Deflection of Beams
(Credit for many illustrations is given to
McGraw Hill publishers
and an array of internet search results)
Parallel Reading
Chapter 7
Section 7.1
Section 7.2
Section 7.3
Section 7.6
(Do Chapter 7 Reading Assignment Work)
Previously on Engr 350
Bending moments deflect beams
along the arc of a circle.
The degree of curvature depends on
the magnitude of the Bending Moment,
Young’s Modulus, and the beams
Moment of Inertia
But as the Stomach Turns We Recently Learned
that Bending Moments Vary over the Length of
Beams
We learned how to do bending moment
Diagrams.
But if the bending moment changes
Over the length x, then the degree of
Beam curvature must also be changing!
How Can We Get the Deflection of
a Beam Under These Conditions?
That curvature is the
second derivative of
vertical displacement
at a distance x down
the beam
Time for some algebraic substitution!
By algebra
By substitution
Let Me See
y
And I want this.
If I have this
Oh Yes! How about integrating twice
(Oh Gosh no – not calculus please)
Ok it May Suck But Lets Get to It
If we integrate once
We get the angle on the deflection
curve.
Or with substitution
If We Integrate Twice
We can get the displacement of the beam
Dare We Try This?
This is a cantilever beam with
a simple end load.
P
Value of x
M
Our Moment Diagram
L
x
L
M=-P*x
Lets Integrate Once
= -P*x
dy
1
2
EI
  P x  c1
dx
2
Houston – We Have a Problem
dy
1
2
EI
  P x  c1
dx
2
I don’t see what c1 is
The first integration gives us the bend angle
Off the Wall Boundary Conditions
The angle of bend at the
wall is 0.
L
X
EI
dy
1
2
  P x  c1
dx
2
So when x = L
1
2
0   P L  c1
2
1
2
c1  2 P L
Anyone going to give
me genius credit for
this?
Integrate Twice
EI
dy
1
1
2
2
 Px  PL
dx
2
2
Slope
Angle
Integrate
1
1
3
2
EIy   P x  P L x  c2
6
2
I get the equation for the elastic curve
We have another unknown c
(I feel a boundary condition
coming on).
Checking Out the Wall
L
x
1
1
3
2
EIy   P x  P L x  c2
6
2
What is the displacement at the Wall
We better hope y=0 at x=L
1
1
3
3
0   P L  P L  c2
6
2
c2 
1
3
PL
3
Finishing Up Our Answer
1
1
1
3
2
3
EIy   P x  P L X  P L
6
2
3
Get y by itself
P
3
2
3
y
( x  3 L x  2 L )
6 EI
And when x = 0
3
PL
y0  6EI
y0
Assignment 23
Do Problem 7.3.5
Do Problem 7.3.16
Remember – Show and explain your work step by step. Scribbles with a circled
Answer at the end – even if the answer is right, will be marked wrong.
We Can Well Imagine Some of the Loadings and
Bending Moments We Might Face Could be
Unpleasant
Are We Always Going to Face
Double Integrations with Boundary
Conditions?
Not Necessarily
As with most
common integrals
there is always
someone making
a standard table.
Lets Try Doing One With a Table
Once upon a time there was a
Uniformly loaded Cantilever Beam
100 lbs/ft
5.5 in
δ
12 ft
5.5 in
(6X6 wood
Beam)
We want the deflection distance δ
We Look Up the Solution On a
Table
The displacement at the end
Of the beam.
Plug In Time
Load = 100 lbs/ft = 8.33 lbs/inch
Length 12 ft = 144 inches
4
3.35 in =
I b
12
Young’s Modulus for
Wood 1,750,000 psi
(From Table F.2 in your
Book)
76.26
But Lets Face It
You Can’t Make a Table of Everything
There are ways of extending basic tables
to complex situations using the method
of superposition. (Linear functions can be
added one atop another to make some
pretty complex stuff.
How would
you find the
answer if you
could?
The Need for Method of
Superposition
What are the chances I will find
This situation in a table?
I’ve got a bad feeling about that.
I Did See Some Stuff in the Tables
that Relates
Here is a set for a uniformly loaded beam
Here is a point loaded
beam
The Method of Superposition
Lets Try a Cantilever Beam With A
Uniform Load Only Over the Outer Half
Tables Give Me a Uniformly
Loaded Cantilever
So How Do I Get a Half Loaded
Cantilever?
We apply superposition
Of Course There Is a Problem
If the tables don’t have
half loaded cantilevers, how
do I get a half loaded cantilever?
The Bending Moment Diagram
Gives Us a Clue
M
6 ft
12 ft
How much of a bending moment
did the maid in the parlor put on
the outer end of that beam?
You say None!
So that outer part of the beam is
not bent at all and only the inner
nalf has anything to do with the
problem.
Therefore We Can Start to Get One of Our
Superposition Components from a half
length cantilever.
Displacement halfway out to the beam end is

w( L
4
2)
8EI
4
wL

128 EI
Now for the Rest of the Story
Because there is no bending the
remaining half of the beam just
keeps going straight up.
But remember – we got the slope
Angle from our first integration so
That is bound to be in the table.
Going up for a distance
Of L/2
We Are Now Able to Give the Displacement
for the cantilever with loading on the inner
half
Now Back to Applying
Superposition
How Might the FE Test Our
Mastery of Beam Deflections?
What is going to happen with this load?
A- That eccentric 50N load is going to twist
the main rod
We need to recognize that with
Superposition we can solve each
problem separately without
interference from the other.
B- The shear diagram will show that the 50N
load will transfer and become a 50 N point
load on the main rod which will act as
a cantilever.
This is Just a Cantilever with a
Point Load at the End
50 N
2 meters
From a deflection table
(or double integration like we did in class)
Going for a Quick Plug
50 N
2M
We just need I for a
Circular rod
3 cm
Get Young’s Modulus
From the problem
Moment of Inertia for a Circular
Rod
From tables such as those at the back of your
Book.
I
r
4
4
=3.976X10-8
And Finishing
And Pick D!
Assignment 24
Do problems
7.3-18
7.3-21
Remember to explain and show step by step how you are solving these
Problems. Putting down and answer with some scribbled steps does not
Cut it and will get you marked wrong even if you chosen number or
Expression is correct.