Is the Suspension Bridge Shaped like a Parabola or a Catenary?

Last edited: 2021-05-14 19:25:34


What makes the suspension wire of the Golden Gate Bridge curve like a parabola? Lets get into the mechanics of the shape of the suspension bridge's cables and what determines them to curve like a parabola or a catenary.

The Assumptions about the Problem

Some assumptions about the problem have to be made first. The main wire that the bridge rests on is assumed to be inextensible, homogeneous and infinitely flexible. The bridge is also assumed to be homogeneous. The main wire is connected at each end of the bridge. The wires that ties together the bridge and the main wire is assumed to be in an infinite amount so the bridge's load function is constant. The origin of coordinate system is located at the bottom of the main wire.

Equilibrium of an Infinitesimal Wire Element

Lets make an free body diagram of an infinitesmal wire element of the main wire:

Image of free body diagram of an infinitesmal wire element

Here T\vec{T} is the tension of the cable, dTd\vec{T} the infinitesmal change in the tension, dsds the infinitesmal length of the wire element and dxdx the length in the xx-direction. qcq_c and qbq_b is respectively the cable's and bridge's linear mass density. The equilibrium of the forces is

T+dTT(qcds+qbdx)gy^=0.\vec{T} + d\vec{T} - \vec{T} - (q_c ds + q_b dx)g \hat{y} = 0.

With some cancellation we get

dT=(qcds+qbdx)gy^.d\vec{T} = (q_c ds + q_b dx)g \hat{y}.

Lastly, the Pythagorean theorem gives us

ds=(dx)2+(dy)2=1+(y)2dx.ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + (y')^2} dx.

So we get

dT=(qc1+(y)2dx+qbdx)gy^.d\vec{T} = (q_c \sqrt{1 + (y')^2} dx + q_b dx)g \hat{y}.

The ODE which Dictates the Shape of The Wire

The tension at the bottom of the main wire has only a xx-component which we'll call T0T_0. So T(0)=T0x^\vec{T}(0) = T_0 \hat{x}. If we do some integration of the equilibrium equation from earlier we get

T=T0x^+(qc0x1+y(x~)2dx~+qbx)gy^,\vec{T} = T_0 \hat{x} + \left(q_c \int_0^x \sqrt{1 + y'(\tilde{x})^2} d\tilde{x} + q_b x \right)g \hat{y},

here we are just using x~\tilde{x} in the integral so the upper limit, xx, of the integral is correct. In the xx-direction we now have

Tcosθ=Tdx(dx)2+(dy)2=T0,T \cos{\theta} = T \frac{dx}{\sqrt{(dx)^2 + (dy)^2}} = T_0,

where θ\theta is the angle between T\vec{T} and the xx-axis. So

T=T0(dx)2+(dy)2dx.T = T_0 \frac{\sqrt{(dx)^2 + (dy)^2}}{dx}.

In the yy-direction we have

Tsinθ=Tdy(dx)2+(dy)2=T0dydx=(qc0x1+y(x~)2dx~+qbx)g.T \sin{\theta} = T \frac{dy}{\sqrt{(dx)^2 + (dy)^2}} = T_0 \frac{dy}{dx} = \left(q_c \int_0^x \sqrt{1 + y'(\tilde{x})^2} d\tilde{x} + q_b x \right)g.

We now have an expression for yy':

y=(qc0x1+y(x~)2dx~+qbx)gT0y' = \left(q_c \int_0^x \sqrt{1 + y'(\tilde{x})^2} d\tilde{x} + q_b x \right) \frac{g}{T_0}

Lets do an additional derivation to get rid of the integral:

y=qbgT0+qcg1+(y)2T0.y'' = \frac{q_b g}{T_0} + \frac{q_c g \sqrt{1+(y')^2}}{T_0}.

This is the differential equation that dictates the curve of suspension bridge.

Is the Cable Shaped as a Parabola or Catenary?

So what shape is the cable? To answer this lets look at two extremes, neglecting either the cable's mass or the bridge's mass. If we neglect the cable's load function i.e. its mass we get the differential equation

y=qbgT0.y'' = \frac{q_b g}{T_0}.

With some simple integration we get

qbg2T0x2+bx+c,\frac{q_b g}{2 T_0} x^2 + bx + c,

where bb and cc are constants. The origin is loacted at the bottom of the cable so c=0c=0 and the cable is symmetric so b=0b=0. Here we have our parabola. Neglecting the cable's mass is a pretty reasonable approximation since the bridge's mass is likely much larger than the cable's mass. Can you guess what we get if we do the opposite? If we neglect the bridge's mass we get the following differential equation:

y=qcg1+(y)2T0.y'' = \frac{q_c g \sqrt{1+(y')^2}}{T_0}.

Since we already have pretty good hunch that the solution is in the shape of a catenary, lets check if y=acoshxay = a \cosh{\frac{x}{a}} works. Then the left side is

y=coshxaay'' = \frac{\cosh{\frac{x}{a}}}{a}

and the right side is

qcg1+(y)2T0=qcg1+sinh2xaT0=qcgcoshxaT0.\frac{q_c g \sqrt{1+(y')^2}}{T_0} = \frac{q_c g \sqrt{1+\sinh^2{\frac{x}{a}}}}{T_0} = \frac{q_c g \cosh{\frac{x}{a}}}{T_0}.

For this to work aa needs to be

a=T0qcg.a = \frac{T_0}{q_c g}.

That is the solution if the bridge's mass can be neglected. I don't know what sort of bridge that is... But hey now you know. I'll leave the harder task of solving the ODE with both masses present for you.

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