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Grades of g₁ = -2.50% and 9₂= +1.50 %, VPI elevation 856.00 ft at station 22 +00, and a fixed elevation 860.00 ft at station 20 + 50. Part A Field conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent vertical curve and full-station elevations. Give the elevations in order of increasing X Express your answers in feet to five significant figures separated by commas. VG| ΑΣΦ.Η. vec 4 → O Elev = 861.68,861.08,859.24,858.3,858.24,859.08,859.43 ft Submit Provique Apeware Request Answer Review ?

Grades of g₁ = -2.50% and 9₂= +1.50 %, VPI elevation 856.00 ft at station 22 +00, and a fixed elevation 860.00 ft at station 20 + 50. Part A Field conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent vertical curve and full-station elevations. Give the elevations in order of increasing X Express your answers in feet to five significant figures separated by commas. VG| ΑΣΦ.Η. vec 4 → O Elev = 861.68,861.08,859.24,858.3,858.24,859.08,859.43 ft Submit Provique Apeware Request Answer Review ?

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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25.11 Grades of g₁ = -2.50% and g2 = +1.50%, VPI elevation 860.00 ft at station 22 +00, and a fixed elevation 864.00 ft at station 20 + 50. L=450.00 ft from quadratic: 0.5L² − 3.25L + 4.50 = 0 2.5L X = (-1.5) sta; YBvc = 860 + 2 864 = 860 + = 2.5L 2 BVC Station = 19+75.000 BVC Elevation 865.625 Minimum elevation -2.5 (1-1.5) + (1.5 +2.5) (²1.5)* 2 2L r = 1.5 +2.5 4.5 = 0.889 Station x (Sta) gl*x r/2*x*x Elev 1st Diff 9.000 863.375 8.028 863.028 24+25.000 4.500 -11.250 24+00.000 4.250 -10.625 23+00.000 3.250 -8.125 22+00.000 2.250 -5.625 21+00.000 1.250 -3.125 20+50.000 0.750 -1.875 20+00.000 0.250 -0.625 0.028 865.028 19+75.000 0 865.625 4.694 862.194 -0.834 2.250 862.250 0.694 863.194 0.250 864.000 0 0 2 1.834 0.056 0.890 0.944 0.888 = 862.109 @ station 22+56.250 2nd Diff 0.890
Transcribed Image Text:25.11 Grades of g₁ = -2.50% and g2 = +1.50%, VPI elevation 860.00 ft at station 22 +00, and a fixed elevation 864.00 ft at station 20 + 50. L=450.00 ft from quadratic: 0.5L² − 3.25L + 4.50 = 0 2.5L X = (-1.5) sta; YBvc = 860 + 2 864 = 860 + = 2.5L 2 BVC Station = 19+75.000 BVC Elevation 865.625 Minimum elevation -2.5 (1-1.5) + (1.5 +2.5) (²1.5)* 2 2L r = 1.5 +2.5 4.5 = 0.889 Station x (Sta) gl*x r/2*x*x Elev 1st Diff 9.000 863.375 8.028 863.028 24+25.000 4.500 -11.250 24+00.000 4.250 -10.625 23+00.000 3.250 -8.125 22+00.000 2.250 -5.625 21+00.000 1.250 -3.125 20+50.000 0.750 -1.875 20+00.000 0.250 -0.625 0.028 865.028 19+75.000 0 865.625 4.694 862.194 -0.834 2.250 862.250 0.694 863.194 0.250 864.000 0 0 2 1.834 0.056 0.890 0.944 0.888 = 862.109 @ station 22+56.250 2nd Diff 0.890
Item 2 Grades of g₁ = -2.50% and 9₂ = +1.50%, VPI elevation 856.00 ft at station 22 +00, and a fixed elevation 860.00 ft at station 20 + 50. Part A Field conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent vertical curve and full-station elevations. Give the elevations in order of increasing X. Express your answers in feet to five significant figures separated by commas. AΣo↓ vec Submit Elev = 861.68,861.08,859.24,858.3,858.24,859.08,859.43 ft ? Previous Answers Request Answer ore chant 2 of 5 ........ ..... Review
Transcribed Image Text:Item 2 Grades of g₁ = -2.50% and 9₂ = +1.50%, VPI elevation 856.00 ft at station 22 +00, and a fixed elevation 860.00 ft at station 20 + 50. Part A Field conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent vertical curve and full-station elevations. Give the elevations in order of increasing X. Express your answers in feet to five significant figures separated by commas. AΣo↓ vec Submit Elev = 861.68,861.08,859.24,858.3,858.24,859.08,859.43 ft ? Previous Answers Request Answer ore chant 2 of 5 ........ ..... Review
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