Consider the 2-D model for truck mounted crane used to save people in high floors. The crane uses one hydraulic pump and one tank to operate the mass M in 3 DOF: 1. Linear actuator for displacement (Y) 2. Hydraulic motor for the rotation about vertical axis (R) 3. Hydraulic motor for the rotation about horizontal axis (θ) Assumptions: – Consider that the two hydraulic motors (R and θ ) and the mass M are located at the same point at a height equals ( Y− initially at 1.5 m ). – Assume return pipe similar to pressure pipes. – Neglect the masses of the hydraulic motors. The load is a lumped mass with: M= The load mass held by the crane =1000 kg J= The moment of inertia =10,000 kg⋅m2 L=5 m. K=5×105 N/m (for linear actuator) C=10 N.s/m (for linear actuator) K=5×105 N/ radian (for hydraulic actuator) C=10 N.s/radian (for hydraulic actuator) Z=10 cm (Internal radius for the vane of hydraulic motors) d=20 mm (for all pipes) Viscosity =100cSt S.G =0.9 Pump output power =5HP Pump max. flow rate (Q)=60 liter /min Volumetric efficiency of pump =80%

Question:

Consider the 2-D model for truck mounted crane used to save people in high floors. The crane uses one hydraulic pump and one tank to operate the mass M in 3 DOF: 1. Linear actuator for displacement (Y) 2. Hydraulic motor for the rotation about vertical axis (R) 3. Hydraulic motor for the rotation about horizontal axis (θ) Assumptions: – Consider that the two hydraulic motors (R and θ ) and the mass M are located at the same point at a height equals ( Y− initially at 1.5 m ). – Assume return pipe similar to pressure pipes. – Neglect the masses of the hydraulic motors.

The load is a lumped mass with:

M= The load mass held by the crane =1000 kg

J= The moment of inertia =10,000 kg⋅m2 L=5 m. K=5×105 N/m (for linear actuator)

C=10 N.s/m (for linear actuator)

K=5×105 N/ radian (for hydraulic actuator)

C=10 N.s/radian (for hydraulic actuator)

Z=10 cm (Internal radius for the vane of hydraulic motors)

d=20 mm (for all pipes) Viscosity =100cSt S.G =0.9

Pump output power =5HP Pump max.

flow rate (Q)=60 liter /min Volumetric efficiency of pump =80%

ANSWER

here is an analysis of the 2-D model for a truck-mounted crane used to save people in high floors:

Objective:

Analyze the performance of a truck-mounted crane used for rescue operations in high-rise buildings.

Assumptions:

  1. Two hydraulic motors (R and θ) and the mass M are located at the same point at a height of (Y – 1.5 m).
  2. Return pipes are similar to pressure pipes.
  3. Masses of hydraulic motors are negligible.

Given Parameters:

  • Load mass (M) = 1000 kg
  • Moment of inertia (J) = 10,000 kg·m²
  • Crane arm length (L) = 5 m
  • Linear actuator stiffness (K) = 5 × 10⁵ N/m
  • Linear actuator damping coefficient (C) = 10 N·s/m
  • Hydraulic actuator stiffness (K) = 5 × 10⁵ N/rad
  • Hydraulic actuator damping coefficient (C) = 10 N·s/rad
  • Internal radius of hydraulic motor vane (Z) = 10 cm
  • Pipe diameter (d) = 20 mm
  • Viscosity (μ) = 100 cSt
  • Specific gravity (S.G.) = 0.9
  • Pump output power = 5 HP
  • Pump maximum flow rate (Q) = 60 L/min
  • Pump volumetric efficiency (η_v) = 80%

Analysis:

  1. Calculate the pump actual flow rate:Actual flow rate (Q_act) = η_v × Q_max = 0.8 × 60 L/min = 48 L/min
  2. Determine the pump’s theoretical pressure and flow rate:Using the pump power and actual flow rate, determine the theoretical pressure:P_th = (5 HP × 746 W/HP) / (Q_act × 60 L/min)P_th = 131000 PaCalculate the theoretical flow rate using the pump’s theoretical pressure:Q_th = V_d × η_v / (T × P_th)where:V_d = displacement volume per revolution (π × Z² × L)T = cycle time (assumed to be 1 second)Substituting values, we get:Q_th = (π × 0.1² × 0.5) × 0.8 / (1 × 131000) = 1.52 × 10⁻⁵ m³/sSince Q_act > Q_th, the pump operates in the overload region.
  3. Analyze the linear actuator:The linear actuator is responsible for vertical movement of the load. Its equation of motion is:M(dY²/dt²) + C(dY/dt) + KY = P_A/Awhere:Y = vertical displacement (m) P_A = pressure in the linear actuator (Pa) A = area of the piston in the linear actuator (m²)Assuming a sinusoidal input with amplitude Y_m and frequency ω:Y = Y_m sin(ωt)Substituting into the equation of motion and solving for P_A:P_A = (MY²ω² sin(ωt) + CY²ω cos(ωt) + KY²) / AThe maximum pressure occurs at t = π/2:P_A_max = (MY²ω² + CY²ω + KY²) / ASubstituting values, we get:P_A_max = (1000 × 0.5² × ω² + 10 × 0.5² × ω + 5 × 10⁵ × 0.5²) / (π × 0.1²) = 4.03 × 10⁵ + 1.00 × 10⁴ω + 1.59 × 10⁵ω²
  4. Analyze the hydraulic motors:The hydraulic motors are responsible for rotation about the vertical and horizontal axes. Their equations of motion are:J(d²θ/dt²) + C(dθ/dt) + KT = P_A/Awhere:θ = angular displacement (rad) P_A = pressure in the hydraulic motor (Pa) A = area of the piston in the hydraulic motor (m²)Assuming a sinusoidal input with amplitude θ_m and frequency ω

The coefficient of friction between steel and oil is approximately 0.1. Plugging in the values, we get:

F_friction = (0.1)(0.1) = 0.01 N

The total force required to move the load is:

F = m*g + F_friction = (1000 kg)(9.81 m/s²) + (0.01 N) = 9810.1 N

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