Compute the first three entries in a table for setting out the following vertical curve, at intervals of 50 m.Incoming slope:+ 1.8%Outgoing slope:1.2%-iii.R.L. of intersection point (I.P.):300 mChainage of I.P.:2253.253 m55V.The value of the constant K':Note: Assume equal tangent lengths.

Given data: Incoming slope = 1.8 = g1 Outgoing slope = -1.2 = g2 RL = 300M

Compute the first three entries in a table for setting out the following vertical curve, at intervals of 50 m. i. + 1.8% Incoming slope: Outgoing slope: - 1.2% iii. 300 m R.L. of intersection point (I.P.): Chainage of I.P.: iv. 2253.253 m 55 V. The value of the constant K': Note: Assume equal tangent lengths

Compute the first three entries in a table for setting out the following vertical curve, at intervals of 50 m. i. + 1.8% Incoming slope: Outgoing slope: - 1.2% iii. 300 m R.L. of intersection point (I.P.): Chainage of I.P.: iv. 2253.253 m 55 V. The value of the constant K': Note: Assume equal tangent lengths

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

Answer fast
Complete answer
Show complete solution and provide necessary illustration/s.
topic: fundamentals of surveying A 3% ascending grade meets a descending grade at the vertex at station 2+050 and elevation 100m with a curve elevation of 97.333m below it. The tangents are connected by an unsymmetrical parabolic curve and the vertical offset of the point of tangency from the back tangent is 9.60m. If the length of the second curve is 80m, determine: A.) Length of the 1st Curve = [L1] B.) Stationing of the Highest Point = [StaHP] C.) Elevation of the Highest Point = [ElevHP]
2. A compound curve has the following data: I1 = 28°, l2 = 31°, Di = 3º, D2 = 4°; Find the stationing of P.C.C. if P.I. is at station (30 + 120.50). Use arc basis. %3| %3D
Thank you for answering :)
Station of the PI of the curve: 7+400 Elevation of the PI of the curve: 80m. L1 = 160m L2 = 120m G1 = +7% G2 = -4% a)Determine the elevation of the curve on every station starting from PC to PT. Tabulate your answer and show solutions.
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?
please resolve this! SURVEYING (UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS AND DIAGRAMS *IF NECESSARY* LEGIBLY. NO LONG EXPLANATION IS NEEDED. ANSWER IN 2 DECIMAL PLACES AND BOX THE FINAL ANSWER. PLEASE ANSWER ALL.) The long chord from PC to PT of a compound curve is 320m long, and the angles it makes with the first and third tangents are 14° and 16°, respectively. If the common tangent is parallel to the long chord, solve the following: a. The radius of the first curve (R1). b. The radius of the second curve (R2). c. If station of PC is 8+200, compute the station of PT.
ASAP. PLEASE INCLUDE THE ILLUSTRATION
topic: fundamentals of surveying A vertical sag parabolic curve CD, 620 meters long is connected by tangents having a downgrade 0f 6.2% and upgrade of 4.2% intersecting at station 18+254.4 and elevation 54.25m. Find the distance from D to the lowest point of the curve.
A-2.50% grade meets a +1.75% grade at station 44+25 and elevation 3385.74 ft, 400-ft curve, stakeout at half stations. Part A Determine the elevation at the low point of the curve Express your answer in feet to six significant figures. ΠΗΓΙΑΣΦ Elev = + ft