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  • 1. 1 ! Development: The Slope-Deflection Equations ! Stiffness Matrix ! General Procedures ! Internal Hinges ! Temperature Effects ! Force & Displacement Transformation ! Skew Roller Support BEAM ANALYSIS USING THE STIFFNESS METHOD
  • 2. 2 Slope – Deflection Equations settlement = ∆j Pi j k w Cj Mij Mji w P θj θi ψ i j
  • 3. 3 • Degrees of Freedom L θΑ A B M 1 DOF: θΑ P θΑ θΒΒΒΒ A B C 2 DOF: θΑ , θΒΒΒΒ
  • 4. 4 L A B 1 • Stiffness Definition kBAkAA L EI kAA 4 = L EI kBA 2 =
  • 5. 5 L A B1 kBBkAB L EI kBB 4 = L EI kAB 2 =
  • 6. 6 • Fixed-End Forces Fixed-End Forces: Loads P L/2 L/2 L w L 8 PL 8 PL 2 P 2 P 12 2 wL 12 2 wL 2 wL 2 wL
  • 7. 7 • General Case settlement = ∆j Pi j k w Cj Mij Mji w P θj θi ψ i j
  • 8. 8 w P settlement = ∆j (MF ij)∆ (MF ji)∆ (MF ij)Load (MF ji)Load + Mij Mji θi θj + i jw P Mij Mji settlement = ∆j θj θi ψ =+ ji L EI L EI θθ 24 ji L EI L EI θθ 42 += ,)()() 2 () 4 ( Loadij F ij F jiij MM L EI L EI M +++= ∆θθ Loadji F ji F jiji MM L EI L EI M )()() 4 () 2 ( +++= ∆θθ L
  • 9. 9 Mji Mjk Pi j k w Cj Mji Mjk Cj j • Equilibrium Equations 0:0 =+−−=Σ+ jjkjij CMMM
  • 10. 10 + 1 1 i jMij Mji θi θj L EI kii 4 = L EI k ji 2 = L EI kij 2 = L EI k jj 4 = iθ× jθ× • Stiffness Coefficients L
  • 11. 11 [ ]       = jjji ijii kk kk k Stiffness Matrix )() 2 () 4 ( ij F jiij M L EI L EI M ++= θθ )() 4 () 2 ( ji F jiji M L EI L EI M ++= θθ         +            =      F ji F ij j iI ji ij M M LEILEI LEILEI M M θ θ )/4()/2( )/2()/4( • Matrix Formulation
  • 12. 12 [D] = [K]-1([Q] - [FEM]) Displacement matrix Stiffness matrix Force matrix w P(MF ij)Load (MF ji)Load + + i jw P Mij Mji θj θi ψ ∆j Mij Mji θi θj (MF ij)∆ (MF ji)∆ Fixed-end moment matrix ][]][[][ FEMKM += θ ]][[])[]([ θKFEMM =− ][][][][ 1 FEMMK −= − θ L
  • 13. 13 L Real beam Conjugate beam • Stiffness Coefficients Derivation MjMi L/3 θi L MM ji + EI M j EI Mi θι EI LM j 2 EI LMi 2 )1(2 0) 3 2 )( 2 () 3 )( 2 (:0' −−−= =+−=Σ+ ji ji i MM L EI LML EI LM M )2(0) 2 () 2 (:0 −−−=+−=Σ↑+ EI LM EI LM F ji iy θ ij ii L EI M L EI M andFrom θ θ ) 2 ( ) 4 ( );2()1( = = L MM ji + i j
  • 14. 14 • Derivation of Fixed-End Moment Real beam 8 ,0 16 2 22 :0 2 PL M EI PL EI ML EI ML Fy ==+−−=Σ↑+ P M M EI M Conjugate beam A EI M B L P A B EI M EI ML 2 EI M EI ML 2 EI PL 16 2 EI PL 4 EI PL 16 2 Point load
  • 15. 15 L P 841616 PLPLPLPL =+ − + − 8 PL 8 PL 2 P 2 P P/2 P/2 -PL/8 -PL/8 PL/8 -PL/16 -PL/8 - PL/4 + -PL/16-PL/8 -
  • 16. 16 Uniform load L w A B w M M Real beam Conjugate beam A EI M EI M B 12 ,0 24 2 22 :0 23 wL M EI wL EI ML EI ML Fy ==+−−=Σ↑+ EI wL 8 2 EI wL 24 3 EI wL 24 3 EI M EI ML 2 EI M EI ML 2
  • 17. 17 Settlements M M L ∆ MjMi = Mj L MM ji +L MM ji + Real beam 2 6 L EI M ∆ = Conjugate beam EI M A B EI M ∆ ,0) 3 2 )( 2 () 3 )( 2 (:0 =+−∆−=Σ+ L EI MLL EI ML MB EI M EI ML 2 EI MEI ML 2 ∆
  • 18. 18 • Typical Problem 0 0 0 0 A C B P1 P2 L1 L2 w CB 8 0 24 11 11 LP L EI L EI M BAAB +++= θθ 8 0 42 11 11 LP L EI L EI M BABA −++= θθ 128 0 24 2 222 22 wLLP L EI L EI M CBBC ++++= θθ 128 0 42 2 222 22 wLLP L EI L EI M CBCB − − +++= θθ P 8 PL 8 PL L w 12 2 wL 12 2 wL L
  • 19. 19 MBA MBC A C B P1 P2 L1 L2 w CB B CB 8 0 42 11 11 LP L EI L EI M BABA −++= θθ 128 0 24 2 222 22 wLLP L EI L EI M CBBC ++++= θθ BBCBABB forSolveMMCM θ→=−−=Σ+ 0:0
  • 20. 20 A C B P1 P2 L1 L2 w CB Substitute θB in MAB, MBA, MBC, MCB MAB MBA MBC MCB 0 0 0 0 8 0 24 11 11 LP L EI L EI M BAAB +++= θθ 8 0 42 11 11 LP L EI L EI M BABA −++= θθ 128 0 24 2 222 22 wLLP L EI L EI M CBBC ++++= θθ 128 0 42 2 222 22 wLLP L EI L EI M CBCB − − +++= θθ
  • 21. 21 A B P1 MAB MBA L1 A C B P1 P2 L1 L2 w CB ByR Cy Ay ByL Ay Cy MAB MBA MBC MCB By = ByL + ByR B C P2 MBC MCB L2
  • 22. 22 2EI EI 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 1 2 1 2 31 2 3 7 8 9 4 5 6 1 2 3 7 8 9 4 5 6
  • 23. 23 i j E, I, A, L Beam-Member Stiffness Matrix d4 = 1 AE/LAE/L AE/L AE/L 6 4 5 1 2 3 [k] = 1 2 3 4 5 6 3 2 1 4 6 5 d1 = 1 k11 = = k44 0 0 0 0 0 0 0 0 - AE/LAE/L -AE/L AE/L k41 k14
  • 24. 24 d2 = 1 [k] = 1 2 3 4 5 6 3 2 1 4 6 5 0 0 0 0 0 0 0 0 - AE/LAE/L -AE/L AE/L 6EI/L2 6EI/L2 d5 = 1 6EI/L2 6EI/L2 12EI/L3 12EI/L3 12EI/L3 12EI/L3 6EI/L2 12EI/L3 - 6EI/L2 - 12EI/L3 0 0 6EI/L2 -12EI/L3 0 0 - 6EI/L2 12EI/L3 k22 = = k55 = k52 k62 = = k32 = k65 = k25 = k35 i j E, I, A, L 6 4 5 1 2 3
  • 25. 25 [k] = 1 2 3 4 5 6 3 2 1 4 6 5 0 0 0 0 0 0 0 0 - AE/LAE/L -AE/L AE/L 6EI/L2 12EI/L3 - 6EI/L2 - 12EI/L3 0 0 6EI/L2 -12EI/L3 0 0 - 6EI/L2 12EI/L3 4EI/L 6EI/L2 6EI/L2 2EI/L 0 0 2EI/L -6EI/L2 0 0 -6EI/L2 4EI/L d6 = 1 d3 = 1 2EI/L 4EI/L 6EI/L26EI/L26EI/L2 4EI/L 2EI/L 6EI/L2 k33 = k23 = = k53 = k63 = k36 = k56k26 = = k66 i j E, I, A, L 6 4 5 1 2 3
  • 26. 26 FF xi FF yi MF i FF xj FF yj MF j + MF i MF j FF xjFF xi FF yi FF yj 1 6EI/L2 6EI/L2 12EI/L3 12EI/L3 + x ∆i 12EI/L3 6EI/L2 6EI/L2 12EI/L3 1 + 4EI/L 2EI/L 6EI/L26EI/L2 x θi 6EI/L2 4EI/L 2EI/L 6EI/L2 x δj AE/L 1 AE/L + AE/L AE/L 6EI/L2 6EI/L2 1 12EI/L3 12EI/L3 x ∆j 6EI/L2 6EI/L2 12EI/L3 12EI/L3 1 + 2EI/L 6EI/L26EI/L2 4EI/L x θj 6EI/L2 2EI/L 6EI/L2 4EI/L Fxi Fyi Mi Fxj Fyj Mj • Member Equilibrium Equations Mj i j Fxj Fxi Fyi Fyj = Mi E, I, A, L x δi AE/L AE/L 1 AE/LAE/L
  • 27. 27                     +                                         − −−− − − − − =                     F j F yi F xj F i F yi F xi j j j i i i j yj xj i yj xi M F F M F F θ ∆ δ θ ∆ δ LEILEILEILEI LEILEILEILEI LAELAE LEILEILEILEI LEILEILEILEI LAELAE M F F M F F /4/60/2/60 /6/120/6/120 00/00/ /2/60/4/60 /6/120/6/120 00/00/ 22 2323 22 2323 F jjjjiiij F yjjjjiiiyj F xijjjiiixj F ijjjiiixi F yijjjiiiyi F xijjjiiixi MθLEI∆LEIδθLEI∆LEIδM FθLEI∆δθLEI∆LEIδF Fθ∆δLAEθ∆δLAEF MθLEI∆LEIδθLEI∆LEIδM FθLEI∆LEIδθLEI∆LEIδF Fθ∆δLAEθ∆δLAEF )/4()/6()0()/2()/6()0( )/6()0()0()/6()/12()0( )0()0()/()0()0()/( )/2()/6()0()/4()/6()0( )/6()/12()0()/6()/12()0( )0()0()/()0()0()/( 22 223 22 2323 −= −−−= −= −= −+= +++−++= [q] = [k][d] + [qF] Fixed-end force matrix End-force matrix Stiffness matrix Displacement matrix
  • 28. 28 6x6 Stiffness Matrix [ ]                     − −−− − − − − =× LEILEILEILEI LEILEILEILEI LAELAE LEILEILEILEI LEILEILEILEI LAELAE k /4/60/2/60 /6/120/6/120 00/00/ /2/60/4/60 /6/120/6/120 00/00/ 22 2323 22 2323 66 θi ∆j θj∆i δjδi Mi Vj Mj Vi Nj Ni 4x4 Stiffness Matrix [ ]             − −−− − − =× LEILEILEILEI LEILEILEILEI LEILEILEILEI LEILEILEILEI k /4/6/2/6 /6/12/6/12 /2/6/4/6 /6/12/6/12 22 2323 22 2323 44 θi ∆j θj∆i Mi Vj Mj Vi
  • 29. 29 Comment: - When use 4x4 stiffness matrix, specify settlement. - When use 2x2 stiffness matrix, fixed-end forces must be included. 2x2 Stiffness Matrix θi θj Mi Mj [ ]       =× LEILEI LEILEI k /4/2 /2/4 22
  • 30. 30 21 31 2 1 2 3 4 5 6 General Procedures: Application of the Stiffness Method for Beam Analysis w P P2y M2M1 Local 1´ 2´ 3´ 4´ 1 1´ 2´ 3´ 4´ 2 Global
  • 31. 31 Member 1 2 3 4 1 3 4 5 6 2 1 2 3 4 5 6 31 2 1 2 Global w P P2y M2M1 Local 1´ 2´ 3´ 4´ 1 1´ 2´ 3´ 4´ 2
  • 32. 32 (q´F 4 )2 (q´F 3 )2 (q´F 1)2 (q´F 2)2 w 2 P (q´F 4)1 (q´F 3)1 (q´F 2)1 (q´F 1 )1 1[FEF] 1 2 3 4 5 6 31 2 1 2 Global Local 1´ 2´ 3´ 4´ 1 1´ 2´ 3´ 4´ 2 w P P2y M2M1
  • 33. 33 [q] = [T]T[q´] 1 2 3 4 5 6 31 2 1 2 Global Local 1´ 2´ 3´ 4´ 1 [k] = [T]T[q´] [T]                         =             '4 '3 '2 '1 4 3 2 1 1000 0100 0010 0001 q q q q q q q q 1´ 4´2´ 3´ 1 2 3 41
  • 34. 34 [q] = [T]T[q´] 1 2 3 4 5 6 31 2 1 2 Global Local 1´ 2´ 3´ 4´ 2 [k] = [T]T[q´] [T]                         =             '4 '3 '2 '1 6 5 4 3 1000 0100 0010 0001 q q q q q q q q 1´ 4´2´ 3´ 3 4 5 62
  • 35. 35 Stiffness Matrix: 21 31 2 1 2 3 4 5 6 [k]1 = 1 2 3 4 1 2 3 4 [k]2 = 3 4 5 6 3 4 5 6 Member 1 [K] = 1 2 3 4 5 6 1 2 3 4 5 6 Member 2 Node2 1 2
  • 36. 36                     +                                         =                     F F F F F F Q Q Q Q Q Q D D D D D D Q Q Q Q Q Q 6 5 1 4 3 2 6 5 1 4 3 2 6 5 1 4 3 2 2 3 4 1 5 6 5 612 3 4 0 0 0 M1z -P2y M3z KBA KAB KBB KAA 21 31 2 1 2 3 4 5 6 Reaction Qu Joint Load Qk Du=Dunknown Dk = Dknown QF B QF A [ ] [ ][ ] [ ]F AuAAk QDKQ += [ ] [ ] [ ] [ ])( 1 F AkAAu QQKD −+= − [ ] [ ][ ] [ ]F qdkqForceMember +=:
  • 37. 37 +                     6 5 4 3 2 1 D D D D D D Global: 21 31 2 1 2 3 4 5 6 1 2 2 3 4 51 6 (q´F 6 )2 (q´F 5 )2 (qF 4)2 (qF 3)2 w 2 P (qF 4)1 (qF 3)1 (qF 2)1 (qF 1 )1 1[FEF]           = −= = 24 23 12 MQ PQ MQ y           4 3 2 D D D           + F F F Q Q Q 4 3 2 2 3 4 Node2           = 2 3 4                     Member 1 1 2 3 4 5 6 1 2 3 4 5 6 Member 2 Node2=                     6 5 4 3 2 1 Q Q Q Q Q Q                     += += F F F 4 F 4 F F 3 F 3 F F F Q Q qqQ qqQ Q Q 6 5 214 213 2 1 )()( )()( M1 -P2y M2 0 0 0
  • 38. 38 1 2 31 2 1 2 3 4 5 6           4 3 2 D D D 2 3 4 Node2           = 2 3 4                     −           = −= = F F F Q Q Q MQ PQ MQ 4 3 2 24 2y3 12
  • 39. 39 Member 1: 21 31 2 1 2 3 4 5 6 P qF 4 qF 3 qF 2 qF 1 1 [FEF]             4 3 2 1 q q q q             = = = = 44 33 22 1 0 Dd Dd Dd d               + F F F F q q q q 4 3 2 1 1 2 3 4 1 2 3 4             = 1k
  • 40. 40 21 31 2 1 2 3 4 5 6 Member 2: qF 6 qF 5 qF 4 qF 3 w 2 [FEM]             6 5 4 3 q q q q             = = = = 0 0 6 5 44 33 d d Dd Dd               + F F F F q q q q 6 5 4 3 1 2 3 4 1 2 3 4             = 1k
  • 41. 41 A C B 9 m 3 m 10 kN 1 kN/m 1.5 m 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.
  • 42. 42 21Global 123 1 Members 2 3 2 12 A C B 9 m 3 m 10 kN 1 kN/m 1.5 m wL2/12 = 6.75wL2/12 = 6.75 PL/8 = 3.75PL/8 = 3.75 10 kN 1.5 m1.5 m 2 1 kN/m 9 m 1 [FEF]
  • 43. 43 4/3 2/3 2/3 21 123 9 m 3 m [k]1= EI 4/9 2/9 3 4/9 2/9 2 3 2 [k]2 = EI 4/3 4/3 2/3 2/3 2 1 2 1 [K] = EI 2 1 2 1 (4/9)+(4/3) 21 123 Stiffness Matrix: θi θj Mi Mj [ ]       =× LEILEI LEILEI k /4/2 /2/4 22
  • 44. 44 A C B 10 kN 1 kN/m 123 Equilibrium equations: MCB = 0 MBA + MBC = 0 Global Equilibrium: [Q] = [K][D] + [QF] = 2.423/EI 0.779/EI θC θB 9 m 1.5 m1.5 m 6.756.75 + -3.75 -6.75 + 3.75 = -3MBA + MBC = 0 MCB = 01 2 θC θB 4/3 2/3 2/3 = EI 2 1 2 1 (4/9)+(4/3) 3.753.756.75
  • 45. 45 = -6.40 6.92 Member 1 : [q]1 = [k]1[d]1 + [qF]1 MBA MAB 2 3 = 0.779/EIθB θA 0 + -6.75 6.75 Substitute θB and θC in the member matrix, 1 2 3 wL2/12 = 6.75wL2/12 = 6.75 1 kN/m 9 m 1 [FEF] 4.56 kN 4.44 kN 6.92 kN•m 6.40 kN•m wL2/12 = 6.75wL2/12 = 6.75 = EI 4/9 2/9 3 4/9 2/9 2 1 kN/m 19 m
  • 46. 46 Member 2 : [q]2 = [k]2[d]2 + [qF]2 Substitute θB and θC in the member matrix, MCB MBC 1 2 = EI 4/3 4/3 2/3 2/3 2 1 = 0 6.40 θC θB = 2.423/EI = 0.779/EI 2 12 [FEF] PL/8 = 3.75PL/8 = 3.75 10 kN 1.5 m1.5 m 2 7.13 kN 2.87 kN 6.40 kN•m 10 kN 2 PL/8 = 3.75PL/8 = 3.75 + -3.75 3.75
  • 47. 47 A C B 10 kN1 kN/m 9 m 1.5 m1.5 m 6.92 kN•m 11.57 kN 2.87 kN4.56 kN 4.56 kN 4.44 kN 7.13 kN 2.87 kN 1 kN/m 6.92 6.40 6.40 10 kN M (kN•m) x (m) -6.92 -6.40 3.48 4.32 V (kN) x (m) 4.56 -4.44 -2.87 4.56 m 7.13 θB = +0.779/EI θC = +2.423/EI
  • 48. 48 20 kN A CB EI2EI 4 m 4 m 9kN/m 40 kN•m 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.
  • 49. 49 1 2 3 0.5625 3 -0.375 -0.375 1 2 3 4 5 6 [K] = EI 3 4 3 4 Use 4x4 stiffness matrix, [k]1 0.375EI -0.75EI2EI 0.75EI - 0.375EI 0.375EI 0.75EI EI -0.75EI 2EI 0.75EI 0.75EI -0.375EI EI -0.75EI -0.75EI 1 2 3 4 1 2 3 4 [k]2 0.375EI 0.375EI -0.1875EI 0.5EI -0.375EI -0.375EI 0.1875EI -0.375EIEI 0.375EI - 0.1875EI 0.1875EI 0.375EI 0.5EI -0.375EI EI 3 4 5 6 3 4 5 6 2EI EI 1 2 1 2 5 6 θi ∆j θj Mi ∆i Vj Mj Vi [ ]             − −−− − − = LEILEILEILEI LEILEILEILEI LEILEILEILEI LEILEILEILEI k /4/6/2/6 /6/12/6/12 /2/6/4/6 /6/12/6/12 22 2323 22 2323
  • 50. 50 20 kN 9kN/m 40 kN•m 4 m 4 m 12 kN•m 18 kN 12 kN•m 18 kN 18 kN D4 D3 Global: D4 D3 = 9.697/EI -61.09/EI Q4 = 40 Q3 = -203 4 + -12 18 3 4 20 kN 40 kN•m 12 kN•m 1 2 1 2 3 4 5 6 1 2 3 2EI EI [Q] = [K][D] + [QF] 0.5625 3 -0.375 -0.375 = EI 3 4 3 4
  • 51. 51 Member 1: 1 9 kN/m A B 12 kN•m [qF]1 12 kN•m 18 kN 18 kN 1 9 kN/m A B [q]1 53.21 kN•m 48.18 kN 67.51 kN•m 12.18 kN 1 1 2 3 4 1 2 2EI 12 kN•m 12 kN•m 18 kN 18 kN q2 q1 q4L q3L 67.51 48.18 53.21 -12.18 12 18 -12 18 12(2EI)/43 -0.75EI2EI 0.75EI - 0.375EI 0.375EI 0.75EI EI -0.75EI 2EI 0.75EI 0.75EI -0.375EI EI -0.75EI -0.75EI 1 2 3 4 1 2 3 4 [q]1 = [k]1[d]1 + [qF]1 d3 = -61.09/EI d4 = 9.697/EI d2 = 0 d1 = 0
  • 52. 52 2 B C [q]2 Member 2: [q]2 = [k]2[d]2 + [qF]2 2 3 4 5 6 2 3 EI q4R q3R q6 q5 0.375EI 0.375EI -0.1875EI 0.5EI -0.375EI -0.375EI 0.1875EI -0.375EIEI 0.375EI - 0.1875EI 0.1875EI 0.375EI 0.5EI -0.375EI EI 3 4 5 6 3 4 5 6 0 0 0 0 d3 =-61.09/EI d4 = 9.697/EI d6 = 0 d5 = 0 18.06 kN•m 7.82 kN 13.21 kN•m 7.82 kN -13.21 -7.818 -18.06 7.818
  • 53. 53 V (kN) x (m) -7.818 48.18 -7.818 - M (kN•m) x (m) 13.21 -18.08 - -67.51 - 1 9 kN/m A B [q]1 53.21 kN•m 48.18 kN 67.51 kN•m 12.18 kN 2 B C [q]2 18.06 kN•m 7.82 kN 13.21 kN•m 7.82 kN 12.18+ 53.21 + 20 kN 4 m 4 m 9kN/m 40 kN•m 7.818 kN 18.06 kN•m67.51 kN•m 48.18 kN D3 = ∆B = -61.09/EI D4 = θB = +9.697/EI
  • 54. 54 20 kN A CB EI2EI 4 m 2 m 9kN/m 40 kN•m 10 kN 2 m 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.
  • 55. 55 Global 1 2 1 2 3 4 5 6 1 2 3 20 kN A C B EI2EI 9kN/m 40 kN•m 10 kN 4 m 2 m 2 m 10 kN 5 kN•m 5 kN 5 kN•m 5 kN 2 12 kN•m 18 kN 12 kN•m 18 kN 9 kN/m 1[FEF] θi ∆j θj Mi ∆i Vj Mj Vi [ ]             − −−− − − =× LEILEILEILEI LEILEILEILEI LEILEILEILEI LEILEILEILEI k /4/6/2/6 /6/12/6/12 /2/6/4/6 /6/12/6/12 22 2323 22 2323 44
  • 56. 56 1 2 1 2 3 4 5 6 1 2 3 4 m 2 m 2 m 1 2 5 1 2 0.375 0.5 0.375 0.5 1 [K] = EI 3 4 6 3 4 6 0.5625 3 -0.375 -0.375 12(2EI)/43 [k]1 = -0.75EI2EI 0.75EI - 0.375EI 0.375EI 0.75EI EI -0.75EI 2EI/L 0.75EI 0.75EI -0.375EI EI -0.75EI -0.75EI 1 2 3 4 1 2 3 4 [k]2 = 0.375EI 0.375EI -0.1875EI 0.5EI -0.375EI -0.375EI 0.1875EI -0.375EIEI 0.375EI - 0.1875EI 0.1875EI 0.375EI 0.5EI -0.375EI EI 3 4 5 6 3 4 5 6
  • 57. 57 MBA+MBC = 40 VBL+VBR = -20 MCB = 0 θB ∆B θC -12 + 5 = -7 -5 θB ∆B θC = -7.667/EI -116.593/EI 52.556/EI Global: [Q] = [K][D] + [QF] = EI 0.5625 3 -0.375 -0.375 0.375 0.5 0.375 0.5 1 3 4 6 3 4 6 20 kN A C B 9kN/m 40 kN•m 10 kN 21 1 2 3 4 5 6 1 2 3 10 kN 5 kN•m 5 kN 5 kN•m 5 kN 2 12 kN•m 18 kN 12 kN•m 18 kN 9 kN/m 1[FEF] 20 kN 40 kN•m 5 kN•m 5 kN•m 5 kN 12 kN•m 18 kN + 18 + 5 = 23
  • 58. 58 0.375EI = -0.75EI2EI 0.75EI - 0.375EI 0.375EI 0.75EI EI -0.75EI 2EI/L 0.75EI 0.75EI -0.375EI EI -0.75EI -0.75EI 1 2 3 4 1 2 3 4 = 91.78 55.97 60.11 -19.97 91.78 kN•m 55.97 kN Member 1: [q]1 = [k]1[d]1 + [qF]1 MAB VA MBA VBL + 12 18 -12 18=-116.593/EI =-7.667/EI 0 0 θB ∆B 1 1 2 3 4 1 2 12 kN•m 18 kN 12 kN•m 18 kN 9 kN/m 1 [FEF] 19.97 kN 60.11 kN•m 4 m 9kN/m 1 12 kN•m 18 kN 12 kN•m 18 kN18 kN
  • 59. 59 VC MBC VBR MCB + 5 5 5 -5 = -20.11 -0.0278 0 10.03 =52.556/EI =-116.593/EI =-7.667/EIθB ∆B θC 0 10 kN 5 kN•m 5 kN 5 kN•m 5 kN 2 [FEM] 10.03 kN 2 m 10 kN 2 m 2 = 0.375EI 0.375EI -0.1875EI 0.5EI -0.375EI -0.375EI 0.1875EI -0.375EIEI 0.375EI - 0.1875EI 0.1875EI 0.375EI 0.5EI -0.375EI EI 3 4 5 6 3 4 5 6 0.0278 kN 20.11 kN•m Member 2: [q]1 = [k]1[d]1 + [qF]1 22 3 5 6 3 4 5 kN•m 5 kN 5 kN•m 5 kN
  • 60. 60 20 kN 4 m 2 m 9kN/m 40 kN•m 10 kN 2 m 10.03 kN 91.78 kN•m 55.97 kN V (kN) x (m) 55.97 -0.0278 19.97 + -10.03 -10.03 - 91.78 kN•m 55.97 kN 19.97 kN 60.11 kN•m4 m 9kN/m 1 10.03 kN0.0278 kN 20.11 kN•m 2 m 10 kN 2 m 2 91.78 kN•m 60.11 kN•m 20.11 kN•m M (kN•m) x (m) -91.78 - + 20.11 60.11 θB = -7.667/EI θC = 52.556/EI ∆B = -116.593/EI
  • 61. 61 40 kN 8 m 4 m 4 m 2EI EI 6 kN/m A B C ∆B = -10 mm 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.
  • 62. 62 [k]1 = 4 8 8 4 1 2 1 2 EI 8 1 2 1 2 3 2EI EI 1 2 42 2 3 2 3 [K] = EI 8 12 2 4 4 2 2 3 2 3 [k]2 = EI 8 Use 2x2 stiffness matrix: θi θj Mi Mj [ ]       =× LEILEI LEILEI k /4/2 /2/4 22
  • 63. 63 75 kN•m 6(2EI)∆/L2 = 75 kN•m 10 mm 1 37.5 kN•m 6(EI)∆/L2 = 37.5 kN•m 10 mm 2 [FEM]∆∆∆∆ 6(EI)∆/L2 = 37.5 kN•m D2 D3 = -61.27/EI rad 185.64/EI rad 40 kN 2EI EI 6 kN/m A B C ∆B = -10 mm 1 2 1 2 3 [FEM]load 32 kN•mwL2/12 = 32 kN•m 6 kN/m 1 40 kN•mPL/8 = 40 kN•m 2 40 kN 8 m 4 m 4 m 32 kN•m 40 kN•mPL/8 = 40 kN•m 75 kN•m 37.5 kN•m + -32 + 40 + 75 -37.5 = 45.5Q2 = 0 Q3 = 0 -40 - 37.5 = -77.5 D2 D3 2 42 2 3 2 3 12 8 EI = Global: [Q] = [K][D] + [QF]
  • 64. 64 6(2EI)∆/L2 = 75 kN•m [FEM]load 32 kN•mwL2/12 = 32 kN•m 6 kN/m 1 75 kN•m 6(2EI)∆/L2 = 75 kN•m 10 mm 1 [FEM]∆∆∆∆ 6 kN/m 18 m 1 2 2EI 1 q1 q2 d1 = 0 d2 = -61.27/EI = 76.37 kN•m -18.27 kN•m Member 1: [q]1 = [k]1[d]1 + [qF]1 32 kN•mwL2/12 = 32 kN•m 75 kN•m + 32 + 75 = 107 -32 + 75 = 43 31.26 kN 76.37 kN•m 16.74 kN 18.27 kN•m 4 8 8 4 1 2 1 2 8 EI =
  • 65. 65 Member 2: [q]2 = [k]2[d]2 + [qF]2 2 2 3 40 kN•mPL/8 = 40 kN•m 2 40 kN 37.5 kN•m 6(EI)D/L2 = 37.5 kN•m 10 mm 2 [FEM]load [FEM]∆∆∆∆ 40 kN•m 37.5 kN•m 6(EI)D/L2 = 37.5 kN•m PL/8 = 40 kN•m = 18.27 kN•m 0 kN•m q2 q3 + 40 - 37.5 = 2.5 -40 - 37.5 = -77.5 d2 = -61.27/EI d3 = 185.64/EI 17.72 kN 18.27 kN•m 22.28 kN 40 kN 2 2 4 4 2 2 3 2 3 8 EI =
  • 66. 66 Alternate method: Use 4x4 stiffness matrix 2EI EI 1 2 2 5 6 3 4 1 8 EI = 1.5 1.5 -0.375 12(2)/82 4 -1.5 8 1.5 -1.5 -1.5 -0.375 0.375 1.5 4 -1.5 8 1 2 3 4 1 2 3 4 [k]1 8 EI = 0.75 0.75 -0.1875 12/82 2 -0.75 4 0.75 -0.75 -0.75 -0.1875 0.1875 0.75 2 -0.75 4 3 4 5 6 3 4 5 6 [k]2 2 -0.75 -0.1875 -0.75 0.1875 -0.75 0.75 2 -0.75 4 -0.1875 0.75 1.5 4 1.5 4 0.375 -0.375 1.5 1.5 8 -1.5 -0.375 -1.5 0 0 0 0 0 0 0 0 -0.75 12 0.5625 -0.758 EI =[K] 1 2 3 4 5 6 1 4 5 62 3
  • 67. 67 Global: [Q] = [K][D] + [QF] 40 kN 8 m 4 m 4 m 2EI EI 6 kN/m A B C ∆B = -10 mm 1 2 2 5 6 3 4 1 40 kN 40 kN•m 20 kN 40 kN•m 20 kN 2 32 kN•m 24 kN 32 kN•m 24 kN 6 kN/m 1 [FEF] 53 2 1 D4 D6 = -61.27/EI = -1.532x10-3 rad 185.64/EI = 4.641x10-3 rad + (200x200/8)(-0.75)(-0.01) = 37.5 (200x200/8)(0.75)(-0.01) = -37.5 D4 D6 Q4 = 0 Q6= 0 2 4 12 2 + 8 -40 4 6 4 68 EI = -0.75 -0.75 D5 = -0.01) 8 200200 ( × + 4 6 5 D4 D6 Q4 = 0 Q6= 0 2 4 12 2 4 6 4 68 EI = + 8 -40
  • 68. 68 + 32 24 -32 24 q1 q2 q4 q3 1.5 1.5 -0.375 12(2)/82 4 -1.5 8 1.5 -1.5 -1.5 -0.375 0.375 1.5 4 -1.5 8 1 2 3 4 1 2 3 4 = (200x200) 8 q1 q2 q4 q3 = 76.37 kN•m 31.26 kN -18.27 kN•m 16.74 kN Member 1: [q]1 = [k]1[d]1 + [qF]1 1 2 3 4 1 ∆B = -10 mm 32 kN•m 24 kN 32 kN•m 24 kN 6 kN/m 1 [FEF]load d2 = 0 d1 = 0 d4 = -1.532x10-3 d3 = -0.01 6 kN/m 18 m 31.26 kN 76.37 kN•m 16.74 kN 18.27 kN•m
  • 69. 69 Member 2: + 40 20 -40 20 q3 q4 q6 q5 d4 = -1.532x10-3 d3 = -0.01 d6 = 4.641x10-3 d5 = 0 0.75 0.75 -0.1875 12/82 2 -0.75 4 0.75 -0.75 -0.75 -0.1875 0.1875 0.75 2 -0.75 4 3 4 5 6 3 4 5 6 = (200x200) 8 q3 q4 q6 q5 = 18.27 kN•m 22.28 kN 0 kN•m 17.72 kN 17.72 kN 18.27 kN•m 22.28 kN 40 kN 2 2 5 6 3 4 -10 mm = ∆B 40 kN 40 kN•m 20 kN 40 kN•m 20 kN 2 [FEF]
  • 70. 70 40 kN 8 m 4 m 4 m 2EI EI 6 kN/m A B C ∆B = -10 mm 31.26 kN 76.37 kN•m 16.74 + 22.28 kN 17.72 kN V (kN) x (m) + + -- 31.26 -16.74 22.28 -17.72 5.21 m M (kN•m) x (m) -76.37 5.06 -18.27 70.85 + - - D6 = θC = 4.641x10-3 rad d4 = θB = -1.532x10-3 rad Deflected Curve ∆B = -10 mm
  • 71. 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 8 m 4 m 4 m 2EI EI 0.6 kN/m A B C ∆C = -10 mm 20 kN•m
  • 72. 72 2EI EI 1 2 -10 mm 2 5 6 3 4 1 2 5 1 Q3 Q6 Q4 0.75 2 0.75 2 4 1.5 1.5 -0.375 12(2)/82 4 -1.5 8 1.5 -1.5 -1.5 -0.375 0.375 1.5 4 -1.5 8 1 2 3 4 1 2 3 4 [k]1 8 EI = 8 EI = 0.75 0.75 -0.1875 12/82 2 -0.75 4 0.75 -0.75 -0.75 -0.1875 0.1875 0.75 2 -0.75 4 3 4 5 6 3 4 5 6 [k]2 D3 D6 D4 -0.1875 -0.75 -0.75 D5 = -0.01 • Global: [Q] = [K][D] + [QF] •Member stiffness matrix [k]4x4 0.5625 -0.75 -0.75 12 3 4 6 3 4 6 8 EI = ) 8 200200 ( × + 3 4 6 5 QF 3 QF 6 QF 4+
  • 73. 73 4 kN 4 kN•m 2 kN 4 kN•m 2 kN 2 3.2 kN•m 2.4 kN 3.2 kN•m 2.4 kN 0.6 kN/m 1 [FEF] 4 kN 8 m 4 m 4 m 2EI EI 0.6 kN/m A B C 10 mm 20 kN•m 2EI EI 1 2 6 3 4 2 5 1 3 4 6 0.5625 0.75 -0.75 -0.75 12 2 0.75 2 3 4 6 4 8 EI = D3 D6 D4 + 9.375 37.5 37.5 2.4+2 = 4.4 -4.0 -3.2+4 = 0.8+ D3 D6 D4 -377.30/EI = -9.433x10-3 m +74.50/EI = +1.863x10-3 rad -61.53/EI = -1.538x10-3 rad= Global: [Q] = [K][D] + [QF] Q3 = 0 Q6 = 20 Q4 = 0
  • 74. 74 + 3.2 2.4 -3.2 2.4 q1 q2 q4 q3 q1 q2 q4 q3 = 43.19 kN•m 8.55 kN 6.03 kN•m -3.75 kN Member 1: [q]1 = [k]1[d]1 + [qF]1 d2 = 0 d1 = 0 d4 = -1.538x10-3 d3 = -9.433x10-3 0.6 kN/m 18 m 8.55 kN 43.19 kN•m 3.75 kN 6.03 kN•m 1 2 3 4 1 3.2 kN•m 2.4 kN 3.2 kN•m 2.4 kN 0.6 kN/m 1 [FEF]load 8 m 1.5 1.5 -0.375 12(2)/82 4 -1.5 8 1.5 -1.5 -1.5 -0.375 0.375 1.5 4 -1.5 8 1 2 3 4 1 2 3 4 8 200200× =
  • 75. 75 Member 2: + 4 2 -4 2 q3 q4 q6 q5 d4 = -1.538x10-3 d3 = -9.433x10-3 d6 = 1.863x10-3 d5 = -0.01 q3 q4 q6 q5 = -6.0 kN•m 3.75 kN 20.0 kN•m 0.25 kN 0.25 kN 6.0 kN•m 3.75 kN 4 kN 2 10 mm 2 5 6 3 4 4 kN 4 kN•m 2 kN 4 kN•m 2 kN 2 [FEF] 8 m 0.75 0.75 -0.1875 12/82 2 -0.75 4 0.75 -0.75 -0.75 -0.1875 0.1875 0.75 2 -0.75 4 3 4 5 6 3 4 5 6 8 200200× = 20 kN•m
  • 76. 76 4 kN 8 m 4 m 4 m 2EI EI 0.6 kN/m A B C 10 mm 20 kN•m Deflected Curve 0.6 kN/m 18 m 8.55 kN 43.19 kN•m 3.75 kN 6.03 kN•m V (kN) x (m) + 8.55 3.75 -0.25 0.25 kN 6.0 kN•m 3.75 kN 4 kN 2 20 kN•m M (kN•m) x (m) -43.19 6 21 + - 20 D3 = -9.433 mm D6 = +1.863x10-3 radD4 = -1.538x10-3 rad D5 = -10 mm
  • 77. 77 30 kN A C B EI2EI 4 m 2 m 9kN/m Hinge 2 m 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. Internal Hinges
  • 78. 78 [K] = EI 1 2 1 2 0.0 0.0 1.0 2.0 A C B 2EI 4 m 2 m 2 m EI 1 2 43 1 2 [k]1 = EI 2EI 2EI EI 3 1 3 1 [k]2 = 0.5EI 1EI 1EI 0.5EI 2 4 2 4 3 4 Use 2x2 stiffness matrix, θi θj Mi Mj [ ]       =× LEILEI LEILEI k /4/2 /2/4 22
  • 79. 79 0.0 0.0 D1 D2 15 kN•m B C 15 kN•m 30 kN 2 12 kN•m A B 12 kN•m 9kN/m 1 [FEF] = EI 1 2 1 2 0.0 0.0 1.0 2.0 D1 D2 = 0.0006 rad -0.0015 rad -12 15 + Global matrix: 30 kN 9kN/m Hinge A C B 1 2 43 12 kN•m 15 kN•m 1 2
  • 80. 80 Member 1: q3 q1 12 -12 += EI 2EI 2EI EI 3 1 3 1 18 0.0 = d3 d1 = 0.0 = 0.0006 12 kN•m A B 12 kN•m 9kN/m 1 [FEF]A B 1 3 1 18 kN•m 22.5 kN 13.5 kN A B 9 kN/m 1
  • 81. 81 B C 30 kN C B 2 4 2 15 kN•m B C 15 kN•m 30 kN 2 q2 q4 = -0.0015 = 0.0 d2 d4 15 -15 += 0.5EI 1EI 1EI 0.5EI 2 4 2 4 0.0 -22.5 = Member 2: 22.5 kN•m 20.63 kN9.37 kN
  • 82. 82 30 kN A C B EI2EI 4 m 2 m 9kN/m Hinge 2 m -20.63 V (kN) x (m) 22.5 9.37 -13.5 + 2.5 m M (kN•m) x (m) -18 10.13 -22.5 18.75 + - + - - θBL= 0.0006 rad θBR= -0.0015 rad 13.5 kN22.5 kN 22.5 kN•m B C 30 kN 20.63 kN9.37 kN A B 18 kN•m 9 kN/m 9.37 kN 13.5 kN 22.87 kN
  • 83. 83 20 kN A CB EI2EI 4 m 4 m 9kN/m Hinge 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.
  • 84. 84 A CB 1 2 2EI EI 6 7 4 5 4 5 6 7 0.375 0.375 1.0 -0.75 -0.75 2.0 0.0 0[K] = EI 1 2 3 1 2 3 0.5625 [k]1 = 0.75EI EI -0.75EI 2EI/L 0.375EI 0.75EI 0.75EI -0.375EI 2EI 0.75EI EI -0.75EI -0.75EI - 0.375EI 0.375EI -0.75EI 4 5 1 2 4 5 1 2 0.375EI 0.375EI -0.1875EI 0.5EI -0.375EI -0.375EI [k]2 = 0.1875EI -0.375EIEI 0.375EI - 0.1875EI 0.1875EI 0.375EI 0.5EI -0.375EI EI 1 3 6 7 1 3 6 7 1 2 3
  • 85. 85 Q1 = -20 Q3 = 0.0 Q2 = 0.0 D1 D3 D2 18 0.0 -12+ D1 D3 D2 = -0.02382 m 0.008933 rad -0.008333 rad [FEM] 12 kN•m A B 12 kN•m 18 kN18 kN 1 9kN/m 20 kN A C B 9kN/m A CB 1 2 6 7 4 5 2EI EI Global: = EI 1 2 3 1 0.5625 0.375 -0.75 2 -0.75 2.0 0.0 3 0.375 0 1.0 12 kN•m 18 kN 20 kN 1 2 3
  • 86. 86 A B 1 2EI 1 24 5 [FEF] 12 kN•m A B 12 kN•m 18 kN18 kN 1 9kN/m + 12 18 -12 18 Member 1: q4 q5 q1 q2 = 0.75EI EI -0.75EI 2EI/L 0.375EI 0.75EI 0.75EI -0.375EI 2EI 0.75EI EI -0.75EI -0.75EI - 0.375EI 0.375EI -0.75EI 4 5 1 2 4 5 1 2 = 0.0 = 0.0 = -0.02382 = -0.00833 d5 d4 d2 d1 = 107.32 44.83 0.0 -8.83 A B 107.32 kN•m 8.83 kN 44.83 kN 9kN/m 1
  • 87. 87 Member 2: CB 2 EI 1 3 6 7 44.66 kN•m B C 11.16 kN 11.16 kN 2 + 0 0 0 0 = -0.02382 = 0.008933 = 0.0 = 0.0 d3 d1 d7 d6 q1 q3 q7 q6 0.375EI 0.375EI -0.1875EI 0.5EI -0.375EI -0.375EI = 0.1875EI -0.375EIEI 0.375EI - 0.1875EI 0.1875EI 0.375EI 0.5EI -0.375EI EI 1 3 6 7 1 3 6 7 = 0.0 -11.16 -44.66 11.16
  • 88. 88 20 kN A CB 4 m 4 m 9kN/m V (kN) x (m) 44.83 8.83 -11.16 -11.16 + - M (kN•m) x (m) -107.32 -44.66 - - A B 107.32 kN•m 8.83 kN44.83 kN 9kN/m 1 44.66 kN•m B C 11.16 kN 11.16 kN 2 θBL= -0.008333 rad θBR= 0.008933 rad
  • 89. 89 30 kN A C B EI2EI 4 m 2 m 9kN/m Hinge 40 kN•m 2 m 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.
  • 90. 90 [K] = EI 1 2 1 2 0.0 0.0 1.0 2.0 A C B 2EI 4 m 2 m 2 m EI 1 2 43 [k]1 = EI 2EI 2EI EI 3 1 3 1 [k]2 = 0.5EI 1EI 1EI 0.5EI 2 4 2 4 3 4 1 2 Use 2x2 stiffness matrix: θi θj Mi Mj [ ]       =× LEILEI LEILEI k /4/2 /2/4 22
  • 91. 91 15 kN•m B C 15 kN•m 30 kN 2 12 kN•m A B 12 kN•m 9kN/m 1 [FEM] Global matrix: A C B 1 2 43 1 2 12 kN•m 15 kN•m 30 kN A C B EI2EI 9kN/m Hinge 40 kN•m Q1 = 40 Q2 = 0.0 D1 D2 -12 15 + D1 D2 = 0.0026 rad -0.0015 rad = EI 1 2 1 2 0.0 0.0 1.0 2.0 40 kN•m
  • 92. 92 12 kN•m A B 12 kN•m 9kN/m 1 [FEF]A B 1 3 1 40 kN•mA B 38 kN•m 9 kN/m 1.5 kN37.5 kN Member 1: 38 40 = q3 q1 = 0.0 = 0.0026 d3 d1 12 -12 += EI 2EI 2EI EI 3 1 3 1
  • 93. 93 C B 2 4 2 15 kN•m B C 15 kN•m 30 kN 2 Member 2: q2 q4 = -0.0015 = 0.0 d2 d4 15 -15 += 0.5EI 1EI 1EI 0.5EI 2 4 2 4 0.0 -22.5 = 22.5 kN•m B C 30 kN 20.63 kN9.37 kN
  • 94. 94 9.37 kN 1.5 kN 30 kN A C B 4 m 2 m 9kN/m Hinge 40 kN•m 2 m -20.63 V (kN) x (m) 37.5 9.371.5+ M (kN•m) x (m) -38 40 -22.5 + - 18.75 + -- θBL= 0.0026 rad θBR=- 0.0015 rad 7.87 kN 40 kN•mA B 38 kN•m 9 kN/m 1.5 kN37.5 kN 22.5 kN•m B C 30 kN 20.63 kN9.37 kN
  • 95. 95 20 kN A CB EI2EI 4 m 4 m 9kN/m Hinge40 kN•m 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
  • 96. 96 A C B 1 2 6 7 4 5 1 2 3 20 kN A CB EI2EI 4 m 4 m 9kN/m Hinge40 kN•m 12 kN•m 18 kN 12 kN•m 18 kN 9 kN/m 1[FEM] θi ∆j θj Mi ∆i Vj Mj Vi [ ]             − −−− − − =× LEILEILEILEI LEILEILEILEI LEILEILEILEI LEILEILEILEI k /4/6/2/6 /6/12/6/12 /2/6/4/6 /6/12/6/12 22 2323 22 2323 44
  • 97. 97 1 22A C B 6 7 4 5 [K] = EI 1 2 3 1 2 3 6 7 4 5 0.5625 0.0 0 0.375 0.375 1.0 -0.75 -0.75 2.0 [k]1 = 0.75EI EI -0.75EI 2EI 0.375EI 0.75EI 0.75EI -0.375EI 2EI 0.75EI EI -0.75EI -0.75EI - 0.375EI 0.375EI -0.75EI 4 5 1 2 4 5 1 2 1 0.375EI 0.375EI -0.1875EI 0.5EI -0.375EI -0.375EI [k]2 = 0.1875EI -0.375EIEI 0.375EI - 0.1875EI 0.1875EI 0.375EI 0.5EI -0.375EI EI 1 3 6 7 1 3 6 7 2EI EI 1 2 3
  • 98. 98 Q1 = -20 Q3 = 0.0 Q2 = 40 D1 D3 D2 18 0.0 -12+ D1 D3 D2 = -0.01316 m 0.0049333 rad -0.002333 rad A C B 1 2 6 7 4 5 2EI EI 20 kN EI 9kN/m 40 kN•m 20 kN 40 kN•m 12 kN•m 18 kN 12 kN•m 18 kN 9 kN/m 1[FEF] = EI 1 2 3 1 2 3 0.5625 0.0 0 0.375 0.375 1.0 -0.75 -0.75 2.0 12 kN•m 18 kN Global: 1 2 3
  • 99. 99 1 2EI [q]1 1 1 2 1 2EI4 5 + 12 18 -12 18 q4 q5 q1 q2 = 0.75EI EI -0.75EI 2EI 0.375EI 0.75EI 0.75EI -0.375EI 2EI 0.75EI EI -0.75EI -0.75EI - 0.375EI 0.375EI -0.75EI 4 5 1 2 4 5 1 2 = 87.37 49.85 40 -13.85 12 kN•m 18 kN 12 kN•m 18 kN 9 kN/m 1[FEF] Member 1: [q]1 = [k]1[d]1 + [qF]1 13.85 kN 40 kN•m87.37 kN•m 49.85 kN = 0.0 = 0.0 = -0.01316 = -0.002333 d5 d4 d2 d1 12 kN•m 18 kN 12 kN•m 18 kN 1
  • 100. 100 22 C B 1 3 EI 6 7 Member 2: [q]2 = [k]2[d]2 + [qF]2 + 0 0 0 0 q1 q3 q7 q6 0.375EI 0.375EI -0.1875EI 0.5EI -0.375EI -0.375EI = 0.1875EI -0.375EIEI 0.375EI - 0.1875EI 0.1875EI 0.375EI 0.5EI -0.375EI EI 1 3 6 7 1 3 6 7 = 0.0 -6.18 -24.69 6.18 24.69 kN•m 6.18 kN6.18 kN = -0.01316 = 0.004933 = 0.0 = 0.0 d3 d1 d7 d6 22 [q]2
  • 101. 101 20 kN A C B EI2EI 4 m 4 m 9kN/m 40 kN•m 24.69 kN•m B C 6.18 kN6.18 kN A B 87.37 kN•m 13.85 kN49.85 kN 40 kN•m M (kN•m) x (m) - -87.37 + 40 - -24.69 θBL= -0.002333 rad θBR = 0.004933 rad 49.85 -6.18 V (kN) x (m) -6.18 - 13.85+
  • 102. 102 • Fixed-End Forces (FEF) - Axial - Bending • Curvature Temperature Effects
  • 103. 103 Tm> TR • Thermal Fixed-End Forces (FEF) Tl Tt 2 bt m TT T + = Tt Tb > Tt β F AF F BF B F BMF AM Rmaxial TTT −=∆ )( tbbending TTT −=∆ )( )()( d T yTy ∆ =∆ y β )(tan d T d TT tb ∆ = − =β A A B σaxialσaxial A B σy y TR Tm Tt Tb Tb - Tt Room temp = TR ct cb d TR NA
  • 104. 104 - Axial A B σaxialσaxial F AF F BF dAF A axial F A ∫= σ dAE axial∫= ε dATE axial∫ ∆= )(α axialTEA )(∆= α AETTF RmA F axial )()( −=α dATE axial ∫∆= )(α Tt Tb > Tt Tm> TR Rmaxial TTT −=∆ )( TR Tm
  • 105. 105 - Bending A B σy y dAyM A y F A ∫= σ dATyE y∫ ∆= )(α dA d T yyE∫ ∆ = )(α ∫ ∆ = dAy d T E 2 )(α EI d TT F ul A F bending )()( − =α F BMF AM dAyE∫= ε tbbending TTT −=∆ )( )()( d T yTy ∆ =∆ y β Tt Tb
  • 106. 106 O´ ds´ dx After deformation dθ dθ Before deformation y dx ds = dx y ρ )()( θρ ddx = )( θdy )( 1 dx dθ ρ = • Elastic Curve: Bending
  • 107. 107 dx d T yyd )()( ∆ =αθ dx d T d )()( ∆ =αθ Tb Tt y y d T∆ • Bending Temperature dx Tt Tb Tl > Tu y Tl > Tu Tu O dθ 2 bt m TT T + = dθ MM Tb cb ct Tt ∆T = Tb - Tt d T∆ =β EI M d T dx d = ∆ == )( 1 )( α ρ θ
  • 108. 108 A CB EI2EI 4 m 4 m T2 = 40oC T1 = 25oC 182 mm 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.
  • 109. 109 1 2 3 1 2 Mean temperature = (40+25)/2 = 32.5 Room temp = 32.5oC FEM +7.5 oC -7.5 oC a(∆T /d)EI19.78 kN•m = 19.78 kN•m A CB EI2EI 4 m 4 m T2 = 40oC T1 = 25oC 182 mm mkNEI d T F F bending •=×× − ×= ∆ = − 78.19)502002)( 182.0 2540 )(1012()2)(( 6 α
  • 110. 110 [M1] = 3EIθ1 + (1.978x10-3)EI 0 Element 1: M1 M2 q1 q2 [q] = [k][d] + [qF] + 1.978 -1.978 (10-3) EI= 1 2 2 1 2 1 2 1 EI Element 2: M3 M1 q3 q1 + 0 0 = 0.5 1 1 0.5 1 3 1 3 EI θ1 = -0.659x10-3 rad 1 2 3 A C B EI2EI 4 m 4 m 1 2 19.78 kN•m19.78 kN•m 182 mm
  • 111. 111 Element 2: M3 M1 q3 q1 = + 0 0 0.5 1 1 0.5 1 3 1 3 EI = -0.659x10-3 = 0 = 6.59 kN•m -26.37 kN•m = -0.659x10-3 = 0 = -3.30 kN•m -6.59 kN•m Element 1: M1 M2 = q1 q2 + 19.78 -19.78 1 2 2 1 2 1 2 1 EI 4.95 kN4.95 kN 26.37 kN•m 6.59 kN•m A B 6.59 kN•m 3.3 kN•m B C 2.47 kN2.47 kN 1 2 3 A C B EI2EI 4 m 4 m 1 2 19.78 kN•m19.78 kN•m 182 mm
  • 112. 112 A CB 4 m 4 m +7.5 oC -7.5 oC 3.3 kN•m 26.37 kN•m 4.95 kN 2.47 kN 2.47 kN M (kN•m) x (m) 26.37 6.59 -3.30 + - V (kN) x (m) -4.95 -2.47 - - θB = -0.659x10-3 radDeflected curve x (m)
  • 113. 113 A CB EI, AE2EI, 2AE 4 m 4 m T2 = 40oC T1 = 25oC 182 mm 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
  • 114. 114 A CB EI2EI 4 m 4 m T2 = 40oC T1 = 25oC 1 2 3 7 8 9 1 2 4 5 6 Element 1: mkN m mkNm L AE /)10(2 )4( )/10200)(1020(2)2( 6 2623 = ×× = − mkN m mmkN L EI •= ××× = − )10(20 )4( )1050)(/10200(24)2(4 3 4626 mkN m mmkN L EI •= ××× = − )10(10 )4( )1050)(/10200(22)2(2 3 4626 kN m mmkN L EI )10(5.7 )4( )1050)(/10200(26)2(6 3 2 4626 2 = ××× = − mkN m mmkN L EI /)10(75.3 )4( )1050)(/10200(212)2(12 3 3 4626 3 = ××× = −
  • 115. 115 Fixed-end forces due to temperatures Mean temperature(Tm) = (40+25)/2 = 32.5 oC , TR = 28 oC A CB EI, AE2EI, 2AE 4 m 4 m T2 = 40oC T1 = 25oC 182 mm mkNEI d T F F bending •=×× − ×= ∆ = − 78.19)502002)( 182.0 2540 )(1012()2)(( 6 α 19.78 kN•m19.78 kN•m 432 kN432 kN A B +12 oC -3 oC kNmkNmAETF F axial 432)/10200)(310202)(285.32)(1012()( 2626 =×−××−×=∆= − α
  • 116. 116 FEM +12 oC -3 oC 19.78 kN•m A B 19.78 kN•m 432 kN432 kN Element 1: [q] = [k][d] + [qF] 0 0 0 d1 d2 d3 q4 q6 q5 q1 q2 q3 = 4 5 6 1 2 3 4 1 - 2x106 0.00 0.00 2x106 0.00 0.00 2 0.00 -3750 -7500 0.00 3750 -7500 3 0.00 7500 10x103 0.00 -7500 20x103 2x106 0.00 0.00 -2x106 0.00 0.00 5 0.00 3750 7500 0.00 -3750 7500 6 0.00 7500 20x103 0.00 -7500 10x103 432 -19.78 0.00 -432 0.00 19.78 + A CB EI, AE2EI, 2AE 4 m 4 m T2 = 40oC T1 = 25oC 182 mm 1 2 3 7 8 9 1 2 4 5 6
  • 117. 117 Element 2: [q] = [k][d] + [qF] d1 d3 d2 0 0 0 0 0 0 0 0 0 + q1 q3 q2 q7 q8 q9 - 1x106 0.00 0.00 1x106 0.00 0.00 0.00 -1875 -3750 0.00 1875 -3750 0.00 3750 5x103 0.00 -3750 10x103 1x106 0.00 0.00 -1x106 0.00 0.00 0.00 1875 3750 0.00 -1875 3750 0.00 3750 10x103 0.00 -3750 5x103 7 8 91 2 3 = 1 2 3 7 8 9 A CB EI, AE2EI, 2AE 4 m 4 m T2 = 40oC T1 = 25oC 182 mm 1 2 3 7 8 9 1 2 4 5 6
  • 118. 118 D1 D3 D2 = 0.000144 m -719.3x10-6 rad -0.0004795 m Q1 = 0.0 Q3 = 0.0 Q2 = 0.0 D1 D3 D2 -432 19.78 0.0+= 1 2 3 1 3x106 0.0 0.0 2 0.0 5625 -3750 3 0.0 -3750 30x103 Global: A CB EI, AE2EI, 2AE 4 m 4 m T2 = 40oC T1 = 25oC 182 mm 1 2 3 7 8 9 1 2 4 5 6
  • 119. 119 = 144x10-6 = -479.5x10-6 = -719.3x10-6 Element 1: 0 0 0 d1 d2 d3 q4 q6 q5 q1 q2 q3 = 4 5 6 1 2 3 4 1 - 2x106 0.00 0.00 2x106 0.00 0.00 2 0.00 -3750 -7500 0.00 3750 -7500 3 0.00 7500 10x103 0.00 -7500 20x103 2x106 0.00 0.00 -2x106 0.00 0.00 5 0.00 3750 7500 0.00 -3750 7500 6 0.00 7500 20x103 0.00 -7500 10x103 432 -19.78 0.00 -432 0.00 19.78 + = q4 q6 q5 q1 q2 q3 144.0 kN -23.38 kN•m -3.60 kN -144.0 kN 3.60 kN 9.00 kN•m A B 1 2 3 4 5 6 A B 144 kN 3.60 kN 23.38 kN•m 144 kN 3.60 kN 9 kN•m
  • 120. 120 = 144x10-6 = -479.5x10-6 = -719.3x10-6 144 kN -9 kN•m -3.6 kN -144 kN 3.6 kN -5.39 kN•m = q1 q3 q2 q7 q8 q9 B C 7 8 9 1 2 3 144 kN B C 3.60 kN 9 kN•m 144 kN 3.60 kN 5.39 kN•m Element 2: d1 d3 d2 0 0 0 0 0 0 0 0 0 + q1 q3 q2 q7 q8 q9 = - 1x106 0.00 0.00 1x106 0.00 0.00 0.00 -1875 -3750 0.00 1875 -3750 0.00 3750 5x103 0.00 -3750 10x103 1x106 0.00 0.00 -1x106 0.00 0.00 0.00 1875 37500 0.00 -1875 3750 0.00 3750 10x103 0.00 -3750 5x103 7 8 91 2 3 1 2 3 7 8 9
  • 121. 121 A CB EI2EI 4 m 4 m T2 = 40oC T1 = 25oC Isolate axial part from the system RC RA RA RA = RC + Compatibility equation: dC/A = 0 RA = 144 kN RC = -144 kN 0)4)(285.32(1012 )4( 2 )4( 6 =−×++ − AE R AE R AA
  • 122. 122 A B 144 kN 3.60 kN 23.38 kN•m 144 kN 3.60 kN 9 kN•m 144 kN B C 3.60 kN 9 kN•m 144 kN 3.60 kN 5.39 kN•m V (kN) x (m) -3.6 - M (kN•m) x (m) 23.38 8.98 -5.39 + - D3 = -719.3x10-6 radDeflected curve x (m) A CB EI2EI 4 m 4 m T2 = 40oC T1 = 25oC D2 = -0.0004795 mm
  • 123. 123 Skew Roller Support - Force Transformation - Displacement Transformation - Stiffness Matrix
  • 124. 124 m i j m i j x´ y´ x y • Displacement and Force Transformation Matrices 1 2 3 4 5 6 θy θx 4´ 5´ 6´ 1´ 2´ 3´
  • 125. 125 λx λy i j θy θx y´ x´ 4´5´ 6´ 1´ 2´ 3´ θy θx m i j x y 1 2 3 4 5 6 Force Transformation L xx ij x − =λ L yy ij y − =λ yx qqq θθ coscos 5'4'4 −= xy qqq θθ coscos 5'4'5 −= 6'6 qq =           6 5 4 q q q           − = 100 0 0 xy yx λλ λλ           6' 5' 4' q q q                                         − − =                     6' 5' 4' 3' 2' 1' 6 5 4 3 2 1 100000 0000 0000 000100 0000 0000 q q q q q q λλ λλ λλ λλ q q q q q q xy yx xy yx [ ] [ ] [ ]'qTq T =
  • 126. 126 x y θy θx x y θy θx d4 d4 cos θx d4 cos θy θy θx d5 θx θy d5 cos θx d5 cos θy x´ Displacement Transformation j 4´ 5´ 6´j 4 5 6 λx λy yx θdθdd coscos' 544 += xy θdθdd' coscos 545 +−= 66' dd =           6' 5' 4' d d d           −= 100 0 0 xy yx λλ λλ           6 5 4 d d d                                         − − =                     6 5 4 3 2 1 6' 5' 4' 3' 2' 1' 100000 0000 0000 000100 0000 0000 d d d d d d λλ λλ λλ λλ d d d d d d xy yx xy yx [ ] [ ][ ]dTd ='
  • 127. 127 [ ] [ ] [ ]'qTq T = [ ] [ ][ ] [ ])'( FT qd'k'T += [ ] [ ][ ] [ ] [ ]FTT qTd'k'T '+= [ ] [ ] [ ][ ][ ] [ ] [ ] [ ][ ] [ ]FFTT qdkqTdTk'Tq +=+= ' [ ] [ ] [ ][ ]TkTkTherefore T ', = [ ] [ ] [ ]FTF qTq '= [ ] [ ] [ ]'qTq T = [ ] [ ][ ]dTd =' [ ] [ ] [ ][ ]TkTk T '=
  • 128. 128 i j θj θi 4* 5* 6* 1* 2* 3* λix = cos θi λiy = sin θi λjx = cos θj λjy = sin θj [q*] = [T]T[q´] 1 Stiffness matrix i j 4´ 5´ 6´ 1´ 2´ 3´                     *6 5* *4 *3 2* *1 q q q q q q                     − − = 100000 0000 0000 000100 0000 0000 jxjy jyjx ixiy iyix λλ λλ λλ λλ 1 4 5 62 3 1* 2* 3* 4* 5* 6* [ T ]T                     6' 5' 4' 3' 2' 1' q q q q q q
  • 129. 129 [ ]                     − − = 100000 0000 0000 000100 0000 0000 jxjy jyjx ixiy iyix λλ λλ λλ λλ T 1 4 5 62 3 1 2 3 4 5 6 [ ]                     − −−− − − − − = LEILEILEILEI LEILEILEILEI LAELAE LEILEILEILEI LEILEILEILEI LAELAE k /4/60/2/60 /6/120/6/120 00/00/ /2/60/4/60 /6/120/6/120 00/00/ ' 22 2323 22 2323 1 3 4 6 2 5 1 2 3 4 5 6
  • 130. 130 [ k ] = [ T ]T[ k´ ][T] = Ui Vi Mi Uj Vj Mj Vj Mj - λiy 6EI L2 λix 6EI L2 2EI L λjy 6EI L2 - λjx 6EI L2 4EI L Ui Vi Mi - λiy 6EI L2 λix 6EI L2 4EI L λjy 6EI L2 λjx 6EI L2 - 2EI L Uj AE L - λixλiy)( 12EI L3 AE L λiy 2 + 12EI L3 λix 2 )( λix 6EI L2 λix 6EI L2 AE L λiyλjx - 12EI L3 λixλjy)-( AE L λixλjx + 12EI L3 λiyλjy)-( λjy 6EI L2 AE L λjx 2 + 12EI L3 λjy 2 )( λjy 6EI L2 λjx 6EI L2 - AE L - λjxλjy)( 12EI L3 - λjx 6EI L2 AE L λixλjy - 12EI L3 λiyλjx)-( AE L - λixλiy)( 12EI L3 - λiy 6EI L2 - λiy 6EI L2 AE L λixλjx + 12EI L3 λiy λjy)-( )( AE L λix 2 + 12EI L3 λiy 2 AE L λiyλjy + 12EI L3 λix λjx )-( AE L λiyλjy + 12EI L3 λixλjx)-( 12EI L3 λjx 2 ) AE L - ( λixλjy- λiyλjx ) 12EI L3 AE L λiyλjx - 12EI L3 λixλjy)-( AE L - λjxλjy)( 12EI L3 AE L λjy 2 +(
  • 131. 131 40 kN 4 m4 m 22.02 o 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.
  • 132. 132 40 kN 40 kN•m 20 kN 40 kN•m 20 kN [ q´F ] [FEF] 1Global i j 1 i j Local 40 kN 4 m4 m 22.02o 20sin22.02=7.5 20cos22.02=18.54 1´ 2´ 3´ 4´ 5´ 6´ 4 5 6 x* x´1* 2* 3* λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271, λjy = cos 67.98o = 0.3749 22.02 o
  • 133. 133 x* x´ 1 4 5 6 1* 2* 3* • Transformation matrix Member 1: [ q ] = [ T ]T[q´] 1 i j1´ 2´ 3´ 4´ 5´ 6´ Global Local λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271, λjy = cos 67.98o = 0.3749                     *3 2* *1 6 5 4 q q q q q q                     6' 5' 4' 3' 2' 1' q q q q q q                     − = 100000 09271.03749.0000 03749.09271.0000 000100 000010 000001 1´ 4´ 5´ 6´2´ 3´ 4 5 6 1* 2* 3* [ ]                     − − = 100000 0000 0000 000100 0000 0000 jxjy jyjx ixiy iyix T T λλ λλ λλ λλ
  • 134. 134 [k´]1 = 103 1´ 2´ 3´ 4´ 5´ 6´ 1´ 4´ 5´ 6´2´ 3´ -150.0 0.000 0.000 150.0 0.000 0.000 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 0.000 3.750 10.00 0.000 -3.750 20.00 150.0 0.000 0.000 -150.0 0.000 0.000 0.000 0.9375 3.750 0.000 -0.9375 3.750 0.000 3.750 20.00 0.000 -3.750 10.00 1 i j1´ 2´ 3´ 4´ 5´ 6´ Local • Local stiffness matrix [ ]                     − −−− − − − − =× LEILEILEILEI LEILEILEILEI LAELAE LEILEILEILEI LEILEILEILEI LAELAE k /4/60/2/60 /6/120/6/120 00/00/ /2/60/4/60 /6/120/6/120 00/00/ 22 2323 22 2323 66 θi ∆j θj∆i δjδi Mi Vj Mj Vi Nj Ni
  • 135. 135 Stiffness matrix [k´]: [k´]1 = 103 1´ 2´ 3´ 4´ 5´ 6´ 1´ 4´ 5´ 6´2´ 3´ -150.0 0.000 0.000 150.0 0.000 0.000 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 0.000 3.750 10.00 0.000 -3.750 20.00 150.0 0.000 0.000 -150.0 0.000 0.000 0.000 0.9375 3.750 0.000 -0.9375 3.750 0.000 3.750 20.00 0.000 -3.750 10.00 Stiffness matrix [k*]: [ k* ] = [ T ]T[ k´ ][T] [k*]1 = 103 4 5 6 1* 2* 3* 4 1* 2* 3*5 6 -139.0 0.351 1.406 129.0 1.406 51.82 -56.25 -0.869 -3.750 51.82 21.90 -3.476 0.000 3.750 10.00 1.406 -3.476 20.00 150.0 0.000 0.000 -139.0 -56.25 0.000 0.000 0.9375 3.750 0.351 -0.869 3.750 0.000 3.750 20.00 1.406 -3.750 10.00 x* x´ 1 4 5 6 1* 2* 3* 1 i j1´ 2´ 3´ 4´ 5´ 6´ Global Local 4 5 6 2*
  • 136. 136 Global Equilibrium: Q1 = 0.0 Q3 = 0.0 D1* D3* = 36.37x10-6 m 0.002 rad [Q] = [K][D] + [QF] = 103 1* 3* 1* 129 1.406 3* 1.406 20.0 D1* D3* -7.5 -40+ x* x´1 4 5 6 1* 2* 3* 4 m4 m 22.02 o 40 kN x* x´ 4 5 6 2* 40 kN 40 kN•m 20 kN 40 kN•m 20 kN [ q´F ] 20sin22.02=7.5 20cos22.02=18.54
  • 137. 137 Member Force : [q] = [k*][D] + [qF] 0 40.0 20 -7.54 18.54 -40.0 + q4 q6 q5 q1* q2* q3* = 103 4 5 6 1* 2* 3* 4 1* 2* 3*5 6 -139.0 0.351 1.406 129.0 1.406 51.82 -56.25 -0.869 -3.750 51.82 21.90 -3.476 0.000 3.750 10.00 1.406 -3.476 20.00 150.0 0.000 0.000 -139.0 -56.25 0.000 0.000 0.9375 3.750 0.351 -0.869 3.750 0.000 3.750 20.00 1.406 -3.750 10.00 0 0 0 36.4x10-6 0 2x10-3 -5.06 60 27.5 0 13.48 0 = 40 kN 13.48 kN 60 kN•m 27.5 kN 5.06 kN 4 m4 m 22.02 o 40 kN x* x´ 1 4 5 6 1* 2* 3* 40 kN 40 kN•m 7.5 18.54 40 kN•m 20 kN 4 5 6 2*
  • 138. 138 4 m4 m 22.02 o 40 kN 5.05 kN 27.51 kN 60.05 kN•m 13.47 kN 40 kN + Bending moment diagram (kN•m) Deflected shape 50.04 - -60.05 0.002 rad
  • 139. 139 40 kN 22.02o 8 m 4 m 4 m 2EI, 2AE EI, AE 6 kN/m 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.
  • 140. 140 40 kN 22.02 o 8 m 4 m 4 m 2EI, 2AE EI, AE 6 kN/m 1 21 2 3 4 5 6 7* 8* 9* Global 21´ 2´ 3´ 4´ 5´ 6´ 11´ 2´ 3´ 4´ 5´ 6´ Local mkN m mkNm L AE •×= × = 3 262 10150 )8( )/10200)(006.0( mkN m mmkN L EI •×= × = 3 426 1020 )8( )0002.0)(/10200(44 mkN L EI •×= 3 1010 2 kN m mmkN L EI 3 2 426 2 1075.3 )8( )0002.0)(/10200(66 ×= × = mkN m mmkN L EI /109375.0 )8( )0002.0)(/10200(1212 3 3 426 2 ×= × =
  • 141. 141 [k]1 = [k´]1 = 2x103 4 5 6 1 2 3 4 1 2 35 6 -150.0 0.000 0.000 150.0 0.000 0.000 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 0.000 3.750 10.00 0.000 -3.750 20.00 150.0 0.000 0.000 -150.0 0.000 0.000 0.000 0.9375 3.750 0.000 -0.9375 3.750 0.000 3.750 20.00 0.000 -3.750 10.00 1 21 2 3 4 5 6 7* 8* 9* Global Local : Member 1 14 5 6 1 2 3 [ q ] 11´ 2´ 3´ 4´ 5´ 6´ [ q´] [q] = [q´] Thus, [k] = [k´]
  • 142. 142 Member 2: Use transformation matrix, [q*] = [T]T[q´] q1´ q3´ q2´ q4´ q5´ q6´ q1 q3 q2 q7* q8* q9* 0.9271 0.3749 -0.3749 0.9271 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 = 1 2 3 7* 8* 9* 1´ 4´ 5´ 6´2´ 3´ 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 21 2 3 4 5 6 7* 8* 9* λix = cos 0o = 1, λiy = cos 90o = 0 λjx = cos 22.02o = 0.9271, λjy = cos 67.98o = 0.3749 21 2 3 x* x´7* 8* 9*[ q* ] 21´ 2´ 3´ 4´ 5´ 6´ [ q´ ]
  • 143. 143 Stiffness matrix [k´]: [k´]2 = 103 1´ 2´ 3´ 4´ 5´ 6´ 1´ 4´ 5´ 6´2´ 3´ -150.0 0.000 0.000 150.0 0.000 0.000 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 0.000 3.750 10.00 0.000 -3.750 20.00 150.0 0.000 0.000 -150.0 0.000 0.000 0.000 0.9375 3.750 0.000 -0.9375 3.750 0.000 3.750 20.00 0.000 -3.750 10.00 Stiffness matrix [k*]: [k*] = [T]T[k´][T] [k*]2 = 103 1 2 3 7* 8* 9* 1 7* 8* 9*2 3 -139.0 0.351 1.406 129.0 1.406 51.82 -56.25 -0.869 -3.477 51.82 21.90 -3.476 0.000 3.750 10.00 1.406 -3.476 20.00 150.0 0.000 0.000 -139.0 -56.25 0.000 0.000 0.9375 3.750 0.351 -0.869 3.750 0.000 3.750 20.00 1.406 -3.477 10.00 21 2 3 x* x´7* 8* 9*[ q* ] 21´ 2´ 3´ 4´ 5´ 6´ [ q´ ]
  • 144. 144 1 21 2 3 4 5 6 7* 8* 9* 1 2 3 4 5 6 8* [k]1= 2x103 4 5 6 1 2 3 4 1 2 35 6 -150.0 0.000 0.000 150.0 0.000 0.000 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 0.000 3.750 10.00 0.000 -3.750 20.00 150.0 0.000 0.000 -150.0 0.000 0.000 0.000 0.9375 3.750 0.000 -0.9375 3.750 0.000 3.750 20.00 0.000 -3.750 10.00 [k*]2 = 103 1 2 3 7* 8* 9* 1 7* 8* 9*2 3 -139.0 0.351 1.406 129.0 1.406 51.82 -56.25 -0.869 -3.477 51.82 21.90 -3.476 0.000 3.750 10.00 1.406 -3.476 20.00 150.0 0.000 0.000 -139.0 -56.25 0.000 0.000 0.9375 3.750 0.351 -0.869 3.750 0.000 3.750 20.00 1.406 -3.477 10.00
  • 145. 145 Global: = -509.84x10-6 rad 18.15x10-6 m 0.00225 rad 58.73x10-6 m D3 D1 D9* D7* 0 0 0 0 + -32 + 40 = 8 0 -40 -7.5 D3 D1 D9* D7* 32 kN•m 24 kN 32 kN•m 24 kN 6 kN/m 1 40 kN 6 kN/m 1 21 2 3 4 5 6 7* 8* 9* 1 2 3 4 5 6 8* = 103 0 0 -139 450 10 1.406 60 0 1.406 1.406 -139 129 0 10 1.406 20 1 3 7* 9* 1 3 7* 9* 40 kN 40 kN•m 7.5 18.54 40 kN•m 20 kN 40 kN•m 7.5 40 kN•m 32 kN•m
  • 146. 146 Member 1: [ q ] = [k ][d] + [qF] 0 32 24 0 24 -32 + 0 0 0 d1=18.15x10-6 0 d3=-509.84 x10-6 q4 q6 q5 q1 q2 q3 = 2x103 4 5 6 1 2 3 4 1 2 35 6 -150.0 0.000 0.000 150.0 0.000 0.000 0.000 -0.9375 -3.750 0.000 0.9375 -3.750 0.000 3.750 10.00 0.000 -3.750 20.00 150.0 0.000 0.000 -150.0 0.000 0.000 0.000 0.9375 3.750 0.000 -0.9375 3.750 0.000 3.750 20.00 0.000 -3.750 10.00 -5.45 kN 21.80 kN•m 20.18 kN 5.45 kN 27.82 kN -52.39 kN•m = q4 q6 q5 q1 q2 q3 6 kN/m 5.45 kN 20.18 kN 21.80 kN•m 5.45 kN 27.82 kN 52.39 kN•m 32 kN•m 24 kN 32 kN•m 24 kN 6 kN/m 114 5 6 1 2 3
  • 147. 147 q1 q3 q2 q7* q8* q9* 21 3 x* x´7* 9* = 103 1 2 3 7* 8* 9* 1 7* 8* 9*2 3 -139.0 0.351 1.406 129.0 1.406 51.82 -56.25 -0.869 -3.750 51.82 21.90 -3.476 0.000 3.750 10.00 1.406 -3.476 20.00 150.0 0.000 0.000 -139.0 -56.25 0.000 0.000 0.9375 3.750 0.351 -0.869 3.750 0.000 3.750 20.00 1.406 -3.750 10.00 0 40 20 -7.5 18.54 -40 + 18.15x10-6 -509.84x10-6 0 58.73x10-6 0 0.00225 q1 q3 q2 q7* q8* q9* -5.45 kN 52.39 kN•m 26.55 kN 0 kN 14.51 kN 0 kN•m = 5.45 kN 26.55 kN 52.39 kN•m 14.51 kN 40 kN 20sin22.02=7.5 20cos22.02=18.54 40 kN 40 kN•m 20 kN 40 kN•m 20 kN [ q´F ] 2 2 8* Member 2: [ q ] = [k ][d] + [qF]
  • 148. 148 5.45 kN 26.55 kN 52.39 kN•m 14.51 kN 40 kN 6 kN/m 5.45 kN 27.82 kN 52.39 kN•m 5.45 kN 20.18 kN 21.80 kN•m 40 kN 22.02o 8 m 4 m 4 m 6 kN/m 5.45 kN 20.18 kN 21.80 kN•m 14.51 kN 54.37 kN 3.36 m M (kN•m) x (m) 53.81 -21.8 - V (kN) x (m) 26.55 + 20.18 + - - -27.82 14.51cos 22.02o =13.45 kN -52.39 - 12.14 + -13.45