CHAPER MECHANICS OF MAERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John. DeWolf orsion Leture Notes: J. Walt Oler exas eh University 006 he MGraw-Hill Companies, In. All rights reserved.
Contents Introdution orsional lloads on Cirular Shafts Net orque Due to Internal Stresses Axial lshear Components Shaft Deformations Shearing Strain Stresses in Elasti Range Normal lstresses orsional Failure Modes Sample Problem 3.1 Angle of wist in Elasti Range Statially Indeterminate Shafts Sample Problem 3.4 Design of ransmission Shafts Stress Conentrations ti Plasti Deformations Elastoplasti l ti Materials Residual Stresses Example 3.08/3.09 09 orsion of Nonirular Members hin-walled WlldHll Hollow Shafts Shft Example 3.10 006 he MGraw-Hill Companies, In. All rights reserved. 3 -
orsional Loads on Cirular Shafts Interested in stresses and strains of irular shafts subjeted to twisting ouples or torques urbine exerts torque on the shaft Shaft transmits the torque to the generator Generator reates an equal and opposite torque 006 he MGraw-Hill Companies, In. All rights reserved. 3-3
Net orque Due to Internal Stresses Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque, ( τ ) ρ df ρ da Although the net torque due to the shearing stresses is known, the distribution of the stresses is not. Distribution of shearing stresses is statially indeterminate must onsider shaft deformations. Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads an not be assumed uniform. 006 he MGraw-Hill Companies, In. All rights reserved. 3-4
Axial Shear Components orque applied to shaft produes shearing stresses on the faes perpendiular to the axis. Conditions i of equilibrium i require the existene of equal stresses on the faes of the two planes ontaining the axis of the shaft. he existene of the axial shear omponents is demonstrated by onsidering a shaft made up of axial slats. he slats slide with respet to eah other when equal and opposite torques are applied to the ends of the shaft. 006 he MGraw-Hill Companies, In. All rights reserved. 3-5
Shaft Deformations From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length. φ φ L When subjeted to torsion, every ross-setion setion of a irular shaft remains plane and undistorted. Cross-setions for hollow and solid irular shafts remain plain and undistorted beause a irular shaft is axisymmetri. Cross-setions of nonirular (nonaxisymmetri) shafts are distorted when subjeted to torsion. 006 he MGraw-Hill Companies, In. All rights reserved. 3-6
Shearing Strain Consider an interior setion of the shaft. As a torsional load is applied, an element on the interior ylinder deforms into a rhombus. Sine 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 φφ and γ L ρ γ 006 he MGraw-Hill Companies, In. All rights reserved. 3-7
Stresses in Elasti Range J J 1 4 π Multiplying the previous equation by the shear modulus, ρ G γ Gγγ ρ From Hooke s Law, τ τ τ Gγ, so he shearing stress varies linearly with the radial position in the setion. 4 4 ( ) 1 1 π Reall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the setion, τ da τ ρτ ρ da ρ τ and τ J J he results are known as the elasti torsion formulas, 006 he MGraw-Hill Companies, In. All rights reserved. 3-8 J
Normal Stresses Elements with faes parallel and perpendiular to the shaft axis are subjeted to shear stresses only. Normal stresses, shearing stresses or a ombination of both may be found for other orientations. Consider an element at 45 o to the shaft axis, ( τ A0 ) os45 0 F τ A F τ A σ τ 0 o 45 A A0 Element a is in pure shear. Element is subjeted to a tensile stress on two faes and ompressive stress on the other two. Note that all stresses for elements a and have the same magnitude 006 he MGraw-Hill Companies, In. All rights reserved. 3-9
orsional Failure Modes Dutile materials generally fail in shear. Brittle materials are weaker in tension than shear. When subjeted to torsion, a dutile speimen breaks along a plane of imum shear, i.e., a plane perpendiular to the shaft axis. When subjeted to torsion, a brittle speimen breaks along planes perpendiular to the diretion in whih tension is a imum, i.e., along surfaes at 45 o to the shaft axis. 006 he MGraw-Hill Companies, In. All rights reserved. 3-10
Sample Problem 3.1 SOLUION: Shaft BC is hollow with inner and outer diameters of 90 mm and 10 mm, respetively. Shafts AB and CD are solid of diameter d. For the loading shown, determine (a) the minimum and imum 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. Cut setions through shafts AB and BC and perform stati equilibrium analyses to find torque loadings. Apply elasti torsion formulas to find minimum and imum stress on shaft BC. Given allowable shearing stress and applied torque, invert the elasti torsion formula to find the required diameter. 006 he MGraw-Hill Companies, In. All rights reserved. 3-11
Sample Problem 3.1 SOLUION: Cut setions through shafts AB and BC and perform stati equilibrium analysis to find torque loadings. M AB x 0 ( ) ( ) ( ) 0 6kN m M 0 6kN m + 14kN m 6kN m CD AB BC x 0kN m BC 006 he MGraw-Hill Companies, In. All rights reserved. 3-1
Sample Problem 3.1 Apply elasti torsion formulas to find minimum and imum stress on shaft BC. Given allowable shearing stress and applied torque, invert the elasti torsion formula to find the required diameter. π J τ τ τ τ 4 4 π 4 4 ( ) [( 0.060) ( 0.045) ] 1 13.9 10 τ 6 m BC J 86.MPa 4 τmin 86.MPa min 1 min 64.7 MPa ( 0kN m)( 0.060 m) 13.9 10 45mm 60mm 6 m 4 τ τ min τ J 38.9 10 86.MPa 64.7 MPa π 3 m 4 65MPa d 6kN m 3 π 77.8mm 006 he MGraw-Hill Companies, In. All rights reserved. 3-13
Angle of wist in Elasti Range Reall that the angle of twist and imum shearing strain are related, φ γ L In the elasti range, the shearing strain and shear are related dby Hooke s Law, τ γ G JG Equating the expressions for shearing strain and solving for the angle of twist, L φ JG If the torsional loading or shaft ross-setion hanges along the length, the angle of rotation is found as the sum of segment rotations il φ i J G i i i 006 he MGraw-Hill Companies, In. All rights reserved. 3-14
Statially Indeterminate Shafts Given the shaft dimensions and the applied torque, we would like to find the torque reations at A and B. From a free-body analysis of the shaft, A + B 90lb ft whih is not suffiient to find the end torques. he problem is statially ti indeterminate. i t Divide the shaft into two omponents whih must have ompatible deformations, AL BL φ φ 1 1 + φ 0 B J G J G 1 L 1 J 90lb ft L J A + A 1 L1 J L J Substitute into the original equilibrium equation, 1 A 006 he MGraw-Hill Companies, In. All rights reserved. 3-15
Sample Problem 3.4 SOLUION: Apply a stati equilibrium analysis on the two shafts to find a relationship between CD and 0. Apply a kinemati analysis to relate the angular rotations of the gears. wo solid steel shafts are onneted by ygears. Knowing that for eah shaft G 11. x 10 6 psi and that the allowable shearing stress is 8 ksi, determine (a) the largest torque 0 that may be applied to the end of shaft AB, (b) the orresponding angle through whih end A of shaft AB rotates. Find the imum allowable torque on eah shaft hoose the smallest. Find the orresponding angle of twist for eah shaft and the net angular rotation of end A. 006 he MGraw-Hill Companies, In. All rights reserved. 3-16
Sample Problem 3.4 SOLUION: Apply a stati equilibrium analysis on the two shafts to find a relationship between CD and 0. Apply a kinemati analysis to relate the angular rotations of the gears. M M CD B C ( 0.875in. ) 0 F 0 F.8 0 (.45in. ) 0 CD r φ r φ φ B B B B r r C B C φ C φ. 8 φ C C.45in. φ 0.875in. C 006 he MGraw-Hill Companies, In. All rights reserved. 3-17