Shaft Machining Error Coursework

Course work · Technological Assurance of Quality in Machine-Building Production

Calculation of Machining Errors of a Shaft

• Author: Tanuj Namboodri
• Approved by: I. I. Kozar
• MSc coursework, 2021
• English version (original in Russian)

5.1 Task

1. For a shaft machined between centres, select the cutting conditions t, s, v from standard tables and calculate the radial (normal) component of the cutting force P_y.
2. Calculate the errors caused by elastic deformation of the machine for 10 sections x along the shaft and plot the resulting longitudinal form error.
3. Calculate the errors caused by elastic deformation of the workpiece for the same sections and plot the shape of the shaft in the longitudinal section.
4. Calculate the errors caused by dimensional wear of the cutting tool for the same coordinates and plot them.
5. Calculate the form errors caused by thermal deformation of the cutting tool and plot them.
6. Calculate the total error in the longitudinal section, taking the sign of each component into account, and plot it against the machining coordinate.
7. Compare the resulting error with the IT6 tolerance of the shaft.

Figure 1. Working drawing of the shaft (title block translated). Overall length 247 mm, controlled diameter Ø50 mm.

5.2 Initial data

Operation	finish turning between centres
Tool	turning tool with T15K6 carbide insert, shank b×h = 16×25 mm, overhang lp = 20 mm
Material	steel 45, HB 260–285 (approx. C45 / AISI 1045), σB = 750 MPa assumed
Controlled diameter	D = 50 mm
Shaft length	L = 247 mm
Depth of cut	t = 0.5 mm
Feed	s = 0.5 mm/rev
Cutting speed	v = 150 m/min
Coefficients for Py	steel σB = 750 MPa, carbide tool: Cp = 243, x = 0.9, y = 0.6, n = −0.3; Kp = 1
Compliance	headstock, tailstock and carriage: Whs = Wts = Wcar = 0.02 µm/N
Elastic modulus	E = 200 000 MPa
Second moment of area	I ≈ 0.05·D4 = 312 500 mm4
Relative tool wear	U0 = 8 µm/km
Sections	10 sections, x = 24.7, 49.4 … 247 mm, measured from the tailstock end where the cut starts

5.3 Radial component of the cutting force

\[ P_y = 10 \, C_p \, t^{x} s^{y} v^{n} K_p = 10 \cdot 243 \cdot 0.5^{0.9} \cdot 0.5^{0.6} \cdot 150^{-0.3} \cdot 1 = 191\ \text{N} \]

The errors caused by elastic deflection of the machine and of the workpiece are calculated (in µm) with:

\[ \Delta_y^{\text{m}} = 2 P_y \left( W_{\text{car}} + W_{\text{hs}} \left(\frac{x}{L}\right)^2 + W_{\text{ts}} \left(\frac{L-x}{L}\right)^2 \right) \]

\[ \Delta_y^{\text{w}} = 10^3 \cdot 2 P_y \, \frac{L^3}{3EI} \left(\frac{x}{L}\right)^2 \left(\frac{L-x}{L}\right)^2 \]

Superscript m = machine, w = workpiece. The factor 2 turns the radial deflection into an error on the diameter.

5.4 Errors from elastic deflection of the workpiece

Example for the first section, x = 24.7 mm:

\[ \Delta_y^{\text{w}} = 10^3 \cdot 2 \cdot 191.1 \cdot \frac{247^3}{3 \cdot 200\,000 \cdot 312\,500} \cdot \left(\frac{24.7}{247}\right)^2 \left(\frac{247-24.7}{247}\right)^2 = 0.249\ \mu\text{m} \]

x, mm	24.7	49.4	74.1	98.8	123.5	148.2	172.9	197.6	222.3	247
Δw, µm	0.249	0.786	1.355	1.769	1.920	1.769	1.355	0.786	0.249	0.000

Figure 2. Error caused by elastic deflection of the workpiece. It is largest at mid-length.

5.5 Errors from elastic deflection of the machine

Example for x = 24.7 mm:

\[ \Delta_y^{\text{m}} = 2 \cdot 191.1 \cdot \left( 0.02 + 0.02 \left(\frac{24.7}{247}\right)^2 + 0.02 \left(\frac{247-24.7}{247}\right)^2 \right) = 13.91\ \mu\text{m} \]

x, mm	24.7	49.4	74.1	98.8	123.5	148.2	172.9	197.6	222.3	247
Δm, µm	13.91	12.84	12.08	11.62	11.47	11.62	12.08	12.84	13.91	15.29

Figure 3. Error caused by elastic deflection of the machine. It is smallest at mid-length and largest near the centres.

5.6 Errors from dimensional wear of the tool

The tool wears in proportion to the length of the cutting path l. With U₀ in µm/km:

\[ l = \frac{\pi D x}{1000 \, s}\ \text{(m)}, \qquad \Delta_u = 2 \cdot \frac{U_0 \, l}{1000} \]

Example for x = 24.7 mm:

\[ l = \frac{3.1416 \cdot 50 \cdot 24.7}{1000 \cdot 0.5} = 7.76\ \text{m}, \qquad \Delta_u = 2 \cdot \frac{8 \cdot 7.76}{1000} = 0.124\ \mu\text{m} \]

x, mm	24.7	49.4	74.1	98.8	123.5	148.2	172.9	197.6	222.3	247
Δu, µm	0.124	0.248	0.372	0.497	0.621	0.745	0.869	0.993	1.117	1.242

Figure 4. Error caused by dimensional wear of the cutting tool. It grows linearly with the cutting path.

5.7 Errors from thermal deformation of the tool

\[ \Delta_{T} = -2 \, \zeta_{\text{st}} \left( 1 - e^{-\tau/T} \right) \]

where

ζ_st

elongation of the tool at thermal equilibrium, µm: \( \zeta_{\text{st}} = C \, \frac{l_p}{F} \, \sigma_B \, (t \, s)^{0.75} \sqrt{v} \), with σ_B in kgf/mm²

C

constant, 4.5

l_p

tool overhang, 20 mm

F

cross-section of the tool shank, 16 × 25 = 400 mm²

σ_B

750 MPa = 76.5 kgf/mm²

τ

cutting time, min: \( \tau = \frac{\pi D x}{1000 \, s \, v} \)

T

thermal time constant of the tool, 4 min (as in the original)

\[ \zeta_{\text{st}} = 4.5 \cdot \frac{20}{400} \cdot 76.5 \cdot (0.5 \cdot 0.5)^{0.75} \cdot \sqrt{150} = 74.5\ \mu\text{m} \]

Example for x = 24.7 mm:

\[ \tau = \frac{3.1416 \cdot 50 \cdot 24.7}{1000 \cdot 0.5 \cdot 150} = 0.0517\ \text{min}, \qquad \Delta_T = -2 \cdot 74.5 \cdot \left(1 - e^{-0.0517/4}\right) = -1.91\ \mu\text{m} \]

x, mm	24.7	49.4	74.1	98.8	123.5	148.2	172.9	197.6	222.3	247
ΔT, µm	−1.91	−3.81	−5.67	−7.51	−9.33	−11.13	−12.90	−14.65	−16.37	−18.08

Figure 5. Error caused by thermal elongation of the tool. The tool grows towards the axis, so the diameter becomes smaller (negative sign).

5.8 Total form error in the longitudinal section

\[ \Delta_\Sigma = \Delta_y^{\text{m}} + \Delta_y^{\text{w}} + \Delta_u + \Delta_T \]

x, mm	Machine Δm	Workpiece Δw	Tool wear Δu	Thermal ΔT	Total ΔΣ
24.7	13.91	0.25	0.12	−1.91	12.37
49.4	12.84	0.79	0.25	−3.81	10.07
74.1	12.08	1.35	0.37	−5.67	8.13
98.8	11.62	1.77	0.50	−7.51	6.37
123.5	11.47	1.92	0.62	−9.33	4.67
148.2	11.62	1.77	0.74	−11.13	3.01
172.9	12.08	1.35	0.87	−12.90	1.40
197.6	12.84	0.79	0.99	−14.65	−0.03
222.3	13.91	0.25	1.12	−16.37	−1.10
247	15.29	0.00	1.24	−18.08	−1.55

All values in µm on the diameter.

Figure 6. The four error components and the total error along the shaft.

6 Comparison with the tolerance

The form error in the longitudinal section is the difference between the largest and the smallest diameter error along the shaft:

\[ \Delta_{\text{form}} = \Delta_{\Sigma,\max} - \Delta_{\Sigma,\min} = 12.37 - (−1.55) = 13.92\ \mu\text{m} \]

IT6 tolerance, D = 50 mm (ISO 286-1)	16 µm on the diameter	13.9 µm < 16 µm: within
Profile of longitudinal section, level A (normal), GOST 24643-81	5 µm (on the radius)	7.0 µm > 5 µm: exceeded

Conclusion: the form error of 13.9 µm is within the IT6 tolerance of 16 µm, but on the radius (7.0 µm) it is larger than the normal form tolerance of 5 µm for this size and grade.

The largest contribution is the thermal elongation of the tool (down to −18.1 µm at the end of the pass). Without it, the form error would be only 2.7 µm. Running the tool until it reaches thermal equilibrium before the finishing pass, or compensating the elongation, would bring the shaft within both limits. The constant part of the error shifts the diameter as a whole and is removed when the tool is set to size.