Pure Shear The stresses on ab and cd (making 45 with xaxis) are
represented by point D and stresses on ad and bc are represented by
point D 1 Normal stress on these planes is zero This element is in
a state of stress called Pure Shear. It can be said that pure shear
is produced by tension and equal amount of compression working in
perpendicular directions.
Slide 3
Deformation in Pure Shear No change in length along ab, bc, cd,
ad Change in length along bd and ac The angle adc and angle abc
decreases & angle dab & dcb increases by
Slide 4
How things can fail? Torsion
Slide 5
Torsion We have covered two types of external loading so far,
1. Tension 2. Compression The third type of loading is TORSION (In
the other part of the course the fourth type of external loading
bending will be covered)
Slide 6
Circular section remains circular during the twist. The plane
cross sections perpendicular to the axis of the bar remain plane
after the application of a torque: points in a given plane remain
in that plane after twisting. Furthermore, expansion or contraction
of a cross section does not occur, nor does a shortening or
lengthening of the bar. Thus all normal strains are zero. The bar
is in shear.
Torsion The torsion angle (angle of twist) is proportional with
the length and torsional moment. angle of twist
Slide 9
Shapes other than cylindrical They are not so simple and will
not be covered in this course.
Slide 10
The element abcd on the surface of the disc is in pure shear.
Hence,
Slide 11
The element abcd on the surface of the disc is in pure shear.
Lets consider an element inside the cylinder. Hence, Shearing
stress varies directly as the distance from The axis of the shaft.
Hence, the stress is maximum at the surface. The twisting couple
(torque) applied is For each small area of the shaft, the moment
about the axis of the shaft = From equilibrium,
Slide 12
Torsion Equation Maximum shear stress which occurs at the
surface is, This is known as the torsion formula for circular
shafts. As the centre of a solid cube carries very little shear
stress, it is more efficient to use hollow shafts. For hollow tube,
internal diameter, d and outside diameter, D
Slide 13
Example 1
Slide 14
Example 2 A shaft consisting of two prismatic circular parts is
in equilibrium under the torques applied to the pulleys fastened to
it, as shown in figure. Find: The maximum shearing stress in the
shaft. Assumption: Stresses are at points somewhat away from the
pulleys.