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Matrix frame
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Matrix frame

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  • 1. FRAME ANALYSIS USING THE STIFFNESS METHOD ! Simple Frames ! ! ! ! Frame-Member Stiffness Matrix Displacement and Force Transformation Matrices Frame-Member Global Stiffness Matrix Special Frames ! Frame-Member Global Stiffness Matrix 1
  • 2. Simple Frames 2
  • 3. Frame-Member Stiffness Matrix y´ = d 1´ 6´ 5´ j 3´ 2´ i x´ 4´ 6EI/L2 d 2´ = 4EI/L 2´ 3´ 4´ 5´ 6´ 1´ AE/L 0 0 - AE/L 0 0 2´ 0 12EI/L3 6EI/L2 0 - 12EI/L3 6EI/L2 3´ 0 6EI/L2 4EI/L 0 - 6EI/L2 0 0 5´ 0 -12EI/L3 6´ 0 6EI/L2 12EI/L3 = d 3´ 2EI/L AE/L 0 6EI/L2 0 12EI/L3 -6EI/L2 2EI/L 0 - 6EI/L2 1 AE/L 6EI/L2 AE/L 1 6EI/L2 d 5´ = 12EI/L3 12EI/L3 0 -6EI/L2 = d 4´ 2EI/L 1 6EI/L2 1´ -AE/L 1 12EI/L3 [k´] 4´ 6EI/L2 AE/L m 1´ AE/L 1 4EI/L 2EI/L = d 6´ 1 6EI/L2 4EI/L 6EI/L2 3
  • 4. 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 4
  • 5. λx Force Transformation x´ q4 = q4´ cos θx - q5´ cos θy q5 = q4´ cos θy + q5´ cos θx 6´ θy 5´ y´ θy 2´ 3´ λx = 1´ i 4´ q6 = q6´ j θx θx λy = y L y j − yi L 5 4 m 3 2 i 1 x q 4   λ x q  =  λ  5  y q 6   0    − λy λx 0 0 0  1  q 4'  q   5'  q 6'    q 1   λ x q    2   λy q 3   0  = q 4   0 q 5   0    q 6   0    x j − xi 6 j λy − λy λx 0 0 0 0 0 0 1 0 0 0 0 0 0 λx λy 0 0 0 0 − λy λx 0 0 0  0  0 0  1  q 1'  q   2'  q 3'    q 4'  q 5'    q 6'    [q ] = [T ]T [q '] 5
  • 6. [q] = [T]T[q´] = [T]T ( [k´][d´] + [q´F] ) = [T]T [k´][d´] + [T]T [q´F] [q] = [T]T [k´][T][d] + [T]T [q´F] = [k][d] + [qF] Therefore, [k] = [T]T [k´][T] [qF] = [T]T [q´F] [q] = [T]T[q´] [d´] = [T][d] [k] = [T]T [k´][T] 6
  • 7. Frame Member Global Stiffness Matrix [q] = [T]T[q´]= [T]T ( [k´][d´] + [q´F] ) = [T]T[k´][d´] + [T]T[q´F] = [T]T [k´][T][d] + [T]T [q´F] [k] [qF] [ 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 λix λjx ) - 6EI L2 6EI L2 λjy 6EI L2 λjx ( AE L ( AE L L3 12EI λiyλjx - L λixλjy) 12EI λixλjx + 4EI λix 12EI 6EI L3 L3 L3 6EI L L2 AE λixλjy ) AE -( L L λjy λixλjy - λiyλjy + 6EI - 12EI 12EI 2EI λiyλjy ) -( λjy λjx2 + - Vj λjy2 ) ) λjxλjy ( AE L2 - L Mj 12EI L3 12EI L3 6EI L2 λiy L2 L2 λix 2EI L 6EI ) λjxλjy λjx 6EI 6EI λixλjx ) AE 12EI λjx2 ) ( L λjy2 + 3 L - - λjx 12EI L3 λiyλjx ) L2 - λjy 6EI L2 λjx 4EI L 7
  • 8. Example 1 For the frame 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. E = 200 GPa, I = 60(106) mm4, A = 600 mm2 6m A 5 kN B 6m C 8
  • 9. 6m A 5 kN B 6m 12 EI = 3 L C Global : 6 A 5 4 2 1 AE = L (600 × 10 −6 m 2 )(200 × 10 6 6m 12(200 × 10 6 6EI = L2 3 1 B 4 EI = L kN ) 2 m = 20000 kN/m kN )(60 × 10 −6 m 4 ) m2 = 666.667 kN/m 3 (6m) kN )(60 × 10 −6 m 4 ) 2 m = 2000 kN (6m) 2 4(200 × 10 6 kN )(60 × 10 −6 m 4 ) 2 m = 8000 kN • m 6m 2(200 × 10 6 kN )(60 × 10 −6 m 4 ) 2 m = 4000 kN • m 6m 2 6(200 × 10 6 8 C 9 7 2EI = L 9
  • 10. Using Transformation Matrix: Global : 6 A Local : 5 4 2 1 1 3´ 2´ 3 1´ A B 6´ 5´ 1 4´ B 3´ 1´ 2 2 2´ 8 C 9 7 6´ 5´ • Member Stiffness Matrix δi ∆i Ni  AE/L 0 Vi  0 12 EI/L3  Mi  0 6 EI/L2 [k'] =  0 Nj − AE/L − 12 EI/L3 Vj  0  6 EI/L2 Mj  0  θi 0 6 EI/L2 4 EI/L 0 − 6 EI/L2 2 EI/L δj − AE/L 0 0 AE/L 0 0 ∆j 4´ θ j 0 − 12 EI/L3 − 6 EI/L2      0  − 6 EI/L2   4 EI/L   0 12 EI/L3 − 6 EI/L2 0 6 EI/L2 2 EI/L 10
  • 11. Stiffness Matrix: Member 1 6 Local : 5 A 4 3 2 1 1 3´ 2´ 1´ A B 6´ 5´ B 3 2 1 4´ Global: [q] = [q´] 9 8 C 4 7 -> [k]1 = [k´]1 6 5 1 2 4 0 0 -20000 0 0 6 [k]1 = 20000 0 666.667 2000 0 -666.667 2000 5 0 2000 8000 0 -2000 4000 0 0 20000 0 0 1 -20000 2 0 -666.667 -2000 0 666.667 -2000 3 0 2000 4000 0 -2000 8000 11
  • 12. Stiffness Matrix: Member 2 5 6 A 3 2 1 4 3´ 1´ 1 λix = cos (-90o) = 0 o 2´ λiy = sin (-90 ) = -1 o B 2 2 Local: Global: 6´ 9 8 C 7 q1 q2 q3 q7 q8 q9 = 1 2 3 7 8 9 2´ 1 0 0 0 0 0 λjx = cos (-90o) = 0 λjy = sin (-90o) = -1 5´ 4´ [q]2 = [ T ]T[ q´]2 1´ 0 -1 0 0 0 0 90 3´ 0 0 1 0 0 0 4´ 0 0 0 0 -1 0 5´ 0 0 0 1 0 0 6´ 0 0 0 0 0 1 q1´ q2´ q3´ q4´ q5´ q6´ [T]T 12
  • 13. 1´ 2´ 4´ 5´ 0 1´ 20000 3´ 6´ 0 -20000 0 0 2´ [k´]2 = 0 666.667 2000 0 -666.667 2000 3´ 0 2000 8000 0 -2000 4000 0 0 20000 0 0 4´ -20000 5´ 0 -666.667 -2000 0 666.667 -2000 6´ 0 2000 4000 0 -2000 8000 2 3 7 8 9 0 2000 0 2000 [k]2 = [ T ]T[ k´ ]2[ T ] 1 1 666.667 -666.667 2 [k]2 = 0 20000 0 0 -20000 0 3 2000 0 8000 -2000 0 4000 7 -666.667 0 -2000 666.667 0 -2000 8 0 -20000 0 0 20000 0 9 2000 0 4000 -2000 0 8000 13
  • 14. Global Stiffness Matrix: [k]1 5 6 A 2 4 4 3 6 2 Global: 0 5 B 0 9 2 5 0 1 -20000 2 0 3 0 8000 0 -2000 4000 0 20666.667 0 2000 -2000 0 3 0 4000 2000 0 -666.667 2000 -2000 4000 8000 0 0 20000 2000 0 0 -666.667 -2000 1 1 -20000 0 -20000 3 4000 0 0 0 666.667 -2000 0 -2000 8000 [k]2 5 0 2 0 0 0 [K] 4 4 20000 2 2000 0 3 7 1 666.667 2000 1 -20000 8 C 5 0 4 20000 1 1 6 20666.667 -2000 -2000 16000 2 1 666.667 0 2 0 3 2000 20000 0 7 -666.667 0 8 0 9 2000 -20000 0 3 7 8 2000 666.667 0 0 0 9 2000 -20000 0 0 4000 -2000 666.667 0 -2000 8000 -2000 0 0 4000 2000 20000 0 0 8000 14
  • 15. 6m A 5 kN 1 A B 4 4 Q5 = 0 5 0 8000 1 -20000 0 20666.667 Q2 = 0 2 0 -2000 0 Q3 = 0 3 0 4000 2000 D4 D5 D1 D2 D3 3 1 B 2 2 Q4 = 0 = 1 8 C [Q] = [K][D] + [QF] 4 20000 Q1 = 5 2 6m C Global: 5 6 9 7 5 0 1 -20000 2 0 3 0 D4 0 0 -2000 4000 D5 0 2000 D1 20666.667 -2000 D2 0 -2000 16000 D3 + 0 0 0 0.01316 m 9.199(10-4) rad = 0.01316 m -9.355(10-5) m -1.887(10-3) rad 15
  • 16. 6m A Member 1 5 kN 1 6 B 2 A 5 3 2 4 B 1 1 6m 1.87 kN C 11.22 kN•m 1.87 kN 1 A B [q]1 = [k]1[d]1 + [qF]1 4 q4 4 20000 q6 6 0 q5 5 0 q1 1 -20000 q2 2 0 q3 3 0 6 5 1 2 0 0 -20000 0 666.667 2000 0 3 0 -666.667 2000 2000 8000 0 -2000 4000 0 0 20000 0 0 -666.667 -2000 2000 4000 0 0 666.667 -2000 -2000 8000 D4 = 0.01316 D6 = 0 0 -1.87 D5 = 9.199(10-4) 0 D1 = 0.01316 0 D2 = -9.355(10-5) 1.87 D3 = -1.887(10-3) -11.22 16
  • 17. 6m A 1 Member 2 5 kN 2 B 2 2 8 9 18.77 kN•m 7 q1 1 666.667 0 q2 2 q3 3 2000 q7 7 -666.667 0 q8 8 q9 9 2000 20000 0 -20000 0 3 7 8 2000 -666.667 0 0 0 11.22 kN•m 5 kN 2 6m [q]2 = [k]2[d]2 + [qF]2 1 2 0 1.87 kN 1 C 0 3 5 kN 1.87 kN 9 2000 D1 = 0.01316 5 -20000 0 D2 = -9.355(10-5) -1.87 0 4000 D3 = -1.887(10-3) 11.22 -2000 666.667 0 -2000 D7 = 0 -5 20000 0 D8 = 0 1.87 0 8000 D9 = 0 18.77 8000 -2000 0 0 4000 -2000 17
  • 18. 1.87 kN A 11.22 kN•m 1.87 kN 1 5 kN A B 11.22 kN•m 1.87 kN 1.87 kN B 6m 5 kN 2 18.77 kN•m 5 kN C 18.77 kN•m 5 kN 6m 1.87 kN 1.87 kN A 11.22 B - A -11.22 Bending moment diagram B -1.87 5 + Shear diagram + C 18.77 C 5 18
  • 19. D4=13.16 mm A D1=13.16 mm D5=0.00092 rad D3=-0.00189 rad B 11.22 B - A -11.22 D3=-0.00189 rad Bending moment diagram Deflected shape C C 6 D4 D5 D3 = 0.01316 m -9.355(10-5) m -1.887(10-3) rad A 5 4 2 1 18.77 3 1 B 2 D1 D2 0.01316 m 9.199(10-4) rad + Global : 8 C 9 7 19
  • 20. 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 (c) Draw the quantitative shear and bending moment diagrams. E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member. 3 kN/m B C 4.5 m A 6m 6m 20
  • 21. 3 kN/m Global 1 6 4.5 m 2 Members 6 3 2 1 1 7 9 8 2 7 1 5 4 [FEM] 2 4 6m 3 9 8 1 5 6m 3 2 3 kN/m 9 kN•m 9 kN•m 2 9 kN 9 kN 21
  • 22. 3 kN/m m 7. 5 4.5 m AE (600 ×10 −6 m 2 )(200 ×106 kN / m 2 ) = L 7.5 m = 16000 kN / m 6m 6m Member 1: 3 2 θy 6 1 θx 5 1 4 λx = cos θx = 6/7.5 = 0.8 λy = cos θy = 4.5/7.5 = 0.6 12 EI 12(200 ×106 kN / m 2 )(60 ×10 −6 m 4 ) = L3 (7.5 m) 3 = 341.33 kN / m 6 EI 6( 200 × 10 6 kN / m 2 )(60 × 10 −6 m 4 ) = 2 L (7.5 m) 2 = 1280 kN 4 EI 4(200 × 106 kN / m 2 )(60 ×10 −6 m 4 ) = L 7.5 m = 6400 kN • m 2 EI 2(200 × 106 kN / m 2 )(60 ×10 −6 m 4 ) = L 7.5 m = 3200 kN • m 22
  • 23. Member m: λx = cos θx λy = cos θy ]T[ [ km ] = [ T Ui ( AE Vi L ( AE Vj - ( Mj - L Mi Uj -( λix2 + k´ ][T ] = - AE L AE L 12EI L3 6EI L2 λixλjx + λixλjy- - L3 6EI L2 λiy2 ) ) λixλiy ( ( AE L AE L 12EI L3 12EI L3 λiy - -( λiyλjx ) -( AE L AE L ) λixλiy L3 λiy2 + L2 λiy λjy) θx λiyλjx - λiyλjy + 6EI L2 Uj Mi 12EI 6EI λiy Uj Vi Vi 12EI θy Mi Ui 12EI L3 - λix2 ) 12EI L3 12EI L3 λix 6EI λiy -( L2 6EI L2 λix -( AE L AE L λix λjx ) - 6EI L2 6EI L2 λjy 6EI L2 λjx ( AE L ( AE L 12EI λiyλjx - L λixλjy) L3 L3 L3 6EI L L2 λixλjy ) AE -( L L λjy λixλjy - λiyλjy + 6EI - 12EI 12EI 2EI AE λjy λjx2 + - L3 λiyλjy ) -( λjy2 ) ) λjxλjy ( m Vj Ui 12EI λixλjx + 4EI λix Mj Vj AE L2 - L Mj 12EI L3 12EI L3 6EI L2 λiy L2 L2 λix 2EI L 6EI ) λjxλjy λjx 6EI 6EI λixλjx ) AE 12EI λjx2 ) ( L λjy2 + 3 L - - λjx 12EI L3 λiyλjx ) L2 - λjy 6EI λjx L2 4EI L 23
  • 24. 3 kN/m 7. 5 4.5 m Member 1: θy m 6 1 θx 5 6m 3 2 1 4 6m λx = cos θx = 6/7.5 = 0.8 λy = cos θy = 4.5/7.5 = 0.6 4 5 6 1 2 3 4 7516.162 -768 -10362.879 -7516.162 -768 5 [k1] = 10362.879 7516.162 5978.451 1024 -7516.162 -5978.451 1024 6 -768 1024 6400 768 -1024 3200 768 10362.879 7516.162 768 1 -10362.879 -7516.162 2 -7516.162 -5978.451 -1024 7516.162 5978.451 -1024 3 -768 1024 3200 768 -1024 6400 24
  • 25. 3 kN/m 4.5 m 7. 5 AE (600 × 10 −6 m 2 )(200 × 106 kN / m 2 ) = L 6m = 20000 kN / m m 6m 6m Member 2: 3 2 1 9 8 2 7 λx = cos 0o = 1.0, λy = cos 90o = 0 12 EI 12(200 ×10 6 kN / m 2 )(60 × 10 −6 m 4 ) = L3 (6 m) 3 = 666.667 kN / m 6 EI 6( 200 × 10 6 kN / m 2 )(60 × 10 −6 m 4 ) = 2 L (6 m) 2 = 2000 kN 4 EI 4(200 × 106 kN / m 2 )(60 ×10 −6 m 4 ) = L 6m = 8000 kN • m 2 EI 2(200 × 106 kN / m 2 )(60 ×10 −6 m 4 ) = L 6m = 4000 kN • m 25
  • 26. 3 2 1 2 1 2 3 7 8 1 AE/L 0 0 - AE/L 0 2 0 12EI/L3 6EI/L2 0 - 12EI/L3 6EI/L2 3 0 6EI/L2 4EI/L 0 - 6EI/L2 2EI/L 7 -AE/L 0 0 AE/L 0 0 8 0 9 0 9 8 7 [k2] = 1 3 0 1 0 2 [k2] = 20000 0 666.667 3 0 2000 7 -20000 0 8 0 -666.667 9 0 2000 12EI/L3 -6EI/L2 0 2EI/L 2 0 0 -12EI/L3 -6EI/L2 6EI/L2 9 - 6EI/L2 7 8 4EI/L 9 - 20000 0 0 2000 0 - 666.667 2000 8000 0 - 2000 4000 0 20000 0 -2000 0 666.667 4000 0 - 2000 0 -2000 8000 26
  • 27. 5 4 6 1 2 3 4 7516.162 -768 -10362.879 -7516.162 -768 5 [k1] = 10362.879 7516.162 5978.451 1024 -7516.162 -5978.451 1024 6 -768 1024 6400 768 -1024 3200 768 10362.879 7516.162 768 1 -10362.879 -7516.162 2 -7516.162 -5978.451 -1024 7516.162 5978.451 -1024 3 -768 1024 3200 768 -1024 6400 1 2 3 0 1 0 2 [k2] = 20000 0 666.667 3 0 2000 7 -20000 0 8 0 -666.667 9 0 2000 7 8 9 0 - 20000 0 2000 0 - 666.667 2000 8000 0 - 2000 4000 0 20000 0 -2000 0 666.667 4000 0 - 2000 0 -2000 8000 27
  • 28. 3 kN/m 3 2 7. 4.5 m 5m 9 kN•m 1 9 kN 9 kN 6 5 6m 9 8 7 2 1 4 6m Global: 1 2 3 1 30362.9 7516.16 768 D1 2 7516.16 6645.12 976 D2 3 768 976 14400 D3 0 Q1 0 Q2 0 Q3 = D1 D2 D3 0 + 9 9 4.575(10-4) m = -1.794(10-3) m -5.278(10-4) rad 28
  • 29. Member 1: 3 2 θy 6.75 kN 0.50 kN•m 1 θx 5 1.19 kN•m 9.15 kN 1 6 1.19 kN•m 11.37 kN 1 0.50 kN•m 6.75 kN 4 9.15 kN λx = cos θx = 6/7.5 = 0.8 11.37 kN 1 0.09 kN 0.09 kN λy = cos θy = 4.5/7.5 = 0.6 4 q4 q1 q2 q3 6 5 = 6 1 2 3 1 2 3 k1 0 0 9.15 0 4 q5 q6 5 0 6.75 0 0 0.50 D1 = 4.575(10-4) + 0 = -9.15 D2 = -1.794(10-3) 0 -6.75 D3 = -5.278(10-4) 0 -1.19 29
  • 30. Member 2: 3 2 9 8 2 1 1.19 kN•m 3 kN/m 7 9.15 kN 3 kN/m 9 kN•m 2 9.15 kN 6.75 kN 9 kN•m 14.70 kN•m 11.25 kN 2 9 kN [FEM] 9 kN 1 2 3 7 8 9 q1 1 D1 = 4.575(10-4) 0 9.15 q2 2 D2 = -1.794(10-3) 9 6.75 q3 3 D3 = -5.278(10-4) 9 1.19 q7 q8 q9 = k2 + = 7 0 8 0 9 11.25 9 0 -9 -14.70 0 -9.15 30
  • 31. 1.19 kN•m 3 kN/m 6.75 kN 9.15 kN 0.50 kN•m 14.70 kN•m 1.19 kN•m 9.15 kN 6.75 kN 1 2 9.15 kN 11.25 kN 6.75 kN 9.15 kN 3 kN/m 14.70 kN•m 9.15 kN 0.50 kN•m 11.25 kN All Reactions 9.15 kN 6.75 kN 31
  • 32. 1.19 kN•m 3 kN/m 11.37 kN 1.19 kN•m 9.15 kN 0.09 kN 1 11.37 kN D1 D2 D3 6.75 kN 9.15 kN 11.25 kN 6.75 0.50 kN•m 0.09 kN 4.575(10-4) = 2 14.70 kN•m + -0.09 m -1.794(10-3) m -5.278(10-4) rad -0.09 - Shear diagram (kN) -11.25 D1 = 0.46 mm D2 = -1.79 mm D3 =-5.278(10-4) -1.19 rad -14.70 Deflected shape 0.5 Bending-moment diagram (kN•m) 32
  • 33. 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. (c) Draw the quantitative shear and bending moment diagrams. E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member. 10 kN 20 kN•m 15 kN B m N/ 3k 4.5 m A C 6m 3m 3m 33
  • 34. 10 kN 20 kN•m 15 kN B m N/ 3k 4.5 m A C θy = 53.13 6 Members 1 6 3 2 3 2 4 3m 3m 1 1 5 1 1 λy = cos θy = 4.5/7.5 = 0.6 9 8 2 7 [FEM] wL2/12 = 14.06 kN•m 11.25(0.8) = 9 kN 10 kN 7.5 kN•m 7.5 kN•m m N/ 3k 1 7 2 14.06 kN•m 4 9 8 θx =36.87 λx = cos θx = 6/7.5 = 0.8 5 6m 3 2 Global 6.75 kN 11.25 kN 9 kN 2 5 kN 5 kN 11.25(0.6) = 6.75 kN wL/2 = 11.25 kN 34
  • 35. 10 kN 20 kN•m 15 kN 1 6 5 A 1 14.06 kN•m 10 kN 7.5 kN•m 7.5 kN•m 6.75 kN 11.25 kN m N/ 9 kN 1 3k 5 kN [FEM] 11.25(0.6) = 6.75 kN 14.06 kN•m 9 kN 2 5 kN 11.25 kN 1 = 2 3 1 Q1 = 15 Q3 = 20 7 2 4 Global: Q2 = 0 9 8 C B m N/ 3k 3 2 30362.9 7516.16 768 D1 2 7516.16 6645.2 976 D2 3 768 14400 D3 976 -6.75 + 9+5 -14.06 + 7.5 35
  • 36. 3 2 1 5 6 9 8 2 7 1 4 D1 D2 D3 1.751(10-3) m = -4.388(10-3) m 2.049(10-3) rad 36
  • 37. Member 1: 2 θy 14.06 kN•m 3 1 m N/ 3k 1 6 θx 5 14.06 kN•m 4 9 kN λx = cos 0.8, λy = cos 53.13o = 0.6 36.87o = 4 q4 5 q6 6 6 4 q5 5 q1 q2 q3 = 1 2 6.75 kN 11.25 kN 9 kN [FEM] 3 1 11.25(0.6) = 6.75 kN 11.25 kN -6.75 6.51 0 k1 0 9 24.17 0 14.06 26.46 + = -20.01 1 D1 = 1.751(10-3) 2 D2 = -4.388(10-3) 9 -6.17 3 D3 = 2.049(10-3) -14.06 4.89 -6.75 37
  • 38. 3 2 θy = 53.13o 1 1 6 θx = 36.87o 5 q4 6.51 q5 24.17 q6 26.46 = q1 4 q2 q3 6.17 kN -20.01 -6.17 4.89 4.89 kN•m 19.71 kN 4.89 kN•m 20.01 kN 26.46 kN•m m N/ 3k 7. 5m 6.51 kN 24.17 kN N/ 3k 26.46 kN•m 19.71 kN m 7.07 kN 1 m .5 7 15.43 kN 38
  • 39. Member 2: 3 2 2 1 9 8 15.12 kN•m 7 35.02 kN 10 kN 7.5 kN•m 7.5 kN•m 10 kN 2 8.08 kN•m 35.02 kN 6.17 kN 3.83 kN 2 5 kN 1 [FEM] 2 5 kN 3 7 8 9 q1 1 D1 = 1.751(10-3) 0 35.02 q2 2 D2 = -4.388(10-3) 5 6.17 q3 3 D3 = 2.049(10-3) 7.5 15.12 q7 q8 q9 = k2 + = 7 0 8 0 5 3.83 9 0 -7.5 -8.08 0 -35.02 39
  • 40. 20.01 kN 26.46 kN•m m N/ 3k 7. 15.12 kN•m 4.89 kN•m 6.17 kN 35.02 kN 10 kN 2 6.17 kN 5m 8.08 kN•m 35.02 kN 3.83 kN 6.51 kN 24.17 kN 10 kN 20 kN•m C 15 kN m N/ 3k 26.46 kN•m A 8.08 kN•m 35.02 kN B 3.83 kN 6.51 kN 24.17 kN 6 m 3m 3m 40
  • 41. 4.89 kN•m 19.71 kN m N/ 3k 7.5 26.46 kN•m 15.12 kN•m 35.02 kN 7.07 kN 1 m 10 kN 2 35.02 kN 6.17 kN 3.83 kN 6.17 15.43 kN 19.71 kN 8.08 kN•m -7.07 D1 1.751(10-3) m D2 -4.388(10-3) m = D3 15.43 -3.83 Shear diagram (kN) 2.049(10-3) rad D1 =1.75 mm 4.89 D2 = -4.39 mm -15.12 D3 = 2.05(10-3) rad Deflected shape -8.08 Bending-moment diagram (kN•m) -26.46 41
  • 42. Special Frames 42
  • 43. 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´ ] 1´ q 1*  1*  λix q   2*  λiy 2*   q 3*  3*  0  =  q 4 *  4*  0  q 5*  5*  0    q 6 *  6*  0    2´ 3´ 4´ 5´ 0 − λiy 0 0 0 0 0 λix 0 1 0 0 0 0 λ jx − λ jy 0 0 λ jy λ jx 0 0 0 0 6´ 0 0  0  0 0  1  q 1'  q   2'  q 3'    q 4'  q 5'    q 6'    [ T ]T 43
  • 44. 1´ 2´ [T ] = 3´ 4´ 5´ 6´ 1* 2* λiy  λix − λ  iy λix  0 0  0  0  0 0  0  0  3* 4* 0 0 0 0 1 0 0 λ jx 0 − λ jy 0 0 5* 0 0 0 λ jy λ jx 0 6* 0 0  0  0 0  1  • Member Stiffness Matrix 1´ 1´ 2´ 3´ [k'] = 4´ 5´ 6´ 2´ 0  AE/L  0 12 EI/L3   0 6 EI/L2  0 − AE/L  0 − 12 EI/L3  6 EI/L2  0  3´ 4´ 5´ 0 − AE/L 0 6 EI/L2 0 − 12 EI/L3 4 EI/L 0 0 AE/L − 6 EI/L2 0 − 6 EI/L2 0 12 EI/L3 2 EI/L 0 − 6 EI/L2 6´  6 EI/L2   2 EI/L   0  − 6 EI/L2   4 EI/L   0 44
  • 45. [ k ] = [ T ]T[ k´ ][T] = Vi Ui Ui ( AE Vi L ( AE Vj - ( Mj - L Mi Uj -( λix2 + - AE L AE L L3 12EI L3 6EI L2 λixλjx + λixλjy- - 12EI 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 - Mi λiyλjx - λiyλjy + 6EI L2 12EI L3 - λix2 ) L3 12EI L3 λix λiy -( L2 L2 λix -( AE L AE L λix λjx ) - 6EI L2 6EI L2 λjy 6EI L2 λjx ( AE L ( AE L L3 12EI λiyλjx - L λixλjy) 12EI λixλjx + 4EI λix 12EI 6EI 6EI Vj Uj L3 L3 6EI L L2 λixλjy ) AE -( L L λjy λixλjy - λiyλjy + 6EI - 12EI 12EI 2EI AE λjy λjx2 + - L3 λiyλjy ) -( λjy2 ) ) λjxλjy ( AE L2 - L Mj 12EI L3 12EI L3 6EI L2 λiy L2 L2 λix 2EI L 6EI ) λjxλjy λjx 6EI 6EI λixλjx ) AE 12EI λjx2 ) ( L λjy2 + L3 - - λjx 12EI L3 λiyλjx ) L2 - λjy 6EI λjx L2 4EI L 45
  • 46. 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 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 3m 20 kN 200 kN•m 6 kN /m 22.02 o 4m 4m 7.416 m 46
  • 47. 40 kN 3m 22.02 o 4m 2* Global 20 kN 200 kN•m 6 kN /m 4m 3* 1* 5 7.416 m 6 4 1 22.02 9 8 2 o 7 3´ Local 5´ 2´ 1´ 1 6´ 4´ 2´ 3´ 1´ 2 5´ 6´ 4´ 47
  • 48. δi ∆i Ni  AE/L 0 Vi  0 12 EI/ L3  Mi  0 6EI/ L2 [k'] =  0 Nj − AE/ L Vj  0 − 12 EI/ L3  6EI/ L2 Mj  0  θi 0 6EI/ L2 δj ∆j − AE/ L 0 − 12 EI/ L3 0 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  2 − 6EI/ L  4 EI/ L   AE (0.006 m 2 )(200 ×106 kN / m 2 ) = = 150 ×103 kN / m L 8m 4 EI 4(200 ×106 kN / m 2 )(0.0002 m 4 ) = = 20 ×103 kN • m L 8m 2 EI = 10 × 103 kN • m L 6 EI 6(200 × 106 kN / m 2 )(0.0002 m 4 ) = = 3.75 × 103 kN 2 2 L (8 m) 12 EI 12(200 × 10 6 kN / m 2 )(0.0002 m 4 ) = = 0.9375 × 103 kN / m 3 3 L (8 m) 48
  • 49. 3´ 2´ 5´ 8m 6´ 4´ 1 1´ 2´ 3´ 1´ 8m 5´ 2 Local 6´ 4´ δi Ni [ k´]1 = [ k´]2 = Vi Mi ∆i θi 150000 0 0 0 937.5 3750 0 3750 20000 Nj -150000 0 0 Vj 0 -937.5 -3750 Mj 0 3750 10000 δj ∆j θj 0 0 0 - 937.5 3750 0 - 3750 10000 0 937.5 0 -3750 - 3750 20000 - 150000 150000 0 0 49
  • 50. Member 1: 2* 3* 5 1 Global o 1 * θi =22.02 λix = cos (22.02o) = 0.927, λiy = sin (22.02o) = 0.375 3´ 6 2´ 4 θj = 0o Local 1´ 5´ 6´ 4´ 1 40 kN λjx = cos (0o) = 1, λjy = sin (0o) = 0 40 kN•m 40 kN•m 1 [ q* ] = [ T ]T[ q´ ] 1´ 20 kN 2´ 3´ 4´ 5´ [FEM] 20 kN 6´ 1* 0.927 - 0.375 0 0 0 0 q*2 2* 0.375 0.927 0 0 0 0 q*3 3* 0 0 1 0 0 0 q3´ 4* 0 0 0 1 0 0 q4´ 5* 0 0 0 0 1 0 q5´ 6* 0 0 0 0 0 1 q6´ q*1 q4 q5 q6 = q1´ q2´ [ T ]1T 50
  • 51. 3* 2* 5 1 Global o 1 * θi =22.02 3´ 6 2´ 4 θj = 0o Local 1´ 1 5´ 6´ 4´ [ k* ]1 = [ T ]1T[ k´ ]1[ T ]1 1* 2* 3* 4 5 6 1* 129.046 51.811 -1.406 -139.058 0.351 -1.406 21.892 3.476 -56.240 -0.869 3.476 0 3* -1.406 3.476 20.00 -3.75 10.00 3 10 150 4 -139.058 -56.240 0.00 0 0 0 5 0.351 -0.869 -3.75 0.938 -3.75 2* 51.811 [ k * ]1 = 6 -1.406 3.476 10.00 0 -3.75 20 51
  • 52. 2* Global 40 kN 3* 5 1 o 1 * θi =22.02 6 40 kN•m 4 θj = 0o 40 kN•m 1 20 kN [FEM] 20 kN [ qF* ] = [ T ]T[ qF´] 0 20 [ qF*] = [ T ]1 T 40 0 20 -40 -7.50 18.54 40 = 0 20 -40 1* 2* 40 kN•m 3* 4 40 kN 18.54 40 kN•m 7.5 1 20 kN 5 6 52
  • 53. λix = λjx = cos (-22.02o) = 0.927, λiy = λjy = sin (-22.02o) = -0.375 6 5 3´ 2´ Member 2 4 22.02o 1´ 5´ 2 8 2 9 7 [q] 32 k 22.02o N•m 24 k [q]=[T 4´ 6 kN /m N 2 24 k ]T[ 6´ q´ ] [ q´] [ q´F] N 32 k 1´ 2´ 3´ 4´ 5´ 6´ 4 0.927 0.375 0 0 0 0 q5 5 -0.375 0.927 0 0 0 0 q6 6 0 0 1 0 0 0 q3´ 7 0 0 0 0.927 0.375 0 q4´ 8 0 0 0 -0.375 0.927 0 q5´ 9 0 0 0 0 0 1 N•m q6´ q4 q7 q8 q9 = q1´ q2´ [ T ]2T 53
  • 54. 6 5 2´ 4 22.02o 8 2 1´ 9 7 [q] 3´ 5´ 2 22.02o 6´ 4´ [ q´ ] [ k ]2 = [ T ]2T[ k´]2[ T ]2 4 5 6 7 8 9 4 [ k ]2 = 1.406 5 -51.811 3.476 6 103 129.046 -51.811 1.406 -129.046 51.811 1.406 7 -129.046 21.892 3.476 3.476 20 51.811 -21.892 -1.406 -3.476 10 51.811 -1.406 129.046 -51.811 -1.406 8 51.811 -21.892 -3.476 -51.811 21.892 -3.476 9 1.4056 3.476 10 -1.406 -3.476 20 54
  • 55. 5 6 32 k 4 22.02o 8 2 9 7 [q] N•m 24 k 6 kN /m N 2 24 k 22.02o [ q´F ] N 32 k N•m [ qF* ] = [ T ]T[ qF´ ] 0 24 [ qF ] = [T ]2T 32 0 24 -32 8.998 22.249 32 = 8.998 22.249 -32 4 5 6 7 8 32 kN•m 8.998kN [ qF ] 6 kN /m 22.249 kN 2 8.998 kN 32 kN•m 22.249 kN 9 55
  • 56. 3* 2* Global Stiffness: 1* 5 6 4 1 8 2 9 7 1* [ k* ]1 = 2* 3* 4 5 6 1* 129.046 51.811 -1.406 -139.058 0.351 -1.406 2* 51.811 21.892 3.476 -56.240 -0.869 3.476 0 3* -1.406 3.476 20.00 -3.75 10.00 3 10 150 4 -139.058 -56.240 0.00 0 0 5 0.351 -0.869 -3.75 0 0.938 -3.75 6 -1.406 4 3.476 5 10.00 6 0 -3.75 8 7 4 5 [ k ]2 = 129.046 -51.811 1.406 -129.046 -51.811 21.892 6 103 1.406 3.476 3.476 20 20 9 51.811 1.406 51.811 -21.892 3.476 -1.406 -3.476 10 7 -129.046 51.811 -1.406 129.046 -51.811 -1.406 8 51.811 -21.892 -3.476 9 1.4056 3.476 10 -51.811 21.892 -3.476 -1.406 -3.476 20 56
  • 57. Global: 40 kN 20 kN 200 kN•m 6 kN /m 40 kN 40 kN•m 18.54 kN 40 kN•m 5 7 [ qF ] 6 kN /m 22.249 kN 2 8.998 kN [ Q ] = [ K ][ D ] + [ QF ] 1* 129.046 Q3* = 0.0 Q4 = 0.0 3* = 103 4 32 kN•m 22.249 kN 1* Q1* = 0.0 9 8 2 32 kN•m 20 kN [ q*F ] 6 4 1 1* 8.998 kN 1 7.5 3* 2* 3* 4 -1.406 -139.058 -1.406 20 -139.058 0 0 5 6 0.351 -1.406 D1* -7.5 -3.75 10 D3* 40 279.046 -51.811 1.406 D4 22.829 -0.274 D5 20 +22.249 -0.274 D6 -40 + 32 Q5 = -20 5 0.352 -3.75 -51.811 Q6 = -200 6 -1.406 10 1.406 40 + 8.998 57
  • 58. Global: 40 kN 20 kN 200 kN•m 6 kN /m 2* 3* 1* 5 1 6 4 2 8 9 7 D1* -0.0205 m D3* -0.0112 rad -0.0191 m D5 -0.0476 m D6 -0.0024 rad D4 = 58
  • 59. 2* 3* 5 1 [q] 4 1* 18.54 Member 1: [ q*] = [ k*][ d*] + [ 1* q1* q2* q3* q4 = 103 7.5 3* 4 5 1 20 kN [ qF* ] qF*] 40 kN•m 6 1* 129.046 51.811 -1.406 -139.058 0.351 -1.406 D1*=-0.0205 2* 51.811 21.892 3.476 -56.240 -0.869 3.476 D2*= 0.0 0 3* -1.406 3.476 20.00 -3.75 10.00 D =-0.0112 4 -139.058 -56.240 0.00 150 3* 0 0 D4= -0.0191 -7.50 18.54 + 40 0 5 = 0.351 -0.869 -3.75 0 0.938 -3.75 D5= -0.0476 20 6 q5 q6 q1* q2* q3* q4 q5 q6 2* 40 kN 40 kN•m 6 -1.406 3.476 10.00 0 -3.75 D6=-0.0024 -40 0 22.63 0 -8.49 19.02 7.87 40 kN kN kN kN kN•m 20 7.87 kN•m 8.49 kN 22.6 3 kN 19.02 kN 59
  • 60. 32 kN•m 6 5 4 8.998 kN 9 8 2 22.249 kN 2 8.998 kN 7 [q] [ qF ] 6 kN /m Member 2 : [ q ] = [ k ][ d ] + [ qF ] 4 5 6 7 8 4 129.046 -51.811 1.406 -129.046 51.811 q5 5 -51.811 q6 6 1.406 q4 q7 =103 3.476 20 22.249 kN 9 1.406 D4= -0.0191 8.998 51.811 -21.892 3.476 D5= -0.0476 22.249 -1.406 -3.476 10 D6=-0.0024 129.046 -51.811 -1.406 D7 = 0 51.811 -21.892 -3.476 1.4056 + -51.811 21.892 -3.476 D8 = 0 9 = 51.811 -1.406 8 q8 q9 q4 q5 q6 q7 q8 q9 7 -129.046 21.892 3.476 32 kN•m 8.49 -39.02 -207.87 9.51 83.51 -248.04 3.476 kN kN kN•m kN kN kN•m 10 -1.406 -3.476 32 8.998 22.249 20 D9 = 0 -32 207.87 kN•m 6 kN /m 8.49 kN 39.02 kN 248.04 kN•m 9.51 kN 83.51 kN 60
  • 61. 22.5 207.87 kN•m 7.87 kN•m kN 40 kN 8.49 kN 22.6 20.98 kN 3 kN 8.49 kN 19.02 kN 83.56 + 7.87 + 6 kN 8.49 kN /m 248.04 kN•m 33 kN 39.02 kN 9.51 kN 22.5 207.87 81 kN kN 83.51 kN D1* Bending-moment diagram (kN•m) D1*=- 20.5 mm - D4=-19.1 mm -248.04 -0.0205 m D3* -0.0112 rad D4 = -0.0191 m D5 -0.0476 m D6 -0.0024 rad D5=-47.6 mm D3*=-0.0112 rad D6 =0.00 24 r ad Deflected shape 61
  • 62. 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 bending moment diagrams and qualitative deflected shape. Take I = 400(106) mm4 , A = 60(103) mm2, and E = 200 GPa for all members. 20 kN/m A B 80 kN•m 50 kN 3m C 4m 20o 2m 62
  • 63. 20 kN/m A B 80 kN•m 50 kN 3m C 4m 3 1 6 5 2´ 5´ 6´ 4´ 3´ 2 2 9* 7* Local 5´ 8* 1 1´ 4 Global 3´ 1´ 1 2m 2´ 2 76.31o 20o 6´ 4´ 63
  • 64. 20 kN/m A B 80 kN•m 50 kN 3m 2 3 1 5 1 C 4m 6 4 20o 2m Member 1: AE (60 × 10 −3 m 2 )(200 × 10 6 kN/m 2 ) = 4m L = 3000 × 10 3 kN/m 4 EI 4(200 × 10 6 kN/m 2 )(400 × 10 −6 m 4 ) = 4m L = 80 × 10 3 kN • m 2EI 2(200 × 10 6 kN/m 2 )(400 × 10 −6 m 4 ) = 4m L = 40 × 10 3 kN • m 6EI 6(200 × 10 6 kN/m 2 )(400 × 10 −6 m 4 ) = 2 (4 m) 2 L = 30 × 10 3 kN 12 EI 12(200 × 10 6 kN/m 2 )(400 × 10 −6 m 4 ) = L3 (4 m) 3 = 15 × 10 3 kN/m 64
  • 65. 20 kN/m A B 80 kN•m 50 kN 2 3m 5 1 1 56.31o 20o C 4m 3 2m 6 4 20 kN/m 76.31o 26.67 kN•m 26.67 kN•m 1 40 kN 40 kN Member 1: [ q ] = [ k ][ d ] + [ qF ] 1 2 3 4 5 6 q1 1 3000 0 0 -3000 0 0 d1 0 q2 2 0 15 30 0 -15 30 d2 40 q3 3 0 30 80 0 -30 40 d3 26.67 4 -3000 0 0 3000 0 0 d4 5 0 -15 -30 0 15 -30 d5 40 6 0 30 40 0 -30 80 d6 -26.67 q4 q5 q6 = 103 + 0 65
  • 66. 2´ 6 5 3´ 4 i Member 2: 1´ 6EI 6(200 × 10 6 kN/m 2 )(400 × 10 −6 m 4 ) = 2 (3.61 m) 2 L 2 2 8* = 36.83 × 10 3 kN 5´ 9* * 7 6´ 4´ j AE (60 × 10 −3 m 2 )(200 × 10 6 kN/m 2 ) = 3.61 m L = 20.41 × 10 3 kN/m = 3324 × 10 3 kN/m 4 EI 4(200 × 10 6 kN/m 2 )(400 × 10 −6 m 4 ) = 3.61 m L = 88.64 × 10 3 kN • m [ k´ ]2 1´ = 1´ 2´ 3´ 3 10 4´ 5´ 6´ 12 EI 12(200 × 10 6 kN/m 2 )(400 × 10 −6 m 4 ) = 3 L (3.61 m) 3 2EI 2(200 × 10 6 kN/m 2 )(400 × 10 −6 m 4 ) = 3.61 m L = 44.32 × 10 3 kN • m 2´ 3324 0 0 20.41 0 36.83 3´ 0 36.83 88.64 -3324 0 0 0 -20.41 -36.83 0 36.83 44.32 4´ 5´ 6´ -3324 0 0 -20.41 0 -36.83 0 36.83 44.32 0 3324 0 0 20.41 -36.83 0 -36.83 88.64 66
  • 67. 6 4 56.31o i i 3´ (-56.31o) 2 1´ 9* 2 λix = cos = 0.55, λiy = sin (-56.31o) = -0.83 2´ 5 5´ 7* 76.31o j q4 q5 q6 q7* q8* q9* 4´ [ q* ] = [ T ]T[ q´ ] 6´ * λjx = cos = 0.24,8 λjy = sin (-76.31o) = -0.97 j (-76.31o) = 1´ 4 0.55 5 -0.83 0 6 7* 0 0 8* 0 9* 4 5 6 * ] = [ T ]T[ k´] [ T ] = 103 [k 2 2 7* 8* 9* 2´ 0.83 0.55 0 0 0 0 3´ 0 0 1 0 0 0 4 5 1036.923 -1524.780 -1524.780 2307.582 30.646 20.431 -452.884 643.585 1787.474 -2689.923 30.646 20.431 4´ 0 0 0 0.24 -0.97 0 6 30.646 20.431 88.643 -35.786 -8.717 44.321 5´ 0 0 0 0.97 0.24 0 7* -452.884 643.585 -35.786 205.452 -759.668 -35.786 6´ 0 0 0 0 0 1 8* 1787.474 -2689.923 -8.717 -759.668 3139.053 -8.717 q1´ q2´ q3´ q4´ q5´ q6´ 9* 30.646 20.431 44.321 -35.786 -8.717 88.643 67
  • 68. 2 3 1 6 5 1 4 2 9* 8* [k]1 = 103 [k*]2 = 1 2 3 4 5 6 4 5 6 3 10 7* 8* 9* 1 3000 0 0 -3000 0 0 2 0 15 30 0 -15 30 3 0 30 80 0 -30 40 4 -3000 0 0 3000 0 0 7* 5 0 -15 -30 0 15 -30 6 0 30 40 0 -30 80 4 5 6 7* 8* 9* 1036.923 -1524.780 30.646 -452.884 1787.474 30.646 -1524.780 2307.582 20.431 643.585 -2689.923 20.431 20.431 88.643 -35.786 30.646 -8.717 44.321 -452.884 643.585 -35.786 205.452 -759.668 -35.786 1787.474 -2689.923 -8.717 -759.668 3139.053 -8.717 30.646 20.431 44.321 -35.786 -8.717 88.643 68
  • 69. 20 kN/m A B 80 kN•m 50 kN 3 2 1 6 5 1 26.67 kN•m 20 kN/m 4 2 3m 9* 8* C 4m D4 D5 D6 D7* D9* 20o 2m Global: Q4 = -50 Q5 = 0 Q6 = -80 Q7* = 0 Q9* = 0 1 = = 4 5 103 6 7 9 40 kN 26.67 kN•m 40 kN 4 5 6 7* 9* 4036.923 -1524.780 30.646 -452.884 30.646 -1524.780 232.582 -9.569 643.585 -9.569 30.646 -9.569 168.643 -35.786 44.321 -452.884 643.585 -35.786 205.452 -35.786 30.646 20.431 44.321 -35.786 88.643 -2.199x10-5 -3.095x10-4 -2.840x10-4 m rad D4 D5 D6 D7* D9* 0 40 + -26.67 0 0 m m rad 0.979x10-3 6.161x10-4 7* 69
  • 70. Member 1: 1 2 3 4 5 6 q1 1 3000 0 0 -3000 0 0 0 0 q2 2 0 15 30 0 -15 30 0 40 q3 3 0 30 80 0 -30 40 0 26.67 4 -3000 0 0 3000 0 0 D4 5 0 -15 -30 0 15 -30 D5 40 6 0 30 40 0 -30 80 D6 -26.67 q4 = q5 q6 103 q1 65.97 kN q2 36.12 kN q3 24.59 -65.97 kN•m kN 43.88 kN -40.11 kN•m 2 q4 q5 q6 = 3 5 1 1 65.97 kN 36.12 kN 0 6 4 20 kN/m 24.59 kN•m + 1 40.11 kN•m 65.97 kN 43.88 kN 70
  • 71. Member 2: 4 5 1036.923 -1508.14 q4 4 q5 5 -1508.14 30.57 6 q6 q7* = q8* q9* 103 6 7* 8* 9* 30.57 -455.21 1769.34 30.57 D4 2296.15 20.26 651.27 -2678.93 20.26 D5 20.26 88.64 -35.73 -8.84 44.32 D6 651.27 7* -455.21 8* 1769.34 -2678.93 -35.73 210.67 -769.1 -35.73 D*7 9* 30.57 -43.88 kN q6 -39.89 0 kN•m kN 46.69 kN 0 kN•m -35.73 kN q5 44.32 q7* q8* q9* = -8.84 88.64 6 5 -8.84 D*9 39.89 kN•m 43.88 kN 15.97 kN 4 i 0 2 15.97 -769.1 3128.82 2 q4 20.26 -8.84 9* 8* j 7* 46.69 kN 71
  • 72. 2 3 1 6 5 1 4 2 65.97 kN 9* 8* 20 kN/m 24.59 kN•m 7* 40.11 kN•m 65.97 kN 1 36.12 kN 43.88 kN 39.89 kN•m 43.88 kN 15.97 kN + -24.59 39.89 - + -40.11 Bending-moment diagram (kN•m) = -2.199x10-5 -3.095x10-4 -2.840x10-4 0.979x10-3 6.161x10-4 m m rad m rad 2 D4 D5 D6 D7* D9* 46.69 kN D4=-2.2x10-5 m D5=-3.1x10-4 m D6 = -2.84x10-4 rad D7*=0.979x10-3 m Deflected shape D9*=6.161x10-4 rad 72

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