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2 years ago

13

What is the difference Plastic vs elastic deformation.

1 answer:

Reika [66]2 years ago

8 0

Answer:

What is the difference Plastic vs elastic deformation

Explanation:

The elastic deformation occurs when a low stress is apply over a metal or metal structure, in this process, the stress' deformation is temporary and it's recover after the stress is removed. In other words, this DOES NOT affects the atoms separation.

The plastic deformation occurs when the stress apply over the metal or metal structure is sufficient to deform the atomic structure making the atoms split, this is a crystal separation on a limited amount of atoms' bonds.

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Differentiate between isohyetal method and arithmetical average method of rainfall

prohojiy [21]

Answer:

Explained below

Explanation:

The isohyetal method is one used in estimating Rainfall whereby the mean precipitation across an area is gotten by drawing lines that have equal precipitation. This is done by the use of topographic and other data to yield reliable estimates.

Whereas, the arithmetic method is used to calculate true precipitation by the way of getting the arithmetic mean of all the points or arial measurements that will be considered in the analysis.

7 0

3 years ago

Reverse masking forms a soft edge on the panel.

valina [46]

True I think I’m not sure?

7 0

2 years ago

Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adia

harina [27]

Answer:

Power = 371.28 kW

Explanation:

Initial pressure, P1 = 5 bar

Final pressure, P2 = 1 bar

Initial temperature, T1 = 320°C

Final temperature, T2 = 160°C

Volume flow rate, V = 0.65m³/s

From steam tables at state 1,

h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

v1 = 0.5416 m³/kg

Mass flow rate, m = V/v1

m = 1.2 kg/s

From steam tables, at state 2

h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

Power developed, P = m(h1 - h2)

P = 1.2(3105.6-2796.2)

P = 371.28 kW

8 0

3 years ago

A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri

babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are 73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

$T_{BE}-W=0$ hence

$T_{BE}=W=20*9.81=196.2 N$

Therefore, tension in the cable, $T_{BE}=196.2 N$

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

$196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0$

$24.525-39.24+0.2D_x=0$

$D_x=73.575 N$

Similarly,

$A_x-D_y=0$

$A_x=73.575 N$

Therefore, both reactions at A and D are 73.575 N

7 0

3 years ago

A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.

sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c) $Stress\ ratio =-0.65$

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

$\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}$

$2\sigma_m={\sigma_1+\sigma_2}$

2 x 46.2 = σ₁ + σ₂

σ₁ + σ₂ = 92.4 MPa --------1

The amplitude stress given as

$\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}$

$2\sigma_a={\sigma_1-\sigma_2}$

2 x 219 = σ₁ - σ₂

σ₁ - σ₂ = 438 MPa --------2

By adding the above equation

2 σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

$Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}$

$Stress\ ratio =\dfrac{-172.8}{265.2}$

$Stress\ ratio =-0.65$

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0

3 years ago