Symmetrical Vertical Curves - CONTINUED - 14071_255
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Figure 11-18.Table of computations of elevations on a symmetrical vertical curve. - 14071_254
Unsymmetrical Vertical Curves - 14071_256
Engineering Aid 2 - Intermediate Structural engineering guide book
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233
234
235
236
237
238
239
240
241
242
243
The elevation on the tangent at station 29
230.12 + 4.50 = 234.62 feet.
The elevation on the tangent at station 30
234.62 + 4.50 = 239.12 feet.
+
50 is
+ 00 is
In this problem, to find the elevation on the tan-
gent at any 50-foot station starting at the
PVC,
add
4.50 to the elevation at the preceding station until you
reach the
PVI.
At this point use a slightly different
method to calculate elevations because the curve
slopes downward toward the
PVT.
Think of the eleva-
tions as being divided into two groups—one group
running from the
PVC
to the
PVI;
the other group
running from the
PVT
to the
PVI.
Going downhill on a gradient of –7 percent from
the
PVI
to the
PVT,
there will be a drop of 3.50 feet
for every 50 feet of horizontal distance. To find the
elevations at stations between the
PVI
to the
PVT
in
this particular problem, subtract 3.50 from the eleva-
tion at the preceding station. The elevation on the
tangent at station 30 + 50 is
239.12-3.50, or 235.62 feet.
The elevation on the tangent at station 31 + 50 is
235.62-3.50, or 232.12 feet.
The elevation on the tangent at station 31 + 50 is
232.12-3.50, or 228.62 feet.
The elevation on the tangent at station 32+00 (PVT) is
228.62-3.50, or 225.12 feet,
The last subtraction provides a check on the work you
have finished. List the computed elevations under col-
umn 2.
STEP 4: Calculate
(e),
the middle vertical offset
at the
PVI.
First, find the
(G),
the algebraic difference of the
gradients using the formula
G = g
_{2}
– g
_{1}
G =
-7 –(+9)
G =
–16%
The middle vertical offset
(e)
is calculated as follows:
e
=
LG/8
= [(4)(–16) ]/8 = -8.00 feet.
The negative sign indicates e is to be subtracted from
the
PVI.
STEP 5: Compute the vertical offsets at each
50-foot station, using the formula
(x/l)
^{2}
e.
To find
the vertical offset at any point on a vertical curve,
first find the ratio
x/l;
then square it and multiply
by
e;
for example,
x/l
=
50/200
=
1/4.
at station 28 + 50, the ratio of
Therefore, the vertical offset is
(1/4)
^{2 }
e
= (1/16)
e.
The vertical offset at station 28 + 50 equals
(1/16)(–8) = –0.50 foot.
Repeat this procedure to find the vertical offset at
each of the 50-foot stations. List the results under
columns 3, 4, and 5.
STEP 6: Compute the grade elevation at each of
the 50-foot stations.
When the curve is on a crest, the sign of the offset
will be negative; therefore, subtract the vertical offset
(the figure in column 5) from the elevation on the
tangent (the figure in column 2); for example, the
grade elevation at station 29 + 50 is
234.62 – 4.50 = 230.12 feet.
Obtain the grade elevation at each of the stations in a
similar manner. Enter the results under column 6.
Note:
When the curve is in a dip, the sign will be
positive; therefore, you will
add
the vertical offset
(the figure in column 5) to the elevation on the tangent
(the figure in column 2).
STEP 7: Find the turning point on the vertical
curve.
When the curve is on a crest, the turning point is
the highest point on the curve. When the curve is in a
dip, the turning point is the lowest point on the curve.
The turning point will be directly above or below the
PVI
only when both tangents have the same percent of
slope (ignoring the algebraic sign); otherwise, the
turning point will be on the same side of the curve as
the tangent with the least percent of slope.
The horizontal location of the turning point is
either measured from the
PVC
if the tangent with the
lesser slope begins there or from the
PVT
if the tangent
with the lesser slope ends there. The horizontal loca-
tion is found by the formula:
Where:
x
_{t}
=
distance of turning point from
PVC
or
PVT
_{g}
= lesser slope (ignoring signs)
L
= length of curve in stations
G
= algebraic difference of slopes.
11-17
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