4.139 Load on timber beams. Timber beams are widely used inhome construction. When the load (measured in pounds) perunit length has a constant value over part of a beam, the loadis said to be uniformly distributed over that part of the beam.Uniformly distributed beam loads were used to derive thestiffness distribution of the beam in the American Institute of

Aeronautics and Astronautics Journal(May 2013). Considera cantilever beam with a uniformly distributed load between100 and 115 pounds per linear foot.

a. What is the probability that a beam load exceeds110 pounds per linearfoot?

b. What is the probability that a beam load is less than102 pounds per linear foot?

c. Find a value Lsuch that the probability that the beamload exceeds Lis only .1.

You should consider a cantilever beam with a uniformly distributed load between 100 and 115 pounds per linear foot.

Let x be the random variable uniformly distributed between 100 and 115 pounds.

The probability density function random variable x is given by

$f\left(x\right)=\frac{1}{d-c};c<x<d$

Here c=100 and d=115

So the pdf of x is:

$$\begin{gathered} f\left(x\right)=\frac{1}{115-100} \\ =\frac{1}{15} \\ =0.066667 \\ f\left(x\right)=\left\{\begin{array}{l}0.066667;100<x<115\\ 0;otherwise\end{array}\right. \end{gathered}$$

$$\begin{gathered} F\left(x\right)=P\left(X\le x\right) \\ ={\int }_{100}^{x}f\left(t\right)dt \\ ={\int }_{100}^{x}0.66667dt \\ =0.66667{\int }_{100}^{x}dt \\ =0.66667{\left[T\right]}_{100}^x \\ =0.66667\left(x-100\right) \\ F\left(x\right)=0.66667\times \left(x-100\right) \end{gathered}$$

For continuous random variable x

$P\left(X<x\right)=P\left(X<x\right)$

The probability that a beam load exceeds110 pounds per linear foot is,

$$\begin{gathered} P\left(x>110\right)=1-P\left(x\le 110\right) \\ =1-F\left(110\right) \\ =1-066667\times \left(110-100\right) \\ =0.333333 \end{gathered}$$

Thus, the required probability is 0.333333.

The probability that a beam load is less than102 pounds per linear foot is,

$$\begin{gathered} P\left(x<102\right)=F\left(102\right) \\ =0.066667\times \left(102\times 100\right) \\ =0.133333 \end{gathered}$$

Thus, the required probability is 0.133333.

To find a value Lsuch that the probability that the beamload exceeds Lis only .1, we require following calculation.

$$\begin{gathered} P\left(x>L\right)=0.10 \\ 1-P\left(x\le L\right)=0.10 \\ P\left(x\le L\right)=0.90 \\ F\left(L\right)=0.90 \\ 0.0666667\times \left(L-100\right)=0.90 \\ L=100+\frac{0.90}{0.0666667} \\ L=113.5 \end{gathered}$$

Thus, the required value is 113.5 pounds.