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The intent here is not to cover all aspects of beam bending. In particular, topics such as determining the neutral axis, the parallel axis theorem, and computing beam deflections are not covered.

We will eventually need an analytical expression for radius of curvature, so it will be developed here. Start with any function, \(y = y(x)\), as shown in the figure. Recall that arc length, \(s\), is related to \(\rho\) through \(\rho \, \theta = s\), where \(\theta\) is the angle of the arc. In differential form, this is \(\rho \, d \theta = ds\). Now divide both sides by \(dx\).

\[ \rho {d \theta \over dx} = {ds \over dx} \]

We need expressions for \(d \theta \over dx\) and \(ds \over dx\). Starting with \(d \theta \over dx\), recall that

\[ \tan ( \theta ) = {dy \over dx} \]

so

\[ \theta = \text{Tan}^{-1} ( y' ) \]

Differentiate with respect to \(x\) to get \(d \theta \over dx\).

\[ {d \theta \over dx} = {y'' \over 1 + (y')^2} \]

To obtain \(ds \over dx\), start with

\[ ds^2 = dx^2 + dy^2 \]

Divide both sides by \(dx^2\)

\[ \left( ds \over dx \right)^2 = 1 + \left( dy \over dx \right)^2 \]

and take the square root of both sides

\[ {ds \over dx} = \sqrt{1 + (y')^2} \]

Substituting all this into \(\rho {d \theta \over dx} = {ds \over dx}\) gives

\[ \rho \, {y'' \over 1 + (y')^2} = \sqrt{1 + (y')^2} \]

And rearrange a little to obtain the popular expression.

\[ {1 \over \rho} = {y'' \over \left[ 1 + (y')^2 \right]^{3/2} } \]

Interestingly, any expression involving the radius of curvature seems to always have it appear in the denominator. And this is no exception, even when it is a defining equation.

Also interesting is the fact that many mechanics applications involve bending, but on a small scale. The beam bending discussed here is no exception. In such cases, the best approach is to define the x-axis along the beam such so that the \(y\) deflections, and more importantly the deformed slope, \(y'\), will both be small. If \(y' \lt \lt 1\), then \(y'\) can be neglected in the above equation. This produces a much simpler expression.

\[ {1 \over \rho} \approx y'' \]

\[ {1 \over \rho} = { | {\bf v'} \times {\bf v''} | \over | {\bf v'} |^3} \]

where \({\bf v}\) is a vector defined parametrically as \({\bf v} = {\bf v}(x(t), y(t), z(t))\), \({\bf v'}\) is its first derivative, \({\bf v''}\) is its second derivative, and \(|...|\) represents the length of a vector, i.e., the square root of the sum of the squares of its components.

Things become more complex when thickness is considered. In the figure below, an object of initial length, \(L_o\), is bent as shown. Since it has finite thickness, different portions of it are stretched, or compressed, different amounts. The outer portion of a beam is stretched the most because it is farthest from that center. Mathematically, all portions are bent the same angle, \(\theta\), but \(\rho\) varies through the thickness, so the quantity \(\rho \, \theta\) varies too, and therefore \(L\) varies as well.

The next step is to make a conscious decision to avoid the confusion of having many different radii of curvature through the thickness of a bent object. This is accomplished in two steps.

First, find the one \(\rho\) that satisfies \(\rho \, \theta = L_o\). Note that \(\rho\) is the computed result here and \(\theta\) and \(L_o\) are the inputs. Note also that the length in the equation is \(L_o\), the original, undeformed length, not the deformed one. This step establishes one unique value of \(\rho\) for a cross-section, rather than having multiple values, which could lead to much confusion.

The "due to bending" caveat in the above paragraph is present because the object may also be simultaneously loaded in tension (or compression) that stretches it until every point in its cross-section is longer (or shorter if compressed) than the original length, \(L_o\).

\[ L = (\rho - y) \theta \]

Recall that \(L_o = \rho \, \theta\). Now the strain, \(\epsilon\), can be expressed as

\[ \epsilon_x = {L - L_o \over L_o} = { (\rho - y) \theta - \rho \, \theta \over \rho \, \theta } \]

which simplifies to

\[ \epsilon_x = - {y \over \rho} \]

This is a key result for the strain in the object. It shows that the strain is zero at \(y=0\), the neutral axis, and varies linearly from it. If the object is thick, then \(y\) can take on large values, but in thin objects, it does not. This is fundamentally why thick objects have more bending stiffness than thin objects.

Also, the radius of curvature in the denominator accounts for many effects of bending. When the object is not bent, then \(\rho\) is infinite, and the strains are naturally zero. As the object bends, \(\rho\) decreases and the equation shows that the strains will increase.

Finally, note that the strain is a

\[ \sigma_x = - {E \, y \over \rho} \]

Although this step was increadibly simple, it is in fact rather profound in what has been neglected by the simplicity. Recall from the page on Hooke's Law that each normal stress component is dependent on all three normal strain components. But here, we have simply multiplied the strain by \(E\) to obtain stress. This step has a key assumption built into it... that there are no lateral loads/stresses acting on the beam. In such cases, the equations "work out" so that \(\sigma_x = E \, \epsilon_x\) as in uniaxial tension. This occurs in most beams because they are thin relative to their length.

\[ \sigma_x = {E \over (1 - \nu^2) } \epsilon_x \]

where \(\nu\) is the materials Poisson ratio. It is clear that the stress for a given strain is higher in this case than for classical uniaxial tension.

The bending moment, \(M_z\), on the cross-section due to the stress field is computed by

\[ M_z \; = \; \int_A r \times dF \; = \; - \int_A y \, \sigma_x dA \]

where \(y\) is the moment arm and \(\sigma_x dA\) is the force.

Recall that \(\sigma_x = - E \, y / \rho\) and insert into the moment equation to obtain

\[ M_z \; = \; - \int_A y \, \left[ {- E \, y \over \rho} \right] dA \; = \; {E \over \rho} \int_A y^2 dA \]

The remaining integral is very important. Note that it is entirely geometric. It is called the

\[ I_{zz} \; = \; \int_A y^2 dA \]

Always keep in mind that the \(y\) values are measured from the neutral axis, which corresponds to \(y = 0\).

The bending moment equation can now be written as

\[ M_z = {E \, I_{zz} \over \rho} \]

which is a very important equation.

\[ M_z = E \, I_{zz} y'' \qquad (\text{when} \;\; y' \lt \lt 1) \]

which is another very important and useful equation.

\[ M_z = {E \, I_{zz} \over \rho} \qquad \qquad \text{and} \qquad \qquad \sigma_x = - {E \, y \over \rho} \]

and recognize that both contain \( E / \rho\). Solving each equation for this ratio gives

\[ {E \over \rho} = {M_z \over I_{zz} } \qquad \qquad \text{and} \qquad \qquad {E \over \rho} = -{y \over \sigma_x} \]

So equate the two to obtain one of the best known equations in Mechanical Engineering.

\[ \sigma_x = - {M_z \, y \over \; I_{zz} } \]

Full 3-D bending involves deflections in two directions, \(y\) and \(z\). The strain, \(\epsilon_x\), now depends on both coordinates.

\[ \epsilon_x = {z \over \rho_y} - {y \over \rho_z} \]

Multiply through by \(E\) to obtain stress.

\[ \sigma_x = {E \, z \over \rho_y} - {E \, y \over \rho_z} \]

The bending moment, \(M_y\), is calculated by integrating the stress over the cross-section with \(z\) as the moment arm.

\[ \begin{eqnarray} M_y & = & \int_A z \, \sigma dA & = & {E \over \rho_y} \int_A z^2 dA - {E \over \rho_z} \int_A y \, z \, dA \\ \\ \\ & & & = & {E \over \rho_y} I_{yy} - {E \over \rho_z} I_{yz} \end{eqnarray} \]

And the other moment, \(M_z\), is calculated by integrating the stress over the cross-section with \(y\) as the moment arm.

\[ \begin{eqnarray} M_z & = & -\int_A y \, \sigma dA & = & {E \over \rho_z} \int_A y^2 dA - {E \over \rho_y} \int_A y \, z \, dA \\ \\ \\ & & & = & {E \over \rho_z} I_{zz} - {E \over \rho_y} I_{yz} \end{eqnarray} \]

The equations can be written compactly in matrix form as

\[ \left\{ \matrix { M_y \\ \\ M_z } \right\} = \left[ \matrix { \; I_{yy} & -I_{yz} \\ \\ -I_{yz} & \; I_{zz} } \right] \left\{ \matrix { E / \rho_y \\ \\ E / \rho_z } \right\} \]

This reveals that the moment of inertia is actually a tensor.

It is more useful to invert the equations to obtain the radii of curvature as functions of the moments. Doing so gives

\[ {E \over \rho_y} = {M_y I_{zz} + M_z I_{yz} \over I_{yy} I_{zz} - (I_{yz})^2 } \]

\[ {E \over \rho_z} = {M_z I_{yy} + M_y I_{yz} \over I_{yy} I_{zz} - (I_{yz})^2 } \]

And recall that

\[ \sigma_x = {E \over \rho_y} z - {E \over \rho_z} y \]

Inserting the above equations for \(E / \rho_y\) and \(E / \rho_z\) gives

\[ \sigma_x = \left[ {M_y I_{zz} + M_z I_{yz} \over I_{yy} I_{zz} - (I_{yz})^2 } \right] z - \left[ {M_z I_{yy} + M_y I_{yz} \over I_{yy} I_{zz} - (I_{yz})^2 } \right] y \]

\[ \sigma_x = {1 \over \text{det} ({\bf I})} \matrix { \left\{ -y \;\;\; z \; \right\} \\ \\ \\ } \left[ \matrix { I_{yy} & I_{yz} \\ \\ I_{yz} & I_{zz} } \right] \left\{ \matrix { M_z \\ \\ M_y } \right\} \]

where \( \text{det} ({\bf I}) = I_{yy} I_{zz} - (I_{yz})^2 \).

Finally, if \(I_{yz} = 0\) things greatly simplify to

\[ \sigma_x = {M_y \, z \over I_{yy} } - {M_z \, y \over I_{zz} } \]

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