This document discusses determining the deflection of beams under load. It introduces the concepts of bending moment (M), modulus of elasticity (E), and moment of inertia (I) in determining curvature and deflection. The maximum deflection can be obtained by solving the second order differential equation that governs the elastic curve of the beam, using the boundary conditions of the beam's supports and applying any loads. Examples are provided to demonstrate how to set up and solve the differential equations to find the deflection at any point on beams with various load configurations.

1. Deflection of Beams Chapter 9

2. Introduction : In this chapter we learn how to determine the deflection of beams (the maximum deflection) under given load . 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 : 1/ρ = M/EI

3. Where , M is the bending Moment , E is the modulus of elasticity , I is the moment of inertia of the cross section about it’s neutral axis .

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” . We write 1/ρ = M(x)/EI

5. We notice that the deflection at the ends y = 0 due to supports dy/dx = 0 at A,B ,also dy/dx = 0 at the max. deflection Deflection

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 So the deflection (y) can be obtained through the boundary conditions .

7. In the next fig. two differential equations are required due to the effective force (p) at point (D) . One for the portion (AD) ,the other one for the portion (DB) . P D

8. The first eq. holds Q1 ,y1 The second eq. holds Q2 ,y2 So ,we have four constants C1 ,C2 ,C3 ,C4 due to the integration process . Two will be determined through that deflection (y=0) at A,B . The other constants can be determined through that portions of beam AD and DB have the same slope and the same deflection at D

9. If we took an example M(X) = -Px We notice that the radius of curvature “ρ” = ∞ ,so that M = 0 P B A L P B A

10. P2 P1 C Also we can conclude from the next example , P1>P2 We notice that +ve M so that the elastic curve is concaved downward . A D B +ve M -ve M Elastic curve Q(x,y)

11. Equation of elastic curve: We know the curvature of a plane curve at point Q(x,y) is expressed as 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

12. Is the governing differential equation for the elastic curve . N.B.: ``EI`` is known as the flexural rigidity . In case of prismatic beams (EI) is constant .