1. GUIDED BY :- ANKIT PATEL
SVMIT BHARUCH DEPARTMENT OF MECHENICAL ENGINEERING A Presentation On Shear force banding moment
Prepared by:-
Parmar Amit R
Lunat Salman
Singer Vikas
Patel Dipal
Parmar Keyur
Patel Afzal
GUIDED BY :- ANKIT PATEL

2. index 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.

3. 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.

4. 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.

5. Types of beams 5.Continuous beam:-
A Beam Supported on more then two support is known as continuous beam.

6. 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.

7. 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.

8. 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.

9. 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.

10. 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.

11. 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.

12. 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.

13. Shear Force And Bending Moment
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

14. 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.)

15. 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.

16. 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.

17. 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 = = 66kN

18. 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= = 0kN
B.M Calculation
MA = 0
MC = (6640) – (2042)
= ( )
= 104kN.m
MD =(66 8)-(20 4 6)
=( )
=48kN.m
MB = 0kN.m

19. 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.

20. 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

21. 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 = = -24kN

22. 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.

23. 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

24. S.F Calculation
FA Left = 0kN
FA Right = -10kN
FC = -10kN
FB Left = = 25kN
FD Right = 25kN
FD Right = = 0kN
B.M Calculation
MA = 0kN.m
MC =(-102) =-20kN.m
MB = = -100kN.m
MD = 0kN.m

25. 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.

26. 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

27. 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

28. -: GUIDED BY :- ANKIT PATEL Reference
Mechanics of solids (Atul Prakashan)
Images – Created By Amit Parmar ( )
-: GUIDED BY :-
ANKIT PATEL

29. THE END