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Cohesion of Soil given Loading Intensity by Terzaghi's Analysis Formula

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Cohesion is the ability of like particles within soil to hold onto each other. It is the shear strength or force that binds together like particles in the structure of a soil. Check FAQs

C = \frac{q - ((\frac{2 \cdot P p}{B}) - (\frac{γ \cdot B \cdot \tan (\frac{φ \cdot π}{180})}{4}))}{\tan (\frac{φ \cdot π}{180})}

C - Cohesion?q - Load Intensity?P_p - Passive Earth Pressure?B - Width of Footing?γ - Unit Weight of Soil?φ - Angle of Shearing Resistance?π - Archimedes' constant?

Cohesion of Soil given Loading Intensity by Terzaghi's Analysis Example

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Here is how the Cohesion of Soil given Loading Intensity by Terzaghi's Analysis equation looks like with Values.

Here is how the Cohesion of Soil given Loading Intensity by Terzaghi's Analysis equation looks like with Units.

Here is how the Cohesion of Soil given Loading Intensity by Terzaghi's Analysis equation looks like.

4.2301 = \frac{26.8 - ((\frac{2 \cdot 26.92}{2}) - (\frac{18 \cdot 2 \cdot \tan (\frac{82.57 \cdot 3.1416}{180})}{4}))}{\tan (\frac{82.57 \cdot 3.1416}{180})}

4.2301kPa = \frac{26.8kPa - ((\frac{2 \cdot 26.92kPa}{2m}) - (\frac{18kN/m³ \cdot 2m \cdot \tan (\frac{82.57° \cdot 3.1416}{180})}{4}))}{\tan (\frac{82.57° \cdot 3.1416}{180})}

4.2301 = \frac{26.8 - ((\frac{2 \cdot 26.92}{2}) - (\frac{18 \cdot 2 \cdot \tan (\frac{82.57 \cdot 3.1416}{180})}{4}))}{\tan (\frac{82.57 \cdot 3.1416}{180})}

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Cohesion of Soil given Loading Intensity by Terzaghi's Analysis Solution

Follow our step by step solution on how to calculate Cohesion of Soil given Loading Intensity by Terzaghi's Analysis?

FIRST Step Consider the formula

C = \frac{q - ((\frac{2 \cdot P p}{B}) - (\frac{γ \cdot B \cdot \tan (\frac{φ \cdot π}{180})}{4}))}{\tan (\frac{φ \cdot π}{180})}

Next Step Substitute values of Variables

∴

C = \frac{26.8kPa - ((\frac{2 \cdot 26.92kPa}{2m}) - (\frac{18kN/m³ \cdot 2m \cdot \tan (\frac{82.57° \cdot π}{180})}{4}))}{\tan (\frac{82.57° \cdot π}{180})}

Next Step Substitute values of Constants

∴

C = \frac{26.8kPa - ((\frac{2 \cdot 26.92kPa}{2m}) - (\frac{18kN/m³ \cdot 2m \cdot \tan (\frac{82.57° \cdot 3.1416}{180})}{4}))}{\tan (\frac{82.57° \cdot 3.1416}{180})}

Next Step Convert Units

∴

C = \frac{26800Pa - ((\frac{2 \cdot 26920Pa}{2m}) - (\frac{18000N/m³ \cdot 2m \cdot \tan (\frac{1.4411rad \cdot 3.1416}{180})}{4}))}{\tan (\frac{1.4411rad \cdot 3.1416}{180})}

Next Step Prepare to Evaluate

∴

C = \frac{26800 - ((\frac{2 \cdot 26920}{2}) - (\frac{18000 \cdot 2 \cdot \tan (\frac{1.4411 \cdot 3.1416}{180})}{4}))}{\tan (\frac{1.4411 \cdot 3.1416}{180})}

Next Step Evaluate

∴

C = 4230.06315111881Pa

Next Step Convert to Output's Unit

∴

C = 4.23006315111881kPa

LAST Step Rounding Answer

∴

C = 4.2301kPa

Cohesion of Soil given Loading Intensity by Terzaghi's Analysis Formula Elements

Variables

Constants

Functions

Cohesion

Cohesion is the ability of like particles within soil to hold onto each other. It is the shear strength or force that binds together like particles in the structure of a soil.

Symbol: C

Measurement: PressureUnit: kPa

Note: Value should be between 0 to 50.

Formulas to find Cohesion

Load Intensity

Load Intensity is defined as load applied per unit area.

Symbol: q

Measurement: PressureUnit: kPa