# How to calculate the mid-span lifting capability of a single-beam wooden gantry crane?

The explanations already submitted (ex: "How to calculate deflection of a simple beam with a load in the center") didn't help me much. I'm building a single beam gantry crane out of Doug Fir and need to know how to calculate the needed dimensions for the beam itself; not just the answer. I see all kinds of wood beam calculators on line but they all ask about the beam spacing of the floor joists, or the rafters or the roof trusses which is irrelevant in my case. I must not be using any of these calculators correctly because I can't get these to show a single-beam calculation. Anyone know how to calculate this long-hand or is there a table available somewhere? I know it can be done because I've lifted many a diesel truck engine block with a quality straight-grained 4 x 10 x 8 foot long Doug Fir beam using block and tackle with nary a problem. The application for what I'd like to calculate and eventually build "Norm Abrams" style is 3000 lbs with perhaps a 10% safety factor. Again I just want to know HOW to calculate the mid-span single-beam lifting capacity based on Wood species, cross dimension, and beam-span. I gotta believe this is much easier than I'm trying to make it. Your thoughts?

• use the formula: $\delta = \frac{PL^3}{48EI}$, where $P$ is the load in the center, in Newtons; $L$ is the length of the beam, in mm; $E$ is the modulus of elasticity, in MPa; $I$ is the centroidal moment of inertia with respect to the axis of bending, in $mm^4$. There are also bending formulas, $f_b = \frac{Mc}{I}$. But I think, deflection will dictate most of its strength. – Jem Eripol Feb 27 '18 at 0:51

The most basic way is to simplify the effects of dynamic loading caused by the start and stop of the crane or the winch , they could introduce a factor of 2.8 for load magnification, is to basically multiply you load by three and calculate as follows:

$$M = \dfrac{P\ell}{4}$$

Where $P$ is your factored load, $\ell$ is span.

And $f(b) = \dfrac{mc}{I}$, where the second moment of inertia is $I= \dfrac{bh^3}{12}$ for rectangular cross sections and $c=\dfrac{h}{2}$. Depending on the species of timber $f(b)$ ranges between 900 and 1250 psi.

And finally you need to check for allowable deflection as has been mentioned in other answers, and shear assuming the load is just very near to one support (worst case scenario).

You will find out many sizes of timber satisfying your calculations, the optimal choice is the fat beam rather slender one $h/b < 2$. And all rails and pulleys need to have damped stops.

Edit:

Lifting and lowering of loads has dynamic unplanned impacts like off-center torque due to the initial dislodging the load off its cart, so the beam has to resist torsional stress. Imagine the load going back and forth in pendulum-like motion in all directions. Best cross sections for torsion are tubular or at least not slender members, more near square cross sections.

• Can you expand on why a non-slender beam is better in this situation? – Wasabi Feb 27 '18 at 16:42
• Lifting and lowering of loads has dynamic unplanned impacts like off-center torque due to the initial dislodging the load off its cart, so the beam has to resist torsional stress. Imagine the load going back and forth in pendulum-like motion in all directions. Best cross sections for torsion are tubular or at least not slender members, more near square cross sections.. – kamran Feb 27 '18 at 18:24
• I'd recommend editing that into the answer. – Wasabi Feb 27 '18 at 18:33