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Chapter 3 Torsion Introduction -- Analyzing the stresses and strains in machine parts which are subjected to torque T Circular -- Cross-section Non-circular Irregular shapes -- Material(1) Elastic (2) Elasto-plastic -- Shaft(1) Solid (2) Hollow
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3.1 Introduction T is a vector Two ways of expression -- Applications: a. Transmission of torque in shafts, e.g. in automobiles
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Assumptions in Torque Analysis: a. Every cross section remains plane and undistorted. b. Shearing strain varies linearly along the axis of the shaft.
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3.2 Preliminary Discussion of the Stresses in a Shaft Free-body Diagram Where = distance (torque arm) Since dF = dA The stress distribution is Statically Indeterminate.
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-- Must rely on “deformation” to solve the problem. Analyzing a small element:
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3.3 Deformations in a Circular Shaft = (T, L) -- the angle of twist (deformation) Rectangular cross section warps under torsion
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A circular plane remains circular plane
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(in radians) Determination of Shear Strain The shear strain
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= c = radius of the shaft Since
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3.4 Stresses in the Elastic Range Hooke’s Law Therefore,(3.6)
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(3.1) (3.9) But Therefore,Or, (3.6)
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Substituting Eq. (3.9) into Eq. (3.6) (3.10) (3.9) These are elastic torsion formulas. For a solid cylinder: For a hollow cylinder:
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Since (3-13)
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Mohr’s Circle (Sec. 7.4) -- Pure Shear Condition
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Ductile materials fail in shear (90 o fracture) Brittle materials are weaker in tension (45 o fracture)
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3.5 Angle of Twist in the Elastic Range (3.3) (3.15) Eq. (3.3) = Eq. (3.15) Therefore, Hence,
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For Multiple-Section Shafts:
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Shafts with a Variable Circular Cross Section
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3.6 Statically Indeterminate Shafts -- Must rely on both (1)Torque equations and (2) Deformation equation, i.e. Example 3.05
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3.7 Design of Transmission Shafts -- Two Parameters in Transmission Shafts: a. Power P b. Speed of rotation where = angular velocity (radians/s) = 2 = frequency (Hz) [N.m/s = watts (W)](3.21)
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(3.9) For a Solid Circular Shaft: Therefore,
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3.8 Stress Concentrations in Circular Shafts
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3.9 Plastic Deformation sin Circular Shafts (3.4) c = radius of the shaft
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c dd dA = 2 d c (3.1) Knowing dF = dA Where = ( ) (3.26)
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(3.9) If we can determine experimentally an Ultimate Torque, T U, then by means of Eq. (3.9), we have R T = Modulus of Rupture in Torsion
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3.10 Circular Shafts Made of an Elasto-Plastic Material Case I: < Y Hooke’s Law applies, < max Case I Case II Case II: < Y Hooke’s Law applies, = max T Y = max elastic torque
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Since Case III: Entering Plastic Region Case III 0 Y : Y – region within the plastic range Y c: (3-29)
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(3.26) By evoking Eq. (3.26) (3.31)
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Case IV Case IV -- Fully Plastic Y 0: = Plastic Torque (3-33)
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3.11 Residual Stresses in Circular Shafts
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3.12 Torsion of Noncircular Members A rectangular shaft does not axisymmetry.
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From Theory of Elasticity:
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3.13 Thin-Walled Hollow Shafts
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