Non-Traditional Manufacturing Processes (NTMP) Assignment - ElectroChemical Machining (ECM) - MRR

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Question: Prove that Volumetric Material Removal Rate (MRR) for an alloy with ‘n’ elements by Electrochemical Machining is given by: 1 MRR 100 n i i i i I F x M      where, xi is weight percent composition of ith element in the alloy, νi is the valency of ith element, Mi is the atomic weight of ith element in the alloy, ρ is the density of the material, F is the Faraday’s constant, I is the current in the machining process. (Hint: Use Faraday’s Laws) Solution: Electrochemical Machining is governed by Faraday’s laws. The first law states that the amount of electrochemical dissolution is proportional to amount of charge passed through the electrochemical cell, which may be expressed as: m  Q The second law states that the amount of material deposited or dissolved proportional to the Electrochemical Equivalence (ECE) of the material. Thus, m ECE A    m QA ItA m ItA F         where, F = Faraday’s constant Now, MRR m t MRR IA F      Now for passing a current of I for a time t, the mass of material dissolved for any element ‘i’ is given by 100 a i i m x    where Γa is the total volume of alloy dissolved. 100 i i i i i i i i a i i i i m QM F Q F m M Q F x M           1 1 1 1 1 1 100 100 100 Now, 100 100 n T i i n n a i i i i T a i i i i a n i i i i a n i i a i i n i i i i Q It Q Q It F x F x M M It F x M MRR t It F x MRR M t t MRR I F x M                                       

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Assignment: (Dead Line – 13th March 2015) Question: Prove that Volumetric Material Removal Rate (MRR) for an alloy with ‘n’ elements by Electrochemical Machining is given by: 1 MRR 100 n i i i i I F x M      where, xi is weight percent composition of ith element in the alloy, νi is the valency of ith element, Mi is the atomic weight of ith element in the alloy, ρ is the density of the material, F is the Faraday’s constant, I is the current in the machining process. (Hint: Use Faraday’s Laws) Solution: Volumetric Material removal rate (MRR) is the volume of material removed from the job material per unit time during the machining. In ECM, material removal takes place due to atomic dissolution of work material. Electrochemical dissolution is governed by Faraday’s laws of electrolysis. The first law states that the amount of chemical change i.e., the amount of electrochemical dissolution or deposition is proportional to amount of charge passed through the electrochemical cell that is again the product of current and the time passed, which may be expressed as: m  Q  It …. eqn (1) where m = mass of material dissolved or deposited; and Q = amount of charge passed The second law states that the amount of material deposited or dissolved further depends on Electrochemical Equivalence (ECE) of the material that is again the ratio atomic weight (M) and valency (ν). Thus, m ECE M    …. eqn (2) From equations (1) & (2), we have, m ItM m ItM F      where, F = Faraday’s constant = 96500 coulombs Now, Volume removed / Time taken MRR m m t t ItM MRR F IM t F MRR IM F                     where, I = current, ρ = density of the material Let us assume there are ‘n’ elements in an alloy. The atomic weights are given as Mi (M1, M2,…Mn) with valency during electrochemical dissolution as νi (ν1, ν2,…νn). The weight percentages of different elements are xi (x1, x2,…xn). Now for passing a current of I for a time t, the mass of material dissolved for any element ‘i’ is given by 100 a i i m V x   where Va is the total volume of alloy dissolved. Each element present in the alloy takes a certain amount of charge to dissolve.  /100 100 i i i i i i i i i a i i i a i i i i m QM F Q F m M F V x Q M Q FV x M              The total charge (QT) passed is the algebraic sum of charge passed of individual elements in the alloy. So, we can write, 1 1 1 1 1 1 1 100 100 100 100 Now, Total volume dissolved Total time 100 100 100 n T i i n n a i i i i T a i i i i n i i a i i a n i i i i a n i i a i i n i i i i Q It Q Q It FV x V F x M M It V F x M V It F x M MRR V t It F x MRR V M I t t F x M MRR                                             1 n i i i i I F x M     Thus, Volumetric Material Removal Rate (MRR) for an alloy with ‘n’ elements by Electrochemical Machining is given by: 1 MRR 100 n i i i i I F x M     
Module 9 Non conventional Machining Version 2 ME, IIT Kharagpur Lesson 38 Electro Chemical Machining Version 2 ME, IIT Kharagpur Instructional Objectives (i) Identify electro-chemical machining (ECM) as a particular type of non-tradition processes (ii) Describe the basic working principle of ECM process (iii) Draw schematically the basics of ECM (iv) Draw the tool potential drop (v) Describe material removal mechanism in ECM (vi) Identify the process parameters in ECM (vii) Develop models for material removal rate in ECM (viii) Analyse the dynamics of ECM process (ix) Identify different modules of ECM equipment (x) List four application of ECM (xi) Draw schematics of four such ECM applications 1. Introduction Electrochemical Machining (ECM) is a non-traditional machining (NTM) process belonging to Electrochemical category. ECM is opposite of electrochemical or galvanic coating or deposition process. Thus ECM can be thought of a controlled anodic dissolution at atomic level of the work piece that is electrically conductive by a shaped tool due to flow of high current at relatively low potential difference through an electrolyte which is quite often water based neutral salt solution. Fig. 1 schematically shows the basic principle of ECM. In ECM, the workpiece is connected to the positive terminal of a low voltage high current DC generator or power source. The tool is shaped and shape of the tool is transferred to the workpiece. The tool is connected to the negative terminal. Machining takes place due to anodic dissolution at atomic level of the work material due to electrochemical reaction. A gap between the tool and the workpiece is required to be maintained to proceed with steady state machining. W O R K T O O L WO R K T O O L electrolyte Fig. 1 Schematic principle of Electro Chemical Machining (ECM) 2. Process During ECM, there will be reactions occurring at the electrodes i.e. at the anode or workpiece and at the cathode or the tool along with within the electrolyte. Let us take an example of machining of low carbon steel which is primarily a ferrous alloy mainly containing iron. For electrochemical machining of steel, generally a neutral salt solution of sodium chloride (NaCl) is taken as the electrolyte. The electrolyte and water undergoes ionic dissociation as shown below as potential difference is applied NaCl ↔ Na+ + Cl- H2O H↔+ + (OH)- Version 2 ME, IIT Kharagpur As the potential difference is applied between the work piece (anode) and the tool (cathode), the positive ions move towards the tool and negative ions move towards the workpiece. Thus the hydrogen ions will take away electrons from the cathode (tool) and from hydrogen gas as: 2H+ + 2e- = H2↑ at cathode Similarly, the iron atoms will come out of the anode (work piece) as: Fe = Fe+ + + 2e- Within the electrolyte iron ions would combine with chloride ions to form iron chloride and similarly sodium ions would combine with hydroxyl ions to form sodium hydroxide Na+ + OH- = NaOH In practice FeCl2 and Fe(OH)2 would form and get precipitated in the form of sludge. In this manner it can be noted that the work piece gets gradually machined and gets precipitated as the sludge. Moreover there is not coating on the tool, only hydrogen gas evolves at the tool or cathode. Fig. 2 depicts the electro-chemical reactions schematically. As the material removal takes place due to atomic level dissociation, the machined surface is of excellent surface finish and stress free. W O R K T O O L Fe++ Fe++ OH−OH−OH−OH− Na+Na+Cl-Cl- H+H+OH−Fe++Fe(OH)2 FeCl2 e e H2 Fig. 2 Schematic representation of electro-chemical reactions The voltage is required to be applied for the electrochemical reaction to proceed at a steady state. That voltage or potential difference is around 2 to 30 V. The applied potential difference, however, also overcomes the following resistances or potential drops. They are: • The electrode potential • The activation over potential • Ohmic potential drop • Concentration over potential • Ohmic resistance of electrolyte Fig. 3 shows the total potential drop in ECM cell. Version 2 ME, IIT Kharagpur Anodic overvoltage Anode potential Activation over potential ohmic potential concentration potential concentration potential Ohmic drop activation overpotentialanode cathode cathodic potential cathodic overpotential Voltage Voltage Fig. 3 Total potential drop in ECM cell 3. Equipment The electrochemical machining system has the following modules: • Power supply • Electrolyte filtration and delivery system • Tool feed system • Working tank Fig. 4 schematically shows an electrochemical drilling unit. Flow control valve Pressure relief valve Pump Filters sludge centrifuge Spent electrolyte Tool Flow meter Pressure gauge Constant feed to the tool Low voltage high current power supply +ve -ve PS O M M O P Fig. 4 Schematic diagram of an electrochemical drilling unit Version 2 ME, IIT Kharagpur 4. Modelling of material removal rate Material removal rate (MRR) is an important characteristic to evaluate efficiency of a non-traditional machining process. In ECM, material removal takes place due to atomic dissolution of work material. Electrochemical dissolution is governed by Faraday’s laws. The first law states that the amount of electrochemical dissolution or deposition is proportional to amount of charge passed through the electrochemical cell, which may be expressed as: mQ∝, where m = mass of material dissolved or deposited Q = amount of charge passed The second law states that the amount of material deposited or dissolved further depends on Electrochemical Equivalence (ECE) of the material that is again the ratio atomic weigh and valency. Thus of thet νααAECEm Thus ναQAm where F = Faraday’s constant = 96500 coulombs νFItAm=∴ ρνρFIAtmMRR==∴ where I = current ρ= density of the material The engineering materials are quite often alloys rather than element consisting of different elements in a given proportion. Let us assume there are ‘n’ elements in an alloy. The atomic weights are given as A1, A2, ………….., An with valency during electrochemical dissolution as ν1, ν2, …………, νn. The weight percentages of different elements are α1, α2, ………….., αn (in decimal fraction) Now for passing a current of I for a time t, the mass of material dissolved for any element ‘i’ is given by iaimραΓ= where Γa is the total volume of alloy dissolved. Each element present in the alloy takes a certain amount of charge to dissolve. iiiiFAQmν= iiiiAFmQν=⇒ iiiaiAFQνραΓ=⇒ The total charge passed Version 2 ME, IIT Kharagpur Σ==iTQItQ Σ==∴iiiaTAFItQναρΓ Now Σ=Γ=iia IFtMRRναρ.1 5. Dynamics of Electrochemical Machining ECM can be undertaken without any feed to the tool or with a feed to the tool so that a steady machining gap is maintained. Let us first analyse the dynamics with NO FEED to the tool. Fig. 5 schematically shows the machining (ECM) with no feed to the tool and an instantaneous gap between the tool and workpiece of ‘h’. job electrolyte tool dh h Fig. 5 Schematic representation of the ECM process with no feed to the tool Now over a small time period ‘dt’ a current of I is passed through the electrolyte and that leads to a electrochemical dissolution of the material of amount ‘dh’ over an area of S rhVssrhVRVI===∴ then ⎟⎠⎞⎜⎝⎛=s1.rhVsA.F1dtdhxxρν Version 2 ME, IIT Kharagpur rhV.A.F1xxρν= for a given potential difference and alloy hch1.rFVAdtdhxx==ρν where c = constant rFVAxxρν= Σ=iiiArFVcναρ hcdtdh=∴ cdthdh= At t = 0, h = ho and at t = t1 and h = h1 ∫∫=∴1hoht0dtchdh ct2hh2o21=−∴ That is the tool – workpiece gap under zero feed condition grows gradually following a parabolic curve as shown in Fig. 6 h t ho Fig. 6 Variation of tool-workpiece gap under zero feed condition As hcdtdh= Thus dissolution would gradually decrease with increase in gap as the potential drop across the electrolyte would increase Version 2 ME, IIT Kharagpur Now generally in ECM a feed (f) is given to the tool fhcdtdh−=∴ Now if the feed rate is high as compared to rate of dissolution, then after sometime the gap would diminish and may even lead to short circuiting. Under steady state condition the gap is uniform i.e. the approach of the tool is compensated by dissolution of the work material. Thus with respect to the tool, the workpiece is not moving Thus fhc0dtdh=== hcf=∴ or h* = steady state gap = c/f Now under practical ECM condition it is not possible to set exactly the value of h* as the initial gap. Thus it is required to be analysed if the initial gap value would have any effect on progress of the process Now fhcdtdh−= Now chf*hh'h== And ctf*hft't2== dtdh.f1dtdh.c/fc/f'dt'dh2==∴ Thus fhcdtdh−= fc'hcff*h'hc'dt'dhf−=−=⇒ ⎟⎠⎞⎜⎝⎛−=⇒'h'h1f'dt'dhf 'h'h1'dt'dh−=⇒ 'dh'h1'h'dt−=∴ Now integrating between t’ = 0 to t’ = t’ when h’ changes from ho’ to h1’ 'dh'h1'h'dt't0'1h'oh∫∫−=∴ ()()()∫−+∫−−−=∴'1h'oh'1h'oh'h1d'h1'h1d't 1h1hlnhh't'1'o'1'o−−+−= now for different value of ho’, h1’ seems to approach 1 as shown in Fig. 7 Version 2 ME, IIT Kharagpur h1' t' h0= 0 h0= 0.5 1 Simulation for ho'= 0, 0.5, 1, 2, 3, 4, 5 Fig. 7 Variation in steady state gap with time for different initial gap Thus irrespective of initial gap 1cfh1*hh'h=⇒== fch=∴ or h1.rFVAhcfxxρν== si.FArhV.FAfxxxxρνρν==∴ s/mminMRRsI.FAfxx==∴ρν Thus it seems from the above equation that ECM is self regulating as MRR is equal to feed rate. 6. Applications ECM technique removes material by atomic level dissolution of the same by electrochemical action. Thus the material removal rate or machining is not dependent on the mechanical or physical properties of the work material. It only depends on the atomic weight and valency of the work material and the condition that it should be electrically conductive. Thus ECM can machine any electrically conductive work material irrespective of their hardness, strength or even thermal properties. Moreover Version 2 ME, IIT Kharagpur as ECM leads to atomic level dissolution, the surface finish is excellent with almost stress free machined surface and without any thermal damage. ECM is used for • Die sinking • Profiling and contouring • Trepanning • Grinding • Drilling • Micro-machining Die sinking 3D profiling Work Tool Fig. 8 Different applications of Electro Chemical Machining drilling (drilling) work tool Version 2 ME, IIT Kharagpur trepanning toolwork Fig. 9 Drilling and Trepanning by ECM 7. Process Parameters Power Supply Type direct current Voltage 2 to 35 V Current 50 to 40,000 A Current density 0.1 A/mm2 to 5 A/mm2 Electrolyte Material NaCl and NaNO3 Temperature 20oC – 50oC Flow rate 20 lpm per 100 A current Pressure 0.5 to 20 bar Dilution 100 g/l to 500 g/l Working gap 0.1 mm to 2 mm Overcut 0.2 mm to 3 mm Feed rate 0.5 mm/min to 15 mm/min Electrode material Copper, brass, bronze Surface roughness, Ra 0.2 to 1.5 μm Version 2 ME, IIT Kharagpur Quiz Test 1. For ECM of steel which is used as the electrolyte (a) kerosene (b) NaCl (c) Deionised water (d) HNO3 2. MRR in ECM depends on (a) Hardness of work material (b) atomic weight of work material (c) thermal conductivity of work material (d) ductility of work material 3. ECM cannot be undertaken for (a) steel (b) Nickel based superalloy (c) Al2O3 (d) Titanium alloy 4. Commercial ECM is carried out at a combination of (a) low voltage high current (b) low current low voltage (c) high current high voltage (d) low current low voltage Problems 1. In electrochemical machining of pure iron a material removal rate of 600 mm3/min is required. Estimate current requirement. 2. Composition of a Nickel superalloy is as follows: Ni = 70.0%, Cr = 20.0%, Fe = 5.0% and rest Titanium Calculate rate of dissolution if the area of the tool is 1500 mm2 and a current of 2000 A is being passed through the cell. Assume dissolution to take place at lowest valency of the elements. ANi = 58.71 ρNi = 8.9 νNi = 2 ACr = 51.99 ρCr = 7.19 νCr = 2 AFe = 55.85 ρFe = 7.86 νFe = 2 ATi = 47.9 ρTi = 4.51 νTi = 3 3. In ECM operation of pure iron an equilibrium gap of 2 mm is to be kept. Determine supply voltage, if the total overvoltage is 2.5 V. The resistivity of the electrolyte is 50 Ω-mm and the set feed rate is 0.25 mm/min. Version 2 ME, IIT Kharagpur Answers Answers to Quiz Test 1 – (b) 2 – (b) 3 – (c) 4 – (a) Solution to Prob. 1 νFAItmmMRR.=== ρνρΓFAItmMRR.===∴ MRR = 600 mm3/min = 600/60 mm3/s = 10 mm3/s = 10x10-3cc/s 2x8.7x96500xI5610x103=∴− As AFe = 56 νFe = 2 F = 96500 coulomb ρ = 7.8 gm/cc 562x8.7x10x10x96500I3−=∴ I = 268.8 A Answer Solution of Problem 2 Now, Σ=iialloy1ραρ TiTiFeFeCrCrNiNi1ραραραρα+++= cc/gm07.851.405.086.705.019.72.09.87.01=+++= Now Σ==iiiAFItmMRRναρρ ⎭⎬⎫⎩⎨⎧+++=9.473x05.085.552x05.099.512x2.071.582x75.0x07.8x965001000 = 0.0356 cc/sec = 2.14 cc/min = 2140 mm3/min Version 2 ME, IIT Kharagpur min/mm43.115002140AreaMRRndissolutioofRate===∴ answer Solution to Prob. 3 fc*h= where FeFeFerFVAcνρ= ()2x50x10x8.7x9650085.55x5.2VC3−−= ()7.13475.2V−= ()6025.0x13475.2Vfc2*h−=== 615.55.2V2−= .Volt73.8V=∴ Answer Version 2 ME, IIT Kharagpur

Design Of Machine Elements (ME30602) - Class test Question Bank with answers

INDIAN INSTITUTE OF TECHNOLOGY, KHARAGPUR

End/Mid Semester Examination

DESIGN OF MACHINE ELEMENTS

ME30602

Class Test
1. Answer all questions. All questions carry equal marks
2. Make suitable assumptions wherever necessary
3. Put √ on your chosen answer.
4. Any data, if not furnished, may be assumed.

1.     A new model of a car is a typical example of

a)     Adaptive design

b)    Developmental design

c) New design

2.     Adaptive design is based on

a)     Existing design adapted for new application

b)    New inventions for new application

c)     Conceptual design for new application

d)    New Technology for new application

3.     Which of the following design is based on based on experience and experiments?

a)     Rational design

b)    Empirical design

c)     Industrial design

4.     Use of factor of safety in designing the elements is to

a)     optimize the design to avoid over-design for reliability

b)    prevent loss of material and premature failure

c)     have good aesthetics of the designed product

d)    increase efficiency of the element

5.     A stainless steel is described as 18/8 steel. This indicates that it contains:

a)     8% Nickel and 18% Chromium

b)    18% Nickel and 8% Chromium

c)     18% Chromium and 8% Silicon

d)    8% Chromium and 18% Silicon

6.     Resilience is

a)     the property of the material that enables it to resist shock and impact by storing energy

b)    the property of the material that enables it to be twisted, bent or stretched under impact load or high stress before rupture

c)     the property of the material that enables it to be drawn out or elongated to an appreciable extent before rupture

d)    the property of the material that enables it to resist permanent deformation, penetration, indentation

7.     Resilience is the property of a material:

a)     that enables it to resist indentation

b)    that enables it to resist wear

c)     that enables it to resist shock and impact

d)    that enables it to resist scratching

8.     A standard alloy steel used for making engineering components is 20Cr18Ni2. The composition of this steel is:

a)     20% carbon, 18% chromium and 2% nickel

b)    0.2% chromium, 18% nickel and 2% carbon

c)     0.2% carbon, 18% chromium and 2% nickel

d)    20% chromium, 18% nickel and 2% carbon

9.     The composition of 20Cr18Ni2 is:

a)     20% Cr, 18%Ni, 2%Carbon

b)    0.2% carbon, 18% chromium and 2% nickel

c)     2% Cr, 0.18%Ni, 0.2%Carbon

d)    2% carbon, 0.18% chromium and 0.2% nickel

10.                        Allowance of a shaft-hole system is the difference between

a)     maximum dimension of shaft and minimum dimension of hole

b)    maximum dimension of shaft and maximum dimension of hole

c)     minimum dimension of shaft and minimum dimension of hole

d)    minimum dimension of shaft and maximum dimension of hole

11.                         Tolerance is the difference between:

a)     Maximum and minimum dimensions of a component

b)    Maximum dimension of the shaft and maximum dimensions of the hole

c)     Minimum and minimum dimensions of a component

d)    Minimum dimension of the shaft and the maximum dimension of the hole

12.                        In a unilateral system of tolerance, the tolerance is allowed on

a)     one side of the actual size

b)    one side of the nominal size

c)     both sides of the actual size

d)    both sides of the nominal size

13.                        Nominal size of a hole and shaft is 40 mm, the tolerance grade for making the hole is IT6 and for shaft is IT5. The fundamental deviation of shaft basis is positive and for hole is zero. Which of the following designation suits the above hole-shaft system?

a)     40M6/h5

b)    40M5/h6

c)     40H5/m6

d)    40H6/m5

14.                        Which of the following standard series is used for Capacities of hydraulic cylinder?

a)     R5

b)    R10

c)     R20

d)    R40

15.                         What is the bearing stress developed at the faces of a rectangular key fixing a gear hub on a shaft acted upon by a torque T as shown in figure?

                                                                                                       

16.                         Let the cross-sectional area of the plate, the larger hole H₁ and the smaller holes H₂ be A, a₁, a₂ respectively. If 2a₂ > a₁ , the critical section in the loaded plate is:

a)     A-A

b)    B-B

c)     C-C

d)    D-D

17.                         A 3m long cantilever beam of solid rectangular cross-section of 100mm width and 150mm depth is subjected to an end loading P as shown in the figure. If the allowable shear stress in the beam is 150 MPa, the safe value of P based on shear alone is

a)     1 MN

b)    1.5 MN

c)     2.3 MN

d)    3.5 MN

18.                         A vertical load P_y = 20 kN is applied at the free end of a cylindrical bar of radius 50 mm as shown in figure. The principal stress at the point A is

a)     56MPa, 7.4MPa

b)    84MPa, −8.9MPa

c)     105MPa, 9.3MPa

d)    112.5MPa, −14.72MPa

19.                         An axially loaded brass (E_brass = 90GPa) strut hinged at both ends is 1m long and is of a square cross-section of sides 20mm. The dimension of steel (E_steel = 200 GPa) strut of the same length and subjected to the same axial loads is:

a)     24.40 mm

b)    13.42 mm

c)     16.38 mm

d)    32.62 mm

20.                         The theoretical stress concentration factor at hole in a plate with transverse elliptical hole and subjected to a tensile load as shown in figure is:

                                                                                                             

21.                         Notch sensitivity factor, q is

                                                                                                     

22.                         The endurance or fatigue limit is defined as the maximum value of the stress which a polished standard specimen can withstand without failure, for infinite number of cycles, when subjected to

a)     static load

b)    dynamic load

c)     static as well as dynamic load

d)    completely reversed load

23.                         In the plate shown below it is required that the stress concentration at hole does not exceed that at the fillet. The hole diameter is

a)     21.5 mm

b)    35 mm

c)     41.9 mm

d)    53 mm

Data Provided:

24.                         Dowel pins are used for

a)     proper alignment

b)    loosening

c)     both for proper alignment and loosening

25.                         Dowel Pins are used

a)     To get a better grip

b)    To transmit torque smoothly

c)     For proper alignment

26.                         Studs are

a)     single headed fasteners

b)    double headed fasteners

c)     headless fasteners

d)    twin headed fasteners

27.                         A heat treated steel shaft of tensile yield strength of 350 MPa has a diameter of 50mm. The shaft rotates at 1000 rpm and transmits 100 kW through a gear. An appropriate key set of dimensions is:

a)     Width = 16 mm, Length = 45 mm, Depth = 10 mm

b)    Width = 10 mm, Length = 35 mm, Depth = 8 mm

c)     Width = 18 mm, Length = 55 mm, thickness = 11 mm

d)    Width = 20 mm, Length = 65 mm, thickness = 12 mm

Data Provided:

Shaft Diameter (mm)	30-38	38-44	44-50	50-58	58-65	65-75	75-85
Key width, w (mm)	10	12	14	16	18	20	22
Key depth, t (mm)	8	9	9	10	11	12	14
Key length, L (mm)	22-110	28-140	36-160	45-180	50-200	56-220	63-250

28.                         In a steam engine the steam pressure is 2 MPa and the cylinder diameter is 250 mm. The contact surfaces of the head and cylinder are ground and no packing is required. Assuming that the number of bolts to be used is 12, a suitable bolt for the joint to be leak proof would be:

a)     M20 x 1.5

b)    M16 x 1.5

c)     M5 x 0.8

d)    M10 x 1.25

Data provided:

Designation	Pitch (mm)	Minor Diameter		Stress area (mm2)
		Bolt (mm)	Nut (mm)
M2	0.40	1.509	1.567	207
M5	0.8	4.019	4.134	14.2
M10	1.25	8.466	8.647	61.6
M16	1.5	14.160	14.376	167
M20	1.5	18.160	18.376	272
M24	2	21.546	21.835	384

29.                         Shear stress developed in the weld in the figure shown below is

a)     16.7 MPa

b)    21.9 MPa

c)     26.2 MPa

30.                         Considering unit weld width the polar moment of inertia of the weld group in above question is

a)     0.00365 m³

b)    0.000198 m³

c)     0.0000325 m³

31.                         If the yield strength of the weld material is 150MPa, the safe value of F is

a)     108 kN

b)    213 kN

c)     269 kN

d)    313 kN

32.                         In a triple riveted butt joint with two unequal straps 30 mm diameter rivets and 10mm plates (both main and straps) are used. The pitch at the outer row is 250mm. Considering that the joint fails only due to shearing of all the rivets and taking the working stresses to be σ_t = 75MPa, τ_y = 60MPa and σ_c = 130MPa, the joint efficiency is

a)     76.3%

b)    81.3%

c)     98.3%

33.                         Two plates of 7 mm thickness are connected by a double riveted lap joint of zigzag pattern. If the rivet diameter is 19mm and σ_t = 90MPa, σ_c = 120MPa, τ = 60 MPa, the rivet pitch is

a)     19 mm

b)    54 mm

c)     73 mm

d)    102 mm

34.                         According to Unwin's formula, the diameter of the rivet hole (in mm) can be obtained by:                   (t = thickness of the plate in mm)

                                                                                                         

35.                         Wahl correction factor takes care of:

a)     curvature effect

b)    shear stress correction

c)     both curvature effect and shear stress correction

36.                         In spring design, Wahl correction factor

a)     takes care of the direct shear stress in springs

b)    takes care of the curvature effect in springs

c)     takes care of the direct stress together with the curvature effect in springs

37.                         In a close coiled helical spring, the spring index (C) is given by D/d where D and d are the mean coil diameter and wire diameter respectively. For considering the effect of curvature, the Wahl’s stress factor K is given by

                                                                                       

38.                         Which of the following threads is used for transmission of power only in one direction?

a)     Acme thread

b)    Knuckle thread

c)     Buttress thread

d)    Square thread

39.                         V-threads are normally used for

a)     power transmission

b)    for securing two components

c)     not normally used

40.                         A bolt of M 24 × 2 means that

a)     the pitch of the thread is 24 mm and depth is 2 mm

b)    the cross-sectional area of the threads is 24 mm² and number of starts is 2

c)     the nominal diameter of bolt is 24 mm and the pitch is 2 mm

d)    the effective diameter of the bolt is 24 mm and there are two threads per cm

41.                         The shear stress developed in the centre bolt (no-2) is

a)     35.3 MPa

b)    43.6 MPa

c)     6.4 MPa

42.                         For zero collar friction, the efficiency of a square-thread screw is:

(λ is lead angle and f is coefficient of friction of the screw)

                                                                                     

43.                         A single square thread power screw is to raise a load of 50 kN. A screw thread of major diameter of 34 mm and a pitch of 6 mm is used. The coefficient of friction at the thread and collar are 0.15 and 0.1 respectively. If the collar frictional diameter is 100 mm and the screw turns at a speed of 1 rev per second. The power input to the screw is

a)     1549 watts

b)    2300 watts

c)     2614 watts

d)    3167 watts

44.                         In the above question, the combined efficiency of the screw and collar is

a)     11.5%

b)    22.3%

c)     33.5%

d)    44.3%

45.                         The machine shown in the figure can be used for a tension test but not for a compression test. Should both screws have the same hand?

a)     Yes

b)    No, screws should always have different hands

c)     No, screws can have either same or different hands

46.                         In the machine shown below, both the screws have the same hand

a)     the machine can be used for a compression test only

b)    the machine can be used for a tension test only

c)     the machine can be used for both tension and compression tests

47.                         The purpose of Rebound clips in Laminated semi-elliptic spring is:

a)     to share the load from the master leaf to the graduated leaf

b)    to hold the leaves of the spring

c)     to prevent the deflected spring from touching the attached machine member

d)    to attach the spring with another machine member

48.                         The purpose of central clamp in Laminated semi-elliptic spring is:

a)     to share the load from the master leaf to the graduated leaf

b)    to hold the leaves of the spring

c)     to prevent the deflected spring from touching the attached machine member

d)    to attach the spring with another machine member

49.                         The purpose of camber in Laminated semi-elliptic spring is:

a)     to share the load from the master leaf to the graduated leaf

b)    to hold the leaves of the spring

c)     to prevent the deflected spring from touching the attached machine member

d)    to attach the spring with another machine member

50.                         Which of the following methods reduces additional stresses in laminated spring hinged at supports?

a)     Master leaf is made of stronger material than the other leaves

b)    Master leaf is made thinner than the other leaves

c)     Increase the radius of curvature of the master leaf than the next leaf

d)    All the above

51.                         Nipping of leaf spring

a)     Increases the stress in master leaf

b)    Decreases the stress in master leaf

c)     Doesn’t affect the stress in master leaf

52.                         A helical spring of wire diameter 6mm and spring index 6 is acted by an initial load of 800N. After compressing it further by 10mm the stress in the wire is 500MPa. Taking G = 84000MPa the number of active coils is

a)     13 turns

b)    21 turns

c)     27 turns

d)    32 turns

53.                         For a simple cantilever type leaf spring with uniform strength the leaf thickness can be given by

a)     σ_max L^3/2/(E δ_max^1/2)

b)    σ_max L³/(E δ_max²)

c)     σ_max L²/(E δ_max)

d)    σ_max L⁴/(E δ_max³)

Where δ_max is the spring deflection, L is the characteristic length, σ_max is the design stress or maximum stress and E is the Modulus of elasticity

54.                         Thermoplastic materials are those materials which

a)     are formed into shape under heat and pressure and results in a permanently hard product

b)    do not become hard with the application of heat and pressure and no chemical change occurs

c)     are flexible and can withstand considerable wear under suitable conditions

d)    are used as a friction lining for clutches and brakes

55.                         A cotter joint is used to transmit

a)     axial tensile load only

b)    axial compressive load only

c)     combined axial and twisting loads

d)    axial tensile or compressive loads

56.                         To ensure self-locking in a screw jack, it is essential that the helix angle is

a)     larger than friction angle

b)    smaller than friction angle

c)     equal to friction angle

d)    such as to give maximum efficiency in lifting

57.                         The figure shows a clamp coupling. If the effective number of bolts in each shaft is n/2, the allowable tensile stress for the bolt material is σ_t and the root diameter of the bolt is d_c, then the clamping pressure between the shaft and the sleeve is

a)     nπd_c²σ_t/(dL)

b)    nπd_c²σ_t/(4dL)

c)     3nπd_c²σ_t/(4dL)

58.                         In a typical clamp coupling the clamping pressure between the shaft and the sleeve is given by

a)     nπd_c²σ_t/(dL)

b)    0.5nπd_c^3/2σ_t/(dL^1/2)

c)     0.75nπd_c⁴σ_t/(dL)²

d)    0.25nπd_c²σ_t/(dL)

Where d_c is the core diameter of the bolts, d is the clamp diameter, L the clamp length and σ_t is the tensile yield stress.

59.                         In a disc clutch, maximum pressure occurs

a)     at the inner radius

b)    at the centre

c)     at the outer radius

60.                         In which of the following, uniform pressure theory is more approximate?

a)     a new clutch

b)    an old clutch

c)     friction clutch

61.                         For a self-locking brake

a)     A large actuating force is needed

b)    A small actuating force is needed

c)     No actuating force is needed

62.                         If in the band brake shown below, drum diameter = 400mm, lining width = 75mm, rotational speed is 200 rpm, a = 250mm, m = 75mm, α = 2700, μ = 0.2 and the maximum lining pressure is 0.5 MPa, the torque is

a)     632 Nm

b)    915 Nm

c)     1150 Nm

d)    3160 Nm

63.                         Hydrodynamic Lubrication occurs at:

a)     Very low Sommerfeld Number

b)    Low Sommerfeld Number

c)     Medium Sommerfeld Number

d)    High Sommerfeld Number

64.                         Hydrodynamic Lubrication occurs at:

a)     At zero Sommerfeld number

b)    At relatively low Sommerfeld number

c)     At relatively high Sommerfeld number

65.                         Hydrodynamic lubricated bearings are

a)     Thick film bearings

b)    Thin film bearings

c)     Zero film bearings

66.                         When a shaft rotates in anticlockwise direction at slow speed in a bearing, then it will

a)     have contact at the lowest point of bearing

b)    move towards right of the bearing making metal to metal contact

c)     move towards left of the bearing making metal to metal contact

d)    move towards right of the bearing making no metal to metal contact

67.                         If Z = Absolute viscosity of the lubricant in kg/m-s, N = Speed of the journal in rpm, and p = Bearing pressure in N/mm², then the bearing characteristic number is

a)     ZN/p

b)    Zp/N

c)     Z/Pn

d)    pN/Z

68.                         A ball bearing subjected to a radial load of 5 kN is expected to have a life of 8000 hours at 1450 rpm with a reliability of 99%. The dynamic load capacity of the bearing so that it can be selected from the manufacturer's catalogue based on a reliability of 90% is

a)     86.5 kN

b)    68.5 kN

c)     58.6 kN

d)    1 kN

69.                         The ball bearings are usually made from

a)     low carbon steel

b)    medium carbon steel

c)     high speed steel

d)    chrome nickel steel

70.                         The tapered roller bearings can take

a)     radial load only

b)    axial load only

c)     both radial and axial loads

71.                         The listed life of a rolling bearing, in a catalogue, is the

a)     minimum expected life

b)    maximum expected life

c)     average life

72.                         Which of the following are called as Anti-friction bearings?

a)     Fluid Film bearings

b)    Rolling contact bearings

c)     Sliding contact bearings

d)    Journal bearings

73.                         The load rating of a bearing is given for

a)     Axial loads only

b)    Radial loads only

c)     Both axial and radial loads

74.                         If two groups of identical ball bearings are tested under loads P₁ and P₂ for respective lives of L₁ and L₂ , then L₁/L₂ is

a)     (P₁/P₂)^1/3

b)    (P₂/P₁)^1/3

c)     (P₁/P₂)³

d)    (P₂/P₁)³

75.                         Which of the following is best suitable for high speeds of shaft and high loads?

a)     Journal bearing

b)    Ball bearing

c)     Roller bearing

76.                         Can a block moving over a constant height fluid film carry load?

a)     Yes

b)    No

c)     Depends upon viscosity of the fluid

77.                         If Sommerfeld number increases, then the Eccentricity ratio

a)     Increases

b)    Decreases

c)     Remains same

78.                         If Sommerfeld number increases, then the minimum film thickness

a)     Increases

b)    Decreases

c)     Remains same

79.                         If Sommerfeld number increases, then the ratio of side flow to total flow

a)     Decreases

b)    Increases

c)     Remains same

80.                         If Sommerfeld number increases, then the coefficient of friction

a)     Decreases

b)    Increases

c)     Remains same

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