# Exact anlytical solution for beam-lifting problem constrained by crane capacity

I have been trying to calculate the analytical solution for the following problem but have not succeeded to reach an exact solution. Can anyone propose a way?

A concrete beam is to be lifted by two cranes. The beam is 20 m long and weights 900 kN. The cranes have lifting capacities of 500 kN and 400 kN. What are distances $$a$$, $$b$$, and $$c$$ in the scheme below so the cranes are not overloaded? (position of cranes are marked with | |)

            <---- a----> <----- b -------> <--- c --->
===========================================  <-- the beam
||                ||
A = 500 kN        B = 400 kN

• A & B: reaction forces of cranes
• distributed self weight q = 900 kN / 20 m = 45 kN/m
• $$a + b + c = L = 20\text{ m}$$

The beam is statically determinate.

My attempted solutions:

I started from the equilibrium equations of the sum of bending moments about the left and right ends which are equal to zero:

\begin{align} \sum M_{left} &= 0 = Aa + B(a + b) = \dfrac{qL^2}{2} \\ \sum M_{right} &= 0 = Bc + A(b + c) = \dfrac{qL^2}{2} \end{align}

A third equation is needed; I have tried $$a + b + c = L$$, but I arrive at the "apple = apple" situation, which implies that the three equations are not independent.

I also tried to add an additional constrained by putting any of the $$a$$, $$b$$ or $$c$$ values equal to a constant value. The issue then is that each of the equilibrium equations become dependent on only one of the two remaining unknowns, as in the following case, where I have put $$b = L/3$$:

\begin{align} \sum M_{left} &= 0 = \dfrac{qL^2}{2} = Aa + B(a + L/3) \\ \sum M_{right} &= 0 = \dfrac{qL^2}{2} = Bc + A(c + L/3) \end{align}

And thus, each solved unknown will not satisfy the other equilibrium equation.

I have run out of tricks in my sleeve ...

Let's call the distance from the anchor to 500 kN crane from the middle of the beam $$D_1$$ and the distance to 400 kN, $$D_2$$.

To do exact analysis we get the moment of cranes' hook forces about the midpoint of the beam at equilibrium configuration.

$$P_1 D_1= P_2 D_2$$

denoting $$P_1$$ and $$P_2$$ as the tributary weight of the beam on 500 kN crane and 400 kN, respectively.

Substituting $$P_2$$ with $$4/5 P_1$$ in above equation we get

$$P_1 D_1 = 4/5 P_1 D_2$$

Dividing both sides by $$P_1$$:

$$D_1= \dfrac{4}{5}D_2$$

One of the safest ways to lift the beam is to attach the 400 kN crane to one end of concrete beam and 500 kN to a point at the other end 90% the length of the beam.

• Actually this isn't the "safest" way to lift a beam which is likely to crack in tension (e.g. concrete or stone). You want to position the ropes to minimize the absolute value of the maximum bending moment at any point along the beam. The ropes will be somewhere near the 1/4 and 3/4 positions along the length (but with the 4:5 ratio distances from the center) to do that. – alephzero Mar 9 '19 at 13:01
• @alephzero, by the safest way I mean as far as actual job site requirements. Because of control of the undesirable wobbling and angular momentum of the beam while cranes' are having the beam raise. What if there is a gust of wind or if they are not operating in good synchrony. The farther the hooks the more control. – kamran Mar 9 '19 at 15:13

There is not a unique solution to the problem, which explains why you can't find one!

To see why, think about a different problem: two people who weigh 50kg and 40kg want to balance each other on a see-saw. How far should they sit from the middle of the see-saw, to make it balance?

It should be obvious that the ratio of their distances from the center is the only thing that matters.