Problem 3.20 Effect of restrained warping

A cantilever I beam (L = 3.0m) is subjected to t = 4 kNm/m uniformly distributed torque load. At the fixed end there is no twist or warping. The thickness of the wall is t_w = t_f = 12 mm, width of the cross section is b = 260 mm, the height is h = 480 mm. The material properties are E = 210 GPa = 210 ×10³ N/mm², G = 80.8 GPa = 80.8 ×10³ N/mm². Calculate the rotation at the free end
a) neglecting warping stiffness (EI_ω = 0),
b) neglecting torsional stiffness (EI_t = 0),
c) taking both the warping and torsional stiffness into account.

Solve Problem

SolveHide solve

Rotation of the free end, ψ [rad]=

Do you need help?

Steps

Step by stepHide steps

Problem a)

Step 1. Give the torsional stiffness of the cross section.

Check torsional stiffnessHide torsional stiffness

Eq.(3-100)

G I_{t} = G \sum_{k = 1}^{3} \frac{b_{k} t_{k}^{3}}{3} = G (2 \frac{b t_{f}^{3}}{3} + \frac{h t_{w}^{3}}{3}) = 80.8 \times 10^{3} (2 \frac{260 \times 12_{}^{3}}{3} + \frac{480 \times 12_{}^{3}}{3}) = 4.598 \times 10^{10} {Nmm}^{2}

Step 2. Write the differential equation of Saint-Venant torsion.

Check equationHide equation

Eq.(3-129), where the warping stiffness is neglected

t = - \frac{d^{2} ψ_{SV}}{d x^{2}} G I_{t}

Step 3. Give the rotation function as the general solution of the differential equation.

Check general solutionHide general solution

$ψ_{SV} = - \frac{t}{2 G I_{t}} x^{2} + C_{1} x + C_{2}$

See Example D.5 and Eq.(D-31)

Step 4. Determine constants from boundary conditions.

Check boundary conditionsHide boundary conditions

Boundary conditions are given in Table 3.11.

At the built-in end:

$ψ_{SV} (0) = 0 \to C_{2} = 0$

At the free end:

$T (L) = \frac{d ψ_{SV}}{d x} G I_{t} = 0 \frac{d ψ_{SV}}{d x} = - \frac{t}{G I_{t}} L + C_{1} = 0 \to C_{1} = \frac{t L}{G I_{t}}$

Step 5. Give the rotation function. Calculate end rotation.

Check end rotationHide end rotation

$ψ_{SV} = \frac{t L}{G I_{t}} - \frac{t x^{2}}{2 G I_{t}} ψ_{SV} (L) = \frac{t L^{2}}{2 G I_{t}} = \frac{4 \times 3^{2}}{2 \times 45.98} = 0.391 rad = 22 . 4^{\circ}$

Problem b)

Step 1. Give the warping stiffness of the cross section.

Check warping stiffnessHide warping stiffness

Table 3.12

E I_{ω} = E \frac{b_{}^{3} t_{f}}{24} d^{2} = 210 \times 10^{3} \frac{260^{3} \times 12}{24} {(480 - 12)}^{2} = 4.042 \times 10^{17} {Nmm}^{2}

Step 2. Write the differential equation of restrained warping.

Check equationHide equation

Eq.(3-129), where the torsional stiffness is neglected

t = \frac{d^{4} ψ}{d x^{4}} E I_{ω}

Step 3. Give the rotation function as the general solution of the differential equation.

Check general solutionHide general solution

$ψ = \frac{t}{24 E I_{ω}} x^{4} + C_{1} \frac{x^{3}}{6} + C_{2} \frac{x^{2}}{2} + C_{3} x + C_{4}$

See Example D.14 and Eq.(D-79)

Step 4. Determine constants from boundary conditions.

Check boundary conditionsHide boundary conditions

Boundary conditions are given in Table 3.11.

At the built-in end:

$ψ (0) = 0 \to C_{4} = 0 ϑ (0) = \frac{d ψ_{SV}}{d x} = 0 \to C_{3} = 0$

At the free end:

$T (L) = - \frac{d^{3} ψ}{d x^{3}} E I_{ω} = 0 \to - \frac{t L}{E I_{ω}} - C_{1} = 0 \to C_{1} = \frac{t L}{E I_{ω}} M_{ω} (L) = E I_{ω} \frac{d^{2} ψ}{d x^{2}} = 0 \to \frac{t L^{2}}{2 E I_{ω}} + C_{1} L + C_{2} = 0 \to C_{2} = - \frac{t L^{2}}{2 E I_{ω}}$

Step 5. Give the rotation function. Calculate end rotation.

Check end rotationHide end rotation

$ψ = \frac{t x^{4}}{24 E I_{ω}} - \frac{t L x^{3}}{6 E I_{ω}} + \frac{t L^{2} x^{2}}{4 E I_{ω}} ψ (L) = \frac{t L^{4}}{24 E I_{ω}} - \frac{t L^{4}}{6 E I_{ω}} + \frac{t L^{4}}{4 E I_{ω}} = \frac{t L^{4}}{8 E I_{ω}} = \frac{1}{8} \frac{4 \times 3^{4}}{404.2} = 0.100 rad = 5 . 74^{\circ}$

Problem c)

Step 1. The torsional and warping stiffnesses of the cross section are calculated in Problem a) and b) respectively.

Step 2. Write the differential equation of torsion.

Check equationHide equation

Eq.(3-129), where the torsional stiffness is neglected

t = - \frac{d^{3} ψ}{d x^{3}} G I_{t} + \frac{d^{4} ψ}{d x^{4}} E I_{ω}

In the solution follow steps of Example 3.18.

Step 3. Give the solution of the homogeneous equation.

Check homogeneous solutionHide homogeneous solution

See Example D.10 and Eq.(D-62)

ψ_{hom} = C_{1} + C_{2} x + C_{3} \cos h μ x + C_{4} \sinh μ x

where

$μ = \sqrt{\frac{G I_{t}}{E I_{ω}}}$

Step 4. Choosing a particular solution give the general solution.

Check particular solutionHide particular solution

$ψ_{part} = - \frac{t x^{2}}{2 G I_{t}} ψ = ψ_{hom} + ψ_{part} = C_{1} + C_{2} x + C_{3} \cosh μ x + C_{4} \sinh μ x - \frac{t x^{2}}{2 G I_{t}}$

where

$μ = \sqrt{\frac{G I_{t}}{E I_{ω}}}$

Step 5. Determine constants from boundary conditions.

Check boundary conditionHide boundary conditions

Boundary conditions are given in Table 3.11.

At the built-in end:

$ψ (0) = 0 \to C_{1} + C_{3} = 0 ϑ (0) = \frac{d ψ}{d x} = 0 \to C_{2} + C_{4} μ = 0$

At the free end:

$T (L) = \frac{d ψ}{d x} G I_{t} - \frac{d^{3} ψ}{d x^{3}} E I_{ω} = 0 \to G I_{t} (C_{2} + C_{3} μ \sinh μ L + C_{4} μ \cosh μ L) - t L - E I_{ω} (C_{3} μ^{3} \sinh μ L + C_{4} μ^{3} \cosh μ L) = 0 \to C_{2} + C_{3} (\underset{0}{\underset{⏟}{μ \sinh μ L - \frac{E I_{ω}}{G I_{t}} μ^{3} \sinh μ L}}) + C_{4} (\underset{0}{\underset{⏟}{μ \cosh μ L - \frac{E I_{ω}}{G I_{t}} μ^{3} \cosh μ L}}) - \frac{t L}{G I_{t}} = 0 \to C_{2} = \frac{t L}{G I_{t}} M_{ω} (L) = E I_{ω} \frac{d^{2} ψ}{d x^{2}} = 0 \to C_{3} μ^{2} \cosh μ L + C_{4} μ^{2} \sinh μ L - \frac{t}{G I_{t}} = 0 \to C_{3} = \frac{t}{G I_{t} μ^{2} \cosh μ L} (1 + μ L \sinh μ L) \to C_{1} = - \frac{t}{G I_{t} μ^{2} \cosh μ L} (1 + μ \sinh μ L) \to C_{4} = - \frac{t L}{G I_{t} μ}$

Step 5. Give the rotation function. Calculate end rotation.

Check equationHide equation

Substituting the constants the rotation function is obtained:

$ψ = \frac{t}{E I_{ω} μ^{2}} (\frac{1}{\cosh (μ L)} ((1 + L μ \sinh (μ L)) (\cosh (μ x) - 1) + L μ^{2} x - μ^{2} \frac{x^{2}}{2} - L μ \sinh (μ x)))$

$where μ = \sqrt{\frac{G I_{t}}{E I_{ω}}} = \sqrt{\frac{45.98}{404.2}} = 0.337$

The end rotation is

$ψ (L) = \frac{t}{E I_{ω} μ^{2}} (\frac{1}{\cosh (μ L)} ((1 + L μ \sinh (μ L)) (\cosh (μ L) - 1) + μ^{2} \frac{L}{2} - L μ \sinh (μ L))) = = \frac{4}{404.2 \times 0 . 337^{2}} (\frac{1}{\cosh (0.337 \times 3)} ((1 + (0.337 \times 3) \sinh (0.337 \times 3)) (\cosh (0.337 \times 3) - 1) + 0 . 337^{2} \frac{3}{2} - (0.337 \times 3) \sinh (0.337 \times 3))) = = 0.072 rad = 4 . 12^{\circ}$

Results

Show worked out solutionHide solution

Problem a)

The torsional stiffness of the thin walled cross section is calculated as

Eq.(3-100)

G I_{t} = G \sum_{k = 1}^{3} \frac{b_{k} t_{k}^{3}}{3} = G (2 \frac{b t_{f}^{3}}{3} + \frac{h t_{w}^{3}}{3}) = 80.8 \times 10^{3} (2 \frac{260 \times 12_{}^{3}}{3} + \frac{480 \times 12_{}^{3}}{3}) = 4.598 \times 10^{10} {Nmm}^{2}

The differential equation of Saint-Venant torsion is

Eq.(3-129), where the warping stiffness is neglected

t = - \frac{d^{2} ψ_{SV}}{d x^{2}} G I_{t}

The rotation function is given as the general solution of the differential equation:

See Example D.5 and Eq.(D-31)

$ψ_{SV} = - \frac{t}{2 G I_{t}} x^{2} + C_{1} x + C_{2}$

where the constants are determined from the boundary conditions.

Boundary conditions are given in Table 3.11.

At the built-in end:

$ψ_{SV} (0) = 0 \to C_{2} = 0$

At the free end:

$T (L) = \frac{d ψ_{SV}}{d x} G I_{t} = 0 \frac{d ψ_{SV}}{d x} = - \frac{t}{G I_{t}} L + C_{1} = 0 \to C_{1} = \frac{t L}{G I_{t}}$

The rotation function and the end rotation are given below.

$ψ_{SV} = \frac{t L}{G I_{t}} - \frac{t x^{2}}{2 G I_{t}} ψ_{SV} (L) = \frac{t L^{2}}{2 G I_{t}} = \frac{4 \times 3^{2}}{2 \times 45.98} = 0.391 rad = 22 . 4^{\circ}$

Problem b)

The warping stiffness of the cross section is calculated as

Table 3.12

E I_{ω} = E \frac{b_{}^{3} t_{f}}{24} d^{2} = 210 \times 10^{3} \frac{260^{3} \times 12}{24} {(480 - 12)}^{2} = 4.042 \times 10^{17} {Nmm}^{2}

The differential equation of restrained warping is

Eq.(3-129), where the torsional stiffness is neglected

t = \frac{d^{4} ψ}{d x^{4}} E I_{ω}

The rotation function is given as the general solution of the differential equation.

See Example D.14 and Eq.(D-79)

$ψ = \frac{t}{24 E I_{ω}} x^{4} + C_{1} \frac{x^{3}}{6} + C_{2} \frac{x^{2}}{2} + C_{3} x + C_{4}$

The constants are determined from the boundary conditions.

Boundary conditions are given in Table 3.11.

At the built-in end:

$ψ (0) = 0 \to C_{4} = 0 ϑ (0) = \frac{d ψ_{SV}}{d x} = 0 \to C_{3} = 0$

At the free end:

The rotation function and the calculated end rotation are given below

Problem c)

The torsional and warping stiffnesses of the cross section are calculated in Problem a) and b) respectively.

The differential equation of torsion is

Eq.(3-129), where the torsional stiffness is neglected

t = - \frac{d^{3} ψ}{d x^{3}} G I_{t} + \frac{d^{4} ψ}{d x^{4}} E I_{ω}

In the solution follow steps of Example 3.18.

First the solution of the homogeneous equation is derived.

See Example D.10 and Eq. (D-62)

ψ_{hom} = C_{1} + C_{2} x + C_{3} \cos h μ x + C_{4} \sinh μ x

where

$μ = \sqrt{\frac{G I_{t}}{E I_{ω}}}$

Then a particular solution is chosen. The general solution is shown below.

$ψ_{part} = - \frac{t x^{2}}{2 G I_{t}} ψ = ψ_{hom} + ψ_{part} = C_{1} + C_{2} x + C_{3} \cosh μ x + C_{4} \sinh μ x - \frac{t x^{2}}{2 G I_{t}}$

The constants are determined from the boundary conditions.

Boundary conditions are given in Table 3.11.

At the built-in end:

$ψ (0) = 0 \to C_{1} + C_{3} = 0 ϑ (0) = \frac{d ψ}{d x} = 0 \to C_{2} + C_{4} μ = 0$

At the free end:

Substituting the constants the rotation function is obtained:

$ψ = \frac{t}{E I_{ω} μ^{2}} (\frac{1}{\cosh (μ L)} ((1 + L μ \sinh (μ L)) (\cosh (μ x) - 1) + L μ^{2} x - μ^{2} \frac{x^{2}}{2} - L μ \sinh (μ x)))$

$where μ = \sqrt{\frac{G I_{t}}{E I_{ω}}} = \sqrt{\frac{45.98}{404.2}} = 0.337$

The end rotation is $ψ (L) = \frac{t}{E I_{ω} μ^{2}} (\frac{1}{\cosh (μ L)} ((1 + L μ \sinh (μ L)) (\cosh (μ L) - 1) + μ^{2} \frac{L}{2} - L μ \sinh (μ L))) = = \frac{4}{404.2 \times 0 . 337^{2}} (\frac{1}{\cosh (0.337 \times 3)} ((1 + (0.337 \times 3) \sinh (0.337 \times 3)) (\cosh (0.337 \times 3) - 1) + 0 . 337^{2} \frac{3}{2} - (0.337 \times 3) \sinh (0.337 \times 3))) = = 0.072 rad = 4 . 12^{\circ}$