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 beamsalong 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 momentDiagrams.

But if the bending moment changesOver the length x, then the degree ofBeam curvature must also be changing!

How Can We Get the Deflection of a Beam Under These Conditions?

That curvature is thesecond derivative ofvertical displacementat a distance x downthe beam

Time for some algebraic substitution!

By algebraBy substitution

Let Me See

y

If I have this

And I want this.

Oh Yes! How about integrating twice (Oh Gosh no – not calculus please)

Ok it May Suck But Lets Get to It

Or with substitution

If we integrate once

We get the angle on the deflectioncurve.

If We Integrate Twice

We can get the displacement of the beam

Dare We Try This?This is a cantilever beam witha simple end load.

Value of x

Lx

M

L

Our Moment Diagram

P

M=-P*x

Lets Integrate Once

= -P*x

cxPdx

dyEI

1

2

2

1

Houston – We Have a Problem

cxPdx

dyEI

1

2

2

1

I don’t see what c1 is

The first integration gives us the bend angle

Off the Wall Boundary Conditions

cLP 1

2

2

10

cxPdx

dyEI

1

2

2

1

The angle of bend at thewall is 0.

X L

So when x = L Lc P2

1 2

1

Anyone going to giveme genius credit for this?

Integrate Twice

Lx PPdx

dyEI

22

2

1

2

1

cLx xPPEIy2

23

2

1

6

1

SlopeAngle

Integrate

I get the equation for the elastic curve

We have another unknown c(I feel a boundary condition coming on).

Checking Out the Wall

cLx xPPEIy2

23

2

1

6

1

cLL PP2

33

2

1

6

10

What is the displacement at the Wall

We better hope y=0 at x=L

xL

Lc P3

2 3

1

Finishing Up Our Answer

LLx PXPPEIy323

3

1

2

1

6

1

)23(6

323

LLx xEI

Py

EI

PLy6

3

0

Get y by itself

And when x = 0

y0

Assignment 23

Do Problem 7.3.5Do Problem 7.3.16

Remember – Show and explain your work step by step. Scribbles with a circledAnswer 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 FaceDouble Integrations with BoundaryConditions?

Not Necessarily

As with mostcommon integralsthere is alwayssomeone makinga standard table.

Lets Try Doing One With a Table

Once upon a time there was aUniformly loaded Cantilever Beam

12 ft

100 lbs/ft

5.5 in

5.5 in

(6X6 woodBeam)

δ

We want the deflection distance δ

We Look Up the Solution On a Table

The displacement at the endOf the beam.

Plug In TimeLoad = 100 lbs/ft = 8.33 lbs/inch

Length 12 ft = 144 inches

Young’s Modulus forWood 1,750,000 psi(From Table F.2 in yourBook)

12

4

bI

76.26

3.35 in =

But Lets Face ItYou Can’t Make a Table of Everything

There are ways of extending basic tablesto complex situations using the methodof superposition. (Linear functions can beadded one atop another to make somepretty complex stuff.

How wouldyou find theanswer if youcould?

The Need for Method of Superposition

What are the chances I will findThis 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 loadedbeam

The Method of Superposition

Lets Try a Cantilever Beam With A Uniform Load Only Over the Outer Half

Tables Give Me a UniformlyLoaded Cantilever

So How Do I Get a Half Loaded Cantilever?

We apply superposition

Of Course There Is a Problem

If the tables don’t havehalf loaded cantilevers, howdo I get a half loaded cantilever?

The Bending Moment Diagram Gives Us a Clue

6 ft

12 ft

M

How much of a bending momentdid the maid in the parlor put onthe outer end of that beam?

You say None!

So that outer part of the beam isnot bent at all and only the innernalf has anything to do with theproblem.

Therefore We Can Start to Get One of Our Superposition Components from a half

length cantilever.

Displacement halfway out to the beam end is

EI

w

EI

Lw L1288

2(4

4

)

Now for the Rest of the Story

Because there is no bending theremaining half of the beam justkeeps going straight up.

But remember – we got the slopeAngle from our first integration soThat is bound to be in the table.

Going up for a distanceOf 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 twistthe main rod

B- The shear diagram will show that the 50Nload will transfer and become a 50 N pointload on the main rod which will act asa cantilever.

We need to recognize that withSuperposition we can solve eachproblem separately withoutinterference from the other.

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 N2M

Get Young’s ModulusFrom the problem

3 cm

We just need I for aCircular rod

Moment of Inertia for a Circular Rod

From tables such as those at the back of yourBook.

4

4

rI

=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 theseProblems. Putting down and answer with some scribbled steps does notCut it and will get you marked wrong even if you chosen number orExpression is correct.