Question

Mechanics of Materials :

Beams: 4b The cantilever beam shown below, of length L and elasticity modulus E, is subjected to load W (weights) at its free end. A dial indicator (considered weightless) and other measurements concluded that the radius of curvature at distance s from the fixed end is p. Considering that the cross section of the beam is square of side b, find, in terms of E, b, L, s, p (a) the weight W; (b) the maximum tensile stress in the beam; (c) the maximum shear stress in the beam. Note that the moment of inertia of a square cross section of side b with respect to its center is I- b4 /12 175 Dial indicator Weights Experimental setup:

Answer 1

Radius of curvature at section X - X = Rho From simple bending theory, M/I = sigma/y = E/R Therefore M/I = E/P = > M = EI/P (at section x - x) Also, Bending t Section X -X = W times (L - s) Therefore W(L - s) = EI/rHO = > W = E/Rho(L - s) times b^4/12 = > weight(W) = Eb^4/12 Rho(L-s) The maximum tensile stress occurs where bending moment is maximum, i.e., at fixed joint. bending moment at A = W times L Therefore M/I = sigma/y = > M/I y where, y = Distance of extreme fibre from neutral - axis = b/2 Therefore = W times L/(b^4/12) times (b/2) = 6wL/b^3 = 6L/b^3 times E b^4/12 Rho (L - S) = > maximum tensile Stress (delta) = ELb/2P(L-S) Maximum shear stress occurs where Shear force is maximum, i.e., at fixed support Shear force at support A = W kN Maximum Shear Force = 1.5 times Average Shear Stress = 1.5 times W/Area afcs Maximum Shear Force = 1.5 times W/b^2 = 1.5 times Eb^4/12 Rho(L-s) times 1/b^2 Therefore Maximum Shear Force = 0.125 Eb^2/Rho(L - S) KN