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fenix001 [56]
3 years ago
10

A steel bar with a diameter of .875 inches and a length of 15.0 ft is axially loaded with a force of 21.6 kip. The modulus of el

asticity of the steel is 29 *106 psi. Determine
Engineering
1 answer:
svetoff [14.1K]3 years ago
3 0

Answer:

35.92 kpsi

Explanation:

Given data:

diameter of the steel bar d = 0.875 in

Area A = πd^2/4 = π(0.875)^2/4

length L = 15.0 ft

Load P = 21.6 kip

Modulus of elesticity E = 29×10^6 Psi

Assume we are asked to determine axial stress in the bar which is given as

\sigma  = Load, P/ Area, A

\sigma  = 4P/\pi d^2

substitute the value

\sigma = \frac{4\times 21.6}{\pi \times (0.875)^2} \\=35.92\ kpsi

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Answer:

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Explanation:

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  R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

  R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

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  R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

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The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

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so, the fire brick interface temperature at the common brick is ...

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For the next layers, the interface temperatures are ...

  common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

  magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

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To find the reactance XLXLX_L of an inductor, imagine that a current I(t)=I0sin(ωt)I(t)=I0sin⁡(ωt) , is flowing through the indu
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Answer:

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