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An analysis of the simple theory of bending, including moments of area(inertia)

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Bending Theory

If a length of beam is acted upon by a constant bending moment (zero shear force), the stress set up on any cross section must constitue a pure couple equal and opposite to the bending moment. Hence it can be deduced that one part of the cross section is in tension whilst the other part is in compression. It is clear that for an initially straight beam the inside edge will be in compression and the outside in tension. There will be an intermediate surface where the stress will be zero and this is called the neutral axis

In the following analysis it has been assumed that:

1) The material is homogeneous, isotropic and has the same value for Youngs Modulus in both tension and compression.
2) The beam is initialy straight and all longitudinal filaments bend into circular arcs with a common centre of curvature.
3) Transverse cross-sections remain plane and perpendicular to the neutral surface after bending.
4) The radius of curvature is large compared the dimensions of the cross section.
5) The stress is purely longitudinal and local effects near concentrated loads are neglected.
6) That the stresses are within the elastic limit of the material and no permanent distortion of the material takes place.

The above diagram shows a beam which is subject to a bending force. If the beam is to bend, there is only one way in which it can happen. If we examine a small bit under stress it will look like this.

Of course it can still be joined up to the sections next to it because the plane sections remain plane.

Now consider a small section of the beam under a bending stress.

Note that all distance are measured from the "Neutral Axis"

Consider what has happened to an element of the beam which is a distance y from the neutral axis. From the above diagram it can be seen that the length has been "stretched" so the strain at y from the neutral axis is given by:-

$Strain\;\;=\;{\frac{length\;-\;original\;length}{original\;length}$

(1)

$=\;\frac{\theta (R\;+\;y)\;-\;\theta \, R}{\theta \, R}$

(2)

$=\;\frac{y}{R}$

(3)

$But\;\;\;\frac{stress}{strain}\;=\;E\;\;\;(Young's\;Modulus)$

(4)

$\therefore\;\;\;\;stress\;=\;\frac{y}{R}\;E$

(5)

From this it is apparent that $\frac{E}{R}$ is constant for any given bending moment. Stress is directly proportional to the distance from the Neutral Axis. |Therefore in the interests of both weight reduction and economy, material is concentrated at the greatest possible distance from the neutral axis. Hence the universal adoption of "I" section steel girders.

If $\delta \, A$ is an element of cross section at a distance y from the neutral axis XX then for pure bending thee net normal force on the cross section must be zero.

$Thus\;\;\;\;\;\int f\,.\delta \, A\;=\;0$

(6)

$or\;\;\;\;\;\frac{E}{R}\int \: y\,.\delta \, A\;=\;0$

(7)

This is the condition that XX passes through the centroid of the section.

Taking moments about the Neutral axis.

$M\;=\;\int f\,y\,\delta \, A$

(8)

$\therefore\;\;\;M\;=\;\frac{E}{R}\int y^2\,\delta \, A$

(9)

This is usually written as in terms of I where $I\;=\;\int \, y^2\: \delta \, A$ and is called the moment of inertia of the section or second moment of area.

$Hence\;\;\;\;\M\;=\;\frac{E}{R}\;I$

(10)

Equations (5) and (10) can now be combined into:-

$\frac{f}{y}\;\;=\;\;\frac{E}{R}\;\;=\;\;\frac{M}{I}$

(11)

Moments of inertia

The moment of Inertia "I" which is also called the second moment of area is defined as either $\int y^2\: da\;\;or\int x^2\: da$ depending upon which axis the moment is taken. It can be shown that the sum of $I_x\;and\;I_Y$ equals J ( The polar Moment of Inertia). This relationship is called the" perpendicular axis theorum" which states that thesum of the moments of area about any two axis in the plane is equal to the moment of inertia about the axis perpendicular to the plane, the three axis being concurrent. . It follows that the sum of the moments of inertia about any two perpendicular axes through a given point in the plane is constant.

Circular Section

To calculate the polar moment of inertia J about the centre of the section O

$\delta \, A\;=\;2\pi r\: dr$

(12)

$\therefore\;\;\;\;\;J\;=\;\int_{0}^{\frac{d}{2}}\: r^2\: 2\pi r\: dr$

(13)

$=\;2\pi \left[\frac{r^4}{4} \right]_0}^{\frac{d}{2}}$

(14)

$\therefore\;\;\;\;J\;=\;\frac{\pi \, d^4}{32}$

(15)

but $J\;=\;I_X\;+\;I_Y$ and by the parallel axis theory ( For proof of theorem see next section) and since $I_X\;and\;I_Y \;are equal$ being moments of inertia about a diameter.

$I_{diam}\;=\;\frac{1}{2}\: J\;=\;\frac{\pi \, d^4}{64}$

(16)

for a hollow section with external diameters and internal diameters of D and d

$J\;=\;\frac{\pi }{32}\left(D^4\;-\;d^4 \right)\;\;\;and\;\;\;I_{diam}.\;=\;\frac{\pi }{64}\left(D^4\;-\;d^4 \right)$

(17)

Section Modulus Z and Moment of resistance.

The ratio I/y max is called the section modulus Z, so that the maximum stress $f_{max}$ equals M/Z. The bending moment which can be carried by a given section for a limiting stress ic called the Moment of resistance

The parallel axis theorum

${I_z\;=\;\int (y\;+\;h)^2\;dA}$

(18)

by definition

$=\;\int \, y\, .\, dA\;+\;2h\int y\, .\, dA\;+\;h^2\int \, dA$

(19)

$\therefore\;\;\;I_z\;=\;I_x\;+\;A\, \;\;\;\; since \int\,y\,da\;=\;0$

Stated in words: The moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the centroid plus the area times the square of the distance between the axes.

It should be noticed that the moment of inertia through the centroid is the minimum value for any axis in that particular direction.

If it is required to transfer from one axis zz to a parallel axis VV, neither being through the centroid, then the operation must be done in two stages.

$i.e.\;\;\;\;\;\;\;\;I_Z\;=\;I_G\;+\:Ah^2$

(20)

$and\;\;\;\;\;\;\;\;I_V\;=\;I_G\;+\:Ak^2$

(21)

$from\;which\;\;\;\;\;\;\;\;I_V\;=\;I_Z\;-\;Ah^2\;+\;Ak^2$

(22)

Calculations of moments of inertia

a) rectangular section.

From the diagram

$I_X\;=\;\int_{-\, \frac{d}{2}}^{\frac{d}{2}}\: y^2\: b\, dy$

(23)

$=\;b\: \left[\frac{y^3}{3} \right]_{-\frac{d}{2}}^\frac{d}{2}$

(24)

$=\;\frac{b\, d^3}{12}$

(25)

For a hollow section of outside dimensions B:D: and inside dimensions b;d;

$I_X\;=\;\frac{1}{12}\left(BD^3\;-\;bd^3 \right)$

(26)

b) "I" section

In the case of standard sections the moments of inertia are computed graphically from the actual shape of the cross section as rolled. However a reasonable approximation can be made by estimating the mean flange thickness and working from a series of rectangles.

Using the dimensions shown, the moment of inertia about XX may be obtained by subtracting that for rectangles $(B\;-\;t_2)$ wide and d deep from the overall figure for B ties D.

$i.e.\;\;\;\;\;\;\;\;\;I_X\;=\;\frac{1}{12}\left[BD^3\;-\;(B\;-\;t_2)d^3 \right]$

(27)

Blank

Alternatively, for greater accuracy of calculation, the web and flanges may be treated separately using the parallel axis theorem for the flanges.

$Hense\;\;\;\I_X\;=\;2\left(\left[\frac{Bt_1^3}{12} \right]\;+\;Bt_1\left[\frac{(d\,+\,t_1)}{2} \right]^2 \right)\;+\;\frac{t_2d^3}{12}$

(28)

Where $\frac{(d\;+\;t_1)}{2}$ is the distance between the centroid axis of the flange itself and the principle axis of the whole cross section XX. The term $\frac{B\,t_1^3}{12}$ is very small and can usually be neglected