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How to calculate the energy required to shake an entire bridge in a clash?
- Thread starter GraveDigger84
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Here's my very rough attempt to calc this. Based on the calcs I looked at, the correct method is to use the Mercalli scale? With the formula:
This link says that 1 cubic meter of concrete equals 2350 kilograms so we have (46,000)(2350)=108100000 kg of concrete.
Then for the timber, this is a little complicated since they don't state what dimensions the timber had besides the feet. The wikipedia link does make mention of timber planks however.
This link states wood planks can come at a thickness and width of 38 mm × 286 mm. So solving for the volume of the total wood gives us: (.038m)(.286m)(2000000m)=21736 m^3 of wood.
The aforementioned wikipedia article mentions the use of pine wood for the pillings so we'll use this link that states 1 cubic meter of pine, wood equals 450 kilograms. Therefore we have (21736)(450)= 9781200 kg of wood.
For the masonry, it is mentioned to be limestone and 99,000 m^3 of it so we will use this link to say 1 cubic meter of limestone equals 2500 kilograms therefore we have (2500)(99,000)= 247500000 kg of masonry
Then we are given a minimum 40,000 short tons of steel. Using this we get 36287389.6 kg of steel.
Adding this up we get 108100000+9781200+247500000+36287389.6= 401668589.6 kg as our mass for the bridge.
Now to find the velocity on this chart: https://en.wikipedia.org/wiki/Peak_ground_acceleration#Correlation_with_the_Mercalli_scale
Since the clash explicitly shook the whole bridge I'll use two ends, one at intensity V and one at intensity VI
Intensity V:
(0.5)(401668589.6)(0.0465^2) = 434253.9539 Joules, Wall Level
Intensity VI:
(0.5)(401668589.6)(0.0964^2) = 1866345.068 Joules, Wall Level
Did I do this right or did I mess this up somewhere along the line cause this seems way too low.
First, we must get the mass of the bridge. The wikipedia page for the Williamsburg bridge says this:
First for the concrete.
This link says that 1 cubic meter of concrete equals 2350 kilograms so we have (46,000)(2350)=108100000 kg of concrete.
Then for the timber, this is a little complicated since they don't state what dimensions the timber had besides the feet. The wikipedia link does make mention of timber planks however.
This link states wood planks can come at a thickness and width of 38 mm × 286 mm. So solving for the volume of the total wood gives us: (.038m)(.286m)(2000000m)=21736 m^3 of wood.
The aforementioned wikipedia article mentions the use of pine wood for the pillings so we'll use this link that states 1 cubic meter of pine, wood equals 450 kilograms. Therefore we have (21736)(450)= 9781200 kg of wood.
For the masonry, it is mentioned to be limestone and 99,000 m^3 of it so we will use this link to say 1 cubic meter of limestone equals 2500 kilograms therefore we have (2500)(99,000)= 247500000 kg of masonry
Then we are given a minimum 40,000 short tons of steel. Using this we get 36287389.6 kg of steel.
Adding this up we get 108100000+9781200+247500000+36287389.6= 401668589.6 kg as our mass for the bridge.
Now to find the velocity on this chart: https://en.wikipedia.org/wiki/Peak_ground_acceleration#Correlation_with_the_Mercalli_scale
Since the clash explicitly shook the whole bridge I'll use two ends, one at intensity V and one at intensity VI
Intensity V:
(0.5)(401668589.6)(0.0465^2) = 434253.9539 Joules, Wall Level
Intensity VI:
(0.5)(401668589.6)(0.0964^2) = 1866345.068 Joules, Wall Level
Did I do this right or did I mess this up somewhere along the line cause this seems way too low.
