Chapter 7 Trusses, Frames, And Machines

MEM202 Engineering Mechanics - StaticsMEMChapter 7Trusses, Frames, and Machines1

MEMMEM202 Engineering Mechanics - Statics7.2 Plane TrussesBefore this chapterR1F1F2Determine the reactions, R1and R2, of a rigid bodysubjected to a pair of forcesF1, and F2.In this chapterR2R1F1F2R2Determine the reactions, R1and R2, and the forces innine rigid members that arejoined together with six pinjoints, subjected to a pair offorces F1, and F2.2

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesIdealized trusses1. Members are connected together at their ends only.2. Members are connected together by frictionless pins.3. Loads are applied only at the joints. (Thus, allmembers are two-force members.)4. Weights of members are neglected.An actual riveted truss joint, which transmits both forcesand moments among connecting membersAn idealized frictionless pin connection, which transmitsforces among connecting members, but not moments.This assumption can be justified so long as the membersare long.3

MEMMEM202 Engineering Mechanics - Statics7.2 Plane TrussesA triangle is the building block of all plane trusses1ABB343 512CA2CDF8796m 2j 31211 1310HGEm NUMBER OF MEMBERS; j NUMBER OF JOINTSTotal unknowns: m (one for each member) 3 (3 support reactions).Each joint yields two equations (ΣFx 0, ΣFy 0)Is a truss always “stable” and “solvable” when m 2j -3 is satisfied?4

MEMMEM202 Engineering Mechanics - Statics7.2 Plane TrussesA triangle is the building block of all plane trusses1ABB343 512CA2CDF8796m 2j 31211 1310HGEm NUMBER OF MEMBERS; j NUMBER OF JOINTSTotal unknowns: m (one for each member) 3 (3 support reactions).Each joint yields two equations (ΣFx 0, ΣFy 0)Is a truss always “stable” and “solvable” when m 2j -3 is satisfied?5

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesRigidity and Solvability of A Trussm 2j rm: # of Membersj: # of Jointsr: # of Reactionsm 2j 3m 2j rj 9 m 152 j 3 15 mj 9 m 15 r 32 j r 15 mj 9 m 142 j 3 15 mj 9 m 14 r 42 j r 14 mj 6 m 92j 3 9 mj 6 m 9 r 32j r 9 m6

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane Trussesm 2j rj 8 m 12 r 32 j r 13 m 12j 8 m 14 r 32 j r 13 m 14j 10 m 16 r 32 j r 17 m 16j 8 m 13 r 32 j r 13 mYet, it is unstable!7

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesMethod of Joints1. Draw a free-body diagram of the entire structureand determine the reactions (if r 3).2. Draw free-body diagrams for all members (assumetensile forces in all members) and all joints.3. Set up the equilibrium equations for each joint andsolve them one joint at a time, begin with those thathave at most two unknowns.4. Check the results at the last joint.8

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesMethod of JointsReactions M (8)B (4)(1000) (8)(2000) 0 M (8)A (4)(1000) 0 F A 1000 0AyBxyxAx 1000 lbAy 500 lbB y 2500 lb9

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesMethod of Joints FA: 2000 lb FDDTCDCTCDTAD TCD TCDTACTADCDCABTADTACATADTACA1000 lb A TAB TAB500 lbTBCTAB TABy TAD TAC sin θ 500 0 Fx TCD TAC cos θ 0C: Fy TBC TAC sin θ 2000 0 Fx TCD 1000 0D: Fy TAD 0TBCB TAB TAC cos θ 1000 0 Fx TAB 0CB: TBC Fy TBC 2500 0TBCTACxBD (TAD , TCD ) C (TAC , TBC ) A(TAB , TAC ) B (TAB , TBC )2500 lb10

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesMethod of Joints? m 2j r ?j 4 m 5 r 32j r 5 mj 4 m 5 r 42j r 4 m 5j 4 m 5 r 32j r 5 mj 4 m 5 r 42j r 4 m 511

MEMMEM202 Engineering Mechanics - Statics7.2 Plane TrussesZero-Force TBC Fy 0 TBC 0 Fy′ 0 TCD 012

MEMMEM202 Engineering Mechanics - Statics7.2 Plane TrussesZero-Force MembersDBDBAZero-force member:BC, CDBACEDACEZero-force member:AB, ACCEZero-force member:AB, AC, BC13

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesMore About Zero-Force MembersZero force members cannot simply be removed from the truss anddiscarded, as they are needed to guarantee stability of the truss.If members BD and AD are removed, then a slight disturbance would cause joint D tobuckle outward. The equilibrium at joint D (see the free-body diagram) requires that F F TDE cos φ TCD cos φ 0 TDE TCD TDE TCD 0y TDE sin φ TCD sin φ 0 TDE TCD Yet equilibrium at joint C requires that TCD 0. Thus, joint D will continue to buckleoutward and the truss is no longer stable.x14

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesMore About Zero-Force MembersZero-force members may become non-zero-force members, orvice versa, as the load moves from one location to another.Consider a “Warren” truss with vertical supports.(See http://www.geocities.com/Baja/8205/truss.htm for other types of bridge truss)15

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesOther Bridge TrussesWarren Trussw/o s16

MEM202 Engineering Mechanics - StaticsMEM7.2 Plane TrussesMore About Zero-Force MembersWashington Crossing BridgeTension-only members17

MEMMEM202 Engineering Mechanics - Statics7.2 Plane TrussesMore About Zero-Force MembersCross members are slender tension-only members. Anycompressive force will buckle the member, render it useless.TCTCTTTCCCTC18

MEM202 Engineering Mechanics - Statics MEM 7.2 Plane Trusses Method of Joints 1. Draw a free-body diagram of the entire structure and determine the reactions (if r 3). 2. Draw free-body diagrams for all members (assume tensile forces in all members) and all joints. 3. Set up the equilibrium equations for each joint and