9 torsion 2015 bach khoa

33 4 0
  • Loading ...
1/33 trang

Thông tin tài liệu

Ngày đăng: 27/01/2019, 14:56

Đây là tài liệu của các bạn sinh viện hiện tại đang học tại Đại học Bách Khoa TP HCM. Đồng thời cũng là giáo án của giảng viên tại Đại học Bách Khoa. Nó sẽ rất hữu ích cho công việc học tập của các Bạn. Chúc Bạn thành công. 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 Mz > (counter clockwise) Eye’s direction z x y 9.1 INTRODUCTION 9.1 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’ 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS NET TORQUE DUE TO INTERNAL STRESSES • Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque, T    dF     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 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS AXIAL SHEAR COMPONENTS • Torque applied to shaft produces shearing stresses on the faces perpendicular to the axis • Conditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft • 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 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  T L • When subjected to torsion, every cross-section 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 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 • Since the ends of the element remain planar, the shear strain is equal to angle of twist • It follows that L   or    L • Shear strain is proportional to twist and radius  max  c  and    max L c 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS STRESSES IN ELASTIC RANGE • Multiplying the previous equation by the shear modulus, G   c G max From Hooke’s Law,   G , so   c  max The shearing stress varies linearly with the radial position in the section J  12  c • Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section,   T    dA  max   dA  max J c c  J  12  c24  c14  • The results are known as the elastic torsion formulas,  max  Tc T and   J J 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 • Consider an element at 45o to the shaft axis, F  2 max A0 cos 45   max A0  45o  F  max A0    max A A0 • Element a is in pure shear • Element c is subjected to a tensile stress on two faces and compressive stress on the other two • Note that all stresses for elements a and c have the same magnitude 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  M B   F 0.875 in.  T0 • Apply a kinematic analysis to relate the angular rotations of the gears rB B  rCC rC 2.45 in C  C rB 0.875 in  M C   F 2.45 in.  TCD B  TCD  2.8 T0  B  2.8C 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS EXAMPLE 9.02 • Find the T0 for the maximum • Find the corresponding angle of twist for each allowable torque on each shaft – shaft and the net angular rotation of end A choose the smallest A / B   max  T 0.375 in. TAB c 8000 psi   0.375 in.4 J AB T0  663 lb  in  max  TCD c 2.8 T0 0.5 in. 8000 psi   0.5 in.4 J CD T0  561lb  in T0  561lb  in 561 lb  in.24in. TAB L  J AB G  0.375 in.4 11.2  10 psi    0.387 rad  2.22 o C / D  TCD L 2.8 561 lb  in.24in.  J CDG  0.5 in.4 11.2  10 psi   0.514 rad  2.95o     B  2.8C  2.8 2.95o  8.26 o  A   B   A / B  8.26 o  2.22 o  A  10.48o 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS DESIGN OF TRANSMISSION SHAFTS • Principal transmission shaft performance specifications are: - power - speed • Designer must select shaft material and cross-section to meet performance specifications without exceeding allowable shearing stress • Determine torque applied to shaft at specified power and speed, P  T  2fT T P   P 2f • Find shaft cross-section which will not exceed the maximum allowable shearing stress,  max  Tc J J  T  c  c  max  solid shafts  J  4 T  c2  c1  c2 2c2  max hollow shafts 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS STRESS CONCENTRATIONS • The derivation of the torsion formula,  max  Tc J assumed a circular shaft with uniform cross-section loaded through rigid end plates • The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, and cross-section discontinuities can cause stress concentrations • Experimental or numerically determined concentration factors are applied as  max  K Tc J 9.2 TORSION OF NONCIRCULAR MEMEBERS TORSION OF NONCIRCULAR MEMBERS • Previous torsion formulas are valid for axisymmetric or circular shafts • Planar cross-sections of noncircular shafts not remain planar and stress and strain distribution not vary linearly • For uniform rectangular cross-sections,  max  T c1ab2  TL c2ab3G • 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 9.2 TORSION OF NONCIRCULAR MEMEBERS SUMMARY FOR TORSION OF RECTANGULAR MEMBERS  max 1 Mz h    max 1 z max 1 b Mz  hb Mz    hb , ,  : 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,  Fx    A t Ax    B t B x   At A  Bt B   t  q  shear flow shear stress varies inversely with thickness • Compute the shaft torque from the integral of the moments due to shear stress dM  p dF  p t ds   q pds   2q dA T   dM   2q dA  2qA  T 2tA • Angle of twist (accepted, from energy method) TL ds   4A G t 9.2 TORSION OF NONCIRCULAR MEMEBERS EXAMPLE 9.03 Extruded aluminum tubing with a rectangular cross-section has a torque loading of 24 kipin 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 • Find the corresponding shearing stress with each wall thickness with a uniform wall thickness,  q 1.335 kip in  t 0.160 in   8.34 ksi with a variable wall thickness A  3.84 in.2.34 in.  8.986 in.2 q T 24 kip - in kip   335 A 8.986 in.2 in    AB   AC  1.335 kip in 0.120 in  AB   BC  11.13 ksi  BD   CD  1.335 kip in 0.200 in  BC   CD  6.68 ksi 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 T F F D 9.3 HELICAL SPRING UNDER AXIAL LOAD BASIC FORMULAS FOR SPRING d : Wire diameter Gd k D : Mean coil diameter 8nD n : Active coils L : Free length 8PD  K 𝝀 : Deflection d P : Load k : Spring constant (stiffness) PDn G : Shear modulus  Gd τ : Torsional corrected stress K : Application correction factor OR D    0.25 d K  D   1 d
- Xem thêm -

Xem thêm: 9 torsion 2015 bach khoa, 9 torsion 2015 bach khoa

Gợi ý tài liệu liên quan cho bạn

Nhận lời giải ngay chưa đến 10 phút Đăng bài tập ngay