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Cohesion of Soil for Steady Seepage along Slope Formula

GoCohesion of Soil given Saturated Unit Weight 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 = h c \cdot ((γ saturated \cdot \tan (\frac{i \cdot π}{180}) \cdot {(\cos (\frac{i \cdot π}{180}))}^{2}) - (y S \cdot \tan (\frac{φ \cdot π}{180}) \cdot {(\cos (\frac{i \cdot π}{180}))}^{2}))

C - Cohesion in Soil as Kilopascal?h_c - Critical Depth?γ_saturated - Saturated Unit Weight of Soil?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 for Steady Seepage along Slope Example

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Here is how the Cohesion of Soil for Steady Seepage along Slope equation looks like with Values.

Here is how the Cohesion of Soil for Steady Seepage along Slope equation looks like with Units.

Here is how the Cohesion of Soil for Steady Seepage along Slope equation looks like.

0.1633 = 1.01 \cdot ((11.89 \cdot \tan (\frac{64 \cdot 3.1416}{180}) \cdot {(\cos (\frac{64 \cdot 3.1416}{180}))}^{2}) - (5 \cdot \tan (\frac{46 \cdot 3.1416}{180}) \cdot {(\cos (\frac{64 \cdot 3.1416}{180}))}^{2}))

0.1633kPa = 1.01m \cdot ((11.89kN/m³ \cdot \tan (\frac{64° \cdot 3.1416}{180}) \cdot {(\cos (\frac{64° \cdot 3.1416}{180}))}^{2}) - (5kN/m³ \cdot \tan (\frac{46° \cdot 3.1416}{180}) \cdot {(\cos (\frac{64° \cdot 3.1416}{180}))}^{2}))

0.1633 = 1.01 \cdot ((11.89 \cdot \tan (\frac{64 \cdot 3.1416}{180}) \cdot {(\cos (\frac{64 \cdot 3.1416}{180}))}^{2}) - (5 \cdot \tan (\frac{46 \cdot 3.1416}{180}) \cdot {(\cos (\frac{64 \cdot 3.1416}{180}))}^{2}))

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Cohesion of Soil for Steady Seepage along Slope Solution

Follow our step by step solution on how to calculate Cohesion of Soil for Steady Seepage along Slope?

FIRST Step Consider the formula

C = h c \cdot ((γ saturated \cdot \tan (\frac{i \cdot π}{180}) \cdot {(\cos (\frac{i \cdot π}{180}))}^{2}) - (y S \cdot \tan (\frac{φ \cdot π}{180}) \cdot {(\cos (\frac{i \cdot π}{180}))}^{2}))

Next Step Substitute values of Variables

∴

C = 1.01m \cdot ((11.89kN/m³ \cdot \tan (\frac{64° \cdot π}{180}) \cdot {(\cos (\frac{64° \cdot π}{180}))}^{2}) - (5kN/m³ \cdot \tan (\frac{46° \cdot π}{180}) \cdot {(\cos (\frac{64° \cdot π}{180}))}^{2}))

Next Step Substitute values of Constants

∴

C = 1.01m \cdot ((11.89kN/m³ \cdot \tan (\frac{64° \cdot 3.1416}{180}) \cdot {(\cos (\frac{64° \cdot 3.1416}{180}))}^{2}) - (5kN/m³ \cdot \tan (\frac{46° \cdot 3.1416}{180}) \cdot {(\cos (\frac{64° \cdot 3.1416}{180}))}^{2}))

Next Step Convert Units

∴

C = 1.01m \cdot ((11890N/m³ \cdot \tan (\frac{1.117rad \cdot 3.1416}{180}) \cdot {(\cos (\frac{1.117rad \cdot 3.1416}{180}))}^{2}) - (5000N/m³ \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 = 1.01 \cdot ((11890 \cdot \tan (\frac{1.117 \cdot 3.1416}{180}) \cdot {(\cos (\frac{1.117 \cdot 3.1416}{180}))}^{2}) - (5000 \cdot \tan (\frac{0.8029 \cdot 3.1416}{180}) \cdot {(\cos (\frac{1.117 \cdot 3.1416}{180}))}^{2}))

Next Step Evaluate

∴

C = 163.320006077776Pa

Next Step Convert to Output's Unit

∴

C = 0.163320006077776kPa

LAST Step Rounding Answer

∴

C = 0.1633kPa

Cohesion of Soil for Steady Seepage along Slope 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.

Formulas to find Cohesion in Soil as Kilopascal

Critical Depth

The Critical Depth is defined as the depth of flow where energy is at a minimum for a particular discharge.

Symbol: h_c

Measurement: LengthUnit: m