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# Indeterminate Structures

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• 1. ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE FORCE METHOD ! ! ! ! ! ! Force Method of Analysis: Beams Maxwell’s Theorem of Reciprocal Displacements; Betti’s Law Force Method of Analysis: Frames Force Method of Analysis: Trusses Force Method of Analysis: General Composite Structures 1
• 2. Force Method of Analysis : Beams 1 Degree of freedom • Compatibility of displacement M1 P 1 2 C A M1 B R1 P 1 2 C A R1 R2 L R2 C A = P ∆´2 B L = P • Compatibility of slope B θ ´1 + + 1 α11 f22 × R2 × M1 1 ∆´2 + f22 R2 = ∆2 = 0 θ ´1 + α11M1 = θ 1= 0 2
• 3. 2 Degree of freedom Ax P A 1 B Ay R1 P 2 C R2 B A D ∆´2 + f21 f11 f12 Dy C ∆´1 A D D 1 x R1 R1 + f22 A D x R2 R2 1 ∆´1 + f11 R1 + f12 R2 ∆´2 + f21 R1 + f22 R2 = ∆1 = 0 = ∆2 = 0 3
• 4. Maxwell’s Theorem of Reciprocal Displacements; Betti’s Law f21 1 1 2 A B f11 f21 m2 M 1 mm dx = ∫ 2 1 dx EI EI L L 1 • f 21 = ∫ m1 m2 m1 dx EI L f 21 = ∫ 1 A f12 B f22 m2 4
• 5. 1 f12 m2 m1 dx EI L 2 f 21 = ∫ 1 A B f22 f12 m1M 2 mm dx = ∫ 1 2 dx EI EI L L 1 • f12 = ∫ m1m2 dx EI L f12 = ∫ m2 f 21 = f12 Maxwell’s Theorem: 1 A B f ij = f ji f21 f11 m1 5
• 6. f11, f22 1 2 1 A B f11 f21 m1M 1 mm dx = ∫ 1 1 dx EI EI L L 1 • f11 = ∫ m2 M 2 mm dx = ∫ 2 2 dx EI EI L L 1 • f 22 = ∫ m1 In general, 1 A B 1 • f ij = f = ∫ mi m j L f22 f12 1 • f ji = f ji = ∫ L EI m j mi EI dx dx m2 6
• 7. 1 A 2 P1 D d11 = f11 P1 d21 = f21 P1 P2 A D d12 = f12 P2 d22 = f22 P2 7
• 8. Force Method of Analysis: General 1 Compatibility Eq. 2 w ∆´1 + f11R1 + f12R2 = ∆1 = 0 ∆´2 + f21R1 + f22R2 = ∆2 = 0 = ∆´1 w ∆´1 ∆´2 ∆´2 + f12 f22 R 2 xR1 1 + f12 f12 f22 R 2 f11 f12 R1 f21 f11 + f11 f12 R1 f22 1 xR2 = =- 0 ∆1 0 ∆2 ∆´1 ∆´2 General form: f11 f12 f1n f21 f22 f2n . . . fn1 fn2 fnn R1 R2 . . . Rn ∆´1 ∆´2 . =- . . ∆´n 8
• 9. Example 9-1 Determine the reaction at all supports and the displacement at C. 50 kN B C A 6m 6m 9
• 10. SOLUTION Use compatibility of displacement for find reaction • Principle of superposition 50 kN MA A 6m 6m RB = RA B C 50 kN ∆´B + RB fBB x RB Compatibility equation : ∆ ' B + f BB RB = 0 -----(1) 1 10
• 11. • Use formulation for ∆´B and fBB 50 kN A 6m C P(6) 3 P ( 6) 2 ∆'B = + ( 6) 3EI 2 EI ∆´B 50(6)3 (50)(6) 2 9000 = + (6) = ,↓ EI 3EI 2 EI C f BB 6θ´C θ´C ∆´B = ∆´C+ (6 m)θ´C A B ∆´C 6m PL3 (1)(12) 3 576 = = = ,↑ 3EI 3EI EI B fBB 1 11
• 12. • Substitute ∆´B and fBB in Eq. (1): ∆ 'B + f BB RB = 0 + ↑: − ∆'B = 9000 576 +( ) RB = 0 EI EI RB = 15.63 kN, 9000 ,↓ EI f BB = 576 ,↑ EI 50 kN 6m MA A 6m C B 15.63 kN RA Equilibrium equation : + ΣMA = 0: + ΣFy = 0: M A − 50(6) + 15.63(12) = 0, + RA − 50 + 15.63 = 0, MA = 112.4 kN, + Ra = 34.37 kN, 12
• 13. • Quantitative shear and bending diagram and qualitative deflected curve 50 kN 6m 112.4 kN•m A 6m B C 15.63 kN 34.37 kN 34.37 V (kN) x (m) -15.63 -15.63 93.78 M (kN•m) x (m) 3.28 -112.44 6 12 13
• 14. Or use compatibility of slope to obtain reaction 50 kN • Principle of superposition 6m 6m C A RA B = MA RB 50 kN A B C θ ´A 1 + fAA A B C x MA MA Compatibility equation : θ ' A + f AA M A =θ A= 0 -----(2) 14
• 15. • Use the table on the inside front cover for θ´B and fBB 50 kN A C B C B θ ´A PL2 θ A'= 16 EI 1 fAA A f CC = L 3EI Substitute the values in equation: θ ' A + f AA M A =θ A= 0 +: PL2 L − + MA =0 16 EI 3EI MA = 3PL 3(50)(12) = 16 16 = 112.5 kN•m, + 15
• 16. Or use Castigliano least work method 50 kN x1 x2 12RB - 300 = MA B C A 6m 50 - RB = RA M 12RB - 300 diagram L ∆B = 0 = ∫ ( 0 6m RB M2 = RBx2 M1 = (12RB - 300) + (50 - RB)x1 x (m) ∂M M ) dx ∂RB EI 6 6 1 1 0= (12 − x1 )(12 RB − 300 + 50 x1 − RB x1 ) dx1 + x2 ( RB x2 ) dx2 EI ∫ EI ∫ 0 0 2 2 3 3 3 6 900 x1 24 x1 50 x1 x1 x 0 = (144 RB x1 − 3600 x1 + − RB − + RB ) 0 + 2 RB 2 2 3 3 3 RB =15.63 kN, 6 0 16
• 17. Use conjugate beam for find the displacement 50 kN 6m 112 kN•m 6m C B A ∆C 34.4 kN Real Beam 15.6 kN 93.6 M (kN•m) x (m) 12 6 3.28 93.6/EI -112 Conjugate Beam -112/EI 281 223 776 M 'C = ( 2) − (6 ) = − EI EI EI ∆ C = M 'C = − 776 ,↓ EI 281/(EI) M´C V´C 2 m 4m 223/(EI) 223/(EI) 17
• 18. Use double integration to obtain the displacement 50 kN 6m 112 kN•m 6m C B A ∆C 34.4 kN 93.6 Real Beam 15.6 kN M (kN•m) x (m) 3.28 -112 6 12 d 2υ EI 2 = −112 + 34.4 x1 dx dυ x12 EI = −112 x1 + 34.4 + C1 dx 2 2 x1 x13 EIυ = −112.4 + 34.4 + C1 x1 + C2 2 6 1 62 63 778 (−112( ) + 34.4( ) + 0 + 0) = − ∆C = ,↓ 2 6 EI EI 18
• 19. Example 9-2 Draw the quantitative Shear and moment diagram and the qualitative deflected curve for the beam shown below.The support at B settles 5 mm. Take E = 200 GPa, I = 60(106) mm4. 16 kN B A C ∆B = 5 mm 2m 2m 4m 19
• 20. Use compatibility of displacement to obain reaction • Principle of superposition 16 kN B A C ∆B = 5 mm 2m 2m 16 kN 4m = SOLUTION + ∆´B fBB × RB 1 kN Compatibility equation : ∆ B = −0.005m = ∆ 'B + f BB RB -----(1) 20
• 21. • Use conjugate beam method for ∆´B 16 kN Real A beam 12 kN B ∆´B 2m 2m 24 M´ diagram 24 Conjugate EI beam 56 EI C 4 3 24 EI 4m 4 kN 16 16 EI 72 EI M´´B 2 3 2m 4m 40 EI V´´B 4 3 32 EI 40 EI 2 4 3 32 4 40 + ΣMB = 0: − M ' ' B + ( )− (4) = 0 EI 3 EI 117.33 ∆' B = M ' ' B = − ,↓ 21 EI
• 22. • Use conjugate beam method for fBB fBB Real A beam 0.5 kN B C 1 kN 4m 4m 0.5 kN m´ diagram m´´B vB´B -2 Conjugate beam 4 EI 4 3 − + ΣMB = 0: 4 − EI − 2 EI 4 − EI 4 EI 2 4 3 4 EI 4 EI 4 4 4 ( )+ (4) = 0 EI 3 EI 10.67 f BB = mB ' ' = ,↑ EI − mB ' '− 22
• 23. • Substitute ∆´B and fBB in Eq. (1): + ↑: − 0.005 = − ∆ B = −0.005 m = ∆ ' B + f BB RB 117.33 10.67 RB + EI EI 117.33 ,↓ EI 10.67 = mB ' ' = ,↑ EI ∆'B = M ' 'B = − (−0.005) EI = −117.33 + 10.67 RB f BB (−0.005)(200 × 60) = −117.33 + 10.67 RB RB = 5.37 kN, + 16 kN 16 kN A 1 kN 0.5 kN 0.5 kN = 4 kN 12 kN xRB = 5.37 B RA = 9.31 kN 5.37 kN C RC = 1.32 kN 23
• 24. • Quantitative shear and bending diagram and qualitative deflected curve 16 kN B A C RA = 9.31 kN 5.37 kN 2m 2m ∆B = 5 mm RC = 1.32 kN 4m 9.31 V diagram -6.69 M diagram Deflected Curve -1.32 18.62 5.24 ∆B = 5 mm 24
• 25. Or use Castigliano least work method • Principle of superposition 16 kN RB RA = 12 - 0.5RB 16 kN 12 0.5RB 4m = 2m + 2m RC = 4 - 0.5RB RB 4 0.5RB 25
• 26. 16 kN x1 x2 x3 RB RA = 12 - 0.5RB 2m x1 2m RC = 4 - 0.5RB 4m M1 = (12 - 0.5RB)x1 V1 x3 (4 - 0.5RB)x3 = M3 M2 = 0.5x2RB + 16 - 2RB + 4 x2 12 - 0.5RB V2 L ∆ B = −0.005 = ∫ ( 0 ∂M i M i ) dx ∂RB EI RB x2 2 V3 4 - 0.5RB 4 - 0.5RB 4m 2 1 1 − 0.005 = ( −0.5 x1 )(12 x1 − 0.5 x1 RB )dx1 + (0.5 x2 − 2)(0.5 x2 RB + 16 − 2 RB + 4 x2 )dx2 EI ∫ EI ∫ 0 0 4 1 + ∫ (−0.5x3 )(4 x3 − 0.5x3 RB )dx3 EI 0 − 0.005 EI = −117.34 + 10.66 RB , RB = 5.38 kN, 26
• 27. Example 9-3 Draw the quantitative Shear and moment diagram and the qualitative deflected curve for the beam shown below.EI is constant. Neglect the effects of axial load. 5 kN/m A B 4m 4m 27
• 28. Use compatibility of displacement to obtain reaction • Principle of superposition A 5 kN/m θA=0 4m 5 kN/m 4m θB=0 θ´B + θ´A 1 kN•m αBA + αAA αAB Compatibility equation : B = SOLUTION ×MA 1 kN•m × M B αBB θ A = 0 = θ ' A +α AA M A + α AB M B -----(1) θ B = 0 = θ 'B +α BA M A + α BB M B -----(2) 28
• 29. • Use formulation: θ´A, θ´B, αAA, αBA, αBB, αAB, 5 kN/m θ´A θ´B 3wL3 3(5)(8) 3 60 θ 'A = = = EI 128 EI 128EI 7 wL3 7(5)(8) 3 46.67 θ 'B = = = EI 384 EI 384 EI 1 kN•m αAA αBA 1 kN•m αAB αBB α AA = M o L 1(8) 2.67 = = 3EI 3EI EI α BB = M o L 1(8) 2.67 = = 3EI 3EI EI α BA = M o L 1(8) 1.33 = = 6 EI 6 EI EI α AB = M o L 1(8) 1.33 = = 6 EI 6 EI EI Note : Maxwell’s theorem of reciprocal displacement is αAB = αBA 29
• 30. • Substitute θ´A, θ´B, αAA, αBA, αBB, αAB, in Eq. (1) and (2) 5 kN/m θA=0 4m 4m B θB=0 α AA α AB = = A θ A = 0 = θ ' A +α AA M A + α AB M B -----(1) θ B = 0 = θ 'B +α BA M A + α BB M B -----(2) + 0= 60 2.67 1.33 +( )M A + ( )M B EI EI EI 0= + 60 EI 2.67 = EI θ 'A = θ 'B = 1.33 EI 46.67 EI 1.33 EI 2.67 = EI α BA = 46.67 1.33 2.67 +( )M A + ( )M B EI EI EI α BB Solving these equations simultaneously, we haave MA = -18.31 kN•m, + MB = -8.36 kN•m, + 30
• 31. MA = -18.31 kN•m, + MB = -8.36 kN•m, + 5 kN/m 18.31 kN•m A RA + ΣMA = 0: + ΣFy = 0: B 4m 4m 8.36 kN•m RB 18.31 − 20( 2) + RB (8) − 8.36 = 0, 3.76 + R A − 20 + RB = 0, RB = 3.76 kN, Ra = 16.24 kN, 31
• 32. • Quantitative shear and bending diagram and qualitative deflected curve 5 kN/m 18.31 kN•m A B 4m 16.24 kN V diagram 4m 8.36 kN•m 3.76 kN 16.24 3.25 m M diagram 8.08 -3.76 6.67 -8.36 -18.31 Deflected Curve 32
• 33. Force Method of Analysis : Frames • Principle of superposition fCC × C x B ∆´CH C Cx Cy w = w 1 kN + Ax A Ay Compatibility equation : ∆ CH = 0 = ∆'CH + f CC C x 33
• 34. Example 9-4 Draw the quantitative Shear and moment diagram and the qualitative deflected curve for the Frame shown below.EI is constant. 2 kN/m B 6m C 6m A 34
• 35. SOLUTION Use compatibility of displacement to obtain reaction • Principle of superposition fCC × C x B 6m ∆´CH C 2 kN/m Cy = 6m 2 kN/m Cx 1 kN + Ax A Ay Compatibility equation : ∆ CH = 0 = ∆'CH + f CC C x -----(1) 35
• 36. 6m 3m Cy P 3m P V´2 6+P 12 kN -12 - P V´1 2x1 -6 - P Ay x2 6+P M´1 = (12 + P)x1- x12 x1 Ax A M´2 = (6 + P)x2 x2 2 kN/m 2 kN/m • Use Castigliano’s method for ∆´CH ∆´CH 6m B C x1 -12 - P -6 - P 0 0 6 6 1 1 2 ∂M 'i M 'i ( x1 )(12 x1 + x1 P − x1 )dx1 + ) = ∫( dx = ∫ ∫ ( x2 )(6 x2 + x2 P)dx2 EI 0 EI 0 ∂P EI 0 L ∆ 'CH 6 6 1 1 2 3 2 = (12 x1 − x1 ) dx1 + (6 x2 ) dx2 EI ∫ EI ∫ 0 0 3 4 3 1 12 x1 x1 6 1 6 x2 = ( − )0+ ( ) EI EI 3 3 4 6 0 = 972 ,→ EI 36
• 37. • Use Castigliano’s method for fCC fCC 6m B C m´2 = x2P x2 P 1 kN P v´2 Cy P 6m x2 P m´1 = x1P x1 -P Ax A v´1 Ay x1 -P -P 1 1 6 6 1 1 ∂m' m' = ∫ ( i ) i dx = ∫ ( x1 )( x1P)dx1 + EI ∫ ( x2 )( x2 P)dx2 EI 0 ∂P EI 0 0 -P L f CC 3 3 1 x1 6 1 x2 = ( ) + ( ) EI 3 0 EI 3 6 0 = 144 ,→ EI 37
• 38. • Substitute ∆´CH and fCC in Eq. (1) ∆ CH = 0 = ∆'CH + f CC C x +: 0= -----(1) 972 ,→ EI 144 f cc = ,→ EI ∆ CH = 972 144 + Cx EI EI Cx = -6.75 kN, 6 kN 1 kN + × C x = −6.75kN 12 kN = C 6.75 kN B 2 kN/m 2 kN/m 1 kN 0.75 kN 5.25 kN 1 kN A 6 kN 1 kN 0.75 kN 38
• 39. Or use Castigliano least work method: x2 B 6m M2 = (6-Cx)x2 C Cx 2 kN/m Cx A x2 V2 6 - Cx 6 - Cx 6m M1 = (12 - Cx)x1- x12 x1 12 - Cx 6 - Cx V1 2x1 x1 12 - Cx L ∂M i M i ∂U i ) dx = ∆ CH = 0 = ∫( ∂C x 0 ∂C x EI 6 0= 6 - Cx 6 1 1 2 (− x1 )(12 x1 − C x x1 − x1 )dx1 + (− x2 )(6 x2 − C x x2 )dx2 EI ∫ EI ∫ 0 0 3 3 4 3 12 x1 C x x1 x1 6 6x C x 0 = (− + + ) 0 + (− 2 + x 2 3 3 4 3 3 0 = -972 + 144Cx , Cx = 6.75 kN, 3 6 0 39
• 40. • Quantitative shear and bending diagram and qualitative deflected curve B 6m C B 2 kN/m 6.75 kN 0.75 kN - 0.75 - 0.75 - 6.75 6m V, (kN) 5.25 5.25 kN A C 2.63 m A 0.75 kN B B C C -4.5 1.33 m -4.5 Deflected curve A M, (kN•m) 6.90 A 40
• 41. Force Method of Analysis : Truss (Externally indeterminate) E A Ax D Cx C B ∆'CH + f CC C x = ∆ CH = 0 Cy Ay P = E E D D + A C A B C P 1 x Cx B ∆´C fCC 41
• 42. Truss (Internally indeterminate) P 3 D C 6 1 2 5 A B 4 = ∆'6 + f 66 F6 = ∆ 6 = 0 P D ∆´6 A C B D + f66 C 1 A xF6 B 42
• 43. Example 9-5 Determine the reaction at support A, C, E and all the member forces. Take = 200 GPa and A = 500. mm2 . E E 40 kN D 4m A 5m B C 5m 43
• 44. SOLUTION Use compatibility of displacement to obtain reaction • Principle of superposition RE E 40 kN D 4 m Ay A C 5m RC × Cy fCC + 40 kN 5m B = Ax ∆´C 1 kN Compatibility equation : ∆ C = 0 = ∆'C + f CC RC -----(1) 44
• 45. • Use unit load method for ∆´C and fcc 5.39 m 53.85 20 E 50 40 kN +53 .85 A B 5m 53.85 2.5 0 2.5 A (53.85)(-2.69)(5.38) (200x106)(500x10-6) =- 7.81 mm, C ∆´C 0 5m fCC = N ´i (kN) 1 E AiEi 0 0 0 ∆´C = = D 4 m 20 kN 85 -53. 50 5.39 m Σ n´iN´iLi = -2.6 9 D 0 +2.5 0 B n´i (kN) -2.6 9 +2.5 fCC C 1 kN + Σ ni´n´iLi AiEi 2(-2.69)2(5.385) (200x106)(500x10-6) 2(2.5)2(5) (200x106)(500x10-6) = 1.41 mm, 45
• 46. • Substitute ∆´Cv and fCC in Eq. (1): ∆ C = 0 = ∆'C + f CC RC ∆ 'C = 7.81 mm, ↓ + ↑: −7.81 + 1.41RC = 0 f CC = 1.41 mm, ↑ RC = 5.54 kN , ↑ 20 50 +53 .85 A 40 kN 36.15 kN 36.15 kN -2.6 9 D 0 0 C B 0 N ´i (kN) 38.93 kN 0 0 0 0 E 2.5 D 20 kN 5 53.8 50 53.85 + E 1 2.5 A +2.5 14.46 kN E +38 .95 20 kN 53.85 A +13.85 B n´i (kN) = 53.85 xRC = 5.54 kN -2.6 9 C +2.5 1 kN 40 kN D -1 4.90 0 21.8o B +13.85 N i (kN) C 5.54 kN 46
• 47. Or use Castigliano least work method: 5.39 m -2.7 RC + 53.85 = RE 4m Ax = -2.5RC +50 = Ax 5.39 E -2.7 R m 40 kN C +5 3.85 D -2.7 RC .85 -53 0 21.8o A Ay = 20 2.5RC 5m B 2.5RC C RC 5m Castigliano’s Theorem of Least Work : ∆ CV = 0 = ∑ ( 0= ∂N i N i Li ) ∂RC AE 1 [(−2.7)(−2.7 RC + 53.85)(5.39) + ( −2.7)(−2.7 RC )(5.39) + 0 + 0 + 2[(2.5)(2.5 RC )(5)]] AE 0 = 39.3RC − 783.68 + 39.3RC + 62.5 RC RC = 5.55 kN, 47
• 48. Example 9-6 Determine the force in all member of the truss shown : (a) If the horizontal force P = 6 kN is applied at joint C. (b) If the turnbuckle on member AC is used to shorten the member by 1 mm. (c) If (a) and (b) are both accounted. Each bar has a cross-sectional area of 500 mm2 and E = 200 GPa. D C 2m A B 3m 48
• 49. SOLUTION Part (a) : If the horizontal force P = 6 kN is applied at joint C. • Principle of superposition D 6 kN 1 C 6 3 A 5 4 2m B 2 3m D C 6 kN = 3m D ∆´6 1 E´ E E´ A C 2m B Compatibility equation : Note : AE + E ' C = L + 1 E f66 A ∆´6 + f66 F6 = 0 ×F6 B ----------(1) 49
• 50. • Use unit load method for ∆´6 and f66 3m D C 6 kN +6 D ∆´6 +4 6 E E´ +6 -7. 2 A 1 0 B 1 N ´i (kN) 2 m -0.555 0 -14.98 -2 6. 0 3 -14.98 n´iN ´iLi (kN2•m) 0 0.616 6 3. -0.555 3.6 1 1 0.616 2.08 n´i2Li (kN2•m) 3.6 2 1 3 Li (m) ∆ '6 = ∑ 1 2 6 3. 0 0 2.08 3 B -0.832 n´i (kN) 0 -4.44 1 1 A 4 4 C -0.832 n'i N 'i Li Ai Ei = 1 [−4.44 − 26.03 − 2(14.98)] AE = − 60.43 AE 2 n' L 1 12.61 [2(0.616) + 2(2.08) + 2(3.61)] = f 66 = ∑ i i = Ai Ei AE AE 50
• 51. • Substitute ∆´6 and f66 in Eq. (1) − 3m +6 60.43 12.61 + ( F6 ) = 0 AE AE F6 = 4.80 kN, (T) D 6 kN ∆´6 +6 6 1 E E´ -7. 2 0 1 2m + -0.555 0 N ´i (kN) 4 4 -0.832 A n´i (kN) 0 0 .8 +4 2.4 1 +2 +1.34 A 4 Ni (kN) 1 -0.555 x F6 = 4.80 kN B 0 6 kN +2 D 6 1 = +4 C -0.832 C -2.66 B 4 51
• 52. Part (b) : If the turnbuckle on member AC is used to shorten the member by 1 mm. f 66 = 12.61 12.61 = = 1.26(10-4) m = 0.126 mm (500)(200) AE F6 = D C -0.832 1 -0.555 0 1 -0.832 A n´i (kN) 0 1 mm (1 kN ) = 7.94 kN 0.126 mm 1 -6.61 D -0.555 B 0 x F6 = 7.94 kN 4 .9 +7 7 .94 -6.61 -4.41 = 6 A 4 Ni (kN) C -4.41 B 4 52
• 53. Part (c) : If the horizontal force P = 6 kN is applied at joint C and the turnbuckle on member AC is used to shorten the member by 1 mm are both accounted. 4 (Ni)load (kN) B 0 A 4 0 1 -3.07 4 .7 2 4 -4.41 B 0 C 5.5 3 -4.61 A (Ni)short (kN) C 6 kN -4.61 D 6 4 .9 +7 7 .94 -6.61 -4.41 = A C -2.66 -6.61 D + 0 .8 +4 2.4 1 +2 +1.34 6 6 kN +2 D (Ni)total (kN) -7.07 B 4 53
• 54. Or use compatibility equation : ∆´6 + f66 F6 = ∆´6 = 0.001 − 60.43 12.61 + ( F6 ) = 0.001 AE AE F6 = +6 4 ∆´6 -7 . 21 0 + 0 N ´i (kN) 4 12 -3.06 6 -0.832 n´i (kN) 0 2 .7 C 5.5 1 -4.58 A 4 (Ni)total (kN) C 1 x F6 = 12.72 kN f66 -0.555 A -4.58 D -0.832 1 1 -0.555 = 6 D 6 kN +6 +4 0.001AE + 60.43 0.001(500)(200) + 60.43 = = 12.72 kN, (T) 12.61 12.61 B 0 6 kN -7.06 B 4 54
• 55. Composite Structures Example 9-7 Find all reaction and the tensile force in the steel support cable. Consider both bending and axial deformation. Steel cable Ac = 2(10-4) m2 Ec = 200(103) kN/m2 C 2m A 5 kN m2 Ab = 0.06 Ib = 5(10-4) m4 Eb = 9.65(103) kN/m2 B 6m 55
• 56. SOLUTION C m 6.32 x 5 kN 0.316T T M = 0.316Tx - 5x N = -0.949T 0.949T V 5 kN 2m 18.43o A RC = T B 6m Bx MB By By Castigliano’s Theorem of Least Work ; ∂ ∆C = 0 = (U ib + U in ) ∂T L L ∂M M ∂N N ∆C = 0 = ∫ ( dx + ∫ ( ) ) dx ∂T EI ∂T AE 0 0 6 6 1 1 1 0= (0.316 x)(0.316 xT − 5 x)dx + (−0.949)( −0.949T )dx + Eb I b ∫ Ab Eb ∫ Ac Ec 0 0 6 1 0.316 2 x 3 (0.316 × 5) x 3 6 1 1 2 0= [( T) − ]0+ (0.949 xT ) 0 + ( xT ) Eb I b 3 Ab Eb Ac Ec 3 0 = (1.49T - 23.58) + 9.33(10-3)T + 0.158T 6.32 ∫ (1)(T )dx 0 6.32 0 ; T = 14.23 kN, (tension) # 56
• 57. 4.5 kN C m 6.32 2m 18.43o A x 5 kN RC = T = 14.23 kN 13.5 kN Bx B 6m MB By + ΣF = 0: x Bx = Rc cos θ = 13.5 kN, + ΣFy = 0: By = 5 - Rc sin θ = 0.5 kN, + ΣMB = 0: MB = 13.5(2) - 5(6) = -3 kN•m, + 57