Highway Engineering Test 2

Concept:

The space mean speed, V_S = \(\frac{{({f_1} + {f_2} + \ldots + {f_n})\;}}{{\left( {\frac{{{f_1}}}{{{V_1}}} + \frac{{{f_2}}}{{{V_2}}} + \ldots + \frac{{{f_n}}}{{{V_n}}}} \right)}}\)

The time mean speed, V_T = \(\frac{{({f_1}{V_1} + {f_2}{V_2} + \ldots + {f_n}{V_n})}}{{({f_1} + {f_2} + \ldots + {f_n})}}\)

Calculation:

The space mean speed, V_S = \(\frac{{2 + 5 + 7 + 9 + 20 + 7\;}}{{\left( {\frac{2}{3} + \frac{5}{8} + \frac{7}{{13}} + \frac{9}{{18}} + \frac{{20}}{{20}} + \frac{7}{{22}}} \right)}}\)

\({V_S} = \frac{{50}}{{3.65}}\)

\({V_S} = 13.7\;m/s\)

The time mean speed, V_T = \(\frac{{\left( {2 \times 3} \right) + \left( {5 \times 8} \right) + \left( {7 \times 13} \right) + \left( {9 \times 18} \right) + \left( {20 \times 20} \right) + \left( {7 \times 22} \right)}}{{2 + 5 + 7 + 9 + 20 + 7}}\)

\({V_S} = \frac{{853}}{{50}}\)

Concept:

Types of Mix

1) Well-graded mix: Dense mix, bituminous concrete has a good proportion of all constituents and are called dense bituminous macadam, offers good compressive strength and some tensile strength

2) Gap-graded mix: Some large coarse aggregates are missing and have good fatigue and tensile strength.

3) Open-graded mix: Fine aggregate and filler are missing, it is porous and offers good friction, low strength and high speed.

4) Unbounded: Binder is absent and behaves under loads as if its components were not linked together, though good interlocking exists. Very low tensile strength and needs kerb protection.

Hence the correct answer is A-3, B-1, C-2, D-4.

\(5\min PHF = \frac{{hourly\;traffic\;volume}}{{\left( {\frac{{60}}{5}} \right) \times {V_{5\left( {max} \right)}}}}\)

\(= \frac{{1200}}{{\left( {\frac{{60}}{5}} \right) \times 200}}\)

Concept:

Angularity Number:

• This represents the degree of packing.
• Angularity no = 67 - % solid volume
• 67 represents the volume of solids (in %) of most rounded gravels in a well-compacted state, which would have 33% voids. Thus the angularity no measures the voids over 33%.
• Higher the angularity number, more angular is the aggregate.
• The range of angularity number for aggregates used for construction is 0 to 11.

Angularity No. \( = 67 - \frac{{100W}}{{C \times {G_a}}}\)        [The value is expressed as nearest whole number]

Where, W = Weight of aggregate in a cylinder; C = Weight of water in the same cylinder

G_a = Specific Gravity of aggregate.

Calculation:

G_a = 2.7, Volume of cylinder = 1 l

C = weight of 1 litre water = 1 kg

Traffic volume (q) = Traffic Density (k) × Traffic speed (v)                                  

⇒ q = kV.

⇒ q = k × 47.5 (1 – 0.32 k)

⇒ q = 47.5 (k – 0.32 k²)

For the maximum traffic volume

\(\begin{array}{l} \frac{{dq}}{{dk}} = 0\\ \frac{{dq}}{{dk}} = 47.5\left( {1 - 2 \times 0.32\;k} \right) = 0 \end{array}\)

⇒ 1 – 0.64 k = 0

\(\Rightarrow k = \frac{1}{{0.64}}\)

⇒ k = 1.5625 vehicle/km.

Maximum traffic volume

q_max = 47.5 × 1.5625 (1 – 0.32 × 1.5625)

= 37.11 vehicle/hr.

≅ 38 vehicle/hr.

Concept:

The optimum cycle length (C_o) is given by:

\({{\rm{C}}_{\rm{o}}} = \frac{{1.5{\rm{L}} + 5}}{{1 - {\rm{Y}}}}\)

And effective green time for each phase is given by:

\({{\rm{G}}_{\rm{a}}} = \frac{{{{\rm{y}}_{\rm{a}}}}}{{\rm{Y}}} \times \left( {{{\rm{C}}_{\rm{o}}} - {\rm{L}}} \right)\)

Calculation:

Given:

Normal flow on roads A & B: q_a = 400 PCU/hr and q_b = 250 PCU/hr, Saturation flow: S_a = 1250 and S_b­ = 1000 PCU/hr, All – red time (R) = 12 sec, and number of phases (n) = 2

\(\therefore {{\rm{y}}_{\rm{a}}} = \frac{{{{\rm{q}}_{\rm{a}}}}}{{{{\rm{S}}_{\rm{a}}}}} = \frac{{400}}{{1250}} = 0.32{\rm{\;and\;}}{{\rm{y}}_{\rm{b}}} = \frac{{{{\rm{q}}_{\rm{b}}}}}{{{{\rm{S}}_{\rm{b}}}}} = \frac{{250}}{{100}} = 0.25\)

Y = y_a + y_b = 0.32 + 0.25 = 0.57

L = 2n + R = 2 × 2 + 12 = 16 sec

Optimum cycle time,

\({{\rm{C}}_{\rm{o}}} = \frac{{1.5{\rm{L}} + 5}}{{1 - {\rm{Y}}}} = \frac{{1.5 \times 16 + 5}}{{1 - 0.57}} + \frac{{29}}{{0.43}} = 67.4{\rm{\;s}} \approx 67.5{\rm{\;sec}}\)

\({{\rm{G}}_{\rm{a}}} = \frac{{{{\rm{y}}_{\rm{a}}}}}{{\rm{Y}}} \times \left( {{{\rm{C}}_{\rm{O}}} - {\rm{L}}} \right) = \frac{{0.32}}{{0.57}}\left( {67.5 - 16} \right) = 29{\rm{\;sec}}\)

\({{\rm{G}}_{\rm{b}}} = \frac{{{{\rm{y}}_{\rm{b}}}}}{{\rm{Y}}} \times \left( {{{\rm{C}}_{\rm{O}}} - {\rm{L}}} \right) = \frac{{0.25}}{{0.57}}\left( {67.5 - 16} \right) = 22.5{\rm{\;sec}}\)

∴ Effective green time for each phase A and phase B respectively are 29 seconds and 22.5 seconds respectively.

For the observer moving with the stream

q - kv_w = 60

For the observer moving against the stream

q + kv_a = 100

where, q = traffic flow

k = traffic density

v_w = velocity of the test vehicle with the stream

v_a = velocity of the test vehicle against the stream

∴ Solving above two equation,

q - 18k = 60           -----(1)

q + 24k = 180           -----(2)

on solving equations (i) and (ii), we get

k = 2.89 vehicle/km