He probably did-we were just a lot dumber back then : )


"Tom Miller" wrote in message
...
Steve. I wish my Strength of Materials prof. would have explained it
that
well 35 years ago!

Tom Miller
wrote in message
oups.com...
You are making this more complicated than it actually is:

The beam that you are trying to design is static. It is a beam
with a
fixed support and a simple support. A fixed support can only have
force
in the vertical and a simple support can have forces in both the
vertical and horizontal. Therefore, to figure the moments and
forces
all you need to do is use the you three equations of equilibrium.
These
a

Sum of Moments=0 Sum of Vertical Forces=0 Sum of Horizontal
Forces=0

When it comes to moments clockwise reactions are negative and
counter
clockwise reactions are positive.

There is no horizontal forces in this problem so we can ignore it.

Your vertical load is 250lbs. You need to put in a safety factor
of
your choosing. I would use 4 to 1 but it is up to you. For this
problem
I will use a 1000lb load at the end of the beam.

Your beam is 3.75ft long with a fixed support on the left end at
0ft
and a simple support somewhere along its length. We will use
1.875ft
for the location of the simple support. We will first calculate
the
reactions about the supports using moment. We will label the fixed
support R1 and the simple support R2. Moment is equal to force X
distance. So we have:

Sum Moments about R1
1.875R2+3.75(-1000)=0
1.875R2=3750
R2=3750/1.875
R2=2000

Sum Moments about R2
-1.875R1+1.875(-1000)=0
-1.875R1=1875
R1=-1000

Sum All Vertical Forces : -1000(load)+ -1000(R1)+ 2000(R2)=0

All vertical forces are equal to zero as are the moments so the
beam is
in equilibrium.

Now you know your reactions at R1 and R2.

Now if you are using mild steel the Modulus of Elasticity or E=
30x10^6

E=stress/strain Strain is how much the material will "stretch" or
displace under load.
Poissons Ratio is also involved here but we will not complicate
things
with it.

Load (P)=Stress/Area

Simply take your load and the cross sectional area of the tube and
figure your Stress.

As long as your stress is under the yield stress of 36,000psi you
are
fine. Since you already have your safety factor of 4:1 all will be
well
if you are under that number.

You can figure out load at a joint by Using the Method of Joint or
Method of Sections.

Maximum Moment will be where the shear stress crosses zero in the
shear
stress diagram.

If you do not know what these are get any good Statics book and it
will
explain it. You could also look through Machinerys Handbook.

Also, look into free body diagrams

If you need any more help, let me know. -Steve