A beam of span 5 m, fixed at A and B, carries a point load of 50 kN at 2 m from 'A'. The fixed end moments at the supports 'A' and 'B', respectively, are;

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SSC JE CE Previous Year Paper 9 (Held on 30 Oct 2020 Morning)
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  1. 24 kNm clockwise and 36 kNm clockwise
  2. 24 kNm anticlockwise and 36 kNm anticlockwise
  3. 36 kNm clockwise and 24 kNm anticlockwise
  4. 36 kNm anticlockwise and 24 kNm clockwise

Answer (Detailed Solution Below)

Option 4 : 36 kNm anticlockwise and 24 kNm clockwise
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Detailed Solution

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Concept:

The FEM’s when a beam is subjected to point load and clear loading such as uDL is given below:

F2 Abhishek M 11.2.21 Pallavi D5

 

F2 Abhishek M 11.2.21 Pallavi D6

Calculation:

Given:

F2 Abhishek M 11.2.21 Pallavi D7

a = 2m

b = 3m

L = a + b = 5m

P = 50 kN

\(M_{AB}^F = - \frac{{P{b^2}a}}{{{L^2}}} = - \frac{{50\; \times\; {{\left( 3 \right)}^2}\; \times \;2}}{{{{\left( 5 \right)}^2}}} = - 36\;kNm\;\left( \uparrow \right)\)

(-ve sign means that \(M_{AB}^F\) is in Anticlockwises

\(M_{AB}^F = + \frac{{P{a^2}b}}{{{L^2}}} = \frac{{50\; \times \;{{\left( 2 \right)}^2}\;\times\; 3}}{{{{\left( 5 \right)}^2}}} = + 24\;kN\;M\left( \downarrow \right)\)

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