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

Question

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 ...

Answer 1

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.

Answer 2

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.