Shear and Bending Moment Diagrams

Shear and Bending Moment
Diagrams
(Credit for many illustrations is given to McGraw Hill publishers and an array of
internet 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 maximum
stress loads in order to design beams to handle
their task
Shear and Bending Moment
Diagrams are One of the Tools we
Use to Find Stress Maximums
My homework
is due When!
O Ha
Got You!
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 we
like to do shear and
bending moment
diagrams.
Our Mission is to Learn How to Plot
the Shear and Moment Down the
Length of the Beam
V
x
M
x
Sign Conventions
Positive Shear
Newly Elected Congressman
End Conditions Help us Figure Out
Where to Start
Simple Support can push and hold
in place but cannot apply a moment.
Fixed end can supply x
and y forces as well as
moment
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 sections
Of the beam till we puke.
Or There is a Graphical Method
Starting with the shear diagram
we note that the change in shear
is equal to the additional weight
added to the next increment.
If we go over a 1 foot increment, how much
shear 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
A
B
V
20 kN
Now What Happens at B?
We see a jump
up 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 up
By 14 kN
That Last One Wasn’t Bad
Lets Try This One!
A evenly distributed load
results in a linear
slope on the shear
diagram.
We Start at the Left with Our
Distributed Load
8 ft
We accumulate a
Shear load of -24 kips
We Note that from C to D there is
no change
3 ft
Things stay the same
till there is change
Now We Have a Point Load at D
5 ft
We see a drop of
10 kips at the point
load
And Our Reaction at B brings us
back to Neutral
Point load brings us back up
by 34 kips to neutral
Lets Get Even More Exciting
We’ll try an unevenly distributed
distributed load.
Oh Ya – Like Figuring the Area of a
Triangle will Scare Us
We draw a curve down
to the appropriate value.
We know the curve is 2nd
order but we don’t have to
be big on detail because our
critical load point is not going
to 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 area
to get the magnitude
of
This
Now We Have to Deal With
Bending Moment Diagrams
A relationship to the rescue
Semi - English Translation
Integrate the Area of the Shear Diagram to get
total Moment Change for the Interval
6 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 a
rectangle is base X height)
To find the area of 108 positive
We now use this for the magnitude of the
Moment 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 Our
Moment Diagram
We take the 16 kip*ft away from our old value
Of 108 kip*ft and get our new value of
92 Kip*ft
Note also that since the shear line
Has no slope the drop in moment
Is 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
Again note the
Steady sloped line
-48
Our Shear Load Jumps at Point D
and then tapers to zero
Our triangle has an area of
12 * 8 / 2 = 48 kip*ft
Obviously a plus 48 kip*ft
Will bring a minus
48 kip*ft to zero
Rules for Shear and Moment
Diagrams
Concentrated loads produce a jump in the shear diagram and a change in
The linear slope of the moment diagram.
Shear and Moment Diagrams
An evenly distributed load produces and linear slope in the shear diagram
and 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 the
Line in the shear diagram (and the value shear is the slope of the moment
Line)
For Your Sanity
You can go nuts trying to get the
slope and curve right between two
points but it is the value inflextion
points on the moment curve that
controls design.
Critical points
Get the values at the section end
points right and don’t sweat to
much about exact values in
between.
Assignment 21
Do problems 5.5-1, 5.5-11, 5.4-14
(yes I know they are out of order)