Problems in Slope Deflection Method

Problems in Slope Deflection Method

Example : 1

Analyse the statically indeterminate beam shown in the figure

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 $P = 10 \ kN$ .

Note: In the following calcuations, clockwise moments and rotations are positive.

Fixed end moments

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

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

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

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

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

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

Slope deflection equations

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

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

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

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

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

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

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):

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

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

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

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

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

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

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jeetu4 October 2012 at 08:03
i like it
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