This presentation is a continuation of our discussion on the shear stress in beams.
In the previous lecture, we derived the shear stress equation
and illustrated its general use in beam analysis.
A link to that lecture is provided in the video description field.
Suppose our beam has a width of 0.2 meters and a height of 0.4 meters.
Assuming we want to calculate the shear stress at the neutral axis,
Q becomes 0.004 cubic meters.
The moment of inertia of the crosssection is:
And t is 0.2 meters.
Therefore, the shear stress at the neutral axis equals 1.87 MPa.
The shear stress at the neutral axis acts along two orthogonal planes:
the plane of the cross-section,
and the plane that lies along the length of the beam.
Consider a small cube having a face along each of the two planes, as shown here.
The state of the shear stress at the beam's neutral axis
can be represented using the cube, like this:
Note that there are two opposing stress arrows on each plane,
keeping the cube in the state of static equilibrium.
If we label the axes of the Cartesian coordinate system this way,
we can see that all the stress arrows acting on the cube lie on the x-y plane.
So, we can think of the stress element as a square in the x-y plane,
subjected to four stress arrows, like this:
This stress element represents a small segment of the beam.
Under the shear stress, the element deforms like this:
We call this shear deformation.
We can use the change in the 90-degree angles of the square as a measure of the shear deformation.
If we denote the right angle of the square as pi/2,
after the element has gone through the shear deformation,
this angle becomes pi over two minus gamma.
And this angle becomes pi/2 + gamma,
where gamma is a measure of the shear deformation, specified in radians.
This measurement scheme is based on the assumption that angle gamma
remains small enough to equal its tangent.
For practical purposes, this is a valid assumption.
A beam made of steel, concrete, or wood is rigid enough to keep the shear strain relatively small;
Gamma does not go above a few degrees.
For example, as shown here, we can see that for a gamma of 5 degrees,
the tangent of the angle is almost equal to the angle itself.
Let us explore the behavior of this stress element in more detail.
Suppose we have a cantilever beam subjected to a large point load.
Consequently, a significant shear force develops in the beam.
Consider a small stress element near the support, as shown here:
We can draw the state of shear stress for this element, like this:
Note: In this discussion, our focus is on the shear stress in the beam;
we are ignoring the effect of the bending stress on our element.
As we increase the magnitude of the applied load, the shear stress increases,
causing an increase in the shear strain being experienced by the element.
We can draw the stress-strain diagram to visualize
the relationship between the shear stress and the shear strain.
The diagram can be constructed by measuring the strain
and calculating the stress at each load increment,
then graphing strain versus stress, as follows.
We can collect and graph data points until the material fails.
So, this is the stress-strain curve for the material that our cantilever beam is made of.
The lower part of the diagram resembles a straight line.
That is, the material behaves linearly in this region.
Generally speaking, structural materials behave linearly
while the stress remains below a certain point.
Below this point, if the load is removed, the member reverts to its undeformed configuration.
We refer to this point as the yield point of the material.
Past the yield point, as the shear stress continues to increase,
the behavior of the material becomes nonlinear.
In this region, most materials behave inelastically.
Beyond the yield point, if the load is removed,
the member does NOT revert to its undeformed shape;
it retains a permanent deformation.
We refer to this peak stress value as the material's shear strength.
The slope of the line representing the elastic region of the diagram is called shear modulus.
It is a measure of the rigidity of the material.
We denote it using the letter G.
For the hypothetical material in this example,
G is approximately equal to 6 MPa,
divided by,
0.01.
Or, 600 MPa.
The mathematical relationship between tau and gamma
forms a basis for calculating the beam's deflection due to the shear force.
Let us see how the expression for the deflection of a typical beam
due to the shear force can be derived using this equation.
The following presentation assumes that you are familiar with
the Virtual Work Method for calculating deflection in beams.
This method was explained in our previous lectures.
You can find the related video lectures in our online course.
See the video description field for the link.
Consider a cantilever beam of length L subjected to a concentrated load.
Suppose the beam has a rectangular cross-section.
The applied load produces two internal forces: a bending moment, and a shear force.
Each internal force contributes to the overall deflection of the beam.
That is, if we refer to the beam's deflection due to the bending moment as Delta-M,
and call the deflection due to the shear force Delta-V,
then, the total deflection can be written as Delta-M plus Delta-V.
We have discussed the calculation of Delta-M in previous lectures.
Herein, we focus solely on Delta-V.
To determine Delta-V, we can use the virtual work method.
According to the virtual work principle, we can write:
the external virtual work done by the applied loads,
equals,
the internal virtual energy stored in the infinitesimal elements forming the beam.
Imagine the beam is composed of a large number of infinitesimal elements like this:
Each element undergoes shear deformation.
The stored energy in each element, therefore,
can be defined as a product of the shear stress acting on the element,
and the resulting shear strain.
It is here that equation: tau equal G times gamma becomes applicable.
Using this equation, we can express gamma in terms of tau.
So, the internal (stored) energy can be written in terms of tau.
And since tau is a function of the shear force (V) and a few section properties,
we can easily calculate the stored energy in each beam element.
Let's examine these calculations in more detail.
To determine Delta-V using the virtual work method,
we need to analyze the beam twice.
We need to analyze the beam under the applied load.
And, we need to analyze it under a virtual unit load
placed at the free end of the beam where the deflection is to be determined.
Delta-V (the real displacement of the beam under the applied load)
and gamma (the shear strain of a typical infinitesimal beam element)
are associated with this loading scenario.
Tau-star (the shear stress in the infinitesimal element due to the virtual unit load)
is obtained from this loading scenario.
We can write the external virtual work as the product of the virtual unit load
and the real displacement (Delta-V).
The virtual internal energy can be written as the sum of the product
of the virtual shear stress and the real shear strain for the beam elements.
This sum can be expressed as the integral of tau-star times gamma over the volume of the beam.
Dividing both sides of this equation by the virtual unit load, we get:
tau-star can be written using the shear stress equation, like this:
Here, V-star is the shear force in the beam due to the virtual unit load.
For the cantilever beam, V-star is constant; it equals the unit load.
So, tau-star simplifies to: Q divided by I times t.
The shear strain for each beam element can be expressed as:
Replacing tau with V Q over I t, we get:
The shear equation for the beam under the applied load can be written as:
If we replace V with P in the equation for gamma, we get:
Therefore, the expression for Delta-V becomes:
Since the beam has a rectangular cross-section, I and t are constant.
The shear modulus for the material (G) is also constant.
So, the virtual work equation simplifies to:
To perform the integration of Q squared,
we define an infinitesimal element on the cross-section of the beam.
We position the element at an arbitrary distance from the beam's neutral axis.
Let's refer to the distance from the element to the neutral axis as y.
Q is the moment of this gray area about the neutral axis.
The height of the area and the distance from the center of the area to the neutral axis
can be expressed in terms of h and y, as shown here.
Therefore, Q (the moment of the gray area about the neutral axis) can be written as:
Substituting Q in the virtual work equation, we get:
The volume integral can be expanded like this:
Since the x-axis runs along the length of the member,
the limits of the integration in the x-direction are: zero and L.
As shown in this diagram,
we have placed the origin of the coordinate system at the center of the beam's cross-section.
Hence, the limits of the integration in the z-direction are:
negative b/2, and b/2.
And, the integration limits in the y-direction are:
negative h/2, and h/2.
Since the moment of inertia of the beam's cross-section equals b times h cubed divided by 12.
And, t equals b, the constant part of this equation simplifies to:
This integral expression evaluates to:
So, the deflection of the beam due to the shear force is a function of the applied load,
the length of the beam,
the area of the beam's rectangular cross-section,
and the shear modulus of the material.
Suppose the beam's cross-section has a width of 0.2 meters, and a height of 0.4 meters.
And, G equals 77 GPa.
So Delta-V can be written as:
If the beam's length is 2 meters,
and the magnitude of the applied load equals 300 kN,
Delta-V evaluates to 0.117 millimeters.
For comparison, let us determine the beam's deflection under the applied load due to bending.
For this specific beam, Delta-M can be written as:
So, the deflection of the beam due to bending equals 3.74 millimeters.
As you can see, Delta-V is significantly smaller than Delta-M.
That is why we typically ignore the effect of the shear force on the beam's deflection.
In this case, the total deflection equals 3.86 millimeters.
So, if we ignored Delta-V, the result would be off by about 3%.
However, the effect of the shear force on the deflection
becomes more significant as the beam becomes shorter in length.
Suppose the length of our beam is 0.5 meters.
This results in a Delta-V of 0.0293 millimeters,
and a delta-M of 0.0585 millimeters.
In this scenario, about 33% of the total deflection is due to the shear force.
Therefore, Delta-V should not be omitted when analyzing this short beam.
We will continue this discussion in future lectures.
See if you can solve the following exercise problems.