Shear Flow

Beams Subjected to Bending Loads So why did these Beams split down Their 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 moment Diagram, but I don’t See anything horizontal.

Consider a Beam in Bending We all know the top of the beam compresses and the bottom goes into tension And there is a neutral axis in the middle yada yada yada. Expected Not Expected

Lets Grab a Little Piece of that Beam Where Shear is Constant We have nice balancing vertical Equilibrium 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’t Have shear at the Air interface It can’t be even

Ok – So What is Q Lets consider a horizontal plane On a beam so distance y1 away From the neutral axis

And What About I? The moment of inertia of the beam

Lets Do Something With It Obviously the neutral plane is Right through the middle Lets go get the Shear flow on The edge of the Boards!

Round Up Q

Now for I If this were a steel I beam We could just look up I. Unfortunately we are going To have to calculate it. Middle board part is Easy. Of course we’re still missing the contribution Of 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 per Unit of beam length. In our case we are interested in what is trying to shear our nails in two if they are Placed 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 the Web crumpled up.

Designing a Beam This could Go wrong! The beam Could split In axial Tension.

Lets Make Sure That Doesn’t Happen We will use our shear and moment diagrams To find the maximum bending moment Then we will zero in on the required Section modulus

Obviously the Next Thing I Need Is Section Modulus as a Function of Beam Depth Remember – Section Modulus Is Moment of Inertia over c Where c is the distance from The neutral axis to the edge of The beam.

Working Through Our Substitution

Plug it in Given in the problem From Our Moment Diagram Just worked out by our Substitution 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 center Of the beam T allowable is 120 psi

Plug and Chug Yipes! We were Going to use a 4 X 10 We didn’t watch the sheer flow And 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 Element At the Beam Center This element is subject to Strong shear forces, but What about axial force? (assume its on the neutral Axis)

Pure Shear Our worst Case is near The beam edges If we assume we use a 4 X 12

Now to Mohr’s Circle τ σ Plot the clockwise shear At 90 degrees to That we find a Counter clockwise shear Since we have pure Shear there is no Tension or compression On these faces.

Since We Have Pure Shear We Have No Tension or Compression? Right? What is this? What angle Is that on? Is it possible that shear flow could buckle A ductile material in compression on a 45 Degree diagonal plane?