Forklift Jib design

[truckdesigner]

Hoping someone can help me out here. See attached loading for a simple fork jib.

With the steel section having a section modulus of 430 E3 mm^3 and 98.1 kN at free end I calculated Mmax = 276.6 E6 kNm and bending stress fb = 643.2 Mpa. Too high I realise and have since beefed the section up.

Could someone offer any help on a deflection calculation/diagram? I don't think it is as simple as superposition of cantilever deflection equations ie load at free end + uniform loading, or is it?

 

[bpstruct]

Actually, I think it IS that simple. Just add the two up. I think you'll find that the uniform load you have shown will produce a deflection that is negligible when compared to the point load anyway.

 

[grantstructure]

Make sure you handle your unbraced length appropriately as well.

 

[chicopee]

This is an elementary problem that you'll get in strength of material solvable a few way however I prefer to do it by differential equation or by diagrams. Yes superposition will work if within elastic range.

 

[truckdesigner]

Thanks folks for the replies. I will post a complete solution when back in the office on Monday.

Question: do I take the whole length for the cantilever, or just the 2800 mm cantilevered section? Perhaps a stupid question - I would say just the 2800 mm section.

Kind regards.

 

[BAretired]

What you show in the sketch is a propped cantilever. If you want to calculate deflection, you must consider the entire length. To consider only the 2800 mm length will give you the wrong answer because the member is not fixed against rotation at the prop. The deflection in your case will be substantially higher than that of a 2800 mm cantilever.

BA

 

[par060]

a couple of thoughts from someone who has used a similar jib on a fork lift.... the loads won't be "static"....things picked up with the jib will dance around pretty good so give yourself plenty of room for a dynamic factoron the antisipated loads. Also the load gets reversed...alot...(maybe not from positive to negative but the load will go to zero) while the dancing goes on causeing some fatigue if you use welds.

this was always a very handy tool...and took the place of a crane many times!

 

[truckdesigner]

OK. So a little more complex than I first anticipated. Had a feeling it may be. Any chance someone could post the appropriate formulas for this loading case?

The 98.1 kN already has a 2.5 safety factor on it - as per applicable Standard.

Regards.

 

[jimstructures]

I get 378.56 kN up at the reaction point closest to the 98.1 kN down and 277.42 kn down at reaction point furthest from the 98.1 kN down load at the end of the jib.

Since you didn't specify a stiffness for the jib beam I didn't look at deflections. I did assume the jib beam was continuous from the left to the right and had a constant moment of inertia.

Hope this helps
Jim

 

[truckdesigner]

E = 200 Gpa, Fy = 350 MPa. Hope that may help.

Any further assistance greatly appreciated.


Regards.

 

[jimstructures]

You need to pick a section for your steel beam that you are using for the jib beam and thus specify a moment of inertia or stiffness that you will try to use to solve this problem.

Jim

This does not need to be the final choice, just an initial attempt for an initial solution.

 

[BAretired]

I recommend that you obtain the services of a practicing structural engineer to assist you in the design of this beam as there are several aspects other than maximum bending moment and deflection to be considered.

Having said that, M = 98.1*2.8 + 0.8*2.8²/2 = 277.8kN-m or 277.8E6 N-mm. If that was an unfactored moment, the stress would be 646MPa for a beam with S = 430,000mm³, very close to the value you found. The fact that your loads already have a safety factor of 2.5 would suggest you should be considering the plastic modulus, not the section modulus. You should also make allowance for the unbraced length of the beam and the possible overturning of the forklift under the prescribed load.

The deflection calculation can be closely approximated by considering a concentrated load at the tip of the cantilever of 98.1 + 0.8*2.8/2 = 99.2kN. This is a factored load and your deflection calculation should be based on an unfactored load or service load, in your case P = 99.2/2.5 = 39.7kN or 39,700N.

The expression for deflection at the tip of the cantilever is Pa²(L+a)/3EI where P = 39700, a = 2800, L = 1000 and I is the moment of inertia of the beam selected.

BA