beams_1 rev

beams_1 rev - Beams are structures where the internal...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Beams are structures where the internal forces are more complex than trusses. While a truss bar only carries an internal axial force that is constant in the bar, beams can carry internal axial forces, shear forces, and internal moments (couples) that can vary along the length of the beam. A B 3000 lb 4' 2' 4 3 A y A x V (shear force) N (axial force) M (bending moment) P P The directions shown are the ones normally taken for positive internal forces and moments on a "right side" cut P These internal forces and moments are the resultants of distributed internal forces acting on the beam cross section. A y A x V N M P As with all internal forces and moments, these always occur in equal and opposite pairs: B 3000 lb 4 3 P N V M Thus, on a "left side" cut, the positive values for these internal forces and moments are opposite to those on a "right side" cut To determine internal forces, we must first find the external reactions. If the beam is statically determinate, this is always possible. A B 2400 lb 4' 2' A y A x B 1800 lb + + + A B 3000 lb 4' 2' 4 3 ( 29 ( 29 6 2400 4 1600 A M B B lb =- = = ∑ 1800 1800 x x x F A A lb = + = = - ∑ 2400 1600 800 y y y F A A lb =- + = = ∑ A 1600 2400 lb 4' 2' 800 1800 B 1800 lb With these external reactions known, we now can find the internal forces and moment at any cut. For a cut 2' to the right of A, for example: A 800 1800 N M 2' P + + + V 1800 1800 x F N N lb =- = = ∑ 800 800 y F V V lb =- = = ∑ ( 29 ( 29 800 2 1600 P M M M ft lb =- = =- ∑ These internal values depend on the location of the cut. For example, for a These internal values depend on the location of the cut....
View Full Document

This note was uploaded on 10/17/2011 for the course EM 274 taught by Professor Boylan during the Fall '08 term at Iowa State.

Page1 / 17

beams_1 rev - Beams are structures where the internal...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online