Grades of +3.00% and -2.00%, sight distance 520 ft, h₁ = 3.50 ft and h₂ = 2.50 ft. Part A What is the minimum length of a vertical curve to provide a required sight distance for the conditions given? Express your answer in feet to five significant figures. L= Submit VAΣo↓ vec Request Answer → C A ? ft

Grades of +3.00% and -2.00%, sight distance 520 ft, h₁ = 3.50 ft and h₂ = 2.50 ft. Part A What is the minimum length of a vertical curve to provide a required sight distance for the conditions given? Express your answer in feet to five significant figures. L= Submit VAΣo↓ vec Request Answer → C A ? ft

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Item 4 Part A Compute the stopping sight distance available given a 550-ft curve, grades of g₁ = +2.00% and g2 = -2.50%, VPI at station 96 + 80, and elevation 845.26 ft, stakeout at full stations. Assume h₁ = 3.50 ft. and h₂ = 2.00 ft. Express your answer in feet to five significant figures. 197| ΑΣΦ ↓↑ vec S= Submit Provide Feedback Request Answer ? ft
What are the benefits of traditional S-curve?
Given the driver (eye height 1.1m) and the object (height 25cm) are within the length of a vertical curve, find out the length of the vertical curve if the road speed was 100kmph. the connecting slopes were - 3.6% and +2.1% in grades and stopping sight distance provided was 150m. A ▾ B I % $$ * US X₂ x² = = = E E # # I i CX
What is the length of tangent for the horizontal curve shown in the sketch below? A= 24° 33' 15" R= 110.00' Not to Scale Select the correct response: 46.71 ft. 47.14 ft 46.78 ft 23.94 ft
Given the following Information for a.vertical curve • 81 = 4.50% • 82 = -8.00% • PVC = 1+50.00 • L= 800.00' • Elevpyc = 250.00' What is the slope of the curve at station 5+50.00?
For items 1, and 2: A vertical parabolic curve of 260-m length has a back-tangent of -4.2% grade and a forward tangent of +3.0% which intersect at Station 11 + 488.00 with elevations 20.80 meters. a.) If cross-drainage pipes will be installed, determine the pipes' stationing. Provide the answer in kilometers e.g. 11,900 (instead of 11+900). 11+509.67
Please answer number 4 and 5 only. Round off the answer to two decimal places
Horizontal Curve Given: Degree of Curve (Daº) = 03°-40¹-47" Delta (A)= 18°-30¹-00" External (E)=? 20.27' 20.51 20.06' 20.42'
Given a horizontal simple curve of 442 m, an intersection angle of 32 degrees, and the Pl station of 1 + 12 4.2744, compute the station of PT. (My time is limited please help, I will vote it up promise)
I need correct solution
Number 3, handwritten work only. PLEASE MAKE SURE IT IS NEAT. Thank you
A downward grade of - 5.6% and another upward grade of +3.8% were connected with a vertical curve of 300m. The RL of EVC was known as 92.500 above MSL. Calculate the RLs at each 50m interval of the vertical curve and present the calculation in a tabular format. A B I E E $5 * U S X₂ x² = E III # # I ✪ 5 C