These Slides will help you to easily understand shear force and shear force diagrams.

3. Shear Force:
Shear force is an internal force in a material which is
usually caused by an external force acting perpendicular to
the material.

4.  When a beam is subjected to applied forces,internal forces develop in the
beam,since the beam has supports and the system is in equilibrium condition.The
beam should resist these internal forces this type of force is called shear force or
simply shear.
Shear Force in Beam:

5.  . We have to know the maximum value of the shear in order to design the beam
properly so that it can resist overall shear and will not fail when it is loaded.
 To calculate the values of shear along the length of the beam so that we can find
out where the beam fails under bending and where the shear value is zero
Objectives in Finding Shear:

6. How to compute shear Force?
 The magnitude of shear force at any section of the beam is equal to the sum of
reactions minus the sum of the loads to the left of the section.
Shear = Reaction – Load
V = R – L
To compute shear force for any section of the beam we simply call
the upward reaction forces positive and downward load forces negative.

7.  Shear diagrams are graphical representations of the shear force variations over the
length of a beam.Shear diagram is drawn directly underneath the free body
diagram of the beam and will show the variation of shear values over the length of
the beam.The following examples (8.9,8.10,8.11) illustrates the construction of
the shear diagram.
Shear Diagram:

8. Example 8.9:
Construct a shear diagram for simply supported beam shown in figure.
800lb 400lb
4ft 10ft 6ft
A B
Solution:
First we have to calculate reactions at point A and B.
800lb 400lb
RA 4ft 10ft 6ft RB

9. RA + RB – 800 – 400 = 0
RA + RB = 1200 (1)
Moment equation with respect to point B.
RA(20) – 800(16) – 400(6) + RB(0) = 0
20RA = 15200
RA = 760lb
By putting the value of RA in Equation 1
760 + RB = 1200
RB = 440lb
Now we also know that
V = R – L
Vx=1ft = 760 – 0 = 760lb
Vx=5ft = 760 – 800 = – 40lb

10. Vx=15ft = 760 – 800 – 400 = – 440lb
800lb 400lb
760lb 4ft 10ft 6ft 440lb
760lb
– 40lb
– 440lb

11. Example 8.10:
A 14-ft beam is loaded with a uniformly distributed load of 400lb/ft. Construct a shear diagram for
this beam as shown in figure.
Solution:
Total Load = 400 × 14 = 5600lb
400lb/ft
A B
14ft

12. RA + RB = 5600lb (1)
5600lb
RA 7ft 7ft RB
Moment with respect to point “A”
RA(0) – 5600(7) + RB(14) = 0
14RB = 39200
RB = 2800lb

13. By putting the value of RB in equation 1
RA +2800 = 5600
RA = 2800lb
Vx = 1ft = 2800 – (1×400) = 2800lb
Vx = 7ft = 2800 – (7×400) = 0lb
Vx = 14ft = 2800 – (14×400) = – 2800lb
400lb/ft
2800lb 2800lb
2800lb
O
– 2800lb

14. Example 8.11:
Construct a shear diagram for the cantilever beam loaded as shown in figure:
10000lb 1500lb/ft
2m 8m
Total load = 1500 × 8 = 12000lb
10000lb 12000lb
2m 4m 4m
R

15. Reaction = 22000lb
V = R – L = 0 – 10000 = – 10000lb
Vx = 3m = Vx = 2 + Vx = 1 = – 10000 – (1500 × 1) = –11500lb
Vx = 10 = –10000 – (1500 × 8) = – 22000lb
10000lb 12000lb
2m 4m 4m
R
–10000lb
–22000lb