Influence Linesor Bridges Under the Influence

What are They?

At first glance they look like the shear andMoment diagrams we use for finding stressMaximums in beams.

What is Different?

Shear and MomentDiagrams show theImpact of “dead” orFixed loads across anEntire beam orStructure.

We use them to find out where peaks of shear or moment are located

Influence Lines

Influence lines show the impact of live moving loadsAt a single point as the load moves across the beam.

They are important because sometimes we have moving traffic on bridges or beams thatContribute to stress and are not accounted for by loads at fixed positions.

Methods of Producing Influence Lines

Several Methods but theTabulation Method isThe easiest to understand

You take a moving load of oneunit weight.

You pick your point of interest

You place your moving load atVarious points and use staticsPrinciples to find the load atYour point of interest.

Make life easier – for staticallyDeterminate structures you getStraight lines (although the lineSlope may change as the loadPasses over key points.

Why Do I Use a Unit Load of One for an Influence Line?

1 is easy to multiply by the weight of any thingOr any number of things I want.

Influence lines are popular for studying the impact of moving – variable loads on bridgesAnd other such structures.

Sign Conventions

A reaction acting upward is positive.

On Shear

Right Side UpIs Positive.

More Sign Conventions

Moments that bend upward are positive

Lets Do a Simple Tabulated Influence Line

X

Create an influence line for theReaction at B as the unit load movesAcross the beam.

Our controlling statics equation isThat the sum of moments aboutA must be 0.

Try different valuesOf x and computeThe result.

Observations

The weight of our moving load is 1(could be one anything)

The points we picked to try for reactionAt B were fairly arbitrary though theyFollow a consistent pattern.

2.5 M1

Reaction B

M1 = 1*2.5 = 2.5

M2= 10* X = 10XSum of Moments = 0

2.5 = 10 * X X= 0.25

10 M

We Now Use Our Tabulation to Draw the Influence Line

Note the straight line shape typical of statically determinateBeams.

Lets Throw in Another Twist

Using a unit weight of 1We calculate the reactionAt B needed to keep theSum of the moments at AEqual to zero

This time we will cantilever thebeam

And We Plot Our Influence Line

Lets Try Some Shear and Moment Influence Lines

We’ll try a total Cantilever beam.

Lets try first for shear at point A

Known Facts at Point A

Therefore a load of 1Anywhere on the beamMust produce a reactionOf 1 at A

So What is the Shear Influence Line at Point A?

Lets Try for Shear at Point B

B

How much of a shear load is transmitted throughPoint B when our load is between A and B?

What About When the Load is Beyond B?

Now Lets Do the Shear Thing When We Have Support at Both Ends

I want an influence line for shear at point c

What Happens When a Load is Parked Smack Dab on C?

So How Did I Know the Shear was 0.5 in Each Direction?

7.5 M7.5 M

If the load is in the middle how muchSupport comes from each reaction?

I can always refer to my reaction diagramFor point B if I am unsure

Next Consideration

How much load is thereTransmitted through pointC when the load is totally atPoint B?

0.5

Working on our shear line

So What Happens In-between?

0.5

0

We know that for staticallyDeterminate structuresWe should get straight lines

?

So Does Our Straight Line Hypothesis Make Sense?

Remember our reaction at point B increases linearly until it has the full load?

That means the shear that has to pass through C to reach supportA is declining linearly - Yup it makes sense

We now also know why we wanted to know theReactions at A and B (they tell us how muchShear has to pass through our point of interest).

The Moment has Come to Do Moments

Lets think about the moment at A as ourMoving load of 1 unit Cow moves downThe cantilever.

Tabulate and Plot

X Moment

1 1 cowmeter

2 2 cowmeter

3 3 cowmeter

6 6 cowmeter

8 8 cowmeter

10 10 cowmeter

The tabulation and equation

X

-10

010

So Lets Up the Toughness and Make it Point B

What is the bending action at point B when our 1 unit AntelopeIs between point A and point B.

But What Happens When Our Antelope Goes Past B

If we are 1 meter past B, what is theMoment at B?

1 unit Antelope * 1 meter = 1 antelope*meter(the cows and antelope illustrate that we doInfluence lines with a unit weight of 1 – what everThat unit might be).

The Result

Now Lets Try it With Supports on both sides

Here we take advantage of having computed the reactions at AAnd B as our unit load moves across the beam.

The Computation

We knowWe know the reaction at B because we haveAlready calculated it (or could have calculated it).

Let us suppose that X = 2.5 meters and calculate it out.

Reaction at B

The moment at C has to cancel theReaction at B X its lever arm of 7.5 M

Referring to the Tabulation at BAt 2.5 meters the reaction at B is 0.167

Plug that into our equation.

We now solve for the moment at C

We Build Our Table Off of the Principle

We take our tabulated reactions at B and solve forMoment at C at each point as the load moves across

And We Plot the Result

There is One Mystery

I calculated the increasing moment asThe load approached C using the reactionAt B. Then after the load passed I usedThe reaction at A.

How did I know whether to use the reactionAt A or the reaction at B for my momentCalculation at C?

Actually It Didn’t Really Make In Difference to the Answer

What if I calculated Using reaction at A

Reaction at A

I would needThe momentFrom reactionAt A

But I would have toDirectly also computeThe opposing momentFrom my load

The Answer Would Be the Same

The amount of work needed to getThe answer would not be the same.

If you get the point.

Assignment

Plot the influence lines for the reactions at A and B

Plot the influence line for shear at point C

Plot the influence line for the moment at point C