Fundamentals of Structural Analysis Episode 1 Part 4 pps

... shown. (1- a) (1- b)(2) (3) (4- a) (4- b)Problem 2.4m 4@ 3m =12 m 1 kNab4m 4@ 3m =12 m 1 kNab 12 kN 4 @ 4m =16 m2m3mabc6@3m =18 m4m4m 12 kNabc3@3m=9m4m4m 15 kNaABbc3@3m=9m4m4m 15 kNaABbTruss ... form:xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23 1 kN 1 kN 1 kN 1 kN 1 kN 1 kNTruss Analysis: ... direct application of Eq. 14 yields 1 kN (∆o) = jMjjVf∑ =1 = f3 V3 = 0.375 kN (1 mm)and 1 24m3m33m 12 3 1 24m3m33m 12 3 1 kNTruss Analysis: Force Method, Part II by S.T.Mau69force....

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Fundamentals Of Structural Analysis Episode 1 Part 4 ppsx

... shown. (1- a) (1- b)(2) (3) (4- a) (4- b)Problem 2.4m 4@ 3m =12 m 1 kNab4m 4@ 3m =12 m 1 kNab 12 kN 4 @ 4m =16 m2m3mabc6@3m =18 m4m4m 12 kNabc3@3m=9m4m4m 15 kNaABbc3@3m=9m4m4m 15 kNaABbTruss ... form:xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23 1 kN 1 kN 1 kN 1 kN 1 kN 1 kNTruss Analysis: Force ... transfer matrix of Eq. 6 to find the solution.Problem 4- 2.xy 1 24m3m33m 1 23 1. 0 kN0.5 kNxy 1 24m3m33m 1 23 1. 0 kN0.5 kN 1. 0 kNTruss Analysis: Force Method, Part II by S.T.Mau72A...

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Fundamentals of Structural Analysis Episode 1 Part 4 pptx

... form:xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23xy 1 24m3m33m 1 23 1 kN 1 kN 1 kN 1 kN 1 kN 1 kNTruss Analysis: Force ... shown. (1- a) (1- b)(2) (3) (4- a) (4- b)Problem 2.4m 4@ 3m =12 m 1 kNab4m 4@ 3m =12 m 1 kNab 12 kN 4 @ 4m =16 m2m3mabc6@3m =18 m4m4m 12 kNabc3@3m=9m4m4m 15 kNaABbc3@3m=9m4m4m 15 kNaABb ... direct application of Eq. 14 yields 1 kN (∆o) = jMjjVf∑ =1 = f3 V3 = 0.375 kN (1 mm)and 1 24m3m33m 12 3 1 24m3m33m 12 3 1 kNTruss Analysis: Force Method, Part II by S.T.Mau68method,...

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Fundamentals of Structural Analysis Episode 1 Part 3 pps

... joint. 4 F6F3F235 4 3 4 5 1 F3F 1 Ry1Rx13 4 5F 4 F2Ry55Rx535 4 10 kN3@2m=6m 1 32 4 567 1 2 35678 10 11 2m9 4 Truss Analysis: Force Method, Part I by S. T. Mau 43 Problem 1. Use the method of joint ... – F 4 (3/5) –F 1 = 0,F 1 = –6 kN.2 1 236 kN 4 56Ry1Rx1 1 3 4 5Rx5Ry5F53 4 56 kN3F63 4 52F5F 4 F 1 35 4 3 4 5Truss Analysis: Force Method, Part I by S. T. Mau 51 In ... trussesshown. (1- a) (1- b) (1- c)(2-a)(2-b)(2-c)(3-a)(3-b)(3-c) (4- a) (4- b) (4- c)Problem 1. 3m4m3 kN3m4m3m8 kN3m4m3m8 kN3 kN3m4m 4 kN3m4m 4 kN3m2m6 kN2m 1. 5m 1. 2 m 1. 6 m0.9 m2 m2 m 1. 2...

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Fundamentals of Structural Analysis Episode 1 Part 3 pps

... F 4 (3/5) –F 1 = 0,F 1 = –6 kN.2 1 236 kN 4 56Ry1Rx1 1 3 4 5Rx5Ry5F53 4 56 kN3F63 4 52F5F 4 F 1 35 4 3 4 5Truss Analysis: Force Method, Part I by S. T. Mau 49 This particular ... trussesshown. (1- a) (1- b) (1- c)(2-a)(2-b)(2-c)(3-a)(3-b)(3-c) (4- a) (4- b) (4- c)Problem 1. 3m4m3 kN3m4m3m8 kN3m4m3m8 kN3 kN3m4m 4 kN3m4m 4 kN3m2m6 kN2m 1. 5m 1. 2 m 1. 6 m0.9 m2 m2 m 1. 2 ... joint. 4 F6F3F235 4 3 4 5 1 F3F 1 Ry1Rx13 4 5F 4 F2Ry55Rx535 4 10 kN3@2m=6m 1 32 4 567 1 2 35678 10 11 2m9 4 Truss Analysis: Force Method, Part I by S. T. Mau50Example...

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Fundamentals of Structural Analysis Episode 1 Part 5 pps

... -0. 71 -1. 00 -3 .40 -4. 806 -56.56 17 ,680 -3.20 1. 00 0. 94 -3.20 -3.007 40 .00 25,000 1. 60 -0. 71 0.33 -1. 14 0.538 -56.56 17 ,680 -3.20 0.00 -0 .47 0.00 1. 509 -40 .00 25,000 -1. 60 -0. 71 -0.33 1. 14 ... is α=5 (10 -6)/oC.Problem 5-3. 1 24m3m33m 1 23 1. 0 kN0.5 kN 1 23 4 56 1 23 4 567894m3@4m =12 m 1 23 4 56 1 23 4 567894m3@4m =12 mTruss Analysis: Force Method, Part II ... shown, given E =10 GPa, A =10 0 cm2 for all bars.Problem 6-2. 1 23 4 56 1 23 4 567894m3@4m =12 m 10 12 0 kN 10 kN 1 4 32 4 m3 m 1 23 4 56Truss Analysis: Force Method, Part II by S.T.Mau77Example...

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Fundamentals of Structural Analysis Episode 1 Part 7 ppsx

... Frame Analysis: Force Method, Part I by S. T. Mau 11 9 (11 ) (12 ) (13 ) ( 14 ) (15 ) (16 )Problem 2. Frame problems.5 m5 m 10 kN5 m5 m 10 kN-m5 m5 m 10 kN5 m5 m 10 kN-m5 m5 m 10 kN5 m5 m 10 ... diagrams. (1) (2)(3) (4) (5) (6)(7) (8)(9) (10 )Problem 2. Beam Problems.3 m5 m 10 kN3 m5 m 10 kN3 m5 m3 kN/m3 m5 m3 kN/m3 m5 m 10 kN-m3 m5 m 10 kN-m6 m 10 kN-m6 m4 m2 kN/m5@2m =10 m 10 ... BeamPa/2EIPa/EI2aaaReactionsPa/2EIPa/EI 11 Pa2 /12 EI5Pa2 /12 EIShear(Rotation)Diagram( Unit: Pa2 /EI ) 1 1/ 125 /12 1/ 6Moment(Deflection) Diagram( Unit: Pa3 /EI )a/6−27 01/ 7776=−0.35 1/ 3=−0.33P/2P/2Beam...

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Fundamentals of Structural Analysis Episode 1 Part 8 ppsx

... displacedconfiguration of the frame as shown below.Displaced configuration.P2P2PP 1 22 1 1 1/ L1/L 1 1/L 1/ L 1 12PL2L 1 12L2LPBeam and Frame Analysis: Force Method, Part II by S. T. Mau 14 0The computing ... table.. 1 (∆c) = ∫ m(x) EIdxxM )(=(EI 1 )[(3 1 ) (2 1 ) ( 4 L)(2L) +(2 1 ) (2 1 ) ( 4 L) (2L)+(6 1 ) (2 1 ) ( 4 L)(2L) ]∆c = EIL2 16 m. Downward.Example 18 . Find ... ∆dLoadDiagramMomentDiagram(M)(m)(m)(m)(m)a~bEI 1 (3 1 )(2PL)(2L)(2L)=EIPL38300EI 1 (3 1 )(2PL)(2L)(2L)=EIPL383b~cEI2 1 (3 1 )(2PL)(2L)(L)=EIPL323EI2 1 (3 1 )(2PL) (1) (L)=EIPL32EI2 1 (6 1 )(2PL) (1) (L)=EIPL62EI2 1 ...

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Fundamentals of Structural Analysis Episode 1 Part 9 pps

... solutions.P 1 23θ 2’M 1 M 1 θ 11 θ 1 ’θ 3’M 1 θ 21 M 1 θ 31 M 1 M2θ 21 M2θ22M2θ32M3M3θ 31 M3θ23M3θ33θ 2=0θ 1 =0θ 3=0PBeam and Frame Analysis: Force Method, Part ... self-evident.LLwPab5 P /16 11 P /16 3PL /16 L/2L/2Beam and Frame Analysis: Force Method, Part III by S. T. Mau 16 3⎥⎦⎤⎢⎣⎡2 212 12 11 δδδδ⎭⎬⎫⎩⎨⎧2 1 RR= −⎭⎬⎫⎩⎨⎧''2 1 ∆∆Note ... redundantforce.Pab5P /16 11 P /16 3PL /16 VM5PL/32∆ 11 P /16 −5P /16 −3PL /16 Inflection pointL/2L/2L/2L/2PabBeam and Frame Analysis: Force Method, Part III by S. T. Mau 16 5A frame example...

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Fundamentals of Structural Analysis Episode 2 Part 4 ppsx

... Mau 243 ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡9998979695 94 8988878685 84 7978777675 74 6968676665 646 362 61 5958575655 545 352 51 49 4 847 4 645 444 342 41 3635 343 332 31 2625 242 322 21 1 615 1 41 3 1 211 000000000 000000000KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK⎪⎪⎪⎪⎪⎪⎭⎪⎪⎪⎪⎪⎪⎬⎫⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎨⎧000000222θ∆∆yx= ... equation.⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡9998979695 94 8988878685 84 7978777675 74 6968676665 646 362 61 5958575655 545 352 51 49 4 847 4 645 444 342 41 3635 343 332 31 2625 242 322 21 1 615 1 41 3 1 211 000000000 000000000KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK⎪⎪⎪⎪⎪⎪⎭⎪⎪⎪⎪⎪⎪⎬⎫⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎨⎧000000222θ∆∆yx= ... (kG) 1 =⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡60,000022,500-30,000022,50001x10001x10-022,500- 011 ,25022,500- 011 ,250-30,0000 22,500-60,000022,5000 1x10-001x10022,500 011 ,250-22,500 011 ,2509999Member 2: (kG)2=⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡ 44 99 44 99 12 x1090,00006x1090,000-090,00090,000-090,00090,000-0002x10002x10-6x1090,000 012 x1090,000-090,000-90,000-090,000-90,0000002x10-002x10 (4) ...

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