Chapter 20, Problem 4PEB

To determine

The statue that needsto be straightened first from the list of the details shared for creep rate in table.

Answer to Problem 4PEB

Solution:

The statue installed in $1933$ must be straightened first.

Explanation of Solution

Given data:

The statues must be straightened when the displacement reaches $15\text{ cm}$ to prevent toppling.

By measuring the tilt angle on the statues, it is determined.

$\begin{array}{ll}\text{Date Installed}\hfill & \text{Rate}\left(\text{mm/yr}\right)\hfill \\ 1933\hfill & 2.01\hfill \\ 1957\hfill & 2.35\hfill \\ 1963\hfill & 3.1\hfill \\ 1976\hfill & 2.69\hfill \\ 1985\hfill & 2.5\hfill \end{array}$

Formula used:

Writing the expression for weathering rate by dividing the displacement by the age of the statues.

$\text{Rate}=\frac{\text{Displacement}}{\text{age}}$

Explanation:

Recallingtheexpression for weathering rate.

$\text{Rate}=\frac{\text{Displacement}}{\text{age}}$

Rearranging it as,

$\text{Displacement}=\text{Rate}\left(\text{age}\right)$…… (1)

For the installation date of 1933, the age is:

$2005-1933=72\text{ yr}$

Converting the displacement rate to $\frac{\text{cm}}{\text{yr}}$.

$\begin{array}{c}2.01\frac{\text{mm}}{\text{yr}}=2.01\frac{\text{mm}}{\text{yr}}\left(\frac{1\text{ cm}}{10\text{ mm}}\right)\\ =0.201\frac{\text{cm}}{\text{yr}}\end{array}$

Substituting $0.201\frac{\text{cm}}{\text{yr}}$ for rate and $72\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\text{Displacement}}_{1933}=0.201\frac{\text{cm}}{\text{yr}}\left(72\text{ yr}\right)\\ =14.5\text{ cm}\end{array}$

Subtracting the displacement from the $15\text{ cm}$ specification to determine remaining displacement (RD).

$\begin{array}{c}{\text{RD}}_{1933}=15\text{ cm}-14.5\text{ cm}\\ =\text{0}\text{.5 cm}\end{array}$

Dividing RD by the creep rate to determine the number of years left:

$\begin{array}{c}{\text{Year remaining}}_{1933}=\frac{0.5\text{ cm}}{0.201\frac{\text{cm}}{\text{yr}}}\\ =2.5\text{ yr}\end{array}$

For installation date of 1957, the age is:

$2005-1957=48\text{ yr}$:

Converting displacement rate to $\frac{\text{cm}}{\text{yr}}$.

$\begin{array}{c}2.35\frac{\text{mm}}{\text{yr}}=2.35\frac{\text{mm}}{\text{yr}}\left(\frac{1\text{ cm}}{10\text{ mm}}\right)\\ =0.235\frac{\text{cm}}{\text{yr}}\end{array}$

Substituting $0.235\frac{\text{cm}}{\text{yr}}$ for rate and $48\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\text{Displacement}}_{1957}=0.235\frac{\text{cm}}{\text{yr}}\left(48\text{ yr}\right)\\ =11.3\text{ cm}\end{array}$

Subtracting the displacement from the $15\text{ cm}$ specification to determine remaining displacement (RD).

$\begin{array}{c}{\text{RD}}_{1957}=15\text{ cm}-11.3\text{ cm}\\ =\text{3}\text{.7 cm}\end{array}$

Dividing RD by the creep rate to determine the number of years left.

$\begin{array}{c}{\text{Year remaining}}_{1957}=\frac{3.7\text{ cm}}{0.235\frac{\text{cm}}{\text{yr}}}\\ =15.7\text{ yr}\end{array}$

For installation date of 1963 the age is:

$2005-1963=42\text{ yr}$

Converting displacement rate to $\frac{\text{cm}}{\text{yr}}$.

$\begin{array}{c}3.1\frac{\text{mm}}{\text{yr}}=3.1\frac{\text{mm}}{\text{yr}}\left(\frac{1\text{ cm}}{10\text{ mm}}\right)\\ =0.31\frac{\text{cm}}{\text{yr}}\end{array}$

Substituting $0.31\frac{\text{cm}}{\text{yr}}$ for rate and $42\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\text{Displacement}}_{1963}=0.31\frac{\text{cm}}{\text{yr}}\left(42\text{ yr}\right)\\ =13.0\text{ cm}\end{array}$

Subtracting the displacement from the $15\text{ cm}$ specification to determine remaining displacement (RD).

$\begin{array}{c}{\text{RD}}_{1963}=15\text{ cm}-13.0\text{ cm}\\ =\text{2 cm}\end{array}$

Dividing RD by the creep rate to determine the number of years left:

$\begin{array}{c}{\text{Year remaining}}_{1963}=\frac{2\text{ cm}}{0.310\frac{\text{cm}}{\text{yr}}}\\ =6.5\text{ yr}\end{array}$

For installation date of 1976, the age is:

$2005-1976=29\text{ yr}$

Converting displacement rate to $\frac{\text{cm}}{\text{yr}}$.

$\begin{array}{c}2.69\frac{\text{mm}}{\text{yr}}=2.69\frac{\text{mm}}{\text{yr}}\left(\frac{1\text{ cm}}{10\text{ mm}}\right)\\ =0.269\frac{\text{cm}}{\text{yr}}\end{array}$

Substituting $0.269\frac{\text{cm}}{\text{yr}}$ for rate and $29\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\text{Displacement}}_{1976}=0.269\frac{\text{cm}}{\text{yr}}\left(29\text{ yr}\right)\\ =7.8\text{ cm}\end{array}$

Subtracting the displacement from the $15\text{ cm}$ specification to determine remaining displacement (RD).

$\begin{array}{c}{\text{RD}}_{1976}=15\text{ cm}-7.8\text{ cm}\\ =\text{7}\text{.2 cm}\end{array}$

Dividing RD by the creep rate to determine the number of years left.

$\begin{array}{c}{\text{Year remaining}}_{1976}=\frac{7.2\text{ cm}}{0.269\frac{\text{cm}}{\text{yr}}}\\ =26.8\text{ yr}\end{array}$

For installation date of 1985, the age is:

$2005-1985=20\text{ yr}$

Converting displacement rate to $\frac{\text{cm}}{\text{yr}}$.

$\begin{array}{c}2.5\frac{\text{mm}}{\text{yr}}=2.5\frac{\text{mm}}{\text{yr}}\left(\frac{1\text{ cm}}{10\text{ mm}}\right)\\ =0.25\frac{\text{cm}}{\text{yr}}\end{array}$

Substituting $0.25\frac{\text{cm}}{\text{yr}}$ for rate and $20\text{ yr}$ for age in equation (1).

$\begin{array}{c}{\text{Displacement}}_{1985}=0.25\frac{\text{cm}}{\text{yr}}\left(20\text{ yr}\right)\\ =5\text{ cm}\end{array}$

Subtracting the displacement from the $15\text{ cm}$ specification to determine remaining displacement (RD).

$\begin{array}{c}{\text{RD}}_{1933}=15\text{ cm}-5\text{ cm}\\ =\text{10 cm}\end{array}$

Dividing RD by the creep rate to determine the number of years left.

$\begin{array}{c}{\text{Year remaining}}_{1985}=\frac{10\text{ cm}}{0.250\frac{\text{cm}}{\text{yr}}}\\ =40\text{ yr}\end{array}$

Conclusion:

The statue installed in $1933$ must be straightened first.