Giáo trình Strength of Materials - Chapter 9: Torsion
Tóm tắt Giáo trình Strength of Materials - Chapter 9: Torsion: ...weaker in tension than shear. • When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis. • When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction in which tension is...max J M z max exists on the perimeter of the circular section The relative angle of twist (góc xoắn tỷ đối), GJ M z The angle of twist (góc xoắn), GJ LM z 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS STATICALLY INDETERMINATE SHAFTS • Given the shaft dimensions and the ap...o48.10A 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS DESIGN OF TRANSMISSION SHAFTS • Principal transmission shaft performance specifications are: - power - speed • Determine torque applied to shaft at specified power and speed, f PP T fTTP 2 2 • Find shaft cross-se...
CHAPTER 9: TORSION 9.0 Introduction 9.1 Torsional loads on circular shafts 9.2 Torsion of noncircular members 9.3 Helical spring under axial load 9.1 INTRODUCTION A bar subjected to loadings has internal torsion moment only on cross sections: Mz x y z Mz > 0 (counter clockwise) Eye’s direction 9.1 INTRODUCTION 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS • Interested in stresses and strains of circular shafts subjected to twisting couples or torques • Generator creates an equal and opposite torque T’ • Shaft transmits the torque to the generator • Turbine exerts torque T on the shaft 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS NET TORQUE DUE TO INTERNAL STRESSES dAdFT • Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque, • Although the net torque due to the shearing stresses is known, the distribution of the stresses is not • Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform. • Distribution of shearing stresses is statically indeterminate – must consider shaft deformations 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS AXIAL SHEAR COMPONENTS • Torque applied to shaft produces shearing stresses on the faces perpendicular to the axis. • The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats. The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft. • Conditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS SHAFT DEFORMATIONS • From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length. L T • When subjected to torsion, every cross-section of a circular shaft remains plane and undistorted. • Cross-sections of noncircular (non- axisymmetric) shafts are distorted when subjected to torsion. • Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 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. • Shear strain is proportional to twist and radius maxmax and cL c L L or • It follows that • Since the ends of the element remain planar, the shear strain is equal to angle of twist. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS STRESSES IN ELASTIC RANGE J c dA c dAT max2max • Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section, 4 2 1 cJ 414221 ccJ and max J T J Tc • The results are known as the elastic torsion formulas, • Multiplying the previous equation by the shear modulus, max G c G max c From Hooke’s Law, G , so The shearing stress varies linearly with the radial position in the section. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS NORMAL STRESSES • Elements with faces parallel and perpendicular to the shaft axis are subjected to shear stresses only. Normal stresses, shearing stresses or a combination of both may be found for other orientations. max 0 0max 45 0max0max 2 2 245cos2 o A A A F AAF • Consider an element at 45o to the shaft axis, • Element a is in pure shear. • Note that all stresses for elements a and c have the same magnitude • Element c is subjected to a tensile stress on two faces and compressive stress on the other two. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS TORSIONAL FAILURE MODES • Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear. • When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis. • When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft axis. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS EXAMPLES 9.01 Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively. Shafts AB and CD are solid of diameter d. For the loading shown, determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD if the allowable shearing stress in these shafts is 65 MPa. SOLUTION: • Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings • Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter • Apply elastic torsion formulas to find minimum and maximum stress on shaft BC 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS EXAMPLES 9.01 SOLUTION: • Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings CDAB ABx TT TM mkN6 mkN60 mkN20 mkN14mkN60 BC BCx T TM 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS EXAMPLES 9.01 • Apply elastic torsion formulas to find minimum and maximum stress on shaft BC 46 444 1 4 2 m1092.13 045.0060.0 22 ccJ MPa2.86 m1092.13 m060.0mkN20 46 2 2max J cTBC MPa7.64 mm60 mm45 MPa2.86 min min 2 1 max min c c MPa7.64 MPa2.86 min max • Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter m109.38 mkN6 65 3 3 2 4 2 max c c MPa c Tc J Tc mm8.772 cd 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS ANGLE OF TWIST IN ELASTIC RANGE • Recall that the angle of twist and maximum shearing strain are related, L c max • In the elastic range, the shearing strain and shear are related by Hooke’s Law, JG Tc G maxmax • Equating the expressions for shearing strain and solving for the angle of twist, JG TL • If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations i ii ii GJ LT 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS SUMMARY 4 2 1 cJ 414221 ccJ P M0 z x Mz = M0 c J M zmax J M z max exists on the perimeter of the circular section The relative angle of twist (góc xoắn tỷ đối), GJ M z The angle of twist (góc xoắn), GJ LM z 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS STATICALLY INDETERMINATE SHAFTS • Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B. • From a free-body analysis of the shaft, which is not sufficient to find the end torques. The problem is statically indeterminate. ftlb90 BA TT ftlb90 12 21 AA T JL JL T • Substitute into the original equilibrium equation, AB BA T JL JL T GJ LT GJ LT 12 21 2 2 1 1 21 0 • Divide the shaft into two components which must have compatible deformations, 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS EXAMPLE 9.02 Two solid steel shafts are connected by gears. Knowing that for each shaft G = 11.2 x 106 psi and that the allowable shearing stress is 8 ksi, determine (a) the largest torque T0 that may be applied to the end of shaft AB, (b) the corresponding angle through which end A of shaft AB rotates. SOLUTION: • Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0 • Find the corresponding angle of twist for each shaft and the net angular rotation of end A • Find the maximum allowable torque on each shaft – choose the smallest • Apply a kinematic analysis to relate the angular rotations of the gears 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS EXAMPLE 9.02 SOLUTION: • Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0 0 0 8.2 in.45.20 in.875.00 TT TFM TFM CD CDC B • Apply a kinematic analysis to relate the angular rotations of the gears CB CC B C B CCBB r r rr 8.2 in.875.0 in.45.2 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS EXAMPLE 9.02 • Find the T0 for the maximum allowable torque on each shaft – choose the smallest in.lb561 in.5.0 in.5.08.2 8000 in.lb663 in.375.0 in.375.0 8000 0 4 2 0 max 0 4 2 0 max T T psi J cT T T psi J cT CD CD AB AB inlb5610 T • Find the corresponding angle of twist for each shaft and the net angular rotation of end A oo / oo o 64 2 / o 64 2 / 2.2226.8 26.895.28.28.2 95.2rad514.0 psi102.11in.5.0 .24in.lb5618.2 2.22rad387.0 psi102.11in.375.0 .24in.lb561 BABA CB CD CD DC AB AB BA in GJ LT in GJ LT o48.10A 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS DESIGN OF TRANSMISSION SHAFTS • Principal transmission shaft performance specifications are: - power - speed • Determine torque applied to shaft at specified power and speed, f PP T fTTP 2 2 • Find shaft cross-section which will not exceed the maximum allowable shearing stress, shafts hollow 2 shafts solid 2 max 4 1 4 2 22 max 3 max T cc cc J T c c J J Tc • Designer must select shaft material and cross-section to meet performance specifications without exceeding allowable shearing stress. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS STRESS CONCENTRATIONS • The derivation of the torsion formula, assumed a circular shaft with uniform cross-section loaded through rigid end plates. J Tc max J Tc Kmax • Experimental or numerically determined concentration factors are applied as • The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, and cross-section discontinuities can cause stress concentrations 9.2 TORSION OF NONCIRCULAR MEMEBERS TORSION OF NONCIRCULAR MEMBERS • At large values of a/b, the maximum shear stress and angle of twist for other open sections are the same as a rectangular bar. Gabc TL abc T 3 2 2 1 max • For uniform rectangular cross-sections, • Previous torsion formulas are valid for axisymmetric or circular shafts • Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly 9.2 TORSION OF NONCIRCULAR MEMEBERS SUMMARY FOR TORSION OF RECTANGULAR MEMBERS b h Mz z 1 max 1 1 2max hb M z max1 3hb M z , , : Tabulated factors depend on the ratio h/b 9.2 TORSION OF NONCIRCULAR MEMEBERS THIN-WALLED HOLLOW SHAFTS • Summing forces in the x-direction on AB, shear stress varies inversely with thickness flowshear 0 qttt xtxtF BBAA BBAAx tA T qAdAqdMT dAqpdsqdstpdFpdM 2 22 2 0 0 • Compute the shaft torque from the integral of the moments due to shear stress t ds GA TL 24 • Angle of twist (accepted, from energy method) 9.2 TORSION OF NONCIRCULAR MEMEBERS EXAMPLE 9.03 Extruded aluminum tubing with a rectangular cross-section has a torque loading of 24 kip- in. Determine the shearing stress in each of the four walls with (a) uniform wall thickness of 0.160 in. and wall thicknesses of (b) 0.120 in. on AB and CD and 0.200 in. on CD and BD. SOLUTION: • Determine the shear flow through the tubing walls • Find the corresponding shearing stress with each wall thickness 9.2 TORSION OF NONCIRCULAR MEMEBERS EXAMPLE 9.03 SOLUTION: • Determine the shear flow through the tubing walls in. kip 335.1 in.986.82 in.-kip24 2 in.986.8in.34.2in.84.3 2 2 A T q A • Find the corresponding shearing stress with each wall thickness with a uniform wall thickness, in.160.0 in.kip335.1 t q ksi34.8 with a variable wall thickness in.120.0 in.kip335.1 ACAB in.200.0 in.kip335.1 CDBD ksi13.11 BCAB ksi68.6 CDBC 9.3 HELICAL SPRING UNDER AXIAL LOAD INTRODUCTION 9.3 HELICAL SPRING UNDER AXIAL LOAD INTRODUCTION 9.3 HELICAL SPRING UNDER AXIAL LOAD INTRODUCTION 9.3 HELICAL SPRING UNDER AXIAL LOAD INTRODUCTION 9.3 HELICAL SPRING UNDER AXIAL LOAD INTERNAL FORCES F F D F T F 9.3 HELICAL SPRING UNDER AXIAL LOAD d : Wire diameter D : Mean coil diameter n : Active coils L : Free length 𝝀 : Deflection P : Load k : Spring constant (stiffness) G : Shear modulus τ : Torsional corrected stress K : Application correction factor BASIC FORMULAS FOR SPRING 3 4 8nD Gd k 3 8 d PD K 1 25.0 d D d D K 4 8 Gd PDn OR
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