Shear and Bending Moment Diagrams

(Credit for many illustrations is given to McGraw Hill publishers and an array ofinternet search results)

Parallel Reading

Chapter 5 Section 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 (Do Chapter 5 Reading Assignment Problems)

Beams are One of the Most Important Structural Members

And the Load Distribution on them can be Complex

This can mean that we have to find our maximumstress loads in order to design beams to handletheir task

Shear and Bending Moment Diagrams are One of the Tools we

Use to Find Stress Maximums

O HaGot You!

My homeworkis due When!

Interesting Fact About Forces on Beam Cross Sections

If we slice a beam at any cross section point

We will find shear and bending forces there

I bet that is why welike to do shear andbending momentdiagrams.

Our Mission is to Learn How to Plot the Shear and Moment Down the

Length of the Beam

M

x

V

x

Sign Conventions

Positive Shear

Newly Elected Congressman

End Conditions Help us Figure Out Where to Start

Simple Support can push and holdin place but cannot apply a moment.

Fixed end can supply xand y forces as well asmoment

Over-hang configuration

Lets Try to Get a Diagram

Here is a nice timber beam to work with

We can get our reactions at B and D using statics

Two Methods to Get the Diagram

We can do statics up and down cross sectionsOf the beam till we puke.

Or There is a Graphical Method

Starting with the shear diagramwe note that the change in shearis equal to the additional weightadded to the next increment.

If we go over a 1 foot increment, how muchshear will be pick up?

Of Course the Easiest Case is the Jump that Occurs at a Point Load

Lets give it a try moving left to right

V

20 kN

A B

Now What Happens at B?

We see a jumpup of 46 kN

Now Lets Look at C

We take a dip of 40 kN

-14 kN

And at D We Come Back the Neutral

Move back upBy 14 kN

That Last One Wasn’t BadLets Try This One!

A evenly distributed loadresults in a linearslope on the sheardiagram.

We Start at the Left with Our Distributed Load

8 ft

We accumulate aShear load of -24 kips

We Note that from C to D there is no change

3 ft

Things stay the sametill there is change

Now We Have a Point Load at D5 ft

We see a drop of10 kips at the pointload

And Our Reaction at B brings us back to Neutral

Point load brings us back upby 34 kips to neutral

Lets Get Even More Exciting

We’ll try an unevenly distributeddistributed load.

Oh Ya – Like Figuring the Area of a Triangle will Scare Us

We draw a curve downto the appropriate value.

We know the curve is 2nd

order but we don’t have tobe big on detail because ourcritical load point is not goingto be on a down sloping curve.

Now Its Boring Same Old Over to the Reaction at the Wall

Knowing that We Can Get the Total Change in Shear Load Over an Interval by Integrating Under the Load Curve lets Us Solve for Any Crazy Load

Distribution.

Integrate this areato get the magnitudeof

This

Now We Have to Deal With Bending Moment Diagrams

A relationship to the rescue

Semi - English TranslationIntegrate the Area of the Shear Diagram to get

total Moment Change for the Interval6 ft 8 ft 10 ft

8 ft

Note that we integrated 18 kips over 6 feet(ok you don’t have to integrate to know the area of arectangle is base X height)

To find the area of 108 positive

We now use this for the magnitude of theMoment change over that 6 ft.

When the Slope of the Shear Diagram is Zero the Slope of the Moment Diagram is a

Straight Line

The area of the shear diagram from B to C is-2 kips X 8 ft = -16 kip*ft

Working Out the Next Part of OurMoment Diagram

92 Kip*ft

We take the 16 kip*ft away from our old valueOf 108 kip*ft and get our new value of

Note also that since the shear lineHas no slope the drop in momentIs linear.

We Next Drop to -14 kips shear over 10 ft.

Integrating that(ok just getting the area of the rectangle on the shear diagram)

We get -14 kips * 10 ft = -140 kip*ft

Now We Will Put that -140 kip*ft into the Moment Diagram

+ 92 kip*ft – 140 kip*ft = -48 kip*ft

-48Again note theSteady sloped line

Our Shear Load Jumps at Point D and then tapers to zero

Our triangle has an area of12 * 8 / 2 = 48 kip*ft

Obviously a plus 48 kip*ftWill bring a minus48 kip*ft to zero

Rules for Shear and Moment Diagrams

Concentrated loads produce a jump in the shear diagram and a change inThe linear slope of the moment diagram.

Shear and Moment Diagrams

An evenly distributed load produces and linear slope in the shear diagramand the value of the shear is the slope of the line in the moment diagram.

Shear and Moment Diagram Rules

For an unevenly distributed load the value of the load is the slope of theLine in the shear diagram (and the value shear is the slope of the momentLine)

For Your SanityYou can go nuts trying to get theslope and curve right between twopoints but it is the value inflextionpoints on the moment curve thatcontrols design.

Get the values at the section endpoints right and don’t sweat tomuch about exact values in between.

Critical points

Assignment 21

Do problems 5.5-1, 5.5-11, 5.4-14(yes I know they are out of order)