
Concept explainers
The central angles and the corresponding chord lengthsfor a compound curve which connects two tangents.

Answer to Problem 17P
For the first curve central angle and chord length is shown in table below,
Station | Deflection angle | Chord length |
671+35.99 | 0 | 0 |
671 | 1.7185 | 35.99 |
672 | 6.4931 | 99.89 |
673 | 11.268 | 99.89 |
674 | 16.042 | 99.89 |
675 | 20.817 | 99.89 |
675+35.25 | 22.50 | 35.25 |
For the second curve central angle and chord length is shown in table below,
Station | Deflection angle | Chord length |
675+35.25 | 0 | 0 |
676 | 4.122 | 64.69 |
677 | 10.488 | 99.79 |
677+70.87 | 15 | 70.80 |
Explanation of Solution
Given information:
Intersection angle Δ=75∘
Angle of first deflection curve Δ1=45∘
Radius of first curve R1=600ft
Radius of second curve R2=450ft
PCC station (675+35.25)
Calculation:
Deflection for second curve is given by,
Δ=Δ1+Δ2Δ1=45o=Angle of first deflection curveΔ=75o=Intersection angle75o=45o+Δ2Δ2=30o
Tangent length of first curve is given by,
t1=R1tan(Δ12)R1=600ft=Radius of first curveΔ1=45o=Angle of first deflection curvet1=600tan(45o2)t1=248.53ft
Similarly tangent length for second curve is given by,
t2=R2tan(Δ22)R2=450ft=Radius of second curveΔ2=30o=Angle of second deflection curvet2=450tan(30o2)t2=120.58ft
Length of first curve is given by,
L1=R1Δ1π180L1=600×45o×π180L1=471.24ft
Length of second curve is given by,
L2=R2Δ2π180L2=450×30o×π180L2=235.62ft
Station of first curve at PC is given by,
Station of PC=Station of PCC-L1Station of PC=(675+35.25)-(471.24)
The station of PCis the point of curve according to the standards of AASHTOit is calculated by dividing the station when it reaches above 100ft intervals into single decimal.
Station of PC=(675+35.25)-(4+71.24)Station of PC=((675−4)+(35.25−71.24))Station of PC=(671+35.99)
Station at point of tangency is given by,
Station of PT=Station of PCC+L2Station of PT=(675+35.25)+235.62Station of PT=(675+35.25)+(2+35.62)Station of PT=(675+2)+(35.25+35.62)Station of PT=(677+70.87)
For the first curve calculate degree of first curve is given by,
R1=5729.6D1D1=degree of curve600=5729.6D1D1=9.549o
Deflection angle for first full station is given by,
δ1Δ1=δ1L1δ145∘=(671−(670+64.01))471.24δ1=3.437∘
Deflection angle is given by,
Deflection angle=δ12Deflection angle=3.437∘2Deflection angle=1.7185∘
For the first curve chord length is given by,
C1=2R1sinδ12C1=2×600sin3.437o2C1=35.987ft
Deflection angle for last full station is given by,
δ2Δ1=δ2L1δ245∘=35.25471.24δ2=3.366∘
For the first curve central angle and chord length is shown in table below,
Station | Deflection angle | Chord length |
671+35.99 | 0 | 0 |
671 | 1.7185 | 35.99 |
672 | 6.4931 | 99.89 |
673 | 11.268 | 99.89 |
674 | 16.042 | 99.89 |
675 | 20.817 | 99.89 |
675+35.25 | 22.50 | 35.25 |
For the second curve calculate degree of second curve is given by,
R2=5729.6D2450=5729.6D1D2=12.732∘
Deflection angle for first full station is given by,
δ2Δ2=l1L2δ130∘=(676−(675+35.25))235.62δ1=8.244∘
Deflection angle is given by,
Deflection angle=δ12Deflection angle=8.244∘2Deflection angle=4.122∘
For the second curve chord length is given by,
C2=2R2sinδ12C2=2×450sin8.244o2C2=64.69ft
Deflection angle for last full station is given by,
δ2Δ2=l2L2δ230∘=70.87235.62δ2=0.300×30∘δ2=9.023∘
For the second curve central angle and chord length is shown in table below,
Station | Deflection angle | Chord length |
675+35.25 | 0 | 0 |
676 | 4.122 | 64.69 |
677 | 10.488 | 99.79 |
677+70.87 | 15 | 70.80 |
Want to see more full solutions like this?
Chapter 15 Solutions
Traffic And Highway Engineering
- Note:arrow_forwardNote:arrow_forward3. Find the reinforcements for the mid span and supports for an interior 8 in. thick slab (S-2) in the floor from Problem 1. Ignore the beams and assume that the slab is supported by columns only. Sketch the slab and show the reinforcements including the shrinkage and temperature reinforcement steel. Use fc’ = 4,000 psi and fy = 60,000 psi.arrow_forward
- Problem 4 (Apx Method) Determine (approximately) the force in each member of the truss. Assume the diagonals can support both tensile and compressive forces. 3 m 50 kN F 000 40 kN 000 000 000 000 000 000 E 000 000 000 000 000 B 3 m 20 kN D 000 000 000 000 C 3 m Problem 5 (Apx Method) Determine (approximately) the force in each member of the truss in problem 4. Assume the diagonals cannot support compressive forces.arrow_forwardThe single degree of freedom (SDOF) system the acceleration at the base (excitation) and the acceleration at the roof (response) of the SDOF system was recorded with sampling rate 50 Hz (50 samples per second, or dt= 0.02 seconds). The file ElCentro.txt includes the two columns of acceleration data. The first column lists the acceleration at the base of the SDOF system. The second column lists the acceleration at the roof of the SDOF system. (a) Plot the time histories of the recorded accelerations at the base and at the roof of the SDOF system. (b) Compute the acceleration, velocity and displacement time histories of the roof of the SDOF system subjected to the recorded base acceleration using the Central Difference method. Plot the accel- eration, velocity and displacement time histories. Plot the restoring force, the damping force, and the inertia force time histories. Compare the recorded acceleration time history at the roof of the SDOF with the acceleration that you computed…arrow_forwardProblem 2 (Using force method) Determine the force in each member of the truss. E = 29000 ksi 3 k 1.5 in² 4 ft 1.5 in² 1.5 in² 2 in² 6 k D 1.5 in² 3 ft 2 in² Barrow_forward
- The single degree of freedom (SDOF) system that you studied under free vibration in Assignment #3 - Laboratory Component has been subjected to a strong ground motion. The acceleration at the base (excitation) and the acceleration at the roof (response) of the SDOF system was recorded with sampling rate 50 Hz (50 samples per second, or dt= 0.02 seconds). The file ElCentro.txt includes the two columns of acceleration data. The first column lists the acceleration at the base of the SDOF system. The second column lists the acceleration at the roof of the SDOF system. (a) Plot the time histories of the recorded accelerations at the base and at the roof of the SDOF system. (b) Compute the acceleration, velocity and displacement time histories of the roof of the SDOF system subjected to the recorded base acceleration using the Central Difference method. Plot the accel- eration, velocity and displacement time histories. Plot the restoring force, the damping force, and the inertia force time…arrow_forwardPlease explain step by step and show formulaarrow_forwardPlease explain step by step and show formulaarrow_forward
- For an reinforced concrete two-way slab shown in figure under the load (P). (the slab continuous over all edges - all sides are fixed), Determine (By using yield line theory): A- Draw the Yield line Pattern B- Determine the moment m 3BAT C- Find The required flexural steel to resist the loads causing the slab to collapse if P = 200 KN, fc = 28 MPa, fy = 420 MPa d = 120 mm. Use 10 mm bars. (Pmin = 0.002) 6m 8m >2m->) 3marrow_forward3BAT For an reinforced concrete two-way slab shown in figure under the load (P). (the slab continuous over all edges - all sides are fixed), Determine (By using yield line theory): A- Draw the Yield line Pattern B- Determine the moment m KN, fc Please don't solve in Al anco if P = 200 6m 8m 2m-)) 3marrow_forwardPlease explain step by step and show formulaarrow_forward
- Traffic and Highway EngineeringCivil EngineeringISBN:9781305156241Author:Garber, Nicholas J.Publisher:Cengage Learning
