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Load Intensity given Max Deflection for Strut Subjected to Uniformly Distributed Load Formula

GoLoad Intensity for Strut Subjected to Compressive Axial and Uniformly Distributed Load Formula

GoLoad Intensity given Max Bending Moment for Strut Subjected to Uniformly Distributed Load Formula

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Load Intensity is the distribution of load over a certain area or length of a structural element. Check FAQs

q f = \frac{C}{(1 \cdot (ε column \cdot \frac{I}{{P axial}^{2}}) \cdot ((\sec ((\frac{l column}{2}) \cdot (\frac{P axial}{ε column \cdot I}))) - 1)) - (1 \cdot \frac{{l column}^{2}}{8 \cdot P axial})}

q_f - Load Intensity?C - Maximum Initial Deflection?ε_column - Modulus of Elasticity of Column?I - Moment of Inertia?P_axial - Axial Thrust?l_column - Column Length?

Load Intensity given Max Deflection for Strut Subjected to Uniformly Distributed Load Example

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Here is how the Load Intensity given Max Deflection for Strut Subjected to Uniformly Distributed Load equation looks like with Values.

Here is how the Load Intensity given Max Deflection for Strut Subjected to Uniformly Distributed Load equation looks like with Units.

Here is how the Load Intensity given Max Deflection for Strut Subjected to Uniformly Distributed Load equation looks like.

-1.4E-5 = \frac{30}{(1 \cdot (10.56 \cdot \frac{5600}{1500^{2}}) \cdot ((\sec ((\frac{5000}{2}) \cdot (\frac{1500}{10.56 \cdot 5600}))) - 1)) - (1 \cdot \frac{5000^{2}}{8 \cdot 1500})}

-1.4E-5MPa = \frac{30mm}{(1 \cdot (10.56MPa \cdot \frac{5600cm⁴}{{1500N}^{2}}) \cdot ((\sec ((\frac{5000mm}{2}) \cdot (\frac{1500N}{10.56MPa \cdot 5600cm⁴}))) - 1)) - (1 \cdot \frac{{5000mm}^{2}}{8 \cdot 1500N})}

-1.4E-5 = \frac{30}{(1 \cdot (10.56 \cdot \frac{5600}{1500^{2}}) \cdot ((\sec ((\frac{5000}{2}) \cdot (\frac{1500}{10.56 \cdot 5600}))) - 1)) - (1 \cdot \frac{5000^{2}}{8 \cdot 1500})}

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Load Intensity given Max Deflection for Strut Subjected to Uniformly Distributed Load Solution

Follow our step by step solution on how to calculate Load Intensity given Max Deflection for Strut Subjected to Uniformly Distributed Load?

FIRST Step Consider the formula

q f = \frac{C}{(1 \cdot (ε column \cdot \frac{I}{{P axial}^{2}}) \cdot ((\sec ((\frac{l column}{2}) \cdot (\frac{P axial}{ε column \cdot I}))) - 1)) - (1 \cdot \frac{{l column}^{2}}{8 \cdot P axial})}

Next Step Substitute values of Variables

∴

q f = \frac{30mm}{(1 \cdot (10.56MPa \cdot \frac{5600cm⁴}{{1500N}^{2}}) \cdot ((\sec ((\frac{5000mm}{2}) \cdot (\frac{1500N}{10.56MPa \cdot 5600cm⁴}))) - 1)) - (1 \cdot \frac{{5000mm}^{2}}{8 \cdot 1500N})}

Next Step Convert Units

∴

q f = \frac{0.03m}{(1 \cdot (1.1E+7Pa \cdot \frac{5.6E-5m⁴}{{1500N}^{2}}) \cdot ((\sec ((\frac{5m}{2}) \cdot (\frac{1500N}{1.1E+7Pa \cdot 5.6E-5m⁴}))) - 1)) - (1 \cdot \frac{{5m}^{2}}{8 \cdot 1500N})}

Next Step Prepare to Evaluate

∴

q f = \frac{0.03}{(1 \cdot (1.1E+7 \cdot \frac{5.6E-5}{1500^{2}}) \cdot ((\sec ((\frac{5}{2}) \cdot (\frac{1500}{1.1E+7 \cdot 5.6E-5}))) - 1)) - (1 \cdot \frac{5^{2}}{8 \cdot 1500})}

Next Step Evaluate

∴

q f = -14.4030742757908Pa

Next Step Convert to Output's Unit

∴

q f = -1.44030742757908E-05MPa

LAST Step Rounding Answer

∴

q f = -1.4E-5MPa

Load Intensity given Max Deflection for Strut Subjected to Uniformly Distributed Load Formula Elements

Variables

Functions

Load Intensity

Load Intensity is the distribution of load over a certain area or length of a structural element.

Symbol: q_f

Measurement: PressureUnit: MPa

Note: Value can be positive or negative.

Formulas to find Load Intensity

Maximum Initial Deflection

Maximum Initial Deflection is the greatest amount of displacement or bending that occurs in a mechanical structure or component when a load is first applied.

Symbol: C

Measurement: LengthUnit: mm