Lecture 8 -(Bending moment & Shear force Part II)

1. STRUCTURAL DESIGN I
STRUCTURAL MECHANICS
Lecture 8

2. • In Building structures:
– What is bending moment?
– What are its characteristics?
– What is shear force?
– What are its characteristics?
– What is deflection?

3. BENDING MOMENT AND SHEAR FORCE. FURTHER
EXAMPLES
• We have learnt so far that
– A concentrated load system will lead to a BM diagram bound by
straight lines, which change their slope at the load points.
– A distributed load will have some form of a curve for a BM diagram. If
the load is uniformly distributed, a parabolic shape will occur.
– For a concentrated load system the SF diagram will be stepped. A step
will take place at every load point, the vertical drop representing the
load to scale.
– The SF diagram for a uniformly distributed load will be of a uniformly
sloping character. Conventionally the slope will be downward towards
the right.

4. Simply supported beam with uniformly distributed loads
A convenient way of understanding
the bending moment and shear
force of a simply supported beam with
a uniformly distributed load is to
investigate the value of SF and BM
at atypical section of the beam. The
value is expressed in terms of ‘x’, the
distance of the section from A. The
form of the expression obtained gives
the clue to the nature of the BM
(or SF) diagram.

5. BMx= reaction moment – Load moment
Reaction moment = wl/2 × x = wlx/2
Load moment = wx × x/2 = wx²/2
BMx= wlx - wx²
2 2
If this expression is plotted for values of x from 0 to
l a graph in the form of a parabola would be
obtained. The maximum ordinate of the parabola
will occur for x = l/2 i.e. at mid span
BMMAX = (wl/2 × l/2) – (w/2 × l²/4) = wl²/4 - wl²/8
= wl²/8
= Wl/8

6. 2m4m4m4m
2kN8kN6kN2kN
5m
Rl Rr
Find the Maximum Bending moment for the loading system shown in figure.
Draw the BM and SF diagram for this load system

7. OVERHANGING BEAM WITH CONCENTRATED LOAD
SYSTEM
• For construction of
the BM and SF
diagram for a
loading system as
shown in the
figure, first the
reactions should be
ascertained.
2m4m4m4m
2kN8kN6kN2kN
5m
Rl Rr
Taking moments about Rr
Rl x 10 = (8 x 2) + (6 x 6) + (2 x 14) – (2 x 5)
= 16 + 36 + 28 – 10
= 70 or Rl = 7kN
Taking moments about Rl
Rr x 10 = (2 x 15) + (8 x 8) + (6 x 4) – (2 x 4)
= 30 + 64 + 24 – 8
= 110 or Rr = 11kN
Rl + Rr = 2 + 6 + 8 + 2 = 18 = 7 + 11

8. 2m4m4m4m
2kN8kN6kN2kN
5m
7kN 11kN
Bending Moment:
BM at Rl = Moment of load to left of section
= 2 x 4 = -8kNm
BM at 6kN load point = (7 x 4) – (2 x 8) = 12kNm
BM at 8kN load point = (7 x 8) - (6 x 4) – (2 x 12)
= 8kNm
BM at Rl = (7 x 10) – (8 x 2) – (6 x 6) – (2 x 14)
= - 10kNm
Shear Force:
SF at left end = -2
SF at Rl = +5
SF at 8m from left end = -1
SF at 12 m from left end = -9
SF at Rl = +2
SF at right end = +2
-10kNm
8kNm
12kNm
-8kNm
-9kN
+2kN
-2kN
-1kN
+5kN
BM Diagram
SF Diagram

9. 4m 12m 2m
3kN/m
Find the Maximum Bending moment for the loading system shown in figure.
Draw the BM and SF diagram for this load system

10. To solve this question we need to remember the following:
The max BM of a simply supported beam with a continuous load is Wl/8
The max BM of a cantilever beam with a continuous load is Wl/2
BM should be seen as positive when the deflection is with concave upward and
negative when the bending or deflection is convex upward.
Concave upward Convex upward
Positive bending
caused by a positive
BM
Positive SF Negative SF
Negative bending
caused by a negative
BM
Convention of Signs for BM and SF

11. TYPES OF BEAMS
The simply supported beam
i.e. beam supported at two
ends is the most commonly
used. There will be continuous
load on beams and there
could be some point or
concentrated loads. Beams with
a number of supports are also
termed as continuous beams.
The figure shows different beams
and and their deflection under
different loading conditions.

12. C BA
4m
D
12m 2m
Overhanging Beams with uniformly distributed load
The Beam shown in the diagram is loaded with a uniformly distributed load
Of 3 kN per m
To understand the method of algebraic adding of BM we need to divide the
beam in 3 parts AC, AB and BD
Taking section CA
BM at A = (3x4) x (4/2) = 24 kNm (-ve)
Taking section BD
BM at B = (3x2) x (2/1) = 6 kNm (-ve)
Taking section AB
BM at center = Wl/8 = {(12x3) x12)} /8 =
54 kNm (+ve)
3kN/m

13. B DAC
C
24kNm
6kNm
54kNm
0
A
B D
0
0 0
Shear Force diagram; First ascertain the reactions
(Rl×12) + (6 ×1) = 3×16×8 or Rl = 31.5kN
(Rr×12) + (12×2) =3×14×7 or Rr = 22.5 kN
Rl + Rr = 54 kn Load = 3×18 = 54kN
From C to A the load on the beam = 12kN hence the SF diagram must drop by this
amount over this length. At A there is there is a vertical jump of 31.5 kN i.e. Rl.
Between and B there is a load of 3×12 = 36 kN. At B there is a vertical jump of
22.5 kN and then till D there is the fall of 3×2=6 kN till D.

14. 6 kN/m
Find the Maximum Bending moment for the loading system shown in figure.
Draw the BM and SF diagram for this load system

15. Beams with Uniformly Distributed Load Partially covering the Span
AD is a simply supported beam loaded as shown.
Calculating the reactions:
Taking moments about Rr
Rl×12 = (6×4×8) or Rl = 16kN
Taking moments about Rl
Rr×12 = (6×4×4) or Rr = 8kN
BM at A = 0
BM at B = 16×2= 32kNm (+ve)
BM at mid point of BC = (16×4) – (6×2×1) = 52kNm
BM at C = (16×6) – (6×4×2) = 48kNm
BM at D = 0
BM from A to B will be a straight line, a curve from
to C and a straight line from C to D.
SF at A will be 16 till B and then slope to -8 at C and
Will be constant till D.
6kN/m