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INDEX
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Types of beams
Types of loads
Types of support
Shear force Bending Moment
Sign conventions.
Point of contraflexure
Example on Simply supported beam
Example on cantilever beam
Example on overhanging beam
Example on overhanging beam with point of contra
flexure.
BEAMS

Members that are slender and support loads applied
perpendicular to their longitudinal axis.
Distributed Load, w(x)
Concentrated Load, P
Longitudinal
Axis
Span, L
TYPES OF BEAMS

Depends on the support configuration
FH
Pin
FH
FV
FV
Fixed
Roller
M
Fv
Roller
FV
Pin
FV
FH
STATICALLY INDETERMINATE BEAMS
Continuous Beam
Propped Cantilever
Beam

Can you guess how we find the “extra” reactions?
INTERNAL REACTIONS IN BEAMS

At any cut in a beam, there are 3 possible internal
reactions required for equilibrium:
normal force,
 shear force,
 bending moment.

P
a
b
L
INTERNAL REACTIONS IN BEAMS

At any cut in a beam, there are 3 possible internal
reactions required for equilibrium:
normal force,
 shear force,
 bending moment.

M
Left Side of Cut
N
V
Pb/L
x
Positive Directions
Shown!!!
INTERNAL REACTIONS IN BEAMS

At any cut in a beam, there are 3 possible internal
reactions required for equilibrium:
normal force,
 shear force,
 bending moment.

M
V
Positive Directions
Shown!!!
Right Side of Cut
N
Pa/L
L-x
FINDING INTERNAL REACTIONS

Pick left side of the cut:
Find the sum of all the vertical forces to the left of the cut,
including V. Solve for shear, V.
 Find the sum of all the horizontal forces to the left of the
cut, including N. Solve for axial force, N. It’s usually, but
not always, 0.
 Sum the moments of all the forces to the left of the cut
about the point of the cut. Include M. Solve for bending
moment, M


Pick the right side of the cut:

Same as above, except to the right of the cut.
EXAMPLE: FIND THE INTERNAL REACTIONS
AT POINTS INDICATED.
ALL AXIAL FORCE
REACTIONS ARE ZERO. POINTS ARE 2-FT
APART.
P = 20 kips
1
2
3
4
5
8
9
10
6 7
8 kips
12 kips
12 ft
20 ft
Point 6 is just left of P and Point 7 is just right of P.
P = 20 kips
1
2
3
4
5
8
9
10
6 7
8 kips
12 kips
12 ft
20 ft
8 kips
V
x
(kips)
-12 kips
M
(ft-kips)
16
32
48
64
80
96
72
48
24
x
V & M Diagrams
8 kips
P = 20 kips
12 kips
12 ft
20 ft
8 kips
V
x
(kips)
What is the slope
of this line?
96 ft-kips/12’ = 8 kips
M
(ft-kips)
-12 kips
96 ft-kips
b
What is the slope
of this line?
-12 kips
a
c
x
V & M Diagrams
8 kips
P = 20 kips
12 kips
12 ft
20 ft
8 kips
V
x
(kips)
What is the area of the
blue rectangle?
96 ft-kips
-12 kips
96 ft-kips
b
What is the area of
the green rectangle?
-96 ft-kips
M
(ft-kips)
a
c
x
DRAW SOME CONCLUSIONS
The magnitude of the shear at a point equals the
slope of the moment diagram at that point.
 The area under the shear diagram between two
points equals the change in moments between
those two points.
 At points where the shear is zero, the moment is
a local maximum or minimum.

The Relationship Between Load, Shear and Bending
Moment
w( x )  the load function

M( x )   V( x )dx
V( x )  w( x )dx
Common Relationships
0
Constant
Linear
Constant
Linear
Parabolic
Linear
Parabolic
Cubic
Load
Shear
Moment
Common Relationships
Load
0
0
Constant
Constant
Constant
Linear
Linear
Linear
Parabolic
M
Shear
Moment
12 kN
8 kN
A
C
D
B
1m
RA = 7 kN 
3m
1m
RC = 13 kN 
12 kN
8 kN
A
C
D
B
1m
3m
8
7
V
1m
8
7
-15
(kN)
-5
7
M
(kN-m)
2.4 m
-8
TYPES OF BEAMS
1.Simply supported beam:A beam supported freely on the walls or columns at
its both ends is known as simply supported beam.

2.Over hanging beam:A Beam is freely supported on two supports. But its
one end or both the ends are projected beyond the
support.

TYPES OF BEAMS
3.Cantilever Beam:
A Beam Fixed at one end and free at the other end is
known as cantilever Beam.

4.Fixed Beam:A beam whose both ends are rigidly fixed is known as
cantilever beam.

TYPES OF BEAMS
5.Continuous beam:A Beam Supported on more then two support is known
as continuous beam.

TYPE OF LOADS
1. Point load or concentrated load:When a load is acting on a relatively small
area it is considered as point load or concentrated
load.
 W = Point Load
 It is given in N or KN.
TYPES OF LOADS
2. Uniformly Distributed Load:A Load which is spread over a beam in such
a manner that each unit length of beam is loaded to
the same intensity is known as uniformly
distributed load.
 W = U.D.L
 It is given in N/m or KN/m.
TYPES OF LOADS
3.Uniformly Varying Load:A Load which is spread over a beam in such
a manner that its intensity varies uniformly on
each unit is called uniformly varying load.
 When load is zero at one end and increases
uniformly to the other end it is known as
triangular load.

TYPES OF SUPPORT




1. Simple Support:Beam is freely supported on the support.
There is no monolithic construction between the beam
and the support.
Only vertical reaction can devlope at the support.
TYPES OF SUPPORT
2. Hinge Support: Beam is hinged to the support at end
 Beam can rotate about the hinge
 Vertical reaction (V) and horizontal reaction (H)
can developed at the support.

TYPES OF SUPPORT
3. Fixed Support :End of beam is rigidly fixed or built in wall.
Beam can not rotate at the end.
 Vertical reaction (V), Horizontal Reaction (H),
Moment (M) can devlope at support.

TYPES OF SUPPORT
4. Roller Support :Beam is hinged to the support at the end roller
are provided below the support.
 End of beam can move on rollers. Only vertical
reaction (V) normal to the plane of roller can devlope.

SHEAR FORCE AND BENDING MOMENT
Shear Force:It is the algebraic sum of the vertical forces
acting to the left or right of a cut section along the span
of the beam
 Unit of S.F is N or kN

Bending Moment:It is the algebraic sum of the moment of the
forces to the left or to the right of the section taken
about the section
 Unit of B.M is N.m or kN.m

SIGN CONVENTION
•
The Shear Force is positive if it tends to rotate the
beam section clockwise with respect to a point inside
the
beam
section.
•
The Bending Moment is positive if it tends to bend
the beam section concave facing upward. (Or if it
tends to put the top of the beam into compression and
the bottom of the beam into tension.)
POINT OF CONTRAFLEXURE
In B.M diagram, The point at which B.M change
its sign from positive to negative or negative to positive
is called point of contraflexure.
It is a point where the beam tends to bend in
opposite direction. It is the point at which curvature of
beam changes.
EXAMPLE ON SIMPLY SUPPORTED BEAM
Problem 1 :Calculate the value and draw a bending
moment and shear force diagram for following
beam shown in fig.

SOLUTION PROBLEM 1

Solution:-
The beam has two unknown
reaction Ay and By.
Ay + By = 90……… (1)
Taking moment from Ay
By10 = (20 42)+(10 8)
By = 240/10
By = 24kN Ay = 90-24 = 66kN

S.F Calculation
FA left = 0
FA Right = 66kN
FC = 66-(204) = -14kN
FD left= -14kN
FD Right=(-14)-10=-24kN
FB Left = -24 kN
FB Right= -24 + 24 = 0kN

B.M Calculation
MA = 0
MC = (6640) – (2042)
= (264-160)
= 104kN.m
MD =(66 8)-(20 4 6)
=(528-140)
=48kN.m
MB = 0kN.m
EXAMPLE ON CANTILEVER BEAM

Problem 2 :-
Calculate the value and draw a bending
moment and shear force diagram for following
cantilever beam shown in fig.
SOLUTION PROBLEM 2


The support reaction
for given beam can be
easily determined by
following method.
Dy = (42 1/2) + (2
4)
Dy =10 kN

S.F Calculation
FA Left = 0kN
FA Right= -2 kN
FB = -2-0 = -2kN
FC = -2-8 = -10
FD = -10

B.M Calculation
MA = 0
MB = -2  1 = -2kN.m
MC = (-2 3) – (4 2 1)
= -14kN.m
MD = -14-10 = -24kN
EXAMPLE ON OVERHANGING BEAM

Problem 3 :-
Calculate the value and draw a bending
moment and shear force diagram for following
Overhanging beam shown in fig.
SOLUTION PROBLEM 3

Applying equation of
static equilibrium.
Ay - By = 45kN
Taking Moment Of A
By6 =(2510)+(5 4 4)
By = (250+80)/6
= 55kN
Ay = -10Kn

S.F Calculation
FA Left = 0kN
FA Right = -10kN
FC = -10kN
FB Left = -10-35 = 25kN
FD Right = 25kN
FD Right = 25-25 = 0kN

B.M Calculation
MA = 0kN.m
MC =(-102) =-20kN.m
MB = -60-40= -100kN.m
MD = 0kN.m
EXAMPLE ON OVERHANGING BEAM WITH
POINT OF CONTRAFLEXURE

Problem 4 :-
Calculate the value and draw a bending
moment and shear force diagram for following
beam shown in fig also find the point of
contraflexure.
SOLUTION PROBLEM 4

The support reaction
can
be
find
as
following
Ay+Cy=(104)+20
=60kN
Taking moment from A
Cy8=(1042)+(20 10)
Cy=280/8=35kN
Ay=60-35 = 25kN

S.F Calculation
FA left=0
FA right=25kN
FB=25-(10 4)= -15 kN
FC left= -15 kN
FC right=25-5=20kN
FD left=20kN
FD right=20-20=0kN

B.M Calculation
MA = 0
MB=(254)-(10 4 2)=
=20kN.m
MC=(25 8)-(10 4 6)
= -40kN.m
Point of contra-flexure can be determined as writing the
equation for BM and put part BC=0
Mx = 25x – 10*4(x-2) = 5.33 so
X = 5.33 from A
THANK U