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
Middle board part is
Easy.
If this were a steel I beam
We could just look up I.
Unfortunately we are going
To have to calculate it.
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 Plug it in
From Our Moment Diagram
Given in the problem
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
V Max  1.5 * 3000  114.29

1
.
5
*
 Max
A
3.5 X 11.25
If we assume we use a 4 X 12
Now to Mohr’s Circle
τ
Plot the clockwise shear
114.29
σ
At 90 degrees to
That we find a
Counter clockwise
shear
-114.29
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?