Structural Mechanics 6 REACTIONS, SFD,BMD – with UDL’s 20 kN/m 2.00m 3.50m RA RB 1.00m

IMAGINE LOAD CONFIGURATION THUS: 20 kN/m RA RB 3.5/2m 3.5/2m 2.00m 1.00m

Sum of moments about RA = 0 6.5RB = 70 x 3.75 RB = 70 x 3.75 / 6.5 RB = 40.385 kNm Likewise take moments about B: RA [2+3.5 +1] –(20 x 3.5)[3.5/2 +1] = 0 6.5RA = 70[2.75] RA = 70[2.75]/6.5 RA = 29.615 kN - (20kN/m x 3.5m) x [2.0 +3.5/2]m + RB x [2.0 + 3.5 +1.0] = 0 RB x [2.0 + 3.5 +1.0] = (20kN/m x 3.5m) x [2.0 +3.5/2]m Check sum of Vertical forces =zero: 40.385 +29.615 = 70 20 x 3.5 = 70 ok

SHEAR FORCE DIAGRAM x 20 kN/m 2.00m 3.50m 1.00m Space diagram SHEAR FORCE DIAGRAM 20 kN/m RB =40.385 RA =29.615kN 2.00m 3.50m 1.00m 29.615kN 29.615 – [20 x3.5) = 40.385kN x

Find Point of ZERO SHEAR Using Similar Triangles: x 29.615 70.0 x 3.5 X / 29.615 = 3.5 / 70 Hence X = 3.5/70 x 29.615 = 1.481m i.e. Point of zero shear and hence Max BM lies 2+1.481 = 3.481m from support A

BENDING MOMENT DIAGRAM Space Diagram 20 kN/m RB =40.385 RA =29.615kN 2.00m 3.50m 1.00m 1.481m 59.23kNm 81.156kNm 40.385kNm Section 1 Section 2 Section 3

Calculate Bending Moments Section 1 29.615 x 2 = 59.230kNm Section 2 29.615 x [2 +1.481] – (20 x 1.481)x[1.481/2] = 81.156 kNm Section 3 29.615[2 +3.5] –(20x 3.5)[3.5/2] = 40.385kNm Check BM from RHS: 40.385 x 1.0 = 40.385 kNm ok