a 600 ft equal tangent sag vertical curve has the PVC at Station 170+00 and elevation 1000 ft. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationing And elevation of the PVI, the PVT, and the lowest point on the curve.

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
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In this question I got everything correct up to where I have to find elevation of PVT. (Not sure how the textbook got 991.0 ft) and I also don’t understand why the textbook found “a” by not putting G1 and G2 in percentages? Handwritten work only please
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K
Transportation engineering textbook.pdf
88
♫
curve will occur when the first derivative of the parabolic function (Eq. 3.1) is zero because
the initial and final grades are opposite in sign. When initial and final grades are no
opposite in sign, the low (or high) point on the curve will not be where the first derivative is
zero because the slope along the curve will never be zero. For example, a sag curve with an
initial grade of -2.0% and a final grade of -2.0% will have its lowest elevation at the PVT
and the first derivative of Eq. 3.1 will not be zero at any point along the curve. However, in
our example problem the derivative will be equal to zero at some point, so the low point
will occur when
From Eq. 3.3 we have
Substituting for a and b gives
b=G₁ = -3.5
with G₁ in percent. From Eq. 3.6 (with L in stations and G₁ and G₂ in percent),
0.5-(-3.5)
2(6)
IOB RADIO
a=
1.K
dy
dx
7.281
= 2ax+b=0
This gives the stationing of the low point at 175 + 25 (5 + 25 stations from the PVC). For
the elevation of the lowest point on the vertical curve, the values of a, b, c (elevation of the
PVC), and x are substituted into Eq. 3.1, giving
dy
-= 2(0.33333)x+ (-3.5) = 0
dx
x = 5.25 stations
Note that the preceding equations can also be solved with grades expressed as the decimal
equivalent of percent (for example, 0.02 ft/ft for 2%) if x is expressed in feet instead of
stations Care must be taken not to mix unite A dimensional analysis of Ea 31 must ensure
-= 0.33333
y = 0.33333(5.25)²+(-3.5) (5.25)+1000
= 990.81 ft
125
Subscribed ✓
MAR
25
Like C
tv
G
Transcribed Image Text:essages File Edit View Conversations Window Help K Transportation engineering textbook.pdf 88 ♫ curve will occur when the first derivative of the parabolic function (Eq. 3.1) is zero because the initial and final grades are opposite in sign. When initial and final grades are no opposite in sign, the low (or high) point on the curve will not be where the first derivative is zero because the slope along the curve will never be zero. For example, a sag curve with an initial grade of -2.0% and a final grade of -2.0% will have its lowest elevation at the PVT and the first derivative of Eq. 3.1 will not be zero at any point along the curve. However, in our example problem the derivative will be equal to zero at some point, so the low point will occur when From Eq. 3.3 we have Substituting for a and b gives b=G₁ = -3.5 with G₁ in percent. From Eq. 3.6 (with L in stations and G₁ and G₂ in percent), 0.5-(-3.5) 2(6) IOB RADIO a= 1.K dy dx 7.281 = 2ax+b=0 This gives the stationing of the low point at 175 + 25 (5 + 25 stations from the PVC). For the elevation of the lowest point on the vertical curve, the values of a, b, c (elevation of the PVC), and x are substituted into Eq. 3.1, giving dy -= 2(0.33333)x+ (-3.5) = 0 dx x = 5.25 stations Note that the preceding equations can also be solved with grades expressed as the decimal equivalent of percent (for example, 0.02 ft/ft for 2%) if x is expressed in feet instead of stations Care must be taken not to mix unite A dimensional analysis of Ea 31 must ensure -= 0.33333 y = 0.33333(5.25)²+(-3.5) (5.25)+1000 = 990.81 ft 125 Subscribed ✓ MAR 25 Like C tv G
a 600 ft equal tangent sag vertical curve has the PVC at Station 170+00 and elevation
1000 ft. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationing
And elevation of the PVI, the PVT, and the lowest point on the curve.
L=600 ft
T=300 ft
Options
-3.5%
PVC = PVT-L
17000 = PVT -600
17600 = PVT
176 +00= PVT
PVC
170+00
PUI = PVC + T
54°F
Cloudy
Elevation
1000
= 17,000 + 300
- 17300
PVI = 173+00
-
y = 2ax + b
O Search
PVI
173+00
Elevation =
a
989.5'
Elevation PUI = E pvc + 9²/12
= 1000 + (-0.035) (600)
=
989.5'
G₂+0.5%
PVT
1776+00
elevation = 974.5¹
Elevation PVT = EPI-9²/2/2
b=
= 989.5 - (+0.05)(b)
= 974.5'
a=
= G₂-G₁₂₁ = 0.05-(-0.035)
2L
2(600)
= 7.0833•10-5
0.035
4:54 PM
Transcribed Image Text:a 600 ft equal tangent sag vertical curve has the PVC at Station 170+00 and elevation 1000 ft. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationing And elevation of the PVI, the PVT, and the lowest point on the curve. L=600 ft T=300 ft Options -3.5% PVC = PVT-L 17000 = PVT -600 17600 = PVT 176 +00= PVT PVC 170+00 PUI = PVC + T 54°F Cloudy Elevation 1000 = 17,000 + 300 - 17300 PVI = 173+00 - y = 2ax + b O Search PVI 173+00 Elevation = a 989.5' Elevation PUI = E pvc + 9²/12 = 1000 + (-0.035) (600) = 989.5' G₂+0.5% PVT 1776+00 elevation = 974.5¹ Elevation PVT = EPI-9²/2/2 b= = 989.5 - (+0.05)(b) = 974.5' a= = G₂-G₁₂₁ = 0.05-(-0.035) 2L 2(600) = 7.0833•10-5 0.035 4:54 PM
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