* Given: Maximum bending moment the beam can resist = 22 kNm * Span of beam = 5 + 5 + 4 = 14 m * Point loads = 1 + 2 + 1 = 4 kN * To find: Maximum uniform load (w) the beam can carry * Bending moment at center due to point loads = (1 × 2.5) + (2 × 2.5) + (1 × 1.5) = 7.5 kNm * Bending moment at center due to uniform load = wl^2/8 = w × (14)^2/8 = w × 98 kNm * Total bending moment should not exceed 22 kNm 7.5 + w

1. STRUCTURAL DESIGN I
LECTURE 6
BENDING MOMENT AND SHEAR
FORCE

2. 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 their deflection under
different loading conditions.

3. • Consider section XX at the cantilever shown in figure. If this
cantilever is saw cut at XX the portion on the right would
collapse.
• This failure would be because of two reasons:
– The load W produces a moment at section XX which the
cantilever, when cut , can not balance
– The load W is able to cause a vertical sliding at the section
when the shear resistance there is destroyed by the saw
cut.

4. It is important to realize that to prevent failure of beam as at
section XX there is the need to independently counterbalance
these two failure tendencies. The picture shows an experimental
model proving this fact.
 If the pull in the vertical string be reduced the cantilever fails.
The tension in the chains and the thrust in the horizontal bars
do not assist vertical equilibrium.
 If either the chains or the bar be removed collapse takes
place. The pull in the vertical string is not able to maintain
equilibrium.

5. • A further experimental result should
be carefully noted. If an additional
load of W1 is added to the detached
portion of the cantilever , W1
weight(load) will have to be placed in
the scale pan to preserve equilibrium.
But, the exact position in which W1
placed on the cantilever is quite
immaterial provided it is placed on
the right of the vertical string.

6. Presuming the gap at XX to be very small
and at the left of the
detached portion:
• Bending moment at section XX
– The load W and the string tension
S being equal and unlike parallel
forces acting out of line forma
couple. It is the moment of this
couple which causes the bending
off tendency of failure at section
XX
• Shear Force at section XX
– The load W the resultant force
tending to cause vertical shearing
at XX is termed as shear force at
the section.

7. • To understand the complete structural design of
beams it will be worthwhile to first investigate in
detail the magnitude of bending moment and shear
force in different loading conditions.
• Subsequently, it will be necessary to understand the
behavior of the internal fibers or particles of the
materials that make up the beams and how these
resist these two forces that will help in the
understanding of the structural design of beams.
• So, on to a detailed study of bending moments and
shear force.

8. Definitions
• Bending Moment (BM)
– The bending moment at any section of a beam is the
resultant moment about that section of all the forces
acting on one side of the section.
• Shear Force
– The bending moment at any section of a beam is the
resultant moment about that section of all the forces
acting on one side of the section
Note: Firstly it should be decided which side of the given beam Section is
going to be taken. Whichever side is taken the answers will be the same. In
cantilevers it is better to take the free side. SF values are algebraic sum of the
vertical forces therefore their position does not matter provided these are on
one side of the section

9. Concave upward Convex upward
Positive bending
caused by a positive
BM
Positive SF Negative SF
Negative bending
caused by a negative
BM
Shown above are the two types of curvature that a given bending moment
may impose on a beam. These are distinguished by employing ‘plus’ or
‘minus’ signs.
Shear force may cause the section to slide either upward or downward.
If the section to the right sides downward it will be positive SF and if the
section to the right upward it will be negative SF.
Convention of Signs for BM and SF

10. BM and SF Diagrams
It is sometimes necessary to draw a graph showing the variation of the
bending moment along the span of the beam. Such a graph is known as
bending moment diagram. A BM diagram has two different scales:
i) a linear scale for the span(e.g. 1cm = 1m)
ii) A BM scale for the vertical coordinate (e.g. 1cm = 1Knm)
In the same way the SF diagram is constructed to linear and force scales.

11. Ignore this
load
X
X
60cm
3kN 1kN
F
90cm
B
Treat FB as a simple lever
with an imaginary fulcrum at F
Calculation of BM and SF at section XX of the cantilever shown in figure:
Considering the forces on the right of the section XX, completely ignoring the
Forces on the left and presuming an imaginary fulcrum at F, FB can be treated
as a lever hinged at F.
Resultant moment about F = (1kN x 150cm) + (3kN x 60 cm)
= 330 kN
Ending moment values are expressed in the usual units of ‘moments’
The resultant vertical force, considering all the forces
which lie to the right of XX = 3kN + 1 Kn = 4kN
Therefore SF at section XX = 4kN
Shear Force values are expressed as force units.

12. Cantilever with single point load at free end
Assume a cantilever AB , fixed at A and free at B with a point load of 1kN at B
1kN
3mA B
w
l
At B the free end : BM = 1kN x 0m = 0 kNm
At 1 m from B: BM = 1kN x 1m = 1 kNm
At 2 m from B: BM = 1kN x 2m = 2 kNm
At 3 m from B or at A: BM = 1kN x 3m = 3 kNm
So BM diagram can be plotted as:
1kNm
2kNm
3kNm
wlkNm

13. SF Diagram
1kN
3mA B
w
l
Just off the beam at B SF = load to right of section = 1kN
At 1 m from B SF = load to right of section = 1kN
At 2 m from B SF = load to right of section = 1kN
At 3 m from B SF = load to right of section = 1kN
So SF diagram:
Base line
1kN l
The shear force will be 1kN for every section of the cantilever. Expressed in
symbols SF = l. The SF graph will be straight line parallel to the base. The BM
diagram is drawn below the base line and the SF above the base line according
to the normal convention
Shear Force diagram

14. When a beam is loaded by a number of loads or different type of loads then,
it may be simpler to calculate the BM and SF separately and then add these
mathematically
w2
l1
w1
l2
l2
l1
l1
w2
W2 x l1
W1 x l2
(W2 x l1) + (W1 x l2)
w2
l1
w1
l2
W1
W2
W1 + W2
Bending Moment Diagram Shear Force Diagram

15. 1kN
3m
2kN
1m
3kN
Construct the BM and SF diagrams of the cantilever shown below.
Ignore the self weight of the cantilever

16. Cantilever with Uniformly Distributed Load
• The cantilever shown in figure carries a
continuous load of 2 kN/m. To find the
BM at any particular section of the
cantilever imagine a fulcrum at the
section and the portion to the right to be
a simple lever.
• For the purpose of taking simple
moments a uniformly distributed load
may be regarded at the center of its
length i.e. its center of gravity.
• Let x = distance from B of atypical section
of the beam.
• When x = 1m(A – A 1 ) a load of 1 x2 kN
lies to the right of the section. The load
can be presumed to be acting at a
distance of .5m from the section, hence
the BM at the section = 2x.5 kNm =
1kNm.

17. • When x = 2m BM=[(2x2)kN x 1m ] =
4kNm
• When x = 3m BM=[(2x3)kN x 1.5m ] = 9
kNm
• When x = 4m BM=[(2x4)kNx 2m ] =
16kNm
• The x values are in a ration of 1:2:3:4
• The BM values are 1:4:3:16 i.e. 1²: 2²:
3²: 4²
• This means that the BM values varies as
the square of the distance from B. The
BM graph therefore a parabola – a
curve satisfying the given conditions.
• If w = load per unit run of the cantilever
of length l the total load carried by the
cantilever is wl = W. BM MAX = W x l/2 =
Wl/2 or wl²/2.

18. Beams simply supported at each end
• In the case of simply supported beams the definition given
earlier can be simplified to:
Bending moment = Reaction moment – load moments.
• The reaction and loads need to be taken to one side of the
section.
Shear Force = left end reaction – sum of loads up to the
section
• If the left sign of section be always taken for SF calculations
the correct sign will be obtained for SF values

19. Simply supported beam with anon central point load
• The usual first step in all beam
problems is to calculate the support
reactions
• Taking moments about B:
Ra x l = W x b or Ra = Wb/l
• Taking moments about A
Rb x l = W x a or Rb = Wa/b
• Consider portion AC of the beam
At 1 m from end A, BM = (Wb/l) x 1
At 2 m from end A, BM = (Wb/l) x 2
At 3 m from end A, BM = (Wb/l) x 3
• The BM value increases in direct
proportion with the distance from the
end A and at the point C the value is
(Wb/l) x a or Wab/l
W
R b
BA
l
a b
R a
C
a b
Wab/l

20. A
l
a b
R a
C
R b
R a
W
Between points A and C , SF = Ra – 0 = Ra
SF= Wb/l and will be positive as per convention
Between the points C and B
SF = Ra – W = Wb – W = Wb - W = Wb – Wa - Wb
l a + b a + b
SF = -Wa = -Rb
l
So the Sf diagram will be as shown in figure

21. A simply supported beam of 9 m effective span carries a concentrated load
of 6 kN at 3m from the left support. Construct the BM and SF diagrams.

22. 9m
3m
4kN
4kN
12kNm
2kN
6kN
2kN

23. SIMPLY SUPPORTED BEAM WITH SEVERAL POINT LOADS
8kN 3kN
1.5m 3.5m 1m
Taking moments about the right end
Rl x 6 = (8 x 5.5) + (3 x 1) = 39; Rl = 6.5
Taking moments bout the left end
Rr x 6 = (3 x 5) + (8 x 1.5) = 27; Rr = 4.5
BM at left end = 6.5 x 0 = 0
BM at 1.5 m from left end = 6.5 x 1.5 = 9.75
BM at 5 m from left end = (6.5 x 5) - (8 x 3.5)
= 4.5
SF just right of left end = 6.5
SF at 1.5 m from left end = 6.5 – 8 = -1.5
SF at 5 m from left end = -1.5 -3.5 = -4.5
9.75 4.5
BM Diagram
6.5
-1.5
-3SF Diagram

24. SIMPLY SUPPORTED BEAM WITH UNIFORM LOAD
BENDING MOMENT
Taking the case of a beam of effective span l with a
load of w kN per meter run:
Total load = wl. Let this be W
Rl = (wl x l/2) = Wl/2
BM at distance x from left reaction = Wl/2 – (wx x x/2)
If this value is plotted it will be a parabola with max
value at the center
BM Max = Wl/2 – Wl/4 = Wl/8
SHEAR FORCE
The shear force at a typical section
= Left end reaction – Load up to that section
At x m from A
SF = wl/2 – wx.
This expression indicates that SF will be linear in nature
When x = 0 (at A) SF = wl/2 – 0 = wl/2 = W/2
When x = l/2 (at mid section) SF = wl/2 – wl/2 = 0
When x = l (at B) SF = wl/2 – wl = -wl/2 = -W/2

25. Calculate the total uniformly distributed the beam in the figure can carry,
in addition to the point loads shown if the beam were capable of resisting
safely a bending moment of 22kNm.
4m
1kN 2kN 1kN
5m 5m 4m