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Question

The total sustained load on the concrete footing of a planned building is the sum of the dead load plus the occupancy load. Suppose that the dead load

X1X _ { 1 }

has a gamma distribution with

α1=50 and β1=2\alpha _ { 1 } = 50 \text { and } \beta _ { 1 } = 2

, whereas the occupancy load

X2X _ { 2 }

has a gamma distribution with

α2=20andβ2=2\alpha _ { 2 } = 20 \operatorname { and } \beta _ { 2 } = 2

. (Units are in kips.) Assume that

X1 and X2X _ { 1 } \text { and } X _ { 2 }

are independent. Find a value for the sustained load that will be exceeded with probability less than 1/16.

Solution

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Given:\textbf{Given:} The dead load X1X_{1} has a gamma distribution with α1=50\alpha_{1} = 50 and β1=2\beta_{1} = 2 wheareas occupancy load X2X_{2} has a gamma distribution with α1=20\alpha_{1} = 20 and β1=2\beta_{1} = 2.

We have to calculate maximum value of sustained load which will be exceeded with probability less than 1/16.

Let X1+X2=YX_{1} + X_{2} = Y

Results obtained in part (a):\textbf{Results obtained in part (a):}

E(Y)=μ=140E(Y) = \mu = 140

And

V(Y)=σ2=280σ=16.733V(Y) = \sigma^{2} = 280 \Rightarrow \sigma = 16.733

According to Tchebysheff’s Theorem,

P(Yμkσ)1k2=116=142P(|Y - \mu| \geq k\sigma) \leq \dfrac{1}{k^{2}} = \dfrac{1}{16} = \dfrac{1}{4^{2}}

Hence, k=4k = 4. So,

Yμ4σ|Y - \mu| \geq 4\sigma

Yμ4σ And Yμ+4σY \leq \mu - 4 \sigma \ \text{And} \ Y \geq \mu + 4 \sigma

Y1404×16.733 And Y140+4×16.733Y \leq 140 - 4 \times 16.733 \ \text{And} \ Y \geq 140 + 4 \times 16.733

Hence

Y73.067 And Y206.933Y \leq 73.067 \ \text{And} \ Y \geq 206.933

Hence, Required sustained load=206.933 kips\boxed{\text{Required sustained load} = 206.933 \ \text{kips}}

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