MECHANICS THEORY



It is important to understand, and be able to find, the internal moments, shear
force and axial force at any point in a beam structure. These values can then
be used to help design the beam. In this section, the internal loads
(moment, shear and axial) will be calculated for a specific location. The next
section will expand this concept, and develop diagrams showing the internals
at all locations.
Before the internal loads are determined, a few topics need to be reviewed,
specifically, equilibrium equations, boundary conditions, and load types.






Static Equilibrium Equations

Typical Loading on Beam 

Determining internal loads of beam structures requires
a good understanding of static equilibrium equations and how to apply boundary
conditions. For any static equilibrium problem, all forces and moments
in all directions must be equal to zero. This is summarized in equation form
as
Most situations require only the 2D form of these equations, or
ΣF_{x} = 0 ΣF_{y}
= 0 ΣM_{z} = 0






Boundary Conditions

Typical Boundary Conditions 

For typical 2D problems, there are three basic types of
boundary conditions,
fixed, pinned and slider (includes smooth surface). All three have different
reaction forces that need to be used when the support is removed as illustrated
in the diagram at the left.
The Fixed condition has three reactions (2 force and 1 moment) since it cannot rotate, nor displace
in either direction.
The Pinned condition cannot displace but it can freely rotate. This requires
only two reaction forces.
The Slider condition allows the member to move in only one direction but it
can freely rotate. This means there is only one reaction force perpendicular
to the direction of the motion. A special case is when there is a smooth surface
without friction. This allows the member to freely slide in the direction of
the slope.
The diagram uses triangular shapes to represent pinned connections. The triangle
vertex is the pin location. When it is sliding, the triangle has two rollers
to indicate free motion along the surface. Note that the rollers DO NOT indicate
the member can lift off the surface, only sliding motion.






Types of Loads

Types of Loads


Several types of loads can act on beams, such as concentrated loads,
distributed loads, and couples. Concentrated loads are idealized from
loads applied on a very small area. Distributed loads are spread along the
axes of beams. For example, the weight of the beam can be assumed as a
distributed force. Couples are moments applied on the beam.




Equivalent Load for a Distributed Load


Generally, distributed loads are converted into
equivalent forces to make the
solution process easier. For vertical loads, the equivalent force is
The location of the equivalent load is found by

Example: Equivalent Loads for
Uniform Distributed Load 



As illustrated in the example at the left, distributed loads can be split into
separate distributed load. This makes finding the total equivalent force easier
without doing the actual integration.
The location of the equivalent force is through the centroid of the distributed
load. Thus, if the distributed load is a basic shape, the centroid is easy to
determine without integration. In fact, most basic shapes are listed in handbook
tables (see
Sections Appendix).





Sign Conventions

Sign Conventions for Bending Moments,
Shear Forces and Axial Forces 

The sign convention for shear forces and bending moments are not
based on their directions along the coordinates axes. The positive bending moment
tends to compress the upper part of the beam and elongate the lower part. The negative
bending moment tends to elongate the upper part of the beam and compress the lower
part of the beam. The positive shear force
tends to rotate the material clockwise. The negative shear force tends to rotate
the material counterclockwise. And the positive axial force tends to pull the material apart
where as the negative force compresses the material. 




Internal Moments and Forces

Typical Beam with Loads
Support Reactions


The first step in finding the internal loads (moment, shear force, and axial
force) at a point is to determine the reactions at all supports. This is done
by using the three equilibrium equations.
To assist in this task, equivalent forces replace the distributed loads, as
shown in the example at the left. Then, the basic equations give,
ΣM_{A} =
0
F(1.25)  B_{y} (1.5) = 0
[(7.2)(2.5)] (1.25)  B_{y} (1.5) = 0
B_{y} = 15 kN
ΣF_{y} =
0
A_{y} + 15  18 = 0
A_{y} = 3 kN




Beam is Cut, and Internal Loads
(M,V, and A) Replace the
Removed Section 

After the reaction forces are known, the structure is cut at the location where
the internal loads need to be determined. In the example at the left, this is
at the section aa line.
In order to maintain a segment of a beam in equilibrium, there exists an internal
vertical force V at the cut to satisfy the equation in the ydirection. This
internal force, which is vertical to the axis of the beam, is called the shearing
force. It should be noted that the same shear is opposite in direction in the
left and right sections.






Similarly, there exists an internal resisting moment M at the cut to keep a
segment of the beam in equilibrium. This internal moment acts in a direction
opposite to the external moment to satisfy the moment equation.
Sign convention is important at this point. The unknown internal loads
should be drawn as positive loads. Remember, the left and right sections have
opposite positive directions. If the two sections were joined again, M, V, and
A internal loads need to cancel out since there is no external resultant force
at cut location.




Beam is Cut, and Internal Loads
(M, V, and A) Replace the
Removed Section


The internal loads can now be determined by analyzing either beam section using
the basic equilibrium equations. There are three unknowns, M, V, and A, and there
are there equations,
ΣF_{x} = 0
ΣF_{y} = 0
ΣM_{z} =
0






Applying these three equations give,
A = 0
V = 4.2 kN
M = 0.6 mkN
The moment in this example is positive, so the assumed internal moment direction was correct. However, the shear force is negative so the actual shear force is opposite to the direction assumed in the diagram.



