MECHANICS  THEORY



Introduction



It is important to know how the shear forces and bending moments
vary along the length of a beam that is being designed. Graphs are
used to describe the change of shear forces and moments. These graphs
are called shear and moment diagrams. Employing these diagrams, the maximum
and minimum shear and moment are easily identified and located. 





General Method to Develop Shear and Moment Diagrams

Basic Example to Construct
a Shear and Moment Diagram


Constructing shear and moment diagrams is similar to finding the
shear and moment at a particular point
on a beam structure. However, instead of using an exact location, the location
is a variable distance 'x'. This allows the shear and moment to be a function of
the distance, x.
In theory, this appears to be simple, but the equations can be
complex, especially with distributed loads that are also a function of the location,
x. Also, if there are multiple loads or supports, more than one function must
be developed, i.e. one shear and moment function for each section or span of
the beam.
The general steps for shear and moment diagrams are as follows:
 Identify all information and
draw diagram (similar to all static or mechanics problems).
 Construct a freebodydiagram (FBD) and solve for all unknown reaction
forces.
 Cut the beam (one cut for each beam segment), draw a FBD, and solve for the
unknown M and V at the cut in terms of x.
 Plot the moment and shear equations developed in step 3.
It should be noted that this example is fairly simple with one load type
and simple supports. More complex examples are given below with multiple beam
segments and loads. Also, the sign convention is important and is reviewed
below.






Sign Convention

Positive Internal Moment and Shear


When constructing shear and moment diagrams, the sign convention is important
so viewers will know what direction the beam is bending or shearing. Generally,
it is assumed that a positive moment causes a beam to bend downward as shown
in the diagram. A positive shear will skew the beam with the left side going
up and the right going down, as shown.
Each textbook can have different conventions, so it is important that you check
what directions are positive or negative.




Multiple Beam Sections for
Different Loading Configurations


Notice, all internal moments and shears need to cancel out if the cut section
is reassembled. This means the direction is opposite on the right and left faces
of the cut. 



Multiple Beam Sections (or Segments)


One confusing aspect of developing moment and shear diagrams is the need to
have separate moment and shear functions for each beam segment. This is because
a single function cannot model the moment (or shear) change over a load or support
(it is a discontinuous function).
Each new beam section will have its own moment and shear equations as
a function of the location, x. The diagram at the left gives various beams
and typical loadings.
After each section is cut, then a FBD is drawn for either
side of the beam, just like a single section beam discussed above. Then the
M and V is determined for that cut and plotted.





Example: Multiple Beam Sections



Moment and shear diagrams are best understood by examining an example. The simple
support beam at the left has a single point load between the supports.
The first
step is to draw a FBD of the whole beam and solve for the reactions.






ΣM_{A} = 0
(10 ft) R_{B}  (6 ft)(120 lb) = 0
R_{B} = 72 lb
ΣF_{y} =
0
R_{A}  120 lb + 72 lb = 0
R_{A} = 48 lb






Determine Beam Section
Section (1), to the left of the applied load, will
have an expression for the shear force and bending moment that will differ
from the section (2), to the right of the applied load. Therefore, the sections
must be evaluated separately and each will have their own moment and shear equations.






Section (1) (0 ≤ x ≤ 6 ft)
First, cut section (1) a distance x from the left side and form a FBD, as shown. Remember,
the discarded right beam section needs to be replaced by unknown an internal shear
force and moment, V_{1} an M_{1}.
The
left beam section must stay in static equilibrium.
ΣF_{y} = 0
48 lb  V_{1} = 0
V_{1} = 48 lb
ΣM_{cut} =
0
M_{1}  x (48 lb) = 0
M_{1} = 48x ftlb






Section (2) (6≤ x ≤ 10 ft)
Now the next and last section can be cut. The left or right section can be
discarded. For consistency with section (1), the left side will be examined.
Again, form a FBD, as shown. The internal loads are labeled, V_{2} an
M_{2} to distinguish them from section (1) shear and moment equations.
Again, applying the static equilibrium equations gives,
ΣF_{y} = 0
48 lb  120 lb  V_{2} = 0
V_{2} = 72 lb
ΣM_{cut} =
0
M_{2}  x (48 lb) + (x  6)(120 lb) = 0
M_{1} = 720  72x ftlb
Plot Shear and Moment Diagrams
The functions for V and M for both beam sections can be plotted to give the shear and
moment over the length of the beam. The plots are given at the left.
The location for maximum and minimum shear force and bending moment are easily
found and evaluated.






Complex Distributed Load Example



Distributed
loading
is one of the most complex loading when constructing shear
and moment diagrams. This causes higher order polynomial equations for the shear
and moment equations.
Recall,
distributed loads can be converted to equivalent forces which are easier
to work with. Also, complex, nonuniform distributed loads can be split into
simpler distributed loads and treated separately.






An example is the best way to illustrate how to work with nonuniform
distributed loads. Take a simple cantilever beam with a linear varying distributed
load as shown at the left. This example has only one beam segment, so only one
cut will be needed. 





Cut the beam some distance x from the left. If the right part of the cut beam
is used, then the support reactions at A do not need to be determined (this is
a unique situation).
The distributed load intensity, w_{cut},
is a simple linear relationship between A and B, or
w_{cut} = 200 + (x/20)(500200)
= 200
+ 15x N/cm






The distributed load can be split into two parts, a rectangular and triangular
shape. The equivalent loads, F_{1} and F_{2} of each shape are
calculated as
F_{1} = (20  x) (200 + 15x)
= 4,000 + 100x  15x^{2}
F_{2} = 0.5 (20  x) [500 (200 + 15x)]
= 3,000  300x + 7.5x^{2}
The internal moment and shear, M and V, can now be determined using the equilibrium
equations,
ΣF_{y} = 0
V = (4,000 + 100x  15x^{2}) + (3,000  300x +
7.5x^{2})
V = 7,000  200x  7.5x^{2} N






ΣM_{cut} =
0
0 = M + [(1/2) (20x)] (4,000 + 100x  15x^{2})
+
[(2/3) (20x)] (3,000  300x + 7.5x^{2})
M =  (40,000  1,000x  200x^{2} + 7.5x^{3})

(40,000  6,000x + 300x^{2}  5x^{3})
M = 80,000 + 7000x  100x^{2}  2.5x^{3} Ncm
Both the shear and moment can now be plotted as a function of position, x, to
give the moment/shear diagrams. Since the distributed load was changing linearly,
the shear is a quadratic equation and the moment is a cubic.






Moment and Shear Relationship



From the examples given in this section, it is possible to see an interesting
relationship between the shear and the moment. The shear at a given location
is the slope of the moment function. In mathematical terms, this is written as,
V = dM/dx
This relationship may be helpful in determining both the moment and shear diagram
without calculating the actual function. Derivation of this relationship is done
in the
Integration of Load Equation.
