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.

Like this presentation? Why not share!

- Unit 5 shear force and bending mom... by Hareesha N Gowda,... 36500 views
- Lecture 9 shear force and bending m... by Deepak Agarwal 125539 views
- Shear Force And Bending Moment In B... by Amr Hamed 56817 views
- Shear force and bending moment by temoor abbasa 3819 views
- Lesson 04, shearing force and bendi... by Msheer Bargaray 7450 views
- Shear force and bending moment Solv... by Royal University ... 3710 views

No Downloads

Total views

6,799

On SlideShare

0

From Embeds

0

Number of Embeds

14

Shares

0

Downloads

331

Comments

0

Likes

3

No embeds

No notes for slide

- 1. Chapter 5<br />The 1st chapter in 2nd term<br />
- 2. Introduction :<br />In this chapter we care for the analysis and the design of beams ,structural members supporting loads applied at different points ,In most cases ,the loads are perpendicular to the axis of the beam ,when loads aren’t at right angles they also produce axial forces .<br />
- 3. The loading at beams consists of <br />Concentrated loads :<br />P1,P2 expressed in newtons ,pounds.<br />
- 4. Distributedloads(w) :<br />Expressed in (N/m) or (b/ft)<br />The load from (A) to (B) is said to be “uniformly distributed”<br />
- 5. Classification of beams<br />Beams are classified due to the way they are supported .<br />The distance “L” shown called the span .<br />Beams a,b,c are said to be statically determinate beams<br />Beams d,e,f are said to be statically intermediate beams that conclude more than three unknowns .<br />
- 6.
- 7. Other beams are connected by hinges to form a simple continuous structure .It would be noted that the reaction at the supports involves four unknowns .<br />
- 8. “Shear & Bending Moment Diagram”<br />If the force is uniform distributed as Fo for a length of span L ,so the resultant = FoL acts in the medium of length .<br />We notice that Fo is uniformly distributed force .<br />The total force value is FoL<br />
- 9. If the force isn’t uniformly distributed (w=ax) ,as shown .<br />So ,wmax= aL , wmin = 0<br />the total resultant force =<br />area of the force triangle <br />= 0.5L wmax = L (aL) =0.5a L2<br />The resultant force =0.5a L2<br />Point of effect (see in fig.)<br />
- 10. The positive & negative Shear “S”<br />If we cut section through abeam :<br />Positive shear : look (from left to right) <br />Negative shear : look (from right to left)<br />
- 11. The positive & negative Moment “M”<br />Physically : we consider the +ve moment ,if the effect of external forces acting on the beam tend to bend it as shown .<br />
- 12. Mathematically : we consider the +ve moment as shown<br />
- 13. So we can conclude the +ve (shear & moment) <br />We can also conclude the –ve (shear & moment) <br />

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment