= -2.50% and 92 = +1.50%, VPI elevation 854.00 ft at station 22 + 00, and a fixed elevation 858.00 ft atGrades of 91station 20 + 50.Part AField conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent verticalcurve and full-station elevations. Give the elevations in order of increasing X.Express your answers in feet to five significant figures separated by commas.Elev =ΜΕ ΑΣΦ ↑ vecSubmitRequest Answer?ft

Answered: Image /qna-images/answer/e7ab8a45-a946-4f5a-97ff-b34255294ef9.jpg

Answered: = -2.50% and 92 = +1.50%, VPI elevation…

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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L-500 250 250 Figure 3 al Exam - Figure 3.pdf + Calculate the elevation of curve B at station 11+50. 115.90 117.94 118.80 119.80 120.80 29+ /2+50 Ind 13175
A-2.40% grade meets a +1.60% grade at station 44+25 and elevation 3387.96 ft, 400-ft curve, stakeout at half stations. Part A Tabulate station elevations for an equal-tangent parabolic curve for the data given above. Express your answers in feet to six significant figures separated by commas in order of increasing station numbers. {—| ΑΣΦ | 11 | vec A Elev= 3412.46,3432.72,3450.30,3465.20,3477.45 ft Your submission doesn't have the correct number of answers. Answers should be separated with a comma. No credit lost. Try again. Submit Previous Answers Request Answer Provide Feedback Review ? Next >
1. A proposed road bend along a highway construction project is to be laid out such that it will be a circular simple curve tangent to the three reference lines DA, AB, and BC. From the given data below, calculate for the following. a. radius of the simple curve b. stationing of PC if V is at 12 + 340 c. stationing of PT POINT LATITUDE DEPARTURE A - 29.34 - 128.76 B - 177.97 - 58.39 Bearing DA = S65°01'50"E Bearing CB = N54°24'10"E
A-2.60% grade meets a +1.70% grade at station 44+25 and elevation 3387.96 ft, 400-ft curve, stakeout at half stations. Part A Tabulate station elevations for an equal-tangent parabolic curve for the data given above. Express your answers in feet to six significant figures separated by commas in order of increasing station numbers. Elev = Submit ΑΣΦ ↓↑ Request Answer vec ? ft
A compound curve passes thru a common tangent AB having a length of 377.83 m. The radius of the first curve is equal to 283.32 and a central angle of 40.56. If the radius of the second curve is 663.37. Compute the stationing of PT if PC is at 14 +812.81. WRITE YOUR ANSWERS in meters. (For example, the stationing is 10 + 252.3 --> write this as 10252.3)
Given a simple circular curve, in feet, with the following known information: o | = 25°10'45" Da = 8°15'00" o STA P.I. = 30+89.45 What is the stationing of the C.T.? O 32+39.55 32+39.68 36+97.40 O 33+94.65
A A Intersection Problem: A and B are control (known) points. Station N A 500.00 1000.00 B 900.00 600.00 Measured angles at A and B are: Angle at A = 60 30 00 Angle at B = 54 45 00 What is the Easting of point P Use 2 decimals precision.
Given a compound curve 1₁ = 24.20°, 12 = 35.16°, D₁ = 6.08° ; D₂ = 4.43°. Compute the stationing of P.T. if P.I. is at station 6 + 936.25 WRITE YOUR ANSWERS in meters. (For example, the stationing is 10 + 252.3 --> write this as 10252.3)
A500-ft curve, grades of g +2.50% and ₂-2 50% VPI at station 96 00 and elevation 839 26 ft, stakeout at full stations A Part A List station elevations for an equal-tangent parabolic curve for the data given. Give the elevations in order of increasing X Express your answers in feet to five significant figures separated by commas. Elev ΚΕΙ ΑΣΦ | Η vec 5 Ĉ 2 ft
Item 3 A -2.50% grade meets a +1.75% grade at station 44+25 and elevation 3386.92 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. Elev = Submit Part B LE ΑΣΦ Request Answer ↓↑ O 46+95.58 Determine the station at the low point of the curve. 44+60.29 O 39+89.71 O 37+54.42 vec Submit Request Answer ? ft 3 of 5 Review
%3D A horizontal curve has a radius of 980.80 ft.; an I = 23°00'00"; and, PC Station %3D 76+68.50 ft. Determine: • Deflection angle from the PC to station 79+00.00 ft. = Provide the answer in degrees, minutes, and seconds. Provide whole numbers and do not include units. • Chord length from the PC to station 79+00.00 ft.= Provide your answer to two decimal places and do not include units. • Straight line distance from the PC to the PI = Provide %3D your answer to two decimal places and do not include units.
Given the following level section notes for stations 1+120 and 1+160: 1+120 | 6.6 | 0 | 7.6 | 0.8 | 3.7 | 1.3 1+160 | 74 | 0 | 98 | 1.2 | 2.3 | 2.4 The side slope is 2:1. The radius of curvature is 275 m. The center of curvature is on the right. a) What is the roadbed width? b) What is the area of section 1+120? c) What is the area of section 1+160? d) What is the volume between section 1+120 and 1+160 using end area method? e) What is the prismoidal volume correction? g) What is the centroid of excess of section 1+160? h) What is the correction due to horizontal curvature?

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