Revision Term 2. Prof Ahmed Kovacevic

Transcription

1 ME 1110 Engineering Practice 1 Engineering Drawing and Design - Lecture 20 Revision Term 2 Prof Ahmed Kovacevic School of Engineering and Mathematical Sciences Room CG25, Phone: 8780, a.kovacevic@city.ac.uk 1

2 To revise for 2 nd test Test for MEA and EME students:» 3 rd April 9,00 OTLT Review test examples on Moodle Revise lectures Test example 2

3 Let us REVISE Design web 3

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11 Example Shaft Determine the diameter for the solid round shaft 450 mm long, as shown in Figure. The shaft is supported by self-aligning bearings at the ends. Mounted upon the shaft are a V-belt pulley, which contributes a radial load of F 1 =8kN to the shaft, and a gear which contributes a radial load of F2=3kN. The two loads are in the same plane and have the same directions. The allowable bending stress (strength) is S=170 MPa. Assume factor of safety 1.5. F1=8 kn; F2=3 kn a=450 mm b=150 mm c=200 mm S=170 Mpa f=1.5 d=? SOLUTION: Assumptions - the weight of the shaft is neglected - the shaft is designed for the normal bending stress in the location of max. bending moment 11

12 Shaft formulas Bending stress Torsional stress Minimum diameter distortion energy theory σ τ z zy M Mc 32M = = = 3 Z I π d T Tc 16T = = = 3 S J π d 32 f 3 = + π S 4 s 2 2 d 3 M T y c=d/2 I=πd 4 /64 Z=c/I J=πd 4 /32 S=c/J f s - maximum span - second moment of area - section modulus - second polar moment of area - polar section modulus - factor of safety 12

13 Solution f R R 1 2 M = ar + ( a b) F + ( a b c) F c = 6 kn = 5kN max A M = M = b R = 900 Nm Second moment of area (moment of inertia) Section modulus = max span 4 π d d I = c= I π d 3 Z = = 0.1d c 32 Stress = Strain = Bending moment / section modulus = S σ Mc M 900 S = fσ = f = f = 1.5 = I Z 0.1 d d = 3 = m = 20mm

14 Example bearings SOLUTION: 14 Select the bearings and determine their rating life for the driving mechanism shown in the Figure. The shaft is 450 mm long and supported by deep-groove bearing in point O and plane roller bearing in point C. Assume minimum shaft diameter to be 20 mm. Mounted upon the shaft are a V-belt pulley, which contributes a radial load of F 1 =8kN to the shaft, and a gear which contributes a radial load of F 2 =3kN. The two loads are in the same plane and have the same direction. Minimum required bearing life is 2000 h with 90% reliability. Shaft rotates constantly at n=1000 rpm. F1=8 kn a=450 mm c=200 mm F2=3 kn b=150 mm d=20 mm L 10h =(L 10h ) 0 =(L 10h ) C =2000 h n=1000 rpm 6 a 10 C 60n a 10h 6 10h O 1 C 2 L = C = P* L P = R = 6000N P = R = 5000N 60n P 10 C C 60*1000 = 6000* 2000 = 29,595N *1000 = 5000* 2000 = 21, 025N Selected from the catalogue for deep-groove ball bearings: x72x19 mm C=30,700 N Selected from the catalogue for cylindrical roller bearings: NU x47x14 mm C=25,100 N

15 Method of Joints Example Using the method of joints, a) Find is the truss determinate b) the force in each member BD. 15

16 Method of Joints Example Calculate restraint reactions Draw the free body diagram of the truss and solve for the equations: F x = 0: C = 0 x F = 0 : E+ C = 0 E+ C = 3000 lb y y y C d=m+r-2. j = 0 y ( ) ( ) ( ) M = 0 : ft ft E 6 ft E = lb C = = 7000 lb 16

17 Method of Joints Example Joint A F F y AD AB 4 = 0 = FAD 2000 lb 5 = 2500 lb F = 2500 lb C AD ( ) ( ) 3 3 F = 0 = F + F = ( 2500 lb) + F 5 5 F = 1500 lb F = 1500 lb T x AD AB AB AB 17

18 Method of Joints Example Joint D F = 0 = F + F = ( 2500) + F F = 2500 lb F = 2500 lb T DB DE y AD DB DB DB ( ) F = 0 = F + F + F = ( 2500) + ( 2500) + F F = 3000 lb F = 3000 lb C x AD DB DE DE DE ( ) 18

19 Example screws The cover of a pressurised cylinder is attached by a self-energising seal and 6 identical bolts M10x1.5 of Design web class 8.8. The fluid pressure is essentially constant at 6 MPa. A safety factor of three is required. Check if the given bolt can sustain the pressure! P=6MPa 6 class 8.8 M10x1.5 d s =120 mm N d = S t /σ=? SOLUTION: 2 2 Force on the cover ds caused by the pressure: Fc p As p π = = Fc = 6 10 = 67858N = 67.9kN 4 4 F Force on the individual bolt F c 67.9 b = = F b = kn From tables: Tensile stress area 2 A = 58mm Proof strength S = 590 MPa F Stress on each bolt: σ = b = σ = 194MPa At 58 S p Nd σ = 194 = t p 19 Selected number of bolts can sustain the load

20 Load that a bolt can sustain Tensile stress: σ = F b A Shear stress: t τ = P Ar 20 class no St Tensile [Mpa] Sy Yield [Mpa] Sp Proof [Mpa] Elongation % Strength table

21 Metric threads (all dimensions in mm) 21