8.6 Differential equations of static equilibrium

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8.6.1 Axial static equilbrium

• In the axial direction (\(z\)):
  • \(P = P(z)\) – axial force resultant
  • \(p_z = p_z(z)\) – distributed axial load [force/length]
    • e.g., \(p_z = \rho A g_z\)
    • Note \(\rho g_z\) is related to the stress equilibrium equations (body force per unit volume) as \(b_z = \rho g_z\)
  • \(\displaystyle\sum F_z = 0 = P(z + dz) - P(z) + p_z dz\)

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Hence:

• \(0 = \frac{P(z + dz) - P(z)}{dz} + p_z\)
• \(\frac{d P}{d z} = - p_z(z)\)

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• For \(P = {\sigma_{zz}}\cdot A = E A \, {\varepsilon_{zz 0}}= E A \, {{w}_0 {}_{,z}}\)
• \({\frac{d P(z)}{d z}} = {\frac{d }{d z}} \left[ E A {\frac{d w_0}{d z}} \right] = - p_z(z)\)
• For constant \(EA\)
  • \(E A \, {w_0 {}_{,zz}} = - p_z(z)\)
• Note, when \(p_z = 0\), the axial force does not vary (for small deflections)

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8.6.2 Bending static equilibrium

Beam bending: In the lateral directions (\(x\) & \(y\)):

• (\(S_x\) ; \(S_y\)) - shear force resultants
• (\(p_x\) ; \(p_y\)) - distributed lateral loads [force/length]
• (\(M_x\) ; \(M_y\)) - bending moment resultants

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• Summing forces, we find: \[\displaystyle \sum F_x = S_x(z+dz) -S_x(z) + p_x dz = 0\] \[{\frac{d S_x(z)}{d z}} = -p_x\]

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• Using similar arguments, we can find: \[{\frac{d S_y(z)}{d z}} = -p_y\]

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• Now use moment equilibrium and including bending moments:

• Consider first the counterclockwise moments about \(y\) axis (using the the center as reference point): \[\begin{align} \left(M_y(z+dz) - M_y(z)\right)& \\ + \left(S_x(z+dz) +S_x(z)\right)& \frac{d z}{2} = \; 0\\ & d M_y = -(2 S_x +d S_x)\frac{d z}{2} \end{align}\]

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• 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. \[\begin{align} \frac{dM_y}{dz} =& -S_x \\ \end{align}\]

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• The same can be applied to moments about the \(x\) axis:

• Note carefully the difference in the convention for direction of the moments. \[\begin{align} \frac{dM_x}{dz} =& +S_y \\ \end{align}\]

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• In summary, equilibrium arguments require: \[\begin{align} {\frac{d S_x(z)}{d z}} =& -p_x \\ {\frac{d S_y(z)}{d z}} =& -p_y \\ \frac{d M_x}{dz} =& +S_y \\ \frac{d M_y}{dz} =& -S_x \\ \end{align}\]

• Combining these leads to: \[\begin{align} \frac{d^2 M_x}{dz^2} =& -p_y \\ \frac{d^2 M_y}{dz^2} =& +p_x \\ \end{align}\]

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• Recall also: \[\begin{align} M_x =& \, - E{I_{xy}}{u {}_{,zz}} - E{I_{xx}}{v {}_{,zz}} \\ M_y =& \, + E{I_{yy}}{u {}_{,zz}} + E{I_{xy}}{v {}_{,zz}} \\ \end{align}\]

• In matrix form these can be written as: \[\begin{align} \left\{ \begin{array}{c} M_x \\ M_y\\ \end{array} \right\} = \left[ \begin{array}{cc} -EI_{xy} & -EI_{xx} \\ EI_{yy} & EI_{xy} \\ \end{array} \right] \left\{ \begin{array}{c} {u {}_{,zz}} \\ {v {}_{,zz}} \\ \end{array} \right\} \end{align}\]

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• Recall inversion: \[\begin{pmatrix}a & b \\ c & d \\ \end{pmatrix}^{-1} = \begin{pmatrix}\frac{d}{a\,d-b\,c} & -\frac{b}{a\,d-b\,c} \\ - \frac{c}{a\,d-b\,c} & \frac{a}{a\,d-b\,c} \\ \end{pmatrix}\]

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• Simple inversion yields the following: \[\begin{align} \left\{ \begin{array}{c} {u {}_{,zz}} \\ {v {}_{,zz}} \\ \end{array} \right\} = \frac{1}{\det EI} \left[ \begin{array}{cc} EI_{xy} & EI_{xx} \\ -EI_{yy} & -EI_{xy} \\ \end{array} \right] \left\{ \begin{array}{c} M_x \\ M_y\\ \end{array} \right\} \end{align}\] where: \[{\det EI} = + EI_{xx} EI_{yy} - (EI_{xy})^2\]

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• Now, lets examine the stress equation again: \[\begin{equation} \begin{split} {\sigma_{zz}}=& E \left( -{u {}_{,zz}} x - {v {}_{,zz}} y \right) \\ =& \frac{E}{\det EI} \left[\right. \\ & -x \left(EI_{xy} M_x + EI_{xx} M_y \right) \\ & \left. + y \left(EI_{yy} M_x + EI_{xy} M_y\right) \right] \\ \end{split} \end{equation}\]

• It is these equations which allow us to compute deflections, strains, and stresses from geometry, material, and loads.