Introduction Types of Beam Types of Loads acting on beam Types of Supports Instrument used for finding “Support Reactions” How to find “Support Reactions”

1. GUJARAT TECHNOLOGICAL
UNIVERSITY
Subject : Mechanics Of Solids
(2130003)
Sem-3 (Chemical) Aug - 2016
Topic : Support Reactions

3. Contents:
Introduction
Types of Beam
Types of Loads acting on beam
Types of Supports
Instrument used for finding “Support
Reactions”
 How to find “Support Reactions”

4. “Support Reactions”
• Support :- A support prevents translation of
a body in a given direction, a force is
developed on the body in that direction.
• Reactions:- The forces and moments
exerted on an object by its supports are called
reactions.

5. Introduction
• When designing buildings, frames, beams it is
important to know the magnitude and
direction of the load that resides within a
support.
• All of these supports can be located anywhere
along a structural element. They are found at
the ends, at midpoints, or at any other
intermediate points. The type of support
connection determines the type of load that
the support can resist. The support type also
has a great effect on the load bearing capacity
of each element.

6. Types of Beam
Beam :- A beam is a structural element that is
capable of withstanding load primarily
by
resisting bending.

7. In engineering, beams are of several types:
1. Simply supported
Beam –
A beam supported on the ends
which are free to rotate and
have no moment resistance.
2.Fixed beam -a beam
supported on both ends and
restrained from rotation.

8. 3) Cantilever :- it has one end fixed and
other end free.

9. 4) Over hanging :– A simple beam extending
beyond its support on one end.

10. 5) Continuous Beam-
A beam extending over
more than two supports.

11. 8) Propped cantilever
Beam

12. Types of Loads acting on Beams
1) Point load(concentrated load)- w1 and w2
are point loads.
the load concentrated at one point is called
point load.
Unit of point load is n or kn.
eg. 20 kn,100kn,60n,etc
w1 w2

13. 2) Uniformly Distributed
Load(U.D.L)- Load uniformly
distributed on certain length of
beam is called uniformly
distributed load.
it is written as u.d.l
unit of u.d.l is kn/m or n/m.

14. 3) Uniformly Varying Load (U.V.L)-
this type of load is gradually increase Or
decrease on the length of the beam. it is also called triangular
load.

15. Types of supports

16. Types of supports
1) Simple support :-
In this type of support,beam is simply
supported on the support.
There is no connection between beam and
support.
At this type of support,only vertical reaction will
be produced.

17. 2) Fix end support :-
Beam is completely fixed at end in the wall or
support.
Beam cannot rotate at end.
Reactions may be vertical,horizontal,inclined and
moment.

18. 3) Roller support:-
In this type of support,rollers are placed below beam and
beam can slide over the rollers.
Reaction will be perpendicular to the surface on which
rollers are supported.
This type of support is normally provided at the end of a
bridge.
Due to breaking forces of vehicles and temperature
forces,bridge slab can slide over the roller support and damage to
bridge pier can be avoided.

19. 4) hinge support:-
Beam and support are connected
by a hinge.
Beam can rotate about the hinge.
Reactions may be vertical,horizontal
or inclined.

20. Methods to calculate
Support Reactions
Analytical Method Graphical Method

21. Q- Determine the support reactions of the following
examples ?
Taking moment @ A :-
vb X 7+ 6x2 = 8x3x5.5
∴ Vb = 17.14 kN
∴ va + Vb = (8 x 3 - 6)
∴ Va = 0.86 kN
∴ Ha = 12 cos30⁰ = 10.39 kN

22. Q- Determine the support reactions of the following
examples ?
Taking moment @ A :-
vb x6 = (12x2x1)+ (30x3)+30+(22.52x5)
∴ Vb = 42.76 kN
∴ va + Vb =
24+30+22.5
∴ Va = 33.76 kN
∴ Ha = 22 cos 60⁰ = 10.39 kN

23. Q- Determine the support reactions of the following
examples ?
Taking moment @ A :-
Vf = (5.1x2.2x10)+ (25x2)-30+(20x0.8)
∴ Vb = 148.2 kN
∴ Va = (10x2.2)+20+25
∴ Va = 67 kN
∴ Ha = 0 kN

24. Q- Determine the support reactions of the following
examples ?
tanӨ =15/8 ∴ Ө = 61.92⁰
Taking moment @ A :-
vb x 2.5 = 150x2.8
∴ Vb = 168 kN
∴ va + Vb = 150
∴ Va = - 18 kN
∴ Ha = 170 cos61.92⁰ = 80 kN

25. Q- If support reactions of the following example is
equal then find load P ?
Taking moment @ A :-
vb x 6 = (16x4x2)+(px7)
∴ Vb x 6 = 128 + 7p ………( a)
∴ va + Vb = 16 x 4 + p ……………..(b) and va =
vb
On comparing (a)&(b) we get ,
∴ Vb = 32 + 0.5 P
∴ P = 16 kN

26. Graphical Method

27. Instrument used for finding
“Support Reactions”

28. References:-
• Wikipedia
• Engineering Mechanics by Beer and
Johnson
• Mechanics of solids by R.P.Rethaliya ,by
M.N.Patel , N.K. Arora
• By Slideshare and Youtube videos
• Researching in google and many more…