cycle tyre to check whether it has - - - - Unit Properties of matter η R = Shearingstress Angleof shear orshearingstrain The shearing stress is σ s = = ∆ Tangential Areaover whichitisapplied force F t The angle of shear or shearing strain ε s = x h θ Therefore, Rigidity modulus is η R = σ ε θ s s t t F x h F = ∆ = ∆ ( . ) Further, the equation ( . ) implies that a material can be easily twisted if it has small value of rigidity modulus. For example, consider a wire twisted through an angle θ, a restoring torque (τ) developed is τ ∝ θ This means that for a larger torque, wire will twist by a larger amount (angle of shear θ is large).
Since the rigidity modulus is inversely proportional to angle of shear, the modulus of rigidity is small. For the best understanding, the elastic coefficients of some of the important materials are listed in Table . . EXAMPLE .
A metallic cube of side cm is subjected to a uniform force acting normal to the whole surface of the cube. The pressure is pascal. If the volume changes by . × – m , calculate the bulk modulus of the material.
Solution By definition, K = F P V ∆ ∆ K= N m Rigidity modulus or shear modulus: The rigidity modulus is defined as the ratio of the shearing stress to the shearing strain. Table . Elastic coefficient of some materials Material Young’s modulus (Y) ( N m – ) Bulk modulus (K) ( N m – ) Shear modulus (η R ) ( N m – ) Steel . .
enough air. What is checked here is essentially the compressibility of air. The tyre should be less compressible for its easy rolling In fact the rear tyre is less