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- 1. Deflection of Beams<br />Chapter 9<br />
- 2. Introduction :<br />In this chapter we learn how to determine the deflection of beams (the maximum deflection) under given load .<br />A prismatic beam subjected to pure Bending is bent into an arc of a circle in the elastic range ,the curvature of the neutral surface expressed as :<br />1/ρ = M/EI<br />
- 3. Where , M is the bending Moment<br /> , E is the modulus of elasticity<br /> , I is the moment of inertia of the cross section about it’s neutral axis .<br />
- 4. Both the Moment “M” & the Curvature of neutral axis “1/ρ” will vary denoting by the distance of the section from the left of the beam “x” .<br />We write <br />1/ρ = M(x)/EI<br />
- 5. We notice that the deflection at the ends <br />y = 0 due to supports<br />dy/dx = 0 at A,B ,also dy/dx = 0 at the max. deflection<br />Deflection<br />
- 6. To determine the slope and the deflectionof the beam at any given point ,we first drive the following second order differential equations which governs the elastic curve<br />So the deflection (y) can be obtained through the boundary conditions .<br />
- 7. In the next fig. two differential equations are required due to the effective force (p) at point (D) .<br />One for the portion (AD) ,the other one for the portion (DB) .<br />P<br />D<br />
- 8. The first eq. holds Q1 ,y1<br />The second eq. holds Q2 ,y2<br />So ,we have four constants C1 ,C2 ,C3 ,C4 due to the integration process .<br />Two will be determined through that deflection (y=0) at A,B .<br />The other constants can be determined through that portions of beam AD and DB have the same slope and the same deflection at D<br />
- 9. If we took an example<br />M(X) = -Px<br />We notice that the radius of curvature “ρ” = ∞ ,so that M = 0<br />P<br />B<br />A<br />L<br />P<br />B<br />A<br />
- 10. P2<br />P1<br />C<br />Also we can conclude from the next example , P1>P2<br />We notice that +ve M so that the elastic curve is concaved downward .<br />A<br />D<br />B<br />+ve M<br />-ve M<br />Elastic curve<br />Q(x,y)<br />
- 11. Equation of elastic curve:<br />We know the curvature of a plane curve at point Q(x,y) is expressed as<br />Where , are the 1st & 2nd derivatives of a function y(x) represented by a curve ,the slope is so small and it’s square is negligible ,so we get that <br />
- 12. Is the governing differential equation for the elastic curve .<br />N.B.: ``EI`` is known as the flexural rigidity .<br />In case of prismatic beams (EI) is constant .<br />

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