# Shop Crane Length of Moment Calculation

• posted

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!

• posted

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 are:

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.

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

• posted

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

• posted

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?

• posted

Steve. I wish my Strength of Materials prof. would have explained it that well 35 years ago!

• posted

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

forces

Forces=0

counter

1.875ft
• posted

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

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