0: 0 3 kipsx x x xF A C\u3a3 = \u2212 + = =A 0: 5.6 kips 0 5.6 kipsy y yF A\u3a3 = \u2212 = =A 3 kips 5 4 3 BC ACF F = = 5.00 kips C BCF = W 4.00 kips T ACF = W 1.6 kips 8.5 4 ABF = 3.40 kips T ABF = W PROBLEM 6.3 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: Joint FBDs: Joint C: Joint B: ( )( ) ( )0: 6 ft 6 kips 9 ft 0 4 kipsB y yM C\u3a3 = \u2212 = =C 0: 6 kips 0 10 kipsy y y yF B C\u3a3 = \u2212 \u2212 = =B 0: 0x xF\u3a3 = =C 4 kips 17 15 8 AC BCF F = = 8.50 kips T ACF = W 7.50 kips C BCF = W By inspection: 12.50 kips C ABF = W 10 kips 5 4 ABF = PROBLEM 6.4 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: Joint FBDs: Joint D: Joint C: Joint B: ( ) ( )( ) ( )0: 1.5 m 2 m 1.8 kN 3.6 m 2.4 kN 0B yM C\u3a3 = + \u2212 = 3.36 kNy =C 0: 3.36 kN 2.4 kN 0y yF B\u3a3 = + \u2212 = 0.96 kNy =B 20: 2.4 kN 0 2.9y AD F F\u3a3 = \u2212 = 3.48 kN T ADF = W 2.10: 0 2.9x CD AD F F F\u3a3 = \u2212 = 2.1 (3.48 kN) 2.9CD F = 2.52 kN C CDF = W By inspection: 3.36 kN C ACF = W 2.52 kN C BCF = W 40: 0.9 kN 0 5y AB F F\u3a3 = \u2212 = 1.200 kN T ABF = W PROBLEM 6.5 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: Joint FBDs: Joint B: Joint C: Joint A: 0 :xF\u3a3 = 0x =C By symmetry: 6 kNy y= =C D 10: 3 kN 0 5y AB F F\u3a3 = \u2212 + = 3 5 6.71 kN T ABF = = W 20: 0 5x AB BC F F F\u3a3 = \u2212 = 6.00 kN C BCF = W 30: 6 kN 0 5y AC F F\u3a3 = \u2212 = 10.00 kN C ACF = W 40: 6 kN 0 5x AC CD F F F\u3a3 = \u2212 + = 2.00 kN T CDF = W 1 30: 2 3 5 kN 2 10 kN 6 kN 0 check 55y F \uf8eb \uf8f6 \uf8eb \uf8f6\u3a3 = \u2212 + \u2212 =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8\uf8ed \uf8f8 By symmetry: 6.71 kN TAE ABF F= = W 10.00 kN C AD ACF F= = W 6.00 kN C DE BCF F= = W PROBLEM 6.6 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: Joint FBDs: Joint C: Joint D: ( ) ( )( ) ( )( )0: 25.5 ft 6 ft 3 kips 8 ft 9.9 kips 0A yM C\u3a3 = + \u2212 = 2.4 kipsy =C 0: 2.4 kips 9.9 kips 0y yF A\u3a3 = + \u2212 = 7.4 kipsy =A 0: 3 kips 0x xF A\u3a3 = \u2212 + = 3 kipsx =A 2.4 kips 12 18.5 18.5 CD BCF F = = 3.70 kips T CDF = W 3.70 kips C BCF = W or: 0:x BC CDF F F\u3a3 = = 60: 2.4 kips 2 0 18.5y BC F F\u3a3 = \u2212 = same answers ( )17.5 40: 3 kips 3.70 kips 0 18.5 5x AD F F\u3a3 = + \u2212 = 8.125 kipsADF = 8.13 kips T ADF = W ( ) ( )6 30: 3.7 kips 8.125 kips 0 18.5 5y BD F F\u3a3 = + \u2212 = 6.075 kipsBDF = 6.08 kips C BDF = W Joint A: PROBLEM 6.6 CONTINUED ( )4 40: 3 kips 8.125 kips 0 5 5x AB F F\u3a3 = \u2212 + \u2212 = 4.375 kipsABF = 4.38 kips C ABF = W PROBLEM 6.7 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: Joint FBDs: Joint A: Joint B: 0: 480 N 0 480 Ny y yF A\u3a3 = \u2212 = =A ( )0: 6 m 0 0A x xM D\u3a3 = = =D 0: 0 0x x xF A\u3a3 = \u2212 = =A 480 N 6 2.5 6.5 AB ACF F = = 200 N C ABF = W 520 N T ACF = W 200 N 2.5 6 6.5 BE BCF F = = 480 N C BEF = W 520 N T BCF = W Joint C: Joint D: PROBLEM 6.7 CONTINUED By inspection: 520 N T CD CEF F= = W ( )2.50: 520 N 0 6.5x DE F F\u3a3 = \u2212 = 200 N C DEF = W PROBLEM 6.8 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: Joint FBDs: Joint F: Joint C: ( ) ( )( ) ( )( )0: 9 ft 6.75 ft 4 kips 13.5 ft 4 kips 0E yM F\u3a3 = \u2212 \u2212 = 9 kipsy =F 0: 9 kips 0 9 kipsy y yF E\u3a3 = \u2212 + = =E 0: 4 kips 4 kips 0 8 kipsx x xF E\u3a3 = \u2212 + + = =E By inspection of joint :E 9.00 kips T ECF = W 8.00 kips T EFF = W By inspection of joint :B 0ABF = W 0BDF = W 40: 8 kips 0 5x CF F F\u3a3 = \u2212 = 10.00 kips C CFF = W 30: (10 kips) 0 5y DF F F\u3a3 = \u2212 = 6.00 kips T DFF = W ( )40: 4 kips 10 kips 0 5x CD F F\u3a3 = \u2212 + = 4.00 kips T CDF = W ( )30: 9 kips 10 kips 0 5y AC F F\u3a3 = \u2212 + = 3.00 kips T ACF = W Joint A: PROBLEM 6.8 CONTINUED 40: 4 kips 0 5x AD F F\u3a3 = \u2212 = 5.00 kips C ADF = W PROBLEM 6.9 Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: Joint A: Joint B: Joint D: 0: 0x xF\u3a3 = =H By symmetry: 12 kNy= =A H By inspection of joints and ,C G and 0CE AC BCF F F= = W and 0EG GH FGF F F= = W 30: 12 kN 3 kN 0 5y AB F F\u3a3 = \u2212 \u2212 = 15.00 kN C ABF = W ( )40: 15 kN 0 5x AC F F\u3a3 = \u2212 = 12.00 kN T ACF = W ( )4 10 40: 15 kN 0 5 10.44 5x BD BE F F F\u3a3 = \u2212 \u2212 = ( )3 3 30: 15 kN 6 kN 0 5 10.44 5y BD BE F F F\u3a3 = \u2212 \u2212 + = Solving yields 11.93 kN C BDF = W 0.714 kN C BEF = W 0: by symmetryxF\u3a3 =