A pure torsion applied to the longitudinal axis of a thin closed section is referred to the St Venant torsion theory when the ends of a beam are free to carry equal but opposing torques. Here the shear stress in a thin wall and the angular twist between the ends are found from a torsion theory which simplifies to three-part formula: T = GJdθ/dz = 2Aq, when the section is uniform in a single material. There is no requirement to specify the axis of torsion in that A refers to an area enclosed by the mean wall closed perimeter and the St Venant torsion constant $J=4A^2\text{/}{\displaystyle \int \text{d}s}\text{/}t$ is a property of the cross-section. In the case of a thin-walled open section a different theory applies in which the St Venant torque is supplemented by the Wagner Kappus torque. The latter arises with the change that occurs to the natural warping displacements for the closed tube when the section has an opening. A further alteration to the warping occurs when as, in this paper, torsion is applied to an open section cantilever beam in which one end is clamped to prevent any axial warping displacement from occurring at that end. Moreover, the influence of the clamped fixing extends over the full length to modify the ‘free’ warping displacements that would otherwise be found under torsion of an open section beam with unclamped ends. The influence of the axis of torsion upon a clamped-end cantilever beam is investigated here following previous analysis [1] of this beam where the torsion was applied at the free end about the beam’s flexural axis. Where the beam section is uniform throughout the length the flexural axis refers to a straight longitudinal axis through the shear centre at every section of the beam. The analyses are complicated by the fixing involving: (1) the location of the position of the beam section’s shear centre; (2) calculation of the axial stress distribution in the thin walls due to the beam’s fixed-end constraint and (3) accounting for the corresponding increase in torsional stiffness between the ends due to the end fixing. No flexure (bending and transverse shear) occurs with torsion applied to the axis at E provided that axis coincides with the length axis at the section’s centre of twist. The latter axis refers to points in the cross-section that do not rotate. The assumption of coincidence between axes at the centres of twist and shear is often made [2] but has been questioned from a minimal energy condition imposed by an absence of rotation. Other approaches might test the assumption of co-incidence more directly, as here, when the axis of torsion is deliberately shifted away from E. Here, two further torsion axes are considered: one through the section’s centroid G and the other through the web centre O. By implication points along torsion axes passing through G and O will rotate and thereby influence the behaviour listed above in (1)-(3). The many equations required to reveal the trends that each shift involves are summarised below. These are applied to three standard US channel sections mounted with the web in a vertical disposition carrying two, equal length, horizontal flanges.

Equation Summary

The equations listed below are used for the calculations that follow. The derivations of all equations can be found in [1] . Here certain equations have been adapted often in non-dimensional form to the specific US Imperial dimensions of the three channel sections investigated. For example, (GJ/Td ²)w appears as the chosen non dimensional measure of small warping displacements allowing the physical displacement w to be calculated from material and sectional constants: G, J, and d, under a given torque T. Similarly, the primary warping constant Γ_I takes a normalised form: Γ_I/(ta²d ³) in which a, d and t are section dimensions.

$A{X}^{\prime }={\displaystyle \sum A_i{X}_i}$

${\Gamma }_{\text{I}}={\displaystyle \int y^{2}t\text{d}s}-{\stackrel{⌣}{y}}^2{\displaystyle {\int }_{s}t\text{d}s}$

$J=\frac{1}{3}{\displaystyle \sum b{t}^3}$

${\mu }^2=GJ/E{\Gamma }_{\text{I}}$

$T/G\theta =\left[\mu J\left(1+\exp \left(-2\mu L\right)\right)\right]/\left[\left(\mu L-1\right)+\left(\mu L+1\right)\exp \left(-2\mu L\right)\right]$

$GJ{w}_o/T=-\left(y-\stackrel{⌣}{y}\right)=-2A_E$

$w=w_o\left[1-\cosh \mu \left(L-z\right)/\cosh \mu L\right]$

$w=w_o\left(1-1/\cosh \mu L\right)$, for $z=L$

$w_o=-\left(T/GJ\right)\left(y-\stackrel{⌣}{y}\right)=-2A_E\left(T/GJ\right)$

Rate of Twist

$\text{d}\theta /\text{d}z=\left(T/GJ\right)\left[1-\cosh \mu \left(L-z\right)/\cosh \mu L\right]$

Integrating w. r. t. z for θ(z):

$\theta \left(z\right)=\left(T/GJ\right)\left\{z+\left[\sinh \mu \left(L-z\right)-\sinh \mu L\right]/\mu \cosh \mu L\right\}$

For z = L, this will result

$\theta \left(L\right)=\left(T/GJ\right)\left[1-{\left(\mu L\right)}^{-1}\tanh \mu L\right]$

$\begin{array}{c}\theta \left(L\right)=\left(TL/GJ\right)\left[\left(\mu L-1\right)+\left(\mu L+1\right){\text{e}}^{-2\mu L}\right]/\mu L\left(1+{\text{e}}^{-2\mu L}\right)\\ =\left(T/GJ\mu \right)\left[\left(\mu L-1\right)+\left(\mu L+1\right){\text{e}}^{-2\mu L}\right]/\left(1+{\text{e}}^{-2\mu L}\right)\end{array}$

Stress Constraint

${\sigma }_z=E\text{d}w/\text{d}z=-E\times \text{d}\left(2A_E\text{d}\theta /\text{d}z\right)/\text{d}z$

in which, from Equation (8a)

${\text{d}}^2\theta /\text{d}{z}^2=\left(T/GJ\right)\left[\mu \sinh \mu \left(L-z\right)/\cosh \mu L\right]$

$\begin{array}{c}{\sigma }_z=-2A_{E}E\left({\text{d}}^2\theta /\text{d}{z}^2\right)=-2A_{E}E{\theta }^{″}\\ =\left(-2A_{E}ET/GJ\right)\left[\mu \sinh \mu \left(L-z\right)/\cosh \mu L\right]\end{array}$

Alternatively, with ${\mu }^2=GJ/E{\Gamma }_{\text{I}}$:

${\sigma }_z=\left(-2A_{E}ET/\mu {\Gamma }_{\text{I}}\right)\left[\sinh \mu \left(L-z\right)/\cosh \mu L\right]$

${\sigma }_z=\left(-2A_{E}ET/GJ\right)\left[\mu \sinh \mu \left(L-z\right)/\cosh \mu L\right]$

and from Equation (7b):

${\sigma }_z=E{w}_o\left[\mu \sinh \mu \left(L-z\right)/\cosh \mu L\right]$

Equation (8a) applies to $0\le z\le L$

${\sigma }_z=-\left(2A_{E}T/\mu {\Gamma }_{\text{I}}\right)\left[\sinh \mu \left(L-z\right)/\cosh \mu L\right]$

which attains its maximum for z = 0

${\sigma }_z=\left(-2A_{E}T/\mu {\Gamma }_{\text{I}}\right)\tanh \mu L$

${\sigma }_z=\left(-2A_{E}T/\mu {\Gamma }_{\text{I}}\right)\times \left(1-{\text{e}}^{-2\mu L}\right)/\left(1+{\text{e}}^{-2\mu L}\right)$

${\sigma }_z=-\left(y-\stackrel{⌣}{y}\right)T/\mu {\Gamma }_{\text{I}}\times \left(1-{\text{e}}^{-2\mu L}\right)/\left(1+{\text{e}}^{-2\mu L}\right)$

${\sigma }_z=\left(\stackrel{⌣}{y}-y\right)T/\mu {\Gamma }_{\text{I}}\times \left(1-{\text{e}}^{-2\mu L}\right)/\left(1+{\text{e}}^{-2\mu L}\right)$

This section and the next examine the effect of a shift in the axis of torsion, away from the shear centre, upon the unconstrained warping displacements. Firstly, the axis is shifted from the shear centre E to pass through the channel’s centroid G. The position X’ of G from the web centre in the X direction (see Figure 1 ) follows from Equation (1) given above as:

$A{X}^{\prime }={\displaystyle \sum A_i{X}_i}$(1)

$\left(dt+2at\right)X^{\prime }=2\left(at\times a/2\right)$

$X^{\prime }=a^2/\left(d+2a\right)$

Here (twice) the swept areas $A_G$ are required between G and the section’s perimeter points 1, 2, 3, ∙∙∙, 7 in Figure 1(a) .

Three calculations are sufficient to construct the swept area plot given in Figure 1(b) . Starting from the free edge at position 1, the three triangular areas are identified:

$A_{G13}=2\left[\left(a/2\right)\times \left(d/2\right)\right]=ad/2$

$A_{G34}=2\times \left(X^{\prime }/2\right)\times \left(d/2\right)=a^{2}d/2\left(a+d\right)$

$A_{134}=A_{G13}+A_{G34}2=\left(ad/2\right)\left(d+3a\right)/\left(d+2a\right)$

from which the mean co-ordinate $\stackrel{⌣}{y}$ for Figure 1(b) follows:

${\displaystyle {\int }_s\stackrel{⌣}{y}\text{d}s}={\displaystyle {\int }_{s}y\text{d}s}$

$\left(2a+d\right)\stackrel{⌣}{y}=2\left(ad/2\times a/2\right)+2\left[ad/2+\left(ad/2\right)\left(d+3a\right)/\left(d+2a\right)\right]\left(d/4\right)$

$\stackrel{⌣}{y}=\left(ad/4\right)\left(7ad+4a^2+2d^2\right)/{\left(2a+d\right)}^2$

$\stackrel{⌣}{y}/d^2=\left(1/4\right)\left(a/d\right)\left[7\left(a/d\right)+4{\left(a/d\right)}^2+2\right]/{\left[1+2\left(a/d\right)\right]}^2$

The straight-line equations: y = y(s), describing the sides of the plot in Figure 1(b) and the integrals ${\displaystyle {\int }_s{y}^2\text{d}s}$ for the warping constant, in Equation (2) are

evaluated, with t constant, as follows:

1-3 for $0\le s\le a$

$y=ds/2;{\displaystyle {\int }_s{y}^2\text{d}s}={\left(d/2\right)}^2{\displaystyle {\int }_s{s}^2\text{d}s}={\left(d/2\right)}^2\times a^3/3=a^3{d}^3/12$

3-4 for $a\le s\le a+d/2$

$y=ad/2+a^{2}d/2\left(a+d\right)\times \left(2/d\right)\left(s-a\right)=ad/2+a^2\left(s-a\right)/\left(d+2a\right)$

$\begin{array}{c}{\displaystyle {\int }_s{y}^2\text{d}s}={\displaystyle {\int }_s{\left[ad/2+a^2\left(s-a\right)/\left(d+2a\right)\right]}^2\text{d}s}\\ ={\left(ad/2\right)}^2\left(d/2\right)+\left[a^{3}d/2\left(d+2a\right)\right]{\left(d/2\right)}^2+\left(1/3\right){\left[a^2/\left(d+2a\right)\right]}^2{\left(d/2\right)}^3\end{array}$

Hence, for the whole cross-section, the two terms appearing in the primary warping constant Γ_I follow from the summary Equation (2) given above

${\Gamma }_{\text{I}}={\displaystyle {\int }_s{y}^{2}t\text{d}s}-{\stackrel{⌣}{y}}^2{\displaystyle {\int }_{s}t\text{d}s}$(2a)

where:

$\begin{array}{c}{\displaystyle {\int }_s{y}^{2}t\text{d}s}=2t\{a^3{d}^3/12+{\left(ad/2\right)}^2\left(d/2\right)+\left[a^{3}d/2\left(d+2a\right)\right]{\left(d/2\right)}^2\\ +\left(1/3\right){\left[a^2/\left(d+2a\right)\right]}^2{\left(d/2\right)}^3\}\\ =\left(t{a}^2{d}^3/2\right)[\left(1/3\right)\left(a/d\right)+1/2+\left(1/2\right){\left(a/d\right)/\left(1+2a/d\right)}_{}^{}\\ +\left(1/6\right){\left(a/d\right)}^2/{\left(1+2a/d\right)}^2]\end{array}$(2b)

${\stackrel{⌣}{y}}^2{\displaystyle {\int }_{s}t\text{d}s}={\stackrel{⌣}{y}}^2\left(2a+d\right)t={\left\{\left(ad/4\right)\left(7ad+4a^2+2d^2\right)/{\left(2a+d\right)}^2\right\}}^2\times \left(2a+d\right)t$(2c)

Equations 2(a)-(c) are applied to the three channel geometries as follows:

Section A: $a=1/2\text{'}\text{'},d=1\text{'}\text{'},t=1/16\text{'}\text{'},L=300\text{mm}\left(11.81\text{'}\text{'}\right)$ $a/d=1/2,t/d=1/16,d=25.4\text{mm},L=300\text{mm}$. $E=210\text{GPa}$, $G=70\text{GPa}$

$\begin{array}{c}{\displaystyle {\int }_s{y}^{2}t\text{d}s}=\left(t{a}^2{d}^3/2\right)\left[\left(1/3\right)\left(1/2\right)+1/2+\left(1/2\right)\left(1/2\right)/\left(1+1\right)+\left(1/6\right){\left(1/2\right)}^2/{\left(1+1\right)}^2\right]\\ =\left(77/192\right)t{a}^2{d}^4\end{array}$

${\stackrel{⌣}{y}}^2/d^2=\left(1/4\right)\left(1/2\right)\left(7/2+4/4+1/2\right)/{\left(1+1\right)}^2=13/64$

${\stackrel{⌣}{y}}^2\left(2a+d\right)t={\left(13d^2/64\right)}^2\left(2a+d\right)t=169t{a}^2{d}^3/512$

1) Constants

The following constants are calculated for this Section A.

${\Gamma }_{\text{I}}=\left(77/192-169/512\right)t{a}^2{d}^3=\left(109/1536\right)t{a}^2{d}^3=0.5677t{a}^5$(2)

${\Gamma }_{\text{I}}=\left(109/1536\right)\times 1.5875\times {12.7}^2\times {25.4}^3=2.9775\times {10}^5{\text{mm}}^6$

$J=\left(2a+d\right)\left(t^3/3\right)=4a{t}^3/3=\left(2\times 12.7+25.4\right)\left({1.5875}^3/3\right)=67.746{\text{mm}}^4$(3)

$\mu =\sqrt{GJ/E{\Gamma }_{\text{I}}}=\sqrt{4a{t}^3/\left(9\times 0.5677t{a}^5\right)}=0.8848t/a^2$(4)

$\mu =\sqrt{\left(70\times {10}^3\times 67.746\right)/\left(210\times {10}^3\times 2.9775\times {10}^5\right)}=8.71\times {10}^{-3}{\text{mm}}^{-1}$

$\mu L=0.8848\left(t/a\right)\left(L/a\right)=0.8848\left(1/8\right)\left(300/12.7\right)=2.613$

$\exp \left(-2\mu L\right)=\exp \left(-5.226\right)=5.375\times {10}^{-3}$

$\cosh \mu L=\cosh 2.613=6.857$

$\cosh \left(\mu L/2\right)=\cosh 1.307=1.983$

2) Torsional Stiffness

$\begin{array}{c}T/G\theta =\left[\mu J\left(1+\exp \left(-2\mu L\right)\right)\right]/\left[\left(\mu L-1\right)+\left(\mu L+1\right)\exp \left(-2\mu L\right)\right]\\ =\left(0.8848t/a^2\right)\times \left(4a{t}^3/3\right)\times 0.6159=0.7266t^4/a\end{array}$(5)

$\begin{array}{l}\therefore T/\theta =0.7266\times 70\times {10}^3\times 1.58754/12.7\\ =25.436\times {10}^3\text{N}\cdot \text{mm}/\text{rad}=25.436\text{N}\cdot \text{m}/\text{rad}\end{array}$

3) Unconstrained Warping Displacements

These displacements w_o, which are independent of length, follow from

$GJ{w}_o/T=-\left(y-\stackrel{⌣}{y}\right)$ (6)

Equation (6) gives at each median point. 1, 3 and 4:

$\left(GJ/T\right)w_{o1}=\left(GJ/T\right)w_{o7}=\stackrel{⌣}{y}$

$\left(GJ/T\right)w_{o3}=\left(GJ/T\right)w_{o5}=\stackrel{⌣}{y}-ad/2$

$\left(GJ/T\right)w_{o4}=\stackrel{⌣}{y}-\left(ad/2\right)\left(d+3a\right)/\left(d+2a\right)$

At 1, $y_1=0:\left(GJ/T\right)w_{o1}=-\left(0-13d^2/64\right)$

$\therefore \left(GJ/T{d}^2\right)w_{o1}=13/64$

At 3, $y_3=ad/2:\left(GJ/T{d}^2\right)w_{o3}=-\left(ad/2-13d^2/64\right)$

$\left(GJ/T{d}^2\right)w_{o1}=-\left(a/2d-13/64\right)=-\left(1/4-13/64\right)$

$\therefore \left(GJ/T{d}^2\right)w_{o1}=-3/64$

At 4, $y_3=\left(ad/2\right)\left(d+3a\right)/\left(d+2a\right)$

$\left(GJ/T\right)w_{o4}=-\left[\left(ad/2\right)\left(d+3a\right)/\left(d+2a\right)-13d^2/64\right]$

$\begin{array}{c}\left(GJ/T{d}^2\right)w_{o4}=-\left[\left(1/2\right)\left(a/d\right)\left(1+3a/d\right)/\left(1+2a/d\right)-13/64\right]\\ =-\left[\left(1/2\right)\left(1/2\right)\left(1+3/2\right)/\left(1+1\right)-13/64\right]\\ =-\left(5/16-13/64\right)\end{array}$

$\therefore \left(GJ/T{d}^2\right)w_{o4}=-7/64$

4) Constrained Warping Displacements

With one end fully constrained these displacements w = w(z) follow from Equations (7a), (7b):

$w=w_o\left[1-\cosh \mu \left(L-z\right)/\cosh \mu L\right]$(7a)

for which, with z = L at the free-end, the constrained warping displacements become:

$w=w_o\left(1-1/\cosh \mu L\right)$(7b)

in which the multiplication factor $\left(1-1/\cosh \mu L\right)$ in Equation (7b) modifies the unconstrained warping displacement w_o at the free end

$w=w_o\left(1-1/6.857\right)=0.854w_o$

The warping constraint increases towards the fixing. At the fixed-end the cross-section becomes fully constrained in that there are no warping displacements. At the mid-length position Equation (7a) shows for z = L/2

$w=w_o\left[1-\cosh \left(\mu L/2\right)/\cosh \mu L\right]$

$w=w_o\left(1-1.983/6.857\right)=0.711w_o$

Table 1 applies the coefficients 0.854 and 0.711 to w_o giving the constrained warping displacements at the end and centre lengths at points 1, 2 and 3 in this section.

To read from Table 1 if, say, the constrained displacement is required for point 1 at mid length under an axial torque of 1 Nm then the coefficient of 0.1444 appears within the conversion of the normalised displacement as follows

$\begin{array}{c}{w}_1=0.711\times \left(13/64\right)\times T{d}^2/GJ\\ =0.1444\times \left(1\times {10}^3\right)\times {25.4}^2/\left[\left(70\times {10}^3\right)\times 67.746\right]\\ =1.965\times {10}^{-3}\text{mm}\end{array}$

The constrained warping distribution for this section is examined further in the summary that follows in § 3.

Section B: a/d = 4/7, t/d = 1/14, d = 44.45 mm, L = 1 m

$a=1\text{'}\text{'},d=1\frac{3}{4}\text{'}\text{'},t=1/8\text{'}\text{'},L=1\text{m}\left(39.37\text{'}\text{'}\right)$

$\stackrel{⌣}{y}/d^2=\left(1/4\right)\left(4/7\right)\left[7\left(4/7\right)+4{\left(4/7\right)}^2+2\right]/{\left(1+8/7\right)}^2=358/1575$

${\stackrel{⌣}{y}}^2\left(2a+d\right)t={\left(358d^2/1575\right)}^2\left(2a+d\right)t=\left(105/16\right){\left(358/1575\right)}^{2}t{a}^2{d}^3$

$\begin{array}{c}{\displaystyle {\int }_s{y}^{2}t\text{d}s}=\left(t{a}^2{d}^3/2\right)[\left(1/3\right)\left(4/7\right)+1/2+\left(1/2\right)\left(4/7\right)/\left(1+8/7\right)\\ +\left(1/6\right){\left(4/7\right)}^2/{\left(1+8/7\right)}^2]\\ =\left(7897/9450\right)\left(t{a}^2{d}^3/2\right)\end{array}$

${\Gamma }_{\text{I}}=\left[\left(7897/9450\right)-\left(105/8\right){\left(358/1575\right)}^2\right]\left(t{a}^2{d}^3/2\right)=0.07877t{a}^2{d}^3$

1) Constants

${\Gamma }_{\text{I}}=0.07877\times 3.175\times {25.4}^2\times {44.45}^3=14.17\times {10}^6{\text{mm}}^6$

$J=\left(2a+d\right)\left(t^3/3\right)=\left(2\times 25.4+44.45\right)\left({3.175}^3/3\right)=1016.19{\text{mm}}^4$

$\begin{array}{c}\mu =\sqrt{GJ/E{\Gamma }_{\text{I}}}\\ =\sqrt{\left(70\times {10}^3\times 1016.19\right)/\left(210\times {10}^3\times 14.17\times {10}^6\right)}\\ =4.889\times {10}^{-3}{\text{mm}}^{-1}\end{array}$

$\mu L=4.889\times {10}^{-3}\times 1000=4.889$

$\exp \left(-2\mu L\right)=\exp \left(-9.778\right)=56.68\times {10}^{-6}$

$\cosh \mu L=\cosh 4.889=66.414$

$\cosh \left(\mu L/2\right)=\cosh 2.445=5.809$

2) Torsional Stiffness

$\begin{array}{c}T/\theta =\left[\mu GJ\left(1+\exp \left(-2\mu L\right)\right)\right]/\left[\left(\mu L-1\right)+\left(\mu L+1\right)\exp \left(-2\mu L\right)\right]\\ =\left[\left(4.889\times {10}^{-3}\right)\times \left(70\times {10}^3\right)\times 1016.89\left(1+\exp \left(-9.778\right)\right)\right]/\left[3.889+5.889\exp \left(-9.778\right)\right]\\ =89.42\times {10}^3\text{N}\cdot \text{mm}/\text{rad}=89.42\text{N}\cdot \text{m}/\text{rad}\end{array}$

3) Unconstrained Warping Displacements w_o

From Equation (6):

$GJ{w}_o/T=-\left(y-\stackrel{⌣}{y}\right)$

giving at each ‘corner’ position:

$\left(GJ/T\right)w_{o1}=\left(GJ/T\right)w_{o5}=\stackrel{⌣}{y}$

$\left(GJ/T\right)w_{o2}=\left(GJ/T\right)w_{o4}=\stackrel{⌣}{y}-ad/2$

$\left(GJ/T\right)w_{o3}=\stackrel{⌣}{y}-d\left(a+e\right)/2$

At 1, $y_1=0:\left(GJ/T\right)w_{o1}=-\left(0-358d^2/1575\right)$

$\therefore \left(GJ/T\right)w_{o1}=0.2273$

At 3, $y_3=ad/2:\left(GJ/T\right)w_{o3}=-\left(ad/2-358d^2/1575\right)$

$\left(GJ/T{d}^2\right)w_{o3}=-\left(ad/2-358/1575\right)=-\left(2/7-358/1575\right)$

$\therefore \left(GJ/T{d}^2\right)w_{o3}=-0.0584$

At 4, $y_4=\left(ad/2\right)\left(d+3a\right)/\left(d+2a\right)$

$\left(GJ/T\right)w_{o4}=-\left[\left(ad/2\right)\left(d+3a\right)/\left(d+2a\right)-358d^2/1575\right]$

$\left(GJ/T{d}^2\right)w_{o4}=-\left[\left(1/2\right)\left(a/d\right)\left(1+3a/d\right)/\left(1+2a/d\right)-358/1575\right]$

$\begin{array}{c}\left(GJ/T{d}^2\right)w_{o4}=-\left[\left(1/2\right)\left(4/7\right)\left(1+12/7\right)/\left(1+8/7\right)-358/1575\right]\\ =-\left(0.3619-0.2273\right)\end{array}$

$\left(GJ/T{d}^2\right)w_{o4}=-0.1346$

4) Constrained Warping Displacements

Again, these follow from Equation (7a) as:

$w=w_o\left[1-\cosh \mu \left(L-z\right)/\cosh \mu L\right]$(7a)

for which, with z = L at the free-end, the constrained warping displacements become (Equation (7b)):

$w=w_o\left(1-1/\cosh \mu L\right)$(7b)

in which the factor $\left(1-1/\cosh \mu L\right)$ modifies the unconstrained warping displacements at the free end according to:

$w=w_o\left(1-1/66.414\right)=0.9849w_o$

showing the 1 m length almost preserves w_o at the free-end. However, the constraint increases towards the fixing. At the fixed end the cross-section becomes fully constrained in that there are no warping displacements. At the mid-length position Equation (7a) shows, for z = L/2:

$w=w_o\left[1-\cosh \left(\mu L/2\right)/\cosh \mu L\right]=w_o\left(1-5.809/66.414\right)=0.9125w_o$

The coefficients 0.9849 and 0.9125 for the end and centre lengths are applied to the free warping displacements w_o at median points 1, 3 and 4 in the section. Table 2 gives the constrained warping displacements within the cross-section at the end and centre lengths at each position.

To read from Table 2 , say the constrained displacement at point 3 is required at mid-length for a torque of 1 Nm then the coefficient of −0.0533 appears in calculating w₃ as follows:

$w_o=-0.0584\times T{d}^2/GJ$

$w_3=0.9125w_o$

$w_3=-0.9125\times 0.0584\times T{d}^2/GJ$

$w_3=-0.0533\times \left(1\times {10}^3\right)\times {44.45}^2/\left(70\times {10}^3\right)\times 1016.19=-1.48\times {10}^{-3}\text{mm}$

Section C: a/d = 1/3, t/d = 1/40, d = 47.625 mm, L = 340 mm

$a=5/8\text{'}\text{'},d=1\frac{7}{8}\text{'}\text{'},t=3/64\text{'}\text{'},L=340\text{mm}\left(13.39\text{'}\text{'}\right)$

$\stackrel{⌣}{y}/d^2=\left(1/4\right)\left(1/3\right)\left[7\left(1/3\right)+4{\left(1/3\right)}^2+2\right]/{\left(1+2/3\right)}^2=43/300$

${\stackrel{⌣}{y}}^2\left(2a+d\right)t={\left(43d^2/300\right)}^2\left(2a+d\right)t=\left(45/3\right){\left(43/300\right)}^{2}t{a}^2{d}^3$

$\begin{array}{c}{\displaystyle {\int }_s{y}^{2}t\text{d}s}=\left(t{a}^2{d}^3/2\right)\left[\left(1/3\right)\left(1/3\right)+1/2+\left(1/3\right)/\left(10/3\right)+\left(1/9\times 9\right)/\left(6\times 25\right)\right]\\ =\left(323/450\right)\left(t{a}^2{d}^3/2\right)\end{array}$

1) Constants

${\Gamma }_{\text{I}}=\left[323/450-30{\left(43/300\right)}^2\right]t{a}^2{d}^3/2=0.05072t{a}^2{d}^3$

${\Gamma }_{\text{I}}=0.05072\times 1.1906\times {15.875}^2\times {47.625}^3=1.644\times {10}^6{\text{mm}}^6$

$J=\left(2a+d\right)\left(t^3/3\right)=\left(2\times 15.875+47.625\right)\left({1.1906}^3/3\right)=44.654{\text{mm}}^4$

$\begin{array}{c}\mu =\sqrt{GJ/E{\Gamma }_{\text{I}}}=\sqrt{\left(70\times {10}^3\times 44.654\right)/\left(210\times {10}^3\times 1.644\times {10}^6\right)}\\ =3.009\times {10}^{-3}{\text{mm}}^{-1}\end{array}$

$\mu L=3.009\times {10}^{-3}\times 340=1.023$

$\exp \left(-2\mu L\right)=\exp \left(-2.046\right)=0.1292$

$\cosh \mu L=\cosh 1.023=1.5705$

$\cosh \left(\mu L/2\right)=\cosh 0.5115=1.1337$

2) Torsional Stiffness

$\begin{array}{c}T/\theta =\left[\mu GJ\left(1+\exp \left(-2\mu L\right)\right)\right]/\left[\left(\mu L-1\right)+\left(\mu L+1\right)\exp \left(-2\mu L\right)\right]\\ =\left[\left(3.009\times {10}^{-3}\right)\times \left(70\times {10}^3\right)\times 44.654\left(1+0.1292\right)\right]/\left[\left(1.023-1\right)+\left(1.023+1\right)0.1292\right]\end{array}$

$T/\theta =37.35\times {10}^3\text{N}\cdot \text{mm}/\text{rad}=37.35\text{N}\cdot \text{m}/\text{rad}$

3) Unconstrained Warping Displacements

From Equation (6):

$GJ{w}_o/T=-\left(y-\stackrel{⌣}{y}\right)$

giving at each median point 1, 3 and 4:

$\left(GJ/T\right)w_{o1}=\left(GJ/T\right)w_{o7}=\stackrel{⌣}{y}$

$\left(GJ/T\right)w_{o3}=\left(GJ/T\right)w_{o5}=\stackrel{⌣}{y}-ad/2$

$\left(GJ/T\right)w_{o4}=\stackrel{⌣}{y}-d\left(a+e\right)/2$

At 1 $y_1=0:\left(GJ/T\right)w_{o1}=-\left(0-43d^2\right)/300$

$\therefore \left(GJ/T{d}^2\right)w_{o1}=43/300$

At 3, $y_3=ad/2:\left(GJ/T\right)w_{o3}=-\left(ad/2-43d^2/300\right)$

$\left(GJ/T{d}^2\right)w_{o1}=-\left(a/2d-43/300\right)=-\left[\left(1/2\right)\left(1/3\right)-43/300\right]$

$\therefore \left(GJ/T{d}^2\right)w_{o1}=-7/300$

At 4, $y_4=\left(ad/2\right)\left(d+3a\right)/\left(d+2a\right)$

$\left(GJ/T\right)w_{o4}=-\left[\left(ad/2\right)\left(d+3a\right)/\left(d+2a\right)-43d^2/300\right]$

$\begin{array}{c}\left(GJ/T{d}^2\right)w_{o4}=-\left[\left(1/2\right)\left(a/d\right)\left(1+3a/d\right)/\left(1+2a/d\right)-43/300\right]\\ =-\left[\left(1/2\right)\left(1/3\right)\left(1+1\right)/\left(1+2/3\right)-43/300\right]\\ =-\left(1/5-43/300\right)\end{array}$

$\therefore \left(GJ/T{d}^2\right)w_{o4}=-17/300$

4) Constrained Warping Displacements

Equation (7a) provides w from knowing w_o in 3 above

$w=w_o\left[1-\cosh \mu \left(L-z\right)/\cosh \mu L\right]$

for which, with z = L at the free-end, the constrained warping displacement Equation (7b) becomes:

$w=w_o\left(1-1/\cosh \mu L\right)$

in which the factor $\left(1-1/\cosh \mu L\right)$ modifies the unconstrained warping displacements at the free end to

$w=w_o\left(1-1/1.5705\right)=0.3633w_o$

showing that fixing one end of the 340 mm length constrains the free-end warping to 0.3633w_o. The constraint increases further towards the fixing where there is no warping when the cross-section becomes fully constrained. At the mid-length position Equation (7a) shows for z = L/2

$w=w_o\left[1-\cosh \mu \left(L-z\right)/\cosh \mu L\right]$

$w=w_o\left(1-1.1337/1.5705\right)=0.2781w_o$

The coefficients 0.3633 and 0.2781 for the end and centre lengths are applied to the free warping displacements w_o at points 1, 3 and 4 in section C. Table 3 gives the constrained warping displacements within the cross-section at the end and centre lengths at each position.

To read from Table 3 , say if the constrained displacement at point 4 is required at mid-span for a torque of 1 Nm then the coefficient of −0.01576 appears in calculating w₄ as follows

$\begin{array}{c}{w}_4=0.2781w_o=0.2781\times \left(-17/300\right)\times T{d}^2/GJ\\ =-0.01576\times \left(1\times {10}^3\right)\times {47.625}^2/\left(70\times {10}^3\times 44.654\right)\\ =-11.436\times {10}^{-3}\text{mm}\end{array}$