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

Fakultas Teknik

Jurusan Teknik Sipil

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

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

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

• 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 2θ with CX meeting the circle at P

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

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

Proof:

From the geometry of the figure, we find that:

2

2

1

p

p

CP

CA

BC

2

2

2

2

2 1

2 1

2

2 1

2

p

p

p

p

p

p

p

p

BC

Normal Stress

2

cos

2

2

2

cos

2

2 1 2 1 2 1

p

p

p

p

p

CP

p

p

CQ

OC

OQ

p

n n

Tangential Stress

sin

2

2

2

sin

p

1

p

2

CP

PQ

p

_t

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 no tangential 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’s Circle

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 stress

Θ = 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 suitable scale on the opposite sides of O (because the two stresses are unlike)

• 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 2θ with CX meeting the circle at P

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

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

Proof:

From the geometry of the figure, we find that:

2

2

1

p

p

CP

CA

BC

2

2

2

2

2 1

2 2

1

2 2

1

p

p

p

p

p

p

p

p

OB

Normal Stress

2

cos

2

2

2

cos

2

2 1

2 1

2 1

p

p

p

p

p

CP

p

p

CQ

OC

OQ

p

n n

Tangential Stress

2

sin

2

2

sin

p

1

p

2

CP

PQ

A point in a strained material the normal tensile stresses are 60 N/mm2 and 30 N/mm2. Determine by Mohr’s Circle, the resultant intensity of stress on a plane inclined at 40° to the axis of the minor stress. Also check the answer analytically

Example :

Solution :

Given:

 Major tensile stress p₁ = 60 N/mm2

 Minor tensile stress p₂ = 30 N/mm2

• 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

Analytical Check: Let,

p_n = normal stress on the plane p_t = tangential stress on the plane p_R = resultant stress on the plane

Using the relation,

2 2 1 2 1

6

,

47

80

cos

2

30

60

2

30

60

2

cos

2

2

mm

N

p

p

p

p

p

p

n n 2 2 1

8

,

14

80

sin

2

30

60

2

sin

2

mm

N

p

p

p

p

t t 2 2 2 2 2

8

,

49

8

,

14

6

,

47

cm

kg

p

p

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

Example :

Solution :

Given:

 Major stress p₁ = 800 kg/cm2

 Minor stress p₂ = -500 kg/cm2

• 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/cm2

Tangential stress pt = PQ = 510 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

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

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

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

meeting the line X’X at H and G

• 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 normal stresses to the scale.

Proof:

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

2 sin ) 2 cos 1 ( 2 2 sin 2 cos 2 2 2 sin 2 cos 2 cos sin 2 sin cos 2 cos 2 q p p q p p p BE CB p p CE CB CE CE p p n n n n

And tangential stress:

The maximum value of normal stress 2 2 1 2 2 1 1

2

2

2

q

p

p

p

BE

CB

p

p

CE

AC

CG

AC

AG

p

n n n

The minimum value of normal stress

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

Example :

Solution :

Given:

 Compressive stress p = 800 kg/cm2

• Draw a horizontal line X-X and cut off AB equal to 800 to some suitable scale, and bisect 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:

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

• 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 suitable scale 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 shear stress q to the scale.

• 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.

Proof:

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

And tangential stress:

2

cos

2

sin

2

2

cos

2

sin

sin

2

cos

cos

2

sin

sin

2

cos

cos

2

sin

sin

2

cos

cos

2

sin

2

sin

2

1

p

q

p

p

AE

CA

p

CE

CE

p

CJ

CE

CJ

CJ

p

CJ

p

CJ

KJ

p

t t t t t t

The maximum value of normal stress:

The minimum value of normal stress:

2 2

2 1

2 1

2 2

2

2

q

p

p

p

p

p

CE

OC

CQ

OC

OQ

p

n n

2

max

p

_t

CG

p

n1

p

n2

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

Example :

Solution :

Given:

Major stress

 Tensile stress p₁= 60 N/mm2 Minor stress

 Compressive stress p₂ = -40 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 the opposite 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 meeting the line XX’ at Q and P. Also erect a perpendicular at C meeting the circle at G

• By measurement, we find:

A little

knowledge

that

acts

is

worth

infinitely

more

than

much