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PolarNik [594]
3 years ago
10

on the same scale for stress, the tensile true stress-true strain curve is higher than the engineeringstress-engineering strain

curve. Explain whether this condition also holds for a compression test.
Engineering
1 answer:
Bess [88]3 years ago
5 0

Answer:

The condition does not hold for a compression test

Explanation:

For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension.  The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.

<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test

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A small pad subjected to a shearing force is deformed at the top of the pad 0.08 in. The height of the pad is 1.38 in. What is t
Aleksandr-060686 [28]

Answer:

The shear strain is 0.05797 rad.

Explanation:

Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.

Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

tan\phi=\frac{\bigtriangleup l}{h}

tan\phi=\frac{0.08}{1.38}

tan\phi= 0.05797

For small angle of \phi, tan\phi can take as\phi.

\phi = 0.05797 rad.

Thus, the shear strain is 0.05797 rad.

7 0
3 years ago
A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at
DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

dQ = m \times c_p \times (T_2 -T_1)

For ideal gas, c_p = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

3 0
3 years ago
Define extensive and intensive properties of thermodynamic system.
Flura [38]

Answer:

An intense property is a physical attribute of a system that is independent of the size of the system or the quantity of material it contains. An extensive property of a system, on the other hand, is dependent on the size of the system or the amount of material in it.

Explanation:

7 0
3 years ago
Read 2 more answers
Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

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So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0
4 years ago
What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road and (b) on an uphill road?
ElenaW [278]

Answer:

a) zero b) zero

Explanation:

Newton's first law tells us that a body remains at rest or in uniform rectilinear motion, if a net force is not applied on it, that is, if there are no applied forces or If the sum of forces acting is zero. In this case there is a body that moves with uniform rectilinear motion which implies that there is no net force.

4 0
3 years ago
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