2E4: SOLIDS & STRUCTURES Lecture 13 Dr. Bidisha Ghosh Notes: lids & Structures

Pure Shear The stresses on ab and cd (making 45˚ with x–axis) 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.

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 

How things can fail? Torsion

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)

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.

Foam Beams

Torsion The torsion angle (angle of twist) is proportional with the length and torsional moment. angle of twist

Shapes other than cylindrical They are not so simple and will not be covered in this course.

The element abcd on the surface of the disc is in pure shear. Hence,

The element abcd on the surface of the disc is in pure shear. Let’s 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,

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

Example 1

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.

Example 3