Math for Truss Calculation Principles of EngineeringTM Lesson 5.1 - Statics Forging new generations of engineers Project Lead The Way, Inc. Copyright 2007

Math for Truss Calculation

Trusses Trusses are structures made up of beams joined together at their endpoints. Solve for the force in each member of the truss in this example to find whether the members are in tension or compression. Start be replacing the supports with reaction forces: 500 N AX CY AY 500 N B A C 2m 45

Trusses This truss consists of: 500 N AX CY AY 500 N AX CY AY 2 unknown member forces and 1 unknown external force at joint B 2 unknown member forces and 2 unknown reaction forces at joint A 2 unknown member forces and 1 unknown reaction force at joint C Draw the free-body diagrams for each pin: 500 N AX CY AY 500 N AX CY AY

Trusses These three separate free-body diagrams assume that: Member AB is in tension Member BC is in compression Member AC is in tension If these assumptions are incorrect, then our solution will show a negative quantity for force. 500 N AX CY AY 500 N AX CY AY FBC FAB FAC

Trusses To really “see” that Member AB is in tension Member BC is in compression Member AC is in tension We must look at the free-body diagrams of the beams, which show the effects of the pins on the beams.

Trusses Before we can solve for the forces, we must break FBC into its x and y components: FBC FBCX FBCY 45 FBCX = FBC * cos (45) FBCY = FBC * sin (45)

Trusses + FX = 0 ; 500N – FBCcos(45) = 0 FBCcos(45) = 500N cos(45) cos(45) FBC = 707.1 N (C) +FY = 0 ; FBCsin(45) – FAB = 0 707.1*sin(45) – FAB = 0 FAB = 707.1*sin(45) FAB = 500 N (T) The following is the free-body diagram of joint B. The force FBC has been replaced with its x and y components: 500 N FBC cos(45) FAB FBC sin(45)

Trusses + FX = 0 ; 500N – FBCcos(45) = 0 FBCcos(45) = 500N cos(45) cos(45) FBC = 707.1 N (C) +FY = 0 ; FBCsin(45) – FAB = 0 707.1*sin(45) – FAB = 0 FAB = 707.1*sin(45) FAB = 500 N (T) The (T) indicates tension and the (C) indicates compression. Both solutions were positive therefore our initially assumptions were correct. (If the solution had been a negative number, then we would simply reverse our assumption from tension to compression or vice versa)

Trusses + FX = 0 ; – FAC+ FBC cos (45) = 0 – FAC+ 707.1 cos (45) = 0 FAC = 707.1 cos(45) FAC = 500 N (T) +FY = 0 ; CY – FBC *sin(45) = 0 CY – 707.1 *sin(45) = 0 CY = 500 N The following is the free-body diagram of joint C. The force FBC has been replaced with its x and y components: CY FAC FBC 45

Trusses + FX = 0 ; FAC - Ax = 0 500 N - Ax = 0 AX = 500 N +FY = 0 ; FAB - AY = 0 500 N - AY = 0 AY = 500 N The following is the free-body diagram of joint A: AX AY FAB FAC

Trusses Solved!!! 500 N B A C 2m 45 500 N 707.1 N (C) 500 N (T)