Slope deflection method

The slope deflection method is a structural analysis method for beams and frames introduced in 1915 by George A. Maney.Citation|first=George A.|last=Maney|year=1915|title=Studies in Engineering|publisher=University of Minnesota|location=Minneapolis] This method neglects the deformations due to shear and axial forces. The slope deflection method was widely used for more than a decade until the moment distribution method was developed.

Introduction

By forming slope deflection equations and applying joint and shear equilibrium conditions, the rotation angles (or the slope angles) are calculated. Substutituting them back into the slope deflection equations, member end moments are readily determined.

Slope deflection equations

The slope deflection equations express the member end moments in terms of rotations angles. The slope deflection equations of member ab of flexural rigidity $EI\_\{ab\}$ and length $L\_\{ab\}$ are::$M\_\{ab\}\; =\; frac\{EI\_\{ab\{L\_\{ab\; left(\; 4\; heta\_a\; +\; 2\; heta\_b\; -\; 6\; frac\{Delta\}\{L\_\{ab\; ight)$:$M\_\{ba\}\; =\; frac\{EI\_\{ab\{L\_\{ab\; left(\; 2\; heta\_a\; +\; 4\; heta\_b\; -\; 6\; frac\{Delta\}\{L\_\{ab\; ight)$where $heta\_a$, $heta\_b$ are the slope angles of ends a and b respectively, $Delta$ is the relative lateral displacement of ends a and b. The absence of cross-sectional area of the member in these equations implies that the slope deflection method neglects the effect of shear and axial deformations.

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

:$M\_\{ab\}\; =\; 2EK\; left(\; 2\; heta\_a\; +\; heta\_b\; -\; 3\; psi\; ight)$:$M\_\{ba\}\; =\; 2EK\; left(\; heta\_a\; +\; 2\; heta\_b\; -\; 3\; psi\; ight)$

Derivation of slope deflection equations

When a simple beam of length $L\_\{ab\}$ and flexural rigidity $EI\_\{ab\}$ 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 the moment-area theorem.

:$heta\_a\; -\; frac\{Delta\}\{L\_\{ab=\; frac\{L\_\{ab\{3EI\_\{ab\; M\_\{ab\}\; -\; frac\{L\_\{ab\{6EI\_\{ab\; M\_\{ba\}$:$heta\_b\; -\; frac\{Delta\}\{L\_\{ab=\; -\; frac\{L\_\{ab\{6EI\_\{ab\; M\_\{ab\}\; +\; frac\{L\_\{ab\{3EI\_\{ab\; M\_\{ba\}$

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,:$Sigma\; left(\; M^\{f\}\; +\; M\_\{member\}\; ight)\; =\; Sigma\; M\_\{joint\}$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 roations due to sidesway in a frame, additional equilibrium conditions, namely the shear equilibrium conditions need to be taken into account.

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 calcuations, clockwise moments and rotations are positive.

Degrees of freedom

Rotation angles $heta\_A$, $heta\_B$, $heta\_C$ 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::$M\; \_\{AB\}\; ^f\; =\; -\; frac\{P\; a\; b^2\; \}\{L\; ^2\}\; =\; -\; frac\{10\; imes\; 3\; imes\; 7^2\}\{10^2\}\; =\; -14.7\; kNcdot\; m$:$M\; \_\{BA\}\; ^f\; =\; frac\{P\; a^2\; b\}\{L^2\}\; =\; frac\{10\; imes\; 3^2\; imes\; 7\}\{10^2\}\; =\; 6.3\; kNcdot\; m$:$M\; \_\{BC\}\; ^f\; =\; -\; frac\{qL^2\}\{12\}\; =\; -\; frac\{1\; imes\; 10^2\}\{12\}\; =\; -\; 8.333\; kNcdot\; m$:$M\; \_\{CB\}\; ^f\; =\; frac\{qL^2\}\{12\}\; =\; frac\{1\; imes\; 10^2\}\{12\}\; =\; 8.333\; kNcdot\; m$:$M\; \_\{CD\}\; ^f\; =\; -\; frac\{PL\}\{8\}\; =\; -\; frac\{10\; imes\; 10\}\{8\}\; =\; -12.5\; kNcdot\; m$:$M\; \_\{DC\}\; ^f\; =\; frac\{PL\}\{8\}\; =\; frac\{10\; imes\; 10\}\{8\}\; =\; 12.5\; kNcdot\; m$

Slope deflection equations

The slope deflection equations are constructed as follows::$M\_\{AB\}\; =\; frac\{EI\}\{L\}\; left(\; 4\; heta\_A\; +\; 2\; heta\_B\; ight)\; =\; 0.4EI\; heta\_A\; +\; 0.2EI\; heta\_B$ :$M\_\{BA\}\; =\; frac\{EI\}\{L\}\; left(\; 2\; heta\_A\; +\; 4\; heta\_B\; ight)\; =\; 0.2EI\; heta\_A\; +\; 0.4EI\; heta\_B$ :$M\_\{BC\}\; =\; frac\{2EI\}\{L\}\; left(\; 4\; heta\_B\; +\; 2\; heta\_C\; ight)\; =\; 0.8EI\; heta\_B\; +\; 0.4EI\; heta\_C$ :$M\_\{CB\}\; =\; frac\{2EI\}\{L\}\; left(\; 2\; heta\_B\; +\; 4\; heta\_C\; ight)\; =\; 0.4EI\; heta\_B\; +\; 0.8EI\; heta\_C$ :$M\_\{CD\}\; =\; frac\{EI\}\{L\}\; left(\; 4\; heta\_C\; ight)\; =\; 0.4EI\; heta\_C$ :$M\_\{DC\}\; =\; frac\{EI\}\{L\}\; left(\; 2\; heta\_C\; ight)\; =\; 0.2EI\; heta\_C$

Joint equilibrium equations

Joints A, B, C should suffice the equilibrium condition. Therefore:$Sigma\; M\_A\; =\; M\_\{AB\}\; +\; M\_\{AB\}^f\; =\; 0.4EI\; heta\_A\; +\; 0.2EI\; heta\_B\; -\; 14.7\; =\; 0$:$Sigma\; M\_B\; =\; M\_\{BA\}\; +\; M\_\{BA\}^f\; +\; M\_\{BC\}\; +\; M\_\{BC\}^f\; =\; 0.2EI\; heta\_A\; +\; 1.2EI\; heta\_B\; +\; 0.4EI\; heta\_C\; -\; 2.033\; =\; 0$:$Sigma\; M\_C\; =\; M\_\{CB\}\; +\; M\_\{CB\}^f\; +\; M\_\{CD\}\; +\; M\_\{CD\}^f\; =\; 0.4EI\; heta\_B\; +\; 1.2EI\; heta\_C\; -\; 4.167\; =\; 0$

Rotation angles

By solving the above simultaneous equations, rotations angles are given.:$heta\_A\; =\; frac\{40.219\}\{EI\}$:$heta\_B\; =\; frac\{-6.937\}\{EI\}$:$heta\_C\; =\; frac\{5.785\}\{EI\}$

Member end moments

Substitution of these values back into the slope deflection equations yields the member end moments.:$M\_\{AB\}\; =\; 0.4\; imes\; 40.219\; +\; 0.2\; imes\; left(\; -6.937\; ight)\; -\; 14.7\; =\; 0$:$M\_\{BA\}\; =\; 0.2\; imes\; 40.219\; +\; 0.4\; imes\; left(\; -6.937\; ight)\; +\; 6.3\; =\; 11.57$:$M\_\{BC\}\; =\; 0.8\; imes\; left(\; -6.937\; ight)\; +\; 0.4\; imes\; 5.785\; -\; 8.333\; =\; -11.57$:$M\_\{CB\}\; =\; 0.4\; imes\; left(\; -6.937\; ight)\; +\; 0.8\; imes\; 5.785\; +\; 8.333\; =\; 10.19$:$M\_\{CD\}\; =\; 0.4\; imes\; 5.785\; -\; 12.5\; =\; -10.19$:$M\_\{DC\}\; =\; 0.2\; imes\; 5.785\; +\; 12.5\; =\; 13.66$

Notes

References

*cite book |last=McCormac|first=Jack C.|coauthors=James K. Nelson, Jr.|title=Structural Analysis: A Classical and Matrix Approach|edition=2nd |year=1997|publisher=Addison-Wesley|isbn=0-673-99753-7|pages=430-451
*cite book|last=Yang|first=Chang-hyeon|title=Structural Analysis|url=http://www.cmgbook.co.kr/category/sub_detail.html?no=1017|edition=4th|date=2001-01-10|publisher=Cheong Moon Gak Publishers|language=Korean|location=Seoul|isbn=89-7088-709-1|pages=357-389

See also

*Moment distribution method