(a) The network for a construction project is shown in the figure given below. The three time estimates for each activity are…

(a) Expected activity time t_e = (t_o + 4t_m + t_p)/6.

(i) Expected times (days):

• 1–2: (4 + 28 + 16)/6 = 8; 1–3: (4 + 32 + 18)/6 = 9; 1–4: (6 + 36 + 18)/6 = 10
• 2–6: (3 + 24 + 15)/6 = 7; 3–5: (4 + 28 + 16)/6 = 8; 4–5: (5 + 36 + 19)/6 = 10
• 4–7: (6 + 40 + 20)/6 = 11; 5–6: (2 + 24 + 10)/6 = 6; 5–7: (8 + 44 + 20)/6 = 12; 6–7: (3 + 20 + 13)/6 = 6
• Dummy 2–3: 0

(ii) Earliest expected times (forward pass, take the maximum at merge events):

E₁ = 0; E₂ = 8; E₃ = max(9, 8 + 0) = 9; E₄ = 10; E₅ = max(9 + 8, 10 + 10) = 20; E₆ = max(8 + 7, 20 + 6) = 26; E₇ = max(10 + 11, 20 + 12, 26 + 6) = 32 days.

(iii) Latest allowable times (backward pass from L₇ = 32, take the minimum at burst events):

L₇ = 32; L₆ = 26; L₅ = min(32 − 12, 26 − 6) = 20; L₄ = min(20 − 10, 32 − 11) = 10; L₃ = 20 − 8 = 12; L₂ = min(26 − 7, 12 − 0) = 12; L₁ = min(12 − 8, 12 − 9, 10 − 10) = 0.

(iv) Event slack S = L − E: S₁ = 0, S₂ = 4, S₃ = 3, S₄ = 0, S₅ = 0, S₆ = 0, S₇ = 0 days.

(v) Critical path: the zero-slack events form two paths of equal length 32 days: 1–4–5–6–7 (10 + 10 + 6 + 6) and 1–4–5–7 (10 + 10 + 12).

Dotted arrow: it is a dummy activity, with zero time and no resources. It shows only a logical dependency: activity 3–5 cannot start until event 2 (completion of 1–2) as well as event 3 has occurred.

Significance of the critical path: it is the longest path through the network, so it fixes the project duration (32 days). Its activities have zero float; any delay on them delays the whole project, so they get the closest monitoring and first call on resources, and crashing to shorten the project must act on them.

(b) Functions of tie bars in rigid pavements:

• They hold adjacent slabs together at longitudinal joints and prevent the joint from opening.
• They prevent lateral separation and differential vertical movement (faulting) between lanes.
• By keeping the joint tight they maintain aggregate interlock and load transfer and keep out water and incompressibles.
• They are not load-transfer dowels; they are thin deformed bars designed to resist the friction force that pulls the slabs apart.

Design (IRC:58 approach): the steel across one metre of joint must resist the friction force from the slab between the joint and the free edge.

• Force per metre = f W h b = 1.35 × 2400 kg/m³ × 0.30 m × 3.75 m = 3645 kg.
• Using deformed bars (S_s = 2000 kg/cm²): A_s = 3645/2000 = 1.82 cm² per metre. With 12 mm bars (A = 1.131 cm²), spacing = 1.131/1.82 = 0.62 m; provide 12 mm bars at 60 cm c/c.
• Length: each half must develop the bar force by bond, so L = d S_s/(2 S_b) = 1.2 × 2000/(2 × 24.6) = 48.8 cm; provide 50 cm long bars (25 cm on each side of the joint).
• If plain bars are used (S_s = 1250 kg/cm², S_b = 17.5 kg/cm²): A_s = 3645/1250 = 2.92 cm²/m, so 12 mm bars at 38 cm c/c, and L = 1.2 × 1250/(2 × 17.5) = 42.9 cm, say 45 cm.

(c) For B.G. (G = 1750 mm), R = 1750/D = 1750/2 = 875 m.

Superelevation (cant) for the equilibrium speed 90 kmph:

C_a = GV²/(127R) = 1750 × 90²/(127 × 875) = 127.6 mm (below the 165 mm maximum).

Check for the slow goods train (60 kmph): its equilibrium cant = 1750 × 60²/(127 × 875) = 56.7 mm, so cant excess = 127.6 − 56.7 = 70.9 mm < 75 mm (permissible).

Maximum permissible speed, assuming a cant deficiency C_d = 100 mm (allowed on high-speed B.G. routes):

V_max = 0.27√[(C_a + C_d)R] = 0.27√(227.6 × 875) = ≈ 120 kmph. At 130 kmph the cant deficiency would be about 137 mm (> 100 mm), so the sanctioned 130 kmph cannot be run on this curve without a speed restriction.

Transition length (rate of change of cant or cant deficiency = 35 mm/s):

• Based on cant: L = C_a V_max/(3.6 × 35) = 127.6 × 120/126 = ≈ 122 m.
• Based on cant deficiency: L = 100 × 120/126 = 95 m.
• Based on cant gradient 1 in 720: L = 0.72 × 127.6 = 92 m.
• Adopt the largest, L ≈ 122 m (say 125 m).