Machine Design
Short Description
machine design...
Description
GATE SOLVED PAPER - ME MACHINE DESIGN
YEAR 2013
ONE MARK
Q. 1
A metric threads of pitch 2 mm and thread angle 60c is inspected for its pitch diameter using 3-wire method. The diameter of the best size wire in mm is (A) 0.866 (B) 1.000 (C) 1.154 (D) 2.000
Q. 2
Two threaded bolts A and B of same material and length are subjected to identical tensile load. If the elastic strain stored in bolt A times of bolt B and the mean diameter of bolt A is 12 mm, the mean diameter of bolt B in mm is (A) 16 (B) 24 (C) 36 YEAR 2013
(D) 48 TWO MARKS
Common Data For Q.3 and 4 A single riveted lap joint of two similar plates as shown in the figure below has the following geometrical and material details.
width of the plate w = 200 mm , thickness of the plate t = 5 mm , number of rivets n = 3 , diameter of the rivet dr = 10 mm , diameter of the rivet hole dh = 11 mm , allowable tensile stress of the plate sp = 200 MPa , allowable shear stress of the rivet ss = 100 MPa and allowable stress of the rivet sc = 150 MPa . Q. 3
If the plates are to be designed to avoid tearing failure, the maximum permissible load P in kN is (A) 83 (B) 125 (C) 167 (D) 501
Q. 4
If the rivets are to be designed to avoid crushing failure, the maximum permissible load P in kN is (A) 7.50 (B) 15.00 (C) 22.50 (D) 30.00
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GATE SOLVED PAPER - ME
MACHINE DESIGN
YEAR 2012 Q. 5
A fillet welded joint is subjected to transverse loading F as shown in the figure. Both legs of the fillets are of 10 mm size and the weld length is 30 mm. If the allowable shear stress of the weld is 94 MPa, considering the minimum throat area of the weld, the maximum allowable transverse load in kN is
(A) 14.44 (C) 19.93 Q. 6
Q. 7
(A) 100.6
(B) 54.4
(C) 22.1
(D) 15.7
A solid circular shaft needs to be designed to transmit a torque of 50 Nm. If the allowable shear stress of the material is 140 MPa, assuming a factor of safety of 2, the minimum allowable design diameter is mm is (A) 8 (B) 16 (C) 24 (D) 32
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TWO MARKS
Two identical ball bearings P and Q are operating at loads 30 kN and 45 kN respectively. The ratio of the life of bearing P to the life of bearing Q is (A) 81 (B) 27 16 8 (C) 9 (D) 3 2 4 YEAR 2010
Q. 9
(B) 17.92 (D) 22.16
A force of 400 N is applied to the brake drum of 0.5 m diameter in a band-brake system as shown in the figure, where the wrapping angle is 180c. If the coefficient of friction between the drum and the band is 0.25, the braking torque applied, in Nm is
YEAR 2011 Q. 8
TWO MARKS
TWO MARKS
A band brake having band-width of 80 mm, drum diameter of 250 mm, coefficient of friction of 0.25 and angle of wrap of 270 degrees is required to exert a friction torque of 1000 Nm. The maximum tension (in kN) developed in the band is (A) 1.88 (B) 3.56 (C) 6.12 (D) 11.56
GATE SOLVED PAPER - ME
MACHINE DESIGN
A bracket (shown in figure) is rigidly mounted on wall using four rivets. Each rivet is 6 mm in diameter and has an effective length of 12 mm.
Q. 10
Direct shear stress (in MPa) in the most heavily loaded rivet is (A) 4.4 (B) 8.8 (C) 17.6 (D) 35.2 A lightly loaded full journal bearing has journal diameter of 50 mm, bush bore of 50.05 mm and bush length of 20 mm. If rotational speed of journal is 1200 rpm and average viscosity of liquid lubricant is 0.03 Pa s, the power loss (in W) will be (A) 37 (B) 74 (C) 118 (D) 237
Q. 11
YEAR 2009
TWO MARKS
A forged steel link with uniform diameter of 30 mm at the centre is subjected to an axial force that varies from 40 kN in compression to 160 kN in tension. The tensile (Su), yield (Sy) and corrected endurance (Se) strengths of the steel material are 600 MPa, 420 MPa and 240 MPa respectively. The factor of safety against fatigue endurance as per Soderberg’s criterion is (A) 1.26 (B) 1.37 (C) 1.45 (D) 2.00
Q. 12
Common Data For Q. 13 and 14 A 20c full depth involute spur pinion of 4 mm module and 21 teeth is to transmit 15 kW at 960 rpm. Its face width is 25 mm. Q. 13
The tangential force transmitted (in N) is (A) 3552 (B) 2611 (C) 1776 (D) 1305
Q. 14
Given that the tooth geometry factor is 0.32 and the combined effect of dynamic load and allied factors intensifying the stress is 1.5; the minimum allowable stress (in MPa) for the gear material is (A) 242.0 (B) 166.5 (C) 121.0 (D) 74.0
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GATE SOLVED PAPER - ME
MACHINE DESIGN
YEAR 2008
TWO MARKS
Q. 15
A journal bearing has a shaft diameter of 40 mm and a length of 40 mm. The shaft is rotating at 20 rad/s and the viscosity of the lubricant is 20 mPa-s . The clearance is 0.020 mm. The loss of torque due to the viscosity of the lubricant is approximately. (A) 0.040 N-m (B) 0.252 N-m (C) 0.400 N-m (D) 0.652 N-m
Q. 16
A clutch has outer and inner diameters 100 mm and 40 mm respectively. Assuming a uniform pressure of 2 MPa and coefficient of friction of liner material is 0.4, the torque carrying capacity of the clutch is (B) 196 N-m (A) 148 N-m (C) 372 N-m (D) 490 N-m
Q. 17
A spur gear has a module of 3 mm, number of teeth 16, a face width of 36 mm and a pressure angle of 20c. It is transmitting a power of 3 kW at 20 rev/s. Taking a velocity factor of 1.5 and a form factor of 0.3, the stress in the gear tooth is about. (A) 32 MPa (B) 46 MPa (C) 58 MPa (D) 70 MPa
Q. 18
One tooth of a gear having 4 module and 32 teeth is shown in the figure. Assume that the gear tooth and the corresponding tooth space make equal intercepts on the pitch circumference. The dimensions ‘a ’ and ‘b’, respectively, are closest to
(A) 6.08 mm, 4 mm
(B) 6.48 mm, 4.2 mm (D) 6.28 mm, 4.1 mm
(C) 6.28 mm, 4.3 mm Q. 19
Match the type of gears with their most appropriate description. Type of gear
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Description
P
Helical
1.
Axes non parallel and non intersecting
Q
Spiral Bevel
2.
Axes parallel and teeth are inclined to the axis
R
Hypoid
3.
Axes parallel and teeth are parallel to the axis
S
Rack and pinion
4.
Axes are perpendicular and intersecting, and teeth are inclined to the axis.
5.
Axes are perpendicular and used for large speed reduction
6.
Axes parallel and one of the gears has infinite radius
GATE SOLVED PAPER - ME
(A) P-2, Q-4, R-1, S-6 (C) P-2, Q-6, R-4, S-2
MACHINE DESIGN
(B) P-1, Q-4, R-5, S-6 (D) P-6, Q-3, R-1, S-5
Common Data For Q. 20 and 21 A steel bar of 10 # 50 mm is cantilevered with two M 12 bolts (P and Q) to support a static load of 4 kN as shown in the figure.
Q. 20
The primary and secondary shear loads on bolt P, respectively, are (A) 2 kN, 20 kN (B) 20 kN, 2 kN (C) 20 kN, 0 kN
Q. 21
The resultant shear stress on bolt P is closest to (A) 132 MPa (B) 159 MPa (C) 178 MPa (D) 195 MPa YEAR 2007
Q. 22
(D) 0 kN, 20 kN
ONE MARK
A ball bearing operating at a load F has 8000 hours of life. The life of the bearing, in hours, when the load is doubled to 2F is (A) 8000 (B) 6000 (C) 4000 (D) 1000 YEAR 2007
TWO MARKS
Q. 23
A thin spherical pressure vessel of 200 mm diameter and 1 mm thickness is subjected to an internal pressure varying form 4 to 8 MPa. Assume that the yield, ultimate and endurance strength of material are 600, 800 and 400 MPa respectively. The factor of safety as per Goodman’s relation is (A) 2.0 (B) 1.6 (C) 1.4 (D) 1.2
Q. 24
A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/ second carries a load of 2 kN. The lubricant used has a viscosity of 20 mPas. The radial clearance is50 mm . The Sommerfeld number for the bearing is (A) 0.062 (B) 0.125 (C) 0.250 (D) 0.785
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GATE SOLVED PAPER - ME
Q. 25
MACHINE DESIGN
A bolted joint is shown below. The maximum shear stress, in MPa in the bolts at A and B , respectively are
(A) 242.6, 42.5 (B) 42.5, 242.6 (C) 42.5, 42.5 (D) 18.75, 343.64 Q. 26
A block-brake shown below has a face width of 300 mm and a mean co-efficient of friction of 0.25. For an activating force of 400 N, the braking torque in Nm is
(A) 30 (B) 40 (C) 45 (D) 60 Q. 27
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The piston rod of diameter 20 mm and length 700 mm in a hydraulic cylinder is subjected to a compressive force of 10 kN due to the internal pressure. The end conditions for the rod may be assumed as guided at the piston end and hinged at the other end. The Young’s modulus is 200 GPa. The factor of safety for the piston rod is (A) 0.68 (B) 2.75 (C) 5.62 (D) 11.0
GATE SOLVED PAPER - ME
MACHINE DESIGN
Common Data For Q. 28 and 29 A gear set has a pinion with 20 teeth and a gear with 40 teeth. The pinion runs at 30 rev/s and transmits a power of 20 kW. The teeth are on the 20c full-depth system and have a module of 5 mm. The length of the line of action is 19 mm. Q. 28
The center distance for the above gear set in mm is (A) 140 (B) 150 (C) 160 (D) 170
Q. 29
The contact ratio of the contacting tooth is (A) 1.21 (B) 1.25 (C) 1.29
Q. 30
The resultant force on the contacting gear tooth in N is (A) 77.23 (B) 212.20 (C) 2258.1 YEAR 2006
Q. 31
(D) 1.33
(D) 289.43 TWO MARKS
A disc clutch is required to transmit 5 kW at 2000 rpm. The disk has a friction lining with coefficient of friction equal to 0.25. Bore radius of friction lining is equal to 25 mm. Assume uniform contact pressure of 1 MPa. The value of outside radius of the friction lining is (A) 39.4 mm (B) 49.5 mm (C) 97.9 mm (D) 142.9 mm
Q. 32
Twenty degree full depth involute profiled 19 tooth pinion and 37 tooth gear are in mesh. If the module is 5 mm, the centre distance between the gear pair will be (A) 140 mm (B) 150 mm (C) 280 mm (D) 300 mm
Q. 33
A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycles is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will be (A) 1071 cycles (B) 15000 cycles (C) 281914 cycles (D) 928643 cycles
Q. 34
A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN along the weld. The shear strength of the weld material is equal to 200 MPa. The factor of safety is (A) 2.4 (B) 3.4 (C) 4.8 (D) 6.8
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GATE SOLVED PAPER - ME
MACHINE DESIGN
YEAR 2005
ONE MARK
Which one of the following is criterion in the design of hydrodynamic journal bearings ? (A) Sommerfeld number (B) Rating life
Q. 35
(C) Specific dynamic capacity
(D) Rotation factor
YEAR 2005
TWO MARKS
Common Data For Q. 36 and 37 A band brake consists of a lever attached to one end of the band. The other end of the band is fixed to the ground. The wheel has a radius of 200 mm and the wrap angle of the band is 270c. The braking force applied to the lever is limited to 100 N and the coefficient of friction between the band and the wheel is 0.5. No other information is given.
Q. 36
The maximum tension that can be generated in the band during braking is (A) 1200 N (B) 2110 N (C) 3224 N
Q. 37
(D) 4420 N
The maximum wheel torque that can be completely braked is (A) 200 Nm (B) 382 Nm (C) 604 Nm (D) 844 Nm YEAR 2004
ONE MARK
Q. 38
Two mating spur gears have 40 and 120 teeth respectively. The pinion rotates at 1200 rpm and transmits a torque of 20 Nm. The torque transmitted by the gear is (A) 6.6 Nm (B) 20 Nm (C) 40 Nm (D) 60 Nm
Q. 39
In terms of theoretical stress concentration factor (Kt) and fatigue stress concentration factor (K f ), the notch sensitivity ‘q ’ is expressed as (K f - 1) (Kt - 1) (Kt - 1) (C) (K f - 1) (A)
Q. 40
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(K f - 1) (Kt + 1) (K f + 1) (D) (Kt + 1) (B)
The S-N curve for steel becomes asymptotic nearly at (A) 103 cycles (B) 10 4 cycles (D) 109 cycles (C) 106 cycles
GATE SOLVED PAPER - ME
MACHINE DESIGN
YEAR 2004 Q. 41
TWO MARKS
In a bolted joint two members are connected with an axial tightening force of 2200 N. If the bolt used has metric threads of 4 mm pitch, the torque required for achieving the tightening force is
(B) 1.0 Nm
(A) 0.7 Nm (C) 1.4 Nm Q. 42
(D) 2.8 Nm
Match the following Type of gears P.
Bevel gears
1.
Non-parallel off-set shafts
Q.
Worm gears
2.
Non-parallel intersecting shafts
R.
Herringbone gears
3.
Non-parallel, non-intersecting shafts
S.
Hypoid gears
4.
Parallel shafts
(A) (B) (C) (D)
P-4 P-2 P-3 P-1
YEAR 2003 Q. 43
Q. 45
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Q-2 Q-3 Q-2 Q-3
R-1 R-4 R-1 R-4
S-3 S-1 S-4 S-2 ONE MARK
A wire rope is designated as 6 # 19 standard hoisting. The numbers 6 # 19 represent (A) diameter in millimeter # length in meter (B) diameter in centimeter # length in meter (C) number of strands # numbers of wires in each strand (D) number of wires in each strand # number of strands YEAR 2003
Q. 44
Arrangement of shafts
TWO MARKS
Square key of side “d/4 ” each and length ‘l ’ is used to transmit torque “T ” from the shaft of diameter “d ” to the hub of a pulley. Assuming the length of the key to be equal to the thickness of pulley, the average shear stress developed in the key is given by (B) 16T (A) 4T ld ld 2 (C) 8T2 (D) 16T3 ld pd In a band brake the ratio of tight side band tension to the tension on the slack side is 3. If the angle of overlap of band on the drum is 180c, the coefficient of friction required between drum and the band is (A) 0.20 (B) 0.25 (C) 0.30 (D) 0.35
GATE SOLVED PAPER - ME
MACHINE DESIGN
Common Data For Q. 46 and 47 The overall gear ratio in a 2 stage speed reduction gear box (with all spur gears) is 12. The input and output shafts of the gear box are collinear. The counter shaft which is parallel to the input and output shafts has a gear (Z2 teeth) and pinion (Z 3 = 15 teeth) to mesh with pinion (Z1 = 16 teeth) on the input shaft and gear (Z 4 teeth) on the output shaft respectively. It was decided to use a gear ratio of 4 with 3 module in the first stage and 4 module in the second stage. Q. 46
Q. 47
Z2 and Z 4 are (A) 64 and 45 (C) 48 and 60
(D) 60 and 48
The centre distance in the second stage is (A) 90 mm (B) 120 mm (C) 160 mm (D) 240 mm YEAR 2002
Q. 48
(B) 45 and 64
ONE MARK
The minimum number of teeth on the pinion to operate without interference in standard full height involute teeth gear mechanism with 20c pressure angle is (A) 14 (B) 12 (C) 18 (D) 32 YEAR 2002
TWO MARKS
Q. 49
The coupling used to connect two shafts with large angular misalignment is (A) a flange coupling (B) an Oldham’s coupling (C) a flexible bush coupling (D) a Hooke’s joint
Q. 50
A static load is mounted at the centre of a shaft rotating at uniform angular velocity. This shaft will be designed for (A) the maximum compressive stress (static) (B) the maximum tensile (static) (C) the maximum bending moment (static) (D) fatigue loading
Q. 51
Large speed reductions (greater than 20) in one stage of a gear train are possible through (A) spur gearing (B) worm gearing (C) bevel gearing (D) helical gearing
Q. 52
If the wire diameter of a closed coil helical spring subjected to compressive load is increased from 1 cm to 2 cm, other parameters remaining same, the deflection will decrease by a factor of (A) 16 (B) 8 (C) 4 (D) 2
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GATE SOLVED PAPER - ME
MACHINE DESIGN
YEAR 2001 Q. 53
ONE MARK
Bars AB and BC , each of negligible mass, support load P as shown in the figure. In this arrangement,
(A) bar AB is subjected to bending but bar BC is not subjected to bending. (B) bar AB is not subjected to bending but bar BC is subjected to bending. (C) neither bar AB nor bar BC is subjected to bending. (D) both bars AB and BC are subjected to bending. TWO MARKS
YEAR 2001 Q. 54
Two helical tensile springs of the same material and also having identical mean coil diameter and weight, have wire diameters d and d/2. The ratio of their stiffness is (A) 1 (B) 4 (C) 64
(D) 128
*********
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GATE SOLVED PAPER - ME
MACHINE DESIGN
SOLUTION Sol. 54
Option (C) is correct. For 3-wire method, the diameter of the best size wire is given by p d = 2 cos a2 where Hence
Sol. 54
p = pitch = 2 mm , a = 60c 2 = 1.154 mm dD = 2 cos 30c
Option (B) is correct. Stain Energy is given by 2 2 2 U = s V = F 2 # AL = F L 2E 2E 2AE A Given U A = 4U B 2 F A LA = 4F B2 LB 2AA EA 2A B E B Since bolts of same material and length and subjected to identical tensile load, i.e. So that
or or Sol. 54
EA = EB , 1 AA 1 d A2 d B2 dB
LA = LB , FA = FB = 4 AB = 42 dB = 4d A2 = 2dA = 2 # 12 = 24 mm
Option (C) is correct. If the rivets are to be designed to avoid crushing failure, maximum permissible load P = sp # A where for crushing failure A = ^w - 3dhole h # t So that P = ^w - 3dhole h # t # sp
= ^200 - 3 # 11h # 5 # 200 = 167000 N = 167 kN
Sol. 54
Option (C) is correct. For Design against tearing failure, maximum permissible load P = n.drivet # t # sc = 3 # 10 # 5 # 150 = 22500 N = 22.5 kN
Sol. 54
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A = p d2 4
Option (C) is correct. Given : Width of fillets s = 10 mm , l = 30 mm , t = 94 MPa
GATE SOLVED PAPER - ME
MACHINE DESIGN
The shear strength of the joint for single parallel fillet weld is, P = Throat Area # Allowable stress = t#l#t t = s sin 45c = 0.707 s
From figure
P = 0.707 # s # l # t = 0.707 # (0.01) # (0.03) # (94 # 106) = 19937 N or 19.93 kN Sol. 54
Option (B) is correct. T1 = 400 N, m = 0.25, q = 180c = 180c # p = p rad. 180c D = 0.5 m, r = D = 0.25 m 2 For the band brake, the limiting ratio of the tension is given by the relation, T1 = emq T2 400 = e0.25 # p = 2.19 T2 T2 = 400 = 182.68 N 2.19 Given :
For Band-drum brake, Braking Torque is TB = (T1 - T2) # r = (400 - 182.68) # 0.25 = 54.33 Nm , 54.4 Nm Sol. 54
Option (B) is correct. F.O.S = Allowable shear stress Design shear stress Design shear stress for solid circular shaft 3 t = 16T3 = 16 # 50 3# 10 pd pd 3 140 p d # Therefore F.O.S = 16 # 50 # 103 3 or, 2 = 140 # pd 3 16 # 50 # 10 3 d 3 = 2 # 16 # 50 # 10 140 # p d = 15.38 mm , 16 mm
Sol. 54
Option (B) is correct. Given : WP = 30 kN , WQ = 45 kN Life of bearing,
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L = b C l # 106 revolutions W k
From T = t r J
GATE SOLVED PAPER - ME
MACHINE DESIGN
C = Basic dynamic load rating = Constant For ball bearing, k = 3 . Thus L = b C l # 106 revolutions W These are the identical bearings. So for the Life of P and Q. WQ 3 LP 45 3 3 3 27 c LQ m = cWP m = b 30 l = b 2 l = 8 3
Sol. 54
Option (D) is correct. Given : b = 80 mm , d = 250 mm , m = 0.25 , q = 270c, TB = 1000 N-m Let, T1 " Tension in the tight side of the band (Maximum Tension) T2 " Tension in the slack side of the band (Minimum Tension) Braking torque on the drum,
Sol. 54
Sol. 54
TB = (T1 - T2) r ...(i) T1 - T2 = TB = 1000 = 8000 N r 0.125 We know that limiting ratio of the tension is given by, p T1 = emq = ea0.25 # 180 # 270k = 3.246 T2 T2 = T1 3.246 Substitute T2 in equation (i), we get T1 - T1 = 8000 & 3.246T1 - T1 = 25968 3.246 2.246T1 = 25968 & T1 = 25968 = 11.56 kN 2.246 Option (B) is correct. Given : d = 6 mm , l = 12 mm , P = 1000 N Each rivets have same diameter, So equal Load is carried by each rivet. Primary or direct force on each rivet, F = P = 1000 = 250 N 4 4 Shear area of each rivet is, A = p ^6 # 10-3h2 = 28.26 # 10-6 mm2 4 Direct shear stress on each rivet, 250 = 8.84 # 106 - 8.8 MPa t =F = A 28.26 # 10-6 Option (A) is correct. Given : d = 50 mm , D = 50.05 mm , l = 20 mm , N = 1200 rpm , m = 0.03 Pa s Tangential velocity of shaft, -3 u = pdN = 3.14 # 50 # 10 # 1200 = 3.14 m/ sec 60 60 And Radial clearance, y = D - d = 50.05 - 50 = 0.025 mm 2 2 Shear stress from the Newton’s law of viscosity, 3.14 t = m # u = 0.03 # = 3768 N/m2 y 0.025 # 10-3 Shear force on the shaft, F = t # A = 3768 # (p # d # l)
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GATE SOLVED PAPER - ME
MACHINE DESIGN
= 3768 # 3.14 # 50 # 10-3 # 20 # 10-3 = 11.83 N T = F # d = 11.83 # 50 # 10-3 = 0.2957 N-m 2 2 We know that power loss, P = 2pNT = 2 # 3.14 # 1200 # 0.2957 60 60 Torque,
= 37.13 W - 37 W Sol. 54
Option (A) is correct. Given : Su or su = 600 MPa , Sy or sy = 420 MPa , Se or se = 240 MPa , d = 30 mm Fmax = 160 kN (Tension), Fmin =- 40 kN (Compression) 3 Maximum stress, smax = Fmax = 160 # 10 = 226.47 MPa p (30) 2 A 4 3 F min Minimum stress, smin = =- 40 # 10 =- 56.62 MPa p A (30) 2 4# Mean stress, sm = smax + smin = 226.47 - 56.62 = 84.925 MPa 2 2 226.47 - (- 56.62) Variable stress, sv = smax - smin = = 141.545 MPa 2 2 From the Soderberg’s criterion, 1 = sm + sv sy se F .S . 1 = 84.925 + 141.545 = 0.202 + 0.589 = 0.791 240 420 F .S . So, F.S. = 1 = 1.26 0.791
Sol. 54
Option (A) is correct. Given : m = 4 mm , Z = 21, P = 15 kW = 15 # 103 W N = 960 rpm b = 25 mm , f = 20c Pitch circle diameter, D = mZ = 4 # 21 = 84 mm Tangential Force is given by, FT = T r 2p Power transmitted, & T = 60P P = NT 60 2pN Then FT = 60P # 1 r = Pitch circle radius r 2pN
...(i)
3 1 = 60 # 15 # 10 # 2 # 3.14 # 960 42 # 10-3
= 3554.36 N - 3552 N Sol. 54
Option (B) is correct. From Lewis equation FT pd FT = by b#y#m 3552 = 25 # 10-3 # 0.32 # 4 # 10-3 sb = 111 MPa Minimum allowable (working stress) sb =
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pd = p = p = 1 pm m pc
GATE SOLVED PAPER - ME
MACHINE DESIGN
sW = sb # Cv = 111 # 1.5 = 166.5 MPa Sol. 54
Option (A) is correct. Given : d = 40 mm , l = 40 mm , w = 20 rad/ sec Z (m) = 20 mPa - s = 20 # 10-3 Pa - s , c (y) = 0.020 mm From the Newton’s law of viscosity...(i) Shear stress, t = mu y u = rw = 0.020 # 20 = 0.4 m/ sec -3 t = 20 # 10 # -03.4 = 400 N/m2 0.020 # 10 Shear force is generated due to this shear stress, F = t A = t # pdl A = pdl = Area of shaft = 400 # 3.14 # 0.040 # 0.040 = 2.0096 N Loss of torque,
Sol. 54
T = F # r = 2.0096 # 0.020 = 0.040192 N-m - 0.040 N-m
Option (B) is correct. Given : d1 = 100 mm & r1 = 50 mm , d2 = 40 mm & r1 = 20 mm p = 2 MPa = 2 # 106 Pa , n = 0.4 When the pressure is uniformly distributed over the entire area of the friction faces, then total frictional torque acting on the friction surface or on the clutch is given by, (r ) 3 - (r2) 3 T = 2pmp ; 1 E 3 = 2 # 3.14 # 0.4 # 2 # 106 6(50) 3 - (20) 3@ # 10-9 3 = 195.39 N-m - 196 N-m
Sol. 54
Option (B) is correct. Given : m = 3 mm , Z = 16 , b = 36 mm , f = 20c, P = 3 kW N = 20 rev/ sec = 20 # 60 rpm = 1200 rpm , Cv = 1.5 , y = 0.3 Module, m =D Z Power,
D = m # Z = 3 # 16 = 48 mm P = 2pNT 60 3 T = 60P = 60 # 3 # 10 2 # 3.14 # 1200 2pN
= 23.88 N-m = 23.88 # 103 N-mm 3 Tangential load, WT = T = 2T = 2 # 23.88 # 10 = 995 N 48 R D From the lewis equation Bending stress (Beam strength of Gear teeth) p p 1 sb = WT Pd = WT :Pd = PC = pm = m D by bym 995 = 36 # 10-3 # 0.3 # 3 # 10-3 995 sb = = 30.70 # 106 Pa = 30.70 MPa 3.24 # 10-5 Permissible working stress sW = sb # Cv = 30.70 # 1.5 = 46.06 MPa , 46 MPa
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GATE SOLVED PAPER - ME
Sol. 54
MACHINE DESIGN
Option (D) is correct.
Given : m = 4 , Z = 32 , Tooth space = Tooth thickness = a We know that, m =D Z Pitch circle diameter, D = mZ = 4 # 32 = 128 mm And for circular pitch, Pc = pm = 3.14 # 4 = 12.56 mm We also know that circular pitch, Pc = Tooth space + Tooth thickness = a + a = 2a a = Pc = 12.56 = 6.28 mm 2 2 From the figure, or
b = addendum + PR PQ a/2 3.14 = = sin f = 64 64 OQ f = sin-1 (0.049) = 2.81c OP = 64 cos 2.81c = 63.9 mm
PR = OR - OP = 64 - 63.9 = 0.1 mm OR = Pitch circle radius And b = m + PR = 4 + 0.1 = 4.1 mm Therefore, a = 6.28 mm and b = 4.1 mm Sol. 54
Option (A) is correct. Types of Gear
Description
P. Helical
2. Axes parallel and teeth are inclined to the axis
Q. Spiral Bevel
4. Axes are perpendicular and intersecting, and teeth are inclined to the axis
R. Hypoid
1. Axes non parallel and non-intersecting
S.
Rack and pinion 6. Axes are parallel and one of the gear has infinite radius
So, correct pairs are P-2, Q-4, R-1, S-6 Sol. 54
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Option (A) is correct. In given figure WS represent the primary shear load whereas WS1 and WS2 represent the secondary shear loads. Given : A = 10 # 50 mm2 , n = 2 , W = 4 kN = 4 # 103 N We know that primary shear load on each bolt acting vertically downwards, Ws = W = 4 kN = 2 kN n 2 Since both the bolts are at equal distances from the centre of gravity G of the two
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MACHINE DESIGN
bolts, therefore the secondary shear load on each bolt is same. For secondary shear load, taking the moment about point G ,
So,
Sol. 54
Sol. 54
Sol. 54
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Ws1 # r1 + Ws2 # r2 = W # e r1 = r2 and Ws1 = Ws2 2r1 Ws1 = 4 # 103 # (1.7 + 0.2 + 0.1) 2 # 0.2 # Ws1 = 4 # 103 # 2 3 Ws1 = 8 # 10 = 20 # 103 = 20 kN 2 # 0.2
Option (B) is correct. From the figure, resultant Force on bolt P is F = Ws2 - Ws = 20 - 2 = 18 kN Shear stress on bolt P is, 18 # 103 t = F = = 159.23 MPa - 159 MPa p Area (12 # 10-3) 2 # 4 Option (D) is correct. Given : W1 = F , W2 = 2F , L1 = 8000 hr We know that, life of bearing is given by k L = b C l # 106 revolution W 3 For ball bearing, k = 3 , L = b C l # 106 revolution W For initial condition life is, 3 L1 = bC l # 106 F 3 ...(i) 8000 hr = bC l # 106 F 3 3 For final load, L2 = b C l # 106 = 1 # bC l # 106 8 2F F From equation (i) = 1 (8000 hr) = 1000 hr 8 Option (B) is correct. Given : d = 200 mm , t = 1 mm , su = 800 MPa , se = 400 MPa Circumferential stress induced in spherical pressure vessel is, p r p 100 s = # = # = 50p MPa 2t 2#1 Given that, pressure vessel is subject to an internal pressure varying from 4 to 8 MPa. So, smin = 50 # 4 = 200 MPa
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smax = 50 # 8 = 400 MPa sm = smin + smax = 200 + 400 = 300 MPa 2 2 s s 400 200 = 100 MPa min = sv = max 2 2
Mean stress, Variable stress,
Sol. 54
From the Goodman method, 1 = sm + sv = 300 + 100 = 3 + 1 = 5 su se 800 400 8 4 8 F .S . 8 = = 1.6 5 Option (B) is correct. Given : d = 50 mm , l = 50 mm , N = 20 rps , Z = 20 mPa - sec = 20 # 10-3 Pa - sec Radial clearance = 50 mm = 50 # 10-3 mm , Load = 2 kN We know that,
&
F.S.
p = Bearing Pressure on the projected bearing area =
Load on the journal l#d
3 = 2 # 10 = 0.8 N/mm2 = 0.8 # 106 N/m2 50 # 50 2 c = diameteral clearance Sommerfeld Number = ZN b d l p c = 2 # radial clearance -3 2 50 S.N. = 20 # 10 #6 20 # b -3 l 0.8 # 10 100 # 10 -3 2 = 20 # 10 #6 20 # b 1 l # 106 = 0.125 2 0.8 # 10
Sol. 54
Option (A) is correct.
Given : Diameter of bolt d = 10 mm , F = 10 kN , No. of bolts n = 3 Direct or Primary shear load of each rivet 3 FP = F = 10 # 10 N = 3333.33 N n 3 The centre of gravity of the bolt group lies at O (due to symmetry of figure). e = 150 mm (eccentricity given) Turning moment produced by the load F due to eccentricity = F # e = 10 # 103 # 150 = 1500 # 103 N-mm Secondary shear load on bolts from fig. rA = rC = 40 mm and rB = 0
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We know that
MACHINE DESIGN
F # e = FA [(rA) 2 + (rB) 2 + (rC ) 2] rA = FA # [2 (rA) 2] rA 1500 # 103 = FA # [2 (40) 2] = 80FA 40
(rA = rC and rB = 0 )
3
FA = 1500 # 10 = 18750 N 80 FB = 0
(rB = 0)
FC = FA # rC = 18750 # 40 = 18750 N rA 40 From fig we find that angle between FA and FP = qA = 90c FB and FP = qB = 90c FC and FP = qC = 90c Resultant load on bolt A, RA =
(FP ) 2 + (FA) 2 + 2FP # FA cos qA
(3333.33) 2 + (18750) 2 + 2 # 3333.33 # 18750 # cos 90c RA = 19044 N Maximum shear stress at A tA = RA = 19044 = 242.6 MPa p (d) 2 p (10) 2 4 4 Resultant load on Bolt B , =
RB = FP = 3333.33 N Maximum shear stress at B , tB = p RB 2 = p3333.33 2 = 42.5 MPa 4 (d) 4 # (10) Sol. 54
(FB = 0)
Option (C) is correct.
Given : P = 400 N , r = 300 mm = 150 mm , l = 600 mm 2 x = 200 mm , m = 0.25 and 2q = 45c Let, RN " Normal force pressing the brake block on the wheel Ft "Tangential braking force or the frictional force acting at the contact surface of the block & the wheel. Here the line of action of tangential braking force Ft passes through the fulcrum O of the lever and brake wheel rotates clockwise. Then for equilibrium, Taking
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the moment about the fulcrum O , RN # x = P # l RN = P # l = 400 # 0.6 = 1200 N x 0.2 Tangential braking force on the wheel, Braking Torque, Sol. 54
Ft = mRN TB = Ft # r = mRN # r = 0.25 # 1200 # 0.15 = 45 N-m
Option (C) is correct. Given : d = 20 mm , l = 700 mm , E = 200 GPa = 200 # 109 N/m2 = 200 # 103 N/mm2 Compressive or working Load = 10 kN According to Euler’s theory, the crippling or buckling load (Wcr ) under various end conditions is given by the general equation, 2 ...(i) Wcr = cp 2EI l Given that one end is guided at the piston end and hinged at the other end. So, c =2 From equation (i), 2 2 I = p d4 Wcr = 2p 2EI = 2p2E # p d 4 64 64 l l 3 # 10 # 3.14 # (20) 4 = 2 # 9.81 # 200 2 64 (700) = 62864.08 N = 62.864 kN We know that, factor of safety (FOS) Crippling Load FOS = = 62.864 = 6.28 10 Working Load
The most appropriate option is (C). Sol. 54
Sol. 54
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Option (B) is correct. Given : ZP = 20 , ZG = 40 , NP = 30 rev/ sec , P = 20 kW = 20 # 103 W , m = 5 mm Module, m = D = DP = DG Z ZP ZG DP = m # ZP = 5 # 20 = 100 mm or, DG = m # ZG = 5 # 40 = 200 mm Centre distance for the gear set, L = DP + DG = 100 + 200 = 150 mm 2 2 Option (C) is correct. Given : Length of line of action, L = 19 mm Pressure angle, f = 20c Length of path of contact (L) Length of arc of contact = cos f = 19 = 20.21 mm cos 20c
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Contact ratio or number of pairs of teeth in contact, Length of arc of contact = 20.21 = 20.21 = 1.29 pm 3.14 # 5 circular pitch Sol. 54
Option (C) is correct.
Let, T " Torque transmitted in N - m We know that power transmitted is, P = Tw = T # 2pN 60
Sol. 54
3 T = 60P = 60 # 20 # 10 = 106.157 N-m 2 # 3.14 # 1800 2pN T Tangential load on the pinion FT = RP = 106.157 = 2123.14 N 0.05 From the geometry, total load due to power transmitted, F = FT = 2123.14 - 2258.1 N cos f cos 20c Option (A) is correct. Given : P = 5 kW , N = 2000 rpm , m = 0.25 , r 2 = 25 mm = 0.025 m , p = 1 MPa Power transmitted, P = 2pNT 60
3 = 23.885 N-m T = 60P = 60 # 5 # 10 2 # 3.14 # 2000 2pN When pressure is uniformly distributed over the entire area of the friction faces, then total frictional torque acting on the friction surface or on the clutch, (r ) 3 - (r2) 3 T = 2pmp ; 1 E 3
Torque,
23.885 # 3 = 2 # 3.14 # 0.25 # 1 # 106 # [r 13 - (0.025) 3] 23.885 # 3 r 13 - (0.025) 3 = 2 # 3.14 # 0.25 # 106 r 13 - 1.56 # 10-5 = 45.64 # 10-6 = 4.564 # 10-5 r 13 = (4.564 + 1.56) # 10-5 = 6.124 # 10-5 r1 = (6.124 # 10-5) 1/3 = 3.94 # 10-2 m = 39.4 mm Sol. 54
Option (A) is correct. Given : ZP = 19 , ZG = 37 , m = 5 mm Also,
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m =D Z
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MACHINE DESIGN
For pinion, pitch circle diameter is, DP = m # ZP = 5 # 19 = 95 mm And pitch circle diameter of the gear,
Sol. 54
DG = m # ZG = 5 # 37 = 185 mm Now, centre distance between the gear pair (shafts), L = DP + DG = 95 + 185 = 140 mm 2 2 2 Option (C) is correct.
We know that in S-N curve the failure occurs at 106 cycles (at endurance strength) We have to make the S-N curve from the given data, on the scale of log 10 . Now equation of line whose end point co-ordinates are or
6 ^log 10 1000, log 10 490h and ^log 10 10 , log 10 70h ^3, log 10 490h and ^6, log 10 70h,
log 10 70 - log 10 490 y - log 10 490 = 6-3 x-3 y - 2.69 = 1.845 - 2.69 x-3 3
y - y1 y 2 - y1 b x - x1 = x 2 - x1 l
y - 2.69 =- 0.281 (x - 3) Given, the shaft is subject to an alternating stress of 100 MPa So, y = log 10 100 = 2 Substitute this value in equation (i), we get
...(i)
2 - 2.69 =- 0.281 (x - 3) - 0.69 =- 0.281x + 0.843 x = - 0.843 - 0.69 = 5.455 - 0.281 log 10 N = 5.455 N = 105.455 = 285101 The nearest shaft life is 281914 cycles.
And
Sol. 54
Option (B) is correct. Given : l = 60 mm = 0.06 m , s = 6 mm = 0.006 m , P = 15 kN = 15 # 103 N Shear strength = 200 MPa We know that, if t is the allowable shear stress for the weld metal, then the shear strength of the joint for single parallel fillet weld, P = Throat Area # Allowable shear stress = t # l # t P = 0.707s # l # t
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t = s sin 45c = 0.707s
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MACHINE DESIGN
P 15 # 103 = 58.93 MPa = 0.707 # 0.006 # 0.06 0.707 # s # l Shear strength = 200 MPa = 3.39 - 3.4 FOS = 58.93 MPa Allowable shear stress t =
Sol. 54
Option (A) is correct. The coefficient of friction for a full lubricated journal bearing is a function of three variables, i.e. m = f b ZN , d , l l p c d ZN Here, =Bearing characteristic Number, d =Diameter of the bearing p l =Length of the bearing, c =Diameteral clearance 2 Sommerfeld Number = ZN b d l p c It is a dimensionless parameter used extensively in the design of journal bearing. i.e. sommerfeld number is also function of b ZN , d l. Therefore option (A) is p c correct.
Sol. 54
Option (B) is correct.
Given : r = 200 mm = 0.2 m , q = 270c = 270 # p = 3p radian , m = 0.5 180 2 At the time of braking, maximum tension is generated at the fixed end of band near the wheel. Let, T2 " Tension in the slack side of band T1 "Tension in the tight side of band at the fixed end Taking the moment about the point O , & T2 = 200 N T2 # 1 = 100 # 2 For the band brake, the limiting ratio of the tension is given by the relation T1 = emq & T1 = T2 # emq T2 3p
T1 = 200 # e0.5 # 2 = 200 # 10.54 = 2108 N - 2110 N So, maximum tension that can be generated in the band during braking is equal
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MACHINE DESIGN
to 2110 N Sol. 54
Option (B) is correct. Maximum wheel torque or braking torque is given by, TW = (T1 - T2) r = (2110 - 200) # 0.2 = 382 N-m
Sol. 54
Option (D) is correct.
Given : ZP = 40 teeth , ZG = 120 teeth , NP = 1200 rpm , TP = 20 N-m ZP = NG ZG NP Z NG = P # NP = 40 # 1200 = 400 rpm 120 ZG Power transmitted is same for both pinion & Gear. P = 2pNP TP = 2pNG TG 60 60 Velocity Ratio,
NP TP = NG TG TG = NP TP = 1200 # 20 = 60 N-m 400 NG So, the torque transmitted by the Gear is 60 N-m Sol. 54
Option (A) is correct. When the notch sensitivity factor q is used in cyclic loading, then fatigue stress concentration factor may be obtained from the following relation. K f = 1 + q (Kt - 1) K f - 1 = q (Kt - 1) K -1 q = f Kt - 1
Sol. 54
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Option (C) is correct. The S-N curve for the steel is shown below :
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We can easily see from the S-N curve that, steel becomes asymptotic nearly at 106 cycles . Sol. 54
Option (C) is correct. Given : Ft = 2200 N , p = 4 mm = 0.004 m Torque required for achieving the tightening force is, T = Ft # r = Ft # Pitch = 2200 # 0.004 = 1.4 N-m 2p 2 # 3.14
Sol. 54
Option (B) is correct. Type of Gears
Arrangement of shafts
P.
Bevel gears
2.
Non-parallel intersecting shafts
Q.
Worm gears
3.
Non-parallel, non-intersecting shafts
R.
Herringbone gears
4.
Parallel shafts
S.
Hypoid gears
1.
Non-parallel off-set shafts
So, correct pairs are P-2, Q-3, R-4, S-1. Sol. 54
Option (C) is correct. The wire ropes are designated by the number of strands multiplied by the number of wires in each strand. Therefore, 6 # 19 = Number of strands # Number of wires in each strand.
Sol. 54
Option (C) is correct. Diameter of shaft = d Given : Torque transmitted = T Length of the key = l We know that, width and thickness of a square key are equal. i.e. w =t=d 4 Force acting on circumference of shaft F = T = 2T r d Shearing Area, A = width # length = d # l = dl 4 4 2T/d Force Average shear stress, = 8T2 t = = shearing Area dl/4 ld
Sol. 54
Option (D) is correct. Let,
(r = d/2)
T1 " Tension in the tight side of the band, T2 " Tension in the slack side of the band q "Angle of lap of the band on the drum T 1 Given : = 3 , q = 180c = p # 180 = p radian 180 T2 For band brake, the limiting ratio of the tension is given by the relation, T1 = emq or 2.3 log T1 = mq bT2 l T2 2.3 # log (3) = m # p 2.3 # 0.4771 = m # 3.14 m = 1.09733 = 0.349 - 0.35 3.14
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Sol. 54
MACHINE DESIGN
Option (A) is correct.
Let N1 , N2 , N 3 and N 4 are the speeds of pinion 1, gear 2, pinion 3 and gear 4 respectively. Given : Z1 = 16 teeth , Z 3 = 15 teeth and Z 4 = ? , Z2 = ? N1 = Z2 /Z1 Velocity ratio N \ 1/Z N4 Z 3 /Z 4 ...(i) = Z2 # Z 4 = 12 Z1 Z3 N1 = Z 2 = 4 ...(ii) But for stage 1, Z1 N2 So, from eq. (i) 4 # Z 4 = 12 Z3 Z4 = 3, & Z 4 = 3 # 15 = 45 teeth Z3 From equation (ii), Sol. 54
Option (B) is correct. Let centre distance in the second stage is D . D = R4 + R3 = D4 + D3 2 D4 = D3 = 4 But, Z4 Z3 Or, So,
Sol. 54
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Z2 = 4 # Z1 = 4 # 16 = 64 teeth
m = D/Z module
D 4 = 4 # Z 4 = 4 # 45 = 180 D 3 = 4 # Z 3 = 4 # 15 = 60 D = 180 + 60 = 120 mm 2
Option (C) is correct. In standard full height involute teeth gear mechanism the arc of approach is not be less than the circular pitch, therefore Maximum length of arc of approach = Circular pitch ...(i) where Maximum length of the arc of approach Max. length of the path of approach r sin f = = r tan f cos f cos f Circular pitch, m = 2r PC = pm = 2pr Z Z
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MACHINE DESIGN
Hence, from equation (i), we get r tan f = 2pr Z = 17.25 - 18 teeth Z = 2p = 2 p tan f tan 20c Sol. 54
Option (D) is correct. (A) Flange coupling :- It is used to connect two shaft having perfect coaxial alignment and no misalignment is allowed between them. (B) Oldham’s coupling :- It is used to join two shafts which have lateral misalignment. (C) Flexible bush coupling :- It is used to join the abutting ends of shafts when they are not in exact alignment. (D) Hook’s joint :- It is used to connect two shafts with large angular misalignment.
Sol. 54
Option (D) is correct. When the shaft rotates, the bending stress at the upper fibre varies from maximum compressive to maximum tensile while the bending stress at the lower fibres varies from maximum tensile to maximum compressive. The specimen subjected to a completely reversed stress cycle. This is shown in the figure.
When shaft is subjected to repeated stress, then it will be designed for fatigue loading. Sol. 54
Option (B) is correct. For a worm gear the velocity ratio ranges between 10 : 1 to 100 : 1. So, Large speed reductions (greater than 20) in one stage of a gear train are possible through worm gearing.
Sol. 54
Option (A) is correct. For Helical spring, deflection is given by, 3 3 d = 64PR4 n = 8PD4 n Gd Gd where, P = Compressive load d = Wire diameter R = Coil diameter G = Modulus of rigidity From the given conditions d \ 14 d Given d1 = 1 cm and d2 = 2 cm
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MACHINE DESIGN
d2 = d1 4 d1 b d2 l d2 = 1 4 = 1 16 d1 b 2 l d2 = d1 16 So, deflection will decrease by a factor of 16. Sol. 54
Option (C) is correct. Bars AB and BC have negligible mass. The support load P acting at the free end of bars AB and BC . Due to this load P , In bar AB compressive stress and in bar BC tensile stress are induced. However, none of these bars will be subjected to bending because there is no couple acting on the bars.
Sol. 54
Option (C) is correct. Let L1 & L2 are lengths of the springs and n1 & n2 are the number of coils in both the springs. Given : W1 = W2 m1 g = m 2 g rn1 g = rn2 g A 1 # L 1 # rg = A 2 # L 2 # rg p d 2 p D n = p d 2 pD n 1 1 2 2 4 1# 4 2# Given :
d 12 # n1 = d 22 # n2 d1 = d & d 2 = d 2 2
d 2 # n1 = d # n 2 4 or, n1 = n 2 4 The deflection of helical spring is given by, 3 d = 8PD4 n Gd 4 Spring stiffness, k = P = Gd3 d 8D n From the given conditions, we get 4 k\d n k 1 = d1 4 n2 So, k 2 b d 2 l # a n1 k k1 = d 4 n2 k2 c d/2 m # n2 /4 k1 = 16 4 = 64 # k2 ***********
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m = rn L = pDn D1 = D 2
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