Derivation of slope deflection equations

When a simple beam of length ${\displaystyle L_{ab}}$ and flexural rigidity ${\displaystyle E_{ab}I_{ab}}$ is loaded at each end with clockwise moments ${\displaystyle M_{ab}}$ and ${\displaystyle M_{ba}}$, member end rotations occur in the same direction. These rotation angles can be calculated using the unit force method or Darcy's Law.

${\displaystyle \theta _{a}-{\frac {\Delta }{L_{ab}}}={\frac {L_{ab}}{3E_{ab}I_{ab}}}M_{ab}-{\frac {L_{ab}}{6E_{ab}I_{ab}}}M_{ba}}$

${\displaystyle \theta _{b}-{\frac {\Delta }{L_{ab}}}=-{\frac {L_{ab}}{6E_{ab}I_{ab}}}M_{ab}+{\frac {L_{ab}}{3E_{ab}I_{ab}}}M_{ba}}$

Rearranging these equations, the slope deflection equations are derived.

Joint equilibrium

Joint equilibrium conditions imply that each joint with a degree of freedom should have no unbalanced moments i.e. be in equilibrium. Therefore,

${\displaystyle \Sigma \left(M^{f}+M_{member}\right)=\Sigma M_{joint}}$

Here, ${\displaystyle M_{member}}$ are the member end moments, ${\displaystyle M^{f}}$ are the fixed end moments, and ${\displaystyle M_{joint}}$ are the external moments directly applied at the joint.

Shear equilibrium

When there are chord rotations in a frame, additional equilibrium conditions, namely the shear equilibrium conditions need to be taken into account.

Degrees of freedom

Rotation angles ${\displaystyle \theta _{A}}$, ${\displaystyle \theta _{B}}$, ${\displaystyle \theta _{C}}$, of joints A, B, C, respectively are taken as the unknowns. There are no chord rotations due to other causes including support settlement.

Fixed end moments

Fixed end moments are:

${\displaystyle M_{AB}^{f}=-{\frac {Pab^{2}}{L^{2}}}=-{\frac {10\times 3\times 7^{2}}{10^{2}}}=-14.7\mathrm {\,kN\,m} }$

${\displaystyle M_{BA}^{f}={\frac {Pa^{2}b}{L^{2}}}={\frac {10\times 3^{2}\times 7}{10^{2}}}=6.3\mathrm {\,kN\,m} }$

${\displaystyle M_{BC}^{f}=-{\frac {qL^{2}}{12}}=-{\frac {1\times 10^{2}}{12}}=-8.333\mathrm {\,kN\,m} }$

${\displaystyle M_{CB}^{f}={\frac {qL^{2}}{12}}={\frac {1\times 10^{2}}{12}}=8.333\mathrm {\,kN\,m} }$

${\displaystyle M_{CD}^{f}=-{\frac {PL}{8}}=-{\frac {10\times 10}{8}}=-12.5\mathrm {\,kN\,m} }$

${\displaystyle M_{DC}^{f}={\frac {PL}{8}}={\frac {10\times 10}{8}}=12.5\mathrm {\,kN\,m} }$

Slope deflection equations

The slope deflection equations are constructed as follows:

${\displaystyle M_{AB}={\frac {EI}{L}}\left(4\theta _{A}+2\theta _{B}\right)={\frac {4EI\theta _{A}+2EI\theta _{B}}{L}}}$

${\displaystyle M_{BA}={\frac {EI}{L}}\left(2\theta _{A}+4\theta _{B}\right)={\frac {2EI\theta _{A}+4EI\theta _{B}}{L}}}$

${\displaystyle M_{BC}={\frac {2EI}{L}}\left(4\theta _{B}+2\theta _{C}\right)={\frac {8EI\theta _{B}+4EI\theta _{C}}{L}}}$

${\displaystyle M_{CB}={\frac {2EI}{L}}\left(2\theta _{B}+4\theta _{C}\right)={\frac {4EI\theta _{B}+8EI\theta _{C}}{L}}}$

${\displaystyle M_{CD}={\frac {EI}{L}}\left(4\theta _{C}\right)={\frac {4EI\theta _{C}}{L}}}$

${\displaystyle M_{DC}={\frac {EI}{L}}\left(2\theta _{C}\right)={\frac {2EI\theta _{C}}{L}}}$

Joint equilibrium equations

Joints A, B, C should suffice the equilibrium condition. Therefore

${\displaystyle \Sigma M_{A}=M_{AB}+M_{AB}^{f}=0.4EI\theta _{A}+0.2EI\theta _{B}-14.7=0}$

${\displaystyle \Sigma M_{B}=M_{BA}+M_{BA}^{f}+M_{BC}+M_{BC}^{f}=0.2EI\theta _{A}+1.2EI\theta _{B}+0.4EI\theta _{C}-2.033=0}$

${\displaystyle \Sigma M_{C}=M_{CB}+M_{CB}^{f}+M_{CD}+M_{CD}^{f}=0.4EI\theta _{B}+1.2EI\theta _{C}-4.167=0}$

Rotation angles

The rotation angles are calculated from simultaneous equations above.

${\displaystyle \theta _{A}={\frac {40.219}{EI}}}$

${\displaystyle \theta _{B}={\frac {-6.937}{EI}}}$

${\displaystyle \theta _{C}={\frac {5.785}{EI}}}$

Member end moments

Substitution of these values back into the slope deflection equations yields the member end moments (in kNm):

${\displaystyle M_{AB}=0.4\times 40.219+0.2\times \left(-6.937\right)-14.7=0}$

${\displaystyle M_{BA}=0.2\times 40.219+0.4\times \left(-6.937\right)+6.3=11.57}$

${\displaystyle M_{BC}=0.8\times \left(-6.937\right)+0.4\times 5.785-8.333=-11.57}$

${\displaystyle M_{CB}=0.4\times \left(-6.937\right)+0.8\times 5.785+8.333=10.19}$

${\displaystyle M_{CD}=0.4\times -5.785-12.5=-10.19}$

${\displaystyle M_{DC}=0.2\times -5.785+12.5=13.66}$