NOTE:     Although 'q' is constant, the shear stress 't ' may not be if the wall thickness 't'
                 varied with 's'.

To determine the relationship between applied torque and shear flow, apply equilibrium to
the end of the beam. In essence the applied Torque T must equal to the torque generated by
the shear flow.

Look at the end of beam, and a small section d_s. The torque produced by the shear flow on element d_s is pq_{sd s}. Integrating about the whole
section gives:

We have previously defined that:

Therefore:

Often referred to as the 'Bredt-Batho Formula'. Substituting this equation into (4.21) gives the
rate of twist due to the Torque 'T':

And substituting equation (5.1) into equation (4.22), the equation for the warping distribution
becomes:

where:

Start with the rate of twist equation:

Now:

and let:

Giving that the rate of twist is:

Look at the warping equation, but starting from the symmetry line because here w₀ = 0 then:

d and A were calculated above, so to determine the warping we need to define d_o,s and A_0,s,
where:

From 0 to 1, 0 £ s₁ £ b/2, so:

From 1 to 2, 0 £ s₂ £ a, so:

and similarly, from 2 to 3, 0 £ s₃ £ b, so:

and from 3 to 4, 0 £ s₄ £ b, so:

All of which gives a linear equation for the warping on all four sides of the section. Meaning
that we need only look at the corners because here the warping is the greatest. At 1:

Because of symmetry you will find that w₂ = -w₁ = -w₃ = w₄.

TORSION OF OPEN BEAM SECTIONS

To do this analysis we need to consider torsion of a thin rectangular strip, which is bent to form
the open section. This analysis is detailed and complex, for that reason, only the final equations will be given.

The equation for rate of twist is:

where the second polar moment of area J is given by:

The shear stress although constant over the section's length, varies linearly through the
thickness of the section and is given by:

with the maximum shear stress occurring at n = ± ½ t to give:

From this definition for shear stress we can now develop the equation for warping.

The shear strain is given by equation (4.5):

Shifting the axis of the beam to coincide with the shear centre, makes equation (4.9) become:

Shear stress is given by:

Combining these 3 equations give:

At the centre line of the section, n = 0 and t _z,s = 0, giving:

Integrating w.r.t. 's' gives:

where:

and

A_R is the area swept out by a generator, about the shear centre, from the point of zero warping.
 

Note :    A_R is positive if movement of P_R along the tangent to the surface in the direction
             of 's' leads to a counterclockwise rotation of P_R about the shear centre.
 

Example 6: Determine the maximum shear stress t _max, the rate of twist and the warping for
                    the open beam channel section of Figure 50 when a torque of magnitude T = 10 Nm
                    is applied at the shear centre. Assume the material shear modulus to be and G = 25 GPa.

a)      Determining sectional property

b)     Determining maximum shear stress

c)     Determining rate of twist

d)     Determining the sections warping

where:

Because the section is symmetrical, at 0, s₁ = 0, w = 0. Therefore, between 0 and 1:

so the warping equation is linearly varying between 0 and 1, and is:

at s₁ = 25 mm, point 1 : w₁ = -0.253 mm

Between 1 and 2 so

at                                                             s₂ = 25 mm, w₂ = 0.5345 mm

As the section is symmetrical, warping on the lower section will be the reverse of this.

The warping distribution looks like this:

Note :    If the section is unsymmetrical and you don't know where w_s = 0, use the following
              equation.

where:

        A _R,0    =     Swept area by generator rotating about the shear centre from some convenient origin

and

where :

_section = Integral over the entire perimeter of the section

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