Chapter 7

Torsion of Thin-Walled Closed Tubes

1. One-Cell Tube

The figure shows a closed tubular cross-section whose geometry is defined by a middle line (C) having arclength s, outward normal n, and a thickness t, which may be variable.  As seen in the general study of torsion, Prandl's stress function f must be constant on the boundary.  For a tubular section f takes on a different constant value on each boundary curve.  We can set f = 0 on the outer boundary, and let f = f_o on the inner boundary, where f_o is to be determined in terms of the applied Torque.  The shear stress at any point is tangent to the contour line, f = constant, and equal to  the negative of the directional derivative -f,_n .  For a thin tube and a gradually varying thickness, f,_n may be considered constant along n, and thus equal to Df/t  = - f_o/t, and the contour lines along n may all be considered parallel to the s direction.  We may thus write at any point along the normal,

t = t_zs = -f,_n = f_o/t            (1)

The shear flow is

q = tt = f_o                          (2)

and is thus constant around the tube.

The result, q = constant, may also be obtained by considering the equilibrium of a portion of the tube between longitudinal cuts at two arbitrary points A and B.  If L is the length of the tube, the resultant longitudinal force is (q_B - q_A)L, and must be zero.  Thus q_B = q_A

The applied torque T must be equal to the resultant torque of the shear flow.  On a differential area element tds act the shear force qds, whose moment about the origin O is dT = hqds, where h is the orthogonal distance from O to the line of action of qds.  Noting that hds is twice the area dA of the shaded triangle in the figure, we have dT = 2qdA, and, q being constant, integration around the middle line yields

T = 2Aq                           (3)

where A is the area enclosed by the middle line of the tube.

Thus, given a torque T, the shear flow is determined as q = T/2A, and the maximum shear stress is

t_max = q/t_min                    (4)

To describe the deformation, consider the shear strain-displacement relation at a point on the middle line of the tube,

g_zs = w,_s + u_s,_z                (5)

where w is the warping displacement, and u_s is the displacement in the s direction.  We have, u_s = hj, where j is the angle of twist, and  u_s,_z = hq, where q = j,_z  is the rate of twist. Thus

g_zs = w,_s + hq

Integration of this relation around the middle line of the tube yields

The integral of w,_s between two points A and B is (w_B - w_A).  w being single valued, the integral of w,_s around the middle line is zero.  The integral of hds is 2A, as seen earlier.  We thus have, as a condition of single-valuedness of w,

In terms of the shear stress, t = Gg_zs = q/t, 

Substituting q = T/2A, and solving for T,

The torsional constant of the tube is thus

Summary

T = applied torque

A = area enclosed by the middle line of the cross-section

q = T/2A = shear flow

t_max = q/t_min = maximum shear stress.   t_min = minimum thickness.

q = T/GJ  =  rate of twist.  G = shear modulus.  J = torsional consant, Eq. (9).

Example 1

The tubular cross section is subjected to a torque T = 50 k-ft.  The internal corners are actually rounded to avoid stress concentration.  Determine,

a) the maximum shear stress

b) the rate of twist q, assuming G = 11,500 ksi.

a) The area enclosed by the middle line of the cross-section is

A = 12(10) = 120 in²

The shear flow is

q = T/2A = 50(12)/240 = 2.5 k/in

The maximum shear stress occurs in the thinner elements. Thus

t_max = q/t' = 2.5/0.25 = 10 ksi

b) To calculate J, the integral of ds/t is carried out piece-wise over elements of constant thickness.  Thus if L_i is a typical length over which t = t_i, we obtain

SL_i/t_i = (12/0.5)(2) + (10/0.25)(2) =  48 + 80  =  128

Then

J = 4A²/SL_i/t_i = 4(120)²/128 = 450 in⁴

q = T/GJ = 50(12)/(11500 x 450) = 1.159(10^-4) rad/in  or 0.0797 deg./ft

2. Multi-Cell Tube

A two-cell tube is shown in the figure for illustration purposes. Each cell has an external branch along the external boundary of the cross-section, and shares with the other cell branch AB, which is along internal boundaries of the cross-section.    The shear flow q in any branch is constant for the same reason as in a single cell tube.  At a node such as A or B, i.e. a point common to three or more branches, the shear flows must satisfy a longitudinal equilibrium equation, similar to the one used in the preceding section, by which the sum of the shear flows coming into a node, or going out of a node, must add up to zero.  In the figure, the shear flows are defined positive as shown.  The shear flows coming toward node A are q₃, q₂, and -q₁.  The equilibrium equation is thus q₃ +q₂ - q₁ = 0, or

q₃ = q₁ - q₂

There is both a fluid-flow analogy and an electrical analogy to this equation.  If q is considered as a fluid-flow, the sum of the flows into a node must equal the sum of the flows away from the node.  The electrical analogy is similar. 

In general, we can define one independent shear flow, q_i,  per cell i, and use the node equilibrium equations to define the shear flow in a branch common to two cells in terms of the independent shear flows.

Letting C_i be the middle line of cell i,  and A_i be the area enclosed by C_i, the condition of single-valuedness of the warping displacement for cell i is 

where q is piece-wise constant.

The applied torque T must be equal to the resultant torque of the shear flow.  It is convenient to account for a common branch such AB by considering it twice, once as part of cell 1 with the shear flow q₁, and once as part of cell 2 with the shear flow q₂ running in in the opposite sense to q₁. The resultant torque is thus evaluated as the sum over the cells, with a constant shear flow in each cell, or

Procedure for Analysis

Given a torque T, the shear stress distribution and the rate of twist may be found by the following procedure:

a) Define a shear flow q_i for each cell i, positive counter-clockwise, and express the shear flows in common branches in terms of the q_i's.

b) Write Eq.(1) for each cell. The resulting equations form a system of linear simultaneous equations in the q_i's, for a given q.  Note that in Eq. (1) the coefficients of the q_i's are non-dimensional, and the coefficients of Gq on the right-hand side have the dimension of area.  Solve the system for the q_i's in terms of Gq.  The solution has the form

q_i = Gqc_i

where the c_i's are numerical values having the physical dimension of an area. 

c) Substitute the solution into Eq.(2):

T = Gq(2A₁c₁ + 2A₂c₂ + . . .)

The torsional constant J is the coefficient of Gq in the above expression, i.e.

J = 2A₁c₁ + 2A₂c₂ + . . .

d) To compute the shear stresses for a given T, compute Gq = T/J,  then the shear flows q_i = Gqc_i. The shear stress at any point is the ratio q/t at that point.  The maximum q/t yields the maximum shear stress.

Example 2

Determine the torsional constant of the two-cell cross section for

t = 1 inch and t' = 1/2 inch.

Step a)

The shear flows for the two cells are defined, respectively, as q₁ and q₂, positive if counter-clockwise.  In the common branch, q₃ = q₁ - q₂, upward.

Step b)

Cell 1: q₁(24/t + 12/t' + 24/t) + (q₁ - q₂)(12/t') = 2G(24)(12)q

Cell 2: (q₂ - q₁)(12/t') + q₂(12/t + 12/t' + 12/t) = 2G(12)(12)q

Substituting t = 1 and t' = 1/2 yields

 96q₁ - 24q₂ = 576Gq

-24q₁ + 72q₂ = 288Gq

The solution is found as

q₁ = (84/11)Gq

q₂ = (72/11)Gq

Step c)

T = 2A₁q₁ + 2A₂q₂ = [2(288)(84/11) + 2(144)(72/11)]Gq = (69,120/11)Gq

The coefficient of Gq is

J = 6283.64 in⁴