Shear and Bending Moment Diagrams

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

Parallel ReadingChapter 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 MembersAnd the Load Distribution on them can be ComplexThis 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 MaximumsO HaGot You!My homeworkis due When!

Interesting Fact About Forces on Beam Cross SectionsIf we slice a beam at any cross section pointWe will find shear and bending forces thereI 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 BeamMxVx

Sign ConventionsPositive ShearNewly Elected Congressman

End Conditions Help us Figure Out Where to StartSimple 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 DiagramHere is a nice timber beam to work withWe can get our reactions at B and D using statics

Two Methods to Get the DiagramWe can do statics up and down cross sectionsOf the beam till we puke.

Or There is a Graphical MethodStarting 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 LoadLets give it a try moving left to rightV20 kNAB

Now What Happens at B?We see a jumpup of 46 kN

Now Lets Look at CWe take a dip of 40 kN-14 kN

And at D We Come Back the NeutralMove back upBy 14 kN

That Last One Wasnt BadLets Try This One!A evenly distributed loadresults in a linearslope on the sheardiagram.

We Start at the Left with Our Distributed Load8 ftWe accumulate aShear load of -24 kips

We Note that from C to D there is no change3 ftThings stay the sametill there is change

Now We Have a Point Load at D5 ftWe see a drop of10 kips at the pointload

And Our Reaction at B brings us back to NeutralPoint load brings us back upby 34 kips to neutral

Lets Get Even More ExcitingWell try an unevenly distributeddistributed load.

Oh Ya Like Figuring the Area of a Triangle will Scare UsWe draw a curve downto the appropriate value.We know the curve is 2ndorder but we dont 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 DiagramsA relationship to the rescue

Semi - English TranslationIntegrate the Area of the Shear Diagram to get total Moment Change for the Interval6 ft

8 ft10 ft8 ftNote that we integrated 18 kips over 6 feet(ok you dont 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 LineThe 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 Diagram92 Kip*ftWe take the 16 kip*ft away from our old valueOf 108 kip*ft and get our new value ofNote 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 zeroOur triangle has an area of12 * 8 / 2 = 48 kip*ftObviously a plus 48 kip*ftWill bring a minus48 kip*ft to zero

Rules for Shear and Moment DiagramsConcentrated loads produce a jump in the shear diagram and a change inThe linear slope of the moment diagram.

Shear and Moment DiagramsAn 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 RulesFor 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 dont sweat tomuch about exact values in between.Critical points

Assignment 21Do problems 5.5-1, 5.5-11, 5.4-14(yes I know they are out of order)