Mohr’s circle & Formability

Dr.R.Narayanasamy,B.E.,M.Tech.,M.Engg.,Ph.D.,(D.Sc.)Professor,

Department of Production Engineering,National Institute of Technology, Tiruchirappalli – 620 015.

Tamil Nadu, India

σ1σ

3

σ2

σ3

σ1

σ

τ1 τ3

τ2

τ

Mohr’s circle representation of a three-dimensional state of stress.

σ2

Mohr’s circle (three – dimensional) for the state of stress

Uniaxial tension

- σ σ

σ1

τ max=σ1/ 2

τ

σ2 = σ3 = 0

(a)

σ13

2

https://www.youtube.com/watch?v=IIWSgApnIpM

Uniaxial tension (video)

Mohr’s circle (three – dimensional) for the state of stress

Uniaxial compression

σ1 = σ2 = 0

σ3

τ max

σ3

2

1

https://www.youtube.com/watch?v=53QjIamBW3w

Uniaxial compression (video)

Uniaxial tension/Compression

• Tensile/Compressive stress is applied in one direction

• A maximum shear stress is achieved in the positive region for tensile stress and vice versa for the compressive stress

• The maximum shear stress is uniform for tension & compression and which is larger than biaxial tension/compression.

Mohr’s circle (three – dimensional) for the state of stress

biaxial tension

σ3 = 0

σ2

σ1

τ max = τ2

τ1 τ3

3σ

1

σ2

σ3= 0

http://youtube.com/watch?v=FlOjVZis4SA

Mohr’s circle (three – dimensional) for the state of stress biaxial tension

• σ2 - Tensile stress right angle to σ1 (Tensile stress)

• Decrease in principal shear stress on two sets of planes out of three

• Maximum shear stress is decreased from what would be for uniaxial tension

biaxial tension simulation (video)

Mohr’s circle (three – dimensional) for the state of stress

biaxial compression

σ1

σ3 = 0

τ1 τ3

σ1

σ2

σ2

3

τ max = τ2

σ3 = 0

https://www.youtube.com/watch?v=h_bL40T5dl4

biaxial compression simulation (video)

Mohr’s circle (three – dimensional) for the state of stress

triaxial tension (unequal)

σ1

σ2 = σ3

τ max = τ2= τ3

σ1

σ3

σ2

σ1 = 2 σ2 = 2σ3

Mohr’s circle (three – dimensional) for the state of stress

triaxial tension (equal)

σ1

σ2 = σ3

τ max = τ2 =τ3

σ1

σ3

σ2

σ1 = σ2 = σ3

σ1 = σ2 = σ3

Mohr’s circle (three – dimensional) for the state of stress triaxial tension (unequal)&(Equal)

• Tensile stress applied in the three directions• Maximum shear stress is reduced• For equal triaxial tension Mohr’s circle will be

a point and no shear stress acts on the body. This will reduce the ductility, because plastic deformation is produced by shear stress

• Equal triaxial state of stress will lead to brittle failure

Mohr’s circle (three – dimensional) for the state of stress

triaxial compression (unequal)

σ2 = σ3

σ1

τ max = τ2 = τ3

σ1

σ2

σ3

triaxial compression simulation (video)

Mohr’s circle (three – dimensional) for the state of stress

combined tension and compression

σ1

τ max = τ2 = τ3

σ3

σ2

σ1

σ1 = -2 σ2 = -2σ3 σ2 = σ3 σ1

τ max = τ2 = τ3

Example: Wire drawing

Mohr’s circle (three – dimensional) for the state of stress combined tension and compression (Wire drawing)

• Compressive stresses are applied to the lateral surface of tensile stress, which gives large maximum shear stress than uniaxial stress.

• Materials subjected to tension + compression will have good formability before fracture.

• It is applied in plastic working of metals. (Ex: More ductility is achieved in wire drawing through a die. Here applied load is tensile in nature whereas the reaction between the die and metal is compressive in nature).

Drawing operation (video)

Drawing operation (video)

Engineering Stress Strain Curve

• Engineering tension test provide basic design information on strength of material.

• In tension test uniaxial tensile force is applied on the specimen and its elongation was observed.

• Engineering stress strain curve was formed using load vs elongation

• The stress used in this curve is average longitudinal stress = load/original c/s area

Engineering stress strain curve

•Tensile strength & yield strength (strength parameters)

•Percentage elongation and reduction of area (ductility parameters)

Engineering stress strain curve

• The strain used for engineering stress strain is average linear strain. (elongation of gauge length/original length)

• Both stress and strain are obtained by dividing the load and elongation by constant factors the (load-elongation) curve will have same shape as engineering stress strain curve.

Engineering stress strain curve

• The shape and magnitude of stress strain curve depends on:1. chemical composition of material2. heat treatment3. prior history of plastic deformation4. strain rate5. temperature6. state of stress imposed during testing.

Engineering stress strain curve

• Stress is proportional to strain (Elastic region)• As load increases the value corresponding to yield

strength the specimen undergoes plastic deformation• The stress to produce continued plastic deformation

increases with increasing plastic strain (metal strain hardens)

• The volume of specimen remains constant during plastic deformation

Engineering stress strain curve• As the specimen elongates it decreases uniformly along

the gauge length in cross sectional area.• Initially, strain hardening compensates for decrease in

area and the engineering stress in proportional to engineering strain (engineering stress is proportional to load)

• At certain point decrease in cross sectional area is > than increase in deformation load arising from strain hardening (this condition occurs in the specimen which is weaker than the rest portion). The specimen begins to neck/thin down locally.

Reference

• Mechanical Metallurgy by George E.Dieter, McGraw – Hill Publication, London,1988.

Thank You