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Matrix beam Presentation Transcript

  • 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
  • 2. Slope – Deflection Equations P i w j k Cj settlement = ∆j Mij P i w j Mji θi ψ θj 2
  • 3. • Degrees of Freedom M θΑ B A 1 DOF: θΑ C 2 DOF: θΑ , θΒ L θΑ A P B θΒ 3
  • 4. • Stiffness Definition kAA kBA 1 B A L k AA = 4 EI L k BA = 2 EI L 4
  • 5. kBB kAB A B 1 L k BB = 4 EI L k AB = 2 EI L 5
  • 6. • Fixed-End Forces Fixed-End Forces: Loads P L/2 PL 8 L/2 PL 8 L P 2 P 2 w wL2 12 wL 2 L wL2 12 wL 2 6
  • 7. • General Case P i w j k Cj settlement = ∆j Mij P i w j Mji θi ψ θj 7
  • 8. Mij P i j w Mji θi L θj ψ 4 EI 2 EI θi + θj = M ij L L Mji = 2 EI 4 EI θi + θj L L θj θi (MFij)∆ + (MFji)∆ settlement = ∆j + P (MFij)Load M ij = ( settlement = ∆j w (MFji)Load 4 EI 2 EI 2 EI 4 EI )θ i + ( )θ j + ( M F ij ) ∆ + ( M F ij ) Load , M ji = ( )θ i + ( )θ j + ( M F ji ) ∆ + ( M F ji ) Load 8 L L L L
  • 9. • Equilibrium Equations i P w j k Cj Cj M Mji Mji jk Mjk j + ΣM j = 0 : − M ji − M jk + C j = 0 9
  • 10. • Stiffness Coefficients Mij i j Mji L θj θi kii = 4 EI L k ji = 2 EI L ×θ i k jj = 4 EI L ×θ j 1 kij = 2 EI L + 1 10
  • 11. • Matrix Formulation M ij = ( 4 EI 2 EI )θ i + ( )θ j + ( M F ij ) L L M ji = ( 2 EI 4 EI )θ i + ( )θ j + ( M F ji ) L L  M ij  (4 EI / L) ( 2 EI / L) θ iI   M ij F   M  =   + (2 EI / L) ( 4 EI / L) θ j   M ji F    ji      kii k ji [k ] =  kij  k jj   Stiffness Matrix 11
  • 12. P i Mij w j Mji θi L [ M ] = [ K ][θ ] + [ FEM ] θj ψ ∆j ([ M ] − [ FEM ]) = [ K ][θ ] [θ ] = [ K ]−1[ M ] − [ FEM ] Mij Mji θj θi Fixed-end moment Stiffness matrix matrix + (MFij)∆ (MFji)∆ [D] = [K]-1([Q] - [FEM]) + (MFij)Load P w (MFji)Load Displacement matrix Force matrix 12
  • 13. • Stiffness Coefficients Derivation Mj θi Mi Real beam i j L Mi + M j L Mi + M j L/3 M jL 2 EI L Mj EI Conjugate beam Mi EI θι MiL 2 EI M j L 2L MiL L )( ) + ( )( ) = 0 2 EI 3 2 EI 3 M i = 2 M j − − − (1) + ΣM 'i = 0 : − ( + ↑ ΣFy = 0 : θ i − ( M L MiL ) + ( j ) = 0 − − − (2) 2 EI 2 EI From (1) and (2); 4 EI )θ i L 2 EI )θ i Mj =( L Mi = ( 13
  • 14. • Derivation of Fixed-End Moment Point load P Real beam Conjugate beam A A B B L M EI M M EI M P M EI ML 2 EI PL2 16 EI PL 4 EI ML 2 EI 2 PL 16 EI M EI PL ML ML 2 PL2 + ↑ ΣFy = 0 : − − + = 0, M = 2 EI 2 EI 16 EI 8 14
  • 15. P PL 8 PL 8 L P 2 P 2 P/2 PL/8 P/2 -PL/8 -PL/8 - -PL/8 -PL/16 - -PL/16 PL/4 + -PL/8 − PL − PL PL PL + = + 16 16 4 8 15
  • 16. Uniform load w Real beam Conjugate beam A A L B B M EI M M M EI M EI ML 2 EI ML 2 EI w wL3 24 EI wL2 8 EI M EI wL3 24 EI wL2 ML ML 2 wL3 + ↑ ΣFy = 0 : − − + = 0, M = 2 EI 2 EI 24 EI 12 16
  • 17. Settlements Mi = Mj Real beam Mj Conjugate beam L Mi + M j M L A M EI B ∆ ∆ M EI Mi + M j L M EI ML 2 EI ML 2 EI M M EI ∆ + ΣM B = 0 : − ∆ − ( ML L ML 2 L )( ) + ( )( ) = 0, 2 EI 3 2 EI 3 M= 6 EI∆ L2 17
  • 18. • Typical Problem CB w P1 P2 C A B L1 P PL 8 PL 8 L2 wL2 12 L 0 PL 4 EI 2 EI M AB = θA + θB + 0 + 1 1 8 L1 L1 0 EI PL 2 EI 4 M BA = θA + θB + 0 − 1 1 8 L1 L1 0 2 P2 L2 wL2 4 EI 2 EI M BC = θB + θC + 0 + + L2 L2 8 12 0 2 − P2 L2 wL2 2 EI 4 EI θB + θC + 0 + − M CB = 8 12 L2 L2 w wL2 12 L 18
  • 19. CB w P1 P2 C A B L1 L2 CB M BC MBA B M BA = M BC PL 2 EI 4 EI θA + θB + 0 − 1 1 8 L1 L1 P L wL 4 EI 2 EI = θB + θC + 0 + 2 2 + 2 L2 L2 8 12 2 + ΣM B = 0 : C B − M BA − M BC = 0 → Solve for θ B 19
  • 20. P1 MAB CB w MBA A L1 B P2 MBC L2 C M CB Substitute θB in MAB, MBA, MBC, MCB 0 PL 4 EI 2 EI θA + θB + 0 + 1 1 M AB = L1 L1 8 0 EI PL 2 EI 4 M BA = θA + θB + 0 − 1 1 8 L1 L1 0 2 4 EI 2 EI P2 L2 wL2 θB + θC + 0 + + M BC = 8 12 L2 L2 0 2 − P2 L2 wL2 2 EI 4 EI θB + θC + 0 + − M CB = L2 L2 8 12 20
  • 21. P1 MAB A Ay CB w MBA B L1 P2 MCB MBC L2 Cy C By = ByL + ByR B P1 MAB B A Ay L1 MBA ByL P2 MBC ByR L2 C MCB Cy 21
  • 22. 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 14 2EI 1 3 2 21 EI 2 8 3 7 22
  • 23. Beam-Member Stiffness Matrix 3 1 i j 2 E, I, A, L 6 5 k14 AE/L k41 AE/L k11 = AE/L 4 AE/L = k44 d1 = 1 d4 = 1 1 2 3 4 1 0 0 3 0 0 4 -AE/L AE/L 5 0 0 6 0 6 - AE/L 2 [k] = AE/L 5 0 23
  • 24. 3 i 1 6EI/L2 = k32 2 4 6 6EI/L2 = k65 5 6EI/L2 = k35 d5 = 1 12EI/L3 = k52 12EI/L3 1 2 1 AE/L 2 [k] = E, I, A, L k62 = 6EI/L2 d2 = 1 k22 = j 3 12EI/L3 = k25 4 5 0 - AE/L 0 0 12EI/L3 0 - 12EI/L3 3 0 6EI/L2 0 - 6EI/L2 4 -AE/L 0 AE/L 0 5 0 -12EI/L3 0 12EI/L3 6 0 6EI/L2 0 12EI/L3 = k55 6 - 6EI/L2 24
  • 25. 3 i 1 j 2 k33 = 4EI/L d3 = 1 E, I, A, L 2EI/L = k63 4 6 5 4EI/L = k66 2EI/L = k36 d6 = 1 k23 = 6EI/L2 6EI/L2 = k53 k26 = 6EI/L2 6EI/L2 = k56 1 3 4 5 6 1 AE/L 0 0 - AE/L 0 0 2 [k] = 2 0 12EI/L3 6EI/L2 0 - 12EI/L3 6EI/L2 3 0 6EI/L2 4EI/L 0 - 6EI/L2 2EI/L 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
  • 26. • Member Equilibrium Equations Mi Fxi i j E, I, A, L AE/L AE/L x δ i 1 Mj Fyj AE/L AE/L + 6EI/L2 6EI/L2 1 6EI/L2 1 x ∆i 12EI/L3 12EI/L3 + + 12EI/L3 4EI/L x δj 1 + 6EI/L2 = Fyi Fxj 2EI/L 1 x ∆j 12EI/L3 4EI/L 2EI/L x θi x θj 1 6EI/L2 MFi FFxi FFyi + 6EI/L2 6EI/L2 FFyj FFxj 6EI/L2 MFj 26
  • 27. Fxi = ( AE / L)δ i + Fyi = (0)δ i + (0)δ i Mxi = Fxj = (−AE / L)δ i Fyj = (0)δ i Mj = (0)δ i Fxi    Fyj  Mi   = Fxj  Fyj    M j    (0) ∆i (0)θ i + (−AE / L)δ j + (0) ∆j + (0)θ j + F Fxi (12EI / L3 ) ∆i (6EI / L2 )θ i (0)δ j (−12EI / L3 ) ∆j (6EI / L2 )θ j F Fyi (6EI / L2 ) ∆i (0) ∆i (4EI / L)θ i (0)θ i (0)δ j ( AE / L)δ j (−6EI / L2 ) ∆j (0) ∆j (2EI / L)θ j (0)θ j MiF F Fxi (0)δ j (0) ∆j (−6EI / L2 )θ j F Fyj (0)δ j (−6EI / L2 ) ∆j (4EI / L)θ j MjF (−12EI / L3 ) ∆i (−6EI / L2 )θ i (6EI / L2 ) ∆i (2EI / L)θ i 0  AE/ L  0 12 EI/ L3   0 6EI/ L2  0 − AE/ L  0 − 12 EI/ L3  6EI/ L2  0  0 6EI/ L2 4 EI/ L 0 − 6EI/ L2 2EI/ L − AE/ L 0 − 12 EI/ L3 0 − 6EI/ L2 0 0 AE/ L 0 12 EI/ L3 − 6EI/ L2 0 Stiffness matrix F 0  δ i   Fxi     F  6EI/ L2   ∆i   Fyi  2EI/ L  θ i  MiF    +  F  0  δ j   Fxj  F − 6EI/ L2   ∆ j   Fyi     F  4 EI/ L  θ j  M j      Fixed-end force matrix [q] = [k][d] + [qF] End-force matrix Displacement matrix 27
  • 28. 6x6 Stiffness Matrix δi [k ]6×6 = ∆i θi Ni  AE/ L 0 Vi  0 12 EI/ L3  Mi  0 6EI/ L2  Nj − AE/ L 0 Vj  0 − 12 EI/ L3  6EI/ L2 Mj  0  δj ∆j − AE/ L 0 − 12 EI/ L3 0 0 6EI/ L2 4 EI/ L 0 − 6EI/ L2 0 AE/ L 0 − 6EI/ L2 0 12 EI/ L3 2EI/ L 0 − 6EI/ L2 θj 0  6EI/ L2   2EI/ L   0  − 6EI/ L2   4EI/ L   4x4 Stiffness Matrix ∆i [k ]4×4 = Vi  12 EI/ L3  Mi  6EI/ L2 Vj − 12 EI/ L3  2 Mj  6EI/ L θi 6EI/ L2 4 EI/ L − 6EI/ L2 2EI/ L ∆j − 12 EI/ L3 − 6EI/ L2 12 EI/ L3 − 6EI/ L2 θj 6EI/ L2   2EI/ L  − 6EI/ L2   4 EI/ L  28
  • 29. 2x2 Stiffness Matrix θi [k ]2×2 = Mi Mj θj 4 EI / L 2EI / L    2EI / L 4 EI / L Comment: - When use 4x4 stiffness matrix, specify settlement. - When use 2x2 stiffness matrix, fixed-end forces must be included. 29
  • 30. General Procedures: Application of the Stiffness Method for Beam Analysis P2y P M2 M1 2 Global 1 4 1 6 3 2 2 1 3 2´ Local w 4´ 5 2´ 1 1´ 4´ 2 3´ 1´ 3´ 30
  • 31. P2y P M2 M1 2 Global 1 4 1 6 3 2 2 1 3 2 Member w 4 5 4 1 6 2 1 3 3 5 2´ 4´ 2´ 4´ Local 1 1´ 2 3´ 1´ 3´ 31
  • 32. P2y P M2 M1 2 4 1 Global w 1 3 2 2 1 6 3 2´ 4´ Local 5 2´ 1 1´ 4´ 2 3´ 1´ 3´ P (q´F 2)1 [FEF] 1 (q´F1 )1 w (q´F1)2 (q´F4 )2 (q´F4)1 (q´F3)1 2 (q´F2)2 (q´F3 )2 32
  • 33. 2 Global 4 1 1 3 2´ 3 2 2 1 6 5 4´ Local 1 1´ 3´ [q] = [T]T[q´] q1  q   2 = q3     q4  1 1 2 3 4 1´ 1 0  0  0 2´ 0 1 0 0 3´ 0 0 1 0 4´ 0 0  0  1 q1'  q   2'  q3'     q4 '  [k] = [T]T[q´] [T] 33
  • 34. 2 Global 4 1 1 6 2 1 3 2 3 5 2´ Local 4´ 2 1´ 3´ [q] = [T]T[q´] q3  q   4 = q5    q6  2 3 4 5 6 1´ 1 0  0  0 2´ 0 1 0 0 3´ 0 0 1 0 4´ 0 0  0  1 q1'  q   2'  q3'     q4 '  [k] = [T]T[q´] [T] 34
  • 35. 2 4 1 1 6 2 1 3 2 3 5 Stiffness Matrix: 1 2 3 4 1 [k]1 = 1 2 1 3 2 4 3 3 [k]2 = 4 4 5 6 [K] = 3 4 5 6 2 3 4 5 6 Member 1 Node 2 Member 2 5 6 35
  • 36. 2 1 Joint Load Qk 1 6 2 2 1 M Q 2 1z  -P2y Q3    M3z Q 4   = Q1   Q 5    Q6     Reaction Qu 4 3 3 Du=Dunknown 5 2 3 4 1 5 6 2   3  4   1  5    6  KAA KBA   KAB      KBB    D2  Q 2F  D   F   3  Q 3  F D 4  Q 4   0 +  F  D1  Q1  D5 0  Q F   0   5F  D6  Q 6      Dk = Dknown [Q k ] = [K AA ][Du ] + [Q AF ] QFA QFB F [Du ] = [K AA ]−1 + ([Qk ] − [QA ]) [ ] Member Force : [q ] = [k ][d ] + q F 36
  • 37. 2 Global: 4 1 1 3 5 (qF2)1 (qF [FEF]       5   6   1 2 3 4 2 3 5 6 Member 1  Q2 = M 1  2 Q = − P   2y  = 3  3   Q4 = M 2    4  (q´F6 )2 (qF3)2 4 Node 2 Member 2 2 3 4    Node 2   w 2 (qF3)1 (qF1 )1 Q1   M  Q 2 1  Q 3 -P2y   M =  Q 4 2  Q 5    Q6     4)2 (qF4)1 1 1 3 2 2 P 1 6           D1 0   D2   D3   + D4   D5 0  0 D6    D2  Q 2F  D   F   3  + Q 3  F D4  Q 4      (q´F5 )2   Q1F   Q 2F   F F F Q 3 = (q 3 ) 1 + (q 3 ) 2   F  Q 4 = (q 4F ) 1 + (q 4F ) 2     Q 5F   Q 6F     37
  • 38. 2 1 4 1 3 2 2 D2    D3  = 3    D4  4    3 3 2 2 1 6 5 4    Node 2     Q2 = M 1  Q2F      Q3 = − P2y  − Q3F       Q = M  Q F   2   4   4 38
  • 39. 2 4 1 1 6 2 1 3 2 3 5 P qF2 qF4 1 qF1 [FEF] qF3 Member 1: 1 q 1    2 q 2  = q 3  3   4 q 4  1       2 3 k1 4   d1 = 0    d 2 = D2   +  d 3 = D3     d 4 = D4  q 1F   F q 2  F q 3   F q 4    39
  • 40. 2 4 1 1 6 2 1 3 2 3 5 w qF6 qF4 2 qF3 [FEM] qF5 Member 2:  q3  q   4 q5    q6  1 = 2 3 4 1       2 3 k1 4       F q3   d 3 = D3   F d = D  q 4 4  +  4 F q5   d5 = 0   F   d6 = 0   q6     40
  • 41. Example 1 For the beam shown, use the stiffness method to: (a) Determine the deflection and rotation at B. (b) Determine all the reactions at supports. (c) Draw the quantitative shear and bending moment diagrams. 10 kN 1 kN/m B A 9m C 1.5 m 3m 41
  • 42. 10 kN 1 kN/m C B A 1.5 m 3m 9m 3 2 Global 1 2 3 2 Members 1 1 2 2 1 10 kN 1 kN/m 1.5 m [FEF] wL2/12 = 6.75 9m 1 wL2/12 = 6.75 PL/8 = 3.75 1.5 m 2 PL/8 = 3.75 42
  • 43. 3 2 1 1 2 9m 3m Stiffness Matrix: θi Mi Mj [k ]2×2 = 4 EI / L 2EI / L    2EI / L 4 EI / L 3 [k]1= EI θj 2 2 4/9 2/9 3 2/9 4/9 2 2 [K] = EI 4/3 2/3 2 2/3 [k]2 = EI 1 4/3 1 1 (4/9)+(4/3) 2/3 2 2/3 1 4/3 43
  • 44. 3 2 10 kN 1 1 kN/m A 6.75 6.75 9m B 3.75 1.5 m 1.5 m C Equilibrium equations: MCB = 0 MBA + MBC = 0 Global Equilibrium: [Q] = [K][D] + [QF] 2 2 MBA + MBC = 0 1 MCB = 0 θB θC = = EI 1 2 (4/9)+(4/3) 2/3 θB 1 2/3 θC 4/3 + -6.75 + 3.75 = -3 -3.75 0.779/EI 2.423/EI 44
  • 45. 3 2 1 1 kN/m [FEF] 9m wL2/12 = 6.75 1 wL2/12 = 6.75 Substitute θB and θC in the member matrix, Member 1 : [q]1 = [k]1[d]1 + [qF]1 3 3 MAB 2 MBA = EI 2 0 4/9 2/9 θA 2/9 4/9 = θB 0.779/EI 6.75 + -6.75 1 kN/m 4.56 kN = -6.40 6.40 kN•m 6.92 kN•m 9m 6.92 1 4.44 kN 45
  • 46. 10 kN 1 2 1.5 m 2 PL/8 = 3.75 1.5 m 2 PL/8 = 3.75 [FEF] Substitute θB and θC in the member matrix, Member 2 : [q]2 = [k]2[d]2 + [qF]2 2 2 MBC 1 MCB = EI 1 4/3 2/3 θB = 0.779/EI 2/3 4/3 θC = 2.423/EI + 3.75 -3.75 6.40 = 0 10 kN 6.40 kN•m 2 7.13 kN 2.87 kN 46
  • 47. 10 kN 1 kN/m 6.40 6.40 6.92 4.56 kN 4.44 kN 1 kN/m 6.92 kN•m 7.13 kN θB = +0.779/EI 10 kN 2.87 kN θC = +2.423/EI C A 4.56 kN B 9m 11.57 kN 2.87 kN 1.5 m 1.5 m 7.13 4.56 V (kN) x (m) 4.56 m -4.44 M (kN•m) -6.92 -2.87 4.32 3.48 x (m) -6.40 47
  • 48. Example 2 For the beam shown, use the stiffness method to: (a) Determine the deflection and rotation at B. (b) Determine all the reactions at supports. 20 kN 9kN/m 40 kN•m EI 2EI A B 4m C 4m 48
  • 49. 2 4 1 3 2EI 1 1 [k ] = Vi  12 EI/ L3 Mi  6EI/ L2  Vj  − 12 EI/ L3 Mj  6EI/ L2  EI θi ∆j 2 4 EI/ L − 6EI/ L 2EI/ L 2 − 12 EI/ L − 6EI/ L2 12 EI/ L3 − 6EI/ L2 6EI/ L   2EI/ L  − 6EI/ L2   4 EI/ L  3 1 0.375EI 2 0.75EI 3 4 2 3 4 -0.75EI EI -0.375EI -0.75EI 0.375EI -0.75EI 0.75EI EI 3 0.5625 -0.375 [K] = EI 4 -0.375 3 [k]2 0.75EI - 0.375EI 0.75EI 2EI 4 2 [k]1 1 3 θj 3 6EI/ L 5 2 2 Use 4x4 stiffness matrix, ∆i 6 -0.75EI 2EI 3 3 4 5 6 4 5 6 0.1875EI 0.375EI - 0.1875EI 0.375EI 0.375EI EI -0.375EI 0.5EI -0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI 0.5EI -0.375EI EI 49
  • 50. 20 kN 9kN/m 12 kN•m 18 kN 4m 40 kN•m 12 kN•m 18 kN 2 4 1 Global: 4 Q4 = 40 D3 D4 = 2 0.5625 -0.375 4 -0.375 3 4 3 5 EI 2 3 = EI 6 3 2EI 1 1 [Q] = [K][D] + [QF] 3 Q3 = -20 4m 3 D3 D4 18 + 3 -12 4 -61.09/EI 9.697/EI 50
  • 51. 2 4 1 3 2EI 1 1 9 kN/m 12 kN•m A B 1 2 [qF]1 18 kN Member 1: 12 kN•m 18 kN [q]1 = [k]1[d]1 + [qF]1 1 q1 1 q2 2 q3L 3 q4L 4 2 3 4 12(2EI)/430.75EI - 0.375EI 0.75EI 0.75EI 2EI -0.75EI EI -0.375EI -0.75EI 0.375EI -0.75EI 0.75EI EI -0.75EI 2EI 9 kN/m A 67.51 kN•m 48.18 kN 1 [q]1 d1 = 0 18 48.18 d2 = 0 12 67.51 d3 = -61.09/EI 18 -12.18 d4 = 9.697/EI -12 53.21 53.21 kN•m B 12.18 kN 51
  • 52. 4 6 3 5 EI 2 2 3 Member 2: [q]2 = [k]2[d]2 + [qF]2 3 q3R 3 q4R 4 q5 5 q6 6 4 5 6 0.1875EI 0.375EI - 0.1875EI 0.375EI 0.375EI EI -0.375EI 0.5EI -0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI 0.5EI -0.375EI 7.82 kN 0 -7.818 d4 = 9.697/EI 0 -13.21 d5 = 0 0 7.818 d6 = 0 0 -18.06 18.06 kN•m 13.21 kN•m B EI d3 =-61.09/EI 2 [q]2 C 7.82 kN 52
  • 53. 9 kN/m A 67.51 kN•m 48.18 kN 53.21 kN•m B B 1 12.18 kN [q]1 18.06 kN•m 13.21 kN•m C 2 7.82 kN 7.82 kN [q]2 20 kN 9kN/m 40 kN•m D3 = ∆B = -61.09/EI 67.51 kN•m 48.18 kN 48.18 4m D4 = θB = +9.697/EI 4 m + V (kN) 18.06 kN•m 7.818 kN 12.18 x (m) M (kN•m) + - -67.51 -7.818 -7.818 53.21 13.21 - x (m) -18.08 53
  • 54. Example 3 For the beam shown, use the stiffness method to: (a) Determine the deflection and rotation at B. (b) Determine all the reactions at supports. 20 kN 9kN/m 10 kN 40 kN•m EI 2EI A B 4m C 2m 2m 54
  • 55. 20 kN 9kN/m A 40 kN•m 2EI 4m B 4 1 3 1 1 18 kN ∆i = 1 Vi  12 EI/ L3 Mi  6EI/ L2  Vj  − 12 EI/ L3 Mj  6EI/ L2  5 3 10 kN 12 kN•m [FEF] 6 2 2 9 kN/m 12 kN•m [k ]4×4 C EI 2m 2m 2 Global 10 kN θi ∆j 6EI/ L2 − 12 EI/ L3 − 6EI/ L2 12 EI/ L3 − 6EI/ L2 − 6EI/ L2 2EI/ L 5 kN•m 2 18 kN 4 EI/ L 5 kN•m 5 kN θj 6EI/ L2   2EI/ L  − 6EI/ L2   4 EI/ L  5 kN 55
  • 56. 2 4 1 3 1 1 1 1 [k]1 = 2 3 4 0.75EI 0.75EI 0.1875EI 4 0.375EI 6 3 2m 2m 3 4 2EI -0.75EI EI 3 -0.375EI -0.75EI 0.375EI -0.75EI 3 5 2 12(2EI)/43 0.75EI - 0.375EI 0.75EI 3 [k]2 = 5 2 4m 2 6 EI 4 6 3 -0.75EI 2EI/L 5 4 [K] = EI 6 0.5625 -0.375 0.375 4 6 -0.375 0.375 3 0.5 0.5 1 0.375EI - 0.1875EI 0.375EI EI -0.375EI 0.5EI -0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI 0.5EI -0.375EI EI 56
  • 57. 20 kN 9kN/m 10 kN 40 kN•m A C 6 2 B4 1 3 1 1 5 2 2 10 kN 9 kN/m 12 kN•m 12 kN•m [FEF] 1 18 kN 5 kN•m 3 4 5 kN 5 kN 6 3 VBL+VBR = -20 MBA+MBC = 40 = EI MCB = 0 ∆B θB θC 5 kN•m 2 18 kN Global: [Q] = [K][D] + [QF] 3 0.5625 -0.375 0.375 ∆B 4 6 -0.375 0.375 3 0.5 0.5 1 θB θC + 18 + 5 = 23 -12 + 5 = -7 -5 -116.593/EI = -7.667/EI 52.556/EI 57
  • 58. 2 4 1 3 1 1 9 kN/m 12 kN•m 2 12 kN•m 18 kN 1 18 kN [FEF] [q]1 = [k]1[d]1 + [qF]1 Member 1: 1 2 3 4 VA 1 0.375EI 0.75EI - 0.375EI 0.75EI 0 18 55.97 MAB 2 0.75EI 0 12 91.78 VBL MBA = 3 4 2EI -0.75EI EI -0.375EI -0.75EI 0.375EI -0.75EI ∆B =-116.593/EI 0.75EI EI -0.75EI 2EI/L θB =-7.667/EI + 18 -12 = -19.97 60.11 9kN/m 91.78 kN•m 4m 55.97 kN 1 60.11 kN•m 19.97 kN 58
  • 59. 4 6 3 10 kN 5 3 2 2 2 5 kN Member 2: VBR 3 0.1875EI MBC 4 4 0.375EI MCB = 5 6 [FEM] 5 kN [q]1 = [k]1[d]1 + [qF]1 3 VC 5 kN•m 5 kN•m 5 6 0.375EI - 0.1875EI 0.375EI EI -0.375EI 0.5EI -0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI 0.5EI -0.375EI EI ∆B =-116.593/EI 5 -0.0278 θB =-7.667/EI 5 -20.11 0 θC =52.556/EI + 5 -5 = 10.03 0 10 kN 20.11 kN•m 2m 2m 2 0.0278 kN 10.03 kN 59
  • 60. 9kN/m 91.78 kN•m 10 kN 20.11 kN•m 2m 1 4m 19.97 kN 55.97 kN 60.11 kN•m 20 kN 9kN/m 2m 2 0.0278 kN 10.03 kN 10 kN θC = 52.556/EI 40 kN•m 91.78 kN•m 55.97 kN θB = -7.667/EI 10.03 kN ∆B = -116.593/EI 4m 55.97 2m 2m 19.97 + V (kN) 60.11 M (kN•m) -91.78 + - -0.0278 -10.03 x (m) -10.03 20.11 x (m) 60
  • 61. Example 4 For the beam shown: (a) Use the stiffness method to determine all the reactions at supports. (b) Draw the quantitative free-body diagram of member. (c) Draw the quantitative shear diagram, bending moment diagram and qualitative deflected shape. Take I = 200(106) mm4 and E = 200 GPa and support B settlement 10 mm. 40 kN 6 kN/m B A 2EI 8m EI ∆B = -10 mm 4m C 4m 61
  • 62. 1 2 3 2EI EI 2 1 θi Use 2x2 stiffness matrix: [k ]2×2 = 1 [k]1 = EI 8 8 4 2 4 8 4 EI / L 2EI / L    2EI / L 4 EI / L 2 1 Mi Mj θj 2 [k]2 = 2 [K] = EI 8 12 2 4 2 3 2 4 2 3 2 3 2 EI 8 3 4 62
  • 63. 40 kN 6 kN/m 1 B A 2 C EI ∆B = -10 mm 4m 4m 2EI 8m 2 1 40 kN 6 kN/m [FEM]load wL2/12 = 32 kN•m 1 32 kN•m 10 mm 6(2EI)∆/L2 = 75 kN•m 2 Q2 = 0 Q3 = 0 D2 D3 40 kN•m 37.5 kN•m 10 mm 6(EI)∆/L2 = 37.5 kN•m [Q] = [K][D] +2[QF] Global: 2 PL/8 = 40 kN•m 75 kN•m 1 [FEM]∆ 3 3 2 12 2 D2 3 2 4 D3 EI = 8 -61.27/EI = 185.64/EI + -32 + 40 + 75 -37.5 = 45.5 -40 - 37.5 = -77.5 rad rad 63
  • 64. 1 2 6 kN/m [FEM]load 2EI wL2/12 = 32 kN•m 1 1 32 kN•m 75 kN•m 1 [FEM]∆ 10 mm 6(2EI)∆/L2 = 75 kN•m Member 1: [q]1 = [k]1[d]1 + [qF]1 1 q1 q2 EI = 8 2 1 8 4 d1 = 0 2 4 8 d2 = -61.27/EI 76.37 kN•m 6 kN/m 32 + 75 = 107 + -32 + 75 = 43 76.37 kN•m = -18.27 kN•m 18.27 kN•m 8m 1 31.26 kN 16.74 kN 64
  • 65. 2 3 40 kN [FEM]load 2 PL/8 = 40 kN•m 2 40 kN•m 37.5 kN•m 2 [FEM]∆ 10 mm 6(EI)D/L2 = 37.5 kN•m Member 2: [q]2 = [k]2[d]2 + [qF]2 2 q2 q3 = EI 8 3 2 4 2 d2 = -61.27/EI 3 2 4 d3 = 185.64/EI 40 - 37.5 = 2.5 + -40 - 37.5 = -77.5 18.27 kN•m = 0 kN•m 40 kN 18.27 kN•m 2 22.28 kN 17.72 kN 65
  • 66. 2 6 4 2EI EI 2 1 1 Alternate method: Use 4x4 stiffness matrix [k]1 3 4 1.5 -0.375 1.5 8 -1.5 4 -1.5 0.375 -1.5 4 1 -1.5 2 1 12(2)/82 EI 2 1.5 = 8 3 -0.375 4 1.5 5 3 3 = 3 12/82 4 0.75 EI 8 5 4 0.75 2 1 2 0.375 1.5 1.5 8 -0.375 -1.5 EI 3 = 8 4 5 -0.375 -1.5 0.5625 1.5 0 4 0 -0.75 -0.1875 12 -0.75 -0.75 0.1875 6 0 0 0.75 2 -0.75 4 5 6 -0.1875 0.75 -0.75 2 0.1875 -0.75 -0.75 4 2 -0.75 1.5 4 5 2 1 [K] 4 0.75 -0.1875 -0.75 6 8 3 4 [k]2 6 0 0 0 0 -0.75 -0.1875 0.75 66
  • 67. 40 kN 6 kN/m 2 B A C EI ∆B = -10 mm 2EI 8m 4m 1 32 kN•m 40 kN•m Q6= 0 Q4 = 0 Q6= 0 D4 D6 5 D4 4 6 D6 12 2 D4 2 4 D6 12 EI 4 = 8 6 2 4 20 kN 20 kN 2 EI 4 = 8 6 = 40 kN•m 2 1 24 kN 24 kN Global: [Q] = [K][D] + [QF] 4 6 Q4 = 0 5 3 40 kN 32 kN•m [FEF] 2 1 4m 6 kN/m 6 4 200 × 200 4 +( ) 8 6 + -0.75 D = -0.01 5 -0.75 (200x200/8)(-0.75)(-0.01) = 37.5 (200x200/8)(0.75)(-0.01) = -37.5 -61.27/EI = -1.532x10-3 rad + -40 8 -40 rad 185.64/EI = 4.641x10-3 + 8 67
  • 68. 2 4 1 32 kN•m ∆B = -10 mm 1 6 kN/m 32 kN•m [FEF]load 3 1 24 kN 24 kN Member 1: [q]1 = [k]1[d]1 + [qF]1 2 1 1 12(2)/82 2 1.5 q1 q2 3 4 1.5 -0.375 1.5 d1 = 0 24 8 -1.5 4 d2 = 0 32 (200x200) 3 8 -0.375 -1.5 0.375 -1.5 q4 4 1.5 4 -1.5 8 q1 31.26 q3 q2 q3 q4 = = kN 76.37 kN•m 16.74 kN -18.27 kN•m 76.37 kN•m + d3 = -0.01 -32 d4 = -1.532x10-3 6 kN/m 24 18.27 kN•m 8m 1 31.26 kN 16.74 kN 68
  • 69. 40 kN 6 4 40 kN•m 40 kN•m -10 mm = ∆B [FEF] 2 3 5 2 20 kN 20 kN Member 2: 3 3 12/82 4 0.75 q3 q4 q5 = 4 (200x200) 5 8 0.75 4 -0.1875 -0.75 q6 6 q3 22.28 kN q4 18.27 kN•m 17.72 kN•m 6 -0.1875 -0.75 0.75 2 0.1875 -0.75 kN 0 5 q5 q6 = 0.75 2 -0.75 4 d3 = -0.01 20 d4 = -1.532x10-3 40 d5 = 0 + d6 = 4.641x10-3 20 -40 40 kN 18.27 kN•m 2 22.28 kN 17.72 kN 69
  • 70. 40 kN 6 kN/m B 76.37 kN•m EI ∆B = -10 mm 17.72 kN 16.74 + 22.28 kN 8m 4m 4m A 2EI 31.26 kN V (kN) C 31.26 + - 5.21 m 22.28 + -16.74 -17.72 x (m) 70.85 M (kN•m) 5.06 - + - -18.27 x (m) -76.37 Deflected Curve ∆B = -10 mm d4 = θB = -1.532x10-3 rad D6 = θC = 4.641x10-3 rad 70
  • 71. Example 5 For the beam shown: (a) Use the stiffness method to determine all the reactions at supports. (b) Draw the quantitative free-body diagram of member. (c) Draw the quantitative shear diagram, bending moment diagram and qualitative deflected shape. Take I = 200(106) mm4 and E = 200 GPa and support C settlement 10 mm. 4 kN 0.6 kN/m 20 kN•m B A EI 2EI 8m 4m C ∆C = -10 mm 4m 71
  • 72. 2 2EI 1 •Member stiffness matrix [k]4x4 1 1 12(2)/82 EI 2 1.5 = 8 3 -0.375 4 1.5 [k]1 6 4 EI 2 1 5 3 3 4 1.5 -0.375 1.5 8 -1.5 4 -1.5 0.375 -1.5 4 -1.5 2 3 EI = 8 6 8 4 3 12/82 4 0.75 5 -10 mm [k]2 0.75 4 -0.1875 -0.75 0.75 2 5 6 -0.1875 0.75 -0.75 2 0.1875 -0.75 -0.75 4 • Global: [Q] = [K][D] + [QF] 3 Q3 Q4 Q6 3 EI 4 = 8 6 4 6 0.5625 -0.75 -0.75 12 0.75 2 D3 D4 0.75 2 4 D6 5 QF3 3 -0.1875 200 × 200 +( ) 4 -0.75 D5 = -0.01 + QF4 8 QF6 6 -0.75 72
  • 73. 4 kN 0.6 kN/m 20 kN•m B A EI 2EI 8m 4m 10 mm 4m D3 D4 D6 = 5 4 kN•m 2 2 kN 3 Q6 = 20 4 kN•m 1 3 EI 4 = 8 6 EI 3 4 kN 2.4 kN 2.4 kN Global: [Q] = [K][D] + [QF] Q3 = 0 Q4 = 0 6 2 1 1 3.2 kN•m [FEF] 4 2EI C 0.6 kN/m 3.2 kN•m 2 4 6 0.5625 -0.75 -0.75 12 0.75 2 0.75 2 4 2 kN D3 9.375 D4 + 37.5 + D6 37.5 2.4+2 = 4.4 -3.2+4 = 0.8 -4.0 -377.30/EI = -9.433x10-3 m -61.53/EI = -1.538x10-3 rad +74.50/EI = +1.863x10-3 rad 73
  • 74. 2 4 0.6 kN/m 3.2 kN•m 1 1 [FEF]load 3 3.2 kN•m 8m 1 2.4 kN 2.4 kN Member 1: [q]1 = [k]1[d]1 + [qF]1 2 1 q1 q2 q3 q4 1.5 -0.375 1.5 d1 = 0 2.4 8 -1.5 4 d2 = 0 3.2 200× 200 3 8 -0.375 -1.5 0.375 -1.5 d3 = -9.433x10-3 1.5 4 -1.5 8 d4 = -1.538x10-3 q1 q3 4 4 = q4 q2 1 12(2)/82 2 1.5 3 8.55 = kN 43.19 kN•m -3.75 kN 6.03 kN•m 43.19 kN•m 0.6 kN/m + 2.4 -3.2 6.03 kN•m 8m 1 8.55 kN 3.75 kN 74
  • 75. 4 kN 6 4 4 kN•m 4 kN•m [FEF] 2 3 5 8m 10 mm 2 2 kN 2 kN Member 2: 3 q3 q4 q5 200× 200 = 8 q6 4 3 12/82 4 0.75 5 6 0.75 4 -0.1875 -0.75 0.75 q3 3.75 kN q4 -6.0 kN•m 0.25 kN•m 6 -0.75 0.75 d3 = -9.433x10-3 2 2 -0.1875 d4 = -1.538x10-3 4 0.1875 -0.75 -0.75 4 q5 q6 = d5 = -0.01 + d6 = 1.863x10-3 2 -4 4 kN kN 20.0 2 5 6.0 kN•m 20 kN•m 2 3.75 kN 0.25 kN 75
  • 76. 4 kN 0.6 kN/m 20 kN•m B A EI 2EI 8m 0.6 kN/m 43.19 kN•m 4m Deflected Curve 4 kN 20 kN•m 6.0 kN•m 3.75 kN 8.55 V (kN) M (kN•m) 10 mm 4m 6.03 kN•m 8m 1 8.55 kN C + 2 3.75 6 -0.25 21 + -43.19 0.25 kN 3.75 kN D3 = -9.433 mm D4 = -1.538x10-3 rad x (m) 20 x (m) D5 = -10 mm D6 = +1.863x10-3 rad 76
  • 77. Internal Hinges Example 6 For the beam shown, use the stiffness method to: (a) Determine all the reactions at supports. (b) Draw the quantitative shear and bending moment diagrams and qualitative deflected shape. E = 200 GPa, I = 50x10-6 m4. 30 kN 9kN/m Hinge EI 2EI A 4m B C 2m 2m 77
  • 78. 3 4 2 2EI A 1 1 4m EI B 2m 2m θi Use 2x2 stiffness matrix, [k ]2×2 = 3 [k]1 = 2EI EI 1 EI Mi Mj 2EI θj 4 EI / L 2EI / L    2EI / L 4 EI / L  1 3 C 2 2 [k]2 = 2 1EI 4 0.5EI 1 [K] = EI 2.0 0.0 1EI 0.0 2 0.5EI 2 1 4 1.0 78
  • 79. 30 kN 9kN/m Hinge 3 4 2 A 1 [FEF] 1 A = 0.0 EI 15 kN•m C 2 1 2 2.0 0.0 D1 2 0.0 1.0 D2 0.0006 = 30 kN 15 kN•m B 1 D1 D2 B 12 kN•m B Global matrix: 0.0 1 9kN/m 12 kN•m C 2 + -12 15 rad -0.0015 rad 79
  • 80. 3 12 kN•m 1 A [FEF] 1 9kN/m 1 A B 12 kN•m B Member 1: 3 q3 q1 = 1 3 2EI EI 1 EI 2EI d3 = 0.0 d1 = 0.0006 + 12 -12 = 18 0.0 9 kN/m A 18 kN•m 22.5 kN 1 B 13.5 kN 80
  • 81. 4 B C 2 2 15 kN•m B 30 kN 15 kN•m C 2 Member 2: 2 q2 q4 = 2 4 1EI 0.5EI 4 0.5EI 1EI d2 = -0.0015 d4 = 0.0 + 15 -15 = 0.0 -22.5 30 kN C B 9.37 kN 22.5 kN•m 20.63 kN 81
  • 82. 30 kN 9 kN/m A 18 kN•m B 13.5 kN 13.5 kN 22.5 kN 9.37 kN C B 22.87 kN 22.5 kN•m 20.63 kN 9.37 kN 30 kN 9kN/m Hinge EI θBR= -0.0015 rad 2EI θBL= 0.0006 rad A 4m B 2m C 2m 22.5 V (kN) 9.37 + 2.5 m M (kN•m) -13.5 10.13 + - -18 x (m) -20.63 18.75 + - x (m) -22.5 82
  • 83. Example 7 For the beam shown, use the stiffness method to: (a) Determine all the reactions at supports. (b) Draw the quantitative shear and bending moment diagrams and qualitative deflected shape. E = 200 GPa, I = 50x10-6 m4. 20 kN 9kN/m 2EI A B 4m EI Hinge C 4m 83
  • 84. 5 7 4 1 3 2EI 1 A 2 2 B 4 1 5 4 0.375EI 5 [k]1 = 1 0.75EI 2EI -0.375EI -0.75EI 2 0.75EI EI 1 [k]2 = 0.1875EI 3 0.375EI 6 7 C 2 0.75EI - 0.375EI 0.75EI -0.75EI EI 0.375EI -0.75EI -0.75EI 2EI/L 3 6 1 2 3 7 1 0.5625 -0.75 0.375 EI 2 -0.75 2.0 0 3 [K] = 1 6 EI 0.375 0.0 1.0 0.375EI - 0.1875EI 0.375EI EI -0.375EI 0.5EI -0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI 0.5EI -0.375EI EI 84
  • 85. 20 kN B 9kN/m A C 5 4 1 3 2EI 1 A 12 kN•m Global: B 1 [FEM] 1 Q1 = -20 2 2 C 12 kN•m 18 kN 3 1 = 0.5625 -0.75 0.375 D1 EI 2 -0.75 2.0 0 D2 3 Q2 = 0.0 0.375 0.0 1.0 D3 Q3 = 0.0 D1 D2 D3 6 EI 2 B 9kN/m A 18 kN 7 18 + -12 0.0 -0.02382 m = -0.008333 rad 0.008933 rad 85
  • 86. 5 9kN/m 4 2EI 2 1 A 1 12 kN•m A 18 kN B 12 kN•m B 1 [FEF] 18 kN Member 1: 4 1 5 q4 4 0.375EI q5 5 0.75EI 2EI 1 -0.375EI -0.75EI 0.75EI EI q1 q2 = 2 2 0.75EI - 0.375EI 0.75EI -0.75EI EI d4 = 0.0 18 d5 = 0.0 12 0.375EI -0.75EI d1 = -0.02382 -0.75EI d2 = -0.00833 107.32 kN•m A 2EI/L + 18 -12 44.83 = 107.32 -8.83 0.0 9kN/m 1 B 8.83 kN 44.83 kN 86
  • 87. 7 1 B 6 EI 2 3 C Member 2: 1 q1 1 0.1875EI q3 3 0.375EI q6 =6 q7 7 6 3 7 0.375EI - 0.1875EI 0.375EI EI -0.375EI 0.5EI -0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI 0.5EI EI -0.375EI C B 2 11.16 kN d1 = -0.02382 0 d3 = 0.008933 0 d6= 0.0 d7= 0.0 + 0 0 -11.16 = 0.0 11.16 -44.66 44.66 kN•m 11.16 kN 87
  • 88. 20 kN 9kN/m θBL= -0.008333 rad B A θBR= 0.008933 rad C 4m 4m 9kN/m 107.32 kN•m A B 1 44.83 kN C B 2 8.83 kN 44.66 kN•m 11.16 kN 11.16 kN 44.83 V (kN) + 8.83 x (m) -11.16 -11.16 M (kN•m) -107.32 - x (m) -44.66 88
  • 89. Example 8 For the beam shown, use the stiffness method to: (a) Determine all the reactions at supports. (b) Draw the quantitative shear and bending moment diagrams and qualitative deflected shape. 40 kN•m at the end of member AB. E = 200 GPa, I = 50x10-6 m4. 30 kN 9kN/m A Hinge 2EI 4m 40 kN•m EI B 2m C 2m 89
  • 90. 3 4 2 2EI A 1 1 4m C 2 B 2m 2m θj θi Use 2x2 stiffness matrix: [k ]2×2 = 3 [k]1 = EI 2EI EI 1 EI 2EI 4 EI / L 2EI / L    2EI / L 4 EI / L  1 3 Mi Mj 2 [k]2 = 2 1EI 4 0.5EI 1 [K] = EI 2.0 0.0 1EI 0.0 2 0.5EI 2 1 4 1.0 90
  • 91. 30 kN 9kN/m Hinge A 2EI 3 40 kN•m EI B C 4 2 A 1 [FEM] 1 A = Q2 = 0.0 EI 15 kN•m C 2 1 2 2.0 0.0 D1 2 0.0 1.0 D2 0.0026 = 30 kN 15 kN•m B 1 D1 D2 B 12 kN•m B Global matrix: Q1 = 40 1 9kN/m 12 kN•m C 2 + -12 15 rad -0.0015 rad 91
  • 92. 3 12 kN•m 1 A [FEF] 1 9kN/m 1 A B 12 kN•m B Member 1: 3 q3 q1 = 1 3 2EI EI 1 EI 2EI d3 = 0.0 d1 = 0.0026 + 12 -12 = 38 40 9 kN/m A 38 kN•m B 40 kN•m 37.5 kN 1.5 kN 92
  • 93. 4 B C 2 2 15 kN•m B 30 kN 15 kN•m C 2 Member 2: 2 q2 q4 = 2 1EI 4 0.5EI 4 0.5EI 1EI d2 = -0.0015 d4 = 0.0 + 15 -15 = 0.0 -22.5 30 kN C B 9.37 kN 22.5 kN•m 20.63 kN 93
  • 94. 1.5 kN 30 kN 9 kN/m A 38 kN•m B 40 kN•m 9.37 kN 7.87 kN 1.5 kN 37.5 kN C B 22.5 kN•m 20.63 kN 9.37 kN 30 kN 9kN/m 40 kN•m Hinge A θBL= 0.0026 B rad 4m C θBR=- 0.0015 rad 2m 2m 37.5 V (kN) + 9.37 1.5 x (m) -20.63 M (kN•m) 40 + - -38 18.75 + - x (m) -22.5 94
  • 95. Example 9 For the beam shown, use the stiffness method to: (a) Determine all the reactions at supports. (b) Draw the quantitative shear and bending moment diagrams and qualitative deflected shape. 40 kN•m at the end of member AB. E = 200 GPa, I = 50x10-6 m4 20 kN 9kN/m 2EI A 4m 40 kN•m B EI Hinge C 4m 95
  • 96. 20 kN 9kN/m 2EI A 40 kN•m B EI Hinge 4m C 4m 5 7 1 4 A 12 kN•m [FEM] 1 3 6 C 2 2 B 9 kN/m 12 kN•m 1 18 kN 18 kN [k ]4×4 = ∆i Vi  12 EI/ L3 Mi  2  6EI/ L Vj  − 12 EI/ L3 Mj  6EI/ L2  θi ∆j θj 6EI/ L2 − 12 EI/ L3 − 6EI/ L2 12 EI/ L3 − 6EI/ L2 6EI/ L2   2EI/ L  − 6EI/ L2   4 EI/ L  4 EI/ L − 6EI/ L2 2EI/ L 96
  • 97. 5 7 4 2EI A 1 4 4 [k]1 = 1 0.375EI 5 0.75EI 1 5 6 EI C 2 2 B 2 0.75EI - 0.375EI 0.75EI 1 -0.375EI 2EI -0.75EI 0.75EI EI 2 3 -0.75EI EI 0.375EI -0.75EI -0.75EI 1 2EI 2 3 1 1 3 6 0.375 EI 2 -0.75 2.0 0 0.375 0.0 1.0 7 1 [k]2 = -0.75 3 [K] = 0.5625 0.1875EI 3 0.375EI EI -0.375EI 0.5EI -0.1875EI -0.375EI 0.1875EI -0.375EI 6 7 0.375EI 0.375EI - 0.1875EI 0.375EI 0.5EI -0.375EI EI 97
  • 98. 9kN/m 20 kN 5 4 2EI A 1 C 2 2 B 1 18 kN 18 kN Global: 1 2 3 1 = Q3 = 0.0 D1 0.5625 -0.75 0.375 D1 EI 2 -0.75 2.0 0 D2 3 Q1 = -20 D3 6 EI 12 kN•m [FEF] D2 7 9 kN/m 12 kN•m Q2 = 40 EI 40 kN•m 1 3 0.375 0.0 1.0 D3 18 + -12 0.0 -0.01316 m = -0.002333 rad 0.0049333 rad 98
  • 99. 5 4 2 2EI 1 12 kN•m 1 5 4 0.375EI q5 5 0.75EI 2EI 1 -0.375EI -0.75EI 0.75EI EI q2 2 18 kN [q]1 = [k]1[d]1 + [qF]1 q4 q1 1 18 kN 4 = 12 kN•m [FEF] 1 Member 1: 9 kN/m 2 0.75EI - 0.375EI 0.75EI 87.37 kN•m -0.75EI EI d4 = 0.0 18 d5 = 0.0 12 0.375EI -0.75EI d1 = -0.01316 -0.75EI + d2 = -0.002333 2EI 2EI 49.85 kN 18 -12 49.85 = 87.37 -13.85 40 40 kN•m 1 13.85 kN [q]1 99
  • 100. 7 1 B Member 2: q1 1 0.1875EI q3 3 0.375EI =6 q7 7 C 2 3 [q]2 = [k]2[d]2 + [qF]2 1 q6 6 EI 6 3 7 0.375EI - 0.1875EI 0.375EI d1 = -0.01316 EI -0.375EI 0.5EI -0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI 0.5EI -0.375EI EI d3 = 0.004933 d6= 0.0 d7= 0.0 0 + -6.18 0 0.0 0 0 = 6.18 -24.69 24.69 kN•m 6.18 kN 2 6.18 kN [q]2 100
  • 101. 20 kN 9kN/m 40 kN•m EI θBR = 0.004933 rad C 4m 2EI θBL= -0.002333 rad A B 4m B A 87.37 kN•m 49.85 kN 40 kN•m 13.85 kN C B 24.69 kN•m 6.18 kN 6.18 kN 49.85 V (kN) + 13.85 x (m) - -6.18 -6.18 M (kN•m) + -87.37 40 - x (m) -24.69 101
  • 102. Temperature Effects • Fixed-End Forces (FEF) - Axial - Bending • Curvature 102
  • 103. tan β = • Thermal Fixed-End Forces (FEF) Room temp = TR T TR t F MA FBF Tm> TR FAF Tb > Tt A σaxial A d ct Tt β NA Tm = cb F MB B σaxial B σy Tb - Tt TR B Tl (∆T ) axial = Tm − TR y A Tt + Tb 2 Tm Tt y Tb − Tt ∆T =( ) d d ( ∆T y ) = y ( ∆T ) d β Tb (∆T ) bending = Tb − Tt 103
  • 104. - Axial σaxial σaxial Tt Tm> TR FAF A Tb > Tt TR Tm FBF B (∆T ) axial = Tm − TR FAF = ∫ σ axial dA A = ∫ Eε axial dA = ∫ Eα (∆T ) axial dA = Eα (∆T ) axial ∫ dA = EAα (∆T ) axial F ( Faxial ) A = α (Tm − TR ) AE 104
  • 105. - Bending σy Tt y y F MA A B F MB (∆Ty ) = y ( ∆T ) d β Tb (∆T ) bending = Tb − Tt M = ∫ yσ y dA F A A = ∫ yEε dA = ∫ yEα (∆Ty ) dA = ∫ yEαy ( = Eα ( ∆T ) dA d ∆T ) ∫ y 2 dA d F ( Fbending ) A = α ( Tl − Tu ) EI d 105
  • 106. • Elastic Curve: Bending O´ dθ ρ (dx ) = ρ ( dθ ) y ( dθ ) ds = dx y dx Before deformation y ds´ 1 ρ =( dθ ) dx dθ dx After deformation 106
  • 107. • Bending Temperature Tt O dx Tl > Tu Tb dθ Tt + Tb 2 M Tm = Tu ∆T = Tb - Tt Tt dθ y M ∆T β= d ct cb Tl > Tu Tb Tt ∆T y d y Tb ∆T ) dx d ∆T (dθ ) = α ( )dx d dθ M 1 ∆T ( ) = = α( ) = dx d EI ρ (dθ ) y = αy ( 107
  • 108. Example 10 For the beam shown, use the stiffness method to: (a) Determine all the reactions at supports. (b) Draw the quantitative shear and bending moment diagrams and qualitative deflected shape. Room temp = 32.5oC, α = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4. T2 = 40oC 182 mm A T1 = 25oC 4m EI 2EI B C 4m 108
  • 109. T2 = 40oC 182 mm A T1 = 25oC 4m 2 B C 4m 3 1 1 FEM EI 2EI 2 Mean temperature = (40+25)/2 = 32.5 Room temp = 32.5oC 19.78 kN•m a(∆T /d)EI = 19.78 kN•m oC +7.5 F Fbending = α ( -7.5 oC ∆T 40 − 25 )(2 EI ) = (12 × 10 −6 )( )(2 × 200 × 50) = 19.78 kN • m d 0.182 109
  • 110. 2 19.78 kN•m 19.78 kN•m 182 mm A 1 2 2EI 1 3 B 4m 4m [q] = [k][d] + [qF] Element 1: 2 M2 M1 = 1 2 2 1 q2 1 1 2 q1 1 3 1 EI 1 0.5 q1 3 0.5 1 q3 Element 2: M1 M3 = EI EI C + + -1.978 1.978 (10-3) EI 0 0 0 [M1] = 3EIθ1 + (1.978x10-3)EI θ1 = -0.659x10-3 rad 110
  • 111. 2 19.78 kN•m 19.78 kN•m A 182 mm 2EI 1 3 1 2 EI B 4m C 4m Element 1: 2 M2 M1 = 2 EI 1 2 1 q2 = 0 1 1 2 q1 = -0.659x10-3 Element 2: M1 M3 = + 1 0.5 q1 = -0.659x10-3 3 0.5 1 q3 = 0 19.78 = -26.37 kN•m 6.59 kN•m 3 1 EI 1 -19.78 26.37 kN•m A 6.59 kN•m B 4.95 kN 4.95 kN + 0 0 6.59 kN•m B = -6.59 kN•m -3.30 kN•m 3.3 kN•m C 2.47 kN 2.47 kN 111
  • 112. 26.37 kN•m +7.5 oC A 4.95 kN -7.5 oC B C 2.47 kN 4m 4m 2.47 kN V (kN) 3.3 kN•m x (m) -2.47 -4.95 26.37 M (kN•m) Deflected curve + 6.59 - θB = -0.659x10-3 rad x (m) -3.30 x (m) 112
  • 113. Example 11 For the beam shown, use the stiffness method to: (a) Determine all the reactions at supports. (b) Draw the quantitative shear and bending moment diagrams and qualitative deflected shape. Room temp = 28 oC, a = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4, A = 20(10-3) m2 T2 = 40oC 182 mm A T1 = 25oC 4m EI, AE 2EI, 2AE B C 4m 113
  • 114. 6 5 9 T2 = 40oC A 4 T1 = 25oC 1 3 2 2EI 4m 1 B 8 2 EI 7 C 4m Element 1: (2 AE ) 2(20 ×10 −3 m 2 )(200 ×106 kN / m 2 ) = = 2(106 ) kN / m L ( 4 m) 4(2 EI ) 4 × 2(200 ×106 kN / m 2 )(50 ×10 −6 m 4 ) = = 20(103 ) kN • m L ( 4 m) 2(2 EI ) 2 × 2(200 ×106 kN / m 2 )(50 ×10 −6 m 4 ) = = 10(103 ) kN • m L ( 4 m) 6(2 EI ) 6 × 2(200 ×106 kN / m 2 )(50 ×10 −6 m 4 ) = = 7.5(103 ) kN L2 ( 4 m) 2 12(2 EI ) 12 × 2(200 ×106 kN / m 2 )(50 ×10 −6 m 4 ) = = 3.75(103 ) kN / m L3 ( 4 m) 3 114
  • 115. T2 = 40oC 182 mm A T1 = 25oC EI, AE 2EI, 2AE C B 4m 4m Fixed-end forces due to temperatures F Fbending = α ( ∆T 40 − 25 )(2 EI ) = (12 × 10 −6 )( )(2 × 200 × 50) = 19.78 kN • m d 0.182 Mean temperature(Tm) = (40+25)/2 = 32.5 oC , TR = 28 oC F Faxial = α (∆T ) AE = (12 × 10 −6 )(32.5 − 28)(2 × 20 × 10 − 3 m 2 )(200 × 10 6 kN / m 2 ) = 432 kN 432 kN 19.78 kN•m +12 oC A -3 oC 19.78 kN•m B 432 kN 115
  • 116. 6 9 T2 = 40oC 5 182 mm 1 4 A T1 = 3 2 2EI, 2AE 25oC 2 1 EI, AE 4m FEM Element 1: 432 kN [q] = [k][d] + 19.78 kN•m +12 oC -3 A [qF] 19.78 kN•m 432 kN oC 4 5 6 B 1 2 q4 4 2x106 0.00 0.00 - 2x106 0.00 q5 5 0.00 3750 7500 0.00 q6 6 0.00 7500 20x103 q1 q2 q3 = 7 C B 4m 8 1 0.00 3 0.00 0 432 7500 0 0.00 -7500 10x103 0 -19.78 -3750 + 0.00 2x106 0.00 0.00 d1 -7500 0.00 3750 -7500 d2 0.00 7500 10x103 0.00 -7500 20x103 d3 19.78 -2x106 0.00 2 0.00 -3750 3 0.00 -432 116
  • 117. 6 9 T2 = 40oC 5 182 mm 1 4 A T1 = 3 2 2EI, 2AE 25oC 8 2 1 7 C B 4m EI, AE 4m Element 2: [q] = [k][d] + [qF] 1 2 3 7 8 9 0.00 d1 0 -1875 3750 d2 0 0.00 -3750 5x103 d3 0 0.00 1x106 0.00 0.00 0 1875 -3750 0 0 -3750 10x103 0 0 q1 1 1x106 0.00 0.00 - 1x106 0.00 q2 2 0.00 1875 3750 0.00 q3 3 0.00 3750 10x103 q7 q8 q9 = 7 -1x106 0.00 8 0.00 -1875 -3750 0.00 9 0.00 5x103 0.00 3750 + 0 117
  • 118. 6 9 T2 = 40oC 5 182 mm 4 A 1 T1 = 2EI, 2AE 25oC 8 2 1 EI, AE 4m 1 Q2 = 0.0 Q3 = 0.0 D1 D2 D3 3 3x106 0.0 0.0 D1 2 0.0 5625 -3750 D2 3 = 2 1 Q1 = 0.0 0.0 -3750 30x103 D3 0.000144 = 7 C B 4m Global: 3 2 0.0 19.78 m -719.3x10-6 + m -0.0004795 -432 rad 118
  • 119. Element 1: 4 5 2 1 6 3 q4 4 2x106 0.00 0.00 - 2x106 0.00 0.00 0 432 q5 5 0.00 3750 7500 0.00 7500 0 0.00 q6 6 0.00 7500 20x103 -7500 10x103 0 -19.78 q1 = q2 q3 1 -3750 0.00 + 0.00 2x106 0.00 0.00 -7500 0.00 3750 -7500 d2 = -479.5x10-6 0.00 7500 10x103 0.00 -7500 20x103 d3 = -719.3x10-6 19.78 -2x106 0.00 2 0.00 -3750 3 0.00 d1 = 144x10-6 -432 6 q4 144.0 kN q5 -3.60 kN q6 -23.38 kN•m -144.0 kN 3.60 kN 9.00 kN•m q1 q2 q3 = 5 A 4 23.38 kN•m 144 kN A 3.60 kN 3 2 B 1 9 kN•m B 3.60 kN 144 kN 119
  • 120. Element 2: 1 2 3 7 8 9 0.00 d1 = 144x10-6 0 -1875 3750 d2 = -479.5x10-6 0 0.00 -3750 5x103 d3 = -719.3x10-6 0 0.00 1x106 0.00 0.00 0 1875 -3750 0 0 -3750 10x103 3 0 0 q1 1 1x106 0.00 0.00 - 1x106 0.00 q2 2 0.00 1875 3750 0.00 q3 3 0.00 37500 10x103 q7 q8 q9 = 7 -1x106 0.00 8 0.00 -1875 -3750 0.00 9 0.00 5x103 0.00 3750 q1 144 kN q2 -3.6 kN q3 -9 kN•m -144 kN 3.6 kN -5.39 kN•m q7 q8 q9 = 2 B 1 9 kN•m 144 kN B 3.60 kN + 0 9 8 C 7 5.39 kN•m C 3.60 kN 144 kN 120
  • 121. Isolate axial part from the system T2 = 40oC RA A T1 = 25oC EI 2EI B 4m RC RA C 4m RA = RC + Compatibility equation: dC/A = 0 R A (4) RA ( 4) + + 12 × 10 −6 (32.5 − 28)(4) = 0 AE 2 AE RA = 144 kN RC = -144 kN 121
  • 122. T2 = 40oC EI 2EI A T1 = 25oC 4m 4m 9 kN•m 23.38 kN•m 144 kN 144 kN A C B 144 kN 144 kN B 3.60 kN 3.60 kN 5.39 kN•m 9 kN•m B C 3.60 kN 3.60 kN V (kN) x (m) -3.6 23.38 M (kN•m) Deflected curve + 8.98 - D2 = -0.0004795 mm D3 = -719.3x10-6 rad x (m) -5.39 x (m) 122
  • 123. Skew Roller Support - Force Transformation - Displacement Transformation - Stiffness Matrix 123
  • 124. • Displacement and Force Transformation Matrices x´ 5´ y´ 6´ j 2´ 3´ 4´ m 1´ i y 6 5 θy 3 2 i 1 j θx 4 m x 124
  • 125. Force Transformation 6´ x´ θy 5´ y´ θy 2´ 3´ λx = 1´ i λy = y q6 = q6' x j − xi L y j − yi L 5 4 m 3 2 i 1 q 4  q   5 q 6    x − λy λx 0 0  1  q4'  q   5'  q6'    λx 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 λx λy 0 0 − λy λx 0 λx =  λy  0  q 1   λ x    q 2   λy q 3   0  = q 4   0 q 5   0    q 6   0    6 j q4 = q4' cos θ x − q5' cos θ y λy q5 = q4' cos θ y − q5' cos θ x 4´ j θx θx λx 0 − λy [q ] = [T ]T [q '] 0  q 1'    0  q 2'   0  q 3'    0  q 4'  0  q 5'    1  q 6'    125
  • 126. Displacement Transformation y 6 x´ θx os c d4 θx4´ d4 6´ j θy θy os c d4 5´ 5 4 j θy θx d5 d θy θy d 4'  d   5'  d 6'     λx = − λy   0  d 1'   λ x    d 2'  − λy d 3'   0  = d 4'   0 d 5'   0    d 6'   0    θy cos 5 d d ' 4 = d 4 cos θ x + d 5 cos θ y d '6 = d 6 θx os c 5 θx λy d' 5 = −d 4 cos θ y + d 5 cos θ x x y λx θx x 0 0  1  λy λx 0 d 4  d   5 d 6    λx 0 0 0 0 0 0 1 0 0 λx 0 − λy 0 0 λy 0 0 0 0 0 λy λx 0 0  d 1    0  d 2   0  d 3    0  d 4  0  d 5    1  d 6    [d '] = [T ][d ] 126
  • 127. [q ] = [T ]T [q '] [ ] = [T ] [k' ][d' ] + [T ] [q ' ] [q ] = [T ] [k' ][T ][d ] + [T ] [q' ] = [k ][d ] + [q ] = [T ] ([k' ][d' ] + q 'F ) T T T T Therefore, F T F F [k ] = [T ]T [k '][T ] [q ] = [T ] [q ' ] F T F [q ] = [T ]T [q '] [d '] = [T ][d ] [k ] = [T ]T [k '][T ] 127
  • 128. Stiffness matrix 2* 5* 3* 1* 3´ 6* 2´ 1 θi 4* i θj i j λjx = cos θj λiy = sin θi 4´ 1´ j λix = cos θi 6´ 5´ λjy = sin θj [q*] = [T]T[q´] q 1*  q   2*  q 3*    q 4 *  q 5*    q 6 *    1* 2* = 3* 4* 5* 6* 1  λix λ  iy 0  0 0  0  2 − λiy λix 0 0 0 0 3 0 0 1 0 0 0 4 5 0 0 0 0 0 0 λ jx − λ jy λ jy λ jx 0 0 6 0 0  0  0 0  1  q 1'  q   2'  q 3'    q 4'  q 5'    q 6'    [ T ]T 128
  • 129. 1 2 [T ] = 3 4 5 6 1 [k '] = 1 2 3 4 5 6 1 2 λiy  λix − λ  iy λix  0 0  0  0  0 0  0  0  2 0  AE/ L  0 12 EI/ L3   0 6EI/ L2  0 − AE/ L  0 − 12 EI/ L3  6EI/ L2  0  3 4 0 0 0 0 1 0 0 λ jx 0 − λ jy 0 0 3 0 6EI/ L2 4 EI/ L 0 − 6EI/ L2 2EI/ L 5 0 0 0 λ jy λ jx 0 4 6 0 0  0  0 0  1  5 − AE/ L 0 0 − 12 EI/ L3 0 − 6EI/ L2 0 AE/ L 0 12 EI/ L3 0 − 6EI/ L2 6   6EI/ L2  2EI/ L   0  2 − 6EI/ L  4 EI/ L   0 129
  • 130. [ k ] = [ T ]T[ k´ ][T] = Ui Ui ( AE Vi L ( AE Vj - ( Mj - L Mi Uj -( λix2 + - AE L AE L 12EI L3 12EI L3 6EI L2 λixλjx + λixλjy- - Vi 6EI L2 λiy2 ) ) λixλiy ( ( AE L AE L L3 12EI L3 λiy 12EI λiy2 + L2 λiy λjy) -( λiyλjx ) -( AE L AE L ) λixλiy L3 6EI λiy 12EI - λiyλjx - λiyλjy + 6EI L2 Uj Mi 12EI L3 - λix2 ) L3 12EI L3 λix λiy -( L2 6EI L2 λix -( AE L AE L λixλjx + L λixλjy) λix λjx ) - 6EI L2 6EI L2 λjy 6EI L2 λjx ( AE L ( AE L L3 L3 L3 L3 6EI L L2 AE λixλjy ) AE -( L L λjy λixλjy - λiyλjy + 6EI - 12EI 12EI 2EI λiyλjy ) -( λjy λjx2 + - 12EI 12EI λiyλjx - 4EI λix 12EI 6EI Vj λjy2 ) ) λjxλjy ( AE L2 - L Mj 12EI L3 12EI L3 6EI L2 λiy L2 λix L2 2EI L 6EI ) λjxλjy λjx 6EI 6EI λixλjx ) AE 12EI λjx2 ) ( L λjy2 + L3 - - λjx 12EI L3 λiyλjx ) λjy L2 - 6EI L2 λjx 4EI L 130
  • 131. Example 12 For the beam shown: (a) Use the stiffness method to determine all the reactions at supports. (b) Draw the quantitative free-body diagram of member. (c) Draw the quantitative bending moment diagrams and qualitative deflected shape. Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members. Include axial deformation in the stiffness matrix. 40 kN 4m 4m 22.02 o 131
  • 132. 40 kN 4m 4m 22.02o 6 5 i 3´ Global 2* 1 2´ 3* 1* x´ 22.02 o j x* λjx = cos 22.02o = 0.9271, λjy = cos 67.98o = 0.3749 4 λix = cos 0o = 1, λiy = cos 90o = 0 40 kN 40 kN•m [FEF] 20 kN [ q´F ] i 6´ Local 5´ 1 1´ j 4´ 40 kN•m 20sin22.02=7.5 20cos22.02=18.54 20 kN 132
  • 133. • Transformation matrix 6 5 3´ Global 2* 1 4 λix = cos 0o = 1, λiy = cos 90o = 0 Member 1: [ q ] = [ q 4    q5  q 6    q 1*  q 2*    q 3*    3* 1* 2´ x´ x* λjx = cos 22.02o = 0.9271, λjy = cos 67.98o = 0.3749 T ]T[q´] = 4 5 6 1* 2* 3* 1´ 1 0  0  0 0  0  2´ 0 1 0 0 0 0 3´ 4´ 0 0 0 0 1 0 0 0 . 9271 0 0 . 3749 0 0 i 6´ Local 5´ 1 1´ [T ]T λix λ  iy 0 = 0 0  0  − λiy 0 0 0 1 0 0 0 0 λ jx 0 λ jy 0 0 λix 4´ j 0 0 0 0 0 0 − λ jy λ jx 0 0 0  0  0 0  1  5´ 6´ 0 0  q 1'    0 0  q 2'   0 0  q 3'    − 0 . 3749 0  q 4'  0 . 9271 0  q 5'    0 1  q 6'    133
  • 134. 3´ • Local stiffness matrix 2´ i δi Local 5´ 1 1´ ∆i j θi δj 4´ ∆j θj Ni  AE/ L 0 Vi  0 12 EI/ L3  Mi  0 6EI/ L2  Nj − AE/ L 0 Vj  0 − 12 EI/ L3  6EI/ L2 Mj  0  [k ]6×6 = [k´]1 6´ = 0 6EI/ L2 4 EI/ L 0 − 6EI/ L2 2EI/ L 1´ 2´ 1´ 150.0 0.000 2´ 0.000 0.9375 3´ 0.000 3.750 103 4´ -150.0 0.000 5´ 0.000 -0.9375 3´ 4´ 5´ 0.000 -150.0 0.000 3.750 0.000 -0.9375 20.00 0.000 -3.750 0.000 -3.750 150.0 0.000 0.000 0.9375 0.000 -3.750 3.750 10.00 0.000 -3.750 20.00 6´ 0.000 0 − AE/ L 0 − 12 EI/ L3 0 − 6EI/ L2 0 AE/ L 0 12 EI/ L3 0 − 6EI/ L2 0   6EI/ L2  2EI/ L   0  2 − 6EI/ L  4EI/ L   6´ 0.000 3.750 10.00 134
  • 135. 6 5 3´ Global 2* 1 3* 1* 4 Stiffness matrix [k´]: 1´ 2´ 1´ 150.0 0.000 2´ 0.000 0.9375 3´ 0.000 3.750 [k´]1 = 103 4´ -150.0 0.000 5´ 0.000 -0.9375 3.750 2´ x´ i 6´ Local 5´ 1 1´ j x* 3´ 4´ 5´ 0.000 0.000 -150.0 3.750 0.000 -0.9375 20.00 0.000 -3.750 6´ 0.000 3.750 10.00 0.000 -3.750 150.0 0.000 0.000 0.9375 0.000 -3.750 10.00 0.000 -3.750 20.00 4 5 4 5 150.0 0.000 0.000 0.9375 6 1* 0.000 -139.0 3.750 0.351 2* -56.25 -0.869 3* 0.000 3.750 6 0.000 20.00 1.406 -3.750 10.00 1* -139.0 0.351 1.406 2* -56.25 -0.869 -3.750 129.0 51.82 3* 0.000 1.406 4´ 6´ 0.000 Stiffness matrix [k*]: [ k* ] = [ T ]T[ k´ ][T] [k*]1 = 103 3.750 3.750 10.00 51.82 1.406 21.90 -3.476 -3.476 20.00 135
  • 136. 6 40 kN 5 2* 1 4 4m 4m 22.02 o 20 kN x´ x* 40 kN 40 kN•m Global Equilibrium: 3* 1* [ q´F ] 40 kN•m 20sin22.02=7.5 20cos22.02=18.54 20 kN [Q] = [K][D] + [QF] 1* Q1 = 0.0 Q3 = 0.0 D1* D3* = = 1* 103 3* 129 1.406 3* 1.406 D1* 20.0 D3* 36.37x10-6 -40 m 0.002 + -7.5 rad 136
  • 137. 6 40 kN 5 2* 1 4 4m 4m Member Force : 22.02 [q] = [k*][D] + [qF] o 40 kN•m q4 q5 4 5 q6 6 0.000 3.750 20.00 1.406 -3.750 10.00 q1* q2* q3* = x´ x* 40 kN•m 18.54 20 kN 4 5 6 1* 2* 3* 150.0 0.000 0.000 -139.0 -56.25 0.000 0.000 0.9375 3.750 0.351 -0.869 3.750 103 40 kN 3* 1* 7.5 0 0 0 20 -5.06 27.5 0 40.0 60 1* -139.0 0.351 1.406 129.0 51.82 1.406 36.4x10-6 0 2* -56.25 -0.869 -3.750 51.82 21.90 -3.476 2x10-3 * 0.000 3.750 10.00 1.406 -3.476 20.00 3 + -7.54 18.54 -40.0 = 0 13.48 0 40 kN 60 kN•m 5.06 kN 27.5 kN 13.48 kN 137
  • 138. 40 kN 40 kN 4m 4m 60.05 kN•m 22.02 o 5.05 kN 27.51 kN 13.47 kN 50.04 + -60.05 Bending moment diagram (kN•m) 0.002 rad Deflected shape 138
  • 139. Example 13 For the beam shown: (a) Use the stiffness method to determine all the reactions at supports. (b) Draw the quantitative free-body diagram of member. (c) Draw the quantitative bending moment diagrams and qualitative deflected shape. Take I = 200(106) mm4, A = 6(103) mm2, and E = 200 GPa for all members. Include axial deformation in the stiffness matrix. 40 kN 6 kN/m 2EI, 2AE 8m EI, AE 4m 22.02o 4m 139
  • 140. 40 kN 6 kN/m EI, AE 2EI, 2AE 8m 4m 6 22.02 o AE (0.006 m 2 )(200 ×106 kN / m 2 ) = L (8 m) 4m = 150 ×103 kN • m 3 5 2 8* 4 2 1 1 9* 7* Global 3´ 6´ 2´ 5´ 1´ 1 5´ 2´ Local = 20 ×103 kN • m 2 EI = 10 ×103 kN • m L 3´ 4´ 4 EI 4(200 ×106 kN / m 2 )(0.0002 m 4 ) = L (8 m) 1´ 2 6´ 6 EI 6(200 ×106 kN / m 2 )(0.0002 m 4 ) = 2 L (8 m) 2 4´ = 3.75 ×103 kN 12 EI 12(200 ×106 kN / m 2 )(0.0002 m 4 ) = L2 (8 m) 3 = 0.9375 ×103 kN / m 140
  • 141. 6 3 5 2 8* 1 1 4 2 9* 7* Global Local : Member 1 6 3 5 [ q] 4 3´ 2 2´ 1 1 [q] = [q´] Thus, [k] = [k´] 6´ [ q´] 1´ 1 4 5 4 5 6 1 2 0.000 150.0 0.000 0.000 -150.0 0.000 0.9375 3.750 0.000 -0.9375 6 0.000 4´ 3 0.000 3.750 20.00 0.000 -3.750 10.00 -150.0 0.000 0.000 2 0.000 -0.9375 -3.750 3 0.000 3.750 10.00 150.0 0.000 0.000 0.9375 0.000 -3.750 0.000 -3.750 20.00 [k]1 = [k´]1 = 2x103 1 3.750 5´ 141
  • 142. 6 3 5 [ 2 1 λix = cos 0o = 1, λiy = cos 90o = 0 9* 7* 2 3´ 8* q* ] 8* 1 1 4 3 2 2 9* 7* [ q´ ] 2´ x´ 1´ 5´ 6´ 4´ 2 x* λjx = cos 22.02o = 0.9271, λjy = cos 67.98o = 0.3749 Member 2: Use transformation matrix, [q*] = [T]T[q´] 1´ 2´ 3´ 4´ 5´ 6´ q1 q2 1 2 1 0 0 1 0 0 0 0 0 0 0 0 q1´ q2´ q3 3 0 0 1 0 0 0 q3´ 7* 8* 0 0 0 0 0 0 0 0 q4´ q5´ 9* 0 0 0 1 q6´ q7* q8* q9* = 0.9271 -0.3749 0.3749 0.9271 0 0 142
  • 143. 2 3 [ q* ] 1 3´ 8* 9* 7* 2 x´ [ q´ ] 2´ 1´ 2 4´ x* Stiffness matrix [k´]: 3´ 4´ 5´ 0.000 0.000 -150.0 3.750 0.000 -0.9375 20.00 0.000 -3.750 6´ 0.000 3.750 10.00 0.000 -3.750 150.0 0.000 0.000 0.9375 0.000 -3.750 10.00 0.000 -3.750 20.00 1 2 [k´]2 = 5´ 6´ 1´ 2´ 1´ 150.0 0.000 2´ 0.000 0.9375 3´ 0.000 3.750 103 4´ -150.0 0.000 5´ 0.000 -0.9375 3.750 1 2 150.0 0.000 0.000 0.9375 3 7* 0.000 -139.0 3.750 0.351 8* -56.25 -0.869 9* 0.000 3.750 3 0.000 20.00 1.406 -3.477 10.00 7* -139.0 0.351 1.406 8* -56.25 -0.869 -3.477 129.0 51.82 9* 0.000 1.406 6´ 0.000 Stiffness matrix [k*]: [k*] = [T]T[k´][T] [k*]2 = 103 3.750 3.750 10.00 51.82 1.406 21.90 -3.476 -3.476 20.00 143
  • 144. 6 3 5 2 4 8* 1 1 2 9* 7* 4 5 4 5 6 1 2 0.000 150.0 0.000 0.000 -150.0 0.000 0.9375 3.750 0.000 -0.9375 3 0.000 3.750 6 0.000 20.00 0.000 -3.750 10.00 -150.0 0.000 0.000 2 0.000 -0.9375 -3.750 3 0.000 3.750 10.00 150.0 0.000 0.000 0.9375 0.000 -3.750 0.000 -3.750 20.00 1 2 1 2 150.0 0.000 0.000 0.9375 3 7* 0.000 -139.0 3.750 0.351 8* -56.25 -0.869 9* 0.000 3.750 3 0.000 20.00 1.406 -3.477 10.00 7* -139.0 0.351 1.406 8* -56.25 -0.869 -3.477 9* 0.000 3.750 10.00 129.0 51.82 [k]1= 2x103 1 [k*]2 = 103 3.750 3.750 51.82 1.406 1.406 21.90 -3.476 -3.476 20.00 144
  • 145. 6 40 kN 6 kN/m 3 5 2 4 8* 2 1 1 40 kN 6 kN/m 32 kN•m 9* 40 kN•m 32 kN•m 40 kN•m 1 24 kN 24 kN Global: 1 1 3 = 103 7* 0 0 0 0 9* 18.54 20 kN 450 0 -139 0 7* 3 0 60 1.406 10 -139 1.406 129 1.406 D1 18.15x10-6 -509.84x10-6 rad 0 10 1.406 20 D1 D3 D7* D9* + 0 -32 + 40 = 8 -7.5 -40 m 0.00225 9* rad 58.73x10-6 7.5 m D3 7* D7* D9* = 145
  • 146. 6 3 5 2 4 32 kN•m q4 q5 4 5 4 5 6 1 2 0.000 150.0 0.000 0.000 -150.0 0.000 0.9375 3.750 0.000 -0.9375 q6 6 0.000 q1 q2 q3 = 2x103 20.00 0.000 -3.750 1 -150.0 0.000 0.000 2 0.000 -0.9375 -3.750 3 0.000 3.750 10.00 150.0 0.000 0.000 0.9375 0.000 -3.750 q4 q5 -5.45 20.18 kN kN q6 21.80 kN•m 5.45 27.82 kN kN -52.39 kN•m q1 q2 q3 = 24 kN 24 kN Member 1: [ q ] = [k ][d] + [qF] 3.750 32 kN•m 1 1 1 6 kN/m 21.80 kN•m 3 0.000 0 3.750 0 10.00 0 0 24 0.000 d1=18.15x10-6 -3.750 0 + 20.00 d3=-509.84 x10-6 6 kN/m 32 0 24 -32 52.39 kN•m 5.45 kN 5.45 kN 20.18 kN 27.82 kN 146
  • 147. 2 3 1 40 kN 8* 9* 7* 2 40 kN•m x´ 20 kN Member 2: [ q ] = [k ][d] + [qF] 1 2 3 7* q1 1 150.0 0.000 0.000 -139.0 q2 2 0.000 0.9375 3.750 0.351 q3 3 0.000 3.750 20.00 1.406 q7* = 103 7* -139.0 0.351 1.406 129.0 q8* 8* -56.25 -0.869 -3.750 51.82 q9* 9* 0.000 3.750 10.00 1.406 -5.45 26.55 kN kN q3 52.39 kN kN 0 kN•m 8* -56.25 -0.869 9* 0.000 3.750 18.15x10-6 0 0 20 -3.750 10.00 -509.84x10-6 40 1.406 21.90 -3.476 51.82 -3.476 20.00 58.73x10-6 0 + 0.00225 -7.5 18.54 -40 kN•m 0 14.51 20sin22.02=7.5 20cos22.02=18.54 20 kN 2 x* q1 q2 [ 40 kN•m q´F ] q7* q8* q9* = 40 kN 52.39 kN•m 5.45 kN 26.55 kN 14.51 kN 147
  • 148. 6 kN/m 21.80 kN•m 52.39 kN•m 52.39 kN•m 40 kN 5.45 kN 5.45 kN 5.45 kN 26.55 kN 27.82 kN 20.18 kN 40 kN 6 kN/m 21.80 kN•m 14.51 kN 14.51cos 22.02o =13.45 kN 5.45 kN 20.18 kN V (kN) 54.37 kN 4m 8m 26.55 20.18 + 3.36 m 4 m 14.51 kN 22.02o + - - x (m) -13.45 -27.82 53.81 12.14 M (kN•m) -21.8 + -52.39 x (m) 148