This document provides an introduction to bending and shear in simple beams. It discusses the different types of loads, supports, and internal forces that beams experience. Beams can experience bending moments and shear forces from applied loads. Shear and moment diagrams can be constructed to understand the internal forces along the beam. The diagrams are developed using the equilibrium method or semi-graphical method based on the loads, supports, and geometry of the beam. An example problem demonstrates constructing shear and moment diagrams.

1. An Introduction to
Bending and Shear in Simple Beams

2. An Introduction to
Bending and Shear in Simple Beams

3. An Introduction to
Bending and Shear in Simple Beams

4. An Introduction to
Bending and Shear in Simple Beams

5. An Introduction to
Bending and Shear in Simple Beams
Roller Support

6. An Introduction to
Bending and Shear in Simple Beams
Pinned Support

7. An Introduction to
Bending and Shear in Simple Beams
Fixed
Support

8. An Introduction to
Bending and Shear in Simple Beams
Types of
Loads on a
Beam
Concentrat
ed
Uniformly
Distribute
d

9. An Introduction to
Bending and Shear in Simple Beams
Types of
Loads on a
Beam
Nonuniform
ly
Distributed
Load
Pure
Moment

10. Understanding
Bending and Shear in Simple Beams
Internal Forces

11. Understanding
Bending and Shear in Simple Beams
Recap:
Internal Forces

12. Understanding
Bending and Shear in Simple Beams
If the beam is sagging, the top of the beam will get shorter and so
the normal forces in the top of the beam will be compressive. The
bottom of the beam will get longer, and so the normal forces in the
bottom of the beam will be tensile. Each of the normal tensile forces
has a corresponding compressive force which is equal in magnitude
but opposite in direction. As such these forces don’t produce a net
normal force, but they do produce a moment.

13. Understanding
Bending and Shear in Simple Beams
As such these forces don’t produce
a net normal force, but they do
produce a moment. This means
that we can conveniently represent
the internal forces acting on the
beam cross section using just two
resultants, one shear force, which is
a resultant of the vertical internal
forces and one bending moment,
which is a resultant of the normal
internal forces. This is a very
common way of representing
internal forces within a beam.
moment

14. Understanding
Bending and Shear in Simple Beams
Drawing the shear force and
bending moment diagrams is just
figuring out what these internal
forces are at each location of the
beam. These resultant shear forces
and bending moments will depend
on the loads acting on the beam
and the way in which the beam is
supported.
Beams can be loaded in a number
of ways (as previously discussed),
the most common being
concentrated forces, distributed
forces, and concentrated moments.

15. Understanding
Bending and Shear in Simple Beams
Beams can also be supported in a number of
different ways. They can have pinned
supports, roller supports, or fixed supports –
which restrain the beam in different ways.
Pins prevent vertical and horizontal
displacement but allow rotation. Roller
supports prevent vertical displacement but
allow horizontal displacement and rotation.
Fixed supports prevent all displacements and
rotations.
If a certain degree of freedom is restrained at
a support, we will have a corresponding
reaction force or reaction moment at that
location. For example, rotations are permitted
for a pinned support so there is no reaction
moment but displacements in the horizontal
and vertical direction are prevented so we will
have horizontal and vertical reaction forces.

16. Bending and Shear in Simple Beams Summarized
• A beam is a long, slender structural member that
resists loads usually applied transverse
(perpendicular) to its longitudinal axis.
• These transverse forces cause the beam to bend in
the plane of the applied loads.
• Internal stresses are developed in the material as it
resists these loads.
• The design of a beam entails the determination of
size, shape, and material based on the bending
stress, shear stress, and deflection due to the
applied loads.
• Not all beams need to be horizontal; they may be
vertical or inclined.
• Beams may have either one, two, or multiple
reactions.
Overhang roof joist supported by a glu-lam beam

17. Beam Classification Based on Supports - Summary
Beam-Column Framework in Steel

18. 7.2 Shear and Bending Moment
• Beams subjected to a variety of loading conditions, either
singly or in any combination, must resist the loads and
remain in equilibrium.
• For the beam to remain in equilibrium, an internal force
system must exist within the beam to resist the applied
forces and moments.
• Stresses and deflections in beams are functions of the
internal reactions, forces, and moments.
• Shear force in a beam is the internal force produced on a
beam cross section, equal and opposite to the external
force(s) present on that section of beam.
• Bending moment is the internal couple acting on a
beam's cross section equal and opposite to the external
moment acting on the beam section.
• For this reason, it is convenient to “map” these internal
forces and to construct diagrams that give a complete
picture of the magnitudes and directions of the forces
and moments that act throughout the beam length.

19. Load, Shear and Moment Diagrams
• The beam diagrams are referred to as load, shear (V),
and moment (M) diagrams.
• The load diagram is essentially the free-body diagram
of the beam.
• Shear diagrams show the transverse shears along
beam's length.
• A shear diagram is a graph in which the abscissa
(horizontal reference axis) represents distances along
the beam length, and the ordinates (vertical
measurements from the abscissa) represent the
transverse shear at the corresponding beam sections.
• Moment diagrams are graphs of the bending moment
along the beam's length.
• A moment diagram is a graph in which the abscissa
represents distances along the beam, and ordinates
represent the bending moment at the corresponding
sections.

20. Understanding
Bending and Shear in Simple Beams
The internal forces acting on the
beam cross section using just two
resultants, one shear force, which is
a resultant of the vertical internal
forces and one bending moment,
which is a resultant of the normal
internal forces. This is a very
common way of representing
internal forces within a beam.
moment

21. An Introduction to
Bending and Shear in Simple Beams

22. Load, Shear and Moment Diagrams - Sign Convention
• A sign convention is necessary for shear
and moment diagrams if the results
obtained from their use are to be
interpreted conveniently and reliably.
• The shear at a section is considered
positive when the portion of the beam to
the left of the section cut tends to be in the
up position with respect to the portion to
the right of the section cut.

23. Sign Convention - (cont’d)
• The bending moment in a horizontal beam
is positive at sections for which the top
fibers of the beam are in compression and
the bottom fibers are in tension.
• A positive moment generates a curvature
that tends to hold water (concave-upward
curvature), whereas a negative moment
causes curvature that sheds water
(concave-downward curvature).
• The overhang beam exhibits a changing
curvature that results in negative to
positive to negative moments.
• The implication here is that there is a
transverse section(s) in the beam span
where the bending moment
is zero to accommodate the required sign
change.
• Such a section, termed the inflection
point(s) or point of inflection, is almost
always present in overhang and multiple-
span beams.

24. 7.3 Equilibrium Method for Constructing Shear and Moment Diagrams
• In the equilibrium method specific values of V and M
are determined from statics equations that are valid for
appropriate sections of the member.
• A convenient arrangement for constructing shear and
moment diagrams is to draw a free-body diagram
(FBD) of the entire beam and construct shear (V) and
moment (M) diagrams directly below.
• An origin should be selected and positive directions
should be indicated for the coordinate axes.
• Since V and M vary as a function of x along the beam
length, equations for V and M can be obtained from
free-body diagrams of portions of the beam.
• Complete shear and moment diagrams should indicate
values of shear and moment at each section where
they are maximum positive and maximum negative.
• Sections where the shear and/or moment are zero
should also be located.

25. Example Problem 7.1 - Equilibrium Method for V and M

26. Example Problem 7.1 - (cont’d)

27. Basic Curves and Their Properties

28. Load, Shear and Moment Diagrams - Semi-Graphical Method
• General considerations for constructing V and M diagrams.
• 1. When all loads and reactions are known, the shear and moment at the
ends of the beam can be determined by inspection.
• 2. At a simply supported or pinned end, the shear must equal the end
reaction, and the moment must be zero.
• 3. Both shear and moment are zero at a free end of a beam (cantilever beam
or overhang beam).
• 4. At a built-in or fixed-end beam, the reactions are equal to the shear and
moment values.
• 5. Load, shear, and moment diagrams are usually drawn in a definite
sequence with the load diagram on top, followed by the shear diagram
directly beneath it, and the moment diagram below the shear diagram.
• 6. When positive directions are chosen as upward and to the right, a
uniformly distributed load acting down will give a negative slope in the
shear diagram, and a positive distributed load (one acting upward) will
result in a positive slope.

29. Semi-Graphical Method - (cont’d)
• 7. A concentrated force produces an abrupt change in shear.
• 8. The change in shear between any two sections is given by the area
under the load diagram between the same two sections:
• 9. The change of shear at a concentrated force is equal to the
concentrated force.
• 10. The slope at any point on the moment diagram is given by the
shear at the corresponding point on the shear diagram; a positive
shear represents a positive slope and a negative shear represents a
negative slope.
• 11. The rate of increase or decrease in the moment diagram slope is
determined by the increasing or decreasing areas in the shear diagram.
• 12. The change in moment between any two sections is given by the
area under the shear diagram between corresponding sections:

30. Example Problem 7.3
• Construct the V and M diagrams for the girder that
supports three concentrated loads.
• Draw the FBD of the girder and solve for the support
reaction at each end.
• Construction lines should be drawn beneath the FBD at
locations where loads occur.

31. Example Problem 7.3 - (cont’d)
• In plotting the shear diagram, follow the direction for each force
(including the support reactions) and maintain a constant shear
until another concentrated load is encountered.
• The constant shear is a result of having no load areas that can
cause a change in shear.
• Beginning at the left support, the reaction pushes the shear to a
magnitude of +11k.
• No loads occur between A and B so the shear remains constant
until the concentrated load at B pushes the shear down by 5k to -
6k.
• Between B and C, the shear is constant until changed by the
concentrated load of -10k.
• The resulting shear goes from +6k to -4k, reflecting a change in
shear equal to 10k (the concentrated load).
• The shear remains constant at -4k between C and D until the -8k
concentrated load pushes the shear down to -12k.
• No load area exists between D and E, so the shear remains
constant until the support reaction at E brings the shear diagram
back to zero (a check on the condition of equilibrium).

32. Example Problem 7.3 - (cont’d)
• Since the supports for the girder are a hinge and roller, the
moment at the ends will be zero.
• Again, beginning from the left support, the moment
changes from zero at A to the area contained under the
shear diagram between A and B.
• The area is equal to: A=(11k)x(4')=44k-ft.
• The shear area is a zero degree-curve, with a positive area.
• This results in a 1st degree-curve, with a positive slope in
the moment diagram.
• Between B and C, the moment changes by the area
A=(6k)x(6')=36k-ft., with a 1st degree-curve, positive slope.
• The area under the shear diagram between C and D is
equal to 20k-ft., generating a 1st degree-curve with a
negative slope (the shear area is negative).
• The last segment of shear area is 60k-ft., producing a
negative slope (1°) which brings the moment diagram back
to zero (which it needs to be for a roller support).