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

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

3. 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

4. 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

5. 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

6. 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

7. 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