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Cohesion of Soil given Saturated Unit Weight Formula

GoCohesion of Soil for Steady Seepage along Slope Formula

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Cohesion in Soil as Kilopascal 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 = (F s \cdot γ saturated \cdot z \cdot \cos (\frac{i \cdot π}{180}) \cdot \sin (\frac{i \cdot π}{180})) - (y S \cdot z \cdot \tan (\frac{φ \cdot π}{180}) \cdot {(\cos (\frac{i \cdot π}{180}))}^{2})

C - Cohesion in Soil as Kilopascal?F_s - Factor of Safety in Soil Mechanics?γ_saturated - Saturated Unit Weight of Soil?z - Depth of Prism?i - Angle of Inclination to Horizontal in Soil?y_S - Submerged Unit Weight in KN per Cubic Meter?φ - Angle of Internal Friction?π - Archimedes' constant?

Cohesion of Soil given Saturated Unit Weight Example

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Here is how the Cohesion of Soil given Saturated Unit Weight equation looks like with Values.

Here is how the Cohesion of Soil given Saturated Unit Weight equation looks like with Units.

Here is how the Cohesion of Soil given Saturated Unit Weight equation looks like.

1.7365 = (2.8 \cdot 11.89 \cdot 3 \cdot \cos (\frac{64 \cdot 3.1416}{180}) \cdot \sin (\frac{64 \cdot 3.1416}{180})) - (5 \cdot 3 \cdot \tan (\frac{46 \cdot 3.1416}{180}) \cdot {(\cos (\frac{64 \cdot 3.1416}{180}))}^{2})

1.7365kPa = (2.8 \cdot 11.89kN/m³ \cdot 3m \cdot \cos (\frac{64° \cdot 3.1416}{180}) \cdot \sin (\frac{64° \cdot 3.1416}{180})) - (5kN/m³ \cdot 3m \cdot \tan (\frac{46° \cdot 3.1416}{180}) \cdot {(\cos (\frac{64° \cdot 3.1416}{180}))}^{2})

1.7365 = (2.8 \cdot 11.89 \cdot 3 \cdot \cos (\frac{64 \cdot 3.1416}{180}) \cdot \sin (\frac{64 \cdot 3.1416}{180})) - (5 \cdot 3 \cdot \tan (\frac{46 \cdot 3.1416}{180}) \cdot {(\cos (\frac{64 \cdot 3.1416}{180}))}^{2})

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Cohesion of Soil given Saturated Unit Weight Solution

Follow our step by step solution on how to calculate Cohesion of Soil given Saturated Unit Weight?

FIRST Step Consider the formula

C = (F s \cdot γ saturated \cdot z \cdot \cos (\frac{i \cdot π}{180}) \cdot \sin (\frac{i \cdot π}{180})) - (y S \cdot z \cdot \tan (\frac{φ \cdot π}{180}) \cdot {(\cos (\frac{i \cdot π}{180}))}^{2})

Next Step Substitute values of Variables

∴

C = (2.8 \cdot 11.89kN/m³ \cdot 3m \cdot \cos (\frac{64° \cdot π}{180}) \cdot \sin (\frac{64° \cdot π}{180})) - (5kN/m³ \cdot 3m \cdot \tan (\frac{46° \cdot π}{180}) \cdot {(\cos (\frac{64° \cdot π}{180}))}^{2})

Next Step Substitute values of Constants

∴

C = (2.8 \cdot 11.89kN/m³ \cdot 3m \cdot \cos (\frac{64° \cdot 3.1416}{180}) \cdot \sin (\frac{64° \cdot 3.1416}{180})) - (5kN/m³ \cdot 3m \cdot \tan (\frac{46° \cdot 3.1416}{180}) \cdot {(\cos (\frac{64° \cdot 3.1416}{180}))}^{2})

Next Step Convert Units

∴

C = (2.8 \cdot 11890N/m³ \cdot 3m \cdot \cos (\frac{1.117rad \cdot 3.1416}{180}) \cdot \sin (\frac{1.117rad \cdot 3.1416}{180})) - (5000N/m³ \cdot 3m \cdot \tan (\frac{0.8029rad \cdot 3.1416}{180}) \cdot {(\cos (\frac{1.117rad \cdot 3.1416}{180}))}^{2})

Next Step Prepare to Evaluate

∴

C = (2.8 \cdot 11890 \cdot 3 \cdot \cos (\frac{1.117 \cdot 3.1416}{180}) \cdot \sin (\frac{1.117 \cdot 3.1416}{180})) - (5000 \cdot 3 \cdot \tan (\frac{0.8029 \cdot 3.1416}{180}) \cdot {(\cos (\frac{1.117 \cdot 3.1416}{180}))}^{2})

Next Step Evaluate

∴

C = 1736.5208132043Pa

Next Step Convert to Output's Unit

∴

C = 1.7365208132043kPa

LAST Step Rounding Answer

∴

C = 1.7365kPa

Cohesion of Soil given Saturated Unit Weight Formula Elements

Variables

Constants

Functions

Cohesion in Soil as Kilopascal

Cohesion in Soil as Kilopascal 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.

Factor of Safety in Soil Mechanics

Factor of Safety in Soil Mechanics expresses how much stronger a system is than it needs to be for an intended load.

Symbol: F_s

Measurement: NAUnit: Unitless