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FIXED BEAM INTRODUCTION

Fixed beam introduction

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FIXED BEAM INTRODUCTION

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  1. D= R - E = 6 - 3 = 3 (GENERAL CONDITION)

  2. D = R - E = 4 - 2 = 2 (vertical load condition) W There is no horizontal load,therefore horizontal reactions are not developed

  3. To analyse the fixed beam we need four (4) equations,but we have only two equations,those are1) Σ Fv= 02) Σ M= 0We need two extra equations to solve the fixed beamThose equations are “compatability equations”

  4. A B elastic curve or deflection curve

  5. From figure before and after application of load there is no change in the points A and B • Therefore the angle between two tangents drawn at A and B is zero(slope is zero), • 3) θAB= 0 A/EI=0 • The tangents drawn at A will pass through B,therefore the intercept on the vertical at A by the tangents drawn at A and B will be zero. • 4) δAB= 0 A.X/EI=0

  6. In the 3) A/EI = 0 and 4) A.X/EI = 0 equations “A” is present. • Where A is area of bending moment diagram it is unknown, for BMD we need bending moments, for bending moments we need to find out reactions. • But in fixed beam we can not find out reactions directly with the available two equations(ΣFv=0, ΣM=0). without reactions we can not find out bending moment and followed by bending moment diagram. • So, due to this problem, we are consider the fixed beam as equal to simply supported beam with end moments applied.

  7. In this case first add the given load on SSB and calculate the reactions and draw BMD, next remove the given load and apply end moments and draw BMD. By using superposition principle add these two diagrams which is equal to fixed beam BMD. Now we have the area of BMD i.e (now “A” is a known value),substitute the “A” value in 3) and 4) equations we can complete the analysis of fixed beam. It is explained in below diagrams.

  8. Fixed beam with vertical loading condition is equal to simply supported beam with verical loading and end moments applied.

  9. Thank you……..!

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