zunia.vn

Tuyển sinh 2024 dành cho Gen-Z

zunia.vn

» Kỹ Thuật - Công Nghệ

» Cơ khí - Chế tạo máy

Lecture Strength of Materials I: Chapter 6 - PhD. Tran Minh Tu

Chia sẻ: _ _ | Ngày: | Loại File: PDF | Số trang:28

Báo xấu

19
lượt xem 1
download

Download Vui lòng tải xuống để xem tài liệu đầy đủ

Lecture Strength of Materials I - Chapter 6: Torsion. The following will be discussed in this chapter: Introduction, torsional loads on circular shafts, strength condition and stiffness condition, statically indeterminate problem, strain energy, examples.

Chủ đề:

Bình luận(0) Đăng nhập để gửi bình luận!

Lưu

Nội dung Text: Lecture Strength of Materials I: Chapter 6 - PhD. Tran Minh Tu

STRENGTH OF MATERIALS 1/10/2013 TRAN MINH TU - University of Civil Engineering, 1 Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam
CHAPTER 6 TORSION 1/10/2013
Contents 6.1. Introduction 6.2. Torsional Loads on Circular Shafts 6.3. Strength Condition and stiffness condition 6.4. Statically Indeterminate Problem 6.5. Strain Energy 6.6. Examples Home’s works 1/10/2013 3
6.1. Introduction 1/10/2013 4
6.1. Introduction 1/10/2013 5
6.1. Introduction Torsion members – the slender members subjected to torsional loading, that is loaded by couple that produce twisting of the member about its axis Examples – A torsional moment (torque) is applied to the lug-wrench shaft, the shaft transmits the torque to the generator, the drive shaft of an automobile... • Torsional Loads on Circular Shafts: the torsional moment or couple A F x Q2 B C Q1 t z 2 T t T 1 1 2 y 1/10/2013 6
6.1. Introduction  Internal torsional moment diagram • Using method of section • Sign convention of Mz - Positive: clockwise - Negative: counterclockwise Mz > 0 M z 0 Mz = y y z z x x 1/10/2013 7
6.2. Torsion of Circular Shafts 6.2.1. Simplifying assumptions 1/10/2013 8
6.2. Torsion of Circular Shafts => In the cross-section, only shear stress exists 6.2.2. Compatibility • Consider the portion of the shaft shown in the figure • CD – before deformation • CD’ – after deformation - From the geometry DD '   d   dz d => The Shear strain:   dz - d – the angle of twist - Following Hooke’s law: d    G  G  1/10/2013 dz 9
6.2. Torsion of Circular Shafts 6.2.3. Equilibrium d 2 d M z      dA  G   dA  G Ip A dz A dz d M z    – the rate of twist dz GI p 6.2.3. Torsion formulas – Shearing stress Mz – internal torsional moment Mz    Ip – polar moment of inertia Ip  – radial position 1/10/2013 10
6.2. Torsion of Circular Shafts - Maximum shearing stress Mz Mz  max  R Ip Wp - Wp - Section modulus of torsion – Angle of twist c   a b O A B L A L M z dz M dz  rad  From 6.2.3:  AB    z 1/10/2013 B GI p 0 GI p 11
6.2. Torsion of Circular Shafts M zL Mz  const  AB  GI p GI p – Multiply torques GIp – stiffness of torsional shaft If the shaft is subjected to several different torques or cross-sectional area, or shear modulus changes abruptly from one region of the shaft to the next.  Mz     const  GI p i n  Mz   AB    li   i 1  GI p  i 1/10/2013 12
6.2. Torsion of Circular Shafts 1/10/2013 13
6.3. Strength Condition – Stiffness condition Mz  Strength condition:  max     Wp 0    - Determine experimentally 0 n     - Third strength hypothesis 3 2     - Fourth strength hypothesis 4 3  Stiffness condition:  Mz   max      rad / m   GI   p max 1/10/2013 14
6.3. Strength Condition – Stiffness condition  Three main problems: Mz For a circular shaft:  max  3     max  Mz 4    0.2D 0.1GD 1. Investigating the strength condition, (stiffness condition) Mz  max  3    ??? 0.2D 2. Determine the diameter of circular cross-section Mz D 3 0.2  3. Determine the maximum torque M z  0.2  D3 1/10/2013 15
6.4. Statically Indeterminate Problem • Assume that the reactions at the fixed ends MA, MD are shown in the figure. MA M MD • Equilibrium: MA + MD = M (1) 2d d A B D • Compatibility condition: AD = 0 a 2a (2) CD Mz MD AB BD  AD   AB   BD  M a M 2a z  z M BD z  MD AB BD D GI GI p p M AB z  MD  M z  AD   MD  M a  M D 2a 0 M/33 G  0,1  2d  G  0,1 d 4 4 Mz 1 32 MD  M; MA  M 33 33 32M/33 1/10/2013 16
6.5. Strain Energy • For a shaft subjected to a torsional load, 2  xy 2 2 T  U  dV   2 dV 2G 2GJ • Setting dV = dA dx, T 2  2  L L T 2 2 U   dA dx   2  dA dx 2GJ 2 2GJ  A  0A 0  T  xy  L T2 J  dx 2GJ 0 • In the case of a uniform shaft, T 2L U 2GJ 1/10/2013 17
6.6. Example Problem 1 • A Circular shaft made from two segments, each having diameter of D and 2D. The Shaft is subjected to the torques shown in the figure. M 3M 1. Draw the internal torsional 2D D moment diagram B C D 2. Determine the maximum 2a a shearing stress 3. Determine the angle of twist of the end D with respect to B With M=5kNm; a=1m; D=10cm; G=8.103 kN/cm2 1/10/2013 18
6.6. Example Problem 1 M 3M 1. Internal torsional moment diagram 2D D B C D Segment CD 0  z1  a  2a a 3M M CD z  3M  15kNm MzCD Segment BC 0  z2  2a  M z1 3M M BC z  2M  10kNm MzBC z2 a 15 10 Mz kNm 1/10/2013 19
6.6. Example Problem 1 M 3M 2. Maximum shearing stress 2D D B C M zCD 15  102 D  max    7,5(kN / cm2 ) CD 0,2 D 3 0,2 10 3 2a a M zBC 10  102  max    0,625(kN / cm2 ) 15 0, 2  2 D  0, 2  20 BC 3 3 10 Mz kNm   max  7,5(kN / cm2 ) 3. Angle of twist of end D M zCD  a M zBC  2a  D   BC  CD D  CD  GI p GI pBC 15  102 102 10 102  2 102 D    0,02(rad ) 8  10  0,110 8 10  0,1 20 3 4 3 4 1/10/2013 20

CÓ THỂ BẠN MUỐN DOWNLOAD

LV.15: Bộ Đồ Án Tốt Nghiệp Chuyên Ngành Cơ Khí

65 tài liệu

2431 lượt tải

THÔNG TIN

TRỢ GIÚP

HỖ TRỢ KHÁCH HÀNG

Theo dõi chúng tôi

Chịu trách nhiệm nội dung:

Nguyễn Công Hà - Giám đốc Công ty TNHH TÀI LIỆU TRỰC TUYẾN VI NA

LIÊN HỆ

Địa chỉ: P402, 54A Nơ Trang Long, Phường 14, Q.Bình Thạnh, TP.HCM

Hotline: 093 303 0098

Email: support@tailieu.vn

Giấy phép Mạng Xã Hội số: 670/GP-BTTTT cấp ngày 30/11/2015 Copyright © 2022-2032 TaiLieu.VN. All rights reserved.