Problem 8.1. Vibration and deflection limits

A simply supported steel beam is loaded by p = 20 kN/m uniformly distributed load. The prescribed deflection limit is L/250, the minimum allowed eigenfrequency is 5 Hz. Bending stiffness of the beam is EI = 2215 kNm².
a) Check whether the deflection and the eigenfrequency limit is satisfied in case of L = 2 m and L = 3.24 m spans.

The eigenfrequency can be approximated by Eq.(8-57).

b) In case of 5 Hz max. allowed eigenfrequency, what is the typical span range of steel beams where vibration is critical compared to deflection. Draw the span-frequency diagram for cases w = L/250.

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Problem a)

Approximate natural frequency, f_n(L = 2m) [Hz]=

Approximate natural frequency, f_n(L = 3.24m) [Hz]=

Problem b)

Span limit, L_max[m]=

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Steps

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Problem a)

Step 1. Calculate the maximum deflection of the beam. Check deflection limit.

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See Table 3.5.

w (L = 2 m) = \frac{5 p L^{4}}{384 E I} = \frac{5 \times 20 \times 2 . 0^{4}}{384 \times 2215} = 0.00188 m = 1.88 mm \leq \frac{L}{250} = \frac{2000}{250} = 8 mm \to Satisfies the deflection limit . w (L = 3.24 m) = \frac{5 p L^{4}}{384 E I} = \frac{5 \times 20 \times 3 . 24^{4}}{384 \times 2215} = 0.01296 m = 12.96 mm \leq \frac{L}{250} = \frac{3240}{250} = 12.96 mm \to Reaches the deflection limit .

Step 2. Determine the approximate natural frequency. Check frequency limit.

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Eq.(8-57)

f_{n} (L = 2 m) = \frac{18}{\sqrt{w}} = \frac{18}{\sqrt{1.88}} = 13.12 Hz > 5 Hz \to safe f_{n} (L = 3.24 m) = \frac{18}{\sqrt{w}} = \frac{18}{\sqrt{12.96}} = 5.00 Hz = 5 Hz \to safe

Frequency limit is satisfied for both spans.

Problem b)

Step 1. Draw the span-frequency diagram for w = L/250.

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$L [m] \to w = \frac{L \times 1000}{250} \to f_{n} = \frac{18}{\sqrt{w}} = \frac{18}{\sqrt{\frac{L \times 1000}{250}}}$

Step 2. Determine span limit from where vibration becomes critical (assuming that w = L/250).

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$f_{n} \geq \frac{18}{\sqrt{\frac{L \times 1000}{250}}} = 5 Hz \to L \geq \frac{18^{2} \times 250}{5^{2} \times 1000} = 3.24 m$

Vibration becomes relevant above 3.24 m span.

Results

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Problem a)

The maximum deflection of a simply supported beam is:

See Table 3.5.

w (L = 2 m) = \frac{5 p L^{4}}{384 E I} = \frac{5 \times 20 \times 2 . 0^{4}}{384 \times 2215} = 0.00188 m = 1.88 mm \leq \frac{L}{250} = \frac{2000}{250} = 8 mm \to Satisfies the deflection limit . w (L = 3.24 m) = \frac{5 p L^{4}}{384 E I} = \frac{5 \times 20 \times 3 . 24^{4}}{384 \times 2215} = 0.01296 m = 12.96 mm \leq \frac{L}{250} = \frac{3240}{250} = 12.96 mm \to Reaches the deflection limit .

The natural frequency can be approximated as:

Eq.(8-57)

f_{n} (L = 2 m) = \frac{18}{\sqrt{w}} = \frac{18}{\sqrt{1.88}} = 13.12 Hz > 5 Hz \to safe f_{n} (L = 3.24 m) = \frac{18}{\sqrt{w}} = \frac{18}{\sqrt{12.96}} = 5.00 Hz = 5 Hz \to safe

Frequency limit is satisfied for both spans.

Problem b)

The span-frequency diagram for w = L/250 is given in the Figure below.

$L [m] \to w = \frac{L \times 1000}{250} \to f_{n} = \frac{18}{\sqrt{w}} = \frac{18}{\sqrt{\frac{L \times 1000}{250}}}$

Now span range is determined where vibration becomes critical, thus the natural frequency is smaller than the 5 Hz limit. (We assume that the deflection reaches its limit: w = L/250.)

$f_{n} \geq \frac{18}{\sqrt{\frac{L \times 1000}{250}}} = 5 Hz \to L \geq \frac{18^{2} \times 250}{5^{2} \times 1000} = 3.24 m$

Vibration becomes relevant above 3.24 m span.