Theory and Design of Automotive Engines [AU51]: Chapter No:06 - CrankShaft

Theory and Design of Automotive Engines [AU51]<br /> <br /> Chapter No:06, Crank Shaft<br /> Objective: The student will learn that crankshaft design is the estimation of the shaft diameter, crankpin dimensions to ensure satisfactory strength and rigidity when the crankshaft is transmitting power under various operating conditions. Outcomes: The student should be able to design crankshafts for various operating and loading conditions. Prerequisites: This topic requires the student to know about, the fundamentals of Engineering Mathematics, Engg physics, Strength of Materials, Engineering Drawing, Workshop Processes, Theory of Machines, Material Science and fundamentals of Machine Design. Number of Question/s expected in examination: 01 [20Marks] INTRODUCTION: Before studying the actual crankshaft and design details, we shall study briefly the basics of Power Transmission of shafts. 6.1 Power Transmitting Shaft: Shaft Design consists primarily of the determination of the correct shaft diameter to ensure satisfactory strength and rigidity when the shaft is transmitting power under various operating and loading conditions. Shafts are usually circular in cross section, and may be either hollow or solid. Design of shafts of ductile materials, based on strength, is controlled by the maximum shear theory. And the shafts of brittle material would be designed on the basis of the maximum normal stress theory. Various loads subjected on Shafting are torsion, bending and axial loads. 6.1.1 Basics of Design for solving Shaft problems: 6.1.1a. Maximum Principal Stress:(σ1)<br /> <br /> σ1 =<br /> <br /> σ x +σ y<br /> 2<br /> <br /> ⎡σ x − σ y ⎤ 2 + ⎢ ⎥ + τ xy ………………………………..(1.11a/2) ⎣ 2 ⎦<br /> 2<br /> <br /> Here 1.11a and 2 refers to the formula number and page number from Design data handbook by K Mahadevan and K Balaveera Reddy, CBS Publications, INDIA, 1989. Same Data handbook and similar procedure is adopted in further discussion.<br /> <br /> Where, σx --- Stress in x direction, in MPa or N/mm2 σy --- Stress in y direction, in MPa<br /> Krishnaraja G. Kodancha, Assistant Professor, Automobile Engineering Department, B.V.B.College of Engineering and technology, HUBLI-31; Session 24 to 28, Oct 18 to 30 -2007<br /> <br /> 1<br /> <br /> Theory and Design of Automotive Engines [AU51] τxy ---Shear stress, in MPa. 6.1.1b.Minimum Principal Stress: (σ2)<br /> <br /> σ2 =<br /> <br /> σ x +σ y<br /> 2<br /> <br /> ⎡σ x − σ y ⎤ 2 − ⎢ ⎥ + τ xy ………………………………..(1.11b/2) 2 ⎦ ⎣<br /> 2<br /> <br /> 6.1.1c. Maximum Shearing Stress:(τmax)<br /> <br /> τ max<br /> <br /> ⎡σ x − σ y ⎤ 2 =± ⎢ ⎥ + τ xy ………………………………………...(1.12/2) ⎣ 2 ⎦<br /> 2<br /> <br /> 6.1.1d.Torsional stresses: (τ) The Torsional formula is given by,<br /> <br /> T Gθ τ = = ………………………………………………………..(1.15/3) J l r<br /> Here T=torque or Torsional moment, N-mm J=polar moment of inertia, mm4 =<br /> <br /> π<br /> <br /> 32<br /> <br /> d 4 , Where d is the solid shaft diameter.<br /> <br /> =<br /> <br /> (d 32<br /> π<br /> <br /> 4 o<br /> <br /> − d i , Where do and di are outer and inner diameter of the hollow shaft<br /> 4<br /> <br /> )<br /> <br /> respectively. G=Modulus of elasticity in shear or modulus of rigidity, MPa θ=Angle of twist, radians l= Length of shaft , mm r= Distance from the Neutral axis to the top most fibre, mm d (For solid shaft) = 2 d = o (For hollow shaft) 2 6.1.1d.Bending Stresses:(σb) The bending equation is given by<br /> M E σb = = ………………………………………………………..(1.16/3) I R c<br /> <br /> Here M=bending moment, N-mm<br /> Krishnaraja G. Kodancha, Assistant Professor, Automobile Engineering Department, B.V.B.College of Engineering and technology, HUBLI-31; Session 24 to 28, Oct 18 to 30 -2007<br /> <br /> 2<br /> <br /> Theory and Design of Automotive Engines [AU51] I= Second moment of area, mm4 =<br /> <br /> π<br /> <br /> 64<br /> <br /> d 4 (For solid shaft)<br /> <br /> =<br /> <br /> (d 64<br /> π<br /> <br /> 4<br /> o<br /> <br /> − di ,<br /> 4<br /> <br /> )<br /> <br /> (For hollow shaft)<br /> <br /> E=modulus of elasticity or Young’ modulus for the material, MPa θ=Angle of twist, radians R= radius of curvature, mm c= Distance from the Neutral axis to the extreme fibre, mm d (For solid shaft) = 2 d = o (For hollow shaft) 2 6.1.2 Methods of obtaining the Twisting moment and Bending Moment.<br /> 6.1.2a Twisting Moment: i) Power transmitted : 2πnT P= kW 60000 Where T - twisting moment in N-m= (103) N-mm n – speed of the shaft, rpm 3 60000( P)(10 ) 9.55(10 6 )( P) = ……………………………..(3.3a/42) Hence T = n 2πn<br /> <br /> ii)<br /> <br /> In case of belt drives Power transmitted (T − T2 )v P= 1 kW …………………………………………….(14.9a/239) 1000 Where T 1- tension of belt on tight side, N T 2- tension of belt on slack side, N v- velocity of belt, m/s [Student should take care of units here, it is in m/sec not in mm/sec] T1 = e µθ …………………………………………………………(14.6a/238) T2 θ ---arc of contact, rad µ---coefficient of friction between belt and pulley. From equation (14.9a/239) and (14.6a/238) get T 1 and T 2. R Pulley T1 Figure 1 T2 3<br /> <br /> Krishnaraja G. Kodancha, Assistant Professor, Automobile Engineering Department, B.V.B.College of Engineering and technology, HUBLI-31; Session 24 to 28, Oct 18 to 30 -2007<br /> <br /> Theory and Design of Automotive Engines [AU51] Knowing the value of radius of the pulley (R) twisting moment can be found by using the following equation:[Refer Figure 1] T= (T 1 - T 2) R, N-mm. iii) In case of Gear drives. Power transmitting capacity of gears is given by Fv P = t kW…………………………….…………… …..…(12.14a/163) 1000 Ft = driving force or tangential load at pitch line, N d The torque is given by, T = Ft ( ) , N-mm………………………..(12.22/165) 2 Where d is the pitch circle diameter of Gear.<br /> <br /> 6.1.2b Bending Moment. i) Cantilever, end load [Figure 2] W<br /> <br /> Figure 2<br /> <br /> Maximum Bending Moment, M=W(l), N-mm …………………………..[Table 1.4/1/10] Where W is the concentrated load, N l is the length of the beam, mm ii) Simply supported beam [End support, center load] [Figure 3] W l/2 A B<br /> l<br /> <br /> Figure 3<br /> RB<br /> <br /> RA<br /> <br /> To find the reactions RA and RB<br /> <br /> ∑M<br /> <br /> A<br /> <br /> = 0,<br /> <br /> For the convenient of calculations, Clock Wise direction is taken as positive bending moment and Counter Clockwise as negative bending moment.<br /> l W ( ) − RB (l ) = 0 2<br /> <br /> Krishnaraja G. Kodancha, Assistant Professor, Automobile Engineering Department, B.V.B.College of Engineering and technology, HUBLI-31; Session 24 to 28, Oct 18 to 30 -2007<br /> <br /> 4<br /> <br /> Theory and Design of Automotive Engines [AU51] Hence R B =<br /> W ,N 2<br /> <br /> ∑F = 0,<br /> taken as negative. RA+RB – W=0 Hence RA = W – RB<br /> <br /> , Upward force is taken as positive and downward is<br /> <br /> l Wl Maximum bending Moment, M = R A ( ) = ……………………….[Table 1.4/4/10]. 2 4<br /> <br /> iii) Simply supported beam [End support, Intermediate][Figure 4] W b a A B<br /> l<br /> <br /> RA<br /> <br /> RB<br /> <br /> Figure 4 To find the reactions RA and RB<br /> <br /> ∑M<br /> <br /> A<br /> <br /> = 0,<br /> <br /> W (a ) − RB (l ) = 0 W (a) Hence R B = ,N l<br /> <br /> ∑F = 0,<br /> RA+RB – W=0 Hence R A = W − RB , N W (a) a l−a W (b) RA = W − = W (1 − ) = W ( )= ,N l l l l Maximum bending Moment, M = R A (a ) =<br /> W (a )(b) …………………[Table 1.4/5/10]. l For different kinds of loading and support students are advised to refer Table 1.4 from page no 10-12 of the Design Data Book.<br /> <br /> Krishnaraja G. Kodancha, Assistant Professor, Automobile Engineering Department, B.V.B.College of Engineering and technology, HUBLI-31; Session 24 to 28, Oct 18 to 30 -2007<br /> <br /> 5<br /> <br />