Maintenance Fundamentals Episode 1 Part 1 pps

Fundamentals of Structural Analysis Episode 1 Part 3 pps

Fundamentals of Structural Analysis Episode 1 Part 3 pps

... m 5kN 1. 2 m 1m 1m 1 kN 2 m2 m 1m 1m 2 kN 2 m2 m 1m 1m 2 kN 1 kN Truss Analysis: Force Method, Part I by S. T. Mau 49 This particular cut separates the truss into two parts. We shall use the left part ... joint. 4 F 6 F 3 F 2 3 5 4 3 4 5 1 F 3 F 1 R y1 R x1 3 4 5 F 4 F 2 R y5 5 R x5 3 5 4 10 kN 3@2m=6m 1 3 2 4 5 6 7 1 2 3 5 6 7 8 10 11 2m 9 4 Truss Analysis: Force Method,...
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Fundamentals of Structural Analysis Episode 1 Part 3 pps

Fundamentals of Structural Analysis Episode 1 Part 3 pps

... joint. 4 F 6 F 3 F 2 3 5 4 3 4 5 1 F 3 F 1 R y1 R x1 3 4 5 F 4 F 2 R y5 5 R x5 3 5 4 10 kN 3@2m=6m 1 3 2 4 5 6 7 1 2 3 5 6 7 8 10 11 2m 9 4 Truss Analysis: Force Method, Part I by S. T. Mau 50 Example ... F 4 (3/5) –F 1 = 0, F 1 = –6 kN. 2 1 2 3 6 kN 4 5 6 R y1 R x1 1 3 4 5 R x5 R y5 F 5 3 4 5 6 kN 3 F 6 3 4 5 2 F 5 F 4 F 1 3 5 4 3 4 5 Truss Analysis: Force Method, Pa...
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Fundamentals of Structural Analysis Episode 1 Part 4 pps

Fundamentals of Structural Analysis Episode 1 Part 4 pps

... form: x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 1 kN 1 kN 1 kN 1 kN 1 kN 1 kN Truss ... ( ∆ T)L= 1. 2 (10 -5 )/ o C (14 o C) (6,000 mm)= 1mm. (b) V 3 =1mm. (c) V 3 = 16 kN (6 m)/[200 (10 6 ) kN/m 2 (500) (10 -6 )m 2 ]=0.0 01 m= 1mm. Next we ne...
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Fundamentals of Structural Analysis Episode 1 Part 5 pps

Fundamentals of Structural Analysis Episode 1 Part 5 pps

... + 2 1 ∑ = n i i 1 P ∆ i + (1) ( ∆ o ) = 2 1 j M j j vf ∑ =1 + 2 1 j M j j VF ∑ =1 + j M j j Vf ∑ =1 (17 ) Substracting Eq. 17 by Eq. 15 and Eq. 16 yields (1) ( ∆ o ) = j M j j Vf ∑ =1 (14 ) which ... -0. 71 -1. 00 -3.40 -4.80 6 -56.56 17 ,680 -3.20 1. 00 0.94 -3.20 -3.00 7 40.00 25,000 1. 60 -0. 71 0.33 -1. 14 0.53 8 -56.56 17 ,680 -3.20 0.00 -0.47 0.00 1. 5...
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Fundamentals of Structural Analysis Episode 1 Part 7 ppsx

Fundamentals of Structural Analysis Episode 1 Part 7 ppsx

... Force Method, Part I by S. T. Mau 11 9 (11 ) (12 ) (13 ) (14 ) (15 ) (16 ) Problem 2. Frame problems. 5 m 5 m 10 kN 5 m 5 m 10 kN-m 5 m 5 m 10 kN 5 m 5 m 10 kN-m 5 m 5 m 10 kN 5 m 5 m 10 kN-m Beam ... Beam P a/2EI P a/EI 2aaa R eactions P a/2EI P a/EI 11 Pa 2 /12 EI 5Pa 2 /12 EI Shear(Rotation)Diagram ( Unit: Pa 2 /EI ) 1 1/ 12 5 /12 1/ 6 M oment(Deflection) Diagram ( Unit:...
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Fundamentals of Structural Analysis Episode 1 Part 8 ppsx

Fundamentals of Structural Analysis Episode 1 Part 8 ppsx

... below. Displaced configuration. P 2P 2P P 1 2 2 1 1 1/ L1/L 1 1/L 1/ L 1 1 2PL 2L 1 1 2L 2L P Beam and Frame Analysis: Force Method, Part II by S. T. Mau 14 0 The computing is carried out using the ... ∆ d Load Diagram Moment Diagram (M)(m)(m)(m)(m) a~b EI 1 ( 3 1 ) (2PL)(2L)(2L) = EI PL 3 8 3 00 EI 1 ( 3 1 ) (2PL)(2L)(2L) = EI PL 3 8 3 b~c EI2 1 ( 3 1 ) (2PL)(2L)(...
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Fundamentals of Structural Analysis Episode 1 Part 9 pps

Fundamentals of Structural Analysis Episode 1 Part 9 pps

... solutions. P 1 2 3 θ 2 ’ M 1 M 1 θ 11 θ 1 ’ θ 3 ’ M 1 θ 21 M 1 θ 31 M 1 M 2 θ 21 M 2 θ 22 M 2 θ 32 M 3 M 3 θ 31 M 3 θ 23 M 3 θ 33 θ 2 =0 θ 1 =0 θ 3 =0 P Beam and Frame Analysis: Force Method, Part ... ∆ 2 ’ + R 1 δ 21 + R 2 δ 22 = 0 These two equations can be put in the following matrix form. L L w L ∆ 1 ’ ∆ 2 ’ L L w L R 1 δ 11 R 1 δ 21 R...
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Fundamentals of Structural Analysis Episode 2 Part 1 ppsx

Fundamentals of Structural Analysis Episode 2 Part 1 ppsx

... M ab M ba M bc M cb M cd M dc EAM 30 DM +15 +15 COM +7.50 +7.50 DM −4.69 −2. 81 COM −2.35 1. 41 DM +0. 71 +0.70 COM +0.36 0.35 DM −0.22 −0 .14 COM −0 .11 −0.07 DM +0.04 +0.03 COM +0.02 +0.02 DM −0. 01 −0. 01 COM 0.00 0.00 SUM ... distributed loads −2.46 4.92 -14 .27 15 .73 -7.87 I nflection point Beam and Frame Analysis: Displacement Method, Part I by S. T. Mau 17 9 M bc = 2 1...
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Dimensioning and Tolerancing Handbook Episode 1 Part 6 ppsx

Dimensioning and Tolerancing Handbook Episode 1 Part 6 ppsx

... condition is ∅.5 01 + ∅.005 = ∅.506. The hole has an MMC size limit of ∅. 511 and a positional tolerance of ∅.005 at MMC. Since it’s an internal feature of size, its virtual condition is ∅. 511 − ∅.005 ... example, ∅.625 LC5 or 30 f7. 5-50 Chapter Five Eliminate unnecessary trailing zeros. 25 .1 not 25 .10 12 not 12 .0 with not with The exceptions are limit dimensions and bilateral (p...
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Dimensioning and Tolerancing Handbook Episode 1 Part 3 ppsx

Dimensioning and Tolerancing Handbook Episode 1 Part 3 ppsx

... subcommittees, including Y14 Main Committee, Y14.3 Multiview and Sectional View Drawings, Y14.5 Dimensioning and Tolerancing, Y14 .11 Molded Part Drawings, Y14.35 Drawing Revisions, Y14.36 Surface Texture, ... Displacement from MMC Allowable Position Tolerance 2.74 0.00 0 .14 2.76 0.02 0 .16 2.78 0.04 0 .18 2.80 0.06 0.20 2.82 0.08 0.22 2.84 0 .10 0.24 2.86 0 .12 0.26 Tolerancing Optimi...
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