Problem 7.3. Displacement magnification factor

Approximate the percentage increase of the moment at the bottom of the column given in the Figure taking into account the second order effect of the normal force. In the approximation use the displacement magnification factor (or moment magnification factor). Bending stiffness of the steel profile is: EI = 1.46×10⁷ kNcm².

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Increase of the moment due to second order effect [%]=

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Steps

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Step 1. Calculate the critical load of the column.

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Eq.(7-29)

N_{cr} = \frac{π^{2} E I}{l_{0}^{2}} = \frac{π^{2} \times 1.46 \times 10^{7}}{{(2 \times 260)}^{2}} = 532.9 kN

Step 2. Determine the displacement magnification factor.

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Eq.(7-35)

ψ = \frac{1}{1 - \frac{N}{N_{c r}}} = \frac{1}{1 - \frac{100}{532.9}} = 1.23

Step 3. Give the percentage increase of the moment.

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See Eq.(7-36)

M = ψ M_{0} = 1.23 M_{0}

The moment is increased due to second order effect by the displacement magnification factor. The percentage increase is 23%.

Results

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The critical load of the column is

Eq.(7-29)

N_{cr} = \frac{π^{2} E I}{l_{0}^{2}} = \frac{π^{2} \times 1.46 \times 10^{7}}{{(2 \times 260)}^{2}} = 532.9 kN

The moment is increased due to second order effect by the displacement magnification factor:

Eq.(7-35)

ψ = \frac{1}{1 - \frac{N}{N_{c r}}} = \frac{1}{1 - \frac{100}{532.9}} = 1.23

See Eq.(7-36)

The increased moment is
$M = ψ M_{0} = 1.23 M_{0}$

The percentage increase is 23%.