Last Updated on October 5, 2026 by Maged kamel
The theory of pure bending.
Lecture objectives.
Our first item is the theory of pure bending assumptions and the relation. The second Point is why we estimate the Product of Inertia. Next, we will discuss Mohr’s circle of Inertia and how it can be used to calculate the maximum and minimum values of the Moment of Inertia.
Assumptions to follow for the theory of pure bending.
First, we make some assumptions: when a Beam is subjected to pure bending, it deforms.
But first, let’s look at the Beam before deformation. We have two axes, X and Y, and we consider the Beam’s section rectangular before deformation.
We have a section that is vertical and parallel to the neutral axis. We assume the plane section that was parallel to the neutral axis before deformation remains parallel to the neutral axis after deformation.
After deformation, the Beam takes on a curved shape and is no longer vertical. Nonetheless, the section remains perpendicular to the neutral axis after deformation.
The second assumption is that bending causes minimal deformation.
The third assumption is that the Beam is initially straight, with all longitudinal filaments bending into a circular arc with a shared center of curvature.
This means all these perpendicular sections lie in one plane, and any other section is a plane section perpendicular to the neutral axis. The third part will be perpendicular to the neutral axis. If you expand all of these portions, they will also come together in the same place.
The line formed will be a circular arc.
The fourth assumption of pure bending theory is that vertical deformation, or transverse strain (εyy), can be omitted.
Compression is always followed by compression or transverse strain. In our situation, the transverse strain (εyy=dh/h) will be negative. The longitudinal deformation equals the change in length divided by the initial length.
In εyy, dl/l replaces length extension and is positive. We assume it is small and neglect it.
The fifth assumption states that the radius of curvature is larger than the cross-sectional dimension. The sixth assumption states that Young’s Modulus of elasticity is the same in tension and compression.
Pure bending: what is it? Pure bending results in a Shear Value of 0 when the Beam experiences equal moments on both sides.
Why will there be no Shear Value? Because there will be a pair produced by this bending Moment on the right. The couple will be upward because the Shear is anti-clockwise, which also causes a downward Shear force on the right side. At the same Time, it produces an anti-clockwise resisting couple.
As a result, the Shear forces will oppose each other—downward on one side and upward on the other—and they should cancel out.
For the bending Moment diagram, the Shear force will therefore equal zero. This is saggy if we wish to sketch it. A positive Value will be allocated to the appropriate assistance, and this will be closed. That’s this, left to right.
The Beam will experience bending with a positive bending Moment. The left- and right-side moments cause the Beam to bend into an arc, and, as the diagram shows, the enclosed angle between the two perpendicular lines is α.
In the second scenario, which involves pure bending, you have a supported Beam that is hinged and acted upon by two Shear forces of equal Value, P and P, operating on the Beam at L/3 and L/3, respectively.
In the Shear force diagram, w is a positive Shear force that rises and falls until it reaches B, and a negative Shear that moves from right to left and closes at Point D.
Since we have a positive bending Moment (P*L/3), the intermediate section will have a bending Moment with no Shear force between C and D = 0. The pure bending Moment is present in this second instance.
Stress formula for the case of pure bending.
The third supposition states that the Beam is originally straight. As shown, the section is rectangular, and the x- and y-axes were there before deformation.
Consider a section where a distance y separates a small element before deformation, and the section’s distance apart is (dx).
After deformation, the Beam bends, and the x, y, and perpendicular axis 3 all show the neutral axis, which was horizontal. Assume an enclosed arc angle (dα) is to be studied, and the arc radius is R.
After equating, we will determine the strain Value before deformation.
Following deformation, there was no strain.
This section, whose length is (dx), will eventually become Rdα. When the element at adistance Y up is deformed, its length becomes (R-Y) ma distancee by (dα), when it was initially dx.
Thus, the length of this red line was initially dx, or (R-y*dα). It’s moving toward the neutral axis if we compare it.
Since nothing has changed, we can interpret dx as = (Rdα), where the deformation is either compression or tension.
In this case, we are above the neutral axis. Thus, the distortion will turn negative.
After subtracting +Rdα and -Rdα from (R-y)dα-Rdα, we are left with (-ydα). Since strain is deformation over the original length, we can calculate its derivative as (-ydα) over the original length R*dα. Consequently, we have (-y/R), and since stress/strain equals Young’s Modulus of elasticity, our strain equals (-y)/R.
Since there is a strain, the strain * Young’s Modulus of elasticity is given by the formula f= (e E). In other words, f=(-y/R)(E).
From assumption #6, Young’s Modulus of elasticity is the same for tension and compression.
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