FormulaDen.com

Physics Chemistry Math Chemical Engineering Civil Electrical Electronics Electronics and Instrumentation Materials Science Mechanical Production Engineering Financial Health

Natural Frequency due to Uniformly Distributed Load Formula

GoNatural Frequency given Static Deflection Formula

Fx Copy

LaTeX Copy

Frequency is the number of oscillations or cycles per second of a system undergoing free transverse vibrations, characterizing its natural vibrational behavior. Check FAQs

f = \frac{π}{2} \cdot \sqrt{\frac{E \cdot I shaft \cdot g}{w \cdot {L shaft}^{4}}}

f - Frequency?E - Young's Modulus?I_shaft - Moment of inertia of shaft?g - Acceleration due to Gravity?w - Load per unit length?L_shaft - Length of Shaft?π - Archimedes' constant?

Natural Frequency due to Uniformly Distributed Load Example

With values

With units

Only example

Here is how the Natural Frequency due to Uniformly Distributed Load equation looks like with Values.

Here is how the Natural Frequency due to Uniformly Distributed Load equation looks like with Units.

Here is how the Natural Frequency due to Uniformly Distributed Load equation looks like.

0.9352 = \frac{3.1416}{2} \cdot \sqrt{\frac{15 \cdot 1.0855 \cdot 9.8}{3 \cdot {3.5}^{4}}}

0.9352Hz = \frac{3.1416}{2} \cdot \sqrt{\frac{15N/m \cdot 1.0855kg·m² \cdot 9.8m/s²}{3 \cdot {3.5m}^{4}}}

0.9352 = \frac{3.1416}{2} \cdot \sqrt{\frac{15 \cdot 1.0855 \cdot 9.8}{3 \cdot {3.5}^{4}}}

Tip: Use pencil ( Pencil

) icon above to edit Input Values and Units.

Calculate

Recalculate

Solution

Copy

Reset

You are here -

Home » Category

Physics » Category

Mechanical » Category

Theory of Machine » fx Natural Frequency due to Uniformly Distributed Load

Natural Frequency due to Uniformly Distributed Load Solution

Follow our step by step solution on how to calculate Natural Frequency due to Uniformly Distributed Load?

FIRST Step Consider the formula

f = \frac{π}{2} \cdot \sqrt{\frac{E \cdot I shaft \cdot g}{w \cdot {L shaft}^{4}}}

Next Step Substitute values of Variables

∴

f = \frac{π}{2} \cdot \sqrt{\frac{15N/m \cdot 1.0855kg·m² \cdot 9.8m/s²}{3 \cdot {3.5m}^{4}}}

Next Step Substitute values of Constants

∴

f = \frac{3.1416}{2} \cdot \sqrt{\frac{15N/m \cdot 1.0855kg·m² \cdot 9.8m/s²}{3 \cdot {3.5m}^{4}}}

Next Step Prepare to Evaluate

∴

f = \frac{3.1416}{2} \cdot \sqrt{\frac{15 \cdot 1.0855 \cdot 9.8}{3 \cdot {3.5}^{4}}}

Next Step Evaluate

∴

f = 0.935192775442116Hz

LAST Step Rounding Answer

∴

f = 0.9352Hz

Natural Frequency due to Uniformly Distributed Load Formula Elements

Variables

Constants

Functions

Frequency

Frequency is the number of oscillations or cycles per second of a system undergoing free transverse vibrations, characterizing its natural vibrational behavior.

Symbol: f

Measurement: FrequencyUnit: Hz

Note: Value can be positive or negative.

Formulas to find Frequency

Young's Modulus

Young's Modulus is a measure of the stiffness of a solid material and is used to calculate the natural frequency of free transverse vibrations.

Symbol: E

Measurement: Stiffness ConstantUnit: N/m