1. 7.2 Shear and Moment
Equations and Diagrams
Beams – structural members designed to support
loadings perpendicular to their axes
Beams – straight long bars with constant cross-
sectional areas
A simply supported beam is pinned at one end
and roller supported at
the other
A cantilevered beam is
fixed at one end and free
at the other

2. 7.2 Shear and Moment
Equations and Diagrams
For actual design of a beam, apply
- Internal shear force V and the bending moment
M analysis
- Theory of mechanics of materials
- Appropriate engineering code to determine
beam’s required cross-sectional area
Variations of V and M obtained by the method of
sections
Graphical variations of V and M are termed as
shear diagram and bending moment diagram

3. 7.2 Shear and Moment
Equations and Diagrams
Internal shear and bending moment
functions generally discontinuous, or their
slopes will be discontinuous at points
where a distributed load changes or where
concentrated forces or couple moments
are applied
Functions must be applied for each
segment of the beam located between any
two discontinuities of loadings
Internal normal force will not be
considered

4. 7.2 Shear and Moment
Equations and Diagrams
Load applied to a beam act
perpendicular to the beam’s axis and
hence produce only an internal shear
force and bending moment
For design purpose, the beam’s
resistance to shear, and particularly to
bending, is more important than its
ability to resist a normal force

5. 7.2 Shear and Moment
Equations and Diagrams
Sign Convention
To define a positive and negative shear
force and bending moment acting on the
beam
Positive directions are denoted by an
internal shear force that causes clockwise
rotation of the member on which it acts
and by an internal moment that causes
compression or pushing on the upper part
of the member

6. 7.2 Shear and Moment
Equations and Diagrams
Sign Convention
A positive moment
would tend to bend the
member if it were
elastic, concave upwards
Loadings opposite to the
above are considered
negative

7. 7.2 Shear and Moment
Equations and Diagrams
Procedure for Analysis
Support Reactions
Determine all the reactive forces and
couple moments acting on the beam’
Resolve them into components acting
perpendicular or parallel to the beam’s
axis

8. 7.2 Shear and Moment
Equations and Diagrams
Procedure for Analysis
Shear and Moment Reactions
Specify separate coordinates x having an origin
at the beam’s left end and extending to regions
of the beams between concentrated force and/or
couple moments or where there is no continuity
of distributed loadings
Section the beam perpendicular to its axis at
each distance x and draw the FBD of one of the
segments

9. 7.2 Shear and Moment
Equations and Diagrams
Procedure for Analysis
Shear and Moment Reactions
V and M are shown acting in their positive sense
The shear V is obtained by summing the forces
perpendicular to the beam’s axis
The moment M is obtained by summing
moments about the sectioned end of the
segment

10. 7.2 Shear and Moment
Equations and Diagrams
Procedure for Analysis
Shear and Moment Diagrams
Plot the shear diagram (V versus x) and the
moment diagram (M versus x)
If computed values of the functions describing V
and M are positive, the values are plotted above
the x axis, whereas negative values are plotted
below the x axis
Convenient to plot the shear and the bending
moment diagrams below the FBD of the beam

11. 7.2 Shear and Moment
Equations and Diagrams
Example 7.7
Draw the shear and bending moments
diagrams for the shaft. The support at A is a
thrust bearing and the support at C is a
journal bearing.

12. 7.2 Shear and Moment
Equations and Diagrams
Solution
Support Reactions
FBD of the shaft

13. 7.2 Shear and Moment
Equations and Diagrams
Solution
+ ↑ ∑ Fy = 0;V = 2.5kN
∑ M = 0; M = 2.5 xkN .m

14. 7.2 Shear and Moment
Equations and Diagrams
Solution
+ ↑ ∑ Fy = 0;2.5kN − 5kN − V = 0
V = −2.5kN
∑ M = 0; M + 5kN ( x − 2m) − 2.5kN ( x) = 0
M = (10 − 2.5 x)kN .m

15. 7.2 Shear and Moment
Equations and Diagrams
Solution
Shear diagram
internal shear force is always
positive within the shaft AB
Just to the right of B, the shear
force changes sign and remains
at constant value for segment
BC
Moment diagram
Starts at zero, increases linearly
to B and therefore decreases to
zero

16. 7.2 Shear and Moment
Equations and Diagrams
Solution
Graph of shear and moment
diagrams is discontinuous at
points of concentrated force
ie, A, B, C
All loading discontinuous are
mathematical, arising from
the idealization of a
concentrated force and
couple moment

17. 7.2 Shear and Moment
Equations and Diagrams
Example 7.8
Draw the shear and bending diagrams for
the beam.

18. 7.2 Shear and Moment
Equations and Diagrams
Solution
Support Reactions
FBD of the beam

19. 7.2 Shear and Moment
Equations and Diagrams
Solution
Distributed loading acting on this
segment has an intensity of 2/3 x at
its end and is replaced by a resultant
force after the segment is isolated as
a FBD
For magnitude of the
resultant force,
½ (x)(2/3 x) = 1/3 x2

20. 7.2 Shear and Moment
Equations and Diagrams
Solution
Resultant force acts through the centroid of the
distributed loading area, 1/3 x from the right
1 2
+ ↑ ∑ Fy = 0;9 − x − V = 0
3
 x2 
V =  9 − kN

 3 
1 2 x 
∑ M = 0; M + x   − 9 x = 0
3 3
 x3 
M =  9 x − kN .m

 9 

21. 7.2 Shear and Moment
Equations and Diagrams
Solution
For point of zero shear,
3
x
V = 9− = 0
3
x = 5.20m
For maximum moment,

M max =  9(5.20 ) −
(5.20) kN .m
3

 9 
 
= 3.12kN .m