Figure P.11.17 shows the cross section of a thin-walled inwardly lipped channel. The lips are of constant thickness while the flanges increase linearly in thickness from 1.27 mm, where they meet the lips, to 2.54 mm at their junctions with the web. The web has a constant thickness of 2.54 mm and

Question

Figure P.11.17 shows the cross section of a thin-walled inwardly lipped channel. The lips are of constant thickness while the flanges increase linearly in thickness from 1.27 mm, where they meet the lips, to 2.54 mm at their junctions with the web. The web has a constant thickness of 2.54 mm and the shear modulus G is 26 700 N/mm² . Calculate the maximum shear stress in the section and also its rate of twist if it is subjected to a torque of 100 Nm.

Accepted Answer

The maximum shear stress in the section is given by Eq. (11.31) in which

[latex] au_{\max }=\pm G t \frac{ d heta}{ d x}=\pm \frac{T t}{J}[/latex] (11.31)

[latex] J=\frac{100 imes 2.54^{3}}{3}+\frac{2 imes 38 imes 1.27^{3}}{3}+\left(\frac{2}{3} ight) \int_{0}^{50}\left[1.27+1.27\left(\frac{s}{50} ight) ight]^{3} d s=854.2 mm ^{4} [/latex]

in which s is measured from a lip/flange junction. Then

[latex] au_{\max }=\pm \frac{100 imes 10^{3} imes 2.54}{854.2}=\pm 297.4 N / mm ^{2} [/latex]

The rate of twist is given by Eq. (11.31) and is

[latex] \frac{ d heta}{ d x}=\frac{100 imes 10^{3}}{26700 imes 854.2}=0.0044 rad / mm [/latex]

Question 11.17: Figure P.11.17 shows the cross section of a thin-walled inwa...

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