SITUATION 3. The perpendicular distance between two parallel tangents of a reversed curve is 8m, and the chord distance from PC to the PT is equal to 30m. Compute the central angle of the reverse curve. 30°56 15°28 30°39 28012

SITUATION 3. The perpendicular distance between two parallel tangents of a reversed curve is 8m, and the chord distance from PC to the PT is equal to 30m. Compute the central angle of the reverse curve. 30°56 15°28 30°39 28012

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: SITUATION 5: A compound curve has the following data: Li = 28° DI = 3° D2 = 4° L2 = 31° Sta. P.I at…

Q: Point E lies on the tangent through PC at a distance of 380m from the PC. The perpendicular distance…

A: Surveying a branch of the construction industry where balancing between different points on the…

Q: Station PT of a compound curve is at 15+480, 11=30°, 12=36°, D1=4°, and D2=5°. is the stationing…

Q: A simple horizontal curve has 35 degrees of curvature and subtends an angle of 110 degrees. If the…

A: Degrees of curvature, D = 35°PC is at station = 3 + 65.00 ft

Q: From a known point V (vertex), the tangent lines of a compound curve are drawn - having azimuths…

A: *Multiple questions* As per our guidelines we are supposed to answer?️ only one question. Kindly…

Q: Two roads AB and BC intersect at a chainage 184+65. These are intersected by a line PQ and the…

A: GIVEN :- Radius of the curve R1 = 330 m R2 = 540 m To Determine :- chainages of the first and the…

Q: SITUATION X: A simple curve has a degree of 6 and the tangent intersects at 42 degrees. 17. What is…

Q: he deflection angles from PC to two intermediate points A and B on a simple curve are 4.5° and 5.6°,…

A: Given, Deflection angle δ A is 4.5° and B is 5.6° Length of chord C is 35.49 m Bearing is S45.5°E…

Q: Find the length of the tangents

Q: Simple Curves: A highway curve was designed to connect intersecting tangents at Pl. Back tangent = N…

Q: A simple curve have tangents AB and BC intersecting at a common point B. AB has an azimuth of 180°…

Q: SITUATION 17: AB and BC are tangents of a simple curve. AB is due north and BC is N50°E. Degree of…

A: From the Above ∆of the curve = 50° Degree of the curve = 4 ° For 20 m chain Degree D = 1146/R R=…

Q: SITUATION 13: A simple curve having a degree of curve of 4° has an angle of intersection of 50°. A…

Q: A simple curve has back tangent AB and forward tangent BC intersecting at B. AB has an azimuth of…

Q: A reverse curve connects two converging tangents that intersect at an angle of 30°. The distance of…

A: Reverse Curve: The road surface planning mandates that the road connecting the two locations take…

Q: Determine the common tangent of a reversed curve given the following properties:

A: Given: I1=27.09deg, I2=32.92deg, D1=5.29deg, D2=4.85deg Find: Common tangent for the reversed curve

Q: The bearing of a back tangent of a simple curve is N 65°00' E, while the forward tangent has a…

Q: the stationing of the PT.

Q: Compute the stationing of PCC of the compound curve if the following data are given: The stationing…

Q: a. Elements of the curve b. Stationing of P.C. and P.T. c. Lengths of sub-chords and its…

Q: Problem 2. The common tangent BC of a reverse curve is 280.5 m and has a bearing of S 47°31′ E. AB…

Q: 3 Four tangents having an angle of intersections shown is to be connected by three simple curves…

Q: The tangent thru the PC has a direction due to North; and the tangent through the PT has a bearing…

Q: Choose the letter of the answer. Point A and B on a simple curve have perpendicular distances from…

A: since no clue is given about angle and radius of the curve so i have used the basic geometry…

Q: simple curve has back and forward tangent bearings of N20°E and N80°E respectively. The radius of…

Q: Problem 1 A simple curve connects two tangents AB and BC with bearings N 75°12' E and $ 78°36' E,…

A: "Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for…

Q: The common tangent CD of a reversed curve is 280.5m and has an azimuth from the south (s)of 312°…

Q: The forward and the backward tangents of a simple curve have an azimuth of 249°15’ and 190°45’…

A: The curve is of two type like horizontal curve and vertical curve, the type of horizontal curve is…

Q: The perpendicular distance between tangents of a reversed curve is 10 m. The chord length from the…

Q: Situation 1: A simple curve have tangents AB and BC intersecting at a common point B. AB has an…

Q: An unsymmetrical parabolic curve has a backward tangent of +5% and a forward tangent of-3%. The…

A: Answer We are given Ascending grade G1 =5% Descending grade G2 = 3% Length of curve on left side =…

Q: Given the following elements of the compound curve shown: h = 37° 30' I2 = 53° 50' R1 = 250 m. %3D…

A: T1=215m

Q: 6. A compound curve has a central angle of the first curve equal to 40° and a radius of 280m. while…

A: R1 = 280m D2 = 5NOTE : As per guidelines, first question is solved. Request you to post remaining…

Q: The common tangent of a compound curve has a length of 70 m. This common tangent makes an angle of…

Q: the stationing of the PT.

Q: Two Tangents intersects at Sta. 0+550. A compound curve laid on their tangents has the following…

Q: Situation 3: The offset distance of the simple curve from the PT is equal to 120.20m. The simple…

A: Angle of intersection (∆)=50o

Q: The long chord of a compound curve is 425 m long and the angles it makes with tangents of the curve…

Q: Two roads AB and BC intersect at a chainage 184+65. These are intersected by a line PQ and the…

Q: Probiem A simple curve connects two tangents AB and BC with bearings N 75012' E and S 78036' E,…

A: "Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for…

Q: Two tangents intersect at chainage 4100 m. Ø 35 35 35". Carry out all the data necessary for setting…

A: A deflection angle is an angle between the onward extension of the previous leg and the line ahead.…

Q: SITUATION 3: A compound curve has the following data: L1 = 28° D1 = 3° L2 = 31° D2 = 4° Sta. P.l at…

Concept explainers

Errors in surveying

An error in surveying is defined as the difference between two or more values of an identical quantity. However, the estimates are never accurate and there will always be a degree of variability regardless of the research tool or method used. These differences are known as errors that need to be reduced or eliminated to maintain certain levels of accuracy. Even if you carefully follow the prescribed test procedures, the considerations may contain errors.

Question

SITUATION 3. The perpendicular distance between two parallel tangents of a reversed curve
is 8m, and the chord distance from PC to the PT is equal to 30m. Compute the central angle of
the reverse curve.
30°56
15°28
30°39
28012

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

At a 25° angle, two tangents AB and BC intersect. A point P point B, with a perpendicular distance of 2.62 meters. (a) Determine the radius of the simple curve that connects the two tangents and passes through the passing point P in m. (b) Determine the chord length between P.C. and point P in m. (c) Determine the area bounded by the curve and tangent lines in m?. 23.97 meters away from
QUESTION 7 Determine the common tangent of a reversed curve given the following properties: = 15.26° D₁ = 4.98° ₁ = 12 = 22.55° D2 = 5.37° NOTE: Solution is not required.
COMPOUND CURVE Station PT of a compound curve is at 15 + 480, I, = 30°, I2 = 36°, D1 = 4°, and D2 = 5°. What is the stationing of PCC? Use arc basis. a. 15 + 319.63 b. 15 + 336.14 C. 15 + 325.28 d. 15 + 342.50
A. Two parallel tangents of a reversed curve are 12m apart. The chord length from the PC to the PT is equal to 150m. If the reversed curve has equal radii. 1) Compute the length of the common radius. 2) Calculate the distance from the V₁ to V2. 3) Calculate the station at PT if station PC= 20+348.
Two roads AB and BC intersect at a chainage 184+65. These are intersected by a line PQ and the angles APQ and PQC are 112° and 102° respectively. The radius of the first arc of the curve is 330m and that of the second arc is 540m. Calculate the chainages of the first and the second tangent points P.C R₂ 2 P.T 185+87.15, 172+18.67 O 170+88.19, 182+14.97 O 176+48.69, 172+18.67 O 185+87.19, 152+14.97 O 190+84.32, 177+12.34 O
SITUATION 4: The distance between two intermediate points A and B on a simple curve is 60 m. Their deflection angles are respectively 4° and 10° measured from tangent through PC. 10. Compute the radius of the simple curve. a. 264.90 m c. 246.16 m d. 224.92 m b. 229.47 m 11. Compute the arc length from PC to A. c. 33.74 m а. 34.37 m b. 37.43 m d. 37.34 m 12. Compute the chord length from PC to B. a. 84.59 m b. 89.54 m c. 85.49 m d. 89.45 m
SITUATION 12: AB and BC are tangents of a simple curve. AB is due north and BC is N50°E. Degree of simple curve is 4° and PC is at station 20 + 130. 33. Compute the length of the long chord. a. 242.14 m c. 386.12 m d. 284.51 m b. 184.52 m 34. Compute the distance from point D on the simple curve to point of intersection of tangents that makes an angle of 54° with the tangent passing through the PC. а. 30.2 m b. 26.8 m С. 23.4 m d. 31.5 m 35. Compute the stationing of point D. a. 20 + 639.3 b. 20 + 739.3 c. 20 + 439.3 d. 20 + 249.3
TOPIC: Surveying (Horizontal Curves)
Answer letter c.
Simple Curve A railway curve is to connect intersecting tangents with the following design data: Back tangent = N 20° 25' E Forward tangent = S 49°35' E Degree of curve = 5⁰ Design speed = 70 Kph Station PI = 5+555.55 1.Determine the length of line from PC to Pl. a) 234.67m b) 244.67m 2.Compute the straight distance from PC to PT. a) 402.34m b) 375.60m c) 245.89m d) 327.42m c) 389.78m d) 423.67m 3. If the width of the land intended for the railway is 10m, determine the area of land from PC to PT. a) 4,456.56 sqm b) 3,456,43 sqm c) 5,112.34 sqm d) 4,399.92 sqm 4.How long would a 300m long train able to leave the curve completely. a) 25.67 sec b) 38.06m sec c) 40.34 sec d) 29.67 sec
Give the correct answer please
SITUATION 3. The perpendicular distance between two parallel tangents of a reversed curve is 8m, and the chord distance from PC to the PT is equal to 30m. Compute the central angle of the reverse curve. 30°56 15°28 30°39 28°12