Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

- Lecture 12 deflection in beams by Deepak Agarwal 42631 views
- 9 beam deflection by Mohamed Yaser 45057 views
- Lesson 05, bending and shearing st... by Msheer Bargaray 5390 views
- Deflection in beams 1 by Satya Narayan Kunwar 11318 views
- 8 beam deflection by Lisa Benson 9523 views
- Structural Mechanics: Deflections o... by Alessandro Palmeri 51001 views

No Downloads

Total views

6,644

On SlideShare

0

From Embeds

0

Number of Embeds

20

Shares

0

Downloads

272

Comments

0

Likes

5

No embeds

No notes for slide

- 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 />

No public clipboards found for this slide

Be the first to comment