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Published byMatthew Souter Modified about 1 year ago

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Structural Mechanics 6 REACTIONS, SFD,BMD – with UDL’s 20 kN/m 2.00m3.50m RARA RBRB 1.00m

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20 kN/m 2.00m 3.50m RARA RBRB 1.00m 3.5/2m IMAGINE LOAD CONFIGURATION THUS:

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

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SHEAR FORCE DIAGRAM Space diagram SHEAR FORCE DIAGRAM 20 kN/m 2.00m3.50m R A =29.615kN R B = m kN x – [20 x3.5) = kN

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

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BENDING MOMENT DIAGRAM Space Diagram 20 kN/m 2.00m3.50m R A =29.615kN 1.00m 1.481m R B = Section 1Section 2Section kNm kNm kNm

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

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