Forklift Jib design
Forklift Jib design
(OP)
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?






RE: Forklift Jib design
RE: Forklift Jib design
RE: Forklift Jib design
RE: Forklift Jib design
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.
RE: Forklift Jib design
BA
RE: Forklift Jib design
this was always a very handy tool...and took the place of a crane many times!
RE: Forklift Jib design
The 98.1 kN already has a 2.5 safety factor on it - as per applicable Standard.
Regards.
RE: Forklift Jib design
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
RE: Forklift Jib design
Any further assistance greatly appreciated.
Regards.
RE: Forklift Jib design
Jim
This does not need to be the final choice, just an initial attempt for an initial solution.
RE: Forklift Jib design
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