\[\begin{align}
{\varepsilon_{zz}}=& \, {w_0 {}_{,z}} - x {u {}_{,zz}} - y {v {}_{,zz}}\\
{\varepsilon_{zz}}=& \, {\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}
\end{align}\]
The symbol \({\varepsilon_{zz 0}}\) is the axial strain of the reference line.
The symbols \({u {}_{,zz}}\) and \({v {}_{,zz}}\) are the
curvatures of the reference line in the \(x-z\) and \(y-z\)
planes.
Stress in a beam
The stress can now be written:
\[\begin{align}
{\sigma_{zz}}=& \, E {\varepsilon_{zz}}\\
{\sigma_{zz}}=& \, E \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right)
\end{align}\]
Beam loads
Axial load in a beam
The axial load in a beam can be expressed as:
\[P = \int_A {\sigma_{zz}}\, dA\]\[dA = dx \cdot dy\]
Therefore:
\[\begin{align}
P =& \, E \int_A \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right) \, dA\\
=& \, E \left({\varepsilon_{zz 0}}\int_A \, dA - {u {}_{,zz}} \int_A x \, dA - {v {}_{,zz}} \int_A y \, dA \right)
\end{align}\]
Definitions
Lets define some terms:
\[\begin{align}
\int_A \, dA =& \, A \\
\int_A x \, dA =& \, x_c A \\
\int_A y \, dA =& \, y_c A \\
\end{align}\]
Now we have:
\[\begin{align}
P =& \, E \left(A {\varepsilon_{zz 0}}- x_c A {u {}_{,zz}} - y_c A {v {}_{,zz}} \right) \\
\end{align}\]
Choice of coordinate system
We choose our coordinate system so that:
\[\begin{align}
x_c = 0 \\
y_c = 0 \\
\end{align}\]
Which leads to:
\[\begin{align}
P =& \, E A \, {\varepsilon_{zz 0}}\\
\end{align}\]
This means that the axial force in the beam is due to the
elongation of the centroidal axis and is not related to
the bending of the beam.
Bending moments about \(x\)
\[\begin{align}
{\sigma_{zz}}=& \, E \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right)
\end{align}\]
The resultant bending moment about the centroid is:\[M_x = \int_A y \cdot {\sigma_{zz}}\, dA\]
\[M_x = E \left( {\varepsilon_{zz 0}}\cdot \int_A y \, dA - {u {}_{,zz}} \cdot \int_A x y \, dA - {v {}_{,zz}} \cdot \int_A y^2 \, dA\right)\]
Recall:
\(\int_A y \, dA = 0\) : By choice of coordinate system
\(\int_A x y \, dA = {I_{xy}}\) : Area product of inertia
\(\int_A y^2 \, dA = {I_{xx}}\) : Area moment of inertia about \(x\)-axis
Note: \({I_{xy}}= 0\) if there is symmetry about the \(x\) or \(y\) axis. (If there
is symmetry about another axis, the computation can be transformed
to be about the principal axis of the cross section.)
If \(E=E(x,y)\), then it must remain inside the integral and we do not
have a clean separation between material and cross-section
properties. (See ME6520 Composite Materials)
Cross section of the rat femur, showing the neutral axis, which goes
through the centroid and has an orientation with respect to the ML
axis. Note that for this example, the bending moment is applied only
about the AP axis. Also shown are the locations of maximum tensile and
compressive bending stress, respectively.
Bending moments about \(y\)
\[\begin{align}
{\sigma_{zz}}=& E \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right)
\end{align}\]
The resultant bending moment about the centroid is:\[M_y = - \int_A x \cdot {\sigma_{zz}}\, dA\]
\[M_y = - E \left( {\varepsilon_{zz 0}}\cdot \int_A x \, dA - {u {}_{,zz}} \cdot \int_A x^2 \, dA - {v {}_{,zz}} \cdot \int_A x y \, dA\right)\]
Recall:
\(\int_A x \, dA = 0\) : By choice of coordinate system
\(\int_A x y \, dA = {I_{xy}}\) : Area product of inertia
\(\int_A x^2 \, dA = {I_{yy}}\) : Area moment of inertia about \(y\)-axis
Note: \({I_{xy}}= 0\) if there is symmetry about the \(x\) or \(y\) axis (If there
is symmetry about another axis, the computation can be transformed
to be about the principal axis of the cross section.)
Using similar arguments, we can find: \[{\frac{d S_y(z)}{d z}} = -p_y\]
Now use moment equilibrium and including bending moments:
Consider first the counterclockwise moments about \(y\) axis (using the
the center as reference point):\[\begin{align}
\left(M_y(z+dz) - M_y(z)\right)& \\
+ \left(S_x(z+dz) +S_x(z)\right)& \frac{d z}{2} = \; 0\\
& d M_y = -(2 S_x +d S_x)\frac{d z}{2}
\end{align}\]
Bringing forward the last equation:
\[\begin{align}
\frac{d M_y}{dz} = - \frac{1}{2} \left(S_x(z+dz) + S_x(z)\right)
\end{align}\]
In the limit as \(d z\) goes to zero.
\[\begin{align}
\frac{dM_y}{dz} =& -S_x \\
\end{align}\]
The same can be applied to moments about the \(x\) axis:
Note carefully the difference in the convention for direction of the
moments.
\[\begin{align}
\frac{dM_x}{dz} =& +S_y \\
\end{align}\]
