(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 4 pptx

Machinery''''s Handbook 27th Episode 2 Part 4 doc

Machinery''''s Handbook 27th Episode 2 Part 4 doc

... 44 19, 4 42 2 , 4 42 7 , 46 15, 4 620 , 4 621 , 4 626 , 47 18, 4 720 , 48 15, 48 17, 4 820 , 5015, 5117, 5 120 , 6118, 8115, 8615, 8617, 8 620 , 8 622 , 8 625 , 8 627 , 8 720 , 8 822 , 94B17 { 125 –175 85 55 f s 16 75 8 140 8 41 0 4 685 26 150 13 160 83 125 20 160 175 22 5 ... 40 9, 42 9 , 43 0, 43 4, 43 6, 4 42 , 44 6, 5 02 135–185 90 f s 7 30 4 80 7...
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Bearing Design in Machinery Episode 2 Part 4 doc

Bearing Design in Machinery Episode 2 Part 4 doc

... The expression for discharge and suction heads are: H d ¼ p d g þ V 2 d 2g þ Z d ð10-51Þ H s ¼ p s g þ V 2 s 2g þ Z s ð10- 52 where H d ¼ head at discharge side of pump ðoutletÞ H s ¼ head at ... equal to the head loss in the loop. The expression for the pump head is H p ¼ p d À p s g þ V 2 d À V 2 s 2g þðZ d À Z s Þð10-53Þ The velocity of the fluid in the discharge and suction can be...
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Fundamentals of Structural Analysis Episode 1 Part 4 pptx

Fundamentals of Structural Analysis Episode 1 Part 4 pptx

... the following 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 ... Method, Part I by S. T. Mau 55 Problem 2. Solve for the force in the marked members in each truss shown. (1-a)(1-b) (2) (3) (4- a) (4- b) Problem 2. 4m 4@ 3m=12m 1 k...
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Fundamentals of Structural Analysis Episode 2 Part 4 ppsx

Fundamentals of Structural Analysis Episode 2 Part 4 ppsx

... = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − +−−−−− −−+−−−−− −− −−−−−−+− −−−−−+ EK L EK C L EK S-EK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L EK L EA CS L EK S L EA C L EK S L EK L EA CS L EK S L EA C EK L EK C L EK SEK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L...
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Fundamentals of Structural Analysis Episode 2 Part 4 doc

Fundamentals of Structural Analysis Episode 2 Part 4 doc

... = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − +−−−−− −−+−−−−− −− −−−−−−+− −−−−−+ EK L EK C L EK S-EK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L EK L EA CS L EK S L EA C L EK S L EK L EA CS L EK S L EA C EK L EK C L EK SEK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L...
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Fundamentals of Structural Analysis Episode 2 Part 5 pptx

Fundamentals of Structural Analysis Episode 2 Part 5 pptx

... maximize F CJ . 1kN 1kN 2 kN 1m 2m 2 kN 10 kN/m 10 kN/m x 6 m 1 .25 1 .25 0. 625 0 .41 F CJ 2 kN 1 .25 1 .25 0. 625 0 .41 F CJ 2 kN 1kN 1kN 1m2m 7 m 9 m Influence Lines by S. T. Mau 25 9 Example 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 kN 10 kN/m 6 m 1 .25 1 .25 0. 625 0 .41 F CJ 10 kN/m 6 m 9 m 1.18 1. 82 1 .25 1 .25 0....
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Fundamentals Of Structural Analysis Episode 2 Part 4 ppt

Fundamentals Of Structural Analysis Episode 2 Part 4 ppt

... = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − +−−−−− −−+−−−−− −− −−−−−−+− −−−−−+ EK L EK C L EK S-EK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L EK L EA CS L EK S L EA C L EK S L EK L EA CS L EK S L EA C EK L EK C L EK SEK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L...
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Fundamentals Of Structural Analysis Episode 2 Part 8 pptx

Fundamentals Of Structural Analysis Episode 2 Part 8 pptx

... 123 D’Alembert force, 29 4 Deflection, 67, 82, 121 , 26 0 transverse, 121 truss, 67 Deflection curve, 144 sketch of, 144 Deformation, 4, 6, 82, 121 axial, 121 flexural, 121 member, 18 shear, 121 Degrees ... 28 8 Global Coordinate, 4, 5, 23 8 Hardening, 28 8 Hinge, 93 Influence lines, 24 9 -27 1 applications of, 25 8 beam, 25 0 -25 9 deflection, 26 0 truss, 26 3 -27 1 Ins...
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Industrial Machinery Repair Part Episode 2 Part 1 pptx

Industrial Machinery Repair Part Episode 2 Part 1 pptx

... 1 728 2 3 /4 20 .80 6 3 /4 308 12 1 /2 1953 3 27 .00 7 343 13 21 97 3 1 /4 34. 33 7 1 /4 381 14 27 44 3 1 /2 42 . 88 7 1 /2 42 2 15 3375 3 3 /4 52. 73 7 3 /4 465 16 40 96 4 64. 00 8 5 12 17 49 13 4 1 /4 76.77 8 1 /4 5 62 ... 125 .0 9 729 1 1 /4 1.95 5 1 /4 145 9 1 /2 857 1 1 /2 3.38 5 1 /2 166 .4 10 1000 1 3 /4 5.36 5 3 /4 190.1 10 1 /2 1157 2 8.00 6...
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