Problem 6.6. Castigliano’s theorem

Applying Castigliano’s theorem derive
a) midspan deflection and
b) end rotation
of a simply supported beam subjected to uniform
load, p = 3 kN/m. Length of the beam is L = 5 m, its bending
stiffness is constant, EI =4.2× 10⁶ Nm.

Hint: Assume a dummy
force, F and a moment, M in the directions
of the unknown displacements in order to
perform the required derivations.

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Problem a)

Midspan deflection, e [mm]=

Problem b)

End rotation, φ [rad]=

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Steps

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Problem a)

Step 1. Place a concentrated “dummy load”, F = 0 at the midspan. Draw moment diagrams from loads, p and F.

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Step 2. Apply Castigliano’s II theorem. Give the derivative of the strain energy parametrically according to the “dummy load”, F.

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The potential energy of the beam subjected to both loads is:

Eq.(6-19)

$U = \frac{1}{2 E I} \int_{L} (M_{p} + M_{F})^{2} d x = \frac{1}{2 E I} \int_{L} (M_{p}^{2} + M_{F}^{2} + 2 M_{p} M_{F}) d x$

Let us substitute $M_{F} = F \times M_{1}$ into the above expression, where $M_{1} = \frac{x}{2}$ is the moment from a unit concentrated force. Perform differentiation with respect to the “dummy load”, F:

$\frac{\partial U}{\partial F} = \frac{\partial}{\partial F} (\frac{1}{2 E I} \int_{L} (M_{p}^{2} + F^{2} M_{1}^{2} + 2 F M_{p} M_{1}) d x) = \frac{1}{2 E I} (2 F \int_{L} M_{1}^{2} d x + \int_{L} 2 M_{p} M_{1} d x)$

According to Castigliagno’s II theorem the above derivative is equal to the midspan deflection, when the dummy load is set equal to zero.

See Eq.(6-93)

$e = \frac{\partial U}{\partial F} = \frac{1}{E I} \int_{L} M_{p} M_{1} d x$

See Footnote j in Section 6.4.

Step 3. Perform integration, calculate the deflection.

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$e = \frac{\partial U}{\partial F} = \frac{1}{E I} \int_{L} M_{p} M_{1} d x = 2 \frac{1}{E I} \int_{0}^{L / 2} p (\frac{L x}{2} - \frac{x^{2}}{2}) \frac{x}{2} d x = = \frac{p}{2 E I} {[\frac{L x^{3}}{3} - \frac{x^{4}}{4}]}_{0}^{L / 2} = \frac{5 p L^{4}}{384 E I} = \frac{5 \times 3000 \times 5^{4}}{4.2 \times 10^{6}} = 0.00581 m = 5.81 mm$

Integration can be also performed visually (see Hint in Problem 6.1):

$e = \frac{1}{E I} \int_{L} M_{p} M_{1} d x = 2 \frac{1}{E I} A h = 2 \frac{1}{E I} (\frac{p L^{2}}{8} \frac{L}{2} \frac{2}{3}) \frac{F L}{4} \frac{5}{8} = \frac{5 p L^{4}}{384 E I} = 5.81 mm$

Problem b)

Step 1. Apply a concentrated “dummy load”, M at the right support. Draw moment diagrams from loads, p and M.

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Step 2. Apply Castigliano’s II theorem. Give the derivative of the strain energy parametrically according to the “dummy load”, M.

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According to Castigliagno’s II theorem (similarly as it was derived above for the deflection) the rotation at the right support results in

$φ = \frac{\partial U}{\partial M} = \frac{1}{E I} \int_{L} M_{p} M_{1} d x$

where M₁ is the moment from a unit moment load acting on the right support.

See Footnote j in Section 6.4.

Step 3. Perform integration, calculate the rotation.

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Perform integration visually:

$φ = \frac{1}{E I} \int_{L} M_{p} M_{1} d x = \frac{A h}{E I} = \frac{1}{E I} (\frac{p L^{2}}{8} L \frac{2}{3}) \frac{1}{2} = \frac{p L^{3}}{24 E I} = \frac{3000 \times 5^{3}}{24 \times 4.2 \times 10^{6}} = 0.0372 rad$

Results

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Problem a)

A concentrated “dummy load”, F (= 0) is placed at midspan. Moment diagrams from loads, p and F are given in the Figure below.

The potential energy of the beam subjected to both loads is:

Eq.(6-19)

$U = \frac{1}{2 E I} \int_{L} (M_{p} + M_{F})^{2} d x = \frac{1}{2 E I} \int_{L} (M_{p}^{2} + M_{F}^{2} + 2 M_{p} M_{F}) d x$

According to Castigliagno’s II theorem the above derivative is equal to the midspan deflection, when the dummy load is set equal to zero.

See Eq.(6-93)

$e = \frac{\partial U}{\partial F} = \frac{1}{E I} \int_{L} M_{p} M_{1} d x$

See Footnote j in Section 6.4.

Performing the integration the deflection results in:

Integration can be also performed visually (see Hint in Problem 6.1):

$e = \frac{1}{E I} \int_{L} M_{p} M_{1} d x = 2 \frac{1}{E I} A h = 2 \frac{1}{E I} (\frac{p L^{2}}{8} \frac{L}{2} \frac{2}{3}) \frac{F L}{4} \frac{5}{8} = \frac{5 p L^{4}}{384 E I} = 5.81 mm$

Problem b)

A concentrated “dummy load”, M is applied at the right support. Moment diagrams from loads, p and M are given below.

According to Castigliagno’s II theorem (similarly as it was derived above for the deflection) the rotation at the right support is

$φ = \frac{\partial U}{\partial M} = \frac{1}{E I} \int_{L} M_{p} M_{1} d x$

where M₁ is the moment from a unit moment load acting on the right support.

See Footnote j in Section 6.4.

Performing the integration visually the rotation results in: