Shear Flow. Beams Subjected to Bending Loads So why did these Beams split down Their length?

Shear Flow

Beams Subjected to Bending Loads

So why did theseBeams split downTheir length?

Maybe they Just Dried Out – They are all Wood

Of course these aren’t wood.

Maybe We Can Find Answers in Our Shear and Moment Diagrams

Hear is a shear and momentDiagram, but I don’tSee anything horizontal.

Consider a Beam in Bending

We all know the top of the beam compresses and the bottom goes into tensionAnd there is a neutral axis in the middle yada yada yada

.

Expected

NotExpected

Lets Grab a Little Piece of that Beam Where Shear is Constant

We have nice balancing verticalEquilibrium

But why doesn’t it spin?

Could it be that we have a mystery force?

What Else Could be Happening as a Beam Bends

Mystery Solved

So What Kinds of Numbers are We Talking?

We know we can’tHave shear at theAir interface

It can’t be even

Ok – So What is Q

Lets consider a horizontal planeOn a beam so distance y1 awayFrom the neutral axis

And What About I?

The moment of inertia of the beam

Lets Do Something With It

Obviously the neutral plane isRight through the middle

Lets go get theShear flow onThe edge of theBoards!

Round Up Q

Now for I

If this were a steel I beamWe could just look up I.Unfortunately we are goingTo have to calculate it.

Middle board part isEasy.

Of course we’re still missing the contributionOf the boards on the ends.

For Our End Boards we Will be rescued by the Parallel Axis Theorem

Getting the Shear Flow

Note that shear flow is shear force perUnit of beam length.

In our case we are interested in what is trying to shear our nails in two if they arePlaced every 25 mm

Nice Spot Check of Shear Stress, but What Does the Stress Profile Look Like?

Note this means the peak stress is

1.5 * Average Shear Stress

Then there are typical Steel Beams

So that’s why theWeb crumpled up.

Designing a Beam

This couldGo wrong!The beamCould splitIn axialTension.

Lets Make Sure That Doesn’t Happen

We will use our shear and moment diagramsTo find the maximum bending moment

Then we will zero in on the requiredSection modulus

Obviously the Next Thing I Need Is Section Modulus as a Function of Beam Depth

Remember – Section ModulusIs Moment of Inertia over cWhere c is the distance fromThe neutral axis to the edge ofThe beam.

Working Through Our Substitution

Plug it in Plug it in

Given in the problem

From Our Moment Diagram

Just worked out by ourSubstitution

Solving the equation for d

Looks Like We Need a 4 X 10 for Our Beam

After all – could anything else go wrong

Yes – We Better Check the Shear Flow

We know the maximum sheer will be at the centerOf the beam

T allowable is 120 psi

Plug and Chug

Yipes! We wereGoing to use a4 X 10

We didn’t watch the sheer flowAnd it nearly bit us in the _ _ _ _

We need a 4 X 12 for this.

Lets Use Mohr’s Circle to Take a Look at the Beam Center

An ElementAt theBeamCenter

This element is subject toStrong shear forces, butWhat about axial force?(assume its on the neutralAxis)

Pure Shear

Our worstCase is nearThe beamedges

29.11425.115.3

3000*5.1*5.1

XAV Max

MaxIf we assume we use a 4 X 12

Now to Mohr’s Circle

τ

σ

Plot the clockwise shear114.29

At 90 degrees toThat we find aCounter clockwiseshear

-114.29

Since we have pureShear there is noTension or compressionOn these faces.

Since We Have Pure Shear We Have No Tension or Compression? Right?

What is this?

What angleIs that on?

Is it possible that shear flow could buckleA ductile material in compression on a 45Degree diagonal plane?