Hi everyone -
I have read through quite a few of the strings on inertia and moments
and am a confused lay-person. I believe the entries were repeatedly
saying that it is not as easy to calculate the inertia/mass per length
of moment, especially adding angle .... but forgive me for asking
again.

After a back surgery, I am building avery simple shop crane from square
tube stock. It is modeled mostly after engine hoists, but without the
ram. It is toned down strength-wise because it simply has to hold
250lbs about 45" out on an arm (rather than the 1 and 2 tons an engine
hoist holds) - and I am probably overbuilding.

There are no hinges, tube stock sleeved 20" into larger tube stock will
allow me to make basic angle/height adjustments with a pulley doing the
rest of the up and down work.

I have seen calculations like I=m(r)squared, but am unsure what length
the "r" is. Have also seeen I=(m)(1)squared/3. Anyway, for a simple
structure as this .... ok ... here it goes .... is there a simple way
to calculate aprox. inertia/mass at the end of the arm to ensure the
tube stock can more than handle it (because of fluctuations from
bounces and jostles when moving about) - thanks!

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

Hey Pumkin,

Can't help you with the theoretical engineering, 'cause I would
probably overbuild it like you anyway.

But a word of warning.

Most "crane" failures, are not from overloading, but from side
loading. Bouncing is nothing compared to a sudden STOP sideways.
They just are not built to take that.

Take care.

Good Luck. Let us know how you made out (pix pix pix)

Brian Lawson,
Bothwell, Ontario.
XXXXXXXXXXXXXXXXXXXXXXX


On 15 May 2005 10:00:49 -0700, wrote:

Hi everyone -
I have read through quite a few of the strings on inertia and moments
and am a confused lay-person. I believe the entries were repeatedly
saying that it is not as easy to calculate the inertia/mass per length
of moment, especially adding angle .... but forgive me for asking
again.

After a back surgery, I am building avery simple shop crane from square
tube stock. It is modeled mostly after engine hoists, but without the
ram. It is toned down strength-wise because it simply has to hold
250lbs about 45" out on an arm (rather than the 1 and 2 tons an engine
hoist holds) - and I am probably overbuilding.

There are no hinges, tube stock sleeved 20" into larger tube stock will
allow me to make basic angle/height adjustments with a pulley doing the
rest of the up and down work.

I have seen calculations like I=m(r)squared, but am unsure what length
the "r" is. Have also seeen I=(m)(1)squared/3. Anyway, for a simple
structure as this .... ok ... here it goes .... is there a simple way
to calculate aprox. inertia/mass at the end of the arm to ensure the
tube stock can more than handle it (because of fluctuations from
bounces and jostles when moving about) - thanks!

On 15 May 2005 10:00:49 -0700, wrote:

Hi everyone -
I have read through quite a few of the strings on inertia and moments
and am a confused lay-person. I believe the entries were repeatedly
saying that it is not as easy to calculate the inertia/mass per length
of moment, especially adding angle .... but forgive me for asking
again.

After a back surgery, I am building avery simple shop crane from square
tube stock. It is modeled mostly after engine hoists, but without the
ram. It is toned down strength-wise because it simply has to hold
250lbs about 45" out on an arm (rather than the 1 and 2 tons an engine
hoist holds) - and I am probably overbuilding.

There are no hinges, tube stock sleeved 20" into larger tube stock will
allow me to make basic angle/height adjustments with a pulley doing the
rest of the up and down work.

Not sure what you mean here. You say "without the ram". Do you mean
the arm is supported only at the end, or that you have replaced the
ram with a sleeved tube, that can be adjusted for gross height? (I
hope the second). The reason I ask is the the sleeved tube _could_
simply slide up and down in the upright portion and you say "no
hinges".

I am also not quite happy about how you are mounting ghe whole thing.
Forces get larger as you move away from the load, in this case. The
base of the upright, or whatever, could be taking more strain than the
arm itself.

Can you do some "ascii art" to show the layout? Or post some rough
drawings to either the metalwork binaries group, or to a website?

I have seen calculations like I=m(r)squared, but am unsure what length
the "r" is. Have also seeen I=(m)(1)squared/3. Anyway, for a simple
structure as this .... ok ... here it goes .... is there a simple way
to calculate aprox. inertia/mass at the end of the arm to ensure the
tube stock can more than handle it (because of fluctuations from
bounces and jostles when moving about) - thanks!

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

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

I had very good Engineering Professors in college who made us work an
endless amount of problems both analytical and practical. We did alot
of bitching and moaning about it as students, but it taught almost all
of us what we needed to know to be successful Engineers in our
respective fields of study. One of the hardest nosed Professors I ever
had was from Ohio State who had a PHD in Materials Science and
Mechanical Engineering. He did alot of research work at Princeton
University in Nano Technology. For the most part he forced us to teach
ourselves and he looked over our shoulders. I learned more in his
classes in Thermodynamics, Statics, Dynamics, and Fluid Power than I
did in any of my other classes and it has stuck with me to this day. So
he proved his point about the importance of his method of teaching. I
see him often and I still joke with him about it.

The funny part is that I am a Police Officer and not a Mechanical
Engineer by profession. Although at this time I am working on my
Masters in Mechanical and Industrial Engineering. It will give me
something to do when I retire from Law Enforcement (-: .

Take Care...-Steve