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venomousdaisy

A 30m section of railway track is installed on a cold winter day at -5°C, each end of the section is rigidly fixed to the sleepers. On a hot summer day at 25°C what force does the rail exert on the sleepers. The coefficient of linear expansion for wrought iron is 10x10^-6K^-1 . And the cross sectional area of the rail is 0.01m². Young 's Modulus for wrought iron E=210GPa

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    30m odcinka linii kolejowej jest zainstalowany na chłodne zimowe w temperaturze -5 ° C, każdy koniec odcinka jest sztywno zamocowany do podkładów. W upalny letni dzień w temperaturze 25 ° C, co życie ma szyny wywierać na podkłady.Współczynnik rozszerzalności liniowej dla kutego żelaza jest 10x10 ^-6K ^ -1. I pole przekroju poprzecznego szyny jest 0,01 m ². Young Moduł dla kutego żelaza E = 210GPa

    zmiana temperatury = 30 stopni razy współczynnik i długość = ile wydłużyły się szyny
    jak każdy koniec sztywno umocowano bez przerw to się rozleciały

Rozwiązania

  • antekL1

    I'm using the following names for variables:
    Delta T = 25 - (-5) = 30 degrees, the temperature difference
    L = 30 m - the section length
    Delta L - the lenght difference, caused by heating
    alpha = 10 * 10^6 1 / K - the linear expansion coeff.
    E = 210 GPA = 210 * 10^9 Pa - the Young's modulus
    S = 0.01 - the crosssectional area.

    F is the force I have to calculate. From the definition of the Young's modulus:
    E = \frac{F}{S}\cdot\frac{L}{\Delta L}
    so the force F equals:
    F = ES\frac{\Delta L}{L}
    The whole ratio delta L / L can be obtained from the definition of alpha:
    \alpha = \frac{\Delta L}{L}\cdot\frac{1}{\Delta T}
    therefore:
    F = ES\alpha \Delta T = 210\cdot 10^9\cdot 0.01\cdot 10\cdot 10^{-6}\cdot 30 = 630000
    The units are: pascals * m^2 * (1 / K) * K = newtons (N).
    Answer: 630 kN on all sleepers. To calculate the force on a particular sleeper one needs to know the number of sleepers but I guess only the total force is needed.
    And important: The force F works in opposite directions on both sides on the ralilway section. NOT the force F/2! But of course if there are 2 sleepers on each end only F/2 per sleeper can be detected.

    The length L is not needed - or I made a mistake in calculations.
    Antek

  • CzaAarNaAa_

    F= 210*10000000000*0.01*10*0,000001*30= 630000

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