1. Sep 15, 2008

### mebuildit

Since I'm new to this site don't razz me to bad since I suck at any type of mathematical equasions.
Anyways, I am building a "Jib" arm and was planning on putting an I beam in the ground with concrete. Although the I beam will have to be free standing and about 10' tall. The "jib" arm is about 8' x 4' tall I will attach a picture.
My question is how big of an I beam will I need before I get any deflection from lifting about 800lbs.?
The I beam will be about 3-4' in the ground with a concrete base. The rest of the I beam will be above ground level and free standing at about 10' tall.

I'm just not sure if a 8" I beam will be strong enough to hold what I'm trying to do.

Here is the idea that I have.

#### Attached Files:

• ###### Jib arm 1.jpg
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2. Sep 16, 2008

### PhanthomJay

You've got to be careful when you are designing something to support a substantial load. You need a qualified engineer. Having said that, the beam will deflect under any load, regardless of how small, because of the overturning moment from the jib arm. You have to specify what is a tolerable deflection. Then be sure the beam can handle both that deflection and the bending and axial loads. And that the concrete foundation is adequate to support the load. Also, your sketch indicates that the jib is attached to the web of the I-beam, creating weak axis bending. You might want to rotate it 90 degrees to reduce deflection and increase beam capacity. And if the jib can pivot (how much?), you've got to consider that case also.

Last edited: Sep 16, 2008
3. Sep 16, 2008

### minger

Well, while I have Roarks out, I'll see what I can find. It seems that you have in section Beams; Flexure of Straight Beams, case 3a - Left End Free, right end cantilevered (where of course your right end is in the ground).

A load on the jig as you're calling it will create a moment on the i-beam. You can do a simple free body to get the moment that will be applied given your load. I find the deflection and slope at the end (top) of the beam as:

$$\theta_A = \frac{-M_0 (l-a)}{EI}$$
$$y_A = \frac{M_0 (l^2 - a^2)}{2EI}$$

Where l is the length, a is the distance from the top of the beam to your load, E and I are material and geometric properties of the beam respectively, and of course M is the moment.

These correspond to the maximum values, where the maximum possible value is when a=0, or the moment is applied at the very top. One consideration, you will have different moment of inertia, I, values depending on which way your bending.

Now, you will also have buckling calculations to run, which are out of the scope of this post, haha.

4. Sep 16, 2008

### mebuildit

Well you bring some information that I didn't think about. As for the arm swing it will rotate 180º and after pricing I beams at \$2400 for a 20 foot length I think that an I beam is out of the equasion.

So what I'm thinking is a 6 X 6 square tubing 3/8" thick with some reinforcements along the sides and front will be enough to support what I"m trying to do.