Slope deflection method

The slope deflection method is a structural analysis method for beams and frames introduced in 1914 by George A. Maney.^[1] The slope deflection method was widely used for more than a decade until the moment distribution method was developed. In the book, "The Theory and Practice of Modern Framed Structures", written by J.B Johnson, C.W. Bryan and F.E. Turneaure, it is stated that this method was first developed,"by Professor Otto Mohr in Germany, and later developed independently by Professor G.A. Maney". According to this book, professor Otto Mohr introduced this method for the first time in his book,"Evaluation of Trusses with Rigid Node Connections" or "Die Berechnung der Fachwerke mit Starren Knotenverbindungen".

Introduction

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.

Slope deflection equations

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

Derivation of slope deflection equations

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

$\theta_a - \frac{\Delta}{L_{ab}}= \frac{L_{ab}}{3E_{ab} I_{ab}} M_{ab} - \frac{L_{ab}}{6E_{ab} I_{ab}} M_{ba}$

$\theta_b - \frac{\Delta}{L_{ab}}= - \frac{L_{ab}}{6E_{ab} I_{ab}} M_{ab} + \frac{L_{ab}}{3E_{ab} I_{ab}} M_{ba}$

Rearranging these equations, the slope deflection equations are derived.

Equilibrium conditions

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,

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

Here, $M_{member}$ are the member end moments, $M^{f}$ are the fixed end moments, and $M_{joint}$ 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.

Example

Example

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

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

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

Degrees of freedom

Rotation angles $\theta_A$, $\theta_B$, $\theta_C$, 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:

$M _{AB} ^f = - \frac{P a b^2 }{L ^2} = - \frac{10 \times 3 \times 7^2}{10^2} = -14.7 \mathrm{\,kN \,m}$

$M _{BA} ^f = \frac{P a^2 b}{L^2} = \frac{10 \times 3^2 \times 7}{10^2} = 6.3 \mathrm{\,kN \,m}$

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

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

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

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

Slope deflection equations

The slope deflection equations are constructed as follows:

$M_{AB} = \frac{EI}{L} \left( 4 \theta_A + 2 \theta_B \right) = 0.4EI \theta_A + 0.2EI \theta_B$

$M_{BA} = \frac{EI}{L} \left( 2 \theta_A + 4 \theta_B \right) = 0.2EI \theta_A + 0.4EI \theta_B$

$M_{BC} = \frac{2EI}{L} \left( 4 \theta_B + 2 \theta_C \right) = 0.8EI \theta_B + 0.4EI \theta_C$

$M_{CB} = \frac{2EI}{L} \left( 2 \theta_B + 4 \theta_C \right) = 0.4EI \theta_B + 0.8EI \theta_C$

$M_{CD} = \frac{EI}{L} \left( 4 \theta_C \right) = 0.4EI \theta_C$

$M_{DC} = \frac{EI}{L} \left( 2 \theta_C \right) = 0.2EI \theta_C$

Joint equilibrium equations

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

$\Sigma M_A = M_{AB} + M_{AB}^f = 0.4EI \theta_A + 0.2EI \theta_B - 14.7 = 0$

$\Sigma M_B = M_{BA} + M_{BA}^f + M_{BC} + M_{BC}^f = 0.2EI \theta_A + 1.2EI \theta_B + 0.4EI \theta_C - 2.033 = 0$

$\Sigma M_C = M_{CB} + M_{CB}^f + M_{CD} + M_{CD}^f = 0.4EI \theta_B + 1.2EI \theta_C - 4.167 = 0$

Rotation angles

The rotation angles are calculated from simultaneous equations above.

$\theta_A = \frac{40.219}{EI}$

$\theta_B = \frac{-6.937}{EI}$

$\theta_C = \frac{5.785}{EI}$

Member end moments

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

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

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

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

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

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

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

See also

• Beam theory

Notes

1. Maney, George A. (1915). "Studies in Engineering". Minneapolis: University of Minnesota.

References

• Norris, Charles Head; John Benson Wilbur; Senol Utku (1976). Elementary Structural Analysis (3rd ed.). McGraw-Hill. pp. 313–326. ISBN 0-07-047256-4.
• McCormac, Jack C.; Nelson, James K. Jr. (1997). Structural Analysis: A Classical and Matrix Approach (2nd ed.). Addison-Wesley. pp. 430–451. ISBN 0-673-99753-7.
• Yang, Chang-hyeon (2001-01-10). Structural Analysis (in Korean) (4th ed.). Seoul: Cheong Moon Gak Publishers. pp. 357–389. ISBN 89-7088-709-1.