Question

The beam has the rectangular cross section shown. A beam of length 6 meters pin-supported 2 meters from the left end and roller-supported 2 meters from the right end. The beam has a rectangular cross section with base length 50 millimeters and height 150 millimeters. Load: w, uniform along beam.

Part A If w = 4 kN/m , determine the maximum bending stress in the beam.

Can you please draw out the moment and shear diagrams for this one using the integration (not area) method? I'm trying to find the formulas for both the shears and moments along each interval (0<=x<=2, 2<=x<=4, 4<=x<=6).

50 mm 150 mm 2 2 2

Answer 1

sigma F_x = 0 H_A = 0 + Sigma Fy = 0 RA + RB = 4 times 6 = 24 bn Sigma M_n = 0 R_B times 2 -(4 times 6 times 1) = 0 R_B = 12 kw R_A = 12 kw shear force and bonding moment For 0 lessthanorequalto x < 2n V(x) = -4x V(0) = 0 V(2) = -8kN m(r) = -4x^2/2 m(0) = 0 m(2) = -8 knm For 2 lessthanorequalto x < 4m V(x) = -4x + RA V(2) = 4 kN V(4) = -4 KN m(x) = -4x^2/2 + R_A(x - 2) m(2) = -8 kNm m(4) = -8 kNm shear force in this span is charging the sign V(x) = -4x + 12 = 0 -4x = -12 x = 3m S.F is becoming '0' at 3m from end Bm is max at this point m(3) = -6 kNm (Local extremum) For 4 lessthanorequalto x < 6 V(x) = -4(x) + R_A + R_B V(4) = 8KN V(6) = 0kN m(x) = -4x^2/2 + R_A(x - 2) + R_B(x - 4) m(4) = -8 kNm m(6) = 0 kNm from bending moment diagram, maximum absolute B.M is 8 kN m moment of inertia I_x = 1/12 times 50 times 150^3