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Q3: For the beam AB acted on by the applied load P of 5 kN. Determine the following: Data: L= 3m, d = 1 m, b= 75 mm, t = 20 mm, h= 100 mm and h1 = 80 mm. a) Free body diagram of the beam AB for the state of equilibrium b) Reactions at A and B c) Centroid of the cross section shown in right view (inverted T section) d) Moment of inertia of cross section e) Maximum stress in beam from bending t P d hi h А, B B H L b
Formula and Materials sheet: Factor of Safety = U.T.S. / 0, o=P/AG = E€ o = Mc/1 8=PL/(AE) 8= a*L* AT T = Tc/ T = P/A T = GY 0 = TL/(JG) 8 = PL/(AE) J = 3.14*c*/2 (Solid Bar) J = 3.14* (c29-C1")/2 (Hollow Bar) I = bh/12 (Rectangle) I=3.14*r*/4 (Circle) lx = Ic+Ad? ΥΣΑ = ΣΥΑ Power= work/time Key: o=Stress, t=Shear Stress, P=Force, A=Area, M=Moment, I=Moment of Inertia, E=Young's Modulus, E=strain T-Torque, Tv=Elastic Torque, J=Polar Moment of Inertia, G=Modulus of Rigidity, y=Shear Strain, =Angle of Twist, L=Length, 8=deformation, b=breadth, h=height, d, y, Y, C=distance or radius, Pr=radius of elastic core, a=coefficient of expansion, U.T.S.= Ultimate Tensile Strength Material Youngs Yield Stress Ultimate Tensile Modulus Coefficient Modulus (MPa) Strength (MPa) of Rigidity of Expansion (GPa) (GPa) (10*/C) Steel 200 250 (Normal) 400 (Normal and 200 12 150 (Shear) Shear) Aluminium 70 95 (Normal) 110 (Normal) 30 23 55 (Shear) 70 (Shear) Copper 120 220 (Normal) 70 (Normal and 17 150 (Shear) Shear) Concrete 12 40 25