HW4_Solution

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Oregon State University, Corvallis *

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392

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Civil Engineering

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Apr 3, 2024

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docx

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Homework 4 February , 2021 CE 392 Introduction to Highway Engineering Chapter 3 Mannering and Washburn (Due Friday 8 th ) Problems Select Answers 3-6 Sta Highpoint = 113+36 Elev Highpoint 245.45ft 3-8 3-9 y= 1201.86 ft 3-13 8.9 ft 3-16 3-21 3-24
Homework 4 February , 2021 3.6  A vertical curve is designed for 55 mi/h and has an initial grade of +2.5% and a final grade of −1.0%. The  PVT  is at station 114 + 50. It is known that a point on the curve at station 112 + 35 is at elevation 245 ft. What is the stationing and elevation of the  PVC ? What is the stationing and elevation of the high point on the curve? GIVEN: K = 114 (@ 55 mi/h), G 1 =2.5, G 2 = -1.0, A= 3.5, PVT = 114+50 Point Station = 112+35, elev point = 245’ Calculate Curve Length L = K*A = 399’ Calculate PVC PVC = PVT-L = 115+51 x:= point-PVC = 184’ Calculate offset of point above curve Y point := (A/200*L)*x 2 = 1.485’ x*(G 1 /100) – Y point = 3.115’ from offset of point calculate elevation of PVC elev PVC := elev point -3.115 = 241.885’ Calculate location of high point x h := KG 1 = 285’ Y h := (A/200*L)*x h 2 = 3.562’ Calculate station of high point elev hp := elev PVC + Y h = 245.45’ STA hp = PVC+x h = 113+36
Homework 4 February , 2021 3.8  An equal-tangent crest vertical curve has a 50-mi/h design speed. The initial grade is +3%. The high point is at station  33 + 40.76  and the  PVT  is at station  37 + 24.66 . What is the elevation difference between the high point and the  PVT ?
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