its explains loads , how to select sections in shear force and bending moment diagrams. what are the techniques in drawing these diagrams.

1. “Some useful techniques for solving
shear force and bending moment
diagrams”
Muhammad Bilal
1st year- CED Uet Peshawar

2. Loads
• Concentrated load
Also called point load
• Uniformly distributed
load
It becomes easy to solve
when we convert UDL to a
point load.

3. Things to keep in mind
• Consider this figure
• The area under the SFD
above the x axis should
equal the area between
the x-axis and the SFD
below the x axis. i.e. the
area should sum to zero.
Check this is true in our
example.
• Any points where the SFD
cross the x-axis, will be a
max or min Bending
Moment
• The SFD should always
equal zero at both ends

4. How to select sections/Cuts
We know when we are solving SFD and BMD we select
different sections of beam and calculate shear force
and moment in each section separately. each section
bear different answer because there is different loading
in each section. Consider this (source meriam statics)

5. Calculate reactions
I started with a FBD and
calculated the two
reactions named as Ra and
Rb.
From categories of
equilibrium It is a parallel
force system so we need to
find moment at any point
and a force equation in the
direction of forces.
Such that
Ra= p/3 and Rb=2p/3

6. Section 1
0<x<2L/3
We see that there is only
one force in this first
section which is actually
the reaction force. We get
the shear force and
bending moment
It should be noted that the
shear force P/3 is same
throughout the section but
moment varies with x
which means that it will be
different at different point
in this section

7. Section 2
Section 1 was before P
and section 2 is after
P.
We can see that now
in section 2 the load P
is included so its shear
and moment will be
different from section
1.

8. Section 2 ( another
possibility)
As the total length of
beam is L we can also
select section 2 like
this
2L/3 <X <L
We can see the
answer is same.in this
section the load P is
not included.

9. Shear force
diagram(explanation)
The diagram can be easily
drawn by looking at the
equations which we have
derived.
Now look at SFD In first
section 1 ‘V was P/3’ as you
see in the diagram then there
is no force so it moves on
horizontally when it reaches
Position 2 it experience a
force P opposite in direction
( P/3-P=-2P/3) again no force
is their in the section 3-4 so
horizontally goes no . At 4 it
experience a upward force
equal in magnitude opposite
in direction which push it to 0.

10. Shear moment diagram(explanation)
• And moment For section 1 it was P/3x when x=2L/3
moment becomes 2/9PL which is actually the
maximum moment
• For section 2 it was 2P/3(L-x) when x=L it reaches
zero which can be seen in the moment diagram

11. Things to remember
In case of uniformly distributed load we will always get
a curve in moment diagram ( mostly parabola). If we
take the clockwise moments positive we will get a
negative parabola and vice versa.

12. In most of the cases
• In case of horizontal
line in shear force
diagram we will get
a linear line in
bending moment
diagram
• And in case of linear
line in shear force
diagram we will get
a curve in moment
diagram

13. Still having problems
Please watch
https://youtu.be/iaS5VLX0NFQ?list=LLrHy-
WQKbvgcLmuS5J4Z9qg
https://youtu.be/IMHits0fs-g?list=LLrHy-
WQKbvgcLmuS5J4Z9qg
https://youtu.be/01ig6qgRj14?list=LLrHy-
WQKbvgcLmuS5J4Z9qg