Solved question of bending moment and shearforce

1. Structural Design I
Structural Mechanics
BM SF Continued
Lecture 9

2. • More Examples

3. 6kN/m
Calculate the magnitude and location of maximum B in the beam shown above and
draw the B M diagram
We need to first draw the FBD:
1m 1m 1m
22kN
10kN
2m
22kN10kN
1m 1m 1m2m
6kN/m

4. To find the location of max BM we need to draw the SF diagram.
For that we need to first calculate the reactions
Rl × 5 = (22 × 1) + (18 × 3.5) + (10 × 4) = 22 + 63 + 40 = 125
Rl = 25kN
Rr × 5 = (10 × 1) + (18 × 1.5) + (22 ×4) = 10 + 27 + 88 = 125
Rl = 25kN
Rl + Rr = 10 + 18 + 22 = 50, Rl + Rr = 25 + 25 = 50
22m10m
1m 1m 1m2m
6kN/m

5. 1m 1m 1m2m
6kN/m
25 kN 25 kN
22 kN10 kN
Just left of Rr SF = 25 kN,
At 1 m from Rr, SF = 25 – 22 = 3 kN
At 2 m from Rr, SF = 25 – 22 = 3 kN
At 3 m from Rr, SF = 3 – 6 = -3kN
At 4 m from Rr, SF = - 3 - 12 = -15 kN
At 5 m from Rr, SF = -15 – 10 = -25
Max BM will lie where SF = 0. From above Sf =0 at 2.5 m from Rr
Max BM at 2.5 m from Rr = (25 × 2.5) – (22 × 1.5) – (6 ×.5 × .25 )
= 62.5 – 33 – .75 = 28.75kNm

6. 1m 1m 1m2m
6kN/m
25 kN 25 kN
10 kN
22 kN
BM at 1 m from Rl = (25 × 1) – ( 6 × 1 × .5) = 25 – 3 = 22 kNm
BM at 2.5 m from Rl = 28.75 kNm
BM at 3 m from Rl = (25 × 3) – (10 × 2) - (6 × 3 × 1.5) = 75 – 20 – 27 = 28kNm

7. Cantilever beam with partial UDL and Point Load
A cantilever beam is loaded as shown.
BM at D = 0
BM at C = 6×4×2=48kNm (-ve)
BM at B = (6×4×4) = 96kNm (-ve)
BM at A = (6×4×6) + (6×2) = 156kNm (-ve)
From D to C BM will be a curve, from C to B it
will be a straight sloping line and from B to A it will
be a straight sloping line.
SF will be 24 kN at C, increase by 6 at B to 30 kN
And remain constant till A.
It will slope from D to C, be parallel to base till B
change vertically at B and be parallel to base till A.

8. Beams carrying Complicated Load Systems
When a beam is carrying a number of
loads both continuous and point loads
then these can be drawn separately
and then added geometrically to plot
the total BM diagram.
In the figure, say the BM produced by the
UDL is a curve that has vertical values
a, b and c and BM by point loads A, B
and C at the load points by the point loads
then the BM diagram can be plotted by
taking values A+a, B+b and C+c at the
load points as vertical ordinates.

9. Position of Maximum Bending Moment in Simply
Supported Beams
When only the maximum BM needs to be calculated without drawing the
BM diagram the following steps can be adopted:
The maximum bending moment occurs at the point where the Shear Force
is zero.
To find this point, first find the left side reaction
Then proceed across from the left end until the load on the beam equals in
total value the left end reaction. This where the shear force is equal to zero.
Calculate the BM at this point by subtracting the total load moments from the
reaction moment around that point.

10. RrRl
4kN/m2kN/m
4m2m
In the beam shown above
Rl×6 = (2×2×5)+(4×6×3) = 20+72 = 92 or Rl = 15.3
Rr×6 = (2×2×1)+(4×6×3) = 4+72 = 76 or Rr = 12.7
Rl + Rr = 28kN Total load = (2×2) + (4×6) = 28kN
SF at right end of 2kN load = 15.3 – 12 = 3.3kN
SF will be zero at a distance of 3.3÷4 = 0.825m right of this point or 2.825m from
left end
To cross check: SF will be zero at a distance from 12.7÷4 from right end
=3.175m
BM at 2.825m from left end = (15.3×2.825) – (2×2×1.825) – (4×2.825×1.412)
=43.2 – 7.3 – 16=19.9kNm or 20kNm
Determine the magnitude and position of the maximum BM in the beam loaded
As shown..

11. Find the Maximum BM for the beam loaded as shown above
1kN/m
2kN/m 4kN
5m
18m4m
8m
First the reactions
Rl×18 = (4×5) + (2×22×11) + (1×8×18)
Rl ×18 = 20 + 484 + 144 = 648 or Rl = 36 kN
Rr×18 = (4×13) + (2×22×7) = 52 + 308 = 360 or Rr = 20kN
Total load = 4 + (2×22) + (1×8) = 4 + 44 + 8 = 56
Rl + Rr = 36 + 20 = 56
Rl Rr

12. SF at A =0, at B = -(1×4) – (2×4) = -12 then moves up by Rl or +24
At C SF slopes down by (4×1)+ ( 4×2) = 12 so becomes (24 – 12) = 12
The load to the right of C is 2kN/m
So SF will be 0 at 6 m to right of C
Max BM will be at 10 m right of Rl
Max BM = (36×10) – (8×10) – (2×14×7) = 360 – 80 – 196 = 84kNm