Fundamentals of Structural Analysis Episode 2 Part 1 pdf

Fundamentals of Structural Analysis Episode 2 Part 1 ppsx

... +7.50DM −4.69 2. 81 COM 2. 35 1. 41 DM +0. 71 +0.70COM +0.36 0.35DM −0 .22 −0 .14 COM −0 .11 −0.07DM +0.04 +0.03COM +0. 02 +0. 02 DM −0. 01 −0. 01 COM 0.00 0.00SUM 2. 46 −4. 92 +4. 92 +14 .27 +15 .73 +7.87In ... 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-mMFcb= 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m(d) Compute DF at b: 2 m 2 m4 macb3 kN/m4 kN 2 m 2 kN 2 m 2 m4 macb3 kN/m4 ... loads 2. 464. 92 -14 .27 15 .73-7.87Inflection point Beam and Frame Analysis: Displacement Method, Part I by S. T. Mau 17 9Mbc = 2 1 ( Mba + Mbc ) = 2 1 (10 0 kN-m) = 50 kN-m.From Eq. 2, ...

Fundamentals of Structural Analysis Episode 2 Part 1 pdf

... 2. 81 COM 2. 35 1. 41 DM +0. 71 +0.70COM +0.36 0.35DM −0 .22 −0 .14 COM −0 .11 −0.07DM +0.04 +0.03COM +0. 02 +0. 02 DM −0. 01 −0. 01 COM 0.00 0.00SUM 2. 46 −4. 92 +4. 92 +14 .27 +15 .73 +7.87In the ... = 2 kN-mMFbc= 2 kN-m 2 m 2 m4 madb4 kNc 2 m2 m4 kN 1. 33 1. 33 1. 33 2. 67 2. 67Inflection point Beam and Frame Analysis: Displacement Method, Part I by S. T. Mau 19 0Solution. (1) ... 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-mMFcb= 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m(d) Compute DF at b: 2 m 2 m4 macb3 kN/m4 kN 2 m 2 kN 2 m 2 m4 macb3 kN/m4...

Fundamentals of Structural Analysis Episode 2 Part 2 pptx

... 0 .22 0.45 0 1 0 .23 0.46 0.31 0MEM MabMbaMbcMbeMebMabMbaMbcMbeMebEAMFEM− 62. 5 62. 5− 62. 5 62. 5DM 62. 5 62. 5COM31.3 31.3DM−31.0 20 .6 − 42. 2 21 .6 −43 .2 29 .0COM0.0 21 .10.0−14.5SUM0.0 ... (2) (3)(4) (5)(6) (7)Problem 1. 2 m 2 m3 macb3 kN/m4 kN2EIEI 2 m 2 m3 macb8 kN2EIEI 2 kN-m50 kNabcd50 kN 2 m 2 m 2 m 2 m 4 m50 kNabc4 m 4 m4 m8 m2EI2EIEIEI2EI2EI50 ... 12 2wL 12 2wL− 20 2 wL 12 2wL− 965 2 wL965 2 wL− b(2a-b)M a(2b-a)MNote: Positive moment acts clockwise.aLwLwLwLwL /2 L /2 PaLaLPPaLbLPPMbLaL Beam and Frame Analysis: ...

Fundamentals of Structural Analysis Episode 2 Part 2 ppt

... MF− 8PL8PL− ab 2 PL − a 2 bPL− a(1-a)PL a(1-a)PL−(6−8a+3a 2 ) 12 22 wLa(4-3a) 12 23wLa− 12 2wL 12 2wL− 20 2 wL 12 2wL− 965 2 wL965 2 wL− b(2a-b)M a(2b-a)MNote: Positive ... 0 .22 0.45 0 1 0 .23 0.46 0.31 0MEM MabMbaMbcMbeMebMabMbaMbcMbeMebEAMFEM− 62. 5 62. 5− 62. 5 62. 5DM 62. 5 62. 5COM31.3 31.3DM−31.0 20 .6 − 42. 2 21 .6 −43 .2 29 .0COM0.0 21 .10.0−14.5SUM0.0 ... kN2EIEI 2 kN-m50 kNabcd50 kN 2 m 2 m 2 m 2 m 4 m50 kNabc4 m 4 m4 m8 m2EI2EIEIEI2EI2EI50 kNabc4 m 4 m4 m8 mEI2EI2EI 2 m 2 m 2 m4 kNabcd 2 m 2 m 2 m4 kNabcddd ...

Fundamentals of Structural Analysis Episode 2 Part 3 potx

... (6)Problem 2. 2 m 2 m 3 macb 3 kN/m4 kN2EIEI 2 m 2 m 3 macb8 kN2EIEI 2 kN-m50 kNabc4 m 4 m4 m8 mEI2EI2EI50 kNabc4 m 4 m4 m8 mEI2EI2EI 2 m 2 m 2 m4 kNabc 2 m 2 m 2 m4 ... kNabcd 2. 77 kN 3. 36 kN-m1 . 23 kN0 .27 kN-m0 .27 kN-m1.64 kN-m0.69 kN0.69 kNcb1.64 kN-m0. 82 kN-m1 . 23 kN1 . 23 kN1 . 23 kN0.69 kN0.69 kN0.69 kN0.69 kN0.69 kN1 . 23 kNc1 . 23 kN1 . 23 kN0.69 ... nodalabc4 kNab 2 kN 2 kN 2 kN-m 2 kN 2 kN 2 kN-m 2 kN-m+abc 2 kN 2 kN-m1 2 3 2 kN 2 kN-mxy 2 kN-m Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau 21 6Σ Mb = 0 Mba...

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Fundamentals of Structural Analysis Episode 2 Part 3 pps

... kNabcd 2. 77 kN 3. 36 kN-m1 . 23 kN0 .27 kN-m0 .27 kN-m1.64 kN-m0.69 kN0.69 kNcb1.64 kN-m0. 82 kN-m1 . 23 kN1 . 23 kN1 . 23 kN0.69 kN0.69 kN0.69 kN0.69 kN0.69 kN1 . 23 kNc1 . 23 kN1 . 23 kN0.69 ... (6)Problem 2. 2 m 2 m 3 macb 3 kN/m4 kN2EIEI 2 m 2 m 3 macb8 kN2EIEI 2 kN-m50 kNabc4 m 4 m4 m8 mEI2EI2EI50 kNabc4 m 4 m4 m8 mEI2EI2EI 2 m 2 m 2 m4 kNabc 2 m 2 m 2 m4 ... nodalabc4 kNab 2 kN 2 kN 2 kN-m 2 kN 2 kN 2 kN-m 2 kN-m+abc 2 kN 2 kN-m1 2 3 2 kN 2 kN-mxy 2 kN-m Beam and Frame Analysis: Displacement Method, Part II by S. T. Mau 22 6Moment and deflection...

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Fundamentals of Structural Analysis Episode 2 Part 4 ppsx

... =⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡−+−−−−−−−+−−−−−−−−−−−−−+−−−−−−+EKLEKCLEKS-EKLEKCLEKSLEKCLEKCLEASLEKLEACSLEKCLEKCLEASLEKLEACSLEKSLEKLEACSLEKSLEACLEKSLEKLEACSLEKSLEACEKLEKCLEKSEKLEKCLEKSLEKCLEKCLEASLEKLEACSLEKCLEKCLEASLEKLEACSLEKSLEKLEACSLEKSLEACLEKSLEKLEACSLEKSLEAC 4 66 2 666 12 ) 12 (6 12 ) 12 (6) 12 ( 126 ) 12 ( 12 266 4 666 12 ) 12 (6 12 ) 12 (6) 12 ( 126 ) 12 ( 12 2 22 22 22 2 22 22 22 22 2 22 22 22 2 22 22 22 22 (26 )The corresponding ... equation.⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡9998979695 94 8988878685 84 7978777675 74 6968676665 646 3 626 15958575655 545 3 525 1 49 4 847 4 645 444 3 42 4 13635 343 3 323 1 26 2 5 24 2 322 211615 141 3 121 1000000000 000000000KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK⎪⎪⎪⎪⎪⎪⎭⎪⎪⎪⎪⎪⎪⎬⎫⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎨⎧000000 2 2 2 θ∆∆yx= ... Mau 24 3⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡9998979695 94 8988878685 84 7978777675 74 6968676665 646 3 626 15958575655 545 3 525 1 49 4 847 4 645 444 3 42 4 13635 343 3 323 1 26 2 5 24 2 322 211615 141 3 121 1000000000 000000000KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK⎪⎪⎪⎪⎪⎪⎭⎪⎪⎪⎪⎪⎪⎬⎫⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎨⎧000000 2 2 2 θ∆∆yx=...

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Fundamentals of Structural Analysis Episode 2 Part 4 doc

... =⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡−+−−−−−−−+−−−−−−−−−−−−−+−−−−−−+EKLEKCLEKS-EKLEKCLEKSLEKCLEKCLEASLEKLEACSLEKCLEKCLEASLEKLEACSLEKSLEKLEACSLEKSLEACLEKSLEKLEACSLEKSLEACEKLEKCLEKSEKLEKCLEKSLEKCLEKCLEASLEKLEACSLEKCLEKCLEASLEKLEACSLEKSLEKLEACSLEKSLEACLEKSLEKLEACSLEKSLEAC 4 66 2 666 12 ) 12 (6 12 ) 12 (6) 12 ( 126 ) 12 ( 12 266 4 666 12 ) 12 (6 12 ) 12 (6) 12 ( 126 ) 12 ( 12 2 22 22 22 2 22 22 22 22 2 22 22 22 2 22 22 22 22 (26 )The corresponding ... equation.⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡9998979695 94 8988878685 84 7978777675 74 6968676665 646 3 626 15958575655 545 3 525 1 49 4 847 4 645 444 3 42 4 13635 343 3 323 1 26 2 5 24 2 322 211615 141 3 121 1000000000 000000000KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK⎪⎪⎪⎪⎪⎪⎭⎪⎪⎪⎪⎪⎪⎬⎫⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎨⎧000000 2 2 2 θ∆∆yx= ... (kG)1=⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡60,000 022 ,500-30,000 022 ,50001x10001x10-0 22 ,500-011 ,25 022 ,500-011 ,25 0-30,0000 22 ,500-60,000 022 ,5000 1x10-001x100 22 ,500011 ,25 0 -22 ,500011 ,25 09999Member 2: (kG) 2 =⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡ 44 99 44 9912x1090,00006x1090,000-090,00090,000-090,00090,000-0002x10002x10-6x1090,000012x1090,000-090,000-90,000-090,000-90,0000002x10-002x10 (4) ...

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Fundamentals of Structural Analysis Episode 2 Part 5 pptx

... −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )-0 .5( 4. 82) (0. 6 25 )(4. 82/ 9)]= −33.0 kN10 kN/m6 m1 . 25 1 . 25 0. 6 25 0.41FCJ6 m9 m1.18 1. 82 Influence Lines by S. T. Mau 26 8(FCJ)max= − [2( 0. 6 25 )+1(0. 6 25 )(7/9)+1(0. 6 25 )(6/9)]= ... maximize FCJ.1kN 1kN 2 kN1m2m 2 kN10 kN/m 10 kN/mx6 m1 . 25 1 . 25 0. 6 25 0.41FCJ 2 kN1 . 25 1 . 25 0. 6 25 0.41FCJ 2 kN1kN 1kN1m2m7 m9 m Influence Lines by S. T. Mau 25 9Example 6. Find the ... =10[0 .5( 1.18)(0.41)+0 .5( 6)(0.41)−0 .5( 1.18)(0.41)(1.18/6)] = 14. 65 kN10 kN/m6 m1 . 25 1 . 25 0. 6 25 0.41FCJ10 kN/m6 m9 m1.18 1. 82 1 . 25 1 . 25 0. 6 25 0.41FCJ6 m9 m1.18 1. 82 Influence Lines by S. T. Mau 25 6Solution....

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Fundamentals of Structural Analysis Episode 2 Part 5 pdf

... maximize FCJ.1kN 1kN 2 kN1m2m 2 kN10 kN/m 10 kN/mx6 m1 . 25 1 . 25 0. 6 25 0.41FCJ 2 kN1 . 25 1 . 25 0. 6 25 0.41FCJ 2 kN1kN 1kN1m2m7 m9 m Influence Lines by S. T. Mau 26 4S = (LaL −) Si ... =10[0 .5( 1.18)(0.41)+0 .5( 6)(0.41)−0 .5( 1.18)(0.41)(1.18/6)] = 14. 65 kN10 kN/m6 m1 . 25 1 . 25 0. 6 25 0.41FCJ10 kN/m6 m9 m1.18 1. 82 1 . 25 1 . 25 0. 6 25 0.41FCJ6 m9 m1.18 1. 82 Influence Lines by S. T. Mau 26 2Problem ... CJ.(FCJ)max=10[0 .5( 1.18)(0.41)+0 .5( 6)(0.41)] = 14.7 kN1 . 25 1 . 25 0. 6 25 0.41FCJ1kN 1kN 2 kN1m2m6 m1 . 25 1 . 25 0. 6 25 0.41FCJ10 kN/m6 m9 m1.18 1. 82 Influence Lines by S. T. Mau 25 9Example 6. Find the maximum...

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Fundamentals of Structural Analysis Episode 2 Part 6 doc

... 17 .63 0 42. 98σ due to axial force (kN/cm 2 )0.0 72 0.0 72 0.088 0.088 0. 125 0. 124 σ due to moment (kN/cm 2 )0 0.0 06 0 0.013 0 0.0 32 Total σ (kN/cm 2 )0.0 72 0.078 0.088 0.101 0. 125 0.1 56 Error ... depth of the beam, h, is the same for both members.The fixed-end moments are obtained from the above table:PL /2 L /2 wLL /2 L /2 L0.2L0.2L 0.6Lhhba0.2L0.8Lhhba0.2L0.2L 0.3Lc0.2L0.8Lhhbahh0.3L100 ... MomentsCabCbaSabSbaMFabMFbaMFabMFba0 .69 1 0 .69 1 9.08EK 9.08EK-0.159PL 0.159PL -0.102wL 2 0.102wL 2 CabCbaSabSbaMFabMFbaMFabMFba0 .69 4 0.475 4.49EK 6. 57EK-0.097PL 0.188PL -0. 067 wL 2 0.119wL 2 Example...

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Fundamentals of Structural Analysis Episode 2 Part 7 potx

... SolutionsMemberForce(MN)Elongation(m)MemberForce(MN)Elongation(m)1 0. 375 0.011 8 0 0 2 0. 375 0.011 9 0. 625 0.0313 0. 375 0.011 10 0 04 0. 375 0.011 11 -0. 625 -0.0315 -0. 625 -0.031 12 -0 .75 0 -0. 023 6 0 0 13 -0 .75 0 -0. 023 7 0. 625 -0.031Case (a): ... kN0.5 kN 1.5 kN−1. 12 kN −1. 12 0 kN 2 kN 2. 24−1 kN0 kN 2. 240 kN1 kN1. 12 2 .24 −1. 12 −1 kN1 kN1 kN14 kN 2 kN1 kN Matrix Algebra Review by S. T. Mau3 02 =1x(−3)−2x(−6)+3x(−3)=0A matrix ... 0.00 2 0.011 -0. 073 6 0.045 -0. 073 3 0. 023 -0. 129 7 0. 023 -0. 129 4 0.034 -0. 073 8 0.000 -0. 073 The reactions are: 0.5 MN upward at both node 1 and node 5.1 2 3456 7 8 12 1314156 7 8911011 12 13xy...

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Fundamentals of Structural Analysis Episode 2 Part 7 pps

... SolutionsMemberForce(MN)Elongation(m)MemberForce(MN)Elongation(m)1 0. 375 0.011 8 0 0 2 0. 375 0.011 9 0. 625 0.0313 0. 375 0.011 10 0 04 0. 375 0.011 11 -0. 625 -0.0315 -0. 625 -0.031 12 -0 .75 0 -0. 023 6 0 0 13 -0 .75 0 -0. 023 7 0. 625 -0.031Case (a): ... 0.00 2 0.011 -0. 073 6 0.045 -0. 073 3 0. 023 -0. 129 7 0. 023 -0. 129 4 0.034 -0. 073 8 0.000 -0. 073 The reactions are: 0.5 MN upward at both node 1 and node 5.1 2 3456 7 8 12 1314156 7 8911011 12 13xy ... SolutionsMemberForce(MN)Elongation(m)MemberForce(MN)Elongation(m)1 0.563 0.0 17 8 0 0 2 0.563 0.0 17 9 0.3 12 0.0163 0.1 87 0.006 10 0 04 0.1 87 0.006 11 -0.3 12 -0.0165 -0.938 -0.041 12 -0. 375 -0.0116 1.000 0.040 13 -0. 375 -0.011 7 -0.3 12 -0.016Case (b):...

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Fundamentals of Structural Analysis Episode 2 Part 8 doc

... kN-mProblem 2. (1) Mba = − 525 kN-m, Mbc = 525 kN-m, Mab = 26 3 kN-m, Mcb = 2, 363 kN-m. (2) Mba = 28 .8 kN-m, Mbc = 28 .8 kN-m, Mab = 28 .8 kN-m, Mcb = 28 .8 kN-m 12 Matrix Algebra ... 29 6Gasset plate, 28 4Gaussian elimination method, 29 3Geometric non-linearity, 28 8 Global Coordinate, 4, 5, 23 8 Hardening, 28 8 Hinge, 93Influence lines, 24 9 -27 1applications of, 25 8 beam, 25 0 -25 9deflection, ... (2) 0. 021 L3 /EI0.5LL 2 /16EI-L 2 /16EI7Pa 2 /6EI7Pa 2 /6EI5Pa 2 /3EI-2Pa/EI8Pa3/3EI4Pa3/3EIabLMb3Mb/2LMb /2 3Mb/2LPabP /2 PL /8 P /2 L /2 L /2 PL /8 Solution to Problems...

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Fundamentals Of Structural Analysis Episode 2 Part 1 pps

... +15 +15 COM +7.50 +7.50DM −4.69 2. 81 COM 2. 35 1. 41 DM +0. 71 +0.70COM +0.36 0.35DM −0 .22 −0 .14 COM −0 .11 −0.07DM +0.04 +0.03COM +0. 02 +0. 02 DM −0. 01 −0. 01 COM 0.00 0.00SUM 2. 46 −4. 92 ... 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-mMFcb= 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m(d) Compute DF at b: 2 m 2 m4 macb3 kN/m4 kN 2 m 2 kN 2 m 2 m4 macb3 kN/m4 ... abcMember ab bcDF 1 0.5 0.5 0MEM MabMbaMbcMcbEAM −4FEM 2+ 2−4+4DM 2 COM 1 DM +1. 5 +1. 5COM +0.8 +0.8DM -0.8COM −0.4DM +0 .2 +0 .2 COM +0 .1 +0 .1 DM −0 .1 COM 0.0Sum −4 +2. 3 2. 3 +4.9The...

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