(a) A two-lane State Highway in plain terrain, with cutting section having camber of 2·0%, negotiates a curve of radius 480 m…

(a) Superelevation for a 480 m curve on a two-lane State Highway

Design speed. For a State Highway in plain terrain IRC:73 gives a ruling design speed of 100 km/h (minimum 80 km/h). Use V = 100 km/h.

Superelevation. IRC provides superelevation to balance the centrifugal force of 75 % of the design speed, neglecting friction:

• e = (0.75V)²/(127R) = V²/(225R) = 100²/(225 × 480) = 0.093
• This exceeds the IRC maximum of 7 % for plain terrain, so e = 0.07 is adopted.

Check of side friction at the full design speed.

• f = V²/(127R) − e = 100²/(127 × 480) − 0.07 = 0.164 − 0.07 = 0.094
• This is below the permissible 0.15, so e = 7 % is adequate and the speed need not be restricted. Provide superelevation of 7 %.

Length of transition (IRC:73, larger of three criteria). Pavement width W = 7.0 m. Extra widening at R = 480 m, W_e = nl²/2R + V/(9.5√R) = 2 × 6²/960 + 100/(9.5 × 21.9) ≈ 0.56 m, so W + W_e ≈ 7.56 m.

• Rate of change of centrifugal acceleration: C = 80/(75 + V) = 0.457 m/s³, L_s = 0.0215V³/(CR) = 0.0215 × 100³/(0.457 × 480) = 98 m
• Rate of introducing superelevation, 1 in 150 for plain terrain, rotating about the inner edge: L_s = e N (W + W_e) = 0.07 × 150 × 7.56 = 79 m
• Empirical formula for plain terrain: L_s = 2.7V²/R = 2.7 × 100²/480 = 56 m
• Provide a transition of 100 m, over which superelevation is introduced.

Attaining the superelevation in a cutting. In a cutting the inner edge is not lowered, because that would put the inner side drain and the pavement edge below the formation and spoil drainage. The pavement is therefore rotated about the inner edge:

1. Eliminating the crown. At the start of the transition the section has 2 % camber. The outer half is raised about the crown until it is level, and then until it slopes towards the inside at 2 %, so the whole carriageway has a uniform 2 % cross-fall towards the inner edge.
2. Rotating to full superelevation. The whole pavement is then rotated about the inner edge, raising the centre line and the outer edge, until the cross-slope reaches 7 % at the start of the circular curve.
3. The outer edge rises at no more than 1 in 150 of the transition length, and the profiles of the outer edge, centre line and inner edge are run out along the transition.

(b)(i) Level book for the five readings

On rising ground the readings fall along the line until the staff can no longer be read, so the instrument is shifted after 1.335 (a foresight on a change point) and 2.980 is the backsight from the new position.

• Point 1: BS 3.815, HI = 99.000 + 3.815 = 102.815, RL 99.000
• Point 2: IS 2.500, RL = 102.815 − 2.500 = 100.315
• Point 3 (change point): FS 1.335, RL = 102.815 − 1.335 = 101.480, then BS 2.980, new HI = 101.480 + 2.980 = 104.460
• Point 4: FS 0.800, RL = 104.460 − 0.800 = 103.660

Checks.

• ΣBS − ΣFS = (3.815 + 2.980) − (1.335 + 0.800) = 6.795 − 2.135 = 4.660 m
• Last RL − first RL = 103.660 − 99.000 = 4.660 m
• Rise and fall: rises of 1.315, 1.165 and 2.180 m, total 4.660 m, with no falls All three agree, so the arithmetic is correct. RL of the last point = 103.660 m.

(b)(ii) Rectangular side drains

• Discharge per drain Q = 1.8/2 = 0.90 m³/s
• With V limited to 0.8 m/s, A = Q/V = 0.90/0.8 = 1.125 m²
• For the most economical rectangular section b = 2y and R = y/2: 2y² = 1.125, so y = 0.75 m, b = 1.50 m, R = 0.375 m
• Manning equation V = (1/n) R^(2/3) S^(1/2), so 0.8 = (1/0.03) × 0.375^(2/3) × S^(1/2) = 17.33 S^(1/2), giving S = 0.00213, about 1 in 470

Provide rectangular drains 1.50 m wide with 0.75 m flow depth on each side, laid at about 1 in 470, with a freeboard of about 0.15 m (total depth 0.90 m).

(c) Updating the road project after 13 weeks

Original network. Paths: 1–2–3–4 = 12 + 18 + 20 = 50 weeks (critical), and 1–3–4 = 9 + 20 = 29 weeks. Normal cost = 90 + 180 + 210 + 360 = ₹ 840 lakh.

Updated network at week 13.

• 1–2 and 1–3 are complete (1–2 finished at week 12, 1–3 at week 9).
• 2–3 started at week 12 and needs 19 more weeks from week 13, so it ends at week 32 (its duration has grown from 18 to 20 weeks).
• 3–4 can start only at week 32 and takes 20 weeks, ending at week 52.
• The new network from week 13 is 2–3 (remaining 19 weeks) followed by 3–4 (20 weeks), and this is the critical path.

Completion by 15 August? 15 August is the 227th day of the year, about 32.4 weeks after 1 January. The project now needs 52 weeks, ending around the end of December, so it will not be completed by 15 August. Even the original 50-week plan could not have met that date.

Crashing to 49 weeks.

• Required reduction = 52 − 49 = 3 weeks, and only 3–4 remains to be crashed.
• 3–4 can be shortened from 20 to 17 weeks, exactly the 3 weeks needed.
• Cost slope = (450 − 360)/(20 − 17) = ₹ 30 lakh per week.
• Crash 3–4 by 3 weeks. Increase in project cost = 3 × 30 = ₹ 90 lakh, raising the total from ₹ 840 lakh to ₹ 930 lakh (assuming the delayed activity 2–3 costs no more than planned).