Question

te a canilever beam loaded s shosn Y gure 4.plot the suay tose and bead monent siag romsaio tined te stactiong at the fixed suport uohat is Ake bench momen t at a distane 0sm tom e ixed sueport

Answer 1

Given that, Span, L = 5m Uniformly distributed load, W = 1.8 KN/m length of UDL = 3m Point load at 'D' = 4.5 kN Relations at the fixed support, R_B = 0 R_A = (1.8 times 3) + 4.5 = 9.9 KN \r\n Shear force calculations shear force at 'B', F_B = 0 Shear force at 'c', F_c = wxx = 1.8 times 3 - = 5.4 KN Shear force at 'D', F_D = 5.4 + 4.5 = 9.9 KN Bending moment calculations Bending moment at B, m_B = 0 Bending moment at C, M_c = -1.8 times 3 (x - 3/2) At C, x = 3, oi.e, M_C = -1.8 times 3(3 - 3/2) = -1.8 times 3 times 1.5 therefore M_C = -8.1 KNm Bending moment at D, m_D = -1.8 times 3 times [x - 3/2] At D, x = 4, i.e., m_D = -1.8 times 3 times (4 - 3/2)