Fakultas TeknikJurusan Teknik SipilUniversitas Brawijaya Malang

Mohr’s Circle for Stresses on an Oblique Section of A Body Subjected to Direct Stresses in Two Mutually Perpendicular

Directions

Another graphical method for the determination of normal tangential and resultant stress is known as Mohr’s Circle

When the two mutually perpendicular principal stresses are unequal and alike

Consider a rectangular body subjected to two mutually perpendicular principal

tensile stresses and an oblique section, on which we are required to find out the

stresses.

Let,

p1 = major tensile stress

p2 = minor tensile stress

Θ = angle which the oblique section makes with the minor tensile stress

• Take some suitable point O and draw a horizontal line OX

• Cut off OA and OB equal to the stress p1 and p2 respectively tosome suitable scale on the same side of O (because the twostresses are alike)

• Bisect BA at C. Now with C as center, and radius equal to CB or CAdraw a circle

Draw the circle of the stresses

• Now through C, draw a line CP making an angle 2θ with CX meeting the circle atP

• Through P, draw PQ perpendicular to OA. Now join OP

• Now OQ and PQ will give the required normal and tangential stresses to thescale. OP will give the required resultant stress to the scale. The angle POA iscalled the angle of obliquity

Proof:

From the geometry of the figure, we find that:

2

21 ppCPCABC

22

2

2

21212

212

ppppp

pppBCOBOC

Normal Stress

2cos22

2cos2

2121

21

ppppp

CPpp

CQOCOQp

n

n

Tangential Stress 2sin2

2sin 21 ppCPPQpt

NOTE:• Since A and B are the ends of the horizontal diameter, therefore maximum

normal stress will be equal to p1 and minimum principal stress will be p2

• On the plane having maximum or minimum principal stresses, there will be notangential stress

• Shear stress on mutually perpendicular planes are numerically equal• The maximum shear stress will be equal to radius of the Mohr’s Circle and will

act on planes inclined at 45° to the principal planes. Mathematically:

• The angle of obliquity will be maximum, when OP is tangential to the Mohr’sCircle

2max 21 pp

pt

When the two mutually perpendicular principal stresses are

unequal and unlike

Consider a rectangular body subjected to two mutually perpendicular unlike

principal tensile stresses and an oblique section, on which we are required to find

out the stresses.

Let,

p1 = major principal tensile stress

p2 = minor principal compressive stressΘ = angle which the oblique section makes with the minor principal stress

• Take some suitable point O and draw a horizontal line X’OX

• Cut off OA and OB equal to the stresses p1 and p2 to some suitablescale on the opposite sides of O (because the two stresses areunlike)

• Bisect BA at C. Now with C as center, and radius equal to CB or CAdraw a circle

Draw the circle of the stresses

• Now through C, draw a line CP making an angle 2θ with CXmeeting the circle at P

• Through P, draw PQ perpendicular to OA. Now join OP

• Now OQ and PQ will give the required normal and tangentialstresses to the scale. OP will give the required resultant stress tothe scale.

Proof:

From the geometry of the figure, we find that:

2

21 ppCPCABC

22

2

2

21221

221

ppppp

ppp

OBBCOC

Normal Stress

2cos22

2cos2

2121

21

ppppp

CPpp

CQOCOQp

n

n

Tangential Stress

2sin2

2sin 21 ppCPPQpt

A point in a strained material the normal tensile stresses are 60 N/mm2 and30 N/mm2. Determine by Mohr’s Circle, the resultant intensity of stress on aplane inclined at 40° to the axis of the minor stress. Also check the answeranalytically

Example :

Solution :

Given:Major tensile stress p1 = 60 N/mm2Minor tensile stress p2 = 30 N/mm2Angle which the plane makes with the axis of minor stress = 40°

• Take some suitable point O and draw a horizontal line OX• Cut off OA equal to 60 and OB equal to 30 to some suitable scale on the

same sides of O• Bisect BA at C. Now with C as center, and radius equal to CB or CA draw a

circle• Now through C, draw a line CP making an angle 2x40 = 80° with CX, meeting

the circle at P• Joint OP. By measurement, we find that the resultant stress,

pR = OP = 49,8 kg/cm2

Analytical Check:

Let,

pn = normal stress on the plane

pt = tangential stress on the plane

pR = resultant stress on the plane

Using the relation,

2

2121

6,4780cos2

3060

2

3060

2cos22

mmNp

ppppp

n

n

2

21

8,1480sin2

3060

2sin2

mmNp

ppp

t

t

22222

8,498,146,47cm

kgppp tnR

A point in a strained material is subjected to a tensile stress of 800kg/cm2 and compressive stress of 500 kg/cm2. Draw the Mohr’s stresscircle and find out the resultant stress at a plane making 64° with thetensile stress. Also find out the normal and tangential stress on theplane.

Example :

Solution :

Given:Major stress p1 = 800 kg/cm2Minor stress p2 = -500 kg/cm2Angle which the plane makes with the tensile stress = 64°

• Take some suitable point O and draw a horizontal line X’OX• Cut off OA equal to 800 and OB equal to 500 to some suitable scale on the

opposite sides of O• Bisect BA at C. Now with C as center, and radius equal to CB or CA draw a circle• Through C, draw a line CP making an angle 2x64 = 128° with CX, meeting the circle

at P• Through P. draw PQ perpendicular to OA. Join OP,• By measurement, we find that:

Normal stress pn = OQ = -250 kg/cm2Tangential stress pt = PQ = 510 kg/cm2Resultant stress pR = OP = 570 kg/cm2

Mohr’s Circle for Stresses on an Oblique Section of A Body Subjected to a Direct Stress in One Plane Accompanied by A Simple

Shear Stress

Consider a rectangular body ABCD subjected to a tensile stress in one plane

accompanied by shear stress, and an oblique section, on which we are required to

find out the stresses.

Let,

p = tensile stress on the faces AD and BC

q = shear stress across the faces DA and BC

Θ = angle which the oblique section EF makes with the normal cross section EG

• draw a horizontal line X’X and cut off AB equal to the stress tosome suitable scale, and bisect it at C

• Now erect a perpendicular at B and cut off BE equal to the shearstress q to the scale

• Now with C as center and radius equal to CE, draw a circlemeeting the line X’X at H and G

Draw the circle of the stresses

• On the base CE, draw a line CF at an angle 2θ meeting the circle at F

• From F, draw FD perpendicular to AD. Join AF

• Now AD and DF will give the required normal and tangential stress to the scale.AF will give the required resultant stress to the scale

• Moreover, AG and AH will give the maximum and minimum values of normalstresses to the scale.

Proof:

From the geometry of the figure, we find that normal stress:

CECFCFCFp

p

CFp

p

CFp

CDACADp

n

n

n

sin2sincos2cos2

sin2sincos2cos2

2cos2

2sin)2cos1(2

2sin2cos22

2sin2cos2

cossin2sincos2cos2

qp

p

qpp

p

BECBp

p

CECBCECEp

p

n

n

n

n

And tangential stress:

2cos2sin2

2cos2sin

cossin2coscos2sin

sin2coscos2sin

sin2coscos2sin

2sin

qp

p

BECBp

CECBCECEp

CECFCFCFp

CFp

CFDFp

t

t

t

t

t

t

The maximum value of normal stress

2

2

1

22

1

1

22

2

qpp

p

BECBp

p

CEACCGACAGp

n

n

n

The minimum value of normal stress

ngan tekansikan tegamengindika hanya minus tanda

22

22

22

2

2

2

2

2

2

2

2

2

2

22

2

2

qpp

p

qpp

p

pq

pp

pBECBp

CACECACHAHp

n

n

n

n

n

A point in a strained material is subjected to a compressive stress of 800kg/cm2 and shear stress of 560 kg/cm2. Determine graphically orotherwise the maximum and minimum intensity of direct stresses.

Example :

Solution :

Given: Compressive stress p = 800 kg/cm2 Shear stress q = 560 kg/cm2

• Draw a horizontal line X-X and cut off AB equal to 800 to some suitable scale, andbisect it at C

• Now erect a perpendicular at B, and cut off BE equal to 560 to the scale• Now with C as centre, and radius equal to CE, draw a circle meeting the line X-X’ at

H and G• By measurement, we find that:

Maximum direct stress AG = 1008,2 kg/cm2 (compressive)Minimum direct stress AH = 288,2 kg/cm2 (tensile)

Mohr’s Circle for Stresses on an Oblique Section of A Body Subjected to a Direct Stress in Two Mutually Perpendicular

Directions Accompanied by A Simple Shear Stress

Consider a rectangular body ABCD subjected to a tensile stress and shear stress

and an oblique section, on which we are required to find out the stresses.

Let,

p1 = tensile stress on the faces AD and BC

p2 = tensile stress on the faces AB and CD

q = shear stress across the faces DA and BC

Θ = angle which the oblique section EF makes with the normal cross section EG

• Take some point O and draw a horizontal line OX

• Cut off OA and OB equal to the stresses p1 and p2 respectively to some suitablescale on the same side of O ( because the two stresses are alike). Bisect BA at C

• Now erect perpendiculars A and B and cut off AE and BF equal to the shearstress q to the scale.

Draw the circle of the stresses

• Now with C as center and radius equal to CE, draw a circle

meeting the line OX at Q and P

• On the base CE, draw a line CJ at an angle 2θ meeting the

circle at J• From J, draw JK perpendicular to AX. Join OJ

• Now OK and KJ will give the required normal and tangential

stresses to the scale. OJ will give the required resultant

stress to the scale.

• Moreover, OQ and OP will give the maximum and minimum

values of normal stresses to the scale and CG will give the

maximum value of shear stress

Proof:

From the geometry of the figure, we find that normal stress:

2sin2cos22

2sin2cos2

sin2sincos2cos2

sin2sincos2cos2

sin2sincos2cos2

2cos

2121

21

21

21

21

qpppp

p

AECApp

p

CECEpp

p

CJCECJCJpp

p

CJpp

p

CJOCCKOCOKp

n

n

n

n

n

n

And tangential stress:

2cos2sin2

2cos2sin

sin2coscos2sin

sin2coscos2sin

sin2coscos2sin

2sin

21 qpp

p

AECAp

CECEp

CJCECJCJp

CJp

CJKJp

t

t

t

t

t

t

The maximum value of normal stress:

2

2

21211

1

22q

ppppp

CEOCCPOCOPp

n

n

The minimum value of normal stress:

2

2

21212

2

22q

ppppp

CEOCCQOCOQp

n

n

2max 21 nn

t

ppCGp

The maximum shear stress:

A point in a strained material is subjected to a tensile stress of 60N/mm2 and a compressive stress of 40 N/mm2, acting on two mutuallyperpendicular planes and a shear stress of 10 N/mm2 on these planes.Determine principal as well as maximum shear stresses. Also find outthe value of maximum shear stress..

Example :

Solution :

Given:Major stress Tensile stress p1= 60 N/mm2Minor stress Compressive stress p2 = -40 N/mm2 Shear stress q = 10 N/mm2

• Take some point O and draw a horizontal line X’OX

• Cut off OA equal to 60 as OB equal to 40 to some suitable scale on theopposite sides of O. Bisect BA at C

• Now erect a perpendicular at A and cut off AE equal to 10 to the scale

• Now with C as center and radius equal to the CE draw a circle meetingthe line XX’ at Q and P. Also erect a perpendicular at C meeting the circleat G

• By measurement, we find:Major principal stress pn1 = OP = 61 N/mm2Minor principal stress pn2 = OQ = -41 N/mm2Maximum shear stress = pt = CG = 51 N/mm2

A little knowledge thatacts is worth infinitely more than much knowledge that is idle.