- #1

themanonthemo

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Here is a problem I have.

QUESTION:

The main span of a bridge has a length of 473 m. On each end there are the expansion joints like the one on the photo below. One day the city changed from -4 to +15 degrees between 6 in the morning and 2 in the afternoon.

At 6 in the morning a piece of tire rubber fell into one of the cracks filling it completely. The rubber was 10 cm long and had a cross section of 4 cm2. What was the stress in the rubber at 2 PM? Clearly state all the assumptions, which you made, while solving this problem.

Data:

Young modulus of steel 200 GPa

Young modulus of rubber 7 kPa

Linear Expansion coefficient of steel 13·10-6 K-1

Linear Expansion coefficient of rubber 77·10-6 K-1

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How I decided to approach it

(dL/L)steel expansion + (dL/L)rubber expansion + (dL/L)steel stress + (dL/L)rubber stress = 0

a_sdT + a_rdt + (F/A)/Ys + (F/A)/Yr = 0

F/A must be equal for both

F/A = -dT(a_s + a_r)/(1/Ys + 1/Yr)

I only get about 6Pa... a number way to low.

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The answer key solves it the following way:

∆L=∆Lbridge+∆Lrubber = 58mm (rubber expansion is much smaller and can be neglected)

F/A = Y(dL)/L

S=7kPa *58mm/100mm = 4kPa

Assumptions:

The span warmed up uniformly to the air temperature

Most construction is steel

The span expanded uniformly in both directions

Only the rubber is compressed by the stress

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Firstly, I there is something wrong with the way the solution key solved it. They obtained that number assuming that all the stress goes into the rubber. It even says so in the assumptions but it still doesn't make sense.

Secondly, I don't see why the number I solved for is wrong. Could someone please explain what incorrect assumption I made?

Thanks so much!