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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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3 years ago
Technician A says that hoods are designed with reinforcements to prevent folding during a collision. Technician B says that some
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Explanation:

hey mate this is the best answer if you're studying engineering!

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3 years ago
Read 2 more answers
Air enters a 28-cm diameter pipe steadily at 200 kPaand 208C with a velocity of 5 m/s. Air is heated as it flows, and leaves the
Alina [70]

Answer:

(a) \dot V_1 = 0.308 m³/s

(b) \dot m = 0.732 kg/m³

(c) v₂ = 5.94 m/s.

Explanation:

(a) The volume flow rate is given by the cross sectional area of the pipe × Velocity of flow of air

Diameter of pipe = 28 cm = 0.28 m

The cross sectional area, A, of the pipe = 0.28²/4×π = 0.0616 m²

Volume flow rate = 5 × 0.0616  = 0.308 m³/s

\dot V_1 = 0.308 m³/s

(b) From the general gas equation, we have;

p₁v₁ = RT₁ which gives;

p₁/ρ₁ = RT₁

ρ₁ = p₁/(RT₁)

Where:

ρ₁ = Density of the air

p₁ = 200 kPa

T₁ = 20 C =

R = 0.287 kPa·m³/(kg·K)

ρ₁ = 200/(0.287 ×293.15) = 2.377 kg/m³

The mass flow rate = Volume flow rate × Density

The mass flow rate, \dot m = 2.377×0.308 = 0.732 kg/m³

\dot m = 0.732 kg/m³

(c) The density at exit, ρ₂, is found from the the universal gas equation as follows;

ρ₂ = p₂/(RT₂)

Where:

p₂ = Pressure at exit = 180 kPa

T₂ = Exit temperature = 40°C = 273.15 + 40 = 313.15 K

∴ ρ₂ = 180/(0.287×313.15) = 2.003 kg/m³

\dot m = ρ₂×\dot V_2

\dot V_2 = \dot m/ρ₂ = 0.732/2.003 = 0.366 m³/s

\dot V_2 = v₂ × A

v₂ = \dot V_2/A = 0.366/0.0616 = 5.94 m/s.

v₂ = 5.94 m/s.

5 0
3 years ago
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