For the propped cantilever beam loaded as shown in Figure, the Ka
![For the propped cantilever beam loaded as shown in Figure, the Ka](http://storage.googleapis.com/tb-img/production/20/10/F1_%20Abhishek.M_01-10-20_Savita_D%201.png)
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