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Elena-2011 [213]
3 years ago
15

Some characteristics of clay products such as (a) density, (b) firing distortion, (c) strength, (d) corrosion resistance, and (e

) thermal conductivity are affected by the extent of vitrification. Will they increase or decrease with increasing degree of vitrification?
1. (a) increase (b) decrease (c) increase (d) decrease (e) increase
2. (a) decrease (b) increase (c) increase (d) increase (e) decrease
3. (a) decrease (b) decrease (c) increase (d) decrease (e) decrease
4. (a) increase (b) increase (c) increase (d) increase (e) increase
5. (a) increase (b) decrease (c) decrease (d) increase (e) decrease
Engineering
1 answer:
Inga [223]3 years ago
6 0

Explanation:

1. increase This due to increase in the pore volume.

2.increase . This is due to the fact that more liquid phase will be present at the firing.

3. Increase. This increase is because of the fact that clay on cooling forms glass.Thus, gaining more strength as the liquid phase formed fills in pore volume.

4. Increase, Rate of corrosion depends upon the surface area exposed.Since, upon vitrification surface area would increase, therefore corrosion increases.

5. Increase , glass has higher thermal conductivity than the pores it fills.

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Problem 3.10 One/half million parts of a certain type are to be manufactured annually on dedicated production machines that run
goldenfox [79]

Answer:

  • 4 machines
  • 4 machines

Explanation:

(a) The production schedule is 4000 hour per year, or 240,000 minutes. In that time, 133,333 1/3 parts can be produced. The production of 500,000 parts in a year will require the use of ...

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The demand can be satisfied by 4 machines.

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(b) If availability is 95% then 3.75/0.95 ≈ 3.95 machines will be required.

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3 years ago
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Ulleksa [173]
The answer is A. Immediately inform her colleague
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Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
raketka [301]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

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3 years ago
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yuradex [85]

Answer:

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

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Explanation:

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there are 7 basic load combination of LRFD that is

1) 1.4(D + F)

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5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

and

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here

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