Question

In: Civil Engineering

1. Compute L, T, E, LC, Δ (delta) and stations of the PC and PT for...

1. Compute L, T, E, LC, Δ (delta) and stations of the PC and PT for the circular curve with the given data of: R= 2200’ and M = 19.5075 and the P.I. station = 28+37.62. Express answers to .01 (ft.).

2. Compute L, T, E, M and Δ (delta) and stations of the and PT for the circular curve with the given data of: R = 2850’ ft. and LC = 985.7122 and the P.C. station = 62+34.17. Express answers to .01 (ft.).

3. Compute L, M, E, LC, Δ (delta) and stations of the PC for the circular curve with the given data of: R= 1250 and T = 237.637 and the P.T. station = 15+14.13. Express answers to .01 (ft.).

4. Tabulate the deflection angles and total chords on full (+00) and half (+50) stations to lay out the circular curve in Problem 1. 5. Tabulate the deflection angles and total chords on full (+00) and half (+50) stations to lay out the circular curve in Problem 3

Solutions

Expert Solution

Given,

Radius of curve (R) = 2200 ft.

Mid ordinate distance (M) = 19.5075 ft

Chainage of point of intersection = 28 + 37.62

As chain length is not given, let us assume 100 ft chain.

Therefore chainage of P.I = 28 * 100 + 37.62 = 2837.62 ft

Mid ordinate distance is given by,

M = R * (1 - cos ( / 2))

Where,

= Deflection angle of the curve.

19.5075 = 2200 * (1 - cos ( / 2))

8.867 * 10^-3 = (1 - cos ( / 2))

cos ( / 2) = 0.99113

= 2 * cos^-1(0.99113)

= 15.274

Hence deflection angle of the curve is 15.274

Tangent length (T) is given by,

T = R * tan ( / 2)

T = 2200 * tan (15.274/ 2)

T = 294.99 ft

Therefore tangent length of the curve is 294.99 ft.

Length of long chord (L) is given by,

L = 2 * R * sin (/ 2)

L = 2 * 2200 * sin (15.274/ 2)

L = 584.74 ft

Therefore length of long chord is 584.74 ft.

External distance (E) is given by,

E = R * (sec (/ 2) -1)

E = 2200 * (sec (15.274/ 2) -1)

E = 19.69 ft

Therefore external distance of the curve is 19.69 ft

Length of curve (LC) is given by,

LC = ( * R * ) / 180

LC = ( * 2200 * 15.274) / 180

LC = 586.48 ft

Therefore length of the curve is 586.48 ft

Chainage of PC = Chainage of PI - Tangent length

Chainage of PC = 2837.62 - 294.99

Chainage of PC = 2542.63 ft = 2542.63/100 = 25.4263 chains = 25 + 0.4263 * 100 = 25 + 42.63

Therefore station of PC is 25 + 42.63

Chainage of PT = Chainage of PC + Length of curve

Chainage of PT = 2542.63 + 586.48

Chainage of PC = 3129.11 ft = 3129.11/100 = 31.2911 chains = 31 + 0.2911 * 100 = 31 + 29.11

Therefore station of PT is 31 + 29.11

Ally Wells answered 1 year ago

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