On a day when the temperature is 22.5°C, a concrete walk is poured in such a way

Question

On a day when the temperature is 22.5°C, a concrete walk is poured in such a waythat its ends are unable to move. a) What is the stress in thecement when its temperature is 42.0°C?
1 Pa
I really don't even know where to start with this one, do Iuse Young's Modulus? If someone could help me that would be great,thanks! On a day when the temperature is 22.5°C, a concrete walk is poured in such a waythat its ends are unable to move. a) What is the stress in thecement when its temperature is 42.0°C?
1 Pa
I really don't even know where to start with this one, do Iuse Young's Modulus? If someone could help me that would be great,thanks!

Explanation / Answer

the Young's modulus of the concrete walk is Y = [(F/A)/(l/lo)] = (F * lo/A *l) = (F/A) * (lo/l) -------------(1) the change in the length of the concrete walk when itstemperature is increased is lt = lo * (1 + * t) or lt = lo + lo * *t or lt - lo = lo * *t or l = lo * * t-------------(2) from equations (1) and (2) we get Y = (F/A) * (lo/lo * *t) = (F/A) * (1/ * t) or (F/A) = Y * * t the Young's modulus for concrete is Y = 30 * 10^9 Pa the coefficient of linear expansion for concrete is =14.5 * 10^-6 m/m K t = (42.0 - 22.5) oC = 19.5 oC =(19.5 + 273) K = 292.5 K

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.