I need some help with #1,2,3, due at midnight! thanks The simply supported beam

Question

I need some help with #1,2,3, due at midnight! thanks

The simply supported beam AB is subjected to a uniform load distribution having an intensity of w0. The length of the beam is 4 m and the cross-sectional dimensions are provided in units of mm. Assume the origin of the coordinate system to be at point A with positive x directed to the right. Figure 1 - Problem Model for questions 1-6 The maximum value of 'Q' the first moment of area about the neutral will be equal to 7.12 times 10-4 m3 1.78 times 10-3 m3 1.78 times 105 mm3 None of the above The moment of inertia about the neutral NA will be equal to 2.91 times 103 mm4 5.13 times 10-3 m4 3.34 times 108 mm4 None of the above Based solely on bending considerations, for a working stress of 14 MPa in bending the maximum allowable intensity of w0 will be to 15.6 kN 15.6 kN/m 12.9 kN/m None of the above

Explanation / Answer

1) Area 1 = 200*50 = 10000 mm^2 Area 2 = 250*50 = 12500 mm^2 For Area 2 : Now First moment of inertia = Integral (y*dA) dA = 50*dy Integral y*50 = (y^2/50 from -125 to 125 ) = 0 ; For Area 1: it is = 2* ( Area 1) *(125-25) = 2*10^6 mm^3 ; So total = 0 + 2*10^6 = 2*10^6 mm^3 2) I =2*( 200*50*100^2) + 2*(50*250^3/12) = 3.302*10^8 mm^4 3)M*y/I = stress M = 14*10^6 *I /y y = 0.125 m M = 36983.33 N-m ; Moment Max = wL^2/4 w = 9245.8 N/m = 9.2458 KN/m

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