To show that the stress function φ = k(r^2 - 4a^2) is applicable to the solution of a solid circular section bar of radius 2a under pure shear, we need to demonstrate that it satisfies the equilibrium equations and boundary conditions of the problem.
In pure shear, the only non-zero stress components are shear stresses, and they are related to the stress function φ through the equations:
τ_x = ∂φ/∂y
τ_y = -∂φ/∂x
Let's calculate these shear stress components using the stress function φ = k(r^2 - 4a^2):
∂φ/∂y = -8ak
∂φ/∂x = 0
From these equations, we can see that the shear stress τ_x is constant and equal to -8ak, while the shear stress τ_y is zero. This means that the stress distribution in the bar is indeed pure shear.
To determine the stress distribution in terms of the applied torque, we can use the relationship between shear stress and torque in a solid circular section bar:
τ = Tc / (2πra^2)
where τ is the shear stress, T is the applied torque, c is the radius of the circular section, and r is the radial distance from the center of the circular section.
Comparing this equation with τ_x = -8ak, we can equate the two and solve for the applied torque:
-8ak = Tc / (2πra^2)
Simplifying the equation, we get:
T = -16πak^2
Therefore, the stress distribution in the bar under pure shear can be expressed as τ_x = -8ak, and the applied torque is related to the stress distribution by T = -16πak^2.
Note: It's important to verify the accuracy of these calculations and ensure they align with the specific problem and assumptions given in the context of your study or reference material.