Equilibrium and Trusses ENGR 221 February 17, 2003
Lecture Goals 6-4 Equilibrium in Three Dimensions 7-1 Introduction to Trusses 7-2Plane Trusses 7-3 Space Trusses 7-4 Frames and Machines
Equilibrium Problem Determine the reactions at A and the force in bar CD due to the loading.
Equilibrium Problem Draw the free-body diagram of the main body. R Ax R Ay 1 6i in. o θ = tan = 26.565 12 in. θ T CD
Equilibrium Problem Look at equilibrium ( o F ) x = RAx + TCDcos 26.565 + 125 lb = 0 R Ax ( o T cos( 26.565 ) 125 lb ) CD = + y Ay CD ( o ) F = R + T sin 26.565 40 lb 60 lb + 80 lb = 0 R Ay CD ( T sin ( 26.565 o ) 20 lb ) =
Equilibrium R Ay Problem R Ax Take the moment about A θ T CD CD M A CD T CD ( o )( ) ( ) = 0 = T cos 26.565 6 in. 40 lb 4 in. 60lb 8in. + 80lb 12in. ( ) ( ) ( o )( ) cos 26.565 6 in. = 320 lb-in T = 59.628 lb
Equilibrium R Ay Problem R Ax Take the moment about A θ T CD (( ) ( o ) ) R Ax = 59.628 lb cos 26.565 + 125 lb = 71.667 lb (( ) ( o ) ) R Ay = 59.628 lb sin 26.565 20 lb = 6.667 lb
Equilibrium in 3-Dimensions In two dimensions, i the equations are solved using the summation of forces in the x, y and z directions and the moment equilibrium includes moment components in the x, y and z directions. F = 0 F = 0 F = 0 x y z M = 0 M = 0 M = 0 x y z
Trusses -Definition Trusses are structures t composed entirely of two force members. They consists generally of triangular sub-element and are constructed and supported so as to prevent any motion.
Frames -Definition Frames are structures t that t always contain at least one member acted on by forces at three or more points. Frames are constructed and supported so as to prevent any motion. Frame like structures that are not fully constrained are called machines or mechanisms.
Planar Trusses - lie in a single plane and all applied loads must lie in the same plane. Truss
Truss Space Trusses - are structures that are not contained in a single plane and/or are loaded out of the plane of the structure.
Truss There are four main assumptions made in the analysis of truss 1 Truss members are connected together at their ends only. 2 Truss are connected dtogether th by frictionless pins. 3 The truss structure is loaded only at the joints. 4 The weights of the members may be neglected.
Simple Truss The basic building block of a truss is a triangle. Large truss are constructed by attaching several triangles together A new triangle can be added truss by adding two members and a joint. A truss constructed in this fashion is known as a simple truss.
Simple Truss It has been observed that the analysis of truss can be done by counting the number member and joints on the truss to determine the truss is determinate, unstable or indeterminate.
Simple Truss A truss is analysis by using m=2*j-3, where m is number of members, j represents the number of joints and 3 represents the external support reactions.
Simple Truss If m< 2j-3, then the truss is unstable and will collapse under load. If m> 2j-3, then the truss has more unknowns than know equations and is an indeterminate i t structure. If m= 2j-3, ensures that a simple plane truss is rigid and solvable, it is neither sufficient nor necessary to ensure that a non-simple plane truss is rigid and solvable.
Simple Truss- Identify Determine type of simple truss is it determinate, indeterminate or unstable.
Method of Joints -Truss The truss is made up of single bars, which are either in compression, tension or no-load. The means of solving force inside of the truss use equilibrium equations at a joint. This method is known as the method of joints.
Method of Joints -Truss The method of joints uses the summation of forces at a joint to solve the force in the members. It does not use the moment equilibrium equation to solve the problem. In a two dimensional i set of equations, F x = 0 F = 0 In three dimensions, y F z = 0
Method of Joints Example Using the method of joints, determine the force in each member of the truss.
Method of Joints Example Draw the free body diagram of the truss and solve for the equations Fx C x = 0 = C = 0 lb x y F = 0 = 2000 lb 1000 lb + E+ C E+ C y = 3000 lb y
Method of Joints Example Solve the moment about C MC = 0 = 2000 lb 24 ft + 1000 lb 12 ft E 6 ft E = 10000 lb y ( ) ( ) ( ) C = 3000 lb 10000 lb = 7000 lb
Look at joint A Method of Joints Example F F y AD 4 = 0 = F AD 2000 lb 5 = 2500 lb F = 2500 lb C AD ( ) 3 3 F = 0 = F + F = ( 2500 lb) + F 5 5 F = 1500 lb F = 1500 lb T AB x AD AB AB AB ( )
Look at joint D Method of Joints Example F = 0 = 4 F + 4 F = 4 ( 2500 lb) + 4 F 5 5 5 5 F = 2500 lb F = 2500 lb T DB y AD DB DB DB ( ) 3 3 Fx = 0 = F AD + FDB + FDE 5 5 3 3 = ( 2500 lb) + ( 2500 lb) + F 5 5 F = 3000 lb F = 3000 lb C DE DE ( ) DE
Method of Joints Example Look at joint B 4 4 Fy = 0 = FBD FBE 1000 lb 5 5 4 4 = ( 2500 lb ) F DE 1000 lb 5 5 F = 3750 lb F = 3750 lb C DE DE ( ) 3 3 Fx = 0 = FBD FBA + FBE + FBC 5 5 3 3 = ( 2500 lb ) 1500 lb + ( 3750 lb ) + 5 5 F = 5250 lb F = 5250 lb T BC DE ( ) F BC
Method of Joints Example Look at joint E 4 4 Fy = 0 = FEB + FEC + 10000 lb 5 5 4 4 = ( 3750 lb ) + F DE + 10000 lb 5 5 F = 8750 lb F = 8750 lb C EC EC ( ) 3 3 Fx = 0 = FEB FED + FEC 5 5 3 3 = ( 3750 lb ) ( 3000 lb ) + F 5 5 F = 8750 lb F = 8750 lb C EC EC ( ) EC
Method of Joints Example Look at joint C to check the solution 4 Fy = 0= FCE 7000 lb 5 4 = ( 8750 lb) 7000 lb = 0 OK! 5 3 Fx = 0 = FCE FCB + Cx 5 3 = ( 8750 lb ) ( 5250 lb ) + 0 = 0 5
Method of Joints Class Problem Determine the forces BC, DF and GE. Using the method of Joints.
Method of Sections -Truss The method of joints is most effective when the forces in all the members of a truss are to be determined. If however, the force is only one or a few members are needed, then the method of sections is more efficient.
Method of Sections -Truss If we were interested in the force of member CE. We can use a cutting line or section to breakup the truss and solve by taking the moment about B.
Method of Sections Example Determine the forces in members FH, GH and GI of the roof truss.
Method of Sections Example Draw a free body diagram and solve for the reactions. F x R = Ax = 0 = R 0 kn Ax R Ax L F y = 0 L + R = Ay 20 kn R Ay
Method of Sections Example Solve for the moment at A. R Ax R Ay L M 6kN( ) 6kN( ) 6kN( ) A = 5 m 10 m 15 m 1 kn 20 m 1 kn 25 m + L 30 m L = 7.5 kn Ay ( ) ( ) ( ) R = 12.5 kn
Method of Sections Example Solve for the member GI. Take a cut between the third and fourth section and draw the free-body diagram. l 15 m 10 m 15 m 8 m HI 10 m = lhi = 8 m ( ) l HI = 5.333 m 8 m α = tan = 28.1 15 m 1 o
Method of Sections Example The free-body diagram of the cut on the right side. M H = 1 kn 5 m + 7.5 kn 10 m F 5.333 m ( ) ( ) ( ) GI F GI = 13.13 kn F = 13.13 kn T GI ( )
Method of Sections Example Use the line of action of the forces and take the moment about G it will remove the F GI and F GH and shift F FH to the perpendicular of G.
Method of Sections Example Take the moment at G M = 1 kn( 5 m) 1 kn( 10 m) + 7.5 kn( 15 m) G +F ( o )( ) FH cos 28.1 8 m F FH = 13.82 kn F = 13.82 kn C FH ( )
Method of Sections Example Use the line of action of the forces and take the moment about L it will remove the F GI and F FH and shift F GH to point G. 5 m β = tan = 133.2 5.333 m 1 o
Method of Sections Example Take the moment at L M L = 1 kn 5 m + 1 kn 10 m + F cos 43.2 15 m ( ) ( ) ( o )( ) GH F GH = 1.372 kn F = 1.372 kn C GH ( )
Method of Sections Class Problem Determine the forces in members CD and CE using method of sections.
Homework (Due 2/24/03) Problems: 6-34, 6-37, 6-38, 6-40, 6-45, 6-6363
Truss Bonus Problem Determine whether the members are unstable, determinate or indeterminate.
Truss Bonus Problem Determine the loads in each of the members.
Truss Bonus Problem Determine the loads in each of the members.
Truss Bonus Problem Determine the loads in each of the members.