llll A horizontal cantilever in the form of an inverted T section (see adjoining

Question

llll

A horizontal cantilever in the form of an inverted T section (see adjoining figure), is used to support a hoist for removing catalyst from a reactor. One end is fixed, and the vertical load is applied 1.0 m along the cantilever. If the maximum allowable normal stress in the material of which the cantilever is made is 330 MN/m2, determine the greatest load that can be lifted from the reactor. Neglect the weight of the cantilever itself. For the greatest load that you determined in the preceding problem, calculate the greatest shear stress in the cantilever beam.

Explanation / Answer

If lower base is considered as y = 0 , we have y-coordinate of centroid of cross-section at [(0.1*0.01)*(0.01/2) + (0.01*(0.1 - 0.01)*(0.01 + (0.1 - 0.01)/2)] / [(0.1*0.01) + (0.01*(0.1 - 0.01))] = 0.02868 m., (This is found using (A1*y1 + A2*y2)/(A1+A2) formula.). MOI of lower flange about its own centroid is 0.1*0.01^3 / 12. By parallel axis theorem, MOI of lower flange about centroid of cross section is = 0.1*0.01^3 / 12 + (0.1*0.01)*(0.02868 - 0.01/2)^2 = 5.6927*10^-7 m^4. MOI of vertical flange about its own axis is 0.01*0.1^3 / 12. By parallel axis theorem, MOI of vertical flange about centroid of cross-section is = 0.01*0.1^3 / 12 + (0.01*(0.1 - 0.01)*(0.01 + (0.1 - 0.01)/2 - 0.02868)^2 = 1.4568*10^-6 m^4. Total MOI of cross-section about its centroid = 5.6927*10^-7 m^4 + 1.4568*10^-6 m^4 = 20.2607*10^-7 m^4. 1) Stress, sigma = My / I 330*10^6 = (P*1)*(0.1 - 0.02868) / (20.2607*10^-7). Load P = 9374.7 kg 2) Shear stress = (9374.7*9.81) / [(0.1*0.01) + (0.01*(0.1 - 0.01))] = 48.4 MPa

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