## Lesson 13. TORSION

 Module 3. Stress Lesson13TORSION13.1 Introduction In solid mechanics, torsion is the twisting of an object due to an applied torque. In circular sections, the resultant shearing stress is perpendicular to the radius. Torsion and bending produce normal and tangential stresses in the same plane simultaneously. The stresses produced by normal and tangential stresses in the same plane simultaneously. The stresses produced by torsion and bending are termed as compound stresses. (Fig. 13.1) Fig. 13.1 Torsion 13.2 Torsion When a member is twisted by a couple, the stress produced is pure shear. However, its intensity on any fiber depends on the distance of the fibre from the center line of the twisted member. Thus in a circular member the shear stress is proportional to the distance from the center line. Thus when a member is subjected to external torsional moment (Fig 13.2) T = Average Torque, M.m. J = Polar moment of inertia , m4 ƒs or Ss= maximum shear stress, N/m2 R= Radius = Distance of outer most fiber from the central line, m G = Shear modulus , N/m2 θ = maximum angular twist , radian L = Length of shaft under twisting , m Where τs-ƒs or Ss is the maximum shear stress developed in the member Zp is the polar section modulus. In case of a solid or hollow circular section, Zp = J / c where J is the polar moment of inertia and c is the distance from the axis to the most remote fibre where the stress is τs. In case of solid round bar, Zp = π D3/16 Hence, ז s = 16 T / (π D3) [some refer Ss as fs] In case of hollow cylindrical bar, D= Outer Diameter d = Inner diameter 13.3 Torsional Stresses and Strains Fig. 13.3 Torsional stresses Consider a shaft fixed at one end, and subjected to a torque at the other as shown above. Let T= Torque in kg-cm l= Length of the shaft, and R= Radius of the shaft. As a result of this torque, every cross-section of the shaft will be subjected to shear stresses. Let the line CA on the surface of the shaft be deformed to CA1 and OA to OA1. Let ∟ACA1=Φ in degrees ∟AOA1=θ in radians fs = Shear stress induced at the outer most surface, and C = Modules of Rigidity, also known as torsional rigidity of the shaft material. We know that shear strain = Deformation per unit length. (AA1)/ I = tan Φ = Φ ( Φ being very small, tanΦ = Φ) We also know that the length of the arc AA1 = Rθ Φ = (AA1 / I) = (Rθ / I) ------------(1) Moreover deformation = (Shear stress / Modulus of rigidity) Φ = (fs / C) -------------(2) Now from equation (1) and (2) we find that (fs / C) = (Rθ / I) (fs / R) = (Cθ / I) if q be the intensity of shear stress, any layer at a distance r be the center of the shaft, then (q /r) = (fs / R) = (Cθ / I) 13.4 Strength of the Solid Shaft It means the maximum torque or power of the shaft can transmit from one to another. Now consider a solid circular shaft subjected to some torque. Let R= Radius of the shaft, and fs= Maximum shear stress developed in the outer most layer of the shaft material Fig. 13.4 Strength of the solid shaft Now consider an elementary ring of thickness dx at a distance from the center. We know that the area of this ring, _________________(1) Shear stress at this section, Where D is the external diameter of the shaft and is equal to 2R Example 1: Find the torque which a shaft of 25 cm diameter can safely transmit, if the shear is not to exceed 460 kg/cm2 Solution: Given. Diameter of shaft, D=25 cm Max. Shear stress, fs = 460 kg/cm2 Let T= Torque transmitted by the shaft. Using the relation, Example 2: A solid steel shaft is to transmit a torque of 10000 kg-m. If the shearing stress is not to exceed 450 kg/cm2, find the minimum diameter of the shaft. Solution: Given Torque, T= 10,000 kg-m = 1000000 kg-cm Max, shear stress, fs= 450 kg/cm2 Let D= Diameter of the shaft. Using the relation, Or 13.5 Strength of the Hallow Shaft It means the maximum torque or power a hallow shaft can transmit from on pulley to another. Now consider a hallow circular shaft subjected to some torque Fig. 13.5 Hallow shaft Let R= Outer radius of the shaft, r = Inner radius of the shaft, and, fs= Maximum shear stress developed in the outer most layer of the shaft material Now consider an elementary ring of thickness dx at a distance x from the center as shown in fig above. We know that the area of this ring, Where D is the external diameter of the shaft and is equal to 2R and 2d is the internal diameter of the shaft and is equal to 2r