For the propped cantilever beam loaded as shown in Figure, the Ka

For the propped cantilever beam loaded as shown in Figure, the Ka
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For the propped cantilever beam loaded as shown in Figure, the Kani’s Rotation moment at the end B is +8.89 kNm. (taking anticlockwise as +). Hence the end moment at fixed end, A is

A. zero

B. + 8.89 kNm

C. + 17.78 kNm

D. – 17.78 kNm

Please scroll down to see the correct answer and solution guide.

Right Answer is: D

SOLUTION

Concept:

Displacement factor in Kani’s Method:

Kani’s method is a similar method as moment distribution method which is often used in analysis of continuous beam or indeterminate frame

In Kani’s method rotation factor for joints is given by,

\({{\rm{R}}_{{\rm{ij}}}} = - 0.5\frac{{{{\rm{k}}_{{\rm{ij}}}}}}{{\sum {{\rm{k}}_{{\rm{ij}}}}}}{\rm{\;where\;}}{{\rm{k}}_{{\rm{ij}}}} = \frac{{{{\rm{I}}_{{\rm{ij}}}}}}{{{{\rm{L}}_{{\rm{ij}}}}}}{\rm{\;}}\)

Where, Iij and Lij is the moment of inertia and length of the member connecting i th and j ih node of the frame.

Calculations:

\({{M}_{AB}}=\frac{ - 20\times {{2}^{2}}\times 4}{{{6}^{2}}}= - 8.889~kN.m\)

\({{M}_{BA}}=\frac{20\times {{4}^{2}}\times 2}{{{6}^{2}}}=17.778~kN.m\)

But as Support B do not carry any moment it is balanced. The moment is carried to Support A

And the carryover factor is 1/2

 ∴ At Support, A Moment is -8.889 + (-17.778/2)

 = - 17.778 kN-m