SFD & BMD

In this presentation you will get knowledge about shear force and bending moment diagram and this topic very useful for civil as well as mechanical engineering department students.

1. JAHANGIRABAD INSTIUTE OF TECHNOLOGY
BARABANKI
Department of Mechanical Engineering
Elements of Mechanical Engineering
RAVI VISHWAKARMA
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2. Index
1.Types of beam
2.Types of load
3.Shear force Bending Moment
4.Sign conventions
5.Point of contraflexure
6.Example on Simply supported beam
7.Example on cantilever beam
8.Example on overhanging beam
9.Example on overhanging beam with point of contra
flexure
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3. Types of beam
Simply supported beam:-
A beam supported freely on the walls or columns at its
both ends is known as simply supported beam.
Over hanging beam:-
A Beam is freely supported on two supports. But its
one end or both the ends are projected beyond the
support.
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4. Cantilever Beam:-
A Beam Fixed at one end and free at the other end is known
as cantilever Beam.
Fixed Beam:-
A beam whose both ends are rigidly fixed is known as
cantilever beam.
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5. Continuous beam:-
A Beam Supported on more then two support is known as
continuous beam.
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6. Type of loads
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.
W
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7. 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.
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8. 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.
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9. 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
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10. 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.)
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11. 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.
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12. Example on simply supported beam
Problem :-
Calculate the value and draw a bending moment
and shear force diagram for following beam
shown in fig.
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13. Solution
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
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14. 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
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15. Example On Cantilever Beam
Problem :-
Calculate the value and draw a bending moment and
shear force diagram for following cantilever beam
shown in fig.
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16. Solution Problem
The support reaction for
given beam can be easily
determined by following
method.
Dy = (4´2 ´1/2) + (2 ´4)
Dy =10 kN
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17. 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
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18. Example On Overhanging Beam
Problem :-
Calculate the value and draw a bending moment and
shear force diagram for following Overhanging beam
shown in fig.
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19. Solution Problem
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
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20. 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
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21. Example On Overhanging Beam with Point of Contraflexure
Problem :-
Calculate the value and draw a bending moment and shear
force diagram for following beam shown in fig also find
the point of contraflexure.
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22. Solution Problem
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
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23. 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
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
B.M Calculation
MA = 0
MB=(25´4)-(10 ´4 ´2)=
=20kN.m
MC=(25 ´8)-(10 ´4 ´6)
= -40kN.m
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