HW_Week01_Solution

# HW_Week01_Solution - 3000 lb 30&quot; 4000 lb 20&quot;...

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60˚ 34.641” 3000 lb 30” 20” 4000 lb 10” 30” Problem 2-4: The structure shown in the figure is an idealization of an engine mount. Find the internal forces in the 5 two-force members. The supports on the right (at C and E) are pinned. B A C D E R cy R cx R ey R ex x y

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ΣM E =(4000)*(40)+(3000)*(90)-R cx (34.641)=0 R cx = 12,413 lb ΣF x =R cx +R ex =0 R ex =-12,413 lb Joint E: 30˚ 2413 lb F ED B A C D E R cy R cx R ey R ex ΣF x =-12,413-F ED cos(30)=0 F ED =-14,338.3 lb ΣF y =R ey -F ED cos(60)=0 R ey =7166.67 lb E 12413 lb R ey
4000 lb 3000 lb B A C D E R cy R cx = 12413 lb R ey = 7166.67 lb R ex = -12413 lb ΣF y =R cy +7166.67 - 4000 - 3000=0 R cy =-166.67 lb C 30˚ 12413 lb 166.67 lb Joint C: F CB F CD ΣF x =-F CB -F CD cos(30)+12413=0 F CB =12413-F CD cos(30) ΣF y =-166.67-F CD cos(60)=0 F CD =-333.33 lb F CB =12701.7 lb

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ΣF x =-F CB -F CD cos(30)+12413=0 F CB =12413-F CD cos(30) ΣF y =-166.67-F CD cos(60)=0 F CD =-333.33 lb F CB =12701.7 lb Joint D: D 30˚ F DB F DA 30˚ 30˚ 333.33 lb ΣF x =-14333.3cos(30)-333.33-F DA cos(30)=0 F DA =-14666.7 lb ΣF y =-333.33sin(30)+(14333.3)sin(30)+F DB +F DA sin(30)=0 F DB =333.33 lb 14,333.3 lb
4000 lb 3000 lb B A C D R cy = 166.7 lb. R cx = 12413 lb 12,702 lb. (T) 333 lb. (C) Final results E R ey = 7166.67 lb R ex = 12,413 lb

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Problem 2-6: Solution A design team insists that a truss structure whose idealization is shown in Figure 2- 6 (b) is the best solution for a space system to be launched to Mars. The two idealizations shown in Problem Figure 2-6(b) and (c) and represent the math models of the two competing designs. Complete the following tasks.
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## This note was uploaded on 01/18/2012 for the course AAE 352 taught by Professor Chen during the Fall '08 term at Purdue University-West Lafayette.

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HW_Week01_Solution - 3000 lb 30&quot; 4000 lb 20&quot;...

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