(a) During turning of a steel rod of 100 mm diameter, at a speed of 600 rpm, feed of 0·32 mm/rev and 3 mm depth of cut by a tool of the following geometry : inclination angle 0°, orthogonal rake angle – 12°, and principal cutting edge angle (φ) 60°,

Question

(a) During turning of a steel rod of 100 mm diameter, at a speed of 600 rpm, feed of 0·32 mm/rev and 3 mm depth of cut by a tool of the following geometry : inclination angle 0°, orthogonal rake angle – 12°, and principal cutting edge angle (φ) 60°, the following have been observed :

Main cutting force component Pz = 1000 N
Radial cutting force component Py = 200 N
Chip thickness = 0·75 mm

Using Merchant's theory, determine the force along the rake surface F, force acting perpendicular to the rake surface N, coefficient of friction between the chip-tool interface, shear force Fs and cutting power consumption. Power consumption due to feed motion may be neglected. (20 marks)

(b) (i) State the sequence and purpose of different types of rolling passes, used for rolling flats and plates. (5 marks)

(ii) A steel plate of 200 mm width and 30 mm thickness is rolled using two-stand rolling mill where diameter of each roller is 400 mm. Rolling is performed at 60 rev/min of roll speed to reduce thickness of steel strip from 30 mm to 26 mm. Average flow stress of metal during rolling is expressed by

δ_f = (K εⁿ)/(1+n)

where

n = strain hardening coefficient

K = strength coefficient, MPa

ε = true strain

Consider strength coefficient of metal (K) is 300 MPa and strain hardening coefficient (n) is 0·2 and coefficient of friction between strip and roll during rolling is 0·15.

Calculate :
(I) Draft in mm
(II) Maximum achievable draft for above condition of rolling in mm
(III) Average flow stress in MPa
(IV) Rolling force in Newton
(Neglect all other possibilities related to rolling) (15 marks)

(c) Write the contributions of Walter A. Shewhart, W. Edwards Deming, Joseph M. Juran, Philip B. Crosby, and K. Ishikawa in the area of quality management. (10 marks)

UPSC Answer Check · Accepted Answer

(a)
Given: D = 100 mm, N = 600 rpm, f = 0.32 mm/rev, d = 3 mm, φ = 60°, α = –12°, P_z = 1000 N, P_y = 200 N, t₂ = 0.75 mm.

Uncut chip thickness t₁ = f sin φ = 0.32 × sin 60° = 0.32 × 0.8660 = 0.2771 mm.
Chip thickness ratio r = t₁/t₂ = 0.2771/0.75 = 0.3695.
Using Merchant's shear-angle relation:
tan φₛ = r cos α / (1 – r sin α).
sin(–12°) = –0.2079, cos(–12°) = 0.9781.
tan φₛ = (0.3695 × 0.9781) / (1 – 0.3695 × (–0.2079)) = 0.3615 / 1.0768 = 0.3357.
φₛ = arctan(0.3357) = 18.56°.

Orthogonal cutting force components: F_c = P_z = 1000 N. For principal cutting edge angle φ, the radial force P_y is the component of the thrust force F_t along the radial direction, so F_t = P_y / cos φ = 200 / cos 60° = 400 N.

Using Merchant's force relations:
F = F_c sin α + F_t cos α = 1000 sin(–12°) + 400 cos(–12°) = –207.9 + 391.2 = 183.3 N.
N = F_c cos α – F_t sin α = 1000 cos(–12°) – 400 sin(–12°) = 978.1 – (–83.2) = 1061.3 N.
Coefficient of friction μ = F/N = 183.3/1061.3 = 0.173.

Shear force F_s = F_c cos φₛ – F_t sin φₛ = 1000 cos 18.56° – 400 sin 18.56° = 948.0 – 127.3 = 820.7 N.

Cutting speed V = π D N / 60 = π × 0.1 × 600 / 60 = π m/s = 3.1416 m/s.
Cutting power P = P_z × V = 1000 × 3.1416 = 3141.6 W = 3.142 kW.

Final answers: F ≈ 183.3 N; N ≈ 1061.3 N; μ ≈ 0.173; F_s ≈ 820.7 N; cutting power ≈ 3.142 kW.

(b)(i)
Sequence of rolling passes for flats and plates:
Breakdown/roughing passes: hot slab is rolled in a reversing mill to reduce thickness, break cast structure, and remove scale.
Broadside/edge passes: slab is rotated 90° and rolled across width to obtain required plate width and sound edges.
Longitudinal/intermediate passes: repeated passes elongate the plate and bring thickness close to final size.
Finishing passes: final passes in finishing train/plate mill give accurate thickness, flatness, surface finish, and desired mechanical properties.
Purpose: gradual reduction avoids excessive roll force and defects, controls width and length, refines grain structure, and meets final dimensional and metallurgical specifications.

(b)(ii)
Given: b = 200 mm, h₀ = 30 mm, h₁ = 26 mm, R = D/2 = 200 mm, μ = 0.15, K = 300 MPa, n = 0.2.

(I) Draft Δh = h₀ – h₁ = 30 – 26 = 4 mm.

(II) Maximum achievable draft Δh_max = μ² R = (0.15)² × 200 = 0.0225 × 200 = 4.5 mm. Since 4 mm < 4.5 mm, rolling is feasible.

(III) True strain ε = ln(h₀/h₁) = ln(30/26) = ln(1.153846) = 0.1431.
Average flow stress δ_f = K εⁿ / (1 + n) = 300 × (0.1431)^0.2 / 1.2.
(0.1431)^0.2 = exp(0.2 ln 0.1431) = exp(0.2 × (–1.9443)) = exp(–0.3889) = 0.6779.
δ_f = 300 × 0.6779 / 1.2 = 169.5 MPa.

(IV) Contact length L = √(R Δh) = √(200 × 4) = √800 = 28.28 mm.
Rolling force F = δ_f × b × L = 169.5 × 200 × 28.28 = 958,600 N = 9.586 × 10⁵ N ≈ 958.6 kN.
(Roll speed is not required for the asked quantities.)

(c)
Walter A. Shewhart: introduced statistical quality control, control charts, distinction between chance and assignable causes, and the Plan-Do-Check-Act (PDCA) cycle.
W. Edwards Deming: promoted statistical quality control and system thinking; gave 14 points for management; emphasized reduction of variation, continuous improvement, and top-management commitment.
Joseph M. Juran: developed quality trilogy (planning, control, improvement); introduced Pareto principle; advocated top-management involvement and cost of quality; defined quality as fitness for use.
Philip B. Crosby: concept of “Quality is Free” and “Zero Defects”; emphasized prevention and doing it right the first time; gave 14 steps to quality improvement; quality management maturity grid.
K. Ishikawa: developed cause-and-effect (fishbone) diagram; advocated company-wide quality control (CWQC), quality circles, and use of seven basic quality tools; introduced internal customer concept.

(a) During turning of a steel rod of 100 mm diameter, at a speed of 600 rpm, feed of 0·32 mm/rev and 3 mm depth of cut by a tool…

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