BEAM ANALYSIS USING THE STIFFNESS METHOD

1 BEAM ANALYSIS USING THE STIFFNESS . METHOD . ! Development: The Slope-Deflection Equations ! STIFFNESS Matrix ! General Procedures ! Internal Hinges ! Temperature Effects ! Force & displacement Transformation ! Skew Roller Support 1. Slope Deflection Equations i P j k w Cj settlement = j i P j Mij w Mji i j 2. Degrees of Freedom M.. A B 1 DOF: . L. P.. B 2 DOF: , . A C.. 3. STIFFNESS Definition kAA 1 kBA. A B. L. 4 EI. k AA =. L. 2 EI. k BA =. L. 4. kAB kBB. A 1 B. L. 4 EI. k BB =. L. 2 EI. k AB =. L. 5. Fixed-End Forces Fixed-End Forces: Loads P. PL L/2 L/2 PL. 8 8. L. P P. 2 2. w wL2 wL2. 12 12. L. wL wL. 2 2. 6. General Case j k i P. w Cj settlement = j i P j Mij w Mji i j 7. i P j Mij w Mji i L settlement = j j 4 EI 2 EI 2 EI 4 EI. i + j = M Mji = i + j L L ij L L. j i +. (MFij) (MFji) . settlement = j +.

2 P. w (MFij)Load (MFji)Load 4 EI 2 EI 2 EI 4 EI. M ij = ( ) i + ( ) j + ( M F ij ) + ( M F ij ) Load , M ji = ( ) i + ( ) j + ( M F ji ) + ( M F ji ) Load 8. L L L L. Equilibrium Equations i P j k w Cj Mji Cj M. jk Mji Mjk j + M j = 0 : M ji M jk + C j = 0. 9. STIFFNESS Coefficients Mij i j Mji L. j i 4 EI. kii = 2 EI. L k ji = i L. 1. +. 2 EI. kij = 4 EI. L k jj = j L. 1. 10. Matrix Formulation 4 EI 2 EI. M ij = ( ) i + ( ) j + ( M F ij ). L L. 2 EI 4 EI. M ji = ( ) i + ( ) j + ( M F ji ). L L. M ij (4 EI / L) ( 2 EI / L) iI M ij F . M = + M F . ji ( 2 EI / L ) ( 4 EI / L ) j ji . kii kij . [k ] = . k ji k jj . STIFFNESS Matrix 11. i P j Mij w Mji i [ M ] = [ K ][ ] + [ FEM ]. L. j j ([ M ] [ FEM ]) = [ K ][ ].. [ ] = [ K ] 1[ M ] [ FEM ]. Mij Mji j i Fixed-end moment + STIFFNESS matrix matrix (MFij) (MFji).

3 [D] = [K]-1([Q] - [FEM]). +. P displacement Force matrix (MFij)Load w (MFji)Load matrix 12. STIFFNESS Coefficients Derivation Mi i Mj Real beam i j L. Mi + M j Mi + M j L L. L/3 M jL Mj 2 EI EI. Conjugate beam Mi EI MiL. 2 EI. MiL L M j L 2L From (1) and (2);. + M 'i = 0 : ( )( ) + ( )( ) = 0. 2 EI 3 2 EI 3 4 EI. Mi = ( ) i M i = 2 M j (1) L. 2 EI. MiL M L Mj =( ) i + Fy = 0 : i ( ) + ( j ) = 0 (2) L. 2 EI 2 EI 13. Derivation of Fixed-End Moment Point load P Real beam Conjugate beam A B. A L B. M M. M EI EI. M. EI ML. M 2 EI. M. ML EI. 2 EI 2. P PL2 PL PL. 16 EI 4 EI 16 EI. ML ML 2 PL2 PL. + Fy = 0 : + = 0, M =. 2 EI 2 EI 16 EI 8 14. P. PL PL. 8 L 8. P P. P/2. 2 2. P/2. PL/8. -PL/8 -PL/8. - -PL/8 -PL/16. - -PL/16. -PL/8. PL/4 PL PL PL PL. + + =. + 16 16 4 8. 15. Uniform load w Real beam Conjugate beam A B.

4 A L B. M M. M EI EI. M. EI ML. M 2 EI. M. ML EI. 2 EI. wL3 wL2 wL3. w 24 EI 8 EI 24 EI. ML ML 2 wL3 wL2. + Fy = 0 : + = 0, M =. 2 EI 2 EI 24 EI 12 16. Settlements M. Mi = Mj Real beam Mj Conjugate beam EI. L. A B. Mi + M j . M. L Mi + M j M EI. L. M. EI ML. ML. 2 EI M. 2 EI. M EI.. ML L ML 2 L. + M B = 0 : ( )( ) + ( )( ) = 0, 2 EI 3 2 EI 3. 6 EI . M= 17. L2. Typical Problem CB P2. P1 w A C. B. L1 L2. wL2. PL P PL w wL2. 12. 8 8 12. L L. 0. 4 EI 2 EI PL. M AB = A + B + 0 + 1 1. L1 L1 8. 0 EI. 2 EI 4 PL. M BA = A + B + 0 1 1. L1 L1 8. 0 2. 4 EI 2 EI P2 L2 wL2. M BC = B + C + 0 + +. L2 L2 8 12. 0 2. 2 EI 4 EI P2 L2 wL2. M CB = B + C + 0 + . L2 L2 8 12. 18. CB P2. P1 w A C. B. L1 L2. MBA CB M. BC. B. 2 EI 4 EI PL. M BA = A + B + 0 1 1. L1 L1 8. 2. 4 EI 2 EI P L wL. M BC = B + C + 0 + 2 2 + 2. L2 L2 8 12.

5 + M B = 0 : C B M BA M BC = 0 Solve for B. 19. CB P2. P1 w MBA. MAB. A C M. CB. B MBC. L1 L2. Substitute B in MAB, MBA, MBC, MCB. 0. 4 EI 2 EI PL. M AB = A + B + 0 + 1 1. L1 L1 8. 0 EI. 2 EI 4 PL. M BA = A + B + 0 1 1. L1 L1 8. 0 2. 4 EI 2 EI P2 L2 wL2. M BC = B + C + 0 + +. L2 L2 8 12. 0 2. 2 EI 4 EI P2 L2 wL2. M CB = B + C + 0 + . L2 L2 8 12. 20. CB P2. P1 w MBA. MAB. MCB. A MBC C. Ay B Cy L1 L2. By = ByL + ByR. B C. P1 P2. MBA MCB. MAB A B MBC. Ay ByL ByR Cy L1 L2. 21. STIFFNESS Matrix Node and Member Identification Global and Member Coordinates Degrees of Freedom Known degrees of freedom D4, D5, D6, D7, D8 and D9. Unknown degrees of freedom D1, D2 and D3. 6 9. 5 2 3 8. 2EI EI. 2 7. 14 1 21 3. 22. Beam-Member STIFFNESS Matrix i j 1 4. 3 6. E, I, A, L. 2 5. k41 k14. k11 = AE/L AE/L AE/L AE/L = k44.

6 D1 = 1 d4 = 1. 1 2 3 4 5 6. 1 AE/L - AE/L. 2 0 0. 3 0 0. [k] =. 4 -AE/L AE/L. 5 0 0. 6 0 0. 23. i j 1 4. 3 6. 6EI/L2 = k32 E, I, A, L. 2 5 6EI/L2 = k65. k62 = 6EI/L2 6EI/L2 = k35. d2 = 1 d5 = 1. 12EI/L3 = k52 12EI/L3 = k25. k22 = 12EI/L3 12EI/L3 = k55. 1 2 3 4 5 6. 1 AE/L 0 - AE/L 0. 2 0 12EI/L3 0 - 12EI/L3. 3 0 6EI/L2 0 - 6EI/L2. [k] =. 4 -AE/L 0 AE/L 0. 5 0 -12EI/L3 0 12EI/L3. 6 0 6EI/L2 0 - 6EI/L2. 24. i j 1 4. 3 6. E, I, A, L. 2 5. k33 = 4EI/L 4EI/L = k66. d3 = 1 2EI/L = k63 2EI/L = k36. d6 = 1. k23 = 6EI/L2 6EI/L2 = k53 k26 = 6EI/L2 6EI/L2 = k56. 1 2 3 4 5 6. 1 AE/L 0 0 - AE/L 0 0. 2 0 12EI/L3 6EI/L2 0 - 12EI/L3 6EI/L2. 3 0 6EI/L2 4EI/L 0 - 6EI/L2 2EI/L. [k] =. 4 -AE/L 0 0 AE/L 0 0. 5 0 -12EI/L3 -6EI/L2 0 12EI/L3 -6EI/L2. 6 0 6EI/L2 2EI/L 0 - 6EI/L2 4EI/L. 25. Member Equilibrium Equations i j Fxi Fxj Mi Mj E, I, A, L.

7 Fyi Fyj =. AE/L AE/L x AE/L AE/L. i x j 1 1. +. +. 6EI/L2. 6EI/L2. 6EI/L2 6EI/L2. 1 1. x i x j 12EI/L3 12EI/L3. 12EI/L3 12EI/L3. +. +. 4EI/L 2EI/L 4EI/L. 1 2EI/L. x i x j 1. 6EI/L2 6EI/L2 6EI/L2 6EI/L2. +. FFyi FFyj FFxi FFxj MFi MFj 26. Fxi = ( AE / L) i + (0) i (0) i + ( AE / L) j + (0) j + (0) j + FxiF. Fyi = (0) i + (12EI / L3 ) i (6EI / L2 ) i (0) j ( 12EI / L3 ) j (6EI / L2 ) j FyiF. Mxi = (0) i (6EI / L2 ) i (4EI / L) i (0) j ( 6EI / L2 ) j (2EI / L) j MiF. Fxj = ( AE / L) i (0) i (0) i ( AE / L) j (0) j (0) j FxiF. Fyj = (0) i ( 12EI / L3 ) i ( 6EI / L2 ) i (0) j (0) j ( 6EI / L2 ) j FyjF. Mj = (0) i (6EI / L2 ) i (2EI / L) i (0) j ( 6EI / L2 ) j (4EI / L) j MjF. Fxi AE/ L i Fxi . F. AE/ L 0 0 0 0.. Fyj 0 12 EI / L3. 6EI/ L2 0 12 EI/ L3 6EI/ L2 i FyiF . Mi 0 6EI/ L2 4 EI/ L 0 6EI/ L2 2EI/ L i MiF.

8 = + F . Fxj AE/ L 0 0 AE/ L 0 0 j Fxj . Fyj 0 12 EI/ L3 6EI/ L2 0 12 EI/ L3 6EI/ L2 j FyiF . F . M j 0 6EI/ L2 2EI/ L 0 6EI/ L2 4 EI/ L j M j . STIFFNESS matrix Fixed-end force matrix [q] = [k][d] + [qF]. End-force matrix displacement matrix 27. 6x6 STIFFNESS Matrix i i i j j j Ni AE/ L 0 0 AE/ L 0 0 . Vi 0 12 EI / L3. 6EI/ L2 0 12 EI/ L3 6EI/ L2 .. Mi 0 6EI/ L2 4 EI/ L 0 6EI/ L2 2EI/ L . [k ]6 6 = . Nj AE/ L 0 0 AE/ L 0 0 . Vj 0 12 EI/ L3 6EI/ L2 0 12 EI/ L3 6EI/ L2 .. Mj 0 6EI/ L2 2EI/ L 0 6EI/ L2 4EI/ L . 4x4 STIFFNESS Matrix i i j j Vi 12 EI/ L3 6EI/ L2 12 EI/ L3 6EI/ L2 .. Mi 6EI/ L2 4 EI/ L 6EI/ L2 2EI/ L . [k ]4 4 =. Vj 12 EI/ L3 6EI/ L2 12 EI/ L3 6EI/ L2 .. Mj 6EI/ L. 2. 2EI/ L 6EI/ L2 4 EI/ L . 28. 2x2 STIFFNESS Matrix i j Mi 4 EI / L 2EI / L . [k ]2 2 = . Mj 2EI / L 4 EI / L . Comment: - When use 4x4 STIFFNESS matrix, specify settlement.

9 - When use 2x2 STIFFNESS matrix, fixed-end forces must be included. 29. General Procedures: Application of the STIFFNESS METHOD for Beam ANALYSIS P P2y w M1 M2. 2 4 6. Global 1. 1 3. 2. 2. 1 3 5. 2 4 2 4 . Local 1 2. 1 3 1 3 . 30. P P2y w M1 M2. 2 4 6. Global 1 3. 1 2. 2. 1 3 5. 2 4 4 6. Member 1 2. 1 3 3 5. 2 4 2 4 . Local 1 2. 1 3 1 3 . 31. P P2y w M1 M2. 2 4 6. Global 1 3. 1 2. 2. 1 3 5. 2 4 2 4 . Local 1 2. 1 3 1 3 . P. w (q F 2)1. (q F1)2. (q F4 )2. [FEF] 1. (q F4)1. 2. (q F1 )1 (q F3)1 (q F2)2 (q F3 )2. 32. 2 4 6. Global 1 3. 1 2. 2. 1 3 5. 2 4 . Local 1. 1 3 . [q] = [T]T[q ]. 1 2 3 4 . q1 1 1 0 0 0 q1' . q 0 q . 2 = 2 1 0 0 2' . q3 3 0 0 1 0 q3' .. q4 1 4 0 0 0 1 q4 ' . [k] = [T]T[q ] [T]. 33. 2 4 6. Global 1 3. 1 2. 2. 1 3 5. 2 4 . Local 2. 1 3 . [q] = [T]T[q ]. 1 2 3 4 . q3 3 1 0 0 0 q1' . q 0 q.

10 4 = 4 1 0 0 2' . q5 5 0 0 1 0 q3' .. q6 2 6 0 0 0 1 q4 ' . [k] = [T]T[q ] [T]. 34. 2 4 6. 1 3. 1 2. 2. 1 3 5. STIFFNESS Matrix: 1 2 3 4. 1. 1 2 3 4 5 6. 2. [k]1 = 3. 1. Member 1. 2. 4 3. [K] = 4. Node 2. 3 4 5 6. 5. 3 Member 2. 6. 4. [k]2 = 5. 6. 35. 2 4 6. 1 3. 1 2. 2. Joint Load 1 3 Du=Dunknown 5. Qk QFA. 2 3 4 1 5 6. M. Q 2 1z 2 D2 Q 2F . -P2y D F . Q 3 3 KAA KAB 3 Q 3 . Q 4 M3z 4 D 4 Q 4F . = 0 + F . Q. 1 1 D1 Q1 . Q 5 5 D5 0 Q F . KBA KBB 0 5F . 6 . Q 6 D6 Q 6 . Reaction Dk = Dknown QFB. Qu [Q k ] = [K AA ][Du ] + [Q AF ]. [Du ] = [K AA ] 1 + ([Qk ] [QAF ]). Member Force : [q ] = [k ][d ] + q F [ ]. 36. Global: 2 4 6. 1 3. 1 2. 2. 1 P 3 5. w (qF2)1 (qF 4)2. (q F6 )2. [FEF] 1 (qF4)1. 2. (qF1 )1 (qF3)1 (qF3)2 (q F5 )2. 1 2 3 4 5 6. Q1 1 D1 0 Q1F . M Member 1 F . Q 2 1 2 D. 2 Q . 2. Q 3 -P 2y 3 D3 Q 3 = (q 3 ) 1 + (q 3 ) 2.