A beam of circular cross-section is subjected to pure bending moment M and the bending stresses are given by the following equation: σ = 32 Mb/πd^3 where d is the diameter of the beam. It has been observed that the diameter (d) of the beam is a normally distributed random variable with a mean of

Question

A beam of circular cross-section is subjected to pure bending moment M and the bending stresses are given by the following equation:

[latex] \sigma=\frac{32 M_{b}}{\pi d^{3}} [/latex]

where d is the diameter of the beam. It has been observed that the diameter (d) of the beam is a normally distributed random variable with a mean of 50 mm and a standard deviation of 0.125
mm. The bending moment [latex] \left(M_{b}\right) [/latex] is also a normally distributed random variable with a mean of 1750 N-m and a standard deviation of 150 N-m.
Determine the mean and standard deviation of the corresponding bending stress variable (σ).
Comment on the analysis.

Accepted Answer

[latex] ext { Given } \mu_{d}=50 mm \quad \hat{\sigma}_{d}=0.125 mm [/latex].

[latex] \mu_{M}=1750 imes 10^{3} N - mm \quad \hat{\sigma}_{M}=150 imes 10^{3} N - mm [/latex].

Step I Population of diameter (d)
d denotes the population of diameters. For this population

[latex] \mu_{d}=50 mm \quad ext { and } \quad \hat{\sigma}_{d}=0.125 mm [/latex].

Step II Population of bending moment (M)
M denotes the population of values of bending moment. For this population,

[latex] \mu_{M}=1750 imes 10^{3} N - mm [/latex].

[latex] ext { and } \hat{\sigma}_{M}=150 imes 10^{3} N - mm [/latex].

[latex] ext { Step III Population of }\left(\pi d^{3} / 32 ight)( Z ) [/latex]

Z denotes a third population. It is obtained by using the expression,

[latex] Z=\frac{\pi}{32} d^{3} [/latex].

In the above expression (π/32) is constant and using the equations in Table 24.7

Table 24.7 Mean and standard deviations of the function Z

[latex] ext { Standard deviation } \hat{\sigma}_{Z}[/latex]	[latex] ext { Mean } \mu_{Z}[/latex]	Function
0	a	Z = a
[latex]a \hat{\sigma}_{X}[/latex]	[latex]a \mu_{X}[/latex]	Z = aX
[latex]\hat{\sigma}_{X}[/latex]	[latex]\mu_{X} \pm a[/latex]	Z = X ± a
[latex]\sqrt{\left(\hat{\sigma}_{X} ight)^{2}+\left(\hat{\sigma}_{Y} ight)^{2}}[/latex]	[latex]\mu_{X} \pm \mu_{Y}[/latex]	Z = X ± Y
[latex]\sqrt{\mu_{X}^{2}\left(\hat{\sigma}_{Y} ight)^{2}+\mu_{Y}^{2}\left(\hat{\sigma}_{X} ight)^{2}+\left(\hat{\sigma}_{X} ight)^{2}\left(\hat{\sigma}_{Y} ight)^{2}}[/latex]	[latex]\mu_{X} \mu_{Y}[/latex]	Z = X Y
[latex]\frac{1}{\mu_{Y}}\left[\frac{\mu_{X}^{2}\left(\hat{\sigma}_{Y} ight)^{2}+\mu_{Y}^{2}\left(\hat{\sigma}_{X} ight)^{2}}{\mu_{Y}^{2}+\left(\hat{\sigma}_{Y} ight)^{2}} ight]^{1 / 2} [/latex]	[latex]\mu_{X} / \mu_{Y}[/latex]	Z = X/Y
[latex] \frac{1}{2}\left(\frac{\hat{\sigma}_{X}}{\mu_{X}} ight)\left[4 \mu_{X}^{2}+\left(\hat{\sigma}_{X} ight)^{2} ight][/latex]	[latex]\mu_{x}^{2}+\hat{\sigma}_{x}^{2}[/latex]	Z = X²
[latex]3 \mu_{X}^{2}\left(\hat{\sigma}_{X} ight)+3\left(\hat{\sigma}_{X} ight)^{3}[/latex]	[latex]\mu_{X}^{3}+3 \mu_{X}\left(\hat{\sigma}_{X} ight)^{2} [/latex]	Z = X³
[latex] \frac{\hat{\sigma}_{X}}{\mu_{X}^{2}}\left[1+\left(\frac{\hat{\sigma}_{X}}{\mu_{X}} ight)^{2} ight][/latex]	[latex] \frac{1}{\mu_{X}}\left[1+\left(\frac{\hat{\sigma}_{X}}{\mu_{X}} ight)^{2} ight][/latex]	Z = 1/X

[latex] \mu_{Z}=\frac{\pi}{32}\left[\mu_{d}^{3}+3 \mu_{d}\left(\hat{\sigma}_{d} ight)^{2} ight] [/latex].

[latex] =\frac{\pi}{32}\left[50^{3}+3(50)(0.125)^{2} ight] [/latex].

= 12 272.08 mm³.

[latex] \hat{\sigma}_{Z}=\frac{\pi}{32}\left[3 \mu_{d}^{2} \hat{\sigma}_{d}+3\left(\hat{\sigma}_{d} ight)^{3} ight] [/latex].

[latex] =\frac{\pi}{32}\left[3(50)^{2}(0.125)+3(0.125)^{3} ight] [/latex].

= 92.04 mm³.

Step IV Population of bending stress (σ)
A fourth population of bending stress is denoted by σ. It is obtained by dividing the population of values of bending moment M by the population Z.
From Table 24.7,

[latex] \mu_{\sigma}=\frac{\mu_{M}}{\mu_{Z}}=\frac{1750 imes 10^{3}}{12272.08}=142.6 N / mm ^{2} [/latex].

[latex] \hat{\sigma}_{\sigma}=\frac{1}{\mu_{Z}}\left[\frac{\mu_{M}^{2} \hat{\sigma}_{Z}^{2}+\mu_{Z}^{2} \hat{\sigma}_{M}^{2}}{\mu_{Z}^{2}+\hat{\sigma}_{Z}^{2}} ight]^{1 / 2} [/latex].

[latex] =\frac{1}{12272.08}\left[\frac{\left(1750 imes 10^{3} ight)^{2} imes(92.04)^{2}+(12272.08)^{2} imes\left(150 imes 10^{3} ight)^{2}}{(12272.08)^{2}+(92.04)^{2}} ight]^{1 / 2} [/latex].

[latex] \hat{\sigma}_{\sigma}=12.27 N / mm ^{3} [/latex].

Step V Comments on analysis
The bending stress population has a mean of 142.6 N/mm² and standard deviation of 12.27 N/mm².
We can predict the mean and standard deviation of this population. However, the population is not a normally distributed random variable.

Question 24.10: A beam of circular cross-section is subjected to pure bendin...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions