UCI ENGRMAE 150 - SolutionProject2 (4 pages)

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SolutionProject2



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SolutionProject2

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Pages:
4
School:
University of California, Irvine
Course:
Engrmae 150 - Mechanics of Strctr
Mechanics of Strctr Documents
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Project 2 Solution FAM Given 8 2 5 ij 2 4 3 ksi 5 3 6 Direction of the element of design weld 1 1 lyn lzn negative lxn 2 2 Design is based on the normal stress acting on the weld Max normal and shear stresses in the whole structure and safe Tables includes several materials Required The selection of an appropriate material to meet design requirements Assumptions uniform stresses isotropic elastic materials Approach 2 2 2 lxn lyn lzn 1 Calculate the normal stress acting on the weld Calculate principal normal and shear stresses Using the value of the normal stress on the weld along with those of the principal normal and shear stresses to select an appropriate material that is ductile enough and that can keep the structure safe Analysis Let s divide the analysis into the following items 1 1 a lzn 2 2 2 lzn 1 lyn lxn 1 1 2 lzn 1 2 4 1 2 lzn 4 1 but since lzn is negative then lzn 12 2 b Normal stress on the weld Use equation 1 14 in FAM s note nr xx lxn lxr yx lyn lxr zx lzn lxr xy lxn lxr yy lyn lyr zy lzn lyr xz lxn lzr yz lyn lzr zz lzn lzr 2 replace r in the above equation by n nn xx lxn lxn yx lyn lxn zx lzn lxn xy lxn lxn yy lyn lyn zy lzn lyn xz lxn lzn yz lyn lzn zz lzn lzn using the values of stresses in the above tensor and the values of direction cosines 8 4 6 2 3 5 nn 2 2 2 4 4 4 2 2 2 nn 7 95 ksi 3 3 4 5 c principal stresses 1 p3 3 I p2 II p III 3 6 Calculate I II III 1 1 10 I x y z 8 4 6 ksi 3 3 3 2 2 2 II 3 x y y z z x xy yz zx 2 2 2 3 8 4 4 6 8 6 2 3 5 II 138 ksi2 2 7 8 2 2 2 II x y z 2 xy yz zx z xy x yz y zx 9 248 ksi3 f p p3 10 p2 46 p 248 10 plot f p against p 1000 500 f p 0 500 1000 1500 10 5 0 5 10 15 ksi p Figure 1 From figure 1 the three roots are p1 12 107 ksi p2 3 59 ksi p1 5 7 ksi 11 The largest principal normal stress is 12 1 ksi The largest principal shear stress 31 5 7 12 1 17 8 8 9 ksi 2 2 12 now using p1 12 1 ksi and 31 8 9 ksi along with nn 7 95 ksi and the data in the table only materials B D E pass the inspection Material E is too expensive Material B is too brittle Therefore we choose M AT ERIAL D 3 4 Appendix From the cosine we know 1 lxn 2 1 lyn 2 lzn 1 2 13 Replace r by n for each component we get 8 4 2 1 1 1 2 2 2 2 1 1 5 5 2 2 2 2 1 1 1 2 2 2 2 4 1 4 1 1 3 3 2 2 4 1 1 5 5 2 2 2 2 1 1 3 3 2 2 4 6 11 6 22 4 xx lxn lxn yx lyn lxn zx lzn lxn xy lxn lyn yy lyn lyn zy lzn lyn xz lxn lzn yz lyn lzn zz lzn lzn 14 so the normal component nn is 7 nn 3 7 95 2 4 15


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