Previously, we introduced the notion of stress as force per unit area.
And we presented two stress types:
axial stress and shear stress.
This lecture focuses on bending stress.
This type of stress plays a key role in the design of beams.
Our objectives in this lecture are:
To explain how bending stress develops in a beam.
How bending stress is related to bending moment.
And how to use the equation for bending stress
to design the beam properly.
Consider a beam subjected to a concentrated load.
We can easily calculate the support reactions
and draw the moment diagram for the beam.
Let’s examine a thin slice of the beam at an arbitrary point, like point C.
According to the moment diagram,
the bending moment at C equals 450 kN.m.
Here is the magnified slice of the beam at point C.
The segment is subjected to a positive bending moment of 450 kN.m.
Since we want to develop a general formulation for bending stress,
let's use variable M as the magnitude of the bending moment.
This bending moment causes the beam segment to bend concave up,
like this:
The deformed shape of the segment follows the arc length of a circle.
We refer to the radius of the circle as the radius of curvature.
We use the Greek letter rho to denote this radius.
Let's refer to the length of the beam segment at point C as dx.
Note that the top part of the beam segment is being compressed,
and its bottom part is being stretched.
Therefore, there must be a fiber within the segment
that is neither compressed nor stretched.
We refer to the length of that fiber as dx.
This is the unchanged length of the beam segment.
The axis that runs along this length is called the neutral axis.
Using basic geometry, we can express dx as:
If we let y represent the distance from the neutral axis to an arbitrary fiber.
Then, the length of that fiber, denoted by dx*, can be written as:
Since dx represents the unchanged length of the beam segment,
the change in the fiber’s length becomes: dx* - dx.
The ratio of the change in length to the original length of the beam segment
is called the tensile strain.
We denote it using the Greek letter epsilon.
For a shortened fiber of length dx*, epsilon equals:
According to Hook’s Law, axial stress sigma
can be expressed as the modulus of elasticity of the material (E)
times axial strain (epsilon).
Therefore, the fiber with length dx* is subjected to a compressive force of:
(E)(y)/(rho).
This is the force per unit area of the cross-section,
which we call axial stress.
We can use this equation to determine the axial stress in the beam fibers.
For example, if the distance from the neutral axis
to the top fiber of the beam is labeled c,
then the axial stress in the fiber equals:
The negative sign indicates that the fiber is in compression.
Let’s denote the axial stress in the top fiber of the segment as sigma_c.
Therefore, the compressive stress at the top of the segment can be shown this way:
The axial stress in the other fibers can be determined in a similar manner.
Since this fiber is closer to the neutral axis,
it carries less axial stress than sigma_c.
The closer we get to the neutral axis, the less axial stress the fiber carries.
The fibers below the neutral axis carry tensile stress.
The stress in the fiber increases as we get closer to the bottom fiber of the segment.
Since this is a linear equation in terms of y,
the stress profile for the beam segment is also linear,
as shown here:
Let us examine the beam segment in 3 dimensions.
Here is the end of a fiber in the compression zone.
The fiber is located at a distance of y from the neutral axis.
Therefore, the axial stress in the fiber can be written as:
-(E)(y)/(rho).
The negative sign indicates that the fiber carries compressive stress.
Similarly, the axial stress in a typical fiber in the tension zone
can be shown this way:
Here is a visualization of the compression and tension zones
on one side of the beam segment.
Since the beam is not subjected to any axial load,
the sum of the axial stresses acting at the beam's cross-section must be zero.
That is, the area under the stress diagram in the compression zone
must be equal to the area under the stress diagram in the tension zone.
We can write this equilibrium condition by integrating the expression
for axial stress over the entire area of the beam's cross-section.
This equation states that the algebraic sum of all the stress elements
acting on the beam cross-section must be zero.
Since E and rho are constant for a given beam segment,
we can conclude that the integral of y dA must be zero.
This integral expression is zero
only if the neutral axis passes through the geometric center of the beam’s cross-section.
Therefore, for a rectangular cross-section,
the neutral axis divides the height of the beam into two equal heights.
If we refer to the overall height of the beam as h,
the distance from the neutral axis to the top and bottom fibers becomes h/2.
We refer to this distance as c.
Let us rearrange this equation so that the ratio:
(E) over (rho) can be expressed as: sigma_c/c.
So we can rewrite the stress equation as:
Note that the axial stress in each beam fiber produces a moment about the neutral axis.
The bending moment in the beam segment (M)
equals the sum of the moments produced by the fiber stresses
in the compression and tension zones.
Mathematically, this can be written as:
The integrand here is the moment of a typical stress element about the neutral axis.
sigma_c/c times (y) is the stress,
and y is the distance from the stress location to the neutral axis.
Since this ratio is constant for the beam segment,
to determine M, we need to integrate: y^2 over the area of the cross-section.
This is called the second moment of area,
or the moment of inertia of the cross-section about the axis of bending.
The moment of inertia of a rectangular cross-section about the neutral axis equals
(1/12)(b)(h^3).
In this equation, b is the width, and h is the height of the cross-section.
We use the letter I to refer to this quantity.
Solving this equation for sigma_c, we get:
This equation gives us the compressive and tensile stresses
at the outer fibers of the beam
as a function of the beam’s bending moment and the dimensions of the beam’s cross-section.
This stress profile, which develops in the beam due to a bending moment,
is called bending stress.
In this equation, if we replace c,
the distance from the neutral axis to the outer fiber of the beam segment,
with y, the distance from the neutral axis to an arbitrary fiber,
we get the general form of the bending stress equation.
This equation enables us to determine the stress magnitude in an arbitrary beam fiber
located at a distance y from the neutral axis.
Let us see how we can use this equation for design purposes.
Consider this bridge structure.
The middle span of the structure consists of two identical simply supported beams.
The maximum load on the bridge, the total weight of two heavy trucks,
can be viewed as four concentrated loads.
Therefore, each beam is responsible for carrying two concentrated loads.
Assume each applied load has a magnitude of 150 kilo-newtons.
The loads, four meters apart, are placed at the center of the beam.
Our objective is to design the beam to carry the applied load safely.
By designing the beam, here,
we mean sizing the member so that the magnitude of the bending stress
at the cross-section of the member remains below the allowable stress.
Let us assume that the allowable stress for the material (wood)
used for the bridge is 35 MPa.
That is, if the stress in any wood fiber exceeds 35 MN/m^2,
the beam will not be able to carry the load effectively.
Therefore, we need to determine the size of the beam's cross-section
so that the bending stress at the outer fibers of the beam is less than 35 MPa.
Here is the bending stress equation for the beam.
If we substitute c for y,
the equation gives us the stress magnitude at the top and bottom fibers of the beam's cross-section.
The moment of inertia of the beam's cross-section about the neutral axis is:
Let's assume the width of the beam is 0.3 meters.
So the only design parameter is h, the height of the beam.
We can write the moment of inertia for the beam’s cross-section as:
Substituting h^3/40 for I in the stress equation, we get:
Replacing c with h/2, we get:
Therefore, the maximum stress in the beam can be expressed in terms of:
M (bending moment in the beam) and h (height of the cross-section).
We want this stress magnitude to be less than 35 MPa.
To calculate h using this equation,
we first need to calculate the maximum bending moment in the beam.
Here is the beam’s bending moment diagram.
As shown here, the bending moment reaches its maximum value at the middle segment of the beam.
The magnitude of the maximum moment is 600 kN.m.
Substituting 600 for M in the stress equation, we get:
Solving this equation for h, we get:
Therefore, to ensure that the bending stress in the beam does not exceed the allowable stress,
the beam’s depth needs to be 0.6 meters.
See if you can solve the following exercise problems.