Transcription
Chapter 3TrussesGermany
Trusses in Building
Trusses Types for Building
٦Chapter 2
Truss Bridge
Truss Bridge Types
Classification of coplanar TrussSimple Truss
CompoundpTruss
Complexp Truss
DeterminacyFor pplane trussIf b r 2jj staticallyy determinateIf b r 2j statically indeterminateWhereb number off barsr number of external support reactionj number off jjoints
StabilityFor pplane trussIf b r 2j unstableThe truss can be staticallyy determinate or indeterminate(b r 2j) but unstable in the following cases:External Stability:y truss is externallyy unstable iff all offits reactions are concurrent or parallel
Internal stability:stabilitInternally stableInternally unstableInternally unstable
Classifyy each of the truss as stable , unstable,,statically determinate or indeterminate.b 19 KK r 3 KK j 11b r 222 j 2(11) 22b 15 KK r 4 KK j 9Statically determinate & StableStatically indeterminate & Stableb r 192 j 2(9) 18
b 9 KK r 3 KK j 6b r 122 j 2(6) 12Statically determinate & Stableb 12 KK r 3 KK j 8b r 152 j 2(8) 16Unstable
Stability of Compound Truss
The Method of JointsSTEPS FOR ANALYSIS1. If the support reactions are not given, draw a FBD of theentire truss and determine all the support reactions usingthe equations of equilibrium.2 Draw the free-body2.free body diagram of a joint with one or twounknowns. Assume that all unknown member forces act intension (pu(pullingg the pin)p )uunless youy u cana determinedbyyinspection that the forces are compression loads.3. Apply the scalar equations of equilibrium, FX 0 and FY 0, to determine the unknown(s). If the answer ispositive, then the assumed direction (tension) is correct,otherwise it is in the opposite direction (compression).4. Repeatp stepsp 2 and 3 at each jjoint in succession until all therequired forces are determined.
The Method of Joints
Example 1Solve the following truss
Fy 0;4 FAG sin 30 0 FAG 8 KN C Fx 0;FAB 8 cos 30 0 FAB 6.93KN T
Fy 0; FGB 3 cos 30 0 FGB 2.6 KN C Fx 0;8 3 sin 30 FGF 0 FGF 6.50 KN C Fy 0;FBF sin 60 2.6 sin 60 0 FBF 2.6 KN T Fx 0;FBC 2.6 cos 60 2.6 cos 60 6.93 0 FBC 4.33KN T
Group WorkDetermine the force in each member
Zero Force Member1- If a joint has only two non-colinear members and there isno external load or support reaction at that joint, then thosetwo members are zero-force members
Zero Force Member2- If three members form a truss joint for which two of themembers are collinear and there is no external load or reaction atthat joint, then the third non-collinear member is a zero forcemember.b
Example 2Find Zero force member of the following truss
Method of Section
The Method of Section
The Method of SectionSTEPS FOR ANALYSIS1. Decide how you need to “cut” the truss. This is based on:a)) where youy need to determine forces,, and,, b)) where the totalnumber of unknowns does not exceed three (in general).2. Decide which side of the cut truss will be easier to workwith (minimize the number of reactions you have to find).3. If required, determine the necessary support reactions byddrawingi ththe FBD off ththe entireti ttruss andd applyingl i ththe EEofE.fE
4. Draw the FBD of the selected part of the cut truss. We needto indicate the unknown forces at the cut members. Initiallyywe assume all the members are in tension, as we did whenusing the method of joints. Upon solving, if the answer ispositive,i i theh memberb iis iin tensioni as per our assumption.iIffthe answer is negative, the member must be in compression.(Please note that you can also assume forces to be either tension orcompression by inspection as was done in the previous exampleabove.))5. Apply the equations of equilibrium (EofE) to the selected cutsection of the truss to solve for the unknown member forces.Please note that in most cases it is possible to write oneequation to solve for one unknown directly.
Example 2Solve the CF & GC members in the truss
ME 0;FCF sin 30(12) 300(6.93) 0 FCF 346.4 Ib C
MA 0;FGC (12) 300(6.93) 346.4 sin 30(12) 0 FGC 346.4 Ib T
Example 3Solve the GF & GD members in the truss
M MD 0 FGF cos 26.6(3) 7(3) 0 FGF 7.83kN Co 0 FGD sin 56.3(6) 7(3) 2(6) 0 FGF 1.8kN C
Example 4Solve the ED & EB members in the truss
Example 5Solve the BC & MC members in the K-truss
ML 0 2900(15) FBC (20) 0 FBC 2175Ib TAt Joint B Fy 0 FMB 1200 Ib TF thForthe totalt t l cuttingtti partt Fy 0 2900 1200 1200 FML FML 2900 Ib T
At Joint M F Fx 0 y 0 2900 1200 313FMC FMK 1532 Ib CFMC 1532 Ib T313FMK 0213FMC 213FMK 0
Group work 1Solve All members
Group work 2Solve All members
Group work 3Solve members CH , CI
Example 4 Compound4 Compound TrussSolve the truss
MC 0 5(4) 4(2) FHG (4 sini 60) 0 FHG 3.46kN C
Joint A FAB & FAIJoint H FHI & FHJJoint I FIJ & FIBJoint B FBC & FBJJoint J FJCJo
Example 5 Compound5 Compound Truss
Example 5 Compound5 Compound TrussSolve the truss
Joint A FAE & FABJoint B FEBAfter Solve FAEJoint A FAF & FAG
Complex TrussSi S xsi'iForce in members'S EC S EC xsEC 0x ?
EComplex Trussesr b 2j,PF3 9D2(6)(6)SADADeterminate Stable CBEEFsECF PF ECD๑ ACBX xS EC x sEC ๐x DS ECS i S i x s isECA FAD๑CB
'S EC S EC xsEC 0x ?Si Si' xsiMemberABACAFFEBEEDFCECSi 'sixsiSi
Example 5 Complex5 Complex Truss
Space Truss Determinacy and Stabilityuunstablestab e trusst ussPb r ๓jstatically determinate-check stabilityb r 3jstaticallyy indeterminate-check stabilityy b r 3j
Space Truss x,x, y, z, Force Componentsl x2 y2 z 2xFx F ( )lyFy F ( )lzFz F ( )lF Fx 2 F y 2 Fz 2
Support ReactionszzyFyshort linkxyxzzyyxxrollerllzzslotted rollery constrainedin a cylinderxyFxxFzzzyxFzball-and -socketFyFxxFzy
Zero Force Member1- If all but one of the members connected to a joint lie on thesame plane,landd providedid d no externall lloadd act on theh jjoint,ithen the member not lying in the plane of the other membersmustsubjected to zero force.Σ Fz ๐ ,FD ๐
Zero Force Member2- If it has been determined that all but two of severalmembersb connectedd at a jointj i support zero fforce, thenh theh tworemaining members must also support zero force, providedthey don’tdon t lie a long the same line and no external load act onthe joint.Σ Fz ๐ ,FB ๐Σ Fy ๐ ,FD ๐
Example 6 Space6 Space Truss
Example 7 Space7 Space Truss
FE BE ED FC EC. Example 5_Complex Truss. Space Truss Determinacy and StabilityDeterminacy and Stability P uustabetussnstable truss b r ๓j statically determinate-check stability b r 3j statically indeterminate-check stability b r 3j. Space Truss x, y, z, Force Componentsx, y, z, Force Components l x2 y2 z2 l x Fx F
