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CivilStructural Mechanics-II
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A beam 4 m long is fixed at it ends ┬╖ It carries a udl of 20 kN/m. find the fixed end moment in the beam.

A

21.78 kNm

B

20.36 kNm

C

24.56 kNm

D

26.66 kNm

Correct Answer

тЪЩя╕П TE тАв Technical Direct FormulaCivilStructural Mechanics-II
Option D

26.66 kNm

Quick Summary:

Given: Length of the beam L = 4 m, Uniformly Distributed Load (UDL) w = 20 kN/m.

ЁЯУРMAMath SolutionDirect Formula
ЁЯУЛ Given

Length of the beam L = 4 m, Uniformly Distributed Load (UDL) w = 20 kN/m.

ЁЯФв Formula Used

MF=wL212M_F = \frac{wL^2}{12}MFтАЛ=12wL2тАЛ

ЁЯУК Diagram / Illustration
Beam length L = 4 mUDL w = 20 kN/m
ЁЯФв Step-by-Step Solution
1

Identify given parameters

The beam length LLL is given as 4┬аm4\text{ m}4┬аm and the intensity of the uniformly distributed load www is 20┬аkN/m20\text{ kN/m}20┬аkN/m.

L=4┬аm,w=20┬аkN/mL = 4\text{ m}, w = 20\text{ kN/m}L=4┬аm,w=20┬аkN/m

2

Select the fixed end moment formula

For a fixed beam subjected to a UDL over the entire span, the fixed end moment MFM_FMFтАЛ at both supports is calculated using the standard structural mechanics formula.

MF=wL212M_F = \frac{wL^2}{12}MFтАЛ=12wL2тАЛ

3

Calculate the fixed end moment

Substitute the given values into the formula: MF=20├Ч4212=20├Ч1612=32012M_F = \frac{20 \times 4^2}{12} = \frac{20 \times 16}{12} = \frac{320}{12}MFтАЛ=1220├Ч42тАЛ=1220├Ч16тАЛ=12320тАЛ.

MF=26.666...┬аkNmтЙИ26.66┬аkNmM_F = 26.666...\text{ kNm} \approx 26.66\text{ kNm}MFтАЛ=26.666...┬аkNmтЙИ26.66┬аkNm

тЬЕ

D is correct because applying the fixed end moment formula for a UDL gives MF=20├Ч4212=26.66┬аkNmM_F = \frac{20 \times 4^2}{12} = 26.66\text{ kNm}MFтАЛ=1220├Ч42тАЛ=26.66┬аkNm.

Core Concepts Used
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Fixed Beam Uniformly Distributed Load (UDL) Fixed End Moment
ЁЯТб EXAM TIP

This formula is derived from the slope-deflection method or the principle of superposition, which is fundamental for solving statically indeterminate structures in advanced structural analysis.

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