(a) (i) Explain any two traffic surveys carried out to decide the geometric design features of a road. (5 marks) (ii) The…

(a)(i) Two traffic surveys for geometric design

Traffic volume survey: vehicles are counted by type and direction over hours, days and seasons, including turning movements at junctions. The peak-hour and design volumes fix the number of lanes, lane width, junction layout and pavement capacity.

Speed survey (spot speed study): spot speeds are measured at a section and the 85th percentile speed is found from the cumulative distribution. It fixes the design speed, curve radii, sight distances, superelevation and speed limits.

(a)(ii) Maximum flow from v = 100 − 1.2k

Flow q = vk = 100k − 1.2k². For maximum flow, dq/dk = 100 − 2.4k = 0, so k = 41.67 veh/km, and d²q/dk² = −2.4 is negative, confirming a maximum.

Critical speed v = 100 − 1.2 × 41.67 = 50 km/h. Maximum flow q_max = 50 × 41.67 = 2083 veh/h. The critical density is 41.67 veh/km.

Other key values: free-flow speed v_f = 100 km/h at k = 0, and jam density k_j = 100/1.2 = 83.3 veh/km at v = 0. The critical values are half of these, v_c = v_f/2 and k_c = k_j/2.

Diagrams (critical point marked on each): the speed–density graph is a straight line from (0, 100) to (83.3, 0). The flow–density graph is a parabola with zeros at k = 0 and 83.3 and its peak at (41.67, 2083). The speed–flow graph is a parabola with zeros at v = 0 and v = 100 and its peak at (50, 2083).

(b) Levels inside the tunnel

Readings taken on the ceiling with an inverted staff are treated as negative. Elevations are in metres.

Setup 1: HI = 100.500 + 1.234 = 101.734. Point B (ceiling, FS −1.782): RL = 101.734 − (−1.782) = 103.516.

Setup 2: BS on B = −1.345, so HI = 103.516 − 1.345 = 102.171. Point C (FS −1.830): RL = 102.171 + 1.830 = 104.001.

Setup 3: BS on C = −1.730, HI = 104.001 − 1.730 = 102.271. Point D (FS −2.000): RL = 104.271.

Setup 4: BS on D = −2.056, HI = 104.271 − 2.056 = 102.215. Point E (FS −2.340): RL = 104.555.

Setup 5: BS on E = −2.156, HI = 104.555 − 2.156 = 102.399. Point F (FS −2.780): RL = 105.179.

Check: ΣBS − ΣFS = (1.234 − 1.345 − 1.730 − 2.056 − 2.156) − (−1.782 − 1.830 − 2.000 − 2.340 − 2.780) = −6.053 + 10.732 = 4.679 m. Last RL − first RL = 105.179 − 100.500 = 4.679 m. The check is satisfied.

(c) Bulking of sand

When dry sand takes up a small amount of moisture, a thin water film forms around each grain. Surface tension in the film holds the grains apart and traps air, so the same mass of sand occupies a larger bulk volume. The bulking rises with moisture to a peak at about 4–6% moisture (up to 25–40% for fine sand, less for coarse sand) and falls back to zero when the sand is fully saturated, because the films break down and the voids fill with water.

The sketches show three containers: dry sand (grains touching), damp sand (higher level with films) and saturated sand (back to the dry level), and a curve of bulking against moisture content.

Effect on concrete: when sand is batched by volume, bulked sand contains less actual sand than intended, so the mix is deficient in fines and is harsh, more porous and lower in strength, and it gives a lower yield of concrete. The moisture on the sand also adds to the mixing water, raising the water–cement ratio if it is ignored. The remedy is to batch sand by weight, or to increase the volume of sand by the bulking percentage, and to reduce the added water.