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Longitudinal Strain Flexure Formula

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Presentation on theme: "Longitudinal Strain Flexure Formula"— Presentation transcript:

1 Longitudinal Strain Flexure Formula
Bending Longitudinal Strain Flexure Formula

2 Bending Theory There are a number of assumptions made in order to develop the Elastic Theory of Bending: The beam has a constant cross-section Is made of a flexible, homogenous material with the same E in tension and compression. The material is linearly elastic. Stress and Strain are proportional (material obeys Hooke’s Law). The beam material is not stressed past its proportional limit.

3 Longitudinal Strains in Beams
Approach: The Longitudinal Strains in beam can be determined by analyzing the curvature of the beam and its deformations. Consider a portion of a beam subjected only to positive bending. Assume that the beam is initially straight and is symmetrical. Draw Picture

4 Longitudinal Strains in Beams
Under the action of the bending moments, the beam deflects in the xy plane and its longitudinal axis is bent into a curve…positive. mn and pq remain plane and normal to the longitudinal axis. This fact is fundamental to beam theory. Even though a plane cross section in pure bending remains plane, there still may be deformations in the plane itself, but they are small and may be neglected.

5 Longitudinal Strains in Beams
With the beam in positive bending, we know that the lower part of the beam in is Tension and the upper part is in compression. Somewhere between the top and bottom of the beam is a surface that does not change in length. This is called the neutral surface of the beam.

6 Longitudinal Strains in Beams
The section mnpq intersect in a line through the center of curvature O, through an angle d, at a radius of . Figure 6.23 At the neutral surface, dx is unchanged, therefore d=dx. All other longitudinal lines between the two planes either lengthen or shorten, creating normal strains x.

7 FG06_23a_b.TIF Notes: Notice that any line segment Dx, located on the neutral surface, does not change its length, whereas any line segment Ds, located at the arbitrary distance y above the neutral surface, will contract and become Ds¢ after deformation.

8 Longitudinal Strains in Beams
To evaluate these normal strains, consider line ∆s, a distance y from the neutral surface. (we are assuming that the x axis lies along the neutral surface of the undeformed beam. When the beam deflects, the neutral surface moves with the beam, but the x axis remains in a fixed position. Yet Δs in the deflected beam is still located at the same distance y from the neutral surface). Therefore the length of ef after bending is:

9 Longitudinal Strains in Beams
Since the original length Δs is dx, it follows that its elongation is Δs-dx or –ydx/ . The longitudinal strain is the elongation divided by the initial length dx. This is called the Strain-Curvature Relation.

10 FG06_24.TIF Notes: Variation in strain over the cross section

11 Longitudinal Strains in Beams
The equation was derived solely from the geometry (without concern for material). Strains in a beam in pure bending vary linearly with distance from the neutral surface regardless of the shape of the stress-strain curve of the material. Longitudinal strains in a beam are accompanied by transverse strains because of Poisson’s ratio.

12 Longitudinal Stresses in Beams
Now we need to find the stresses from the strains, using a stress-strain curve. Note there are no accompanying transverse stresses because beams are free to deform laterally.

13 Normal Stresses in Beams
Stresses act over the entire area cross section of the beam and vary in intensity depending upon the shape of the stress-strain diagram and the dimensions. Fig 6.26 For a linearly elastic material, we can use Hooke’s law

14 FG06_26a_b.TIF Notes: A linear variation of normal strain must then be the consequence of a linear variation in normal stress,

15 Normal Stresses in Beams
By substitution we get: This equation shows that the normal stresses acting on the cross section vary linearly with the distance from y from the neutral surface. Fig 6.26b.

16 Normal Stresses in Beams
In general, the resultant of the normal stresses consists of two stress resultants: A force acting in the x-direction A bending couple acting about the z axis. Since the axial force is zero for a beam in pure bending: The resultant force in the x direction is zero (this gives us the location of the neutral axis) The resultant moment is the bending moment.

17 Moment-Curvature Relationship
Therefore the neutral axis passes through the centroid of the cross-sectional area for our given constraints. And the moment resulting from the normal stresses acting over the cross section is equal to the bending moment.

18 Moment-Curvature Relationship
The integral of all such elemental moments over the entire cross-sectional area is then equal to the bending moment. Equation 6-11 Recall that is the moment of inertia. Then we can express curvature in terms of the bending moment called the Moment Curvature Equation.

19 Moment-Curvature Relationship
This shows that the curvature is directly proportional to the bending moment applied and inversely proportional to EI – called the Flexural Rigidity of the beam. Flexural Rigidity is a measure of the resistance of a beam to bending.

20 Flexure Formula Now that we have located the neutral axis and derived the moment curvature relationship we can find the stresses in terms of bending moment. If we substitute the equation for curvature into our equation for normal stress we get Called the Flexure Formula.

21 Flexure Formula Stresses calculated with the flexure formula are called Bending Stresses or Flexural Stresses. The maximum tensile and compressive bending stresses acting at any given cross section occur at points farthest from the neutral axis.

22 Limitation Reminders The beam is subjected to Pure Bending
The beam has a constant cross-section Is made of a flexible, homogenous material with the same E in tension and compression. The material is linearly elastic. Stress and Strain are proportional. The beam material is not stressed past its proportional limit.

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