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ioda
3 years ago
13

2. Given that a quality-control inspection can ensure that a structural ceramic part will have no flaws greater than 100 µm (100

10-6 m)in size, calculate the maximum service stress available with SiC (KIC = 3 MPa m1/2) and partially stabilized zirconia (PSZ) (KIC = 9 MPa m1/2). (See Example 8.3)
Engineering
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

SiC=169.26 Mpa

Partially stabilized zirconia=507.77 Mpa

Explanation:

<u>SiC </u>

Stress intensity, K is given by

K=\sigma Y\sqrt{\pi a} hence making \sigma the subject where \sigma is applied stress, Y is shape factor, a is crack length

\sigma=\frac {K}{Y\sqrt{\pi a}} substituting the given figures and assuming shape factor, Y of 1

\sigma=\frac {3}{1\sqrt{\pi 100*10^{-6}}}= 169.2569\approx 169.26 Mpa

<u>Stabilized zirconia </u>

Stress intensity, K is given by

K=\sigma Y\sqrt{\pi a} hence making \sigma the subject where \sigma is applied stress, Y is shape factor, a is crack length

\sigma=\frac {K}{Y\sqrt{\pi a}} substituting the given figures and assuming shape factor, Y of 1

\sigma=\frac {9}{1\sqrt{\pi 100*10^{-6}}}= 507.7706\approx 507.77 Mpa

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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

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T = 81.84 Nm

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

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

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torsion = 81.84 / πd³/ 16

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