Area of Circular Bar and Square Bar

To compare the two bars, the area must be the same. Either the circular dimension, r, or the square dimension, s, can be used. This solution will use the r dimension in the calculations. Thus the circular bar cross sectional area, A_c is r²π.

The square bar cross sectional area, A_s is s². Since they need to be question,

   A_c = A_s     =>     r²π = s²

Thus,

   s = π^1/2 r = 1.7772 r

The circular bar maximum stress can be easily determined using the basic shear stress equation derived in the circular bars section.

Rectangular bars is a simple geometry, but the solution is still lengthy since a second order partial differential equation must be solved. Recall, the equation that models all torsional shear stress is,

where G is the shear modulus, θ is the angle (radians) of twist per unit length (not the total twist) and φ is the scalar stress function (used to find shear stress).

To start, set the geometry to a width of 2a and a height of 2b as shown in the diagram. The dimension 'a' and 'b' are different for now, so the derived solution can be used for any rectangular shaped bar. The use of the z-axis pointing to the left is consistent with standard beam coordinates (x is along the beam center, or pointing out of the page in this case).

Notice, the torsional shear stresses are only estimates in the diagram. Unlike a circular bar, the stresses will change for different location around the center. This makes the solution difficult (and lengthy).

Like all solutions to differential equations, a trial solution is proposed, and then substituted back in to find a particular solution. Since the solution should be symmetrical around both the z-axis and y-axis, a cosine function should work. As a starting point, assume the stress function, φ, is

where b_n are constants (will cancel out later) and Y_n are functions of 'y' that are not yet set. Now, substitute this assumed solution form back into the differential equation, giving

All the terms on the left have cosine functions, but the right hand side does not. To make all terms consistent, the -2Gθ constant term can be written as a fourier series (basically just a series using sine and cosine) as

This may seem complicated, but it will work out in the end. If the 2Gθ series is substituted into the previous equation, the cos terms will cancel, giving

Where the ' (prime) marks represent derivatives with respect to y. Rearranging, moving b_n term to the right hand side, and dropping the summation symbol for simplicity (still summing on all terms with n), gives

Remember, the Y_n terms are functions of y that have not yet been set. It will be done now. Basically, the equation above is another differential equation. The general solution is,

Now the boundary conditions can be applied to determine the constants A and B. First, since the solution will be symmetrical about the z-axis, there can be no sinh term (anti-symmetric). Thus A must be zero. Next, B can be found with the boundary conditions φ = 0 at the edge, or when y = ±b, giving

This give Y_n as

Substituting Y_n ,back into the stress function, φ, gives

Notice, the b_n terms cancel. This is the full solution for the stress function at any point (y, z) for a rectangular cross section.

But there are few more steps to make it useful for actual shear stresses. First is to take a derivative with respect to z to find the shear stress on the x-y plane, giving

Likewise, the stress in the x-z plane can be found from τ_xz = dφ/dy. The total shear stress will be a combination of both the x-y and x-z plane stress.

The maximum shear stress will occur when z = ±a or y = ±b (depends if 'a' or 'b' is the long side). But for this problem, a = b, so it will not matter. Taking y = 0, and z = a, gives

Notice, the sin term and the (-1)^(n-1)/2 term will be both negative or both positive for the same n, so the final result will always be positive. This can be further simplified noting that Σ(1/n²) = π²/8 for n = 1, 3, 5, ... This gives,

This expression will converge quickly. Using only three terms (n = 1, 3, and 5), it is accurate to five significant digits. The result is,

   τ_max-s = 1.3506 Gθa

While the maximum is at the sides, it is interesting to note that the corners have zero shear stress as shown in the diagram.

One last issue is the solution is in terms of G or θ, which are not given. The requires another equation relating Gθ with the applied torque, T. This can be done by summing both shear stresses over the cross sectional area. It can be shown (not given here) that summing the stress function, φ, over the domain, also gives the total torque, T.

Completing this integration, and setting a = b, gives,

   T = 0.1406 Gθ (2a)⁴

or

   Gθ = 0.4445 T / a⁴

Substituting this back into the maximum shear stress equation, gives

   τ_max-s = 0.6004 T / a³

Or in terms of the circular bar radius, r, where a = 0.5s and s = 1.7772 r, the maximum shear stress for the bar is

   τ_max-s = 0.8633 T / r³

The ratio of the maximum shear stress for the square bar to the circular bar (but with the same area) is,

This means, the maximum shear stress in the bar is 35.6% higher than the circular bar. Unless the bars are at least 35% cheaper, it would not be wise to use the square bars. This is also why almost all drive shafts are round.