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2 years ago

Which of the two materials (brittle vs. ductile) usually obtains the highest ultimate strength and why?

Engineering

1 answer:

sveta [45]2 years ago

5 0

Answer:

Explanation:

Ductile materials typically have a higher ultimate strength because they stretch absorbing more energy before breaking. While fragile materials snap in half before larger deformations due to larger loads occur.

It should be noted that when ductile materials stretch their section becomes smaller, and in that reduced section the stresses concentrate.

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Solid spherical particles having a diameter of 0.090 mm and a density of 2002 kg/m3 are settling in a solution of water at 26.7C

IrinaVladis [17]

Answer:

Settling Velocity (Up)= 2.048*10^-5 m/s

Reynolds number Re = 2.159*10^-3

Explanation:

We proceed as follows;

Diameter of Particle = 0.09 mm = 0.09*10^-3 m

Solid Particle Density = 2002 kg/m3

Solid Fraction, θ= 0.45

Temperature = 26.7°C

Viscosity of water = 0.8509*10^-3 kg/ms

Density of water at 26.7 °C = 996.67 kg/m3

The velocity between the interface, i.e between the suspension and clear water is given by,

U = [ ((nf/ρf)/d)D^3] [18+(1/3)D^3)(1/2)]

D = d[(ρp/ρf)-1)g*(ρf/nf)^2]^(1/3)

D = 2.147

U = 0.0003m/s (n = 4.49)

Up = 0.0003 * (1-0.45)^4.49 = 2.048*10^-5 m/s

Re=0.09*10^-3*2.048*10^-5*996.67/0.0008509 = 2.159*10^-3

6 0

2 years ago

A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a

vladimir2022 [97]

Answer:

diameter is 14 mm

Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

to find out

minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60 .................1

put here value

15 × $10^{3}$ = 2π×1750×T / 60

T = 81.84 Nm

and

torsion = T / Z ..........2

here Z is section modulus i.e = πd³/ 16

so from equation 2

torsion = 81.84 / πd³/ 16

so torsion = 416.75 / / d³ .................3

so from shear stress theory

torsion = σy / factor of safety

so here σy = 530 for 1020 steel

torsion = σy / factor of safety

416.75 / d³ = 530 × $10^{6}$ / 3

so d = 0.0133 m

so diameter is 14 mm

3 0

2 years ago

Discuss your interpretation of the confidence-precision trade-off, and provide a few examples of how you might make a choice in

LenKa [72]

The given question is incomplete, the complete question is as follows:

Our text describes a trade-off that we must make as engineers between our confidence in the value of a parameter versus the precision with which we know the value of that parameter. That trade-off might be affected by whether we are looking at a two-sided or bounded (one-sided) interval.

Question: Discuss your interpretation of the confidence-precision trade-off, and provide a few examples of how you might make a choice in one direction or the other in an engineering situation.

Answer: A balancing point is required to be reached to obtain a better confidence level in the predicted values.

Explanation:

The confidence interval and precision are the two terms that aims at providing the accurate estimation of the measurability of an object. If the precision increases, we can compromise on the confidence level and if the confidence level increases, then the precision of the predicted value also dilutes.

Thus a balance point is required to be reached between these two variables so that we get better confidence in the values being predicted without losing the correct estimation on precision. Ensuring that both the confidence and precision are maintained.

8 0

2 years ago

A spherical tank for storing gas under pressure is 25 m in diameter and is made of steel 15 mm thick. The yield point of the mat

nalin [4]

Answer:

a) (option B) 230 kPa

b) (option A) 100 N/m

Explanation:

Given:

Diameter, d = 25 m

Thicknesses, t = 15 mm

Yield point = 240 MPa