The rope bridge
A bridge made out of two parallel, high tenacity polyester ropes connected by wooden boards spans a 12m wide canyon.
- What thickness (linear density) of rope is needed to accept a maximum 0.3 m deflection at a total 250 kg weight (bridge included)?
- What will the safety factor for the bridge be?
High tenacity polyester material parameters |
Specific Modulus |
700 - 800 cN/tex |
Specific Tenacity |
78 - 80 cN/tex |
Density |
1.39 g/cm3 |
Extension to break |
13 - 16% |
Tensile Modulus |
9 - 11 GPa |
Shrinkage @ 100oC |
1.5 - 6% |
Tensile Strength |
0.9 - 1.1 GPa |
Solution:
We consider the material as linear elastic. The 0.3 m deflection is connected to a rope elongation that is determined by the Specific modulus and the cross-sectional area of the rope. We have to use the less favourable values of the table. Hence, a Specific Modulus of 700 cN/tex = 7 N/tex.
What will the strain be at 0.3 m deflection? Consider the triangle:
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The hypotenuse compared to the half-length of the bridge, using Pythagoras, gives a 0.125% strain and tan α = 0.3/6 and since α is small tan α = α ⇔ α = 0.05 rad.
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In case the mass m = 250 kg works in the middle of the bridge, the following free body diagram for one of the ropes can be drawn:
Vertical force equilibrium tells that 2F sin α – mg/2 = 0 ⇔ F = mg/4α
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Using the definition of normal stress put into Hooke’s law tells that the linear density of the rope should be F/Eε = mg/4αEε = 1.4 Mtex//
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The second part of the question - to check the safety factor means that we should look at the ratio of the bridge’s ultimate strength minus the actual stress divided by the actual stress. The actual stress is given by Hooke’s law; i.e. σ250 kg = Eε = 7·0.00125 = 0.875 cN/tex.
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Hence the safety factor becomes (78-0.875)/0.875 = 88 times// That sure is comforting!!
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Project
This resource was developed as part of an Erasmus+ project, funded with support from the European Commission under grant agreement 2016-1-SE01-KA203-22064.
The project was a collaboration between:
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