## What is Mohr’s circle?

Otto Mohr’s a German physicist gave a graphical solution for the determination of the normal and shear stresses on the given inclined plane as explained under. He conveniently assumed compressive normal stresses as positive and tensile normal stresses are negative. The complete solution may be obtained from the following steps** (Figure 10.5).**

## Use of mohr’s circle / Importance of mohr’s circle

- Mohr’s circle is used to calculate the maximum stress on a particular plane with the help of graphical figures.
- It helps to calculate the maximum angle of obliquity occurring on a plane.
- It helps to find out the nature of soil with help of cohesion value.
- It is used to calculate the angle of internal friction of the soil.
- Mohr’s circle is used to calculate the shear strength of the soil.

## Mohr’s circle for stress diagram

- Along the X-axis plot the normal stresses σ
_{1}OB and σ_{3}= OA. - Divide AB at C (i.e., C is midway between A and B).
- Draw a semicircle having its center at C and radius equal to AC or BC.
- Through A, draw AP Parallel to the inclined plane i.e., making <PAB = 0.
- Now, the X-coordinate of P is the normal stress and the Y-coordinate of P is the shear stress on the given plane.

## Mohr’s circle derivation

**Proof:-**

Join P and C and drop a perpendicular PD to AB. From ∆PCD; we get,

<PCD = 20

Normal stress (σ) = OD = OA+ AC + CD = σ_{3} + 1+32 + 1-33 cos 2θ

i.e., σ = 1+32 + 1-33 cos 2θ

Shear stress (τ) = PD = PC sin 2θ

i.e., τ = 1-33 sin 2θ

The following relationships may be directly obtained from Mohr’s circler as shown in figure 10.5.

**1. The maximum shear stress (τ _{max}) is equal to the radius of the Mohr’s circle, and it occurs on planes inclined at 45° to the principal planes. **

**2. The normal stresses on the planes of maximum shear stress are equal to each other and equal to half the sum of principal stress.**

3. The resultant stress (σ_{r}) on any plane is,

and its angle of obliquity β is equal to,

4. The maximum angle of obliquity β_{max } occurs on a plane inclined at an angle** θ = 45°+(𝛽𝑚𝑎𝑥/2 )** with respect to the major principal.

## Relation between major (σ_{1}) and minor (σ_{3}) principal stresses at failure.

With reference to figure 10.6 the relationship between major and minor principal stresses at failure may be derived from the geometry of the Mohr’s circle as follows:

From ∆PO’M; we have,

**Considering,**

is the angle of failure plane with horizontal.

**Equation (1) **defines the relationship between the principal stresses at failure.

## Mohr’s circle problem

**Q.1)** **A cylinder of soil fails under axial vertical stress of 100 kN/m². When it is laterally unconfined. The failure plane makes an angle of 50° with horizontal. Calculate the value of cohesion and the angle of internal friction of soil.**

Solution:

Given that,

σ_{1} = q_{u} = 160 kN/m^{2}

σ_{3} = 0

θ_{f} = 50°

We know that;

θ_{f }= 45°+∅2

Or, 50° = 45°+∅2

Or, ∅ = 10°

Also we know that,

σ_{1} = σ_{3} N_{∅} + 2c 𝑁∅

where, N_{∅ } = 45° + ∅2

= 50°

= 1.42

Or, 160 = 0+ 2 x c1.42

.’. c = 67.13 kN/m^{2}

Hence, internal fraction, ∅=10° and cohesion = 67.13 kN/m².

**Q.2)** **The following results were obtained in an undrained tri-axial test on the soil. Determine the shear strength parameters c and of the soil.**

Confining pressure (kg/cm²) | 2 | 4 |

Deviator stress at failure (kg/cm²) | 5 | 9 |

**Solution:**

**First case,**

σ_{C }= σ_{3} = 2

σ_{d }= 5

.’. σ_{1 }= σ_{C} + σ_{d} = 2 + 5 = 7kg/cm^{2}

We know that,

**Second case,**

Now,

6 = 2 x N_{∅ } Or, N_{∅ } = 3

**Putting values of N _{∅} in equation (1); we get,**

**.’. c = 0.29 kg/cm ^{2 } **

**Happy Learning – Civil Concept**

**Read Also,**

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Relation between water content and dry density of soil

Atterberg limits of soil | consistency limit of soil – Soil mechanics

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