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slava [35]
3 years ago
9

A 356 cast aluminum test bar is tested in tension. The initial gage length as marked on the sample is 50mm and the initial diame

ter is 13.1mm. After fracture, the gage length is 53.7mm and the final diameter is 12.6mm. What is the UTS?
Engineering
1 answer:
Andreas93 [3]3 years ago
6 0

Answer

given,

l₀ = 50 mm

d₀ = 13.1 mm

l₁ = 53.7 mm

d₁ = 12.6 mm

Area of cross section

A = \dfrac{\pi}{4}d_0^2

A = \dfrac{\pi}{4} \times 13.1^2

A = 134. 782 mm²

The strain is

\epsilon = \dfrac{l_f-l_0}{l_0}

             = \dfrac{53.7-50}{50}

             = 0.074

the tensile modulus of A356 aluminium is

E = 72.4 GPa

The stress is

σ = εE

  = 72.4 × 10⁹ × 0.074

  = 5357.6 × 10⁶ Pa

σ = 5357.6 MPa

UTS = \dfrac{ultimate\ load}{cross\ sectional\ area}

UTS = \dfrac{ultimate\ load}{134.782}

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Answer:

slenderness ratio = 147.8

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Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

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moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

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buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

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3 years ago
Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes
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Answer:

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Explanation:

Given:

  At the inlet of turbine P=10 MPa  ,T=500 C

 AT the exit of turbine  P=10 KPa   ,x=0.9

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Now by putting the values

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m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

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