Slope deflection method


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


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