Simplified method for suspension bridge

Rankine Theory of Suspension Bridges – Detailed Solved Example and Interactive Calculator

The Rankine Theory of Suspension Bridges provides a simple and powerful way of understanding the fundamental behaviour of the main cable of a suspension bridge.

For the classical simplified analysis, the main cable is considered perfectly flexible and capable of resisting tension only. When the load transferred to the cable is treated as uniformly distributed along the horizontal span, the cable assumes a parabolic profile.

Important: This article deals with the classical simplified parabolic-cable approach. It is intended for understanding, preliminary calculations, teaching and engineering estimation. Detailed bridge design must additionally consider the actual stiffening girder, hangers, cable self-weight, temperature, wind, live-load position, construction stages, seismic effects, anchorage behaviour and the applicable bridge design codes.

1. What is Rankine Theory?

Rankine's approach provides a simplified method of analysing suspension bridges by considering the equilibrium of the flexible main cable.

The central concept is:

\[ \boxed{\text{Uniform horizontal load} \quad\Rightarrow\quad \text{Parabolic cable}} \]

The method is especially useful for understanding how the following parameters influence cable force:

• Span L
• Sag f
• Uniform load w
• Horizontal cable force H
• Vertical support reaction V
• Resultant cable tension T

2. Basic Assumptions

• The main cable is perfectly flexible.
• The cable carries tension only.
• Bending stiffness of the cable is neglected.
• Supports are at the same elevation.
• The loading is symmetrical.
• The load is uniformly distributed along the horizontal span.
• Cable self-weight is neglected in the simplest calculation.
• The cable profile is therefore parabolic.
• The stiffening girder is treated in a simplified manner rather than modelling its complete flexural interaction with the cable.

For a suspension bridge deck supported by many closely spaced hangers, treating the roadway/deck loading as uniformly distributed along the horizontal projection is a standard idealisation for the parabolic cable calculation.

3. Difference Between Parabola and Catenary

This distinction is extremely important.

Loading condition	Ideal cable shape
Uniform load per unit horizontal distance	Parabola
Uniform self-weight per unit cable length	Catenary

Suspension bridge analysis commonly uses the parabolic approximation when the principal load is the bridge deck loading distributed through numerous hangers.

4. Geometry of the Cable

Consider a suspension cable with:

• Span = L
• Central sag = f
• Uniform load = w kN/m

Let x be measured from the left support.

The cable equation is:

\[ \boxed{ y=\frac{4fx(L-x)}{L^2} } \]

At:

\[ x=0 \]

we obtain:

\[ y=0 \]

At:

\[ x=\frac{L}{2} \]

we obtain:

\[ y=f \]

Thus the maximum ordinate is equal to the specified sag.

5. Derivation of the Cable Equation

Consider a small portion of the cable from its lowest point to a section at horizontal distance x.

The uniformly distributed load acting over this portion is:

\[ W_x=wx \]

Let H be the horizontal component of cable tension.

For the free body of the cable segment:

\[ T\sin\theta=wx \]

and:

\[ T\cos\theta=H \]

Dividing:

\[ \tan\theta=\frac{wx}{H} \]

Since:

\[ \tan\theta=\frac{dy}{dx} \]

we obtain:

\[ \frac{dy}{dx}=\frac{wx}{H} \]

Integrating:

\[ y=\frac{wx^2}{2H} \]

This is a parabola. This derivation is also presented in standard engineering-mechanics treatments of parabolic cables.

At the support:

\[ x=\frac{L}{2} \]

and:

\[ y=f \]

Therefore:

\[ f= \frac{w(L/2)^2}{2H} \]

which gives:

\[ \boxed{ H=\frac{wL^2}{8f} } \]

6. Horizontal Component of Cable Tension

The most important Rankine/parabolic cable equation is:

\[ \boxed{ H=\frac{wL^2}{8f} } \]

where:

• H = horizontal component of cable tension
• w = uniformly distributed load per horizontal metre
• L = span
• f = sag

Notice that sag occurs in the denominator.

\[ H\propto\frac{1}{f} \]

Therefore:

• More sag → lower horizontal force.
• Less sag → higher horizontal force.

7. Vertical Reaction at Each Support

The total uniformly distributed load over the span is:

\[ W=wL \]

Because the loading is symmetrical:

\[ V_A=V_B=\frac{W}{2} \]

Therefore:

\[ \boxed{ V_A=V_B=\frac{wL}{2} } \]

8. Maximum Cable Tension

At the support, the cable has:

• Horizontal component H
• Vertical component V=wL/2

Therefore:

\[ T_{max}= \sqrt{H^2+V^2} \]

or:

\[ \boxed{ T_{max}= \sqrt{ \left(\frac{wL^2}{8f}\right)^2+ \left(\frac{wL}{2}\right)^2 } } \]

For symmetrical supports at the same elevation, the maximum cable tension occurs at the supports because the vertical component is greatest there.

9. Minimum Cable Tension

At the lowest point of the cable, the tangent is horizontal.

Therefore:

\[ V=0 \]

Hence:

\[ \boxed{T_{min}=H} \]

Thus:

\[ \boxed{ T_{min}=\frac{wL^2}{8f} } \]

10. Cable Angle at the Support

At the support:

\[ \tan\theta=\frac{V}{H} \]

Substituting:

\[ \tan\theta= \frac{wL/2}{wL^2/(8f)} \]

Therefore:

\[ \boxed{ \tan\theta=\frac{4f}{L} } \]

Hence:

\[ \boxed{ \theta=\tan^{-1}\left(\frac{4f}{L}\right) } \]

11. DETAILED SOLVED EXAMPLE

Problem: A suspension bridge has a main span of 200 m. The main cable has a central sag of 20 m. The cable supports a uniformly distributed load of 50 kN/m over the horizontal span. Determine:

1. Total load on the cable
2. Vertical reaction at each support
3. Horizontal component of cable tension
4. Minimum cable tension
5. Maximum cable tension
6. Angle of cable at the support
7. Equation of the cable
8. Cable ordinate at selected locations
9. Approximate cable length
10. Effect of changing sag

Given Data

Parameter	Symbol	Value
Span	L	200 m
Sag	f	20 m
Uniform load	w	50 kN/m

Step 1 – Calculate Total Load

Total load:

\[ W=wL \]

Substituting:

\[ W=50\times200 \]

\[ \boxed{W=10,000\ kN} \]

Step 2 – Calculate Vertical Reaction

Since loading is symmetrical:

\[ V_A=V_B=\frac{W}{2} \]

\[ V_A=V_B=\frac{10,000}{2} \]

\[ \boxed{V_A=V_B=5,000\ kN} \]

Step 3 – Calculate Horizontal Cable Force

Use:

\[ H=\frac{wL^2}{8f} \]

Substitute:

\[ H= \frac{50(200)^2}{8(20)} \]

\[ H= \frac{50\times40,000}{160} \]

\[ H=12,500\ kN \]

Horizontal cable force:

\[ \boxed{H=12.50\ MN} \]

Step 4 – Minimum Cable Tension

At midspan:

\[ V=0 \]

Therefore:

\[ T_{min}=H \]

\[ \boxed{T_{min}=12,500\ kN=12.50\ MN} \]

Step 5 – Maximum Cable Tension

At the support:

\[ T_{max}=\sqrt{H^2+V^2} \]

Therefore:

\[ T_{max} = \sqrt{12,500^2+5,000^2} \]

\[ T_{max} = \sqrt{156,250,000+25,000,000} \]

\[ T_{max} = \sqrt{181,250,000} \]

\[ \boxed{ T_{max}\approx13,463\ kN } \]

or:

\[ \boxed{T_{max}\approx13.46\ MN} \]

Step 6 – Calculate Cable Angle at Support

Use:

\[ \tan\theta=\frac{4f}{L} \]

Therefore:

\[ \tan\theta= \frac{4(20)}{200} =0.4 \]

Hence:

\[ \boxed{ \theta=\tan^{-1}(0.4) \approx21.80^\circ } \]

Step 7 – Cable Equation

The general equation is:

\[ y=\frac{4fx(L-x)}{L^2} \]

For this example:

\[ y= \frac{4(20)x(200-x)}{200^2} \]

Therefore:

\[ \boxed{ y=\frac{x(200-x)}{500} } \]

This equation gives the cable elevation measured downward from the support chord.

Step 8 – Calculate Cable Sag at Various Locations

x from left support	Calculated y
0 m	0 m
25 m	8.75 m
50 m	15.00 m
75 m	18.75 m
100 m	20.00 m
125 m	18.75 m
150 m	15.00 m
175 m	8.75 m
200 m	0 m

Step 9 – Cable Slope at Any Location

Differentiate the cable equation:

\[ y=\frac{4fx(L-x)}{L^2} \]

Therefore:

\[ \frac{dy}{dx} = \frac{4f(L-2x)}{L^2} \]

The slope is zero at:

\[ x=\frac{L}{2} \]

which corresponds to the lowest point of the cable.

Step 10 – Cable Tension at Any Location

The vertical component of cable tension at a distance x from the left support is:

\[ V(x)=w\left(\frac{L}{2}-x\right) \]

when x is measured from the left support and the sign convention is chosen appropriately.

The magnitude of tension is:

\[ \boxed{ T(x)=\sqrt{H^2+V(x)^2} } \]

At the left support:

\[ V=\frac{wL}{2} \]

At midspan:

\[ V=0 \]

At the right support the vertical component again reaches:

\[ |V|=\frac{wL}{2} \]

12. Cable Length

The horizontal span is 200 m, but the actual cable length is slightly greater than 200 m because the cable is curved.

For the parabolic profile:

\[ y=\frac{4fx(L-x)}{L^2} \]

The exact arc length can be obtained from:

\[ S= \int_0^L \sqrt{ 1+\left(\frac{dy}{dx}\right)^2 } \,dx \]

For the parabolic cable:

\[ \frac{dy}{dx} = \frac{4f(L-2x)}{L^2} \]

For a relatively shallow cable, a useful approximation is:

\[ \boxed{ S\approx L \left[ 1+\frac{8}{3} \left(\frac{f}{L}\right)^2 \right] } \]

For the example:

\[ \frac{f}{L}=\frac{20}{200}=0.10 \]

Therefore:

\[ S\approx200 \left[ 1+\frac{8}{3}(0.1)^2 \right] \]

\[ S\approx205.33\ m \]

The exact parabolic integral can be evaluated numerically and is slightly different from the small-sag approximation.

13. Effect of Changing Sag

Keep:

• L=200\,m
• w=50\,kN/m

Now compare different sag values.

Sag	Horizontal Force H	Approximate Interpretation
10 m	25,000 kN	Very high horizontal force
15 m	16,667 kN	High force
20 m	12,500 kN	Reference case
25 m	10,000 kN	Lower horizontal force
30 m	8,333 kN	Still lower force

This demonstrates directly:

\[ \boxed{ H\propto\frac{1}{f} } \]

Therefore, reducing the sag from 20 m to 10 m doubles the horizontal cable force.

14. Engineering Meaning of the Result

The calculated horizontal force of:

\[ H=12.50\ MN \]

is not simply an internal force that disappears. It has to be resisted by the complete structural system.

Depending on the structural arrangement, the cable force influences:

• Tower forces
• Anchorage forces
• Foundation forces
• Stiffening girder behaviour
• Local cable stresses
• Anchor block stability

This is why sag is a major preliminary design parameter in suspension bridges.

15. Interactive Rankine Suspension Cable Calculator

Input Data

Span L (m)

Sag f (m)

Uniform load w (kN/m)

Profile interval (m)

16. JavaScript Calculation Method

The calculator uses the following fundamental equations:

\[ W=wL \]

\[ V=\frac{wL}{2} \]

\[ H=\frac{wL^2}{8f} \]

\[ T_{max}=\sqrt{H^2+V^2} \]

\[ \theta=\tan^{-1}\left(\frac{4f}{L}\right) \]

\[ y=\frac{4fx(L-x)}{L^2} \]

17. Engineering Interpretation of Span

The horizontal cable force varies with the square of span:

\[ H\propto L^2 \]

Therefore, if span doubles while load and sag remain unchanged:

\[ H_{new}=4H_{old} \]

This shows why long-span suspension bridges require careful control of cable geometry and anchorage forces.

18. Effect of Uniform Load

The horizontal force is directly proportional to the loading:

\[ H\propto w \]

Thus, if the distributed load increases by 20%, the calculated horizontal cable force also increases by 20%, provided span and sag remain unchanged.

19. Dimensionless Form

The equation can also be expressed using the sag-to-span ratio:

\[ r=\frac{f}{L} \]

Then:

\[ H= \frac{wL}{8r} \]

This form is useful for preliminary comparison between bridges of different spans.

It clearly shows that, for a fixed sag/span ratio, horizontal cable force increases approximately in proportion to span:

\[ H\propto L \]

20. Important Limitation – Uniform Horizontal Loading

The equation:

\[ H=\frac{wL^2}{8f} \]

should not be blindly applied when the loading is strongly non-uniform or concentrated.

For example, a single heavy vehicle positioned near one point on the bridge does not produce the same response as a uniform load over the entire span.

In an actual suspension bridge, the stiffening girder redistributes live load through the hanger system. The girder also develops bending moments and shear forces.

Therefore: The simple Rankine/parabolic model is excellent for understanding the fundamental cable force, but it is not a complete analysis of an actual modern suspension bridge under arbitrary traffic loading.

21. Rankine Theory and Elastic Theory

Parameter	Rankine / Simplified Parabolic Approach	Elastic Theory
Main cable	Flexible tension member	Flexible tension member interacting with deck
Load distribution	Idealized uniform horizontal loading	More realistic load distribution
Stiffening girder	Simplified	Flexural stiffness considered
Unsymmetrical traffic loading	Limited	Can be analysed
Girder bending	Not fully represented	Explicitly represented
Construction stages	Generally not represented	Can be incorporated
Preliminary design	Very useful	Useful
Detailed bridge design	Insufficient by itself	More appropriate

22. DOs

• Use horizontal span consistently.
• Check the units of w, L and f.
• Remember that W=wL.
• Use V=wL/2 for each support under symmetrical loading.
• Check that cable sag is measured from the support chord to the lowest cable point.
• Use the parabolic model when the loading assumption is appropriate.
• Check the sensitivity of horizontal force to sag.
• Use a more detailed structural model for final bridge design.

23. DON'Ts

• Do not confuse w with total load W.
• Do not use W=wL/2. The total load is W=wL.
• Do not assume that cable tension is equal to horizontal force everywhere.
• Do not ignore the vertical component at the supports.
• Do not automatically use a parabolic cable for a cable carrying only its own weight.
• Do not use this simplified calculation as a complete modern suspension bridge design.
• Do not ignore the stiffening girder under concentrated or unsymmetrical live loading.

24. Quick Formula Sheet

Quantity	Formula
Total load	W=wL
Vertical reaction	V=wL/2
Horizontal cable force	H=wL^2/(8f)
Minimum tension	T_{min}=H
Maximum tension	T_{max}=\sqrt{H^2+V^2}
Support angle	\theta=\tan^{-1}(4f/L)
Cable profile	y=4fx(L-x)/L^2
Cable slope	dy/dx=4f(L-2x)/L^2
Approximate cable length	S\approx L[1+(8/3)(f/L)^2]

25. Final Engineering Summary

The Rankine/parabolic cable approach demonstrates one of the most important principles of suspension bridge mechanics:

\[ \boxed{ H=\frac{wL^2}{8f} } \]

From this one equation, several important engineering observations follow.

• Horizontal cable force increases with the square of span.
• Horizontal cable force increases directly with load.
• Horizontal cable force decreases as sag increases.
• The cable tension is minimum at the lowest point.
• The cable tension is maximum at the supports for the symmetrical same-level case.
• The support reaction is wL/2.
• The cable profile is parabolic when the load is uniform per unit horizontal span.

For the worked example:

Span: 200 m

Sag: 20 m

Load: 50 kN/m

Total load: 10,000 kN

Vertical reaction: 5,000 kN per support

Horizontal cable force: 12,500 kN

Minimum cable tension: 12,500 kN

Maximum cable tension: approximately 13,463 kN

Support cable angle: approximately 21.80°

The simplified theory therefore provides an excellent bridge between basic engineering mechanics and the more advanced Elastic Theory of Suspension Bridges.

26. Frequently Asked Questions

Q1. Why is the suspension cable parabolic?

When the load is uniformly distributed along the horizontal projection, equilibrium gives a constant horizontal tension component and a linear change in vertical component. Integration of the resulting slope equation produces a parabolic cable profile.

Q2. Where is cable tension maximum?

For symmetrical supports at the same elevation and uniform horizontal loading, the maximum tension occurs at the supports.

Q3. Where is cable tension minimum?

It occurs at the lowest point of the cable, where the vertical component is zero and tension equals H.

Q4. What happens if sag is reduced?

Horizontal cable force increases according to:

\[ H=\frac{wL^2}{8f} \]

Q5. Is this sufficient for designing a real suspension bridge?

No. It is a simplified analytical model. A complete bridge analysis requires consideration of the cable, hangers, stiffening girder, towers, anchorages, foundations, actual load combinations, wind, temperature, seismic actions, construction stages and applicable design standards.