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3 years ago

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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)

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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Where;

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2 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

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Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

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$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

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Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

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4 years ago