A fixed beam is subjected to loading as shown in the figure below
A fixed beam is subjected to loading as shown in the figure below.
The fixed end moments MAB and MBA will be (Assume EI constant)
A. 17.78 kN-m and 39.56 kN-m
B. 17.78 kN-m and 35.56 kN-m
C. 0 kN-m and 4 kN-m
D. None of these
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
When a beam is fixed at the ends such that it cannot rotate, then bending moments are induced at the supports and are called fixed end moments.
When a beam is considered free at the ends to rotate such that no bending moment is developed at the ends, the bending moment diagram so obtained is called free bending moment diagram.
Standard fixed end moments formulae:
\(\therefore {M_{AB}} = \frac{{{M_0}b\left( {3a - l} \right))}}{{{l^2}}}\)
\({M_{BA}} = \frac{{{M_0}a\left( {3b - l} \right)}}{{{l^2}}}\)
\({M_{BA}} = \frac{{Pb{a^2}}}{{{l^2}}}\)
\({M_{AB}} = \frac{{Pa{b^2}}}{{{l^2}}}\)
Calculation:
The fixed end moments generated due to the applied moment and concentrated load can be calculated independently and then added as per the superposition principle.
Using the standard results:-
\({M_{ABI}} = \frac{{12 \times 8 \times \left( {3 \times 4 - 12} \right)}}{{{{\left( {12} \right)}^2}}} = 0\)
\({M_{BAI}} = \frac{{12 \times 4\left( {3 \times 8 - 12} \right)}}{{{{\left( {12} \right)}^2}}} = 4\;kN - m\)
Again, for concentrated load-
\({M_{ABII}} = \frac{{ - Pa{b^2}}}{{{l^2}}} = \frac{{ - 20 \times 8 \times {4^2}}}{{{{\left( {12} \right)}^2}}} = - 17.78\;kN - m\)
\({M_{BAII}} = \frac{{P{a^2}b}}{{{l^2}}} = \frac{{ 20 \times {4} \times {8^2}}}{{{{\left( {12} \right)}^2}}} = 35.56\;kN - m\)
∴ MAB = MABI + MABII = 0 + (-17.78) = -17.78 kN-m
& MBA = MBAI + MBAII = 4 + 35.56 = 39.56 kN-m
Important Point:
Standard fixed end moments:
