Statics and Mechanics of Materials

Transcription

1 Statics and Mechanics of Materials Chapter 7. orsional Loading: Shafts

2 Contents orsional Loads on Circular Shafts Net orque Due to Internal Stresses Axial Shear Components Shaft Deformations Shearing Strain Stresses in Elastic Range Normal Stresses orsional Failure Modes Example Angle of wist in Elastic Range Statically Indeterminate Shafts Example

3 orsional Loads on Circular Shafts Interested in stresses and strains of circular shafts subjected to twisting couples or torques urbine exerts torque on the shaft Shaft transmits the torque to the generator Generator creates an equal and opposite torque

4 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 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.

5 Axial Shear Components orque 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 he existence of the axial shear components is demonstrated by considering a shaft made up of axial slats. he slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.

6 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 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.

7 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 Lγ ρφ or γ ρφ L Shear strain is proportional to twist and radius γ cφ L and γ ρ γ c

8 Stresses in Elastic Range J 1 4 π c Multiplying the previous equation by the shear modulus, ρ G γ From Hooke s Law, τ ρ c τ Gγ c τ Gγ, so he shearing stress varies linearly with the radial position in the section. Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section, τ da τ ρτ ρ da c c J J 4 4 ( c ) 1 c 1 π he results are known as the elastic torsion formulas, c ρ τ and τ J J

9 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 45 o to the shaft axis, F ( τ A ) cos 45 τ A σ o 45 F A 0 τ A 0 A 0 τ Element a is in pure shear. 0 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

10 orsional 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 imum 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 imum, i.e., along surfaces at 45 o to the shaft axis.

11 Example SOLUION: Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings Shaft BC is hollow with inner and outer diameters of 90 mm and 10 mm, respectively. 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. Apply elastic torsion formulas to find minimum and imum stress on shaft BC Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter

12 SOLUION: Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings M AB x 0 6kN m ( 6kN m) CD AB M BC x 0 ( 6kN m) + ( 14kN m) BC 0kN m

13 Apply elastic torsion formulas to find minimum and imum stress on shaft BC Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter J τ τ τ τ π 4 4 π 4 4 ( c c ) [( 0.060) ( 0.045) ] min min τ 6 86.MPa c c 1 m BC c J 64.7 MPa 4 τmin 86.MPa ( 0kN m)( m) m 4 45mm 60mm τ 86.MPa τ min 64.7 MPa τ c J c π 3 c c m 4 65MPa d 6kN m π 3 c c 77.8mm

14 Angle of wist in Elastic Range Recall that the angle of twist and imum shearing strain are related, γ cφ L In the elastic range, the shearing strain and shear are related by Hooke s Law, γ τ c G JG Equating the expressions for shearing strain and solving for the angle of twist, φ L JG If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations il φ i J G i i i

15 Example 7-1

16 Solution: he shaft consists of three portions AB, BC, CD each of uniform cross section and each with a constant internal torque

17 Angle of wist: Recalling that G8- Gpa for the entire shaft, we have

18 Example SOLUION: Apply a static equilibrium analysis on the two shafts to find a relationship between CD and 0 Apply a kinematic analysis to relate the angular rotations of the gears wo solid steel shafts are connected by gears. Knowing that for each 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 corresponding angle through which end A of shaft AB rotates. Find the imum allowable torque on each shaft choose the smallest Find the corresponding angle of twist for each shaft and the net angular rotation of end A

19 SOLUION: Apply a static equilibrium analysis on the two shafts to find a relationship between CD and 0 Apply a kinematic analysis to relate the angular rotations of the gears M M CD B C F F ( 0.875in. ) (.45in. ) 0 CD r φ φ B φ r φ B B B r r C B C φ C.8φ C C.45in. φ 0.875in. C

20 Find the 0 for the imum allowable torque on each shaft choose the smallest Find the corresponding angle of twist for each shaft and the net angular rotation of end A φ A/ B J AB AB L G π ( 561lb in. )( 4in. ) 4 6 ( 0.375in. ) ( psi) τ τ lb in. J J AB AB CD c CD c 561lb in psi 8000 psi 0 π ( 0.375in. ) 4 ( 0.375in. ).8 π 0 ( 0.5in. ) 4 ( 0.5in. ) 0 561lb in φ C / D φ φ B A rad. J.8φ π rad.95 φ CD CD L G C.8.8 o ( 561lb in. )( 4in. ) 4 6 ( 0.5in. ) ( psi) o o (.95 ) 8.6 o o o B + φa/ B o φ A o

21 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, A + B 90lb ft which is not sufficient to find the end torques. he problem is statically indeterminate. Divide the shaft into two components which must have compatible deformations, φ φ φ Substitute into the original equilibrium equation, A 1 B B J1G JG L 1 J 90lb ft L J1 A + A L L L1 J L J 1 A

22 Example aluminum steel

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