KAJI DAYA BAHAN 1 J 3009

JJ310-STRENGTH OF MATERIALTORSION1This topic explains the second polar moment of area for the round solid shafts and hollow shafts.

It also deals with the maximum shear stress and angle of twist subjected to torque and solve problems on the power transmitted by the shaft, simple parallel and series composite shaft2INTRODUCTION-Simple Torsion TheoryWhen a uniform circular shaft is subjected to a torque it can be shown that every section of the shaft is subjected to a state of pure shear.

Consider a bar to be rigidly attached at one end and twisted at the other end by a torque or twisting moment T equivalent to F d, which is applied perpendicular to the axis of the bar, as shown in the figure. Such a bar is said to be in torsion.

3Assumptions-The material is homogeneous, i.e. of uniform elastic properties throughout.The material is elastic, following Hook's law with shear stress proportional to shear strain.The stress does not exceed the elastic limit of proportionality.Circular sections remain circular.Cross sections remain plane.Cross sections rotate as if rigid.

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5Modulus of Rigidity (G)It is a measure of how much deformation the material would undergo subjected to a shear. G is the property which relates to the stiffness of the material, units N/m2.

6TORSIONAL SHEARING STRESS,

7For a solid or hollow circular shaft subject to a twisting moment T, the torsional shearing stress at a distance from the center of the shaft is:-

and

where J is the polar moment of inertia of the section and r is the outer radius.

8Distribution of shearing stress over the shaft cross sectional:Solid shaft:-

Hollow shaft:-

9TORSIONAL SHEARING STRESS, where ; = Maximum shear stress in a circular cross section (N/m2). T = Torque in that section (Nm). R = Radius of the section (m). J = Polar Moment of Inertia (m4). G = Modulus of Rigidity (N/m2). = Angle of twist (radians).

Note: Shear stress linearly with radius.

10After applying the torque point A moves to point B and made an angle .AB subtends an angle at the fixed end, is the shear strain.

ANGLE OF TWIST, 11ANGLE OF TWIST, The angle through which the bar length L will twist is:-

where T is the torque in Nmm, L is the length of shaft in mm, G is shear modulus in MPa, J is the polar moment of inertia in mm4.

in radians12Polar moment of inertia is a quantity used to predict an object's ability to resist torsion, in objects (or segments of objects) with an invariant circular cross section and no significant warping or out-of-plane deformation. It is used to calculate the angular displacement of an object subjected to a torque. It is analogous to the area moment of inertia, which characterizes an object's ability to resist bending and is required to calculate displacement.The larger the polar moment of inertia, the less the beam will twist, when subjected to a given torque.Polar moment of inertia should not be confused with moment of inertia, which characterizes an object's angular acceleration due to a torque.

POLAR MOMENT OF INERTIA13POLAR MOMENT OF INERTIA

For circular solid shaft:

For hollow shaft:

14Section Modulus

15Example 1:A torque of 1000 Nm is acting on a solid cylinder shaft with diameter 50 mm and length 1 m. The shaft is made in steel with modulus of rigidity 79 GN/m2. Calculate the maximum shear stress and angle of twist of the shaft.

Solutions:-

Maximum shear stress can be calculated as= T R / J= T (D/2) / ( D4/32) = (1000 Nm) ((0.05 m)/2) / ( (0.05 m)4/32) = 40.8 MN/m2

16The angle of twist of the shaft can be calculated as:

= L T / J G = L T / ( D4/32) G = (1 m) (1000 Nm) / ( (0.05 m)4/32) (79 GPa) = 0.021 radians= 1.2 o

17A moment of 1000 Nm is acting on a hollow cylinder shaft with outer diameter 50 mm, inner diameter 30 mm and length 1 m. The shaft is made in steel with modulus of rigidity 79 GPa.

Solutions:-

Maximum shear stress can be calculated as = T r / J = T (D/2) / ( (D4 - d4)/32) = (1000 Nm) ((0.05 m)/2) / ( ((0.05 m)4 - (0.03 m)4)/32) = 46.8 Mpa

Example 2:18The angle of twist of the shaft can be calculated as

= L T / J G = L T / ( D4/32) G= (1 m) (1000 Nm) / ( ((0.05 m)4 - (0.03 m)4)/32) (79 GPa) = 0.023 radians= 1.4 o

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Torsion in Composite Shafts1. Series Connection-free end

When two similar or dissimilar shafts of the same or of different materials are connected together to form one composite shaft, the driving torque being applied at one end and the resisting torque at the other, the shafts are said to be connected in series.Consider each component shaft separately, applying the torsion theory to each in turn.The composite shaft will therefore be as weak as its weakest component.24Series composite shaftsAccording to the definition of series connected shaft and making use of torsion equation, the eq. can write:-T = T1 = T2

where:

So:-

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Torsion in Composite ShaftsSeries Connection-one free end and one fixed26

2. Parallel Connection both end fixed

If the driving torque is applied at the junction of the two shafts connected together and the resisting torque at the other ends of the two shafts, then the shafts are said to be connected in parallel.

Torsion in Composite Shafts27Parallel composite barsAccording to the definition of parallel connected shaft and making use of torsion equation, the eq. can write:-T = T1 + T2

where:

So:-

28The torque applied in this case is divided between the two shafts but the angle of twist is the same for each shaft.

29Example 3:A compound shaft consisting of a steel segment and an aluminum segment is acted upon by two torques as shown in below figure. Determine the maximum permissible value of T subject to the following conditions: st = 83 MPa, al = 55 MPa, and the angle of twist of the free end is limited to 6. For steel, G = 83 GPa and for aluminum, G = 28 GPa.

30Solution:-At r = R or (d/2), is maximum.

So,

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The torque for steel and aluminum is:-

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33Example :The compound shaft shown in below figure is attached to rigid supports. For the bronze segment AB, the diameter is 75 mm, = 60 MPa, and G = 35 GPa. For the steel segment BC, the diameter is 50 mm, 80 MPa, and G = 83 GPa. If a = 2 m and b = 1.5 m, compute the maximum torque T that can be applied.

34Solution:-

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38POWER TRANSMITTED BY THE SHAFTA shaft rotating with a constant angular velocity (in radians per second) is being acted by a twisting moment T. The power transmitted by the shaft is:-

where T is the torque in Nm, f is the number of revolutions per second, and P is the power in watts.

39Power TransmissionA torsion member known as shaft is used to transmit power from source to a driven machine. The torque acting on such a member depends upon power transmitted and the speed of rotation.If T be the torque shaft which is rotating with speed of N rpm, then :- work per minute = 2NT The work per minute is the power transmitted by the shaft. So, the Power, P = T where =

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