Mechanical Engineers Handbook Episode 8 potx

Mechanical Engineers Handbook Episode 8 potx

Mechanical Engineers Handbook Episode 8 potx

... Transfer from Extended Surfaces (Fins) 4 68 2.7 Unsteady Conduction Heat Transfer 472 3. Convection Heat Transfer 488 3.1 External Forced Convection 488 3.2 Internal Forced Convection 520 3.3 ... moment for the rotor is J H R  Md 2 R a8kg m 2 ). Compute the inertia moments for shafts J I  M I d 2 I 8 Y J II  M II d 2 II 8 Y J III  M III d 2 III 8 kg m 2 X The corrected inertia...

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Mechanical Engineers Handbook Episode 3 potx

Mechanical Engineers Handbook Episode 3 potx

... 173 3.2 Fluctuating Stresses 1 78 3.3 Constant Life Fatigue Diagram 1 78 3.4 Fatigue Life for Randomly Varying Loads 181 3.5 Criteria of Failure 183 References 187 119 Let us consider an element ... C V A Distance y 1 0 0.2c 0.4c 0.6c 0.8cc Factor C 1.50 1.44 1.26 0.96 0.54 0 Source: J. E. Shigley and C. R. Mischke, Mechanical Engineering Design. McGraw-Hill, New York, 1 989 . Used wit...

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Mechanical Engineers Handbook Episode 7 potx

Mechanical Engineers Handbook Episode 7 potx

... Vibration Vibration One can write dL  dU dV Y 2X 183  and the potential energy is V  1 2 kx 2 X 2X 184  2 .8. 2 SIMPLE HARMONIC VIBRATION The mechanical model for this vibration was shown in Fig. ... the model is m  x  kx  0X 2X 185  Multiplying Eq. (2. 185 ) by x gives mx  x  kx  x  0X 2X 186  This can be written as d dt 1 2 m  x 2  1 2 kx 2   0X 2X 187  Using the...

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Smithells Light Metals Handbook Episode 8 potx

Smithells Light Metals Handbook Episode 8 potx

... Metals Handbook Fe Ti Ga Mg Equilibrium diagrams 143 Mg Mn Mg Na 146 Smithells Light Metals Handbook Mg Sc Mg Si 1 48 Smithells Light Metals Handbook Mg Th Mg Ti Mg Tl 1 38 Smithells Light Metals Handbook Ge ... 144 Smithells Light Metals Handbook Mg Ni Mg Pb Mg Pr Equilibrium diagrams 145 Mg Pu Mg Sb 150 Smithells Light Metals Handbook Mg Zn Mg Zr 154 Smithells Light Metals Ha...

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Mechanical Estimating Manual Episode 8 potx

Mechanical Estimating Manual Episode 8 potx

... 42x 18 10 0.19 3.30 2. 48 3.61 2.69 1.62 6.61 6.72 48x 18 11 0.20 3. 48 2.62 3.67 2 .88 1.64 6.97 7.26 60x 18 13 0.23 4.93 3.71 3.79 3. 28 1.90 9 .88 8. 17 72x 18 15 0.26 5.94 4.45 5.34 3 .89 2.74 11 .87 ... 42x 18 10 0.16 2.75 2.07 3.01 2.24 1.35 5.51 5.60 48x 18 11 0.17 2.90 2. 18 3.06 2.40 1.37 5 .81 6.00 60x 18 13 0.19 4.11 3.09 3.16 2.73 1. 58 8.23 6 .81 72x 18 15 0....

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Mechanical Engineers Handbook Episode 1 pot

Mechanical Engineers Handbook Episode 1 pot

... 189 , 243, 339) Department of Mechanical Engineering, Auburn University, Auburn, Alabama 3 684 9 Dumitru Mazilu, (339) Department of Mechanical Engineering, Auburn University, Auburn, Alabama 3 684 9 Alexandru ... Auburn, Alabama 3 684 9 Bogdan O. Ciocirlan, (1, 51, 119, 559) Department of Mechanical Engi- neering, Auburn University, Auburn, Alabama 3 684 9 Nicolae Craciunoiu, (243, 559...

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Mechanical Engineers Handbook Episode 2 pptx

Mechanical Engineers Handbook Episode 2 pptx

... v 1 2 d dt v 2 Y 3X 18 where v 2  v Á v is the square of the magnitude of v. Using Eq. (3. 18) one may write Eq. (3.17) as F Á dr  1 2 mdv Á v 1 2 mdv 2 X 3X19 Figure 3.6 80 Dynamics Dynamics If ... angle, Figure 2.7 Figure 2 .8 2. Kinematics of a Point 61 Dynamics If we use the function V , the integral de®ning the work is U 12   r 2 r 1 F Á dr   V 2 V 1 ÀdV ÀV 2 À V 1...

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Mechanical Engineers Handbook Episode 4 pptx

Mechanical Engineers Handbook Episode 4 pptx

... 57 16±20 6.4±7 .8 18. 5 235 1.25 50 52.5 164 73 18. 8±22 .8 7.2 8. 0 21.5 262 1.35 50 62.5 187 .5 88 .5 20.4±23.5 7 .8 8. 5 24.5 302 1.50 Source: Joseph E. Shigley and Charles R. Mischke, Mechanical Engineering ... torsion (kpsi) H B K f 20 22 83 26 9.6±14 3.9±5.6 10 156 1.00 25 26 97 32 11.5±14 .8 4.6±6.0 11.5 174 1.05 30 31 109 40 13±16.4 5.2±6.6 14 201 1.10 35 36.5 124 48. 5 14.5±17...

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Mechanical Engineers Handbook Episode 5 pot

Mechanical Engineers Handbook Episode 5 pot

... angular speed n 1  2970 rpm (o 1  o 10  pn 1 a30  311X0 18 radas). Find the absolute Figure 2 .8 Used with permission from Ref. 18. 2. Gears 263 Machine Components The base pitch is like the ... circle (Fig. 1 .8) . The angle l is the helix angle of the thread. The screw is loaded by an axial compressive force F (Figs. 1.7 and 1 .8) . The force diagram for lifting the load is shown in...

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Mechanical Engineers Handbook Episode 6 docx

Mechanical Engineers Handbook Episode 6 docx

... 0. 981 7 0.0256 0.0111 39, 980 19,990 0.0500 0.6 283 0.0270 0.0093 52, 980 31,790 0.0670 0.3499 0.0 288 0.0075 77,450 54,220 0.0900 0.1939 0.0 288 0.00 68 110,540 88 ,430 0.1100 0.12 98 0.0275 0.0062 140,400 ... 25.5 13 .8 20.0 32.5 23.6 45.4 80 14.2 18. 4 28. 0 16.6 22.5 35.5 17.3 26.2 51.6 85 15.0 22.5 30.0 17.2 26.5 38. 5 18. 0 30.7 55.2 90 17.2 25.0 32.5 20.0 28. 0 41.5 37.4 65 .8...

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