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

Universitas 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, p = major tensile stress ₁ p = minor tensile stress ₂ 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 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 Draw the circle of the stresses

• Now through C, draw a line CP making an angle 2θ with CX meeting the circle at
• Through P, draw PQ perpendicular to OA. Now join OP
• Now OQ and PQ will give the required normal and tangential stresses to the

  2

  2 p p p p p p p

  1

  2

  2

  1

  2

  1

  2

  P

  2

  2

  2

  1 p p CP CA BC

  2

  2

  Proof: From the geometry of the figure, we find that:

  scale. OP will give the required resultant stress to the scale. The angle POA is called the angle of obliquity

  BC p OB OC

  Normal Stress p p _{1 2} p OQ OC CQ CP cos _n

  2

  2 p p p p _{1 2 1 2} p _n cos

  2

  2

  2 p p _{1 2} Tangential Stress p PQ CP sin _t 2 sin

  2

  2 NOTE:

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

  normal stress will be equal to p and minimum principal stress will be p _{1 2}

• 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:

  p p _{1 2} max p _t

  2

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

  Circle

  

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, p = major principal tensile stress ₁ p = minor principal compressive stress ₂ Θ = angle which the oblique section makes with the minor principal stress Draw the circle of the stresses

• 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 2
• Through P, draw PQ perpendicular to OA. Now join OP
• Now OQ and PQ will give the required normal and tangential

  2

  2

  2

  1

  2

  2

  1

  2

  2

  θ with CX

  2

  2

  1 p p CP CA BC

  2

  2

  Proof: From the geometry of the figure, we find that:

  stresses to the scale. OP will give the required resultant stress to the scale.

  meeting the circle at P

  1 p p p p p

p

p p

  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 ^{2 1} p p CP PQ p _t Example :

  A point in a strained material the normal tensile stresses are 60 N/mm2 and

  Solution :

  Given: = 60 N/mm2  Major tensile stress p ₁

  = 30 N/mm2  Minor tensile stress p ₂ Angle 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, _R

  p = OP = 49,8 kg/cm2 Analytical Check: Let, p = normal stress on the plane _n p = tangential stress on the plane _t p = resultant stress on the plane _R Using the relation, p p p p _{1 2 1 2} p _n cos

  2

  

2

  2

  60

  30

  60

  30 N p _n cos

  80 47 ,

  6 ₂ mm

  

2

  2 p p _{1 2}

p sin

_t

  2

  2

  60

  

p sin

_t

  80 14 ,

  8 ₂ mm

  2

  2

  2

  2

  2 kg p p p

  47 ,

  6 14 ,

  8 49 ,

  8

  2 R n t cm Example : Solution :

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

  Given:  Major stress p ₁

  = 800 kg/cm2  Minor stress p ₂

  = -500 kg/cm2 Angle 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: _n

  Normal stress p = OQ = -250 kg/cm2 _t Tangential stress p = PQ = 510 kg/cm2 _R

  Resultant stress p = 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 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

  Draw the circle of the stresses

• Now with C as center and radius equal to CE, draw a circle
• 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: p p AD AC CD CF cos _n

  2

  2 p p CF cos _n 2 cos sin 2 sin

  2 p p CF cos _n 2 cos CF sin 2 sin CF CE

  2

  ) 2 sin 2 cos 1 (

  And tangential stress: 2 cos 2 sin

  t t t t t t

  CE CB CE CE p CE CF CF CF p CF p

  2 sin q p p BE CB p

  2 sin sin 2 cos cos 2 sin

  2 sin sin 2 cos cos

  2 cos 2 sin cos sin 2 cos cos

  p p n ^{n n n}

  2 2 sin 2 cos

  BE CB p p

  q p p q p p p

  2

  2 sin cos 2 cos

  2 cos sin

  2 2 sin 2 cos

  2

CF DF p

  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

  2

  2

  2

  2

  2

  2

  2

  2 ² ₂

²

^{2 2 2 2 2 2 2 2 2} q p p p q p p p p q p p p BE CB p

  CA CE CA CH AH p n ^{n n n n} Example :

  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.

  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, 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:

  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, p = tensile stress on the faces AD and BC ₁ p = tensile stress on the faces AB and CD ₂ q = shear stress across the faces DA and BC

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

  Draw the circle of the stresses

• Take some point O and draw a horizontal line OX

  and p respectively to some suitable

• Cut off OA and OB equal to the stresses p
^{1 2} scale on the same side of O ( because the two stresses are alike). Bisect BA at
• 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.

• 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:

  2 sin 2 cos

  2

  2 2 sin 2 cos

  2 sin 2 sin cos

  2 cos

  2 sin 2 sin cos

  2 cos

  2 sin 2 sin cos

  2 cos

  2 2 cos

  2 ^{1 2 1 2 1 2 1 2 1 2 1} q p p p p p

  AE CA p p p

  CE CE p p p

  CJ CE CJ CJ p p p CJ p p p

  CJ OC CK OC OK p n ^{n n n n n}

  And tangential stress: 2 cos 2 sin

  2

  

2

q p p p p p

  2

  1

  1

  1

  2

  1

  2

  2

  2 2 cos 2 sin sin 2 cos cos

  The maximum value of normal stress:

  CJ KJ p t t t t t t

  CE CE p CJ CE CJ CJ p CJ p

  1 q p p p AE CA p

  2

  2 sin

  2 sin sin 2 cos cos 2 sin

  2 sin sin 2 cos cos

  CE OC CP OC OP p n n

  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 2 1 n n t p p CG p

  The maximum shear stress:

  

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 p ₁

  = 60 N/mm2 Minor stress  Compressive stress p ₂

  = -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 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:

  Major principal stress p _n1 = OP = 61 N/mm2 Minor principal stress p _n2 = OQ = -41 N/mm2

  Maximum shear stress = p _t = CG = 51 N/mm2

  

A little knowledge that

acts is worth infinitely

more than much knowledge that is idle.