Question

3. Two vertical forces are applied to a beam of the cross section as shown in Figure 3 Determine the maximum tensile and compressive stresses in portion BC of the beam 20 marks) (CLO3:PLO3:C4) 75 65 kN 65 kN A B 50 mm 50 mm Figure 3

Answer 1

First we calculate the centroid of inverse T section. Area of lower portion A1 = 50*(75 + 75 + 75) = 11250 mm2 A2 = 75*150 = 11250. Y 1 = 25 mm from bottom Y2 = 75 + 50 = 125mm from bottom Y = (Y 1 + Y2)/2 because area are same. Y = 150/2 = 75m m from bottom. Yt = 75mm for tensile strength Yc = (200-75) = 125mm for compressive strength. lxx = (lg1 + A1*r1A2) + (lg2 + A2*r2A2) = ((225*(5QA3)/ 12) + 1125Q*(75-25)A2) + ((75*(150A3)/ 12) + 1125Q*(75-125)A2)\\Maximum moment will be at centre between B and C.\\M = 1.75*75-0.75*65 = 65kN-m = 65000000 N-mm.\\Tensile strength St = (MYt/I) = 65000000*75/79687500 = 61.176 N/mm2 or MPa Compressive strength Sc = (MYc/1) = 65000000*125/7 9687500 = 101.96 N/mm2 orMPa.9: 23 AM

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