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## Q. 5.4

The median dimensions of the two cells shown in the cellular section of Fig. 5.6 are $A_{1} = 20 mm × 40 mm$ and  $A_{2} = 50 mm ×40 mm$ with wall thicknesses $t_{1} = 2 mm$, $t_{2} = 1.5 mm$ and  $t_{3} = 3 mm$. If the section is subjected to a torque of 320 Nm, determine the angle of twist per unit length and the maximum shear stress set up. The section is constructed from a light alloy with a modulus of rigidity G = 30 GN/m².

## Verified Solution

From eqn. (5.18),

$T=2(\tau _{1}t_{1}A{1}+\tau _{2}t_{2}A{2})$       (5.18)

$320=2(\tau _{1} \times 2\times 20\times 40+\tau _{2}\times 1.5\times 50\times 40 )10^{-9}$    (1)

From eqn. (5.19),

$\frac{2G\theta }{L}=\frac{1}{A_{1}}(\tau _{1}s_{1}+\tau _{3}s_{3}) =\frac{1}{A_{2}}(\tau _{2}s_{2}-\tau _{3}s_{3})$     (5.19)

$2\times 30\times 10^{9}\times \theta =\frac{1}{20\times 40\times 10^{-6}}[\tau _{1}(40+2\times 20)10^{-3}+\tau _{3}\times 40\times10^{-3} ]$    (2)

and    $2\times 30\times 10^{9}\times \theta =\frac{1}{50\times 40\times 10^{-6}}[\tau _{2}(40+2\times 50)10^{-3}-\tau _{3}\times 40\times10^{-3} ]$            (3)

Equating (2) and (3),

From eqn. (5.17),

$\tau _{1}t_{1}=\tau _{2}t_{2}+\tau _{3}t_{3}$     (5.17)

$2\tau _{1}=1.5\tau _{2}+3\tau _{3}$         (4)
$\frac{1}{8} [80\tau _{1}+40\tau _{3}]=\frac{1}{20}[140\tau _{2}-40\tau _{3}]$

Multiply through by 40,

$400\tau _{1}+200\tau _{3}=280\tau _{2}-80\tau _{3}$
$40\tau _{1}=28\tau _{2}-28\tau _{3}$                  (5)

(5) × 60/28            $85.7\tau _{1}60\tau _{2}-60\tau _{3}$              (6)

But, from (4), multiplied by 20,

$40\tau _{1}=30\tau _{2}+60\tau _{3}$             (7)

(6) + (7),         $125.7\tau _{1}=90\tau _{2}$        (8)

and from (1),      $320=(3.2\tau _{1}+6\tau _{2})10^{-6}$

$320\times 10^{6}=3.2\tau _{1}+6\tau _{2}$               (9)

substituting for $τ_{2}$ from (8),

$320\times 10^{6}=3.2\tau _{1}+6\times \frac{125.7}{90} \tau _{1}$
$=3.2\tau _{1}+8.4\tau _{1}$
∴               $\tau _{1}=\frac{320\times 10^{6}}{11.6} =27.6\times 10^{6}=27.6 MN/m^{2}$

From (8),

$\tau _{2}=\frac{125.7}{90}\times 27.6=38.6 MN/m^{2}$

From (4),

$\tau _{3}=\frac{1}{3} (2\times 27.6-1.5\times 38.6)$
$=\frac{1}{3} (55.2-57.9)=\frac{1}{3}\times (-2.7) =-0.9 MN/m^{2}$

The negative sign indicates that the direction of shear flow in the wall of thickness $t_{3}$ is reversed from that shown in Fig. 5.6.

The maximum shear stress present in the section is thus 38.6 MN/m² in the 1.5 mm wall thickness.
From (3),

$2\times 30\times 10^{9}\times \theta =\frac{(140t_{2}-40t_{3})}{50\times 40\times 10^{-3}}$
$=\frac{140\times 38.6\times 10^{6}-40(-0.9\times 10^{6})}{50\times 40\times 10^{-3}\times 2\times 30\times 10^{9}}$

$=\frac{(5.40+0.036)}{120}$  radian
$=\frac{5.440}{120}\times \frac{360}{2\pi } =2.6°$

The angle of twist of the section is 2.6°.