ENTC 376 Chapter 6 Lecture Notes 3 Flexure Formula

8/2/2019 ENTC 376 Chapter 6 Lecture Notes 3 Flexure Formula

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ENTC-376 Strength of Materials

• Continue Chapter 6• Work Examples

• Discuss Exam I

Chapter 6: Bending



MM

MM

cyσ

σ

How to determine the orientations of bending stresses?

stress state at a “POINT”

IMy=

Chapter 6: Bending



Chapter 6: Bending

Let’s apply anarbitrary moment,M, in the z-y planemaking an angle θ,with the z-axis.



Chapter 6: Bending

Decompose the moment intothe z-axis component and they-axis component.

Mz = M cos θ

And

My = M sin θ



Chapter 6: Bending

Applying theflexure formula,

To every pointon the surfaceyields the

stress planeshown.

The neutral axis forms an angle,α

with the z-axis.



Chapter 6: Bending

The angle α, of the neutral axiscan be found by the following

equation, noting that the neutralaxis is when stress is zero.

Substituting the previous momentequations yield,



Chapter 6: Bending

Since the definition of the slope of the line is rise over run,

or:

Substituting yields,

NOTE: α ≠ θ

θ = defines the direction of the moment.α = defines the angle of the neutral axis.



Chapter 6: Bending

Composite Beams

Steel re-enforced

concrete

Why is the

steelembedded inthe concrete?



Chapter 6: Bending

Flexure formula only valid for homogeneousmaterial.

Therefore,

We need to “transform” the composite beam into ahomogeneous beam.

Sounds great,but how do we do it?



Chapter 6: Bending

Like a homogeneous beamthe stress will very linearlyfrom zero at the neutral

axis to a max at theoutermost fiber.

If ….material has linear elastic behavior, Hooke’s

Law applies or….

Material 1Material 2



Chapter 6: Bending

If material 1 (blue) is stiffer than material 2 (brown), the

bending stress could look likethis. Notice the disconnect atthe material boundary.

Stiffness changes → Stresschanges



Chapter 6: Bending

Two conditions:

1 Stress distribution produceszero resultant force on the

cross section.

2. Moment of the stress

distribution about the neutralaxis equals to M.

What?



Chapter 6: Bending

Simpler way:

Lets transition the material into onehomogeneous beam.

Let the beam consist entirelyof the less stiff material, but keep

the same height, and hencesame strain distribution.

Therefore, we have to widen thetop stiffer section to accommodatethe stress.



Chapter 6: Bending

Equating the forces of thetwo sections, yield:

or,

Where, n = transformationfactor

Therefore, the original width b must be increased to nb.



Chapter 6: Bending

Now that the material has

been transformed into asingle material, the normalstress distribution will be

linear.

One additional point,

The resultant stress of thetransformed material must

be multiplied by n (or n’).

We will work an example

toward the end of class.



Chapter 6: Bending

Curved Beams

Remember, the flexure formula assumeda linear stress profile, which is accuratefor a straight beam, but not for a curved

one.

Assumptions

1. Uniform X-section

2. Homogeneous

& Isentropic3. Behaves in alinear-elasticmanner



Chapter 6: Bending

Three radius are defined from the center of curvature

1.¯ represents the known location of the centroid.

2. R represents the yet unknown location of the neutral axis.

3. r represents an arbitrary point or area element dA

r



Chapter 6: Bending

Lets start with the strain. Using the arc formula,

Define

Yields

Strains vary in a hyperbolic function of r.



Chapter 6: Bending

Using Hooke’s Law, we get,

To find the neutral axis (i.e. the value or R) we require thestress to equal zero.



Chapter 6: Bending

Lets solve the equation

Doing some algebra, we get,

We can solve for some geometries (Table 6-2)



Chapter 6: Bending

We still need to find the stress as a function of the moment, so:

Integrating over the entire cross section, yields



Chapter 6: Bending

Lets look at this equation

A / R

Area

r A

Therefore,

¯



Chapter 6: Bending

Finally, substituting for Ek yields,

where,σ = normal stress on the member M = internal moment (+ tends to straighten the beam)

A = Cross sectional area.R = Distance from the center of curvature to theneutral axis.

= Distance from the center of curvature to thecentroid.

r = Distance from the center of curvature to the areawhere the stress is determined



Chapter 6: Bending

One last substitution, let,

Then,



Chapter 6: Bending

Homework # 6Due 3/5/09

1. 6-5

2. 6-16

3. 6-17

4. 6-425. 6-49

6. 6-52

7. 6-53

8. 6-569. 6-69

10. 6-72

11. 6-82

12. 6-9413. 6-95

14. 6-120

15. 6-149 (bonus)



Chapter 6: Bending

Exam Discussion

Scheduled for next Thursday (3/5/09)

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ENTC 376 Chapter 6 Lecture Notes 3 Flexure Formula