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  • 1. TRUSSES ANALYSIS ! ! ! ! ! ! ! Fundamentals of the Stiffness Method Member Local Stiffness Matrix Displacement and Force Transformation Matrices Member Global Stiffness Matrix Application of the Stiffness Method for Truss Analysis Trusses Having Inclined Supports, Thermal Changes and Fabrication Errors Space-Truss Analysis 1
  • 2. 2-Dimension Trusses 2
  • 3. Fundamentals of the Stiffness Method y x 4 3 2 (x2, y2) 1 1 2 3 (x3, y3) 1 (x1, y1) • Node and Member Identification 3 6 5 2 8 4 7 (x4, y4) • Global and Member Coordinates • Degrees of Freedom •Known degrees of freedom D3, D4, D5, D6, D7 and D8 • Unknown degrees of freedom D1 and D2 3
  • 4. Member Local Stiffness Matrix x´ y´ q´j j q´i q 'i = i q' j = − x´ y´ d´ i =1 AE/L x di i d´ AE/L d´ j = y´ AE/L AE AE d 'i − d'j L L 1 AE/L AE AE d 'i + d'j L L q 'i  AE  1 − 1  d 'i  q '  = − 1 1  d '  L    j  j [q´] = [k´][d´] ----------(1) x´ x dj j d´ AE  1 − 1 [k ' ] = L − 1 1    4
  • 5. Displacement and Force Transformation Matrices y x´ y´ θy m θx j (xj,yj) x i (xi,yi) λ x = cosθ x = λ y = cosθ y = x j − xi L y j − yi L = = x j − xi ( x j − xi ) 2 + ( y j − yi ) 2 y j − yi ( x j − xi ) 2 + ( y j − yi ) 2 5
  • 6. • Displacement Transformation Matrices y djy x´ d´j y´ j m diy i d´i i Local θy m θx j djx x dix Global λx λy d 'i = d ix cos θ x + d iy cos θ y d ' j = d jx cos θ x + d jy cos θ y  d 'i   λ x d '  =  0  j  λy 0 0 λx [d´] = [T][d]  d ix  0   d iy     d  λ y  jx   d jy   ----------(2) λ x [T ] =  0 λy 0 0 λx 0  λy   6
  • 7. • Force Transformation Matrices qjy y qiy θy m θx i qix j y qjx θy qiy = q 'i cos θ y q jx = q ' j cos θ x q jy = q ' j cos θ y j θx q´i x i Local λx λy m x Global qix = q 'i cos θ x x´ q´j y´  q ix   λ x q    iy  =  λ y  q jx   0     q jy   0    0  0   q 'i   λ x  q' j     λy   [q] = [T]T[q´] where λ x λ y T [T ] =  0  0  0  0   λx   λy   ----------(3) 7
  • 8. Member Global Stiffness Matrix ----------(3) [q] = [T]T[q´] Substitute ( [q´] = [k´][d´] + [q´F] ) into Eq. 3, yields the result, [ q ] = [ T ]T ([k´][d´] + [q´F] ) = [ T ]T [ k´ ][T][d] + [ T ]T [q´F] = [k][d] + [qF] [ k ] = [ T ]T[ k´ ][T] [k] [qF] = [ T ]T [q´F] [ k] = λx λy 0 0 0 0 λx λy 1 -1 λx λy 0 0 -1 AE L 1 0 0 λx λy U V U V λxλx λxλy −λxλx −λxλy AE V λyλx L U −λ λ x x λyλy −λyλx −λyλy −λxλy λxλx λxλy V −λyλx −λyλy λyλx λyλy U [k] = 8
  • 9. Application of the Stiffness Method for Truss Analysis Equilibrium Equation: [Qa] = [K][D] + [QF] Partitioned Form: Joint Load Qk = Qu Unknown Displacement K11 K12 Du K21 K22 Dk Reaction + QFk QFu Boundary Condition [Qk] = [K11][Du] + [K12][Dk] + [QF] [Du] = [Ku] -1 (([Qk] - [QF]) - [K12][Dk]) 9
  • 10. Member Forces x´ q´j y´ y θy m j θx q´i x i q 'i  AE  1 − 1  d 'i   q ' F i  q '  = − 1 1  d '  +  F  L    j  q ' j   j λ x 0  q 'i  AE  1 − 1 λ x q '  = − 1 1   0 L     j λy 0 0 λx λy 0 0 λx  d ix  0   d iy     d  λ y  jx   d jy    Dix  0   Diy   q 'F i    D  +  F  λ y  jx q' j     D jy  10
  • 11. q 'i  AE  λ x q '  =  L  − λx  j λy − λy − λx λx  Dix  − λ y   Diy   q ' F i    D  +  F  λ y  jx q' j     D jy  Dxi q´j = AE L −λx −λy x´ q´j y´ y θy m θx q´i λx λy Dyi Dxj + qj´ F Dyj j x i Member Forces 11
  • 12. Member Forces Dxi qm = AE L −λx −λy λx λy x´ qm y´ y θy m θx Dyi Dxj + qj´F Dyj j x i Member Forces 12
  • 13. Example 1 For the truss shown, use the stiffness method to: (a) Determine the deflections of the loaded joint. (b) Determine the end forces of each member and reactions at supports. Assume EA to be the same for each member. 5m 3m 50 kN 3m 5m 5m 80 kN 4m 4m 13
  • 14. 6 5m 3m (-4,3) 50 kN 3m 5 3 5m 5m 1 80 kN 4m 2 2 (-4,-3) 4m 4 3 1 (0,0) 1 3 8 4 2 ˆ ˆ ˆ = ( x j − xi )i + ( y j − yi ) j λij L L Member λx 7 (4,-3) λy Ui Ui [ k ]m = Vi cosθy = λy Uj Vj λxλx λxλy −λxλx −λxλy #1 -4/5 = -0.8 -3/5 = -0.6 #2 cosθx = λx -4/5 = -0.8 3/5 = 0.6 #3 4/5 = 0.8 -3/5 = -0.6 AE Vi λyλx λyλy −λyλx −λyλy L U −λ λ −λ λ λ λ λ λ j x x x y x x x y Vj −λyλx −λyλy λyλx λyλy 14
  • 15. 6 Member λx 5 3 2 2 1 2 3 0.64 0.48 0.36 - 0.8 0.6 0.64 -0.48 0.36 0.8 0.64 -0.48 0.36 1 0.64 0.48 -0.64 -0.48 AE 2 5 3 0.48 0.36 -0.48 -0.36 -0.64 -0.48 0.64 0.48 4 -0.48 -0.36 1 1 1 [ k ]2 = AE 2 -0.48 5 5 -0.64 6 0.48 0.36 7 6 5 0.36 0.48 -0.36 0.48 0.64 -0.48 0.48 -0.36 -0.48 0.36 8 1 1 0.36 0.48 -0.36 0.48 0.64 -0.48 0.48 -0.36 -0.48 2 0.64 -0.48 -0.64 0.48 0.64 -0.48 -0.64 0.48 AE 2 -0.48 [ k ]3 = 5 7 -0.64 8 2 -0.6 4 1 [ k ]1 = -0.8 -0.6 7 4 1 λy2 #3 8 3 2 λx λy #2 3 4 λx2 #1 1 1 λy 0.36 [K] = AE 5 2 2 1.92 -0.48 -0.48 1.08 15
  • 16. 5m 3m Global 5 2 50 kN 3m 6 5m 5m 1 1 3 4 80 kN 4m 2 7 3 4m 1 1 Q1 = -50 Q2 = -80 8 = AE 5 2 D1 D2 2 1.92 -0.48 D1 -0.48 1.08 D2 + 0 1 0 2 -250.65/AE = -481.77/AE 16
  • 17. 6 Local 5 3 2 2 1 1 [ q ' F ]m = 1 4 3 3 [ AE − λx L − λy 8 4 2 λx λy #1 -0.8 -0.6 #2 -0.8 0.8 -0.6 [ ] ] 7 [q´F]1 = AE 5 0.8 0.6 -0.8 -0.6 0.6 #3 λx  Dxi  D  yi λx   + q'F  Dxj     D yj    Member D4= 0.0 = -97.9 kN (C) [q´F]2 = AE 5 0.8 -0.6 -0.8 0.6 17.7 kN 50 kN 36.87o 97.9 kN 80 kN AE 5 -0.8 +0.6 = -17.7 kN (C) D1= -250.65/AE D2= -481.77/AE D5= 0.0 D6= 0.0 = +17.7 kN (T) 17.7 kN [q´ ] = F 3 D1= -250.65/AE D2= -481.77/AE D3= 0.0 +0.8 -0.6 D1= -250.65/AE D2= -481.77/AE D7= 0.0 D8= 0.0 17
  • 18. 6 5 3 2 2 1 17.7 kN 1 4 3 50 kN 1 3 36.89o 8 4 2 Member λx λy #1 -0.8 -0.6 #2 -0.8 0.6 #3 0.8 -0.6 7 Check : 17.7 kN 97.9 kN 80 kN + ΣF = 0: 17.7 + 17.7 +50cos 36.89 x´ - 97.9cos73.78 - 80cos53.11 = 0, O.K 17.7(0.6)=10.62 kN 17.7(0.8)=14.16 kN 50 kN 97.9(0.6)=58.74 kN 80 kN 97.9(0.8)=78.32 kN 17.7(0.6)=10.62 kN 17.7(0.8)=14.16 kN 18
  • 19. Example 2 For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the deflections of the loaded joint. The support B settles downward 2.5 mm. Temperature in member BD increase 20 oC. Take α = 12x10-6 /oC, AE = 8(103) kN. 8 kN D 4 kN A oC 3m 0 +2 B ∆B = 2.5 mm C 4m 19
  • 20. 8 kN 4 D 3 2 4 kN 1 1 (-4,0) A 2 oC 3m 0 +2 3 4m ∆B = 2.5 mm 1 (0,0) 3 6 C B 2 8 5 4 (-4,-3) 7 (0,-3) L + cosθy = λy Ui [ k ]m = Vi Uj λy -4/4 = -1 0 #2 L λx #1 ( y j − yi ) ˆ j cosθx = λx Ui Member -4/5 = -0.8 -3/5 = -0.6 #3 ˆ λij = ˆ ( x j − xi )i 0 -3/3 = -1 Vj λxλx λxλy −λxλx −λxλy AE Vi λyλx λyλy −λyλx −λyλy L U −λ λ −λ λ λ λ λ λ j x x x y x x x y Vj −λyλx −λyλy λyλx λyλy 20
  • 21. 4 3 2 2 1 1 (-4,0) 2 4 (-4,-3) 2 3 0.25 0 -0.25 -1 0 0.25 0 0 0.128 0.096 0.072 0 0 0.333 2 0 0 0 0 - 0.8 -0.6 0 0 3 -0.25 0 0.25 0 0 1 2 7 [k]2 = 8x103 5 6 2 0.096 0.072 -0.096 -0.072 5 -0.128 -0.096 0.128 0.096 6 -0.096 -0.072 0.096 0.072 8 1 0 0 2 0 0.333 7 0 0 0 0 8 0 -0.333 0 0.333 0 2 1 0.128 0.096 -0.128 -0.096 0 0 -1 1 0 4 [k]3 = λy2/L 4 1 8x103 λx λy/L #3 7 (0,-3) 1 [k]1 = 8x103 λx2/L #2 8 5 λy #1 3 6 3 Member λx 1 (0,0) 1 0 0 -0.333 [K] = 8x103 2 1 0.378 0.096 2 0.096 0.405 21
  • 22. Member 2: [q] = [k]m[d] + [qF] 4 2 1 3 1 2 1 (0,0) (-4,0) 2 3 oC 20 + 6 (-4,-3) 1 q5 q1 q2 q1 q2 7 1.92 kΝ =α(∆T1)AE -6 3 (0,-3) = (12x10 )(20)(8x10 ) 2 5 = 8x103 = 8x103 1.152 kN 6 0.128 0.096 -0.128 -0.096 2 0.096 0.072 -0.096 -0.072 d2 5 -0.128 -0.096 0.128 0.096 -0.096 -0.072 0.096 0.072 1 2 -1.536 1 d1 6 q6 +20oC 1.536 kN 1 = 8x103 1.536 kN 2 +20oC 3 4 ∆B = 2.5 mm q2 2 8 5 q1 1.152 kN 1.92 kN 1 0.128 0.096 d1 2 0.096 0.072 d2 1 0.128 0.096 d1 2 0.096 0.072 d2 + 8x103 d5 = 0 + d6 = -2.5x10-3 5 6 -0.128 -0.096 0 -1.152 2 1.536 5 1.152 6 5 -3 -0.096 -0.072 -2.5x10 6 + 1.92 1.44 + + -1.536 1 -1.152 2 -1.536 1 -1.152222
  • 23. 8 kN D 4 4 kN 3 2 A 1 (-4,0) 2 oC 3m 0 +2 ∆B = 2.5 mm 1 3 4m 1 (0,0) 3 6 C B 2 8 5 4 (-4,-3) 7 (0,-3) [Q] = [K][D] + [QF] Global: 1 Q1 = -4 Q2 = -8 D1 D2 = = 8x103 2 1 0.378 0.096 D1 2 0.096 0.405 D2 -0.8514x10-3 1.44 + -1.536 -1.152 m -2.356x10-3 + 1.92 m 23
  • 24. 4 3 2 1 2 1 1 2 [ q ' F ]m = 3 6 [ AE − λx L − λy λx  Dxi  D  yi λx   + q'F  Dxj     D yj    8 5 3 Local 4 λx λy #1 -1 0 #2 - 0.8 0 -1 [q´F]1 = 8x103 1.0 0.0 -1.0 0.0 4 -0.6 #3 1.92 kN = -1.70 kN (C) 1.92 kN D2= -2.356x10-3 D3= 0.0 D4= 0.0 D1= -0.8514x10-3 [q´F]2 = 8x103 0.8 0.6 -0.8 -0.6 5 = -2.87 kN (C) 2 +20oC [ ] D1= -0.8514x10-3 7 Member ] [q´F]3 = 8x103 0.0 1.0 0.0 -1.0 3 = -6.28 kN (C) D2= -2.356x10-3 + -1.92 D5= 0.0 D6= -0.0025 D1= -0.8514x10-3 D2= -2.356x10-3 D7= 0.0 D8= 0.0 24
  • 25. 4 3 2 1 2 1 2 6 5 3 8 4 7 cosθx cosθy [q´]m -1 0 -1.70 #2 3 Member #1 1 - 0.8 -0.6 -2.87 #3 0 -1 -6.28 8 kN 8 kN 1.70 kN 2.87 kN 4 kN 1 4 kN 1.70 kN 2 3 2.87(0.8) = 2.30 kN 6.28 kN 2.87(0.6) = 1.72 kN 6.28 kN 25
  • 26. Example 3 For the truss shown, use the stiffness method to: (a) Determine the end forces of each member and reactions at supports. (b) Determine the displacement of the loaded joint. Take AE = 8(103) kN. 8 kN 4 kN A ∆ AD =+ ∆ = - 4 mm 3m mm 3 D C B 4m 4m 26
  • 27. 8 kN 4 kN A ∆ AD =+ D mm 3 1 ∆ = - 4 mm 3m 2 4 C B 4m ˆ ˆ ˆ = ( x j − xi )i + ( y j − yi ) j λij L L [ k ]m = 3 4 5 3 5 4 8 (0,-3) 7 (4,-3) Vi Member λx λy #1 -4/5 =-0.8 -3/5 = -0.6 cosθy = λy cosθx = λx Ui 3 6 (-4,-3) 4m Ui 2 1 2 (0,0) 1 #2 0 -3/3 = -1 Vj #3 4/5 = 0.8 -3/5 = -0.6 #4 4/4 = 1 0 #5 4/4 = 1 0 Uj λxλx λxλy −λxλx −λxλy AE Vi λyλx λyλy −λyλx −λyλy L U −λ λ −λ λ λ λ λ λ j x x x y x x x y Vj −λyλx −λyλy λyλx λyλy 27
  • 28. 2 1 2 5m 4 2 3 (-4,-3) 3m 1 4 3 5 4m (0,-3) 4 m 2 4 8 0 -1 0.8 -0.6 λx2/L λxλy/L λy2/L 0.128 0.096 0 0 0.072 0.333 0.128 -0.096 0.072 7 (4,-3) 4 3 0.128 0.096 -0.128 -0.096 2 0.096 0.072 -0.096 -0.072 3 -0.128 -0.096 0.128 0.096 4 -0.096 -0.072 0.096 0.072 [k]3 = 8x103 6 1 2 5 1 [k]2 = 8x103 -0.6 #3 1 [k]1 = 8x103 -0.8 #2 5 1 λy #1 5 m3 6 λx Member (0,0) 1 0 0 0 2 0 0.333 0 -0.333 5 0 0 0 0 -0.333 0 0.333 2 7 8 1 0.128 -0.096 -0.128 0.096 2 -0.096 0.072 0.096 -0.072 7 -0.128 0.096 0.128 -0.096 8 0.096 -0.072 -0.096 0.072 0 6 1 0 28
  • 29. 2 1 2 5m 4 2 3 (-4,-3) 3m 1 4 3 5 4m λx2/L λxλy/L λy2/L 1 0 0.25 0 0 #5 5 1 0 0.25 0 0 4 (0,-3) 4 m 5 6 7 5 0.25 0 -0.25 0 6 0 0 0 0 7 -0.25 0 0.25 0 8 0 0 0 0 8 7 (4,-3) 6 3 4 5 3 0.25 0 -0.25 0 4 0 0 0 0 5 -0.25 0 0.25 0 6 [k]4= 8x103 λy #4 5 m3 6 λx Member (0,0) 1 0 0 0 0 1 Global Stiffness Matrix [k]5= 8x103 2 5 8 7 1 [K] = 8x103 2 5 7 29
  • 30. Global Stiffness Matrix 1 2 1 2 0.096 0.072 -0.096 -0.072 3 -0.128 -0.096 0.128 0.096 5 1 0 0 0 2 0 0.333 0 -0.333 5 0 0 0 0 6 0 -0.333 0 1 [k]3 = 8x103 [k]4= 8x103 0.333 2 0 -0.25 0 4 0 0 0 0 5 -0.25 0 0.25 0 0 0 0 0 [k]5= 8x103 7 0.25 0 -0.25 0 6 0 0 0 0 7 -0.25 0 0.25 0 8 0 0 0 0 8 0.128 -0.096 -0.128 0.096 7 -0.128 0.096 0.128 6 5 0 7 8 5 2 -0.096 0.072 0.096 -0.072 8 0.25 6 2 1 5 6 -0.096 -0.072 0.096 0.072 1 [k]2 = 8x103 4 3 0.128 0.096 -0.128 -0.096 4 [k]1 = 8x103 6 3 4 3 [K] = 8x103 -0.096 0.096 -0.072 -0.096 0.072 1 1 2 5 2 0.256 0.0 5 7 0.0 -0.128 0.0 0.477 0.0 0.096 0.0 0.0 0.50 -0.25 7 -0.128 0.096 -0.25 0.378 30
  • 31. 8 kN 4 kN A ∆ AD =+ 1 1 4 ∆ AD 4.8 kN =+ 4 mm 3 ∆AE/L = 10.67 kN ∆AE/L = 4.8 kN 3.84 kN ∆ = -4 mm 3.84 kN 3 5 5 8 7 4m 2.88 kN 1 2 6 3 C B 4m 2 D ∆ = - 4 mm 3m mm 3 Global Fixed end forces 2 Fixed End 10.67 kN -3.84 1 -2.88 + 10.67 = 7.79 2 5 0.0 7 0.0 2.88 kN 31
  • 32. 8 kN 4 kN A ∆ AD =+ D 1 ∆ = -4 mm 3m mm 3 2 4 4m 3 6 3 C B Global: 2 1 5 4 5 8 7 4m [Q] = [K][D] + [QF] 7 1 Q1 = 4 Q2 = -8 Q5 = 0 1 5 0.256 0.0 0.0 -0.128 = 8x103 0.0 0.477 0.0 0.0 0.50 0.0 0.096 -0.25 D2 D5 0.378 D7 6.4426x10-3 -5.1902x10-3 2.6144x10-3 5.2288x10-3 m m m m -3.84 D1 7 -0.128 0.096 -0.25 = Q7 = 0 D1 D2 D5 D7 2 5 2 + 7.79 0.0 0.0 32
  • 33. Member forces 2  Dxi  D  yi λx   + q'F  Dxj     D yj    2 1 4 5 4 8 5 7 6.4426x10-3 D1 D2 D5 D7 = ∆ AD 4.8 kN =+ m m m m -5.1902x10-3 2.6144x10-3 5.2288x10-3 4.8 kN 1 mm 3 − λy λix -0.8 -0.6 #2 0 -1 10.67 kN ] D1 [q´F]1 = 8x103 0.8 0.6 -0.8 -0.6 5 = -1.54 kN (C) 10.67 kN 2 [ ] λiy λx 3 6 3 [ Member [ q ' F ]m = AE − λx L #1 1 D2 0 + -4.8 0 D1 [q´F]2 = 8x103 0.0 1.0 0.0 -1.0 3 = -3.17 kN (C) D2 D5 + 10.67 0 33
  • 34. 2 1 [ q ' F ]m = 2 1 4 3 6 3 D1 D2 D5 D7 5 4 = 7 m m m m Member λx λy #3 #4 0.8 1 -0.6 0 #5 1 0 − λy λx [ ] ] 8 5 6.4426x10-3 -5.1902x10-3 2.6144x10-3 5.2288x10-3 [ AE − λx L  Dxi  D  yi λx   + q'F  Dxj     D yj    D1 [q´F]3 = 8x103 -0.8 0.6 0.8 -0.6 5 = -6.54 kN (C) D2 D7 0 0 [q´F]4 = 8x103 -1.0 0.0 1.0 0.0 4 = 5.23 kN (T) 0 D5 0 D5 [q´F]5 = 8x103 -1.0 0.0 1.0 0.0 4 = 5.23 kN (T) 0 D7 0 34
  • 35. Member λx λy [q´] #1 -0.8 -0.6 -1.54 #2 2 0 -1 -3.17 #3 #4 0.8 1 -0.6 0 -6.54 5.23 #5 1 0 5.23 1 2 1 4 3 6 3 5 4 4 kN 1.54 kN 1.54 kN 5.23 kN 8 5 7 8 kN 4 kN 3.17 kN 3.17 kN 8 kN 6.54 kN 4 kN 6.54 kN 5.23 kN 0.92 kN 3.17 kN 3.92 kN 35
  • 36. Special Trusses (Inclined roller supports) 36
  • 37. Transformation Matrices 2 1 [ q* ] = [ T ]T[ q´ ] 3 1 4 6 4* 5 2 8 5 7 3* y y* i θi 3* 2 j 1 4* λjx = cos θj 1 θj q1 = q2 λix λiy 0 0 λjx λjy 0 0 λjy = sin θj x [ T ] = [[ λix = cos θi T ]T]T = q´i q´j [T]T λix λiy 0 0 0 0 λjx λjy λiy = sin θi x* q´j 1 q´i q*3 q*4 i j 37
  • 38. [ k ] = [ T ]T[ k´ ][T] [ k ]m = λix λiy 0 0 0 0 λjx λjy Ui Ui [ k ]m = λixλix 1 -1 λix λiy 0 0 -1 AE L 1 0 0 λjx λjy Vi Uj Vj λixλiy −λixλjx −λixλjy AE Vi λiyλix λiyλiy −λiyλjx −λiyλjy L U −λ λ −λ λ λjxλjx λjxλjy j jx ix jx iy Vj −λjyλix −λjyλiy λjyλjx λjyλjy 38
  • 39. Example 5 For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the displacement of the loaded joint. AE is constant. 30 kN 3m 45o 4m 39
  • 40. 2 Member 1: 1 3m 3 4* 6 5 θi = 0, 5 λix = cos 0 = 1, λiy = sin 0 = 0 2 1 3* 45o 4m 4* 6 q´i 3* o 45 o o [q*] θij = -45 = 135 , λix = cos (-45o) = 0.707, λiy = sin(- 45o) = -0.707 1 q´j 1 i j [q*] = [T*]T[q´] + [T*]T[q´F] q5 q6 q3* q4* = 1 0 0 0 0 0 0.707 -0.707 q´i q´j [T*]T 40
  • 41. 2 Member 1: 1 3m 3 4* 6 1 5 θi = 0 ; 5 λix = cos 0 = 1, λiy = sin 0 = 0 2 3* 45o 4m [k*]1 = = q´j 1 i 0 0 0.707 -0.707 5 [k*]1 3* o 45 o o [q*] θij = -45 = 135 , λix = cos (-45o) = 0.707, λiy = sin(- 45o) = -0.707 1 q´i AE  1 − 1 [k *] = [T T ] − 1 1  [T ] L   1 0 0 0 4* 6 5 0.25 6 0 AE 3* -0.1768 4* 0.1768 AE 4 1 -1 -1 1 j 1 0 0 0 0 0 0.707 -0.707 6 3* 4* 0 0 0 0 -0.1768 0 0.125 -0.125 0.1768 0 -0.125 0.125 41
  • 42. 2 Member 2: 2 1 3m 3 4* 6 [k *] = [T T ] 3* 4* 45o 4m AE  1 − 1 − 1 1 [T ] L   [k*]2 = 0 -1 0 0 0 0 -0.707 -0.707 AE 3 3* j 90o+45o θj = -135o = 215o , =135o λix = cos (-135o) = -0.707, λiy = sin(- 135o) = -0.707 1 -1 -1 1 0 -1 0 0 1 [k*]2 = 1 2 AE 3* 4* i 2 5 o o 1 θi = -90 = 270 , 90o λix = cos(-90o) = 0, 2 λiy = sin(-90o) = -1 2 1 q´i 2 3* 0 0.3333 -0.2357 -0.2357 0 -0.2357 0.1667 0.1667 0 -0.2357 0.1667 0.1667 0 -0.707 -0.707 4* 0 0 0 0 0 q´j 42
  • 43. 2 2 Member 3: 1 3m 3 4* 6 [k ] = [T T ] 6 3* 45o 4m AE  1 − 1 − 1 1  [T ] L   [k]3 = = q´j 3 36.87o i 5 q´i θi = θj = 36.87o ; λix = λjx = cos (36.87o) = 0.8, λiy = λjy = sin(36.87o) = 0.6 0 0 0.8 0.6 AE 5 5 [k]3 0.8 0.6 0 0 j 3 2 1 5 1 36.87o 6 1 2 0.096 0.072 -0.096 -0.072 -0.128 -0.096 0.128 0.096 -0.096 -0.072 0.096 0.072 5 0.128 6 0.096 AE 1 -0.128 2 -0.096 1 -1 -1 1 0.8 0.6 0 0 0 0 0.8 0.6 43
  • 44. 2 Global Stiffness: 1 3 3m 4* 6 2 3* 1 5 5 45o 4m [k*]1 = 5 0.25 6 0 AE 3* -0.1768 4* 0.1768 1 [k*]2 = 1 2 AE 3* 4* 2 3* 0 0 0 0 0 0.3333 -0.2357 -0.2357 0 -0.2357 0.1667 0.1667 0 -0.2357 0.1667 0.1667 = 3* 4* 0 0 0 0 -0.1768 0 0.125 -0.125 0.1768 0 -0.125 0.125 4* 5 [k]3 6 5 0.128 6 0.096 AE 1 -0.128 2 -0.096 6 1 2 0.096 0.072 -0.096 -0.072 -0.128 -0.096 0.128 0.096 -0.096 -0.072 0.096 0.072 1 [K] = 2 3* 1 0.128 0.096 0 2 0.096 0.4053 -0.2357 AE 3* 0 -0.2357 0.2917 44
  • 45. 2 Global : 1 30 kN 3m 3m 3 4* 6 1 5 45o 4m 2 3* 45o 4m [Q] = [K][D] + [QF] 1 Q1 = 30 Q2 = 0 Q3*= 0 D1 D2 D3* = = 2 3* 1 0.128 0.096 0 AE 2 0.096 0.4053 -0.2357 3* 0 -0.2357 0.2917 1 AE D1 D2 D3* 352.5 -157.5 -127.3 45
  • 46. 2 Member Forces :  Dxi  D  yi λx   + q'F  Dxj     D yj    1 3m 3 4* 6 1 5 [ q ' F ]m = 2 = − λy [ ] ] λx 3* 0 45o 4m D1 D2 D3* [ AE − λx L 1 AE [q´F]1 = AE -1 0 0.707 -0.707 4 = -22.50 kN, (C) 352.5 -157.5 -127.3 λiy #1 1 0 #2 0 -1 -0.707 -0.707 -0.707 0.8 0.6 0.8 λjy [q´F]2 = AE 0 1 -0.707 3 = -22.50 kN, (C) 0.707 -0.707 #3 λjx 0 D1 λix Member 0 D3* 0.6 D2 D3* 0 0 [q´F]3 = AE -0.8 -0.6 5 = 37.50 kN, (T) 0.8 0.6 0 D1 D2 46
  • 47. 2 Reactions : 1 30 kN 3m 3 3m 4* 6 1 5 45o 4m Member [q´]1 Member Force (kN) -22.50 [q´]2 3* 45o [q´]3 -22.50 4m 2 37.50 22.50 kN 37.50 kN 45o 36.87o 22.50 kN 7.50 kN 45o 31.82 kN 22.50 kN 47
  • 48. Example 6 For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the displacement of the loaded joint. AE is constant. 30 kN 3m 45o 4m 4m 48
  • 49. 2 Member 1: 3m 8 1 3 6 7 4 2 θi = 0o, 5 λix = cos 0o = 1, λiy = sin 0o = 0 5 4* 3* 1 5 4m 4* 6 4m q´i o [q*] θij = -45 λix = cos (-45o) = 0.707, λiy = sin(- 45o) = -0.707 q´j 1 i 3* o 45 1 j [q*] = [T*]T[q´] + [T*]T[q´F] q5 q6 q3* q4* = 1 0 0 0 0 0 0.707 -0.707 q´i q´j [T*]T 49
  • 50. 2 Member 1: 3m 8 1 2 3 6 1 4m 4m [k*]1 = = q´j 1 i 0 0 0.707 -0.707 5 [k*]1 3* o 45 o o [q*] θij = -45 = 315 , λix = cos (-45o) = 0.707, λiy = sin(- 45o) = -0.707 1 q´i AE  1 − 1 [k *] = [T T ] − 1 1  [T ] L   1 0 0 0 4* 6 θi = 0o, 5 λix = cos 0o = 1, λiy = sin 0o = 0 5 4* 3* 5 7 4 5 0.25 6 0 AE 3* -0.1768 4* 0.1768 AE 4 1 -1 -1 1 j 1 0 0 0 0 0 0.707 -0.707 6 3* 4* 0 0 0 0 -0.1768 0 0.125 -0.125 0.1768 0 -0.125 0.125 50
  • 51. 2 Member 2: 3m 8 1 2 3 6 7 4 o 1 θi = -90 , 90o λix = cos(-90o) = 0, 2 λiy = sin(-90o) = -1 5 4* 3* 4m [k *] = [T T ] 4* 4m AE  1 − 1 − 1 1  [T ] L   [k*]2 = 0 -1 0 0 0 0 -0.707 -0.707 AE 3 3* j 90o+45o θj = -135o = 215o, q´j o) = -0.707, =135o λix = cos (-135 λiy = sin(- 135o) = -0.707 1 -1 -1 1 0 -1 0 0 1 [k*]2 = 1 2 AE 3* 4* i 2 1 5 q´i 2 2 3* 0 0.3333 -0.2357 -0.2357 0 -0.2357 0.1667 0.1667 0 -0.2357 0.1667 0.1667 0 -0.707 -0.707 4* 0 0 0 0 0 51
  • 52. 2 Member 3: 3m 8 1 2 3 6 2 7 4 1 5 3 6 4* 3* 4m [k *] = [T T ] AE  1 − 1 − 1 1  [T ] L   i [k]3 = = 0 0 0.8 0.6 AE 5 5 [k]3 q´j 5 q´i θi = θj = 36.87o ; λix = λjx = cos (36.87o) = 0.8, λiy = λjy = sin(36.87o) = 0.6 4m 0.8 0.6 0 0 j 3 36.87o 1 5 36.87o 6 1 2 0.096 0.072 -0.096 -0.072 -0.128 -0.096 0.128 0.096 -0.096 -0.072 0.096 0.072 5 0.128 6 0.096 AE 1 -0.128 2 -0.096 1 -1 -1 1 0.8 0.6 0 0 0 0 0.8 0.6 52
  • 53. 2 Member 4: 3m 8 1 6 5 8 1 1 4m 4m 1 i AE  1 − 1 [k *] = [T T ] − 1 1  [T ] L   j [k]4 = = 1 2 AE 7 8 0 0 1 0 AE 4 1 [k]1 1 0 0 0 7 [q] θi = θij = 0o; λix = λjx = cos 0o = 1, λiy = λjy = sin 0o = 0 q´i q´j 4* 3* 5 2 4 2 3 7 4 2 7 8 0.25 0 -0.25 0 0 0 0 0 -0.25 0 0.25 0 0 0 0 0 1 -1 -1 1 1 0 0 0 0 0 1 0 53
  • 54. 2 Member 5: 3m 8 1 4 6 36.87o , 0 .8 = 7 7o ; .87o ) 0.6 8 36. s (36 7o ) = 5 5 = θ j = co (36.8 λ jx = sin 7 5 2 3 8 8.13o 4* 3* 4* 3* 5 4m [k *] = [T T ] 4m AE  1 − 1 − 1 1 [T ] L   [k*]5 = 0.9899 -0.1414 0 0 0 0 0.8 0.6 3* [k*]5 = 3* 4* AE 7 8 q´j i λ jy 1 j q´i θi = - 8.13o; λix = cos (- 8.13o) = 0.9899, λiy = sin(- 8.13o) = -0.1414 AE 5 4* -1 0.9899 -0.1414 0 0 1 0 0 0.8 0.6 1 -1 7 8 0.196 -0.028 -0.1584 -0.1188 -0.028 0.004 0.02263 0.01697 -0.1584 0.02263 0.128 0.096 -0.1188 0.01697 0.096 0.072 54
  • 55. 2 Global Stiffness: 1 2 3 3m 8 6 5 7 4 [k*]3 5 = 4* 3* 5 6 AE 1 2 6 1 2 0.128 0.096 -0.128 -0.096 0.096 0.072 -0.096 -0.072 -0.128 -0.096 0.128 0.096 -0.096 -0.072 0.096 0.072 1 5 4m 4m 5 [k*]1 = 1 6 4* 0 0 0 0 5 0.25 6 0 AE 3* -0.1768 4* 0.1768 3* -0.1768 0 0.125 -0.125 0.1768 0 -0.125 0.125 [k*]5 1 [k*]2 = 1 2 AE 3* 4* 2 3* 0 0 0 0 0 0.3333 -0.2357 -0.2357 0 -0.2357 0.1667 0.1667 4* [k]4 = 1 0.25 2 0 AE 7 -0.25 8 0 3* = 2 4* 7 8 0 0 0 0 -0.25 0 0.25 0 0 0 0 0 7 8 3* 0.196 -0.028 -0.1584 -0.1188 4* -0.028 0.004 0.02263 0.01697 AE 0.096 7 -0.1584 0.02263 0.128 0.072 8 -0.1188 0.01697 0.096 1 2 3* 0 -0.2357 0.1667 0.1667 [K] = 1 0.378 0.096 0 2 0.096 0.4053 -0.2357 AE 3* 0 -0.2357 0.4877 55
  • 56. Global : 2 30 kN 8 1 2 3 3m 6 4 7 5 4* 3* 1 4m 5 4m [Q] = [K][D] + [QF] 1 Q1 = 30 Q2 = 0 Q3*= 0 D1 D2 D3* = 2 3* 1 0.378 0.096 0 AE 2 0.096 0.4053 -0.2357 3* 0 -0.2357 0.4877 = 1 AE D1 D2 D3* 86.612 -28.535 -13.791 56
  • 57. 2 Member Forces : 3m 8 1 2 3 6 7 4 [ q ' F ]m = 5 [ AE − λx L − λy  Dxi  D  yi λx   + q'F  Dxj     D yj    4* 3* 1 5 D1 D2 D3* 4m = 1 AE [ ] ] λx 0 4m [q´F]1 = AE -1 0 0.707 -0.707 4 = -2.44 kN, (C) 86.612 -28.535 -13.791 0 D1 λix λiy #1 1 0 0.707 -0.707 #2 0 -1 λjx #3 0.8 0.6 0.8 λjy [q´F]2 = AE 0 1 -0.707 3 = -6.26 kN, (C) -0.707 -0.707 -0.707 Member 0 D3* 0.6 D2 D3* 0 0 [q´F]3 = AE -0.8 -0.6 5 = 10.43 kN, (T) 0.8 0.6 0 D1 D2 57
  • 58. 2 Member Forces : 3m 8 1 2 3 6 7 4 [ q ' F ]m = 5 [ AE − λx L − λy 4* 3* 1 5 D1 D2 D3* Member #4 #5 4m = 1 AE λx  Dxi  D  yi λx   + q'F  Dxj     D yj    [ ] ] D1 4m 0 [q´F]4 = AE -1 4 = -21.65 kN, (C) 86.612 -28.535 -13.791 1 0 D2 0 0 D3* λix λiy λjx λjy 1 0 1 0 0.8 [q´F]5 = AE -0.9899 0.141 0.8 0.6 5 = 2.73 kN, (T) 0 0 0.6 0.9899 -0.141 0 58
  • 59. Reactions : 2 30 kN 8 1 2 3 3m 6 4 7 5 4* 3* 1 4m 5 4m Member [q´]1 [q´]2 [q´]3 [q´]4 [q´]5 Member Force (kN) -2.44 -6.26 10.43 -21.65 2.73 21.65 kN 36.87o 10.43 kN 5.90 kN 36.87o 2.44 kN 6.26 kN 6.26 kN 45o 45o 2.44 kN 1.64 kN 19.47 kN 2.73 kN 81.87o 6.54 kN 59
  • 60. Example 7 For the truss shown, use the stiffness method to: (b) Determine the end forces of each member and reactions at supports. (a) Determine the displacement of the loaded joint. Take AE = 8(103) kN. 8 kN 4 kN Ui D Ui 3m [ k *]m = A C B 36.87o 4m λixλix Vi Uj Vj λixλiy −λixλjx −λixλjy AE Vi λiyλix λiyλiy −λiyλjx −λiyλjy L U −λ λ −λ λ λjxλjx λjxλjy j jx ix jx iy Vj −λjyλix −λjyλiy λjyλjx λjyλjy 4m 60
  • 61. 8 kN 4 kN 2 D 1 1 3m 4m 4m 5 2 C B 4 6 4* A 3 5 3* 8 7 36.87o Member 1: λjx = cos 36.87o = 0.8, λjy = sin 36.87o = 0.6 y y* 4* i 2 j 1 36.87o 1 x 1 73.74o 3* q´j x* q´i λix = cos 73.74o = 0.28, λiy = sin 73.74o = 0.96 i j [q*] = [T*]T[q´] + [T*]T[q´F] q3* q4* q1 q2 = 0.28 0.96 0 0 0 0 0.8 0.6 q´i q´j [T*]T 61
  • 62. 8 kN 4 kN 2 D 1 3m A 3 1 6 4* 4m 4m 36.87o 5 2 C B 4 8 5 7 3* AE  1 − 1 [k ] = [T ] [T ] L − 1 1    T [k]1 = 0.28 0.96 0 0 0 0 0.8 0.6 3* [k]1 = 3* 4* 3 8x10 1 2 0.01568 0.05376 -0.0448 -0.0336 8x103 5 4* 1 -1 -1 1 1 0.28 0.96 0 0 0 0 0.8 0.6 2 0.05376 -0.0448 -0.0336 0.18432 -0.1536 -0.1152 0.128 0.096 -0.1536 -0.1152 0.096 0.072 62
  • 63. 8 kN 4 kN 2 D 1 3 1 3m 6 4* A 4m 4m C B 36.87o 4 5 2 8 5 7 3* [q*] = [T*]T[q´] + [T*]T[q´F] Member 2: y y* 4* 2 i 3* 36.87 j x* λix = cos 36.87o = 0.8, λiy = sin 36.87o = 0.6 q´i 6 5 x λjx = cos 0o = 1, λjy = sin 0o = 0 q´j 2 i j [k]2 = q3* q4* q5 q6 = 0.8 0.6 0 0 0 0 1 0 q´i q´j [T*]T -1 [T] [k] = [TT] AE 1 1 L -1 3* 4* 5 6 3* 0.16 4* 0.12 3 8x10 5 -0.2 0 6 0 0 0 0 0.12 0.09 -0.15 0 -0.2 -0.15 0.25 0 63
  • 64. 8 kN 4 kN 2 D 1 3 1 3m 6 4* A 4m 4m B C 4 5 2 8 5 7 3* 36.87o Member 3: y 2 o 270 3 λx = cos 270o = 0, λy = sin 270o = -1 1 Member 4: y 2 λx = cos 323.13o = 0.8, λy = sin 323.13o = -0.6 x 1 x 323.13o 4 6 7 5 [k]3 = 8x103 8 1 2 5 6 1 2 0 0 0 0.333 0 0 0 -0.333 6 5 0 0 0 -0.333 0 0 0 0.333 [k]4 = 8x103 1 2 7 8 1 0.128 -0.096 -0.128 0.096 2 -0.096 0.072 0.096 -0.072 7 -0.128 0.096 0.128 -0.096 8 0.096 -0.072 -0.096 0.072 64
  • 65. 8 kN 4 kN 2 D 1 3 1 3m 6 4* A 4m 4m 36.87o y Member 5: 6 5 2 C B 4 3* 5 8 7 8 5 7 5 x λx = cos 0o = 1, λy = sin 0o = 0 5 [k]5 = 5 6 8x103 7 8 6 7 0.25 0 -0.25 0 0 0 0 0 -0.25 0 0.25 0 8 0 0 0 0 65
  • 66. 2 1 1 3 [k]1 4 6 4* 5 2 8 5 4* 0.12 0.09 -0.15 0 2 -0.096 0.072 0.096 -0.072 [K] = 2 4* 1 0.05376 -0.0448 -0.0336 0.18432 -0.1536 -0.1152 0.128 0.096 -0.1536 -0.1152 0.096 0.072 7 3* 3* 3* 0.16 * 0.12 3 4 [k]2 = 8x10 5 -0.2 6 0 1 1 0.128 [k]4 = 8x103 2 -0.096 7 -0.128 8 0.096 = 3* * 3 4 8x10 1 2 3* 0.01568 0.05376 -0.0448 -0.0336 5 -0.2 -0.15 0.25 0 7 -0.128 0.096 0.128 -0.096 1 2 8x103 * 3 5 6 6 1 2 5 0 0 0 0 1 0 0 -0.333 0 [k]3 = 8x103 2 0 0.333 0 0 0 5 0 0 6 0 -0.333 0 0.333 0 8 5 6 7 8 0 -0.25 0 5 0.25 0.096 0 0 0 0 -0.072 6 [k]5 = 8x103 -0.096 0 0.25 0 7 -0.25 0.072 8 0 0 0 0 5 2 3* 1 0.256 0.0 -0.0448 0 0.0 0.474 -0.0336 0 -0.0448 -0.0336 0.17568 -0.2 0 0 -0.2 0.5 66
  • 67. 8 kN 4 kN 2 D 1 1 3m A 4m 5 2 C B 4 6 4* 4m 3 5 3* 8 7 36.87o Global: [Q] = [K][D] + [QF] 1 Q1 = 4 1 Q2 = -8 Q3*= 0 = Q5 = 0 D1 D2 D3* D5 8x103 0.0 5 3* -0.0448 0 2 0.0 0.474 -0.0336 0 3* -0.0448 -0.0336 0.17568 -0.2 0 0 1.988x10-3 -2.0824x10-3 1.996x10-4 7.984x10-5 m m m m 5 = 0.256 2 -0.2 0.5 D1 D2 D3* D5 67
  • 68. 8 kN 4 kN Member forces D [ q ' F ]m = 3m A 4m 4m C B 36.87o 2 λix λx λiy 3 0.28 0.96 #2 0.8 0.6 [ ] ] λjx λjy 0.8 0.6 1 0 D3* 4 [q´F]1 = 8x103 -0.28 -0.96 8 5 7 = 0.46 kN, (T) 6 4* Member − λy #1 2 1 1 [ AE − λx L  Dxi  D  yi λx   + q'F  Dxj     D yj    5 5 3* 0.8 0.6 0 D1 D2 D3* D1 D2 D3* D5 = 1.988x10-3 -2.0824x10-3 1.996x10-4 7.984x10-5 m m m m [q´F]2 = 8x103 -0.8 4 = -0.16 kN, (C) -0.6 1 0 0 D5 0 68
  • 69. 2 1 3 1 [ q ' F ]m = 4 6 4* 5 2 1.988x10-3 -2.0824x10-3 1.996x10-4 7.984x10-5 = − λy λx 8 5 7 3* D1 D2 D*3 D5 [ AE − λx L m m m m λix λiy λjx λjy #3 0 -1 0 -1 #4 0.8 -0.6 0.8 1 0 1 0 [ ] ] D1 [q´F]3 = 8x103 0 3 = -5.55 kN 1 0 -1 Member D2 D5 0 D1 [q´F]4 = 8x103 -0.8 5 = -4.54 kN 0.6 0.8 - 0.6 -0.6 #5  Dxi  D  yi λx   + q'F  Dxj     D yj    D2 0 0 D5 [q´F]5 = 8x103 -1 4 = -0.16 kN 0 1 0 0 0 0 69
  • 70. 8 kN 4 kN 2 D 1 3m A 3 1 4m 4m B 6 4* 5 2 C 4 5 8 7 3* 36.87o Member Member Force (kN) y* 36.87o [q]1 [q]2 0.46 -0.16 [q]3 [q]4 [q]5 -5.55 -4.54 -0.16 4.54 kN 0.46 kN 5.55 kN 36.87o 0.16 kN 0.16 kN 3.79 kN 36.87o 5.55 kN 0.36 kN x* 2.72 kN 70
  • 71. Space-Truss Analysis 71
  • 72. Member Local Stiffness [k´]: [q´] = [k´][d´] + [q´F] = [k´][T][d] + [q´F]  q 'i  EA  1 − 1  d 'i  q 'iF  q '  = − 1 1  d '  + q ' F    j  j   j L     d ix  d   iy   d iz  EA  1 − 1 F = − 1 1  [T ] d  + q ' L    jx  d jy     d jz    [ ] where, λ x 0 [T ] =  λ y λz 0 0 λx λ y 0 0 0 λz   72
  • 73. Member Global Stiffness [km]: [km]= [T]T[k´] [T] λ x λ  y λ z [km ] =  0 0  0  0 0  0  EA  1 − 1 λ x   λ x  L − 1 1   0   λy   λz   λy λz 0 0 0 λx λ y 0 0 λz   73
  • 74. Global equilibrium matrix: [Q] = [K][D] + [QF] Unknown Displacement Joint Load QI QII Reaction = KI,I KII,I KI,II Du KII,II Dk Fixed End Forces QFI + QFII Support Boundary Condition 74
  • 75. q´j j qix qiy qiz qjx m i q´i qjx EA L −λ λ −λ λ −λ λ λ λ λ λ λ λ x x x y x z x x x y x z −λyλx −λyλy −λyλz λyλx λyλy λyλz −λzλx −λzλy −λzλz λzλx λzλy λzλz dix diy diz djx djy djz qjz qiy i qjy qjz qjy j = λxλx λxλy λxλz −λxλx −λxλy −λxλz λyλx λyλy λyλz −λyλx −λyλy −λyλz λzλx λzλy λzλz −λzλx −λzλy −λzλz m qix  q 'i  EA  1 − 1  d 'i  q 'iF  q '  = − 1 1  d '  + q ' F    j  j   j L    qiz [q' ] j m = [ EA − λx L − λy − λz λx λ y  d ix  d   iy   d iz  λz   + q' F d jx  d jy     d jz    ] [ ] 75
  • 76. Example 6 For the truss shown, use the stiffness method to: (b) Determine the end forces of each member. (a) Determine the deflections of the loaded joint. Take E = 200 GPa, A = 1000 mm2. z 60 kN 80 kN 10 m y 4m 4m O 3m 3m x 76
  • 77. z 60 kN 80 kN 1 3 2 (0, 0, 10) 10 m 1 1 4 y 2 3 2 4m 4m 5 O 3m 3m x (-4, -3, 0) (-4, 3, 0) 4 λm = λxi + λyj + λzk λ1 = (-4/11.18)i + (3/11.18)j + (-10/11.18)k = -0.3578 i + 0.2683 j - 0.8944 k λ2 = (+4/11.18)i + (3/11.18)j + (-10/11.18)k = +0.3578 i + 0.2683 j - 0.8944 k λ3 = (+4/11.18)i + (-3/11.18)j + (-10/11.18)k = +0.3578 i - 0.2683 j - 0.8944 k λ4 = (-4/11.18)i + (-3/11.18)j + (-10/11.18)k = -0.3578 i - 0.2683 j - 0.8944 k 3 Member λx (4, 3, 0) (4, -3, 0) λy λz #1 -0.3578 +0.2683 -0.8944 #2 +0.3578 +0.2683 -0.8944 #3 +0.3578 -0.2683 -0.8944 #4 -0.3578 -0.2683 -0.8944 77
  • 78. λx Member λy λz #1 +0.3578 +0.2683 -0.8944 #3 #4 +0.3578 -0.2683 -0.8944 -0.3578 -0.2683 -0.8944 1 [k11]1 = [k11]3x3 -0.3578 +0.2683 -0.8944 #2 Member Stiffness Matrix [k]6x6 AE L 3 2 +0.320 -0.240 [k22]3x3 [k11]3 = AE L 2 3 [k11]4 = AE L -0.096 +0.072 +0.240 3 -0.320 +0.240 +0.80 3 -0.320 -0.240 +0.80 [KI,I] = AE L 3 2 +0.096 +0.072 +0.240 3 +0.320 +0.240 +0.80 1 +0.096 +0.072 -0.240 2 1 +0.128 +0.096 +0.320 1 +0.128 +0.096 -0.320 2 3 2 1 +0.80 2 1 +0.128 -0.096 -0.320 -0.096 +0.072 -0.240 3 [k21]3x3 1 1 +0.128 -0.096 +0.320 1 AE [k11]2 = L 2 [k]m = [k12]3x3 1 0.512 2 3 0.0 0.0 2 0.0 0.288 0.0 3 0.0 0.0 3.2 78
  • 79. z 60 kN 80 kN 3 1 2 (0, 0, 10) 10 m 1 1 4 y 3 2 2 4m 4m 5 O 3m [Q] = [K][D] + [QF] 60 -80 0.0 = AE L 1 1 0.512 3m x (-4, -3, 0) (-4, 3, 0) 3 4 2 3 0.0 0.0 D1 2 0.0 0.288 0.0 D2 3 0.0 0.0 3.2 D3 (4, 3, 0) (4, -3, 0) 0.0 + 0.0 0.0 79
  • 80. Global equilibrium matrix: [Q] = [K][D] + [QF] Unknown Displacement Joint Load QI QII KI,II = KII,I Du KII,II KI,I Dk Reaction Fixed End Forces QFI + QFII Support Boundary Condition (AE/L) = (1x10-3)(200x106)/(11.18) = 17.89x103 kN 1 60 -80 = 0.0 D1 D2 D3 = AE L L AE 1 0.512 2 3 0.0 0.0 D1 2 0.0 0.288 0.0 0.0 0.0 3.2 + D2 3 0.0 D3 +117.2 -277.8 0.0 0.0 6.551 = 0.0 mm -15.53 mm 0.0 mm 80
  • 81. z 60 kN 80 kN λx Member λy λz #1 #2 y 4m 4m O 3m 3m +0.3578 +0.2683 -0.8944 #3 #4 10 m -0.3578 +0.2683 -0.8944 +0.3578 -0.2683 -0.8944 -0.3578 -0.2683 -0.8944 x dxi Member forces: [q´j]m = AE L −λx −λy −λz D1 D2 D3 λx λy λz dyi dzi dxj + q´F dyj [q´j]1 = AE L +0.3578 -0.2683 +0.8944 = +116.5 kN (T) L AE 117.2 dzj [0] -277.8 0.0 81
  • 82. z 60 kN 80 kN λx Member λy λz #1 #2 y 4m [q´j]2 = [q´j]3 = [q´j]4 = 4m AE L AE L AE L O -0.3578 -0.3578 3m 3m -0.2683 +0.8944 +0.2683 +0.8944 +0.3578 +0.2683 +0.8944 +0.3578 +0.2683 -0.8944 #3 #4 10 m -0.3578 +0.2683 -0.8944 +0.3578 -0.2683 -0.8944 -0.3578 -0.2683 -0.8944 x L AE L AE L AE 117.2 -277.8 0.0 = +32.61 kN (T) 117.2 -277.8 0.0 = -116.5 kN (T) 117.2 -277.8 0.0 = -32.61 kN (T) 82
  • 83. λx Member λy λz [q´j]m #1 -0.3578 +0.2683 -0.8944 116.5 #2 +0.3578 +0.2683 -0.8944 32.6 #3 #4 +0.3578 -0.2683 -0.8944 -116.5 -0.3578 -0.2683 -0.8944 -32.6 60 kN 80 kN 1 4 R5z = (-32.6)(-0.8944) = 29.16 kN 5 32.6 kN 3 2 32.6 kN 116.5 kN 116.5 kN R5y = (-32.6)(-0.2683) = 8.75 kN R5x = (-32.6)(-0.3578) = 11.66 kN 83