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Method of Sections Method of joints works well when you want to know the forces in all the members. If we want to know a force in a particular member, the method of sections is more efficient. Look at the following truss: F1 F2 A F3 B D G E C To solve for the reactions at E and G, we use the entire truss as a free-body. F1 F2 A F3 B Gy D G Gx E C Ey By method of joints we use free-body diagrams of individual nodes. F1 A AB AC We know that the entire truss is in equilibrium. If this is the case, then each joint is also in equilibrium. Well, if this is the case, then a portion of the truss smaller than the entire but larger than the joints must also be in equilibrium. Lecture16.doc 1 For instance: F1 F2 A B BD BE C CE E This section must also be in equilibrium. So, we can cut the truss into sections and solve for the unknowns. What is the only limitation of this method? We can only cut such that there are 3 unknowns because: F y 0 Fx 0 M 0 If I wanted to find BD what would I do? BD is only unknown!! M E 0 What about CE? M B 0 CE only unknown!!! What about BE? Fy 0 BE only unknown!! Lecture16.doc 2 Examples 1). Given: The truss shown below 28 kips A 28 kips C E G I K 16 kips 10 ' B D 8' F J H 8' 8' 8' 8' Find: Determine the force in members EF and GI. FBD 28 kips A 28 kips C E G I K 16 kips y Bx B D F J H By x Jy Fy 0 M B 0 Fx 0 28(8) 28(24) 16(10) J y (32) 0 16 B x 0 28 28 33 B y 0 J y 33 kips B x 16 kips B y 23 kips Member EF 28 kips F y 0 A C FEG 23 28 FEF 0 FEF 5 kips C FEF 16 kips B D FDF 23 kips Lecture16.doc 3 Member GI I FGI K 16 kips FHI H FHJ J 33 kips M H 0 33(8) 16(10) FGI (10) 0 FGI 10.4 kips FGI 10.4 kips C Lecture16.doc 4 2). Given : A P B C D E d F G I J H d K O L M d d N d d Find: Determine the force in members FK and JO FBD P A F B FFG FFK C FGH H D FHI E FIJ J FJO M F 0 M J 0 FJO (4d ) Pd 0 FFK (4d ) Pd 0 FJO 1 P Comp 4 Lecture16.doc FFK 1 P Tension 4 5 3). Given: the truss shown below H 3m I F G D E B 3m K J C L 3m M 3m A 16 kN 16 kN N 6 Panels @ 4m = 24 m Find: Force in members DF,EF, and EG. FBD - whole structure => Ax = 0 Ay = 0 Member DF (enlarged view) F FDF d D B C A tan FEG FEF 16 kN 3 4 36.87 M E 0 16(6) FDF (4 sin 36.87 ) 0 FDF 40 kN T Member EF Member EG M A 0 M F 0 FEF (8) FDF (4 sin 36.87 ) 0 16(9) FEG (4 sin 36.87 ) 0 FEF (8) 40(4 sin 36.87 ) FEG 60 kN FEF 12 kN T Lecture16.doc FEG 60 kN C 6