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Magazine Chapter 5: Q66E (page 242)
If two loads are applied to a cantilever beam as shown in the accompanying drawing, the bending moment at \({\rm{0}}\) due to the loads is \({{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)
a. Suppose that \({{\rm{X}}_{\rm{1}}}{\rm{ and }}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)are independent rv鈥檚 with means \({\rm{2 and 4kip}}\), respectively, and standard deviations \({\rm{.5}}\) and \({\rm{1}}{\rm{.0kip}}\), respectively. If \({{\rm{a}}_{\rm{1}}}{\rm{ = 5ft and }}{{\rm{a}}_{\rm{2}}}{\rm{ = 10ft }}\), what is the expected bending moment and what is the standard deviation of the bending moment?
b. If \({{\rm{X}}_{\rm{1}}}{\rm{ and }}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)are normally distributed, what is the probability that the bending moment will exceed 75 kip-ft?
c. Suppose the positions of the two loads are random variables. Denoting them by \({{\rm{A}}_{\rm{1}}}{\rm{ and }}{{\rm{A}}_{\rm{2}}}{\rm{ }}\), assume that these variables have means of \({\rm{5 and 10ft }}\), respectively, that each has a standard deviation of \({\rm{.5}}\), and that all are independent of one another. What is the expected moment now?
d. For the situation of part (c), what is the variance of the bending moment?
e. If the situation is as described in part (a) except that \({\rm{Corr}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.5}}\) (so that the two loads are not independent), what is the variance of the bending moment?
Short Answer
a. \({{\rm{\mu }}_{{\rm{5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}}}{\rm{ = 50}}\)and \({{\rm{\sigma }}_{{\rm{5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}}}{\rm{ = }}\sqrt {{\rm{106}}{\rm{.25}}} {\rm{\gg 10}}{\rm{.3078}}\)
b. \({\rm{P}}\left( {{\rm{5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}{\rm{ > 75}}} \right){\rm{ = 0}}{\rm{.0075 = 0}}{\rm{.75\% }}\)
c. \({\rm{E}}\left( {{{\rm{A}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{A}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = 50}}\)
d. \({\rm{V}}\left( {{{\rm{A}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{A}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = }}\frac{{{\rm{1785}}}}{{{\rm{16}}}}{\rm{ = 111}}{\rm{.5625}}\)
e. \({\rm{\sigma }}_{{{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}}^{\rm{2}}{\rm{ = 131}}{\rm{.25}}\)
Step by step solution
Definition of standard deviation
The square root of the variance is the standard deviation of a random variable, sample, statistical population, data collection, or probability distribution. It is less resilient in practice than the average absolute deviation, but it is algebraically easier.
Calculating the expected bending moment and the standard deviation of the bending moment
Given:
\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{{{\rm{X}}_{\rm{1}}}}}{\rm{ = 2}}}\\{{{\rm{\mu }}_{{{\rm{X}}_{\rm{2}}}}}{\rm{ = 4}}}\\{{{\rm{\sigma }}_{{{\rm{X}}_{\rm{1}}}}}{\rm{ = 0}}{\rm{.5}}}\\{{{\rm{\sigma }}_{{{\rm{X}}_{\rm{2}}}}}{\rm{ = 1}}{\rm{.0}}}\\{{{\rm{a}}_{\rm{1}}}{\rm{ = 5}}}\\{{{\rm{a}}_{\rm{2}}}{\rm{ = 10}}}\end{array}\)
The mean, variance, and standard deviation for the linear combination \({\rm{W = a}}{{\rm{X}}_{\rm{1}}}{\rm{ + b}}{{\rm{X}}_{\rm{2}}}\)are as follows:
\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{\rm{W}}}{\rm{ = a}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + b}}{{\rm{\mu }}_{\rm{2}}}}\\{{\rm{\sigma }}_{\rm{W}}^{\rm{2}}{\rm{ = }}{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{\sigma }}_{\rm{2}}^{\rm{2}}\left( {{\rm{\;If\;}}{{\rm{X}}_{\rm{ - }}}{\rm{1\;and\;}}{{\rm{X}}_{\rm{ - }}}{\rm{2\;are independent\;}}} \right)}\\{{{\rm{\sigma }}_{\rm{W}}}{\rm{ = }}\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{\sigma }}_{\rm{2}}^{\rm{2}}} \left( {{\rm{\;If\;}}{{\rm{X}}_{\rm{ - }}}{\rm{1\;and\;}}{{\rm{X}}_{\rm{ - }}}{\rm{2}}} \right.{\rm{\;are independent)\;}}}\end{array}\)
After that \({\rm{W = }}{{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}{\rm{ = 5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}\)( with \({\rm{a = 5}}\)and \({\rm{b = 10}}\)), we get:
\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{{{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}}}{\rm{ = }}{{\rm{\mu }}_{{\rm{5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}}}{\rm{ = 5}}{{\rm{\mu }}_{{{\rm{X}}_{\rm{1}}}}}{\rm{ + 10}}{{\rm{\mu }}_{{{\rm{X}}_{\rm{2}}}}}{\rm{ = 5(2) + 10(4) = 50}}}\\{{{\rm{\sigma }}_{{{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}}}{\rm{ = }}{{\rm{\sigma }}_{{\rm{5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}}}{\rm{ = }}\sqrt {{{\rm{5}}^{\rm{2}}}{\rm{\sigma }}_{{{\rm{X}}_{\rm{1}}}}^{\rm{2}}{\rm{ + 1}}{{\rm{0}}^{\rm{2}}}{\rm{\sigma }}_{{{\rm{X}}_{\rm{2}}}}^{\rm{2}}} {\rm{ = }}\sqrt {{\rm{25(0}}{\rm{.5}}{{\rm{)}}^{\rm{2}}}{\rm{ + 100(1}}{{\rm{)}}^{\rm{2}}}} {\rm{ = }}\sqrt {{\rm{106}}{\rm{.25}}} {\rm{\gg 10}}{\rm{.3078}}}\end{array}\)
Calculating the probability that the bending moment will exceed \({\rm{75kip - ft}}\)
The standardized score is the value \({\rm{x}}\) divided by the standard deviation after being reduced by the mean.
\({\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{75 - 50}}}}{{\sqrt {{\rm{106}}{\rm{.25}}} }}{\rm{\gg 2}}{\rm{.43}}\)
Using table A.3, calculate the corresponding probability:
\(\begin{array}{*{20}{c}}{{\rm{P}}\left( {{\rm{5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}{\rm{ > 75}}} \right){\rm{ = P(Z > 2}}{\rm{.43) = 1 - P(Z < 2}}{\rm{.43)}}}\\{{\rm{ = 1 - 0}}{\rm{.9925 = 0}}{\rm{.0075 = 0}}{\rm{.75\% }}}\end{array}\)
Calculating the expected moment now
Assumption:
\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{{{\rm{A}}_{\rm{1}}}}}{\rm{ = 5}}}\\{{{\rm{\mu }}_{{{\rm{A}}_{\rm{2}}}}}{\rm{ = 10}}}\\{{{\rm{\sigma }}_{{{\rm{A}}_{\rm{1}}}}}{\rm{ = 0}}{\rm{.5}}}\\{{{\rm{\sigma }}_{{{\rm{A}}_{\rm{2}}}}}{\rm{ = 0}}{\rm{.5}}}\end{array}\)
We get the following for \({\rm{W = }}{{\rm{A}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{A}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}\)using the conditions for the expected value of independent random variables: \({\rm{E}}\left( {{{\rm{A}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{A}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = E}}\left( {{{\rm{A}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}} \right){\rm{ + E}}\left( {{{\rm{A}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = E}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{E}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ + E}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{E}}\left( {{{\rm{X}}_{\rm{2}}}} \right)\)
\({\rm{ = }}{{\rm{\mu }}_{{{\rm{A}}_{\rm{1}}}}}{{\rm{\mu }}_{{{\rm{X}}_{\rm{1}}}}}{\rm{ + }}{{\rm{\mu }}_{{{\rm{A}}_{\rm{2}}}}}{{\rm{\mu }}_{{{\rm{X}}_{\rm{2}}}}}{\rm{ = 5(2) + 10(4) = 50}}\)
Calculating variance of the bending moment
Assumption:
\(\begin{array}{*{20}{c}}{{{\rm{\mu }}_{{{\rm{A}}_{\rm{1}}}}}{\rm{ = 5}}}\\{{{\rm{\mu }}_{{{\rm{A}}_{\rm{2}}}}}{\rm{ = 10}}}\\{{{\rm{\sigma }}_{{{\rm{A}}_{\rm{1}}}}}{\rm{ = 0}}{\rm{.5}}}\\{{{\rm{\sigma }}_{{{\rm{A}}_{\rm{2}}}}}{\rm{ = 0}}{\rm{.5}}}\end{array}\)
We get \({\rm{W = }}{{\rm{A}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{A}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}\)using the property \({\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ + (E(X)}}{{\rm{)}}^{\rm{2}}}\) for the variance and the properties for the expected value of independent random variables:
Calculating variance of the bending moment
\({\rm{Corr}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.5}}\)
The correlation is calculated by dividing the covariance by the standard deviations of \({{\rm{X}}_{\rm{1}}}\)and \({{\rm{X}}_{\rm{2}}}\)
\({\rm{Corr}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = }}\frac{{{\rm{cov}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right)}}{{{{\rm{\sigma }}_{{{\rm{X}}_{\rm{1}}}}}{{\rm{\sigma }}_{{{\rm{X}}_{\rm{2}}}}}}}\)
Solve the following equation for covariance:
\({\rm{cov}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = }}{{\rm{\sigma }}_{{{\rm{X}}_{\rm{1}}}}}{{\rm{\sigma }}_{{{\rm{X}}_{\rm{2}}}}}{\rm{Corr}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right)\)
The mean, variance, and standard deviation for the linear combination \({\rm{W = a}}{{\rm{X}}_{\rm{1}}}{\rm{ + b}}{{\rm{X}}_{\rm{2}}}\)are as follows:
\({{\rm{\mu }}_{\rm{W}}}{\rm{ = a}}{{\rm{\mu }}_{\rm{1}}}{\rm{ + b}}{{\rm{\mu }}_{\rm{2}}}\)
\({\rm{\sigma }}_{\rm{W}}^{\rm{2}}{\rm{ = }}{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}{\rm{\sigma }}_{\rm{2}}^{\rm{2}}{\rm{ + 2abcov}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right)\)(If \({{\rm{X}}_{\rm{ - }}}{\rm{1}}\) and \({{\rm{X}}_{\rm{ - }}}{\rm{2}}\), are not self-sufficient)
We get the following for \({\rm{W = }}{{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}{\rm{ = 5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}\) (with \({\rm{a = 5}}\)and \({\rm{b = 10}}\)):
\(\begin{array}{*{20}{c}}{{\rm{\sigma }}_{{{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}}^{\rm{2}}{\rm{ = \sigma }}_{{\rm{5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}}^{\rm{2}}{\rm{ = }}{{\rm{5}}^{\rm{2}}}{\rm{\sigma }}_{{{\rm{X}}_{\rm{1}}}}^{\rm{2}}{\rm{ + 1}}{{\rm{0}}^{\rm{2}}}{\rm{\sigma }}_{{{\rm{X}}_{\rm{2}}}}^{\rm{2}}{\rm{ + 2(5)(10)cov}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right)}\\{{\rm{ = }}{{\rm{5}}^{\rm{2}}}{\rm{\sigma }}_{{{\rm{X}}_{\rm{1}}}}^{\rm{2}}{\rm{ + 1}}{{\rm{0}}^{\rm{2}}}{\rm{\sigma }}_{{{\rm{X}}_{\rm{2}}}}^{\rm{2}}{\rm{ + 2(5)(10)}}{{\rm{\sigma }}_{{{\rm{X}}_{\rm{1}}}}}{{\rm{\sigma }}_{{{\rm{X}}_{\rm{2}}}}}{\rm{Corr}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right)}\\{{\rm{ = 25(0}}{\rm{.5}}{{\rm{)}}^{\rm{2}}}{\rm{ + 100(1}}{{\rm{)}}^{\rm{2}}}{\rm{ + 100(0}}{\rm{.5)(1)(0}}{\rm{.5) = 131}}{\rm{.25}}}\end{array}\)
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