Ball Screw Whip on a 300 kg Vertical Lift
A vertical lift raises a 300 kg carriage along a 1.5 m unsupported length of a 25 mm-diameter steel ball screw with a 10 mm lead. The screw is supported at both ends by angular-contact bearings; the nut carries the load in compression. The drive is a 3000 rpm servo connected directly to the screw. The required lift speed is 0.30 m/s. The screw is made of steel with E = 210 GPa and density 7850 kg/m³. The bearing arrangement is claimed to give simple supports. A junior engineer selected the screw by checking only the static buckling load with a safety factor of 2 and concluded it was acceptable. During commissioning, the screw whips violently above about 1200 rpm and the drive trips on following error. The machine builder proposes increasing the motor current limit and adding a support bearing halfway along the screw.
1. Calculate the first critical speed and the Euler buckling load. Is the junior engineer's conclusion correct? Show the intermediate quantities.
Read the worked answer
Area A = πD^2/4 = π(0.025)^2/4 = 4.91 × 10^-4 m². Second moment I = πD^4/64 = π(0.025)^4/64 = 1.92 × 10^-8 m^4. EI = 210 × 10^9 × 1.92 × 10^-8 = 4030 N·m². Mass per length = 7850 × 4.91 × 10^-4 = 3.85 kg/m.
For simple supports, the first natural frequency is f1 = (π/2) × sqrt[EI / (ρA L^4)] = (π/2) × sqrt[4030 / (3.85 × 1.5^4)] = (π/2) × sqrt(4030 / 19.5) = (π/2) × 14.4 = 22.6 Hz. Critical speed = 60 × 22.6 = 1356 rpm. Using a 0.8 safe operating factor gives about 1085 rpm.
Euler buckling for pinned-pinned ends: P_cr = π²EI/L² = 9.87 × 4030 / 2.25 = 17.7 kN. The static load is 300 × 9.81 = 2.94 kN, so the static buckling safety factor is 17.7 / 2.94 = 6.0, not 2; the junior engineer's number is conservative but the check is incomplete.
The required screw speed is 0.30 m/s / 0.010 m/rev = 30 rev/s = 1800 rpm, which is above the 1356 rpm first critical speed. The conclusion is therefore wrong: static buckling is acceptable, but the screw is operating above its critical speed, which explains the whip above 1200 rpm.
2. Evaluate the two proposed fixes - increasing motor current limit and adding a mid-span support - and propose at least one alternative that meets the 0.30 m/s requirement.
Read the worked answer
Increasing the motor current limit does not address whip. Whip is a shaft dynamics problem; more torque will simply drive the screw harder into a whirling condition and increase bearing and nut loads. It may also mask the following-error trip while damaging the screw.
Adding a mid-span support reduces the unsupported length from 1.5 m to 0.75 m. Critical speed scales with 1/L², so it rises by a factor of 4 to about 5420 rpm, which is ample. This is technically effective, but it adds a bearing, a mounting structure and maintenance points inside the lift; it may not fit the existing envelope.
Alternatives:
- Use a larger screw. Critical speed scales with diameter. A 32 mm screw gives (32/25)^2 = 1.64 times the critical speed, about 2220 rpm, which is above 1800 rpm but with limited margin. A 40 mm screw gives about 3470 rpm and a comfortable margin.
- Increase the lead. A 20 mm lead requires only 900 rpm at 0.30 m/s, below the 1085 rpm safe limit, but it doubles the required torque and halves the resolution.
- Keep the 10 mm lead and add a 2:1 belt reduction. The screw then runs at 900 rpm while the motor runs at 1800 rpm. This also doubles reflected torque and adds backlash, so the belt must be sized carefully.
The best choice depends on space and torque; the key judgement is that critical speed, not only static buckling, must be checked.
3. Check the buckling case with a realistic dynamic factor and support condition. What must be verified about the bearing arrangement before accepting the design?
Read the worked answer
The static load is 2.94 kN. For a vertical lift, use a dynamic factor of at least 1.5, giving a design thrust of 4.41 kN.
With pinned-pinned supports, P_cr = 17.7 kN, so the safety factor is 17.7 / 4.41 = 4.0, which is acceptable. However, if one end is effectively free to move laterally or the bearing arrangement behaves as fixed-free, the effective length factor K = 2 and P_cr = π²EI/(2L)^2 = 4.42 kN. The safety factor then becomes 4.42 / 4.41 = 1.0, which is not acceptable and leaves no margin for friction, preload or shock.
The claim that the angular-contact bearings give simple supports must be verified. Check the bearing housings for lateral stiffness, the nut mounting for side load, and whether the screw is in tension or compression over the full stroke. If the bearings cannot be guaranteed to act as simple supports, either use a larger screw, reduce the unsupported length, or add a counterbalance to reduce thrust. Also check the buckling case at the worst position, because the unsupported length changes as the nut travels.