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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?