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Natural Frequencies Calculators







Vibration of a Beam with Simply Supported and Fixed End

In this calculation, a beam of length L with a moment of inertia of the cross-section Ix and own mass m is considered. The one end of the beam is fixed, and the other is simply supported. For the calculation, the elastic modulus E of the beam should be specified. As a result of calculations, the natural vibration frequency of the beam f is determined for the first vibration mode.

Natural frequency of beam with simply supported and fixed ends
Calculation of Natural frequency of beam with simply supported and fixed ends

INITIAL DATA

L - Beam length;


Ix - moment of inertia of the beam in the direction of vibration;


E - Young's modulus;


m - Beam mass.

RESULTS DATA

f - First mode frequency.

Length (L)

Moment of inertia (Ix)

Young's modulus (E)

Beam mass (m)

Natural frequency (f)

BASIC FORMULAS

Natural frequency of the first mode:

f = (2.45 / L2)*(EIxL / m)0.5

INITIAL DATA

L - Beam length;


Ix - moment of inertia of the beam in the direction of vibration;


E - Young's modulus;


m - Beam mass.

RESULTS DATA

f - First mode frequency.

MATERIALS PROPERTIES

Material

Young’s modulus

Pa (psi)

Poisson’s ratio

Steel

1.86÷2.1×1011 (2.7÷3.05×107)

0.25÷0.33

Cast iron

0.78÷1.47×1011 (1.1÷2.1×107)

0.23÷0.27

Copper

1.0÷1.3×1011 (1.45÷1.9×107)

0.34

Tin bronze

0.74÷1.22×1011 (1.1÷1.8×107)

0.32÷0.35

Brass

0.98÷1.08×1011 (1.4÷1.6×107)

0.32÷0.34

Aluminum alloy

0.7×1011 (1.0×107)

0.33

Magnesium alloy

0.4÷0.44×1011 (5.8÷6.4×106)

0.34

Nickel

2.5×1011 (3.6×107)

0.33

Titanium

1.16×1011 (1.7×107)

0.32

Lead

0.15÷0.2×1011 (2.2÷2.9×106)

0.42

Zinc

0.78×1011 (1.1×107)

0.27

Glass

4.9÷5.9×1010 (7.1÷8.5×106)

0.24÷0.27

Concrete

1.48÷2.25×1010 (2.1÷3.3×106)

0.16÷0.18

Wood (along the grain)

8.8÷15.7×1010 (12.8÷22.8×106)

-

Wood (across the grain)

3.9÷9.8×1010 (5.7÷14.2×106)

-

Nylon

1.03×1010 (1.5×106)

-

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