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**Structural Mechanics 6 REACTIONS, SFD,BMD – with UDL’s**

20 kN/m 2.00m 3.50m RA RB 1.00m

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**IMAGINE LOAD CONFIGURATION THUS:**

20 kN/m RA RB 3.5/2m 3.5/2m 2.00m 1.00m

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**Sum of moments about RA = 0**

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

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**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 – [20 x3.5) = kN x

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**Find Point of ZERO SHEAR**

Using Similar Triangles: x 29.615 70.0 x 3.5 X / = / 70 Hence X = /70 x = m 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 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

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**Calculate Bending Moments**

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

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ENGR 220 Section 6.1~6.2 BENDING.

ENGR 220 Section 6.1~6.2 BENDING.

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