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Mechanical Engineering 2024 Paper I 50 marks Calculate

Paper I — Q6

(a) A perfectly plastic material having yield strength of 300 MPa is subjected to extrusion at 400 °C. The extrusion speed is 300…

(a)

A perfectly plastic material having yield strength of 300 MPa is subjected to extrusion at 400 °C. The extrusion speed is 300 mm/s. The extrusion process is done on billet of diameter 50 mm and length 300 mm. Extrusion reduces the diameter of billet from 50 mm to 20 mm. Assuming that the extrusion process is frictionless, determine the true strain, average true strain, ideal extrusion force and work done on billet during extrusion. The initial diameter of billet is significantly greater than the final diameter after extrusion. 20 marks

(b)

The data for the number of dissatisfied customers in a department store observed for 20 samples of size 300 are shown in the table below. Construct a suitable control chart for the collected data. Any out-of-control situation may be treated as the presence of some assignable cause, accordingly revise the control limits:

SampleNo. of Dissatisfied CustomersSampleNo. of Dissatisfied Customers
110116
2121219
381310
49147
56158
611164
7131711
8101810
98196
109207

Management believes that the dissatisfaction rate is 2%, so establish control limits based on this value. Discuss whether the department can meet this goal. What actions would you recommend? 20 marks

(c)

List down different types of wastes indicated in JIT system. How are they connected with flow layout? 10 marks

हिंदी में प्रश्न पढ़ें
(a)

एक पूरी तरह से पराप्रत्यास्थ (प्लास्टिक) सामग्री, जिसका प्रारंभ सामर्थ्य 300 MPa है, 400 °C पर बहिवेधन से गुजरती है। बहिवेधन गति 300 mm/s है। बहिवेधन प्रक्रिया 50 mm व्यास और 300 mm लंबाई के बिलेट पर की जाती है। बहिवेधन, बिलेट का व्यास 50 mm से 20 mm तक घटा देता है। यह मानते हुए कि बहिवेधन प्रक्रिया घर्षण रहित है, वास्तविक विकृति (स्ट्रेन), औसत वास्तविक विकृति, आदर्श बहिवेधन बल और बहिवेधन के दौरान बिलेट पर किया गया कार्य निर्धारित कीजिए। बिलेट का प्रारंभिक व्यास, बहिवेधन के बाद अंतिम व्यास से बहुत अधिक है। (20 अंक)

(b)

एक विभागीय स्टोर में असंतुष्ट ग्राहकों की संख्या के आँकड़े नीचे तालिका में दिखाए गए हैं, जो 300 आमाप (साइज़) के 20 नमूनों के लिए हैं। एकत्रित आँकड़ों के लिए एक उपयुक्त नियंत्रण चार्ट बनाइए। किसी भी नियंत्रण से बाहर की स्थिति को किसी निर्देश्य कारण की उपस्थिति के रूप में माना जा सकता है, तदनुसार नियंत्रण सीमाओं को संशोधित कीजिए:

नमूनाअसंतुष्ट ग्राहकों की संख्यानमूनाअसंतुष्ट ग्राहकों की संख्या
110116
2121219
381310
49147
56158
611164
7131711
8101810
98196
109207

प्रबंधन का मानना है कि असंतोष दर 2% है, इसलिए इस मान के आधार पर नियंत्रण सीमाएँ स्थापित कीजिए। विवेचना कीजिए कि क्या विभाग इस लक्ष्य को पूरा कर सकता है। किन कार्यवाहियों की अनुशंसा आप कर सकते हैं? (20 अंक)

(c)

JIT प्रणाली में बताए गए विभिन्न प्रकार के कचरे की सूची बनाइए। ये सभी प्रवाह अभिन्यास से कैसे जुड़े हुए हैं? (10 अंक)

Q6 of the 2024 UPSC Mains Mechanical Engineering Paper I, as printed
The question as printed in the 2024 Mechanical Engineering paper
Evaluate your answer to this question → See all 2024 Mechanical Engineering questions

The figure this question refers to, in words

The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.

(b) Table: Data for the number of dissatisfied customers in a department store observed for 20 samples of size 300.

Columns: Sample, No. of Dissatisfied Customers, Sample, No. of Dissatisfied Customers

Rows: 1, 10, 11, 6 2, 12, 12, 19 3, 8, 13, 10 4, 9, 14, 7 5, 6, 15, 8 6, 11, 16, 4 7, 13, 17, 11 8, 10, 18, 10 9, 8, 19, 6 10, 9, 20, 7

(c) A two-dimensional Cartesian coordinate system with a grid. The horizontal axis is labeled 'X' with the text 'इकाई' (Unit) below it, and the vertical axis is labeled 'Y' with the text 'इकाई' (Unit) to its left. Both axes have tick marks and labels from 100 to 700 in increments of 100. Four points are plotted on the grid, each marked with a circle and a letter:

  • Point A is located at coordinates (125, 550).
  • Point B is located at coordinates (350, 400).
  • Point C is located at coordinates (450, 125).
  • Point D is located at coordinates (700, 300).

To the right of the graph is a table with the heading '(दूरी निर्देशांक कोष्ठक में हैं)' which translates to '(Distance coordinates are in brackets)'. The table has three columns: 'आपूर्ति बिंदु' (Supply Point), 'निर्देशांक (इकाई)' (Coordinates (Unit)), and 'वार्षिक आपूर्ति (टन)' (Annual Supply (Ton)). The rows contain the following data:

  • Row 1: A, (125, 550), 200
  • Row 2: B, (350, 400), 450
  • Row 3: C, (450, 125), 175
  • Row 4: D, (700, 300), 350

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a) Given D0=50 mm, Df=20 mm, L0=300 mm, Y=300 MPa=300 N/mm². A=πD²/4, so A0=625π mm², Af=100π mm², A0/Af=25/4=6.25, V=A0L0=187500π mm³. Assumptions: incompressible, homogeneous, frictionless, perfectly plastic, no strain hardening; 400 °C fixes flow stress.

True strain by logarithmic strain and volume constancy: ε=ln(A0/Af)=ln(25/4)=1.8326.

Average true strain rate. The given ram speed and the remark that D0 ≫ Df point to the average true strain rate in extrusion: ε̇_avg = 6V D0² ln(A0/Af)/(D0³ − Df³) = 6×300×50²×1.8326/(50³ − 20³) ≈ 70.5 s⁻¹. With D0 ≫ Df this reduces to ε̇_avg ≈ 6V ln(A0/Af)/D0 = 6×300×1.8326/50 ≈ 66.0 s⁻¹.

Ideal frictionless extrusion pressure from the energy/upper-bound method: p=Yε=300×1.8326=549.77 MPa. The ram force is F=pA0=A0Yε=187500π ln(25/4) N=1.0795 MN.

Work by the energy integral: W=∫Y dε dV=YVε=300×187500π×ln(25/4) N mm=56250000π ln(25/4) N mm=56250π ln(25/4) J=323.84 kJ. For complete extrusion the ram stroke is L0=0.300 m, because the ram sweeps the initial billet volume A0L0; hence W=FL0 gives the same value. The speed is used only for the strain rate.

Final: ε=1.8326, ε̇_avg≈66 s⁻¹ (≈70.5 s⁻¹ by the full formula), F=1.0795 MN, W=323.84 kJ.

(b) Use an np-chart: plot the 20 counts against sample number with 3-sigma limits. This is suitable because n=300 is constant, the data are counts of nonconforming customers, observations are independent, and the count is approximately binomial; n=300 supports the normal approximation.

Initial limits: Σd=184, m=20, p̄=184/(20×300)=0.030667. CL=np̄=300×0.030667=9.20 customers. σ=√(np̄(1-p̄))=√(300×0.030667×0.969333)=2.986. UCL=9.20+3×2.986=18.16, LCL=9.20-3×2.986=0.24. Sample 12 has 19, above UCL; treat it as an assignable cause and remove it.

Revised limits: Σd=165, m=19, p̄'=165/(19×300)=0.028947. CL'=165/19=8.68. σ'=√(300×0.028947×0.971053)=2.904. UCL'=8.68+3×2.904=17.40, LCL'=8.68-3×2.904=-0.03, set to 0. All remaining samples lie within 0 to 17.40.

Diagram: np chart dissatisfied customers

For the management target p0=0.02, fixed np-chart limits: CL0=np0=300×0.02=6. σ0=√(300×0.02×0.98)=√5.88=2.425. UCL0=6+3×2.425=13.27, LCL0=6-3×2.425=-1.27, set to 0. Equivalently, the p-chart limits are LCL=0, CL=0.02, UCL=0.04425.

The revised process mean is 8.68 dissatisfied customers per 300, i.e. 2.895%, above the 2% target. The process is in control after removing sample 12, but it does not meet the goal because the target centre line 6 is below the process centre line 8.68. The department cannot currently meet the goal without improvement.

Actions: investigate high-count samples for assignable causes; use Pareto and cause-effect analysis; improve complaint handling, service recovery, waiting times, and billing accuracy; train staff; redesign the service process; monitor with revised limits during improvement; once the average is ≤6 per 300, use the 2% target limits for ongoing control.

Final: revised np limits 0 to 17.40; target np limits 0 to 13.27; current mean 8.68/300, so the 2% goal is not met.

(c) JIT wastes (muda):

  • Overproduction: making more or earlier than needed.
  • Waiting: idle time for material, machines, or information.
  • Transportation: unnecessary movement of material.
  • Over-processing: extra operations not required by the customer.
  • Inventory: excess raw material, WIP, or finished goods.
  • Motion: unnecessary movement of people or equipment.
  • Defects: rework, scrap, or complaints.
  • Unused talent: underuse of employee skills and suggestions.

Connection with flow layout:

  • A flow layout places workstations in the order of value-adding operations, so material moves in one continuous direction.
  • It reduces transportation and motion by short, direct paths.
  • It reduces waiting and inventory by enabling one-piece or small-lot flow and quick changeovers.
  • It exposes defects and over-processing quickly, allowing immediate correction.
  • It supports pull and levelled production, preventing overproduction.

Final: flow layout reduces JIT wastes by making material flow continuous and exposing non-value-adding steps.

How this answer will be evaluated

What earns full marks: Rigorous calculations with clear assumptions; precise statistical analysis with correct chart selection; comprehensive JIT explanation.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Frictionless extrusion of 50 → 20 mm billet: true strain, average true strain (rate), ideal force and work. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • True strain ε = ln(A₀/A_f) = ln 6.25 = 1.833
    • Ideal pressure p = Yε ≈ 550 MPa; force F = pA₀ ≈ 1.08 MN
    • Work W = YVε ≈ 324 kJ (= F × stroke)
    • Average true strain rate using speed: ε̇ ≈ 6V ln R/D₀ ≈ 66 s⁻¹ (≈ 70.5 s⁻¹ by full formula)

    Loses marks

    • Using engineering strain
    • Ignoring the given speed without comment

    Earns more

    • Assumptions (volume constancy, no friction, no hardening)

    Extra mark

    • Comment that real force exceeds ideal due to friction and redundant work
  2. (b) np-chart for dissatisfied customers (n = 300, 20 samples), revised limits, target 2% limits and recommendations. 20 marks

    analyse— intro → causes → effects → stakeholders/linkages → way forward

    Must cover

    • p̄ = 184/6000 = 0.0307; CL = 9.2; UCL ≈ 18.16, LCL ≈ 0.24
    • Sample 12 (19) out of control; revised CL ≈ 8.68, UCL ≈ 17.40, LCL = 0
    • Target limits: CL = 6, UCL ≈ 13.27, LCL = 0
    • Conclusion: goal not met (process ≈ 2.9%); improvement actions

    Loses marks

    • Using c-chart without justification
    • Not revising limits

    Earns more

    • Equivalent p-chart limits

    Extra mark

    • Control chart sketch
  3. (c) Types of waste in JIT and their connection with flow layout. 10 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Seven wastes: overproduction, waiting, transport, over-processing, inventory, motion, defects (+ unused talent)
    • Flow layout: equipment arranged in process sequence
    • How flow layout reduces transport, motion, waiting, inventory
    • Links to one-piece flow, pull, visual control, quick defect detection

    Loses marks

    • Listing wastes without the layout link

    Earns more

    • Cellular manufacturing example

    Extra mark

    • Value-stream mapping
What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

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