Lecture6 - LECTURE Third Edition SHAFTS: TORSION LOADING...

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Unformatted text preview: LECTURE Third Edition SHAFTS: TORSION LOADING AND DEFORMATION • A. J. Clark School of Engineering •Department of Civil and Environmental Engineering 6 Chapter 3.1 - 3.5 by Dr. Ibrahim A. Assakkaf SPRING 2003 ENES 220 – Mechanics of Materials Department of Civil and Environmental Engineering University of Maryland, College Park LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading Slide No. 1 ENES 220 ©Assakkaf Introduction – Members subjected to axial loads were discussed previously. – The procedure for deriving loaddeformation relationship for axially loaded members was also illustrated. – This chapter will present a similar treatment of members subjected to torsion by loads that to twist the members about their longitudinal centroidal axes. 1 Slide No. 2 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf Torsional Loads on Circular Shafts • Interested in stresses and strains of circular shafts subjected to twisting couples or torques • Turbine exerts torque T on the shaft • Shaft transmits the torque to the generator • Generator creates an equal and opposite torque T’ Slide No. 3 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf Net Torque Due to Internal Stresses • Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque, T = ∫ ρ dF = ∫ ρ (τ dA) • Although the net torque due to the shearing stresses is known, the distribution of the stresses is not • Distribution of shearing stresses is statically indeterminate – must consider shaft deformations • Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform. 2 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading Slide No. 4 ENES 220 ©Assakkaf Introduction Cylindrical members Fig. 1 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading Slide No. 5 ENES 220 ©Assakkaf Introduction Rectangular members Fig. 2 3 Slide No. 6 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf Introduction – This chapter deals with members in the form of concentric circular cylinders, solid and hollow, subjected to torques about their longitudinal geometric axes. – Although this may seem like a somewhat a special case, it is evident that many torquecarrying engineering members are cylindrical in shape. Slide No. 7 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf Deformation of Circular Shaft – Consider the following shaft ab cd Fig. 3 l T T l 4 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading Slide No. 8 ENES 220 ©Assakkaf Deformation of Circular Shaft – In reference to the previous figure, the following observations can be noted: • The distance l between the outside circumferential lines does not change significantly as a result of the application of the torque. However, the rectangles become parallelograms whose sides have the same length as those of the original rectangles. LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading Slide No. 9 ENES 220 ©Assakkaf • The circumferential lines do not become zigzag; that is ; they remain in parallel planes. • The original straight parallel longitudinal lines, such as ab and cd, remain parallel to each other but do not remain parallel to the longitudinal axis of the member. These lines become helices*. *A helix is a path of a point that moves longitudinally and circumferentially along a surface of a cylinder at uniform rate 5 Slide No. 10 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf Deformation of Circular Shaft Subjected to Torque T Fig. 4 a T T b Slide No. 11 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 3.5) Torsion Loading ENES 220 ©Assakkaf Deformation of a Bar of Square Cross Section Subjected to Torque T Fig. 5 a T T b 6 Slide No. 12 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf Shaft Deformations • From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length. φ ∝T φ∝L • When subjected to torsion, every crosssection of a circular shaft remains plane and undistorted. • Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric. • Cross-sections of noncircular (nonaxisymmetric) shafts are distorted when subjected to torsion. LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading Slide No. 13 ENES 220 ©Assakkaf An Important Property of Circular Shaft – When a circular shaft is subjected to torsion, every cross section remains plane and disturbed – In other words, while the various cross sections along the shaft rotate through different amounts, each cross section rotates as a solid rigid slab. 7 Slide No. 14 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf An Important Property of Circular Shaft – This illustrated in Fig.4b, which shows the deformation in rubber model subjected to torsion. – This property applies to circular shafts whether solid or hollow. – It does not apply to noncircular cross section. When a bar of square cross section is subjected to torsion, its various sections are warped and do not remain plane (see Fig. 5.b) Slide No. 15 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf Stresses in Circular Shaft due to Torsion – Consider the following circular shaft that is subjected to torsion T Fig. 6 B C A T T 8 Slide No. 16 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf Stresses in Circular Shaft due to Torsion – A section perpendicular to the axis of the shaft can be passed at an arbitrary point C as shown in Fig. 6. – The Free-body diagram of the portion BC of the shaft must include the elementary shearing forces dF perpendicular to the radius ρ of the shaft. Slide No. 17 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading ENES 220 ©Assakkaf Stresses in Circular Shaft due to Torsion B C T ρ T Fig. 7 dF = τρ dA 9 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading Slide No. 18 ENES 220 ©Assakkaf Stresses in Circular Shaft due to Torsion – But the conditions of equilibrium for BC require that the system of these elementary forces be equivalent to an internal torque T. – Denoting ρ the perpendicular distance from the force dF to axis of the shaft, and expressing that the sum of moments of LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsion Loading Slide No. 19 ENES 220 ©Assakkaf Stresses in Circular Shaft due to Torsion – of the shearing forces dF about the axis of the shaft is equal in magnitude to the torque T, we can write T = Tr = ∫ ρ dF = ∫ ρ τ dA area (1) area 10 Slide No. 20 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsion Loading Stresses in Circular Shaft due to Torsion B C T T ρ T = Tr = ∫ ρ τ dA (2) area LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Shearing Strain Slide No. 21 ENES 220 ©Assakkaf If a plane transverse section before twisting remains plane after twisting and a diameter of the the section remains straight, the distortion of the shaft of Figure 7 will be as shown in the following figures (Figs. 8 and 9): 11 Slide No. 22 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Shearing Strain ENES 220 ©Assakkaf Shearing Strain • Consider an interior section of the shaft. As a torsional load is applied, an element on the interior cylinder deforms into a rhombus. • Since the ends of the element remain planar, the shear strain is equal to angle of twist. • It follows that Lγ = ρφ or γ = ρφ L • Shear strain is proportional to twist and radius γ max = ρ cφ and γ = γ max L c Slide No. 23 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Shearing Strain ENES 220 ©Assakkaf Shearing Strain c φ ρ L Fig. 8 12 Slide No. 24 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Shearing Strain Shearing Strain γ a′ a L ρ φ aa′ aa′ ≈ sin L L because γ is very small tan γ = Fig. 9 γ a′ a L Slide No. 25 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Shearing Strain ENES 220 ©Assakkaf Shearing Strain – From Fig. 9, the length aá can be expressed as – But aa′ = L tan γ = Lγ – Therefore, (3) aa′ = ρφ Lγ = ρφ ⇒ γ = (4) ρφ L (5) 13 Slide No. 26 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Shearing Strain Shearing Strain For radius ρ, the shearing strain for circular shaft is ρφ γρ = (6) L For radius c, the shearing strain for circular shaft is γc = cφ L (7) Slide No. 27 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Shearing Strain Shearing Strain Combining Eqs. 6 and 7, gives γ ρ L γ cL φ= = ρ c Therefore γρ = γc c ρ (8) (9) 14 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Slide No. 28 Torsional Shearing Stress ENES 220 ©Assakkaf The Elastic Torsion Formula If Hooke’s law applies, the shearing stress τ is related to the shearing strain γ by the equation τ = Gγ (10) where G = modulus of rigidity. Combining Eqs. 9 and 10, results in τρ G = τc Gc ρ ⇒τρ = τc c (11) ρ LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Slide No. 29 Torsional Shearing Stress ENES 220 ©Assakkaf The Elastic Torsion Formula When Eq. 11 is substituted into Eq. 2, the results will be as follows: ∫ ρ τ dA T = Tr = area τ τ = ∫ ρ c ρ dA = ∫ ρ c ρ dA c c 0 area c = τc c ∫ρ 2 dA (12) area 15 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Slide No. 30 Torsional Shearing Stress ENES 220 ©Assakkaf Polar Moment of Inertia The integral of equation 12 is called the polar moment of inertia (polar second moment of area). It is given the symbol J. For a solid circular shaft, the polar moment of inertia is given by c 4 J = ∫ ρ 2 dA = ∫ ρ 2 (2πρ dρ ) = πc 2 0 (13) LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Slide No. 31 Torsional Shearing Stress ENES 220 ©Assakkaf Polar Moment of Inertia – For a circular annulus as shown, the polar moment of inertia is given by c b c J = ∫ ρ dA = ∫ ρ 2 (2πρ dρ ) 2 b • = πc 2 4 − πb 2 4 = π 2 (r 4 o − ri 4 ) (14) r0 = outer radius and ri = inner radius 16 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Slide No. 32 Torsional Shearing Stress ENES 220 ©Assakkaf Shearing Stress in Terms of Torque and Polar Moment of Inertia – In terms of the polar second moment J, Eq. 12 can be written as T = Tr = τc c ∫ρ 2 dA = area τcJ (15) c – Solving for shearing stress, Tc τc = J (16) LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Slide No. 33 Torsional Shearing Stress ENES 220 ©Assakkaf Shearing Stress in Terms of Torque and Polar Moment of Inertia τ max = τρ = Tc J Tρ J (17a) (18a) τ= shearing stress, T = applied torque ρ = radius, and J = polar moment on inertia 17 Slide No. 34 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Shearing Stress Distribution of Shearing Stress within the Circular Cross Section τ Tc τc = J τ τmax τmin τmax ro c ρ ρ ri Fig. 10 Slide No. 35 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Stresses in Elastic Range ENES 220 ©Assakkaf • Multiplying the previous equation by the shear modulus, Gγ = ρ c Gγ max From Hooke’s Law, τ = Gγ , so τ= ρ c τ max The shearing stress varies linearly with the radial position in the section. J = 1 π c4 2 • Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section, τ τ T = ∫ ρτ dA = max ∫ ρ 2 dA = max J c c J= ( 1 π c4 2 2 4 − c1 ) • The results are known as the elastic torsion formulas, τ max = Tc Tρ and τ = J J 18 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Shearing Stress Slide No. 36 ENES 220 ©Assakkaf Example 1 – A hollow cylindrical steel shaft is 1.5 m long and has inner and outer diameters equal to 40 mm and 60 mm. (a) What is the largest torque which may be applied to the shaft if the shearing stress is not to exceed 120 MPa? (b) What is the corresponding minimum value of the shearing stress in the shaft? LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Shearing Stress Slide No. 37 ENES 220 ©Assakkaf Example 1 (cont’d) T 60 mm 40 mm T 1.2 m Fig. 11 19 Slide No. 38 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Shearing Stress ENES 220 ©Assakkaf Example 1 (cont’d) – (a) Largest Permissible Torque Using Eq.17a τJ Tc ⇒ Tmax = max J c τ max = (19) Using Eq.14 for claculating J , J= π (r 2 4 o ) − ri 4 = [(0.03) − (0.02) ] 2 π 4 4 = 1.021× 10 −6 m 4 Slide No. 39 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Shearing Stress Example 1 (cont’d) Substituting for J and τmax into Eq. 19, we have T= ( )( ) Jτ max 1.021× 10 −6 120 × 106 = = 4.05 kN ⋅ m 0.03 ro (b) Minimum Shearing Stress τρ = Tri 4.05 ×103 (0.02) = 79.3 MPa = J 1.021×10 −6 20 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements Slide No. 40 ENES 220 ©Assakkaf Angle of Twist in the Elastic Range – Often, the amount of twist in a shaft is of importance. – Therefore, determination of angle of twist is a common problem for the machine designer. – The fundamental equations that govern the amount of twist were discussed previously LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements Slide No. 41 ENES 220 ©Assakkaf Angle of Twist in the Elastic Range – The basic equations that govern angle of twist are – Recall Eqs. 6, γρ = ρφ L or γρ = ρ dθ dL (20) 21 Slide No. 42 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements ENES 220 ©Assakkaf Angle of Twist in the Elastic Range τc = Tc J or τ ρ = Tρ J (21) and G= τ γ (22) LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements Slide No. 43 ENES 220 ©Assakkaf Angle of Twist in the Elastic Range – Recall Eq. 17a and 7 τ max = Tc J γ max = cθ L – Combining these two equations, gives θ= γ max L c TL = GJ = τ max L Tc 1 L = G c J G c 22 Slide No. 44 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Displacements Angle of Twist in the Elastic Range The angle of twist for a circular uniform shaft subjected to external torque T is given by TL θ= GJ (22) LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements Slide No. 45 ENES 220 ©Assakkaf Angle of Twist in the Elastic Range – Multiple Torques/Sizes • The expression for the angle of twist of the previous equation may be used only if the shaft is homogeneous (constant G) and has a uniform cross sectional area A, and is loaded at its ends. • If the shaft is loaded at other points, or if it consists of several portions of various cross sections, and materials, then 23 Slide No. 46 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Displacements Angle of Twist in the Elastic Range – Multiple Torques/Sizes • It needs to be divided into components which satisfy individually the required conditions for application of the formula. • Denoting respectively by Ti, Li, Ji, and Gi, the internal torque, length, polar moment of area, and modulus of rigidity corresponding to component i,then n n θ = ∑θ i = ∑ i =1 i =1 Ti Li Gi J i (23) Slide No. 47 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Displacements n i =1 θ = ∑θ i = ∑ E1 E2 L2 Ti Li Gi J i E3 L1 Fig. 12 n i =1 Multiple Torques/Sizes L3 Circular Shafts 24 Slide No. 48 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Displacements Angle of Twist in the Elastic Range The angle of twist of various parts of a shaft of uniform member can be given by n n θ = ∑θ i = ∑ i =1 i =1 Ti Li Gi J i LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Angle of Twist in Elastic Range (24) Slide No. 49 ENES 220 ©Assakkaf • Recall that the angle of twist and maximum shearing strain are related, γ max = cφ L • In the elastic range, the shearing strain and shear are related by Hooke’s Law, γ max = τ max G = Tc JG • Equating the expressions for shearing strain and solving for the angle of twist, φ= TL JG • If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations φ =∑ i Ti Li J i Gi 25 Slide No. 50 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) ENES 220 ©Assakkaf Torsional Displacements Angle of Twist in the Elastic Range If the properties (T, G, or J) of the shaft are functions of the length of the shaft, then L T dx GJ 0 θ =∫ (25) Slide No. 51 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements ENES 220 ©Assakkaf Angle of Twist in the Elastic Range – Varying Properties L T dx GJ 0 θ =∫ L × x Fig. 13 26 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements Slide No. 52 ENES 220 ©Assakkaf Example 2 – What torque should be applied to the end of the shaft of Example 1 to produce a twist of 20? Use the value G = 80 GPa for the modulus of rigidity of steel. LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements Slide No. 53 ENES 220 ©Assakkaf Example 2 (cont’d) T 60 mm 40 mm T 1.2 m Fig. 14 27 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Slide No. 54 ENES 220 ©Assakkaf Torsional Displacements Example 2 (cont’d) Solving Eq. 22 for T, we get JG T= θ (26) L Substituting the given values G = 80 × 109 Pa L = 1.5 m 2π rad −3 = 34.9 ×10 rad 3600 θ = 20 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Slide No. 55 ENES 220 ©Assakkaf Torsional Displacements Example 2 (cont’d) From Example 1, J was computed to give a value of 1.021×10-6 m4. Therefore,using Eq. 26 ( )( )( 1.021×10 −6 80 ×109 JG 34.9 ×10 −3 θ= T= 1.5 L = 1.9 ×103 N ⋅ m = 1.9 kN ⋅ m ) 28 Slide No. 56 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements ENES 220 ©Assakkaf Example 3 What angle of twist will create a shearing stress of 70 MPa on the inner surface of the hollow steel shaft of Examples 1 and 2? T 60 mm 40 mm T 1.2 m Fig. 14 LECTURE 6. SHAFTS: TORSION LOADING AND DEFORMATION (3.1 – 3.5) Torsional Displacements Slide No. 57 ENES 220 ©Assakkaf Example 3 (cont’d) τρ = Jτ ρ Tρ ⇒T = ρ J (1.021×10 -6 )(70 × 106 ) = = 3.5735 kN ⋅ m 0.02 3.5735 × 103 (1.5) TL = = 0.65625 φ= GJ 80 × 109 (1.021× 10 −6 ) To obtain θ in degrees, we write θ = 0.65625 360 = 3.760 2π 29 ...
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