东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A15 Indeterminate Problem（Statically Indeterminate Structures）

Force Method Solution Procedure? Making the indeterminate structure determinate: Obtaining its equations of equilibrium: Analyzing the deformation compatibility that must besatisfied at selected redundant restraints: Substituting the deformation-load relationship intodeformation compatibility, resulting in complementaryequations: Joining of equilibrium equations and complementaryequations: Types of Problems can be addressed include (a) bartension/compression; (b) shaft torsion; and (c) beambending11

• Making the indeterminate structure determinate Force Method Solution Procedure • Obtaining its equations of equilibrium • Analyzing the deformation compatibility that must be satisfied at selected redundant restraints • Substituting the deformation-load relationship into deformation compatibility, resulting in complementary equations • Joining of equilibrium equations and complementary equations • Types of Problems can be addressed include (a) bar tension/compression; (b) shaft torsion; and (c) beam bending 11

Statically Indeterminate Bars: For an axially loaded bar with both ends fixed, determinethe reaction forces. Solution:A(P-R) LRBL0=AL=EAEARPLBLPRRBL12

Statically Indeterminate Bars • For an axially loaded bar with both ends fixed, determine the reaction forces. 2   1 1 2 0 B B B A R L P R L L EA EA L R P L L R P L           • Solution: 12

Sample Problem: Determine the reaction forces at the bar endsAA=250mm. Solution:150mmDRB300kN150mm400×10-6CA=400mm150mm(600- R,)x103K400×10-6150×10-34FL,600kN150mm0=AL=BE(600 R.)×103EAi=1250×10-6AA(900- R.)×103250x10-6300kN300kN= R = 577kNR,=900-577=323kN600kN600kN8=0OROLBBR

• Determine the reaction forces at the bar ends.       6 3 4 3 6 3 1 6 3 6 400 10 600 10 150 10 400 10 0 600 10 250 10 900 10 250 10 577kN 900 577 323kN B B i i i i B B B A R R F L L EA E R R R R                                                      • Solution: Sample Problem

Sample Problem Determine the reaction forces, if any, at the bar ends. E = 200 Gpa? Solution:4.5×10-==AEAi=lA=250mm150mmDR400×10-6300kN150mm(600 R,)×103CA= 400 mm2150×10-3400×10-6150mmK200×10%(600-R,)×103600 kN150mm250×10-6B(900-R.)×1031250×10-6B4.5mm= R, =115.4kN=R,=900-115.4=784.6kN14

• Solution: Sample Problem • Determine the reaction forces, if any, at the bar ends. E = 200 Gpa.       4 3 1 6 3 3 6 9 3 6 3 6 4.5 10 400 10 600 10 150 10 400 10 200 10 600 10 250 10 900 10 250 10 115.4kN 900 115.4 784.6kN i i i i B B B B B A F L L EA R R R R R R                                         14

Sample Problem. What is the deformation of theTube(A2,E,)rod and tube when a force P isRod (Ar,E)exerted on a rigid end plate asshown?Endplate: Solution:P+P=PPLPL△L,ALPE,AE,A2E,A,PE,A,PPE,A + E, A,E,A + E, A2PL=E,A + E, A,15

Sample Problem • What is the deformation of the rod and tube when a force P is exerted on a rigid end plate as shown? • Solution: 1 2 1 2 1 2 1 1 2 2 1 1 2 2 1 2 1 1 2 2 1 1 2 2 1 1 2 2 , P P P PL P L L L E A E A E A P E A P P P E A E A E A E A PL L E A E A                      15