Solution:

Given data:

- Voltage (V) = 120 V
- Armature resistance (Ra) = 0.2 Ω
- Field resistance (Rf) = 60 Ω
- Full-load current (I) = 40 A
- Full-load speed (N) = 1800 rpm

We need to find the speed of the motor when it is operating with half full-load.

Formula used:

- The speed of the DC shunt motor can be calculated using the following formula:

N ∝ (V - Ia Ra) / Φ

where, N = speed of the motor in rpm
V = applied voltage in volts
Ia = armature current in amperes
Ra = armature resistance in ohms
Φ = flux per pole in Weber

- The flux per pole (Φ) of the DC shunt motor can be calculated using the following formula:

Φ ∝ V / (Rf + Ra)

where, Rf = field resistance in ohms

- At half full-load, the current drawn by the motor will be 20 A.

Calculation:

- Flux per pole (Φ) = V / (Rf + Ra)
= 120 / (60 + 0.2)
= 1.98 Wb
- Full-load torque (T) = (I * Φ) / (2 * π)
= (40 * 1.98) / (2 * π)
= 12.56 Nm
- Half full-load torque = T / 2
= 6.28 Nm
- Half full-load current (Ia) = I / 2
= 20 A
- Speed of the motor at half full-load = (V - Ia Ra) / Φ
= (120 - (20 * 0.2)) / 1.98
= 59.60 rpm

Therefore, the speed of the motor when it is operating with half full-load is 59.60 rpm.