Truss nalysis Method of Joints Steven Vukazich San Jose State University
General Procedure for the nalysis of Simple Trusses using the Method of Joints 1. raw a Free Body iagram (FB) of the entire truss cut loose from its supports and find the support reactions using the equations of equilibrium (we will see that for some truss structures this step is not always necessary); 2. raw a FB of a truss joint that has no more than two unknowns and use the two equations of equilibrium to find the two unknown truss member forces;. raw a FB of a truss joint adjacent to the joint analyzed in Step 2 that has no more than two unknowns (using the results from Step 2) use the two equations of equilibrium to find the two unknown truss member forces;. Repeat Step until all truss member forces are found a good check is if the last truss joint is in equilibrium then one has good confidence that the analysis is correct.
nalysis xample Using the Method of Joints m B m m C Consider the idealized truss structure with a pin support at and a roller support at C. The truss is subjected to applied loads at and. The objective of our analysis is to find all seven of the truss member internal forces
1. raw a Free Body iagram (FB) of the entire truss cut loose from its supports and find the support reactions using the equations of equilibrium (we will see that for some truss structures this step is not always necessary) m x B C y y m m
Use quilibrium to Find Support Reactions m x B C! M # = 0 y m m y y = 8 kn
Use quilibrium to Find Support Reactions m x B C! F ) = 0 y m m y y = kn
Use quilibrium to Find Support Reactions m x B C! F * = 0 y m m y x = -
FB Showing Known Support Reactions m C B kn 8 kn m m Joints and C are the only joints with two unknowns 2. raw a FB of a truss joint that has no more than two unknowns and use the two equations of equilibrium to find the two unknown truss member forces;
FB of the Connecting Pin at Joint C F BC F C C 8 kn Note Unknown truss member forces are assumed to act in tension (pulling away form the joint).! F * = 0 F BC = 0
FB of the Connecting Pin at Joint C F BC F C C 8 kn Note Unknown truss member forces are assumed to act in tension (pulling away form the joint).! F ) = 0 F C = 8kN
Return to FB Showing Support Reactions m B C Knowing F C and F BC from the previous analysis, now joints and are joints with two unknowns kn 8 kn m m. raw a FB of a truss joint adjacent to the joint analyzed in Step 2 that has no more than two unknowns (using the results from Step 2) use the two equations of equilibrium to find the two unknown truss member forces;
FB of the Connecting Pin at Joint F F B θ F C = 8 kn Notes Unknown truss member forces are assumed to act in tension (pulling away form the joint); Known force F C is shown acting in compression as was found in the previous step (no need for minus sign!)
FB of the Connecting Pin at Joint F,- cos θ = F,-! F ) = 0 F θ F C = 8 kn F,- sin θ = F,- Notes Unknown truss member forces are assumed to act in tension (pulling away form the joint); Known force F C is shown acting in compression as was found in the previous step (no need for minus sign!) Best to start with equilibrium in the vertical direction. F B = 10 kn
FB of the Connecting Pin at Joint F,- cos θ = F,-! F * = 0 F θ F C = 8 kn F,- sin θ = F,- Notes Unknown truss member forces are assumed to act in tension (pulling away form the joint); Known force F C is shown acting in compression as was found in the previous step (no need for minus sign!) F =
Return to FB Showing Support Reactions m C B kn 8 kn m m Knowing F C, F BC, F, and F B, from the previous analyses, now joints, B, and are joints with two unknowns. Repeat Step until all truss member forces are found a good check is if the last truss joint is in equilibrium then one has good confidence that the analysis is correct.
FB of the Connecting Pin at Joint F = F B θ F B Notes Unknown truss member forces are assumed to act in tension (pulling away form the joint); Known force F is shown acting in compression as was found in the previous step (no need for minus sign!)
FB of the Connecting Pin at Joint F #6 cos θ = F #6! F * = 0 θ F = F B F #6 sin θ = F #6 Notes Unknown truss member forces are assumed to act in tension (pulling away form the joint); Known force F is shown acting in compression as was found in the previous step (no need for minus sign!); Best to start with equilibrium in the horizontal direction. F = kn
FB of the Connecting Pin at Joint F #6 cos θ = F #6 θ F = F B F #6 sin θ = F #6 Notes Unknown truss member forces are assumed to act in tension (pulling away form the joint); Known force F is shown acting in compression as was found in the previous step (no need for minus sign!)! F ) = 0 F B = 8 kn
Return to FB Showing Support Reactions m B C From the previous analysis, joint has one unknown (F B ) and joint B has one unknown (F B ). kn 8 kn m m. Repeat Step until all truss member forces are found a good check is if the last truss joint is in equilibrium then one has good confidence that the analysis is correct.
FB of the Connecting Pin at Joint kn F B θ F = kn Notes Unknown truss member force is assumed to act in tension (pulling away from the joint); Known force F is shown acting in compression as was found in the previous step (no need for minus sign!) Two equations of equilibrium and one unknown one equation is used as a check.
FB of the Connecting Pin at Joint θ kn = kn kn = F B! F ) = 0 kn
FB of the Connecting Pin at Joint θ kn = kn kn = F B! F * = 0 kn F B = 6 kn
Check quilibrium of Connecting Pin at Joint B 10 kn = 8 kn F B = 8 kn B F B = 6 kn F BC = 0 θ F B = 10 kn 10 kn = 6 kn Joint B is in equilibrium!! F * = 0! F ) = 0
Show Results on FB of ntire Truss ( ) m ( 8 kn) (6 kn) B (0) ( 8 kn) C kn 8 kn m m Tension is Positive