Torsional Stress

Initialization

<< Miscellaneous`Units`

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Problem statement

A solid circular shaft has a diameter of 3 in. and a length of 24 in. It is subjected to a torque T=2kp in. Determine the maximum torsional stress in the shaft.

Solution

To determine the state of stress, we select an element of the bar dx and radius ρ. Due to the torque, a straight line parallel to x will be deflected or twisted by ρdθ, and so the shear strain is

γ=ρ ((dθ(
                x))/(dx))

This equation clearly shows that the shear stress is directly proportional to the radius of the element and hence is largest at the surface of the bar.

Next we have to relate the shear stress to the torque which produced it. Any element of the cross section is subjected to a tangential force τ dA due to the external torque T. The torque is the integral over all these moments.

T = ∫ρτ dA

We know that τ = Gγ. Using equation (a), we have

\(T = 
      G \(dθ\/dx\) \(∫\(ρ\^2\) dA\)\)

∫(ρ^2)dA is commonly called the polar-area moment of inertia denoted by J.

Thus the twist per unit length is

d(θ/d)x = (T/GJ)

Thus, for a bar of length l inches,

θ=(Tl/GJ)

where θ is the total angle of twist in radians.

To obtain the shear stress, we note that

\(γ = \(ρ dθ\/dx = Tρ\/GJ\)\)

Then, since τ = Gγ,

τ = T(ρ/J)

And, because of equation (a),

(Subscript[τ, max])= (Tr/( J))

The cross section element dA is ρdρ for circular cross sections, therefore

(Subscript[τ, max])=(Tr/J)=(Tr/(\(∫\_0\%r\( ρ\^3\) dρ\)))

Rewriting this result in Mathematica syntax, we get a function for the stress.

(Subscript[τ, max])[r_, T_]:=((T r)/((∫_0^r) (ρ^3) \[DifferentialD]ρ))

For the above example, we get

(Subscript[τ, max])[3 Inch, 2 Kilopound Inch]

N[%]

\(\(Plot3D[τ\_max[r, \ t], {r, 1, 2}, {t, 1, 2}, 
        PlotRange -> All, \ 
        AxesLabel -> {"r", "T", \*""\!\(τ\_max\)""}];
      \)\)