Problem 3.18. Linearly distributed torque

A beam built-in at both ends is subjected to a linearly varying torque. Present the differential equation of Saint-Venant torsion and give the boundary conditions. The torsional stiffness of the solid circular cross section, GI_t is constant. Give the rotation function.

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Derive the rotation function parametrically.

Check expression

$ψ = \frac{t_{0}}{6 G I_{t} L} (x L^{2} - x^{3})$

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Steps

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Step 1. Give the load function of the beam subjected to linearly distributed torque. Write the differential equation of Saint-Venant torsion.

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$t (x) = t_{0} \frac{x}{L}$

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The geometrical and material equations results in

$T_{SV} = ϑ G I_{t} = \frac{d ψ}{d x} G I_{t}$

See Eqs.(3-83),(3-72) and Table 13 (T = T_SV)

The equilibrium equation is $t (x) = - \frac{d T_{SV}}{d x}$

Eq.(3-81) and Table 13

The differential equation of Saint-Venant torsion is obtained from the above equations:

$t (x) = - \frac{d^{2} ψ}{d x^{2}} G I_{t} where t (x) = t_{0} \frac{x}{L}$

See Eq.(3-129) when restrained warping is neglected.

Note: The differential equation and its solution is equivalent to that of a tension rod (see Problem 3.4)

Step 2. Write the solution of the homogeneous equation.

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$ψ_{hom} = C_{1} + C_{2} x$

Step 3. Find a particular solution.

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The following particular solution satisfies the inhomogeneous equation

$ψ_{part} = - \frac{t_{0}}{G I_{t} L} \frac{x^{3}}{6}$

$- G I_{t} \frac{d^{2} ψ_{part}}{d x^{2}} = - G I_{t} \frac{d^{2} (- \frac{t_{0}}{G I_{t} L} \frac{x^{3}}{6})}{d x^{2}} = - G I_{t} (- \frac{t_{0}}{G I_{t} L} x) = t_{0} \frac{x}{L}$

Step 4. Write the general solution and determine its constants from the boundary conditions.

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$ψ = ψ_{hom} + ψ_{part} = C_{1} + C_{2} x - \frac{t_{0}}{G I_{t} L} \frac{x^{3}}{6}$

Boundary conditions:

At both supports the rotation is zero.

$at x = 0 ψ = 0 \to C_{1} = 0 at x = L ψ = 0 ψ (L) = C_{2} L - \frac{t_{0}}{G I_{t} L} \frac{L^{3}}{6} = 0 \to C_{2} = \frac{t_{0} L}{6 G I_{t}}$

The solution of the differential equation, i.e. the rotation function is

$ψ = \frac{t_{0}}{6 G I_{t} L} (x L^{2} - x^{3})$

Results

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The torque load is linearly distributed along the length of the beam:

$t (x) = t_{0} \frac{x}{L}$

The geometrical and material equations results in

$T_{SV} = ϑ G I_{t} = \frac{d ψ}{d x} G I_{t}$

See Eqs.(3-83), (3-72) and Table 13 (T = T_SV)

The equilibrium equation is $t (x) = - \frac{d T_{SV}}{d x}$

Eq.(3-81) and Table 13

The differential equation of Saint-Venant torsion is obtained from the above equations:

$t (x) = - \frac{d^{2} ψ}{d x^{2}} G I_{t} where t (x) = t_{0} \frac{x}{L}$

See Eq.(3-129) when restrained warping is neglected.

Note: The differential equation and its solution is equivalent to that of a tension rod (see Problem 3.4)

The solution of the homogeneous equation is

$ψ_{hom} = C_{1} + C_{2} x$

The following particular solution satisfies the inhomogeneous equation

$ψ_{part} = - \frac{t_{0}}{G I_{t} L} \frac{x^{3}}{6}$

$- G I_{t} \frac{d^{2} ψ_{part}}{d x^{2}} = - G I_{t} \frac{d^{2} (- \frac{t_{0}}{G I_{t} L} \frac{x^{3}}{6})}{d x^{2}} = - G I_{t} (- \frac{t_{0}}{G I_{t} L} x) = t_{0} \frac{x}{L}$

Constants of the general solution are determined from the boundary conditions:

$ψ = ψ_{hom} + ψ_{part} = C_{1} + C_{2} x - \frac{t_{0}}{G I_{t} L} \frac{x^{3}}{6}$

Boundary conditions:

At both supports the rotation is zero.

$at x = 0 ψ = 0 \to C_{1} = 0 at x = L ψ = 0 ψ (L) = C_{2} L - \frac{t_{0}}{G I_{t} L} \frac{L^{3}}{6} = 0 \to C_{2} = \frac{t_{0} L}{6 G I_{t}}$

The solution of the differential equation, i.e. the rotation function is

$ψ = \frac{t_{0}}{6 G I_{t} L} (x L^{2} - x^{3})$