A circular curve of 500m radius is to be laid to connect two straights APB and CQB intersecting at point B. the intersection point of B is inaccessible due to an intervening river. A traverse APQC has been run and the chainage of points P and Q are 93+78 and 94+73 and angles APQ and PQC are 115°24' and 152°46' respectively. Find out the second .chainage of the tangent point River 115°24 152 46 P.C P.T 98+07.632 88+16.645

A circular curve of 500m radius is to be laid to connect two straights APB and CQB intersecting at point B. the intersection point of B is inaccessible due to an intervening river. A traverse APQC has been run and the chainage of points P and Q are 93+78 and 94+73 and angles APQ and PQC are 115°24' and 152°46' respectively. Find out the second .chainage of the tangent point River 115°24 152 46 P.C P.T 98+07.632 88+16.645

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

A reverse curve of equal radii connects 2 parallel tangents 21 m apart. The length of the chord from PC to PT is 140m. PC is at 3+100 1. Determine the Intersection angle a) 12.57 b) 15.72 c) 17.25 d) 71.52 2. Determine the radius of the curve a) 333.22 b) 222.33 c) 233.33 d) 322.22 3. Determine the length of tangent a) 30.54 b) 35.40 c) 50.34 d) 43.50 4. Determine stationing at PT a) 3+240.50 b) 5+240.50 c) 2+340.50 d) 3+540.30
. A grade line AB having a slope of +5% intersect another grade line BC having a slope of –3% atB. The elevations of pointsA,BandCare 95 m, 100 m and 97 m respectively. Determine the elevation of the summit of the 100 m parabolic vertical curve to connect the grade lines. Resolve using equations for a parabola. (y=Ax^2+Bx+C)
A Compound circular curve have: - R1= 290.75 m, A1= 40° 14', C2=175 m, A2 = 27° 40', Stat. P.C. = 50 + 10, P.C (500, 500), AZ.p.c.C.-P.C. = 275° 22' %3D 1- Stations of main points. 2- Coordinates of P.I., P.T., P.C.C.
1. Two tangents AB & BC are intersected by a line having angles 40 o & 35 o respectively. The radius of 1 st arc is 1000m and of 2 nd arc is 800m. Carry out all the necessary calculation to find the chainage of tangent points points and the point of compound curvature given that the chainage of intersection point is 5000 m. (CLO2; C3; PLO1)
A reverse curve connects three tangent lines with azimuths AB=252⁰16’ BC=314⁰10’and CD=216⁰48’, the degree of the first curve is 6⁰ and the stationing of PC is 2 +410.80. What is the stationing of the PT if line BC is 164. 71m. long.
A vertical parabolic has a forward tangent of 2% and backward tangent of -4% intersecting at STA 2 + 160. Locate the stationing of the lowest point of the curve if the length of curve is 240m. Kindly put the complete solutions. a) 2 + 185.14b) 2 + 200c)2 + 221.4d)2 + 194.19
7. In a railroad layout, the centerline of two parallel tracks is connected with a reversed curve of unequal radi. The central angle of the first and second curve is 16° and the distance between parailel tracks is 27.6 m. Stationing of the PC is 15+420 and the radius of the second curve is 290. a) Compute the length of the long chord from PC to PT. b) Compute the radius of the first curve. c) Compute the stationing of PT.
Provide complete solution and illustration. The bearings of backward and forward tangents are S 65°30’ E and N 75°40’ E, respectively. If the degree of curve is 8° for a full station of 20m and the stationing of the vertex is 5+005.25m, using arc basis determine the: a. Elements of the simple curveb. Station of P.C. and P.T.c. Offset from tangent distance to the 4th station and its distance along the tangent from P.C.d. Offset from the Long chord to the 5th station and its distance along the Long chord from P.C.
The bearings of forward and backward tangent are N 15°30’ E and N 25°40’ W, respectively. If the degree of curve is 2° for a full station of 20m and the stationing of the vertex is 12 + 582.50, determine the ff with illustrations and sketch : a. Elements of the curveb. Stationing of P.C. and P.T.c. Lengths of sub-chords and its corresponding sub-angles
1. The curve of a diversion road has a tangent “AB” which is directed due North and theother tangent “BC” is having a bearing of ?70°?. Point “A” is located at P.C. whosestationing is 12 + 600. The degree of curve using arc-basis is 5°.a. Compute the radius of the curve.b. Compute the long chord of the curve.c. What is the stationing at P.I.?d. Compute for the middle ordinate.e. Compute for the external distance.
Station PT of a compound curve is at 15+480, 1=30°, 12=36°, D1=4°, and D2=5°. is the stationing PCC? Use arc basis.
On a railroad line, two tangents that intersect at station 10+243 so as to form an angle of 36028' are to be connected by a compound curve consisting of two simple curves. The BC which is at station 10 + 163 is to be a 40 curve based on a 20 m chord and is to have a central angle of 170. Compute R2, the stationing of PCC, and the length of tangent from Pl to EC. 10 + 243 PI 36°28' V1. ti t2 V2 PCC EC 10 + 163 BC R2 R1