Matrix beam

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  • 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 Support1

2. Slope Deflection Equations PiwjkCjsettlement = jMijPiwjMjii j2 3. Degrees of Freedom MBA1 DOF: C2 DOF: , LAP B 3 4. Stiffness DefinitionkAAkBA1BA L k AA =4 EI Lk BA =2 EI L4 5. kBBkABAB1 L k BB =4 EI Lk AB =2 EI L5 6. Fixed-End Forces Fixed-End Forces: Loads P L/2PL 8L/2PL 8L P 2P 2w wL2 12 wL 2LwL2 12 wL 26 7. General CasePiwjkCjsettlement = jMijPiwjMjii j7 8. MijPijwMjii Lj 4 EI 2 EI i + j = M ij L LMji =2 EI 4 EI i + j L Lji (MFij)+ (MFji) settlement = j+ P (MFij)LoadM ij = (settlement = jw (MFji)Load4 EI 2 EI 2 EI 4 EI ) i + ( ) j + ( M F ij ) + ( M F ij ) Load , M ji = ( ) i + ( ) j + ( M F ji ) + ( M F ji ) Load 8 L L L L 9. Equilibrium Equations iPwjkCjCj MMji MjijkMjk j+ M j = 0 : M ji M jk + C j = 09 10. Stiffness Coefficients MijijMjiLjikii =4 EI Lk ji =2 EI L ik jj =4 EI L j1kij =2 EI L+ 110 11. Matrix FormulationM ij = (4 EI 2 EI ) i + ( ) j + ( M F ij ) L LM ji = (2 EI 4 EI ) i + ( ) j + ( M F ji ) L L M ij (4 EI / L) ( 2 EI / L) iI M ij F M = + (2 EI / L) ( 4 EI / L) j M ji F ji kii k ji[k ] = kij k jj Stiffness Matrix 11 12. PiMijwjMjiiL[ M ] = [ K ][ ] + [ FEM ]jj([ M ] [ FEM ]) = [ K ][ ][ ] = [ K ]1[ M ] [ FEM ]MijMjijiFixed-end moment Stiffness matrix matrix+ (MFij)(MFji)[D] = [K]-1([Q] - [FEM]) + (MFij)LoadPw(MFji)LoadDisplacement matrixForce matrix12 13. Stiffness Coefficients Derivation MjiMiReal beam ijLMi + M j LMi + M jL/3M jL 2 EILMj EIConjugate beam Mi EIMiL 2 EIM j L 2L MiL L )( ) + ( )( ) = 0 2 EI 3 2 EI 3 M i = 2 M j (1)+ M 'i = 0 : (+ Fy = 0 : i (M L MiL ) + ( j ) = 0 (2) 2 EI 2 EIFrom (1) and (2); 4 EI ) i L 2 EI ) i Mj =( L Mi = (13 14. Derivation of Fixed-End Moment Point loadP Real beamConjugate beam AABBLM EIM M EIMPM EIML 2 EIPL2 16 EIPL 4 EIML 2 EI 2 PL 16 EIM EIPL ML ML 2 PL2 + Fy = 0 : + = 0, M = 2 EI 2 EI 16 EI 8 14 15. P PL 8PL 8L P 2P 2P/2PL/8P/2-PL/8-PL/8 --PL/8-PL/16 --PL/16 PL/4 +-PL/8 PL PL PL PL + = + 16 16 4 815 16. Uniform load wReal beamConjugate beam AALBB M EIMMM EIM EIML 2 EIML 2 EIwwL3 24 EIwL2 8 EIM EIwL3 24 EIwL2 ML ML 2 wL3 + Fy = 0 : + = 0, M = 2 EI 2 EI 24 EI 12 16 17. Settlements Mi = MjReal beamMjConjugate beamL Mi + M jMLAM EIB M EIMi + M j L M EIML 2 EIML 2 EIMM EI+ M B = 0 : (ML L ML 2 L )( ) + ( )( ) = 0, 2 EI 3 2 EI 3 M=6 EI L217 18. Typical ProblemCB wP1P2 CA BL1PPL 8PL 8L2 wL2 12L 0 PL 4 EI 2 EI M AB = A + B + 0 + 1 1 8 L1 L1 0 EI PL 2 EI 4 M BA = A + B + 0 1 1 8 L1 L1 0 2 P2 L2 wL2 4 EI 2 EI M BC = B + C + 0 + + L2 L2 8 12 0 2 P2 L2 wL2 2 EI 4 EI B + C + 0 + M CB = 8 12 L2 L2wwL2 12L18 19. CB wP1P2 CA BL1L2CB M BCMBABM BA = M BCPL 2 EI 4 EI A + B + 0 1 1 8 L1 L1P L wL 4 EI 2 EI = B + C + 0 + 2 2 + 2 L2 L2 8 122+ M B = 0 : C B M BA M BC = 0 Solve for B19 20. P1 MABCB wMBAA L1BP2MBC L2C M CBSubstitute B in MAB, MBA, MBC, MCB 0 PL 4 EI 2 EI A + B + 0 + 1 1 M AB = L1 L1 8 0 EI PL 2 EI 4 M BA = A + B + 0 1 1 8 L1 L1 0 2 4 EI 2 EI P2 L2 wL2 B + C + 0 + + M BC = 8 12 L2 L2 0 2 P2 L2 wL2 2 EI 4 EI B + C + 0 + M CB = L2 L2 8 1220 21. P1 MAB AAyCB wMBA BL1P2 MCBMBC L2CyCBy = ByL + ByRBP1 MABBA AyL1MBAByLP2MBC ByRL2C MCBCy21 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 D3695 142EI 132 21EI 28 3722 23. Beam-Member Stiffness Matrix 31ij2E, I, A, L65k14 AE/Lk41 AE/Lk11 = AE/L4AE/L = k44d1 = 1d4 = 1 12341003004-AE/LAE/L500606- AE/L2[k] =AE/L50 23 24. 3i16EI/L2 = k32246 6EI/L2 = k655 6EI/L2 = k35d5 = 1 12EI/L3 = k5212EI/L3 121AE/L2[k] =E, I, A, Lk62 = 6EI/L2d2 = 1 k22 =j312EI/L3 = k25 450- AE/L0012EI/L30- 12EI/L3306EI/L20- 6EI/L24-AE/L0AE/L050-12EI/L3012EI/L3606EI/L2012EI/L3 = k556- 6EI/L2 24 25. 3i1j2 k33 = 4EI/Ld3 = 1E, I, A, L2EI/L = k63465 4EI/L = k662EI/L = k36 d6 = 1k23 = 6EI/L26EI/L2 = k53k26 = 6EI/L26EI/L2 = k56134561AE/L00- AE/L002[k] =2012EI/L36EI/L20- 12EI/L36EI/L2306EI/L24EI/L0- 6EI/L22EI/L4-AE/L00AE/L0050-12EI/L3-6EI/L2012EI/L3-6EI/L2606EI/L22EI/L0- 6EI/L24EI/L 25 26. Member Equilibrium Equations MiFxiijE, I, A, LAE/LAE/L x i1MjFyj AE/LAE/L+ 6EI/L2 6EI/L216EI/L2 1x i 12EI/L312EI/L3++12EI/L3 4EI/Lx j1+6EI/L2=FyiFxj2EI/L1x j12EI/L3 4EI/L2EI/L x ix j 16EI/L2 MFiFFxiFFyi+6EI/L26EI/L2FFyjFFxj6EI/L2 MFj26 27. Fxi = ( AE / L) i + Fyi =(0) i +(0) i Mxi = Fxj = (AE / L) i Fyj =(0) iMj =(0) iFxi Fyj Mi = Fxj Fyj M j (0) i(0) i +(AE / L) j +(0) j +(0) j +F Fxi(12EI / L3 ) i(6EI / L2 ) i(0) j(12EI / L3 ) j(6EI / L2 ) jF Fyi(6EI / L2 ) i (0) i(4EI / L) i (0) i(0) j ( AE / L) j(6EI / L2 ) j (0) j(2EI / L) j (0) jMiF F Fxi(0) j(0) j(6EI / L2 ) jF Fyj(0) j(6EI / L2 ) j(4EI / L) jMjF(12EI / L3 ) i (6EI / L2 ) i (6EI / L2 ) i(2EI / L) i0 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 0Stiffness matrixF 0 i Fxi F 6EI/ L2 i Fyi 2EI/ L i MiF + F 0 j Fxj F 6EI/ L2 j Fyi F 4 EI/ L j M j Fixed-end force matrix[q] = [k][d] + [qF] End-force matrixDisplacement matrix27 28. 6x6 Stiffness Matrixi[k ]66 =iiNi 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 jj AE/ L 0 12 EI/ L3 00 6EI/ L2 4 EI/ L0 6EI/ L20AE/ L0 6EI/ L2012 EI/ L32EI/ L0 6EI/ L2j 0 6EI/ L2 2EI/ L 0 6EI/ L2 4EI/ L 4x4 Stiffness Matrixi[k ]44=Vi 12 EI/ L3 Mi 6EI/ L2 Vj 12 EI/ L3 2 Mj 6EI/ Li 6EI/ L2 4 EI/ L 6EI/ L2 2EI/ Lj 12 EI/ L3 6EI/ L2 12 EI/ L3 6EI/ L2j 6EI/ L2 2EI/ L 6EI/ L2 4 EI/ L 28 29. 2x2 Stiffness Matrix i[k ]22 =Mi Mjj4 EI / L 2EI / L 2EI / L 4 EI / LComment: - 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 P2yP M2M12 Global14 16 322132 Localw45 2114 231330 31. P2yP M2M12 Global14 16 322132 Memberw45 416 213352424Local112313 31 32. P2yP M2M1241Globalw132216324Local5 2114 2313P (qF2)1[FEF]1(qF1 )1w(qF1)2(qF4 )2(qF4)1 (qF3)12(qF2)2(qF3 )232 33. 2 Global411323221654Local113[q] = [T]T[q] q1 q 2 = q3 q4 11 2 3 41 1 0 0 02 0 1 0 03 0 0 1 04 0 0 0 1q1' q 2' q3' q4 ' [k] = [T]T[q] [T] 33 34. 2 Global4116213235 2Local4 213[q] = [T]T[q] q3 q 4 = q5 q6 23 4 5 61 1 0 0 02 0 1 0 03 0 0 1 04 0 0 0 1q1' q 2' q3' q4 ' [k] = [T]T[q] [T] 34 35. 24116213235Stiffness Matrix: 12341[k]1 =121324 3 3[k]2 =4456[K] =3 4 5 623456Member 1Node 2 Member 25 6 35 36. 2 1Joint Load Qk16 221M Q 2 1z -P2y Q3 M3z Q 4 = Q1 Q 5 Q6 Reaction Qu433Du=Dunknown 52 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 ]QFAQFBF [Du ] = [K AA ]1 + ([Qk ] [QA ])[ ]Member Force : [q ] = [k ][d ] + q F36 37. 2Global:41135(qF2)1(qF[FEF] 5 6 1 2 3 42356Member 1 Q2 = M 1 2 Q = P 2y = 3 3 Q4 = M 2 4 (qF6 )2 (qF3)24Node 2 Member 2 234 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)111322P16 D1 0 D2 D3 + D4 D5 0 0 D6 D2 Q 2F D F 3 + Q 3 F D4 Q 4 (qF5 )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 14 132 2 D2 D3 = 3 D4 4 33221654 Node 2 Q2 = M 1 Q2F Q3 = P2y Q3F Q = M Q F 2 4 438 39. 24116213235P qF2 qF41qF1[FEF]qF3Member 1: 1 q 1 2 q 2 = q 3 3 4 q 4 1 23 k14 d1 = 0 d 2 = D2 + d 3 = D3 d 4 = D4 q 1F F q 2 F q 3 F q 4 39 40. 24116213235w qF6 qF42qF3[FEM]qF5Member 2: q3 q 4 q5 q6 1 =2 3 41 23 k14 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 BA 9mC 1.5 m 3m41 42. 10 kN 1 kN/m CBA1.5 m 3m9m 32Global1232Members112 2110 kN 1 kN/m1.5 m[FEF] wL2/12= 6.759m1wL2/12= 6.75 PL/8 = 3.751.5 m 2PL/8 = 3.75 42 43. 321129m3mStiffness Matrix:i Mi Mj[k ]22 =4 EI / L 2EI / L 2EI / L 4 EI / L3[k]1= EIj224/92/932/94/92 2[K] = EI4/32/322/3[k]2 = EI14/311(4/9)+(4/3) 2/322/314/343 44. 3210 kN11 kN/m A6.756.75 9mB3.75 1.5 m 1.5 mCEquilibrium equations: MCB = 0 MBA + MBC = 0 Global Equilibrium: [Q] = [K][D] + [QF] 2 2MBA + MBC = 01MCB = 0B C== EI12(4/9)+(4/3) 2/3B12/3C4/3+-6.75 + 3.75 = -3 -3.750.779/EI 2.423/EI 44 45. 32 11 kN/m [FEF] 9mwL2/12 = 6.751wL2/12 = 6.75Substitute B and C in the member matrix, Member 1:[q]1 = [k]1[d]1 + [qF]1 33MAB2MBA= EI204/92/9A2/94/9= B 0.779/EI6.75+-6.751 kN/m 4.56 kN=-6.406.40 kNm6.92 kNm 9m6.9214.44 kN 45 46. 10 kN 121.5 m2PL/8 = 3.751.5 m 2PL/8 = 3.75[FEF] Substitute B and C in the member matrix, Member 2 :[q]2 = [k]2[d]2 + [qF]2 22MBC1MCB= EI14/32/3B = 0.779/EI2/34/3C = 2.423/EI+3.75 -3.756.40=010 kN 6.40 kNm27.13 kN2.87 kN 46 47. 10 kN1 kN/m6.40 6.406.92 4.56 kN4.44 kN 1 kN/m6.92 kNm7.13 kNB = +0.779/EI10 kN2.87 kNC = +2.423/EI CA 4.56 kNB 9m11.57 kN 2.87 kN 1.5 m 1.5 m 7.134.56 V (kN)x (m) 4.56 m -4.44M (kNm) -6.92-2.87 4.323.48x (m) -6.4047 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/m40 kNm EI2EI AB 4mC 4m48 49. 24132EI 11[k ] =Vi 12 EI/ L3 Mi 6EI/ L2 Vj 12 EI/ L3 Mj 6EI/ L2 EIij 24 EI/ L 6EI/ L 2EI/ L2 12 EI/ L 6EI/ L2 12 EI/ L3 6EI/ L26EI/ L 2EI/ L 6EI/ L2 4 EI/ L 310.375EI20.75EI3 4234-0.75EIEI-0.375EI -0.75EI 0.375EI -0.75EI 0.75EIEI30.5625-0.375[K] = EI4-0.3753[k]20.75EI - 0.375EI 0.75EI 2EI42[k]1 13j 36EI/ L522Use 4x4 stiffness matrix,i6-0.75EI 2EI3 3 4 5 64560.1875EI 0.375EI - 0.1875EI 0.375EI 0.375EIEI-0.375EI0.5EI-0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI0.5EI-0.375EIEI 49 50. 20 kN 9kN/m 12 kNm 18 kN4m40 kNm 12 kNm 18 kN241Global:4Q4 = 40 D3 D4=20.5625-0.3754-0.3753435EI23= EI632EI1 1 [Q] = [K][D] + [QF]3 Q3 = -204m3D3 D418 +3-124-61.09/EI 9.697/EI50 51. 24132EI 119 kN/m12 kNm AB12[qF]118 kNMember 1:12 kNm18 kN[q]1 = [k]1[d]1 + [qF]1 1 q11q22q3L3q4L423412(2EI)/430.75EI - 0.375EI 0.75EI 0.75EI2EI-0.75EIEI-0.375EI -0.75EI 0.375EI -0.75EI 0.75EIEI-0.75EI 2EI 9 kN/mA 67.51 kNm 48.18 kN1 [q]1d1 = 01848.18d2 = 01267.51d3 = -61.09/EI18-12.18d4 = 9.697/EI-1253.2153.21 kNm B 12.18 kN 51 52. 4635EI 223Member 2: [q]2 = [k]2[d]2 + [qF]2 3 q3R3q4R4q55q664560.1875EI 0.375EI - 0.1875EI 0.375EI 0.375EIEI-0.375EI0.5EI-0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI0.5EI-0.375EI7.82 kN0-7.818d4 = 9.697/EI0-13.21d5 = 007.818d6 = 00-18.0618.06 kNm13.21 kNm BEId3 =-61.09/EI2 [q]2C 7.82 kN 52 53. 9 kN/m A 67.51 kNm 48.18 kN53.21 kNmBB112.18 kN[q]118.06 kNm13.21 kNmC27.82 kN7.82 kN[q]220 kN 9kN/m40 kNm D3 = B = -61.09/EI67.51 kNm 48.18 kN 48.184mD4 = B = +9.697/EI 4 m+V (kN)18.06 kNm 7.818 kN12.18x (m)M (kNm)+ --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/m10 kN40 kNm EI2EI AB 4mC 2m2m54 55. 20 kN 9kN/m A40 kNm2EI 4mB 413 1118 kNi=1Vi 12 EI/ L3 Mi 6EI/ L2 Vj 12 EI/ L3 Mj 6EI/ L2 5 310 kN 12 kNm[FEF]6229 kN/m12 kNm[k ]44CEI 2m2m2Global10 kNij6EI/ L2 12 EI/ L3 6EI/ L2 12 EI/ L3 6EI/ L2 6EI/ L2 2EI/ L5 kNm218 kN4 EI/ L5 kNm5 kNj6EI/ L2 2EI/ L 6EI/ L2 4 EI/ L 5 kN55 56. 2413 111 1[k]1 =2 3 40.75EI 0.75EI0.1875EI40.375EI632m2m342EI-0.75EIEI3-0.375EI -0.75EI 0.375EI -0.75EI35212(2EI)/43 0.75EI - 0.375EI 0.75EI3[k]2 =524m 26EI 463-0.75EI 2EI/L 54[K] = EI 60.5625-0.3750.3754 6-0.375 0.3753 0.50.5 10.375EI - 0.1875EI 0.375EI EI-0.375EI0.5EI-0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI0.5EI-0.375EIEI 56 57. 20 kN 9kN/m10 kN40 kNmAC 62B413 115 2210 kN9 kN/m12 kNm12 kNm[FEF]118 kN5 kNm345 kN5 kN 63VBL+VBR = -20 MBA+MBC = 40= EIMCB = 0B B C5 kNm218 kNGlobal: [Q] = [K][D] + [QF]30.5625-0.3750.375B4 6-0.375 0.3753 0.50.5 1B C+18 + 5 = 23 -12 + 5 = -7 -5-116.593/EI=-7.667/EI 52.556/EI57 58. 2413 119 kN/m12 kNm212 kNm 18 kN118 kN[FEF][q]1 = [k]1[d]1 + [qF]1Member 1: 1234VA10.375EI 0.75EI - 0.375EI 0.75EI01855.97MAB20.75EI01291.78VBL MBA=3 42EI-0.75EIEI-0.375EI -0.75EI 0.375EI -0.75EI B =-116.593/EI 0.75EIEI-0.75EI 2EI/LB =-7.667/EI+18 -12=-19.97 60.119kN/m 91.78 kNm4m 55.97 kN160.11 kNm 19.97 kN58 59. 46310 kN5 32225 kN Member 2:VBR30.1875EIMBC440.375EIMCB=5 6[FEM]5 kN[q]1 = [k]1[d]1 + [qF]1 3VC5 kNm5 kNm560.375EI - 0.1875EI 0.375EI EI-0.375EI0.5EI-0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI0.5EI-0.375EIEIB =-116.593/EI5-0.0278B =-7.667/EI5-20.110C =52.556/EI+5 -5=10.03 010 kN 20.11 kNm2m2m 20.0278 kN10.03 kN59 60. 9kN/m91.78 kNm10 kN20.11 kNm 2m 14m19.97 kN55.97 kN60.11 kNm 20 kN9kN/m2m 20.0278 kN10.03 kN10 kNC = 52.556/EI40 kNm91.78 kNm 55.97 kNB = -7.667/EI10.03 kNB = -116.593/EI4m55.972m2m19.97 +V (kN)60.11 M (kNm) -91.78+ --0.0278-10.03x (m) -10.0320.11 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 kN6 kN/m B A2EI 8mEI B = -10 mm 4mC4m61 62. 1232EIEI 21i Use 2x2 stiffness matrix:[k ]22 =1[k]1 =EI 8842484 EI / L 2EI / L 2EI / L 4 EI / L21Mi Mjj2[k]2 = 2[K] =EI 812 24232423232EI 834 62 63. 40 kN6 kN/m1B A2C EI B = -10 mm 4m 4m2EI 8m2140 kN6 kN/m[FEM]loadwL2/12 = 32 kNm132 kNm10 mm6(2EI)/L2 = 75 kNm2Q2 = 0 Q3 = 0 D2 D340 kNm 37.5 kNm10 mm6(EI)/L2 = 37.5 kNm[Q] = [K][D] +2[QF]Global:2PL/8 = 40 kNm75 kNm1[FEM]332122D2324D3EI = 8-61.27/EI=185.64/EI+-32 + 40 + 75 -37.5 = 45.5 -40 - 37.5 = -77.5rad rad63 64. 126 kN/m[FEM]load2EIwL2/12 = 32 kNm1132 kNm 75 kNm1[FEM]10 mm6(2EI)/L2 = 75 kNm Member 1:[q]1 = [k]1[d]1 + [qF]1 1q1 q2EI = 82184d1 = 0248d2 = -61.27/EI76.37 kNm6 kN/m32 + 75 = 107+-32 + 75 = 4376.37 kNm=-18.27 kNm18.27 kNm8m 1 31.26 kN16.74 kN 64 65. 2340 kN [FEM]load2PL/8 = 40 kNm240 kNm 37.5 kNm2[FEM]10 mm6(EI)D/L2 = 37.5 kNm Member 2:[q]2 = [k]2[d]2 + [qF]2 2q2 q3=EI 83242d2 = -61.27/EI324d3 = 185.64/EI40 - 37.5 = 2.5+-40 - 37.5 = -77.518.27 kNm=0kNm40 kN 18.27 kNm 222.28 kN17.72 kN 65 66. 2642EIEI 211 Alternate method: Use 4x4 stiffness matrix[k]1341.5-0.3751.58-1.54-1.50.375-1.541-1.521 12(2)/82 EI 2 1.5 = 8 3 -0.375 4 1.5533=3 12/82 4 0.75EI 8 540.7521 20.375 1.51.5 8-0.375 -1.5EI 3 = 8 4 5-0.375-1.50.56251.5 04 0-0.75 -0.187512 -0.75-0.75 0.18756000.752-0.75456-0.18750.75-0.7520.1875-0.75-0.7542 -0.751.5 4521[K]40.75-0.1875 -0.7568 34[k]260 00 0-0.75 -0.18750.7566 67. 40 kN6 kN/m2B AC EI B = -10 mm2EI 8m4m132 kNm40 kNmQ6= 0 Q4 = 0 Q6= 0 D4 D65 D44 6D6122D424D612EI 4 = 8 62 420 kN20 kN2EI 4 = 8 6=40 kNm2124 kN 24 kN Global: [Q] = [K][D] + [QF] 4 6 Q4 = 053 40 kN32 kNm[FEF]214m6 kN/m64200 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 = -1.532x10-3rad+-40 8 -40rad185.64/EI = 4.641x10-3+867 68. 24132 kNmB = -10 mm16 kN/m 32 kNm[FEF]load3124 kN24 kN Member 1:[q]1 = [k]1[d]1 + [qF]1 21 1 12(2)/82 2 1.5q1 q2341.5-0.3751.5d1 = 0248-1.54d2 = 032(200x200) 3 8-0.375-1.50.375-1.5q441.54-1.58q131.26q3q2 q3 q4==kN76.37kNm16.74kN-18.27kNm76.37 kNm+d3 = -0.01-32d4 = -1.532x10-3 6 kN/m2418.27 kNm8m 1 31.26 kN16.74 kN 68 69. 40 kN6440 kNm40 kNm-10 mm = B[FEF]235220 kN20 kNMember 2: 3 3 12/82 4 0.75q3 q4 q5=4(200x200) 5 80.75 4-0.1875 -0.75q66q322.28kNq418.27kNm17.72kNm6-0.1875 -0.750.75 20.1875 -0.75kN05q5 q6=0.752-0.754d3 = -0.0120d4 = -1.532x10-340d5 = 0+d6 = 4.641x10-320 -4040 kN 18.27 kNm 222.28 kN17.72 kN 69 70. 40 kN6 kN/m B76.37 kNmEI B = -10 mm 17.72 kN 16.74 + 22.28 kN 8m 4m 4mA2EI31.26 kNV (kN)C31.26 +-5.21 m22.28 + -16.74-17.72x (m)70.85 M (kNm)5.06 -+ --18.27x (m)-76.37 Deflected CurveB = -10 mm d4 = B = -1.532x10-3 radD6 = C = 4.641x10-3 rad70 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 kN0.6 kN/m20 kNmB AEI2EI 8m4mC C = -10 mm4m71 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]164EI 2153341.5-0.3751.58-1.54-1.50.375-1.54-1.523EI = 86843 12/82 4 0.75 5-10 mm[k]20.75 4-0.1875 -0.75 0.75256-0.18750.75-0.7520.1875-0.75-0.754 Global: [Q] = [K][D] + [QF] 3 Q3 Q4 Q63 EI 4 = 8 6460.5625 -0.75-0.75 120.75 2D3 D40.7524D65 QF3 3 -0.1875 200 200 +( ) 4 -0.75 D5 = -0.01 + QF4 8 QF6 6 -0.75 72 73. 4 kN0.6 kN/m20 kNmB AEI2EI 8m4m10 mm 4mD3 D4 D6=5 4 kNm22 kN3Q6 = 204 kNm13 EI 4 = 8 6EI3 4 kN2.4 kN 2.4 kN Global: [Q] = [K][D] + [QF] Q3 = 0 Q4 = 06 2113.2 kNm[FEF]42EIC0.6 kN/m3.2 kNm2460.5625 -0.75-0.75 120.75 20.75242 kND3 9.375 D4 + 37.5 + D6 37.52.4+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. 240.6 kN/m3.2 kNm 11[FEF]load33.2 kNm 8m12.4 kN2.4 kN Member 1:[q]1 = [k]1[d]1 + [qF]1 21 q1 q2 q3q41.5-0.3751.5d1 = 02.48-1.54d2 = 03.2200 200 3 8-0.375-1.50.375-1.5d3 = -9.433x10-31.54-1.58d4 = -1.538x10-3q1 q344=q4q21 12(2)/82 2 1.538.55=kN43.19kNm-3.75kN6.03kNm43.19 kNm0.6 kN/m+2.4 -3.26.03 kNm8m 1 8.55 kN3.75 kN 74 75. 4 kN644 kNm4 kNm [FEF]2358m10 mm22 kN2 kNMember 2: 3 q3 q4 q5200 200 = 8q643 12/82 4 0.75 5 60.75 4-0.1875 -0.75 0.75q33.75kNq4-6.0kNm0.25kNm6-0.750.75d3 = -9.433x10-322-0.1875d4 = -1.538x10-340.1875 -0.75 -0.754q5 q6=d5 = -0.01+d6 = 1.863x10-32 -44 kNkN20.0256.0 kNm20 kNm 23.75 kN0.25 kN 75 76. 4 kN0.6 kN/m20 kNmB AEI2EI 8m0.6 kN/m43.19 kNm4mDeflected Curve4 kN 20 kNm6.0 kNm 3.75 kN8.55V (kN)M (kNm)10 mm 4m6.03 kNm8m 1 8.55 kNC+23.756-0.25 21 +-43.190.25 kN3.75 kND3 = -9.433 mm D4 = -1.538x10-3 radx (m) 20 x (m) D5 = -10 mm D6 = +1.863x10-3 rad76 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 EI2EI A 4mBC 2m2m77 78. 34 22EIA11 4mEIB2m2mi Use 2x2 stiffness matrix,[k ]22 =3[k]1 =2EIEI1EIMi Mj2EIj4 EI / L 2EI / L 2EI / L 4 EI / L 13C22[k]2 =21EI4 0.5EI1[K] =EI2.0 0.01EI0.020.5EI2141.0 78 79. 30 kN 9kN/m Hinge 34 2A1[FEF]1A=0.0EI15 kNm C 2122.00.0D120.01.0D20.0006=30 kN15 kNm B1D1 D2B 12 kNm BGlobal matrix: 0.019kN/m12 kNmC2+-12 15rad-0.0015 rad79 80. 312 kNm1 A[FEF]19kN/m 1AB12 kNm BMember 1: 3 q3 q1=132EIEI1EI2EId3 = 0.0 d1 = 0.0006+12 -12=18 0.09 kN/m A 18 kNm22.5 kN1B 13.5 kN80 81. 4 BC2215 kNm B30 kN15 kNm C2Member 2: 2 q2 q4=241EI0.5EI4 0.5EI1EId2 = -0.0015 d4 = 0.0+15 -15=0.0 -22.530 kN CB 9.37 kN22.5 kNm 20.63 kN81 82. 30 kN9 kN/m A 18 kNmB13.5 kN13.5 kN22.5 kN9.37 kNCB22.87 kN22.5 kNm 20.63 kN9.37 kN 30 kN9kN/m Hinge EIBR= -0.0015 rad2EIBL= 0.0006 radA4mB2mC2m22.5 V (kN)9.37+ 2.5 mM (kNm)-13.5 10.13 +- -18x (m) -20.63 18.75 + -x (m) -22.582 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 AB 4mEIHingeC 4m83 84. 5741 32EI 1A22 B41540.375EI5[k]1 = 10.75EI2EI-0.375EI-0.75EI20.75EIEI1[k]2 =0.1875EI30.375EI6 7C20.75EI - 0.375EI 0.75EI -0.75EIEI0.375EI -0.75EI -0.75EI2EI/L36123710.5625-0.750.375EI 2-0.752.003[K] = 16EI0.3750.01.00.375EI - 0.1875EI 0.375EI EI-0.375EI0.5EI-0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI0.5EI-0.375EIEI 84 85. 20 kN B9kN/m AC 5 41 32EI 1A 12 kNmGlobal:B1[FEM] 1Q1 = -2022C12 kNm 18 kN 31=0.5625-0.750.375D1EI 2-0.752.00D23Q2 = 0.00.3750.01.0D3Q3 = 0.0 D1 D2 D36EI2 B9kN/m A 18 kN718+-12 0.0-0.02382 m=-0.008333 rad 0.008933 rad85 86. 59kN/m42EI21A112 kNm A 18 kNB12 kNmB1[FEF]18 kNMember 1: 415q440.375EIq550.75EI2EI1 -0.375EI-0.75EI0.75EIEIq1 q2=220.75EI - 0.375EI 0.75EI -0.75EIEId4 = 0.018d5 = 0.0120.375EI -0.75EId1 = -0.02382-0.75EId2 = -0.00833107.32 kNm A2EI/L+18 -1244.83=107.32 -8.83 0.09kN/m 1B 8.83 kN44.83 kN 86 87. 7 1 B6EI 23CMember 2: 1 q110.1875EIq330.375EIq6=6q776370.375EI - 0.1875EI 0.375EI EI-0.375EI0.5EI-0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI0.5EIEI-0.375EICB 211.16 kNd1 = -0.023820d3 = 0.0089330d6= 0.0 d7= 0.0+0 0-11.16=0.0 11.16 -44.6644.66 kNm 11.16 kN87 88. 20 kN 9kN/mBL= -0.008333 rad BABR= 0.008933 rad C4m4m9kN/m107.32 kNm AB144.83 kNCB 28.83 kN44.66 kNm 11.16 kN11.16 kN44.83 V (kN)+8.83x (m) -11.16-11.16M (kNm) -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 kNm at the end of member AB. E = 200 GPa, I = 50x10-6 m4. 30 kN 9kN/m AHinge 2EI 4m40 kNmEIB 2mC2m89 90. 34 22EIA114mC2B2m2m jiUse 2x2 stiffness matrix:[k ]22 = 3[k]1 =EI2EIEI1EI2EI4 EI / L 2EI / L 2EI / L 4 EI / L 13Mi Mj2[k]2 =21EI4 0.5EI1[K] =EI2.0 0.01EI0.020.5EI2141.0 90 91. 30 kN 9kN/m HingeA2EI340 kNmEIBC 42 A1[FEM]1A=Q2 = 0.0EI15 kNm C 2122.00.0D120.01.0D20.0026=30 kN15 kNm B1D1 D2B 12 kNm BGlobal matrix: Q1 = 4019kN/m12 kNmC2+-12 15rad-0.0015 rad 91 92. 312 kNm1 A[FEF]19kN/m 1AB12 kNm BMember 1: 3 q3 q1=132EIEI1EI2EId3 = 0.0 d1 = 0.0026+12 -12=38 409 kN/m A 38 kNmB 40 kNm 37.5 kN1.5 kN92 93. 4 BC2215 kNm B30 kN15 kNm C2Member 2: 2 q2 q4=21EI4 0.5EI4 0.5EI 1EId2 = -0.0015 d4 = 0.0+15 -15=0.0 -22.530 kN CB 9.37 kN22.5 kNm 20.63 kN 93 94. 1.5 kN30 kN9 kN/m A 38 kNmB 40 kNm9.37 kN 7.87 kN1.5 kN37.5 kNCB22.5 kNm 20.63 kN9.37 kN 30 kN9kN/m40 kNm HingeABL= 0.0026 B rad 4mCBR=- 0.0015 rad 2m2m37.5 V (kN)+9.371.5x (m) -20.63M (kNm)40 + - -3818.75 +-x (m) -22.594 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 kNm at the end of member AB. E = 200 GPa, I = 50x10-6 m4 20 kN 9kN/m 2EI A 4m40 kNmBEIHingeC 4m95 96. 20 kN 9kN/m 2EI A40 kNmBEIHinge4mC 4m57 14 A12 kNm [FEM]136 C22 B9 kN/m 12 kNm 118 kN18 kN[k ]44=iVi 12 EI/ L3 Mi 2 6EI/ L Vj 12 EI/ L3 Mj 6EI/ L2 ijj6EI/ L2 12 EI/ L3 6EI/ L2 12 EI/ L3 6EI/ L26EI/ L2 2EI/ L 6EI/ L2 4 EI/ L 4 EI/ L 6EI/ L2 2EI/ L96 97. 5742EIA14 4[k]1 =10.375EI50.75EI156EIC22 B 20.75EI - 0.375EI 0.75EI1 -0.375EI2EI -0.75EI0.75EIEI23-0.75EI EI 0.375EI -0.75EI -0.75EI12EI2311360.375EI 2-0.752.000.3750.01.071[k]2 =-0.753[K] =0.56250.1875EI30.375EI EI -0.375EI 0.5EI -0.1875EI -0.375EI 0.1875EI -0.375EI6 70.375EI0.375EI - 0.1875EI 0.375EI0.5EI-0.375EIEI 97 98. 9kN/m20 kN5 42EIA1C22 B118 kN18 kN Global:1231=Q3 = 0.0 D10.5625-0.750.375D1EI 2-0.752.00D23Q1 = -20D36EI12 kNm[FEF]D279 kN/m12 kNmQ2 = 40EI40 kNm 1 30.3750.01.0D318+-12 0.0-0.01316 m=-0.002333 rad 0.0049333 rad98 99. 5 422EI112 kNm1540.375EIq550.75EI2EI1 -0.375EI-0.75EI0.75EIEIq2218 kN[q]1 = [k]1[d]1 + [qF]1q4q1118 kN4=12 kNm[FEF]1Member 1:9 kN/m20.75EI - 0.375EI 0.75EI87.37 kNm-0.75EIEId4 = 0.018d5 = 0.0120.375EI -0.75EId1 = -0.01316-0.75EI+d2 = -0.0023332EI2EI 49.85 kN18 -1249.85=87.37 -13.85 4040 kNm 113.85 kN[q]1 99 100. 7 1 B Member 2:q110.1875EIq330.375EI=6q77C23[q]2 = [k]2[d]2 + [qF]2 1q66EI6370.375EI - 0.1875EI 0.375EI d1 = -0.01316 EI-0.375EI0.5EI-0.1875EI -0.375EI 0.1875EI -0.375EI 0.375EI0.5EI-0.375EIEId3 = 0.004933 d6= 0.0 d7= 0.00+-6.1800.00 0=6.18 -24.6924.69 kNm 6.18 kN26.18 kN [q]2 100 101. 20 kN 9kN/m40 kNm EI BR = 0.004933 rad C 4m2EI BL= -0.002333 radAB4m BA 87.37 kNm 49.85 kN40 kNm13.85 kNCB24.69 kNm 6.18 kN6.18 kN49.85 V (kN)+13.85x (m)--6.18-6.18 M (kNm)+ -87.3740 -x (m) -24.69 101 102. Temperature Effects Fixed-End Forces (FEF) - Axial - Bending Curvature102 103. tan = Thermal Fixed-End Forces (FEF) Room temp = TR TTRtF MAFBFTm> TRFAFTb > TtAaxialAdctTt NATm =cbF MB BaxialByTb - Tt TRBTl(T ) axial = Tm TRy ATt + Tb 2TmTtyTb Tt T =( ) d d( T y ) = y (T ) d Tb(T ) bending = Tb Tt103 104. - AxialaxialaxialTt Tm> TRFAFATb > TtTRTmFBFB(T ) axial = Tm TRFAF = axial dA A= E axial dA = E (T ) axial dA= E (T ) axial dA = EA (T ) axial F ( Faxial ) A = (Tm TR ) AE104 105. - BendingyTt yy F MAABF MB(Ty ) = y (T ) d Tb(T ) bending = Tb TtM = y y dA F AA= yE dA = yE (Ty ) dA= yEy ( = E (T ) dA dT ) y 2 dA dF ( Fbending ) A = (Tl Tu ) EI d 105 106. Elastic Curve: BendingO d(dx ) = ( d ) y ( d )ds = dx ydxBefore deformationyds1=(d ) dxddxAfter deformation106 107. Bending TemperatureTtOdxTl > TuTbd Tt + Tb 2 MTm =TuT = Tb - TtTtd yMT = dct cbTl > TuTbTtT y dy TbT ) dx d T (d ) = ( )dx d d M 1 T ( ) = = ( ) = dx d EI (d ) y = y (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, = 12x10-6 /oC, E = 200 GPa, I = 50x10-6 m4.T2 = 40oC 182 mm AT1 = 25oC 4mEI2EI BC 4m108 109. T2 = 40oC 182 mm AT1 = 25oC 4m2BC 4m 31 1FEMEI2EI2Mean temperature = (40+25)/2 = 32.5 Room temp = 32.5oC 19.78 kNm a(T /d)EI = 19.78 kNm oC +7.5F Fbending = (-7.5 oCT 40 25 )(2 EI ) = (12 10 6 )( )(2 200 50) = 19.78 kN m d 0.182 109 110. 2 19.78 kNm 19.78 kNm 182 mmA1 22EI13B4m4m[q] = [k][d] + [qF]Element 1: 2 M2 M1=1221q2112q1131EI10.5q130.51q3Element 2: M1 M3=EIEIC++-1.978 1.978(10-3) EI0 00 [M1] = 3EI1 + (1.978x10-3)EI1 = -0.659x10-3rad110 111. 2 19.78 kNm 19.78 kNm A182 mm2EI131 2EIB4mC4mElement 1: 2 M2 M1=2 EI121q2 = 0112q1 = -0.659x10-3Element 2: M1 M3=+10.5q1 = -0.659x10-330.51q3 = 019.78=-26.37 kNm 6.59 kNm31 EI1-19.7826.37 kNm A6.59 kNm B4.95 kN4.95 kN+0 06.59 kNm B=-6.59 kNm -3.30 kNm 3.3 kNm C2.47 kN2.47 kN111 112. 26.37 kNm+7.5 oC A4.95 kN-7.5 oCBC2.47 kN 4m4m2.47 kNV (kN)3.3 kNmx (m) -2.47-4.95 26.37 M (kNm)Deflected curve+6.59 -B = -0.659x10-3 radx (m) -3.30x (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 AT1 =25oC 4mEI, AE2EI, 2AE BC 4m113 114. 6 59 T2 = 40oCA 4 T1 = 25oC132 2EI4m1 B8 2EI7 C4mElement 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 AT1 =25oCEI, AE2EI, 2AECB4m4mFixed-end forces due to temperatures F Fbending = (T 40 25 )(2 EI ) = (12 10 6 )( )(2 200 50) = 19.78 kN m d 0.182Mean temperature(Tm) = (40+25)/2 = 32.5 oC , TR = 28 oC F Faxial = (T ) AE = (12 10 6 )(32.5 28)(2 20 10 3 m 2 )(200 10 6 kN / m 2 ) = 432 kN432 kN19.78 kNm +12 oC A-3oC19.78 kNm B432 kN115 116. 69 T2 = 40oC5 182 mm14 AT1 =322EI, 2AE25oC21EI, AE 4mFEMElement 1: 432 kN [q] = [k][d] +19.78 kNm +12 oC -3A[qF]19.78 kNm 432 kNoC456B 12q442x1060.000.00- 2x106 0.00q550.00375075000.00q660.007500 20x103q1 q2 q3=7 CB4m810.003 0.000432750000.00-7500 10x1030-19.78-3750+0.002x1060.000.00d1-75000.003750-7500d20.007500 10x1030.00-7500 20x103d319.78-2x106 0.0020.00 -375030.00-432116 117. 69 T2 = 40oC5 182 mm14 AT1 =322EI, 2AE25oC8 217 CB4mEI, AE 4mElement 2:[q] = [k][d] + [qF] 123789 0.00d10-18753750d200.00-37505x103d300.001x1060.000.0001875-375000-3750 10x10300q111x1060.000.00- 1x106 0.00q220.00187537500.00q330.003750 10x103q7 q8 q9=7 -1x106 0.00 80.00 -1875-37500.0090.005x1030.003750+0117 118. 69 T2 = 40oC5 182 mm 4 A1T1 =2EI, 2AE25oC8 21EI, AE 4m1Q2 = 0.0 Q3 = 0.0 D1 D2 D333x1060.00.0D120.05625-3750D23=21Q1 = 0.00.0-375030x103D30.000144=7 CB4m Global:320.0 19.78m-719.3x10-6+m-0.0004795-432rad118 119. Element 1:452163q442x1060.000.00- 2x106 0.000.000432q550.00375075000.00750000.00q660.007500 20x103-7500 10x1030-19.78q1=q2 q31-37500.00+0.002x1060.000.00-75000.003750-7500d2 = -479.5x10-60.007500 10x1030.00-7500 20x103d3 = -719.3x10-619.78-2x106 0.0020.00 -375030.00d1 = 144x10-6-4326 q4144.0kNq5-3.60kNq6-23.38kNm-144.0kN3.60kN9.00kNmq1 q2 q3=5 A 4 23.38 kNm 144 kNA3.60 kN32 B 19 kNmB 3.60 kN144 kN 119 120. Element 2:123789 0.00d1 = 144x10-60-18753750d2 = -479.5x10-600.00-37505x103d3 = -719.3x10-600.001x1060.000.0001875-375000-3750 10x103 300q111x1060.000.00- 1x106 0.00q220.00187537500.00q330.0037500 10x103q7 q8 q9=7 -1x106 0.00 80.00 -1875-37500.0090.005x1030.003750q1144kNq2-3.6kNq3-9kNm-144kN3.6kN-5.39kNmq7 q8 q9=2 B 1 9 kNm 144 kNB3.60 kN+098 C 75.39 kNmC 3.60 kN144 kN 120 121. Isolate axial part from the system T2 = 40oC RA AT1 = 25oCEI2EI B4mRCRAC 4m RA = RC+ Compatibility equation:dC/A = 0R A (4) RA ( 4) + + 12 10 6 (32.5 28)(4) = 0 AE 2 AERA = 144 kN RC = -144 kN121 122. T2 = 40oC EI2EI AT1 = 25oC4m4m9 kNm23.38 kNm 144 kN144 kNACB144 kN144 kNB 3.60 kN3.60 kN5.39 kNm9 kNm BC 3.60 kN3.60 kNV (kN)x (m) -3.623.38 M (kNm) Deflected curve+8.98 -D2 = -0.0004795 mmD3 = -719.3x10-6 radx (m) -5.39 x (m)122 123. Skew Roller Support- Force Transformation - Displacement Transformation - Stiffness Matrix123 124. Displacement and Force Transformation Matrices x5y6 j2 34m1iy6 5y3 2 i1jx4mx 124 125. Force Transformation 6xy5yy2 3x =1iy =yq6 = q6'x j xi L y j yi L5 4m3 2 i1q 4 q 5 q 6 x y x0 0 1 q4' q 5' q6' x0 00 00 00 0 0 01 0 0 00 x y 00 y x 0x = y 0 q 1 x q 2 y q 3 0 = q 4 0 q 5 0 q 6 0 6jq4 = q4' cos x q5' cos yyq5 = q4' cos y q5' cos x4j xxx0 y[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 y6xx os c d4 x4 d4 6 j y y os c d4 55 4jyxd5dyyd 4' d 5' d 6' x = y 0 d 1' x d 2' y d 3' 0 = d 4' 0 d 5' 0 d 6' 0 y cos 5dd ' 4 = d 4 cos x + d 5 cos yd '6 = d 6x os c 5xyd' 5 = d 4 cos y + d 5 cos xx yxx x0 0 1 y x0d 4 d 5 d 6 x0 00 0 0 01 0 0 x 0 y 0 0y0 00 0 0 y x 00 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 ) TTTTTherefore,FTFF[k ] = [T ]T [k '][T ][q ] = [T ] [q ' ] FTF[q ] = [T ]T [q '] [d '] = [T ][d ][k ] = [T ]T [k '][T ]127 128. Stiffness matrix 2*5*3* 1*36*21i4*ij ijjx = cos jiy = sin i41jix = cos i65jy = sin j[q*] = [T]T[q]q 1* q 2* q 3* q 4 * q 5* q 6 * 1* 2* =3* 4* 5* 6*1 ix iy 0 0 0 0 2 iy ix 0 0 0 03 0 0 1 0 0 04 5 0 0 0 0 0 0 jx jy jy jx 0 06 0 0 0 0 0 1 q 1' q 2' q 3' q 4' q 5' q 6' [ T ]T128 129. 1 2[T ]=3 4 5 61[k ']=1 2 3 4 5 61 2 iy ix iy ix 0 0 0 0 0 0 0 0 20 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 jx 0 jy 0 03 0 6EI/ L2 4 EI/ L 0 6EI/ L2 2EI/ L5 0 0 0 jy jx 046 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/ L26 6EI/ L2 2EI/ L 0 2 6EI/ L 4 EI/ L 0129 130. [ k ] = [ T ]T[ k ][T] = Ui Ui (AEViL (AEVj - (Mj-LMiUj -(ix2 +-AE L AE L12EI L312EI L3 6EI L2ixjx +ixjy--Vi6EI L2iy2 )) ixiy((AE LAE LL3 12EI L3iy12EIiy2 +L2iy jy)-(iyjx )-(AE L AE L) ixiyL36EIiy 12EI-iyjx -iyjy + 6EI L2UjMi12EI L3-ix2 )L3 12EI L3ixiy -(L2 6EI L2ix-(AE L AE Lixjx +Lixjy)ix jx ) -6EI L26EI L2jy6EI L2jx(AE L(AE LL3L3L3L36EILL2AEixjy )AE-(LLjyixjy -iyjy + 6EI-12EI12EI2EIiyjy ) -(jyjx2 +-12EI12EIiyjx -4EIix 12EI6EIVjjy2 )) jxjy(AEL2-LMj 12EI L312EI L36EI L2iyL2ixL22EI L 6EI) jxjyjx6EI6EIixjx )AE 12EI jx2 ) ( L jy2 + L3--jx12EI L3iyjx )jyL2-6EI L2jx4EI L130 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 kN4m4m22.02 o131 132. 40 kN4m4m22.02o6 5 i3 Global2*123* 1*x 22.02 o j x* jx = cos 22.02o = 0.9271, jy = cos 67.98o = 0.37494ix = cos 0o = 1, iy = cos 90o = 040 kN 40 kNm [FEF] 20 kN[ qF ]i6 Local5 11j440 kNm20sin22.02=7.5 20cos22.02=18.54 20 kN 132 133. Transformation matrix 6 53 Global2*14 ix = cos 0o = 1, iy = cos 90o = 0Member 1: [ q ] = [q 4 q5 q 6 q 1* q 2* q 3* 3* 1*2 xx* jx = cos 22.02o = 0.9271, jy = cos 67.98o = 0.3749T ]T[q]=4 5 6 1* 2* 3*1 1 0 0 0 0 0 2 0 1 0 0 0 03 4 0 0 0 0 1 0 0 0 . 9271 0 0 . 3749 0 0i6 Local5 11[T ]Tix iy 0 = 0 0 0 iy 00 0 10 0 00 jx 0 jy 0 0ix4j0 0 00 0 0 jy jx 00 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 iiLocal5 11ijij4jjNi 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 ]66 =[k]16=0 6EI/ L2 4 EI/ L 0 6EI/ L2 2EI/ L1 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 -0.93753 4 5 0.000 -150.0 0.000 3.750 0.000 -0.9375 20.00 0.000 -3.750 0.000 -3.750150.0 0.0000.000 0.93750.000 -3.7503.75010.000.000-3.75020.006 0.0000 AE/ L 0 12 EI/ L3 0 6EI/ L2 0 AE/ L 0 12 EI/ L3 0 6EI/ L20 6EI/ L2 2EI/ L 0 2 6EI/ L 4EI/ L 6 0.000 3.750 10.00134 135. 6 53 Global2*13* 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 3.7502 x i6 Local5 11jx* 3 4 5 0.000 0.000 -150.0 3.750 0.000 -0.9375 20.00 0.000 -3.7506 0.000 3.750 10.000.000 -3.750150.0 0.0000.000 0.93750.000 -3.75010.000.000-3.75020.004 54 5 150.0 0.000 0.000 0.93756 1* 0.000 -139.0 3.750 0.3512* -56.25 -0.8693* 0.000 3.75060.00020.001.406-3.75010.001* -139.0 0.351 1.406 2* -56.25 -0.869 -3.750129.051.823* 0.0001.40646 0.000Stiffness matrix [k*]: [ k* ] = [ T ]T[ k ][T][k*]1= 1033.7503.75010.0051.821.406 21.90 -3.476 -3.47620.00135 136. 6 40 kN52*14 4m4m22.02 o20 kNx x*40 kN40 kNm Global Equilibrium:3* 1*[ qF ]40 kNm20sin22.02=7.5 20cos22.02=18.54 20 kN[Q] = [K][D] + [QF] 1* Q1 = 0.0 Q3 = 0.0D1* D3*==1* 103 3*129 1.4063* 1.406D1*20.0D3*36.37x10-6-40m0.002+-7.5rad 136 137. 6 40 kN52*14 4m4mMember Force :22.02[q] = [k*][D] +[qF]o40 kNmq4 q54 5q660.000 3.750 20.00 1.406 -3.750 10.00q1* q2* q3*=x x*40 kNm18.5420 kN4 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 3.75010340 kN3* 1*7.50 00 20-5.06 27.5040.0601* -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 13.48 040 kN60 kNm 5.06 kN 27.5 kN13.48 kN137 138. 40 kN40 kN4m4m60.05 kNm 22.02 o5.05 kN 27.51 kN13.47 kN50.04 + -60.05 Bending moment diagram (kNm) 0.002 radDeflected 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 kN6 kN/m2EI, 2AE 8mEI, AE 4m22.02o 4m139 140. 40 kN6 kN/mEI, AE2EI, 2AE 8m4m622.02 o AE (0.006 m 2 )(200 106 kN / m 2 ) = L (8 m)4m= 150 103 kN m3528*42119* 7*Global 3625 1152Local= 20 103 kN m 2 EI = 10 103 kN m L344 EI 4(200 106 kN / m 2 )(0.0002 m 4 ) = L (8 m)126 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 / m140 141. 63528*11429* 7*GlobalLocal : Member 1 635[ q] 4322 11[q] = [q] Thus, [k] = [k]6 [ q]114 54 5 6 1 2 0.000 150.0 0.000 0.000 -150.0 0.000 0.9375 3.750 0.000 -0.937560.00043 0.000 3.75020.000.000-3.75010.00-150.0 0.000 0.000 2 0.000 -0.9375 -3.750 3 0.000 3.750 10.00150.0 0.0000.000 0.93750.000 -3.7500.000-3.75020.00[k]1 = [k]1 = 2x103 13.7505141 142. 635[2 1 ix = cos 0o = 1, iy = cos 90o = 09* 7*238*q* ]8*1143 229* 7*[ q ]2x156 42x* jx = cos 22.02o = 0.9271, jy = cos 67.98o = 0.3749Member 2: Use transformation matrix, [q*] = [T]T[q] 123456q1 q21 21 00 10 00 00 00 0q1 q2q33001000q37* 8*0 00 00 00 0q4 q59*0001q6q7* q8* q9*=0.9271 -0.3749 0.3749 0.9271 00142 143. 23 [ q* ] 138*9* 7*2x[ q ]2 124x*Stiffness matrix [k]:3 4 5 0.000 0.000 -150.0 3.750 0.000 -0.9375 20.00 0.000 -3.7506 0.000 3.750 10.000.000 -3.750150.0 0.0000.000 0.93750.000 -3.75010.000.000-3.75020.001 2[k]2 =561 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 3.7501 2 150.0 0.000 0.000 0.93753 7* 0.000 -139.0 3.750 0.3518* -56.25 -0.8699* 0.000 3.75030.00020.001.406-3.47710.007* -139.0 0.351 1.406 8* -56.25 -0.869 -3.477129.051.829* 0.0001.4066 0.000Stiffness matrix [k*]: [k*] = [T]T[k][T][k*]2 = 1033.7503.75010.0051.821.406 21.90 -3.476 -3.47620.00143 144. 6352 48*1129* 7*4 54 5 6 1 2 0.000 150.0 0.000 0.000 -150.0 0.000 0.9375 3.750 0.000 -0.93753 0.000 3.75060.00020.000.000-3.75010.00-150.0 0.000 0.000 2 0.000 -0.9375 -3.750 3 0.000 3.750 10.00150.0 0.0000.000 0.93750.000 -3.7500.000-3.75020.001 21 2 150.0 0.000 0.000 0.93753 7* 0.000 -139.0 3.750 0.3518* -56.25 -0.8699* 0.000 3.75030.00020.001.406-3.47710.007* -139.0 0.351 1.406 8* -56.25 -0.869 -3.477 9* 0.000 3.750 10.00129.051.82[k]1= 2x103 1[k*]2 = 1033.7503.75051.82 1.4061.406 21.90 -3.476 -3.47620.00 144 145. 640 kN6 kN/m352 48* 21140 kN6 kN/m32 kNm9*40 kNm32 kNm40 kNm124 kN24 kN Global:1 1 3 = 103 7*0 0 0 09*18.5420 kN450 0 -139 07*3 0 60 1.406 10-139 1.406 129 1.406D118.15x10-6 -509.84x10-6rad0 10 1.406 20D1 D3 D7* D9*+0 -32 + 40 = 8 -7.5 -40m0.002259*rad58.73x10-67.5mD37*D7* D9*=145 146. 6352 432 kNmq4 q54 54 5 6 1 2 0.000 150.0 0.000 0.000 -150.0 0.000 0.9375 3.750 0.000 -0.9375q660.000q1 q2 q3= 2x10320.000.000-3.7501 -150.0 0.000 0.000 2 0.000 -0.9375 -3.750 3 0.000 3.750 10.00150.0 0.0000.000 0.93750.000-3.750q4 q5-5.45 20.18kN kNq621.80kNm5.45 27.82kN kN-52.39kNmq1 q2 q3=24 kN24 kNMember 1: [ q ] = [k ][d] + [qF]3.75032 kNm 1116 kN/m21.80 kNm3 0.000 0 3.750 0 10.00 00 240.000 d1=18.15x10-6 -3.750 0+20.00 d3=-509.84 x10-6 6 kN/m32 0 24 -3252.39 kNm 5.45 kN5.45 kN 20.18 kN27.82 kN 146 147. 23 140 kN8* 9* 7*240 kNm x 20 kNMember 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 26.55kN kNq352.39kN kN0kNm8* -56.25 -0.8699* 0.000 3.75018.15x10-6 00 20-3.75010.00-509.84x10-6401.406 21.90 -3.476 51.82-3.47620.0058.73x10-6 0+0.00225-7.5 18.54 -40kNm0 14.5120sin22.02=7.5 20cos22.02=18.54 20 kN 2x*q1 q2[40 kNmqF ]q7* q8* q9*=40 kN52.39 kNm5.45 kN 26.55 kN14.51 kN 147 148. 6 kN/m21.80 kNm52.39 kNm52.39 kNm40 kN5.45 kN5.45 kN5.45 kN 26.55 kN27.82 kN20.18 kN40 kN6 kN/m21.80 kNm14.51 kN 14.51cos 22.02o =13.45 kN5.45 kN 20.18 kNV (kN)54.37 kN 4m8m26.5520.18 + 3.36 m4 m 14.51 kN22.02o+ --x (m) -13.45-27.82 53.8112.14 M (kNm) -21.8+ -52.39x (m) 148