TORSION By Prof. Ahmed Amer

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1 ORSION By Prof. Ahmed Amer

2 orque wisting moments or torques are fores ating through distane so as to promote rotation. Example Using a wrenh to tighten a nut in a bolt. If the bolt, wrenh and fore are all perpendiular to one another, the moment is the fore F times the length of the wrenh. Simple torque : F * l

3 Bars subjeted to orsion Let us now onsider a straight bar supported at one end and ated upon by two pairs of equal and opposite fores. hen eah pair of fores P1 and P form a ouple that tend to twist the bar about its longitudinal axis, thus produing surfae trations and moments. hen we an write the moments as 1 P1d1 Pd

4 Cirular 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 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.

5 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 φ L and γ ρ γ

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

7 he relationship between the applied torque and angle of twist Φ may be determined as: φ L GI P he torsion formula ould be arranged in the following form: I P r φg L

8 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, F ( A ) os45 A σ 45 o F A 0 A 0 A 0 Element a is in pure shear. 0 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

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.

10 Sample Problem.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 analysis 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

11 Sample Problem.1 SOLUION: Cut setions through shafts AB and BC and perform stati equilibrium analysis to find torque loadings M AB x 0 ( 6kN m) M 0 ( 6kN m) + ( 14kN m) 6kN m CD AB BC x 0kN m BC

12 Sample Problem.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 4 4 π 4 4 ( ) [( 0.060) ( 0.045) ] min min MPa 1 m BC 64.7 MPa 4 min 86. MPa ( 0kN m)( m) mm 60mm 6 m 4 min 86. MPa 64.7 MPa π 4 m 65MPa d 6kN m π 77.8mm

13 Angle of wist in Elasti Range Reall that the angle of twist and imum shearing strain are related, γ φ L In the elasti range, the shearing strain and shear are related by Hooke s Law, γ G G Equating the expressions for shearing strain and solving for the angle of twist, φ L G 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 G i i i

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 indeterminate. Divide the shaft into two omponents whih must have ompatible deformations, φ φ φ Substitute into the original equilibrium equation, A 1 B B 1G G L 1 90lb ft L A + A 1 L L L1 L 1 A

15 Sample Problem.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 gears. 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

16 Sample Problem.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 F F ( 0.875in. ) (.45in. ) 0 CD r φ φ B φ B B B r φ r r C B C φ C.8φ C C.45in. φ 0.875in. C

17 Sample Problem.4 Find the 0 for 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 φ A / B AB AB L G π ( 561lb in. )( 4in. ) 4 ( ) ( 6 ) 0.75in psi lb in. AB AB CD CD 561lb in psi 8000 psi 0 π ( 0.75in. ) 4 ( 0.75in. ).8 π 0 ( 0.5in. ) 4 ( 0.5in. ) 0 561lb in φ C / D φ φ B A 0.87 rad..8φ π rad.95 φ CD CD B L G C + φ.8 A / B.8 o ( 561lb in. )( 4in. ) 4 ( ) ( 6 ) 0.5in psi o ( o ) o o o φ A o 10.48

18 Design of ransmission Shafts Prinipal transmission shaft performane speifiations are: - power - speed Designer must selet shaft material and ross-setion to meet performane speifiations without exeeding allowable shearing stress. Determine torque applied to shaft at speified power and speed, P ω πf P ω P πf Find shaft ross-setion whih will not exeed the imum allowable shearing stress, π π ( solid shafts) 4 4 ( ) ( hollow shafts) 1

19 orsion of Nonirular Members Previous torsion formulas are valid for axisymmetri or irular shafts Planar ross-setions of nonirular shafts do not remain planar and stress and strain distribution do not vary linearly For non-irular ross-setion, the setion may bulge or warp when the shaft is twisted. Warping is a Longitudinal displaement whih ours in all ross- setion, exept irular setion, that are subjet to twisting moments.

20 Solid Elliptial Setion (at ends of minor axis) Z t Zt Setion modulus Π ab πab a y b x φ.l GI Zt Where orsional stiffness oeffiient Π a a b + b

21 Solid Retangular Setion y φ.l GI Zt Z t b x Where Zt αab Izt βab a a/b α β η

22 hin retangle setions φ. t 1 at. L 1 G at 1 at a t

23 hin Walled Open Cross- Setions, At large values of a/b, the imum shear stress and angle of twist for other open setions are the same as a retangular thin setions. Note:- For Retangular setion, as For, L and other shapes K shape fator 1.00 Angle 1. I-beam beam 1.1 Channels a t maz a 10 t. t k Σat φ α β. L k G Σat 1

24 hin-walled Closed setions One vent (Hollow Shafts) Summing fores in the x-diretion on AB, F 0 ( t Δx) ( t Δx) t A x A t B B A t A q shear flow shear stress varies inversely with thikness Compute the shaft torque from the integral of the moments due to shear stress dm 0 dm tao p df p Angle of twist : L φ 4Ao G ds t 0 B B ( t ds) q( pds) q da qa q da

25 hin Walled losed setions :- (multi vent) A A A O1 O O 1 F F F 1 For eah vent :- f, i i i A L A G o (1) o i φ f ds t φ l 1 A G o fds t () Solving Eqns (1) and () for eah vent get f1, f, f, φ/l

26 Example.10 Extruded aluminum tubing with a retangular ross-setion has a torque loading of 4 kipin. Determine the shearing stress in eah of the four walls with (a) uniform wall thikness of in. and wall thiknesses of (b) 0.10 in. on AB and CD and 0.00 in. on CD and BD. SOLUION: Determine the shear flow through the tubing walls Find the orresponding shearing stress with eah wall thikness

27 Example.10 SOLUION: Determine the shear flow through the tubing walls A (.84in. )(.4in. ) 4kip - in. q A 8.986in in. kip 1.5 ( ) in. Find the orresponding shearing stress with eah wall thikness with a uniform wall thikness, t q 1.5 kip in in. 8.4ksi with a variable wall thikness AB AC BD CD 1.5kip in. 0.10in. AB BC 1.5kip in. 0.00in. 11.1ksi BC CD 6.68ksi

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