Engineering Analysis with Ansys Software Episode 1 Part 3 ppt

... c 3 1 c 1 1 b 1 1 c 2 1 b 2 1 c 3 1 b 3 1 ⎤ ⎥ ⎦ = 1 2 (1) ⎡ ⎢ ⎣ y 2 1 −y 3 1 0 y 3 1 −y 1 1 0 y 1 1 −y 2 1 0 0 x 3 1 −x 2 1 0 x 1 1 −x 3 1 0 x 2 1 −x 1 1 x 3 1 −x 2 1 y 2 1 −y 3 1 x 1 1 −x 3 1 y 3 1 −y 1 1 x 2 1 −x 1 1 y 1 1 −y 2 1 ⎤ ⎥ ⎦ (1. 102) Since ... Element 1. From Equations (1. 73) and (1. 69a)...

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... plate areas 12 3 3.5 .3. 4 Input of boundary conditions 13 3 3.5 .3. 5 Solution procedures 13 5 3. 5 .3. 6 Contour plot of stress 13 6 3. 5.4 Discussion 13 6 3. 5.5 Problems to solve 13 8 References 14 1 Chapter 4 Mode ... tension plate area 11 0 3. 4 .3. 4 Input of boundary conditions 11 3 3.4 .3. 5 Solution procedures 11 4 3. 4 .3. 6 Contour plot of stress 11 5 3. 4.4...

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... Ch 03- H6875.tex 24 /11 /2006 17 : 2 page 51 3 Chapter Application of ANSYS to Stress Analysis Chapter outline 3 .1 Cantilever beam 51 3. 2 The principle of St. Venant 84 3. 3 Stress concentration ... 40 60 80 Cross section 10 mm 5mm 10 0 Point load c f Figure 3. 3 Bending of a cantilever beam to solve. Ch 03- H6875.tex 24 /11 /2006 17 : 2 page 53 3 .1 Cantilever beam 53...

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... shown in Figure 3. 41. A B C D Figure 3. 41 “Contour Nodal Solution Data” window. Ch 03- H6875.tex 24 /11 /2006 17 : 2 page 72 72 Chapter 3 Application of ANSYS to stress analysis A Figure 3. 38 “Solve Current ... =0.005 m,namely node #10 8 in Figure 3. 32. Ch 03- H6875.tex 24 /11 /2006 17 : 2 page 83 A Figure 3. 51 “Subtract Areas” window. Figure 3. 52 The larger rectangula...

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... → Mesh →Areas →Free (1) The Mesh Areas window opens. Ch 03- H6875.tex 24 /11 /2006 17 : 2 page 93 3 .3 Stress concentration due to elliptic holes 93 0 5 10 0 5 10 15 20 x (mm) 200 19 0 18 0 10 0 Longitudinal ... this Ch 03- H6875.tex 24 /11 /2006 17 : 2 page 10 1 3. 3 Stress concentration due to elliptic holes 10 1 Figure 3. 72 Boundary conditions applied to the quarter mode...

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... (3 .11 ) 0 .1 1.0060 1. 0062 1. 0048 1. 0060 0.2 1. 0246 1. 0254 1. 0208 1. 0245 0 .3 1. 0577 1. 0594 1. 0 510 1. 0574 0.4 1. 1094 1. 111 8 1. 1000 1. 1090 0.5 1. 1867 1. 1892 1. 1757 1. 1862 0.6 1. 30 33 1. 30 43 1. 29 21 1 .30 27 0.7 ... 1. 1892 1. 1757 1. 1862 0.6 1. 30 33 1. 30 43 1. 29 21 1 .30 27 0.7 1. 4882 1. 48 41 1.4779 1. 48 73 0.8 1....

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... k  b 2 −x 2 (P3 .17 a) k = 2P’ πb 2 (P3 .17 b) b = 2 √ π  P’R 1 R 2 R 1 +R 2  1 −ν 2 1 E 1 + 1 −ν 2 2 E 2  (P3 .17 c) p 0 = 1 √ π     P’(R 1 +R 2 ) R 1 R 2 · 1  1 ν 2 1 E 1 + 1 ν 2 2 E 2  (P3 .17 d) PROBLEM ... Equation (3 .14 ) below. k = 2P’ πb 2 = 11 5.4N/mm 2 (3 . 13 ) b = 1. 522  P’R E = 2 .34 8 mm (3 .14 ) Ch 03- H6875.tex 24 /11 /2006 17 : 2...

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... 24 /11 /2006 17 : 47 page 16 0 A Figure 4 .30 Window of Solve Current Load Step. B Figure 4. 31 Window of Note. C Figure 4 .32 Window of /STATUS Command. Ch04-H6875.tex 24 /11 /2006 17 : 47 page 16 1 4.2 ... = 1. 042 10 10 m 4 Mass per unit width M =ρAL/L =ρA=7.8 10 3 kg/m 3 ×5 10 −5 m 2 =0 .39 kg/m λ 1 = 1. 875 λ 2 = 4.694 λ 3 = 7.855. For that set of data the following so...

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... 1. 4e 3 5 13 .5e 3 1. 4e 3 6 8.3e 3 1. 4e 3 7 8.3e 3 2 .15 e 3 8 0 0.6e 3 9 8.3e 3 −0.8e 3 10 10 .5e 3 −0.8e 3 11 10 .5e 3 0.8e 3 12 8.3e 3 0.8e 3 13 0 −0.6e 3 0.3e 3 14 8.3e 3 −2 .15 e 3 0.3e 3 15 8.3e 3 2 .15 e 3 ... 24 /11 /2006 17 : 47 page 17 0 17 0 Chapter 4 Mode analysis Table 4 .1 X, Y, and Z coordinates of keypoints for suspension KP No. X Y Z...

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