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MODULE 1. BASIC CONCEPTS

MODULE 2. SYSTEM OF FORCES

MODULE 3.

MODULE 4. FRICTION AND FRICTIONAL FORCES

MODULE 5.

MODULE 6.

MODULE 7.

MODULE 8.

## LESSON 29.

**29.1 MOHR’S CIRCLE**

A German scientist Otto Mohr devised a graphical method for finding the normal and shear stresses on any interface of an element when it is subjected to two perpendicular stresses. This method is explained as follows:

**29.1.1 Mohr’s circle construction for ‘like stresses’**

Steps of construction:

Using some suitable scales, measure *OL* and *OM* equal to *σ _{x}* and

*σ*respectively on the axis

_{y}*OX*.

Bisect

LMatN.With

Nas centre andNLorNMradius, draw a circle.At the centre

Ndraw a lineNPat an angle 2Ө, in the same direction as the normal to the plane makes with the direction ofσ. In Fig.1 which represents the stress system, the normal to the plane makes an angle_{x}Өwith the direction ofσin the anticlockwise direction. The line_{x}NPtherefore, is drawn in the anticlockwise direction.From

P, draw a perpendicularPQon the axisOX.PQwill representτandOQ σ._{n}

Now, from stress diagram

*NP* = *NL* = \[{{{\sigma _{x - {\sigma _y}}}} \over 2}\]

*PQ* = *NP* sin2Ө = \[{{{\sigma _{x - {\sigma _y}}}} \over 2}\] sin2*Ө* = *τ* ----------------------------------------(8.1)

Similarly, *OQ* = *ON* + *NQ* = \[{{{\sigma _{x + {\sigma _y}}}} \over 2}\] + \[{{{\sigma _{x - {\sigma _y}}}} \over 2}\] cos2*Ө* = *σ _{n}*--------------(8.2)

Also, from stress circle, *τ* is maximum when

2*Ө* = 90°, or *Ө* = 45°

And, *τ** _{max}* = \[{{{\sigma _{x - {\sigma _y}}}} \over 2}\] ----------------------------------------------------(8.3)

Sign conventions used:

(i) In order to mark *τ* in stress system, we will take the clockwise shear as positive and anticlockwise shear as negative.

(ii) Positive value of *τ* will be above the axis and negative values below the axis.

(iii) If *Ө* is in the anticlockwise direction, the radius vector will be above the axis and *Ө* will be reckoned positive. If *Ө* is in the clockwise direction, it will be negative and the radius vector will be below the axis.

(iv) Tensile stress will be reckoned positive and will be plotted to the right of the origin *O*. Compressive stress will be reckoned negative and will be plotted to the left of the origin *O*.

**29.1.2 Mohr’s circle construction for ‘Unlike stresses’**

In case *σ _{x}* and

*σ*are not like, the same procedure will be followed except that

_{y}*σ*

_{x}_{ }and

*σ*

_{y}_{ }will be measured to the opposite sides of the origin. The construction is given in Fig.2.It may be noted that the direction of

*σ*

_{n}_{ }will depend upon its position with respect to the point

*O*. If it is to the right of

*O*, the direction of

*σ*will be the same as that of

_{n}*σ*.

_{x}

**29.1.3 Mohr’s circle construction for two perpendicular direct stresses with state of simple shear**

Refer to Fig.3. Following steps of construction are followed if the material is subjected to direct stresses *σ _{x}* and

*σ*

_{y}_{ }along with a state of simple shear.

(a)

(b)

Fig.29.3

Using some suitable scale, measure

OL=σand_{x}OM=σalong the axis_{y}OX.At

LdrawLTperpendicular toOXand equal toτ._{xy}LThas been drawn downward (as per sign conventions adopted) becauseτis acting up with respect to the plane across which_{xy}σ_{x}_{ }is acting, tending to rotate it in the anticlockwise direction and is negative.Similarly, make

MSperpendicular toOXand equal toτ, but above_{xy}OX.Join

STto cut the axis inN.With

Nas centre andNSorNTas radius, draw a circle.At

NmakeNPat angle 2ӨwithNTin the anticlockwise direction.Draw

PQperpendicular to the axis.PQwill giveτwhileOQwill giveσand_{n}OPwill giveσ._{r}

**Proof:** Let the radius of the stress circle be *R*.

Then, *R* = \[\sqrt {N{L^2} + L{T^2}}\] = \[\sqrt {{{\left[ {{{{\sigma _x} - {\sigma _y}} \over 2}} \right]}^2} + {\tau _{xy}}^2}\]

Also, *R* cos *β* = *NL *= \[{{{\sigma _{x - {\sigma _y}}}} \over 2}\]

*R* sin *β* = *LT* = *τ*_{xy}

Now, * * *OQ* = *ON* + *NQ* = *ON* + *R* cos (2*Ө* – *β*)

= *ON* + *R *cos 2*Ө* cos *β* + *R* sin 2*Ө* sin *β*

= \[{{{\sigma _{x + {\sigma _y}}}} \over 2}\] + \[{{{\sigma _{x - {\sigma _y}}}} \over 2}\] cos2*Ө* + *τ*_{xy}_{ }sin 2*Ө* (as per eqn-------)

= *σ _{n}*

Similarly,

*PQ* = *R* sin (2*Ө* – *β*) = *R* sin2*Ө* cos *β* – *R* cos2*Ө* sin *β*

= \[{{{\sigma _{x - {\sigma _y}}}} \over 2}\] sin2*Ө* – *τ*_{xy}_{ }cos 2*Ө* (as per eqn-------)

** ****29.1.4 Mohr’s circle construction for principal stresses**

Refer to Fig.4.The following are the steps of construction:

(a)

(b)

Fig. 29.4

Mark

OLandOMproportional toσ_{x}_{ }andσ._{y}At

LandM, erect perpendicularsLT=MSproportional to in appropriate directions.Join

ST, intersecting the axis inN.

Since *τ* = o, *NV* represents the major principal plane, *P* coinciding with *B*. Similarly *NP*ꞌ represents minor principal plane, *P*ꞌ coinciding with *A*.

*OV* = *ON* + *NV* = \[{{{\sigma _{x + {\sigma _y}}}} \over 2}\] + *R*, where *R* is the radius of the circle.

= \[{{{\sigma _{x - {\sigma _y}}}} \over 2}\] + \[\sqrt {{{\left[ {{{{\sigma _x} - {\sigma _y}} \over 2}} \right]}^2} + {\tau _{xy}}^2}\] = *σ*_{1}

Similarly, *OU* = *ON* – *NU*

= \[{{{\sigma _{x + {\sigma _y}}}} \over 2}\] - \[\sqrt {{{\left[ {{{{\sigma _x} - {\sigma _y}} \over 2}} \right]}^2} + {\tau _{xy}}^2}\] = *σ*_{2}

tan *β* =\[\frac{{LT}}{{LN}}\] = \[\frac{{\tau xy}}{{\frac{{{\sigma _x} - {\sigma _y}}}{2}}}\]

= \[{{2\tau xy} \over {{\sigma _x} - {\sigma _y}}}\] = tan 2*Ө* (where *β* = 2*Ө*)