Slope deflection method

By forming slope deflection equations and applying joint and shear equilibrium conditions, the rotation angles (or the slope angles) are calculated. Substituting them back into the slope deflection equations, member end moments are readily determined. Deformation of member is due to the bending moment.

The slope deflection equations can also be written using the stiffness factor ${\displaystyle K={\frac {I_{ab}}{L_{ab}}}}$ and the chord rotation ${\displaystyle \psi ={\frac {\Delta }{L_{ab}}}}$:

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 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.

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

The statically indeterminate beam shown in the figure is to be analysed.

• Members AB, BC, CD have the same length ${\displaystyle L=10\ m}$.
• Flexural rigidities are EI, 2EI, EI respectively.
• Concentrated load of magnitude ${\displaystyle P=10\ kN}$ acts at a distance ${\displaystyle a=3\ m}$ from the support A.
• Uniform load of intensity ${\displaystyle q=1\ kN/m}$ acts on BC.
• Member CD is loaded at its midspan with a concentrated load of magnitude ${\displaystyle P=10\ kN}$.

In the following calculations, clockwise moments and rotations are positive.

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 are:

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

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

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

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

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

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

The slope deflection equations are constructed as follows:

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

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

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

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

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

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

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

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

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

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

The rotation angles are calculated from simultaneous equations above.

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

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

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

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

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

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

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

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

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

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

• Beam theory