Problem 3.21. Bus stop – Solution Manual

The main girder of a bus stop booth is given in the Figure. The thin walled steel beam has fork supports. Determine the torque in the beam (assume uniform distribution), calculate the stiffnesses of the cross section. Write the differential equation describing the equilibrium of the beam, give the boundary conditions. The vertical load on the roof is uniformly distributed p = 4.0 kN/m² (contains the snow load and the self-weights of the glass roof and of the secondary girders). Geometrical sizes are given in the Figure. Solve the differential equation.

a) neglecting warping stiffness (EI_ω = 0),
b) neglecting torsional stiffness (EI_t = 0),
c) taking both the warping and torsional stiffness into account.

(Bending of the girder due to the vertical load is not discussed.)

Solve Problem

SolveHide solve

Rotation at the middle of the beam, $ψ (\frac{L}{2}) [rad] =$

Do you need help?

Steps

Step by stepHide steps

Step 0. Calculate the torque load acting on the beam.

Check torque loadHide torque load

The torque load is equal to the support bending moment of the secondary girders, which are assumed to be uniformly distributed:

$t = p \frac{1 . 5^{2}}{2} = 4.0 \frac{1 . 5^{2}}{2} = 4.5 kNm / m$

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} = 80.8 \times 10^{3} (\frac{350 \times 10_{}^{3}}{3} + \frac{370 \times 6_{}^{3}}{3} + \frac{100 \times 10_{}^{3}}{3}) = 1.43 \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 x = 0:

$ψ_{SV} (0) = 0 \to C_{2} = 0$

At x = L:

$ψ (L) = - \frac{t L^{2}}{2 G I_{t}} + C_{1} L = 0 \to C_{1} = \frac{t L}{2 G I_{t}}$

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

Check end rotationHide end rotation

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

$ψ (\frac{L}{2}) = - \frac{t L^{2}}{8 G I_{t}} + \frac{t L^{2}}{4 G I_{t}} = \frac{t L^{2}}{8 G I_{t}} = \frac{4.5 \times 4^{2}}{8 \times 14.3} = 0.631 = 36 . 15^{\circ}$

Problem b)

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

Check warping stiffnessHide warping stiffness

Table 3.12

Distance of the shear centre from the bottom of the cross section is

$e = \frac{b_{f 2}^{3} t_{f 2}}{b_{f 1}^{3} t_{f 1} + b_{f 1}^{3} t_{f 1}} d = \frac{100_{}^{3} \times 10}{100_{}^{3} \times 10 + 350_{}^{3} \times 10} \times 370 = 361.6 mm$

The warping stiffness of the cross section is
$E I_{ω} = E \frac{b_{2}^{3} t_{f}}{12} d e = 210 \times 10^{3} \frac{100^{3} \times 10}{12} 370 \times 361.6 = 2.341 \times 10^{16} {Nmm}^{2}$

Step 2. Write the differential equation of restrained warping.

Check equationHide equation

Eq.(3-129)

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-69)

Step 4. Determine constants from boundary conditions.

Check boundary conditionsHide boundary conditions

Boundary conditions are given in Table 3.11.

At x = 0:

$ψ (0) = 0 \to C_{4} = 0 M_{ω} (0) = - E I_{ω} \frac{d^{2} ψ}{d x^{2}} = 0 \to C_{2} = 0$

At x = L:

$\begin{array}{r} ψ (L) = - \frac{t L^{4}}{24 E I_{ω}} + C_{1} \frac{L^{3}}{6 E I_{ω}} + C_{3} \frac{L}{E I_{ω}} = 0 \\ M_{ω} (L) = - E I_{ω} \frac{d^{2} ψ}{d x^{2}} = - \frac{t L^{2}}{2 E I_{ω}} - C_{1} L = 0 \end{array}\} \to \begin{matrix} C_{1} = - \frac{t L}{2 E I_{ω}} \\ C_{3} = - \frac{t L^{3}}{24 E I_{ω}} \end{matrix}$

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

Check end rotationHide end rotation

$ψ = \frac{t x^{4}}{24 E I_{ω}} - \frac{t L x^{3}}{12 E I_{ω}} + \frac{t L^{3} x}{24 E I_{ω}} = \frac{t}{24 E I_{ω}} (x^{4} - 2 L x^{3} + L^{3} x) ψ (\frac{L}{2}) = \frac{t L^{4}}{24 E I_{ω}} (\frac{1}{64} - \frac{2}{8} + \frac{1}{2}) = \frac{5 t L^{4}}{384 E I_{ω}} = \frac{5}{384} \frac{4.5 \times 4^{4}}{23.41} = 0.641 = 36 . 73^{\circ}$

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)

t = - \frac{d^{2} ψ}{d x^{2}} 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 Equation (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. Choose 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 x = 0:

$ψ (0) = 0 \to C_{1} + C_{3} = 0 \to C_{1} = - C_{3} M_{ω} (0) = - E I_{ω} \frac{d^{2} ψ}{d x^{2}} = 0 \to - μ^{2} C_{3} - \frac{t}{G I_{t}} = 0 \to C_{3} = \frac{t}{μ^{2} G I_{t}} = 0.517$

At x = L: $M_{ω} (L) = 0 \to μ^{2} \frac{t}{μ^{2} G I_{t}} \cosh μ L + μ^{2} C_{4} \sinh μ L - \frac{t}{G I_{t}} = 0 \to C_{4} = (\frac{t}{μ^{2} G I_{t}} - \frac{t}{μ^{2} G I_{t}} \cosh μ L) \frac{1}{μ^{2} \sinh μ L} = - 0.474 ψ (L) = - \frac{t}{μ^{2} G I_{t}} + C_{2} L + \frac{t}{μ^{2} G I_{t}} \cosh μ L + C_{4} \sinh μ L - \frac{t L^{2}}{2 G I_{t}} = 0 \to C_{2} = - \frac{t}{μ^{2} G I_{t} L} \cosh μ L - \frac{C_{4}}{L} \sinh μ L + \frac{t L}{2 G I_{t}} + \frac{t}{μ^{2} G I_{t} L} = 0.630$

Step 5. Give the rotation function. Calculate rotation at the middle of the beam.

Check rotationHide rotation

$ψ = - 0.571 + 0.630 x + 0.517 \cosh (0.788 x) - 0.473 \sinh (0.788 x) - \frac{4.5 x^{2}}{2 \times 14.27} ψ (\frac{L}{2}) = 0.321 rad = 18 . 4^{\circ}$

Results

Show worked out solutionHide solution

The torque load is equal to the support bending moment of the secondary girders, which are assumed to be uniformly distributed:

$t = p \frac{1 . 5^{2}}{2} = 4.0 \frac{1 . 5^{2}}{2} = 4.5 kNm / m$

Problem a)

First the torsional stiffness of the cross section is calculated:

Eq.(3-100)

G I_{t} = G \sum_{k = 1}^{3} \frac{b_{k} t_{k}^{3}}{3} = 80.8 \times 10^{3} (\frac{350 \times 10_{}^{3}}{3} + \frac{370 \times 6_{}^{3}}{3} + \frac{100 \times 10_{}^{3}}{3}) = 1.43 \times 10^{10} {Nmm}^{2}

The differential equation of Saint-Venant torsion is

Check equationHide equation

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

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

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

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

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

Constants are determined from the boundary conditions.

Boundary conditions are given in Table 3.11.

At the x = 0:

$ψ_{SV} (0) = 0 \to C_{2} = 0$

At x = L:

$ψ (L) = - \frac{t L^{2}}{2 G I_{t}} + C_{1} L = 0 \to C_{1} = \frac{t L}{2 G I_{t}}$

Substituting the constants the rotation function is obtained.

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

The maximum rotation at the middle of the beam is calculated below as

$ψ (\frac{L}{2}) = - \frac{t L^{2}}{8 G I_{t}} + \frac{t L^{2}}{4 G I_{t}} = \frac{t L^{2}}{8 G I_{t}} = \frac{4.5 \times 4^{2}}{8 \times 14.3} = 0.631 = 36 . 15^{\circ}$

Problem b)

Now the warping stiffness of the cross section is calculated:

Table 3.12

Distance of the shear centre from the bottom of the cross section is

$e = \frac{b_{f 2}^{3} t_{f 2}}{b_{f 1}^{3} t_{f 1} + b_{f 1}^{3} t_{f 1}} d = \frac{100_{}^{3} \times 10}{100_{}^{3} \times 10 + 350_{}^{3} \times 10} \times 370 = 361.6 mm$

The warping stiffness of the cross section is
$E I_{ω} = E \frac{b_{2}^{3} t_{f}}{12} d e = 210 \times 10^{3} \frac{100^{3} \times 10}{12} 370 \times 361.6 = 2.341 \times 10^{16} {Nmm}^{2}$

Write the differential equation of restrained warping is

Eq.(3-129)

t = \frac{d^{4} ψ}{d x^{4}} E I_{ω}

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

See Example D.14 and Eq.(D-69)

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

Constants can be determined from the boundary conditions.

Boundary conditions are given in Table 3.11.

At x = 0:

$ψ (0) = 0 \to C_{4} = 0 M_{ω} (0) = - E I_{ω} \frac{d^{2} ψ}{d x^{2}} = 0 \to C_{2} = 0$

At x = L:

The rotation function and the maximum 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)

t = - \frac{d^{2} ψ}{d x^{2}} 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 shown.

See Example D.10 and Equation (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, which results in the following the general solution:

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

Constants are determined from the boundary conditions.

Boundary conditions are given in Table 3.11.

At x = 0:

$ψ (0) = 0 \to C_{1} + C_{3} = 0 \to C_{1} = - C_{3} M_{ω} (0) = - E I_{ω} \frac{d^{2} ψ}{d x^{2}} = 0 \to - μ^{2} C_{3} + \frac{t}{G I_{t}} = 0 \to C_{3} = \frac{t}{μ^{2} G I_{t}} = 0.517$

At x = L:

$M_{ω} (L) = 0 \to - μ^{2} \frac{t}{μ^{2} G I_{t}} \cosh μ L - μ^{2} C_{4} \sinh μ L + \frac{t}{G I_{t}} = 0 \to C_{4} = (\frac{t}{G I_{t}} - μ^{2} \frac{t}{μ^{2} G I_{t}} \cosh μ L) \frac{1}{μ^{2} \sinh μ L} = - 0.474 ψ (L) = C_{2} x + \frac{t}{μ^{2} G I_{t}} \cosh μ L + C_{4} \sinh μ L - \frac{t L^{2}}{2 G I_{t}} = 0 \to C_{2} = - \frac{t}{μ^{2} G I_{t} L} \cosh μ L - \frac{C_{4}}{L} \sinh μ L + \frac{t L}{2 G I_{t}} = 0.630$

The rotation function and the maximum rotation at the middle of the beam is given below. $ψ = - 0.571 + 0.630 x + 0.517 \cosh (0.788 x) - 0.473 \sinh (0.788 x) - \frac{4.5 x^{2}}{2 \times 14.27} ψ (\frac{L}{2}) = 0.321 rad = 18.39 °$