Statically indeterminate structures MT07 handout

Statically Indeterminate Structures Mt07 Handout-PDF Download

  • Date:11 Jan 2020
  • Views:62
  • Downloads:0
  • Pages:28
  • Size:235.00 KB

Share Pdf : Statically Indeterminate Structures Mt07 Handout

Download and Preview : Statically Indeterminate Structures Mt07 Handout


Report CopyRight/DMCA Form For : Statically Indeterminate Structures Mt07 Handout


Transcription:

Statically Determinate Structures, Reactions and internal forces can be determined solely from. free body diagrams and equations of equilibrium, Results are independent of the material from which the structure. has been made,5 kN reaction forces bar forces,Independent equations. equilibrium in x y directions,at each joint,2 number of joints. Double check structure for,10 kN internal mechanisms etc.
Statically Indeterminate Structures, Reactions and internal forces cannot be found by statics alone. more unknown forces than independent equations of equilibrium. Results are dependent on the material from which the structure. has been made,A 2 unknown forces,Only 1 useful equation of. C L equilibrium,b RA P RB 0,B Need to find another. Flexibility or Force Method,RA A A1 A2 RA redundant. released structure, Equation of compatibility expresses the fact that the change.
in length of the bar must be compatible with the conditions at. the supports,A A1 A 2 0, Write the force displacement relations These take the mechanical. properties of the material into account,A1 and A 2. Substituting into the equation of compatibility gives. A 0 Note that flexibilities b EA and L EA,appear in this equation Hence this. RA approach is called the flexibility method,Substituting into the equilibrium equation gives. Pa Note that we have solved for forces,RB P R A Hence this approach is also called the.
L force method,Stiffness or Displacement Method,Equation of equilibrium forces at C must balance. Equation of compatibility at point C C C1 C 2, Write the force displacement relations and solve for the forces. C1 and C 2 B,R A C1 and RB C 2,Substituting into the equilibrium equation gives. C1 EA C 2 EA,P Note that stiffnesses EA a and EA b. a b appear in this equation Hence this,approach is called the stiffness method.
Using the compatibility condition displacements equal gives. Pab Pab Note that we have solved for displacement,EA a b EAL Hence this approach is also called the. displacement method, Finally substituting into the expressions for forces gives. So both the flexibility method and the stiffness method give the. same result, The choice of approach will depend on the problem being solved. Use of Displacement Diagrams in,Statically Indeterminate Problems. based on Example 3 page 70 Gere Timoshenko,LCD L 2 sin 1.
Bar ADB is supported by two wires CD and CB A load P is. applied at B The wires have axial rigidity EA Disregarding. the weight of the bar find the forces in the wires. RAy TCD TCB,Equilibrium, Horizontal Direction Fx R Ax TCD cos 1 TCB cos 2 0. Vertical Direction Fy R Ay TCD sin 1 TCB sin 2 P 0. Moments about A ccw,M A TCD sin 1 L TCB sin 2 2L P 2L 0. 4 unknown forces only 3 equations,Compatibility, We can relate the tensions in the two wires by considering the. extensions of the wires,D CD Displacement diagram,CD D sin 1. CB B sin 2 2 D sin 2,Force displacement relations,TCD LCD TCB LCB.
Using the equilibrium moment equation the compatibility equation. and the force displacement relations it is possible to solve for the. forces in the wires We find that,TCB 1 125 P and TCD 1 406 P. Elasto plastic Analysis of a Statically Indeterminate Structure. based on Example 2 19 from Gere,Bar 1 L Bar 2, Horizontal beam AB is rigid Supporting bars 1 and 2 are made of an. elastic perfectly plastic material with yield stress Y yield strain Y and. Young s modulus E Y Y Each bar has cross sectional area A. a Find the yield load PY and the corresponding yield displacement BY at point. b Find the plastic load PP and the corresponding plastic displacement BP at. c Draw a load displacement diagram relating the load P to the displacement B. of point B,Moment Equilibrium ccw,M A F1 b F2 2b P 3b 0. F1 2F2 3P 3P,Compatibility,Force displacement,Bar 2 will yield first. 1 and 2 since F2 F1,The corresponding elongation of bar 2 is.
F2 L F2 L Y L, The downward displacement of the bar at point B is. When the plastic load PP is reached both bars will be stretched to. 1 P4 Stress and Strain Dr A B Zavatsky MT07 Lecture 3 Statically Indeterminate Structures Statically determinate structures Statically indeterminate

Related Books