The Falling Chimney

Last edited: 2021-04-11 13:19:36


The falling chimney problem is the perfect mix of mechanical dynamics and beam theory. By just using the fundamentals of those two fields we can accurately predict where the chimney cracks when falling, under some assumptions...

So lets get the assumptions of the falling chimney problem out of the way first. The first assumption is that there is no air resistance. If air resistance had been present the top of the chimney would experience a greater stress from the air than the lower part which makes our calculations much more complex and this is not an article about fluid mechanics. The second assumption is that the chimney is homogeneous, meaning that it has the same diameter across its length and the same material. If it had been inhomogeneous certain parts would handle stress better than others.

Angle Acceleration of the Chimney

Now when that is out of the way, lets make some calculations! The first thing we'll have to determine is the angle acceleration θ¨\ddot{\theta}. To do that we use a beam's, in this case a chimney's, moment of inertia when attached at one of its ends:

I=13mL2.I = \frac{1}{3} m L^2.

Here mm is the chimneys mass and LL is its length. The moment from the gravitation acting on the chimney must equal that of the angle acceleration:

mgL2sinθ=Iθ¨=13mL2θ¨.mg \frac{L}{2} \sin{\theta} = I \ddot{\theta} = \frac{1}{3}mL^2 \ddot{\theta}.

So now we have our angle acceleration θ¨\ddot{\theta} and apparently it is dependent on θ\theta:

θ¨=3g2Lsinθ.\ddot{\theta} = \frac{3g}{2L} \sin{\theta}.

Is this plausible? Well θ¨\ddot{\theta} increases with gg and decreases with LL which makes sense. At θ=0\theta=0 the angle acceleration is zero which also makes sense since at that angle the chimney is "standing up".

The Chimney's Load Function

If the chimney was locked and stationary at an angle θ\theta the load function would be:

w(d)=mgLsinθ.w(d) = -\frac{mg}{L} \sin{\theta}.

If we go to the other extreme and say that the chimney was in free fall it would lead to the load function being zero. So falling negates the load function and in our case the chimney is rotating. So θ¨\ddot{\theta} has a similar effect to free fall. So the load function in our case will be the following:

w(d)=dmLθ¨mgLsinθ=3dmg2L2sinθmgLsinθ.w(d)=\frac{dm}{L} \ddot{\theta} - \frac{mg}{L} \sin{\theta} = \frac{3dmg}{2L^2} \sin{\theta} - \frac{mg}{L} \sin{\theta}.

We can see here that the load function is zero at d=2L3d = \frac{2L}{3}. This makes sense because θ¨\ddot{\theta} is equal to gsinθg \sin{\theta} at d=2L3d = \frac{2L}{3}. In other words, this point is in a "virtual" free fall.

The Chimney's Shear Force Function

The relation between the load function and the shear force function is:


If we integrate that we get:

V(d)=3d2mg4L2sinθ+dmgLsinθ+C,V(d) = -\frac{3d^2mg}{4L^2} \sin{\theta}+ \frac{dmg}{L} \sin{\theta}+ C,

where CC is a constant. We can determine this constant with the boundary condition V(L)=0V(L) = 0. In other words: the shear force is zero at the end of the chimney which is logical. This means:

C=mg4.C = -\frac{mg}{4}.

If we take the absolute value of the shear force function we can check where the force is the greatest:

V(d)=mgsinθ(3d24L2+dL14).|V(d)|=\left|mg\, \sin{\theta} \left(-\frac{3d^2}{4L^2} + \frac{d}{L} - \frac{1}{4}\right) \right|.

Here we can see that the shear force is the greatest when d0d \to 0 on the interval 0<d<L0< d < L, so if the shear force was the cause of the crack in the chimney it would be at the base. But that is not the only place where the chimney can break...

The Chimney's Bending Moment Function

The final step is looking at the bending moment function which has the following relation with the shear force function:


So with a bit of integration we get:

M(d)=mgsinθ(d34L2+d22Ld4)+K,M(d)=mg\, \sin{\theta} \left(-\frac{d^3}{4L^2} + \frac{d^2}{2L} - \frac{d}{4}\right) + K,

where KK is a constant. We'll once again use the boundary condition at the end of the chimney, which in this case is also zero. This leads to K=0K=0. So where is M(d)|M(d)| the greatest value? At d=0d=0 the bending moment is zero so it is not there. Well lets check where its derivative, the shear force function, is zero to get the maximums:

M(d)=V(d)=03d24L2+dL14=0.M'(d)=V(d)=0 \Longrightarrow -\frac{3d^2}{4L^2} + \frac{d}{L} - \frac{1}{4} = 0.

From this we get the roots d=Ld=L and d=L3d=\frac{L}{3}. d=Ld=L does not make much sense since its the end of the chimney, you cant real chop of something that isn't there... So we get d=L3d=\frac{L}{3} and that is why a falling chimney breaks at a third of its length from its base.

The video above shows a demolition of a smokestack, as you can see it breaks close to a third of its length when it falls.

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