Page 709 - 9780077418427.pdf

Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile
tiL12214_appe_643-698.indd Page 686 09/10/10 8:37 AM user-f463
tiL12214_appe_643-698.indd Page 686 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefiles

Case 2: After Weathering the current displacement. This difference is then divided by the
L = 5 m creep rate to determine the number of years left before the
W = 5 m tombstone must be straightened.
H = 2 m
Step 1: Displacement
pieces = 4
date = 1925
A surface = ?
age = 2010 − 1925 = 85 yr
A surface = pieces [2LH + 2WH + 2LW]
mm
= 4[2(5 m)(2 m) + 2(5 m)(2 m) + 2(5 m)(5 m)] rate = 2.14 _
yr
2
2
2
= 4[20 m + 20 m + 50 m ] displacement = ?
2
= 4[90 m ] displacement

= 360 m 2 rate = _ ∴ displacement = age(rate)
age
Difference Convert displacement rate to .
_
cm

A 1 = 280 m 2 yr
mm _
_
1 cm

A 2 = 360 m 2 2.14 ( 10 mm )
yr
_
cm
difference = ? 0.214

2
difference = A − A 1 yr _
cm
(
yr )

2

= 360 m − 280 m 2 displacement = 85 yr 0.214
= 80 m 2 = 18.2 cm

20.2. The weathering rate is determined by dividing the surface Using the same method, determine the displacement for the
reduction by the age of the tombstone. remainder of the values. Results are below:
date = 1906 _
reduction

rate = age displacement 1931 = 17.8 cm
reduction = 0.09 mm displacement 1948 = 18.3 cm
0.09 mm
age = 2000 − 1906 = 94 yr = _

94 yr displacement 1965 = 13.4 cm
rate = ? displacement 1970 = 11.8 cm
−4 mm

= 9.6 × 10 _
yr
Step 2: Remaining Displacement
Using the same method, determine the weathering rate for the
Subtract the displacement from the 20 cm specifi cation to
remainder of the values. Results are shown below:
determine RD:
−3 mm
rate 1912 = 1.5 × 10 _

yr RD 1925 = 20 cm − 18.2 cm = 1.8 cm
−3 mm
rate 1937 = 1.7 × 10 _ RD 1931 = 20 cm − 17.8 cm = 2.2 cm

yr
RD 1948 = 20 cm − 18.3 cm = 1.7 cm
−3 mm
rate 1946 = 1.1 × 10 _ RD 1965 = 20 cm − 13.4 cm = 6.6 cm

yr
RD 1970 = 20 cm − 11.8 cm = 8.2 cm
−4 mm
rate 1955 = 4.4 × 10 _

yr
Step 3: Years Left
Average the results.
Divide RD by the creep rate to determine the number of
−4 mm
−3 mm
−3 mm

9.6 × 10 _ + 1.5 × 10 _ + 1.7 × 10 _ years left :
yr yr yr
1.8 cm

−3 mm
−4 mm

+ 1.1 × 10 _ + 4.4 × 10 _ years remaining 1925 = _ = 8.4 yr
cm
_
yr yr 0.214

rate average = yr
5 _
2.2 cm

−3 mm
_
cm
= 1.1 × 10 _ years remaining 1931 = 0.225 = 9.8 yr

yr yr
1.7 cm

20.3. The creep rate is the distance the rock moved divided by the years remaining 1948 = _ = 5.8 yr

cm
_

time it took to move that distance. 0.295
yr
_
6.6 cm
d = 129 m d _ years remaining 1965 =

= 22.1 yr
cm
_
rate creep = 0.298

t = 2,260 yr t yr
8.2 cm
rate creep = ? Convert distance fromm to cm. years remaining 1970 = _ = 27.8 yr

_
cm

0.295
yr
_ )
100 cm
129 m (
1 m The 1948 stone will need to be straightened fi rst.

4
1.29 × 10 cm

20.5. Steepness in percent is the slope times 100. The units must be
__
4
1.29 × 10 cm

rate creep = the same for Δx and Δy.
2,260 yr
_ Δx = 452 m
cm
= 5.71
yr Δy = 956 m − 823 m = 133 m
slope percent = ?
20.4. This problem requires several steps. First, the amount of Δy
_
displacement of each tombstone needs to be determined by slope percent = × 100%
Δx
multiplying the creep rate by the age of the tombstone. Then _
133 m
the remaining available displacement (RD) must be calculated = 453 m × 100%
by taking the difference between the 20 cm specification and = 29.4%
686 APPENDIX E Solutions for Group A Parallel Exercises E-44