1. Principal Stresses & Strains

2.  Principal Planes:-- Principal Stresses:-- 2 1 pt p p pn 1 2
The planes which carry only Direct Stresses and no Tangential Stresses are called “ Principal Planes ”
Principal Stresses:--
The intensity of stress on the Principal Planes are known as “ Principal Stresses ” 2 1  pt p p pn 1 2

3. All planes parallel to this planes are principal plane with zero principal stress2 1  pt p p pn 1 2
All planes parallel to this planes are Principal plane with principal stress =p

4. Importance of Finding Principal Planes:--
1. The Principal Planes carry the Maximum Direct Stress (Tensile or Compressive).
2. The Principal Planes helps to locate the Planes of Maximum Shear (Inclined at 45 with Principal Planes) 2 1 45 p p 1 2

5. Case:- 1 Element Subjected to Direct Stress & Shear Stresses :--
Consider an element subjected to the direct stresses p & p’ (both tensile) alongwith a shear stress of intensity “ q ”as shown in figure. The shear stress intensity on all the sides will be “q” for equilibrium of the element. Let “ t ” be the thickness of the element.
Consider a plane BE inclined at angle  with BC. A B C p  p' q D E
Force on face BC
= P1 = p * BC * t
= P2 = q * BC * t
Force on face CE
= P3 = q * CE * t
= P4 = p’ * CE * t

6.  A B C p  p' q D P4 B C E P1 Pn Pt P3 P2
Resolving forces parallel & perpendicular to BE,
Pn = (P1+P3)cos  + (P2+P4)sin 
Pt = (P1+P3) sin  - (P2+P4)cos  A B C p  p' q D E
P2sin
P2cos P4 B C E P1 Pn Pt 
P1cos
P1sin
P4cos
P4sin P3 P2
P3cos
P3sin
Force on face BC
= P1 = p * BC * t
= P2 = q * BC * t
Force on face CE
= P3 = q * CE * t
= P4 = p’ * CE * t

7. Resolving forces perpendicular & parallel to BE,
Pn = (P1+P3)cos  + (P2+P4)sin 
Pt = (P1+P3) sin  - (P2+P4)cos 
pn = Pn / (BE.t)
=[(p*BC*t +q*CE*t) Cos+(q*BC*t + p’ *CE*t) Sin] / (BE.t)
= p cos2 + q sin cos + q sin cos + p’ sin2
= p/2 (1+cos2) + p’/2 (1-cos2) + q sin2
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2……….(I)
pt = Pt / (BE.t)
=[(p*BC*t +q*CE*t) Sin - (q*BC*t + p’ *CE*t) Cos] / (BE.t)
= p sin cos + q sin2 - q cos2 - p’ sincos
pt = {(p - p’)/2} * sin2 - q cos2 ……………….…….(II)

8. To locate Principal Planes:--
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2……….(I)
pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)
The Principal Planes are the plane which carry no Tangential stress. Therefore pt = 0
at pt = 0, let  = 
{(p - p’)/2} * sin2 - q cos2 = 0
{(p - p’)/2} * sin2 = q cos2
tan2 = q / {(p - p’)/2 }
tan2 = 2q / (p - p’)
(p - p’)2 + 4q2 2q 2
p - p’
1 = [ tan-1{ 2q / (p - p’) } ]/2
2 = 90 + 1

9. tan2 = 2q / (p - p’) There are two values of  which will satisfy the
above eq. 1 & 2, where 2=90+ 1
sin21=2q/((p-p’)2+4q2) & cos21=(p-p’)/((p-p’)2+4q2)
sin22=-2q/((p-p’)2+4q2)& cos22= -(p-p’)/((p-p’)2+4q2)
Substituting value of sin21 & cos21, in eq.(I),
p1 = {(p + p’)/2} + {(p - p’)/2}* cos21 + q sin21…….(I)
= (p+p’)/2 + {(p-p’)/2}*{(p-p’)/((p-p’)2+4q2)} q* 2q/((p-p’)2+4q2)
= (p+p’)/2+ (p-p’)2*/ 2((p-p’)2+4q2) +2q2/((p-p’)2+4q2)
= (p+p’)/2 +  {(p-p’)/2}2+q2
Similarly, substituting value of cos22 & sin22 in eq.(I),
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 2
p - p’ 2q
(p - p’)2 + 4q2

10. All planes parallel to this plane are principal planes and principal stress = p1, inclination = 90  p q
All planes parallel to this plane are principal planes and principal stress = p2, inclination =  + 90 deg.

11. To locate Planes of Maximum Shear Stress:--
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress
Let “ t ” be the thickness of the element.
Consider a plane AC inclined at angle  with AB. p2 p2
p2cos 
p2sin C B C pt B pt
p1cos  p1 pn p1  p1  pn A
p1sin p2 A

12. Equating all the forces along to AC,
Consider a plane AC inclined at angle  with AB. Now, Force on face AB = P1 = p1*AB*t
Force on face BC = P2 = p2*BC*t
Equating all the forces along to AC,
Pt = P1sin - P2cos
= p1 AB t sin - p2 BC t cos
pt = Pt / (AC. t)
p1 AB t sin - p2 BC t cos P2 A B C P1 Pn Pt 
P1cos
P1sin
P2cos
P2sin =
AC. t =
p1sin (AB/AC) - p2 cos (BC/AC) =
p1sincos - p2 sin cos
pt =
(p1- p2)sincos

13. The Shear stress pt = (p1- p2)sin cos; H E pt = (p1- p2)sin cos; 45
135
pt max. sin2 = max.
sin2 =+/- 1
2 = 90 or 270
 = 45 or 135
When  = 45, (EF) pt max = (p1- p2)/2
When  = 135,(GH) pt max = (p1- p2)/2
Planes of maximum shear are always at Right angles to
each other (Inclined at 45 & 135 with principal planes.) G F

14. p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress
Location of Principal Planes;
tan2 = 2q / (p - p’) (From previous slide)
( anticlockwise from vertical)
pt max = (p1- p2)/2 =  {(p-p’)/2}2 + q2
(at 45 & 135 with principal planes in anticlockwise) p1 p2   p2 p1
pt2
pt1 45

15. Case:- 1 Element Subjected to Direct Stress & Shear Stresses :--
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)
= (p cos2 + q sin cos + q sin cos + p’ sin2 )
pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 (Major Prin. Stress )(III)
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 (Minor Prin. Stress)(IV)
Location of Principal Planes;
tan2 = 2q / (p - p’) ( anticlockwise from vertical) (V)
pt max = (p1- p2)/2 =  {(p-p’)/2}2 + q2 _ _ (VI)
(at 45 & 135 with principal planes in anticlockwise)

16. Q.1 A rectangular bar of cross sectional area 100cm2 is subjected to an axial load of 20kN. Calculate the normal and tangential stresses on a section which is inclined at an angle of 30deg. with normal cross section of the bar.
Ans. Direct stress σ = load / cross sectional area
= 20000/10000 = 2MPa.
Normal stress σn = (p cos2 + q sin cos + q sin cos + p’ sin2 ) as p’= q= zero we have
σn = σcos230 = 2* cos230 = 1.5MPa
and tangential stress
σt = {(p - p’)/2} * sin2 - q cos2, as p’= q= zero
σt =(σ/2)sin2*30 = 0.866MPa

17. Q.2 Calculate diameter of circular bar which is subjected to an axial pull of 160kN, if the permissible shear stress on any cross section is 60MPa.
Ans. Direct stress p = /(π*r2 )
Maximum shear stress q = p/2= 80000/(π*r2)
Max. q = 60 = 80000/ ( π*r2)
Gives r = 20.6mm and diameter = 41.2 mm,
say diameter = 42mm

18. Q.3 A rectangular bar of cross section 200mm x 400mm is subjected to an axial pull P. If permissible normal and tangential stresses are 7 and 3.7MPa, calculate safe value of force P.
Ans. Direct stress p = P/ (200*400) = 7 MPa
hence P = 560kN _ _ _ _(i)
and Maximum shear stress q = 3.7 =
p/2= P/ (2*200*400)
Hence P = 592kN _ _ _ _(ii)
From i and ii safe value of P = 560kN ans.

19. Q.4 Two wooden pieces 100mm X100mm cross section are glued to gather as shown in fig. If permissible shear stress in glue material is 1MPa, calculate maximum axial pull P.
Glued joint P P
30deg
With vertical the inclination of the joint is 60deg.
σt = {(p - p’)/2} * sin2 - q cos2, as p’= q= zero shear stress q = (p/2)*sin 2θ
So that q = 1 = [P/(2*100*100)]sin (2*60)
Give P = kN. Ans.

20. Q. 5 Two mutually perpendicular planes are stresses as shown in fig
Q.5 Two mutually perpendicular planes are stresses as shown in fig. calculate normal and tangential stress on an oblique plane AB.
100MPa
 = 30deg. 
120MPa
120MPa
100MPa
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2 _(I)
σn = {(p - p’)/2} * sin2 - q cos2 _ _ (II) as q = zero
pn = ( )/2 + [(120 – 100)/2 ]cos 2*30
= 115MPa and
σn = [( )/2]sin 2*30 = 8.66MPa Ans

21. Q. 6 Two mutually perpendicular planes are stresses as shown in fig
Q.6 Two mutually perpendicular planes are stresses as shown in fig. calculate normal and tangential stress on an oblique plane AB.
60MPa
 = 30deg. 
120MPa
120MPa
60MPa
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2 _(I)
σn = {(p - p’)/2} * sin2 - q cos2 _ _ (II) as q = zero
pn = ( )/2 +[ ( )/2 ]cos 2*30
= 75 MPa
σn = [( )/2]sin 2*30 = MPa Ans

22. Example:-7 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other are 800N/mm2 and 350N/mm2 (both tensile) and a shear stress of 400N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 50 with vertical as shown in figure.
350N/mm2
400N/mm2 50
800N/mm2
400N/mm2

23. Solution:
Given, p = 800 N/mm2, p’= 350 N/mm2,
q = 400 N/mm  = 50
(i) pn & pt
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I) =
= N/mm2
pt = {(p - p’)/2} * sin2 - q cos2 ………………..……(II)
= ( )
= N/mm2

24. (ii) Principal Planes:--
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress =
= N/mm2
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress =
= N/mm2
(iii) Location of Principal Planes:--
tan2 = 2q / (p - p’)  = [ tan-1{ 2q / (p - p’) } ]/2
1 = [ tan-1{ 2q / (p - p’) } ]/2 = 30.32 (ACW from p)
2 = 90 + 1 = =  (with vertical.)
(iv) pt max = (p1- p2)/2 =  {(p -p’)/2}2 + q2
= N/mm2 (at 45 with principal planes.)

25. Example:-8 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other are 500N/mm2 (tensile) and 100N/mm2 (Compressive) and a shear stress of 200N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 30 with vertical as shown in figure.
100N/mm2
200N/mm2 30
500N/mm2
200N/mm2

26. Solution:
Given, p = 500 N/mm2, p’= -100 N/mm2,
q = 200 N/mm  = 30
(i) pn & pt
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I) =
= N/mm2
pt = {(p - p’)/2} * sin2 - q cos2 ………………..……(II)
= (100)
= N/mm2

27. (ii) Principal Planes:--
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress =
= N/mm2
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress =
= N/mm2
(iii) Location of Principal Planes:--
tan2 = 2q / (p - p’)  = [ tan-1{ 2q / (p - p’) } ]/2
1 = [ tan-1{ 2q / (p - p’) } ]/2 = 16.85 (with vertical.)
2 = 90 + 1 = =106.85 (with vertical.)
(iv) pt max = (p1- p2)/2 =  {(p -p’)/2}2 + q2
= N/mm2 (at 45 with principal planes.)

28. Example:-9 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other are 650N/mm2 (Compressive) and 150N/mm2 (Compressive) and a shear stress of 320N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 60 with vertical as shown in figure.
150N/mm2
320N/mm2 60
650N/mm2
320N/mm2

29. Solution:
Given, p = N/mm2, p’= N/mm2,
q = 320 N/mm  = 60
(i) pn & pt
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I) =
= N/mm2
pt = {(p - p’)/2} * sin2 - q cos2 ………………..……(II)
= ( - 160)
= N/mm2

30. (ii) Principal Planes:--
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress =
= 6.08 N/mm2
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress =
= N/mm2
(iii) Location of Principal Planes:--
tan2 = 2q / (p - p’)  = [ tan-1{ 2q / (p - p’) } ]/2
1 = [ tan-1{ 2q / (p - p’) } ]/2 =  (with vertical.)
2 = 90 + 1 = =  (with vertical.)
(iv) pt max = (p1- p2)/2 =  {(p -p’)/2}2 + q2
= N/mm2 (at 45 with principal planes.)

31. Mohr’s Stress Circle Method:--
When principal stresses & planes at a point in a two dimensional stresses are given, we can use the Mohr’s Stress Circle Method to obtain the normal and tangential stress-intensities across a given plane through the point.
In Mohr’s Circle ;
The Normal Stress (pn) is plotted on X-axis.
The Tangential Stress (pt) is plotted on Y-axis.
Angle on Mohr’s circle is taken as twice the given angle

32. p' q D C  p p A q B p' Sign Conventions:--
Shear Stress (pt) p p A q B p' O
Normal Stress (pn)
Sign Conventions:--
1. Tensile Stresses are taken as Positive (+ve)
2. Compressive Stresses are taken as Negative (-ve)
3. Clockwise Shear is taken as (+ve)
4. Anticlockwise Shear is taken as (-ve)

33. Mohr’s Stress Circle Method:--
Case:-1 If an element is subjected to the direct stresses (p & p’) & shear stress (q) then the principal stresses are (p1 & p2) and the normal and tangential stresses (pn & pt) across a plane inclined at angle  to the plane carrying stress p are,
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)
pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress
Location of Principal Planes; tan2 = 2q / (p - p’)
pt max = (p1- p2)/2 =  {(p-p’)/2}2 + q2

34. A B C p p' D P’(CD) pt q Q pn O Q’ p' C q p P(BC)
Steps to Draw Mohr’s Circle:- A B C p p' q D
P’(CD) pt q Q pn O Q’ p' C q p
P(BC)
1. Select origin O and draw axis OX.
2. Draw OQ = p . On faces BC & DA the shear stress q is anticlockwise, ( -ve) , draw QP = q (downwards)
3. OQ’ = p’ and on faces AB & CD the shear stress q is clockwise, hence it is +ve, so draw Q’P’ = q ( upwards as it is +ve)
4. Join the points P & P’ which will intersect OX at C.

35. pt q Q B A O p2 Q’ C pn p' q p P(BC) p1
(CD)P’ q Q B A O p2 Q’ C pn p' q p p1
P(BC)
5. Point C is the Centre of Mohr’s Circle. With C as centre and radius equal to CP or CP’, draw a circle, which will intersect OX at points A & B.
6. Measure OA which gives maximum direct stress and no Shear stress & it is Major principal stress (p1) OA=p1
Measure OB as Minor principal stress (p2) OB = p2.

36. C D p p pt A q p' B q B Q A p2 O Q’ pn C q p' p P(BC) p1 p p pt
(CD)P’ A q p' B q B Q A p2 O Q’ 2 pn C q p' p
P(BC) p1
7. From plane CP measure angle as 2 anticlockwise
up to CA, therefore the principal plane will be located at  anticlockwise from face BC.

37. T pt q Q B A O p2 Q’ C pn q p' p P(BC) p1
(CD)P’ q Q B A O p2 2 Q’ C pn q p' p
P(BC) p1
8. Measure CT which gives maximum Shear stress . Which is inclined at 90 & 270 in Mohr’s circle, so in actual case it is inclined at 45 & 135 with principal planes.

38. A B C p  p' q D C R pn Q’ q pt P(BC) Q 2 R’ p' p O
(CD)P’ R pn Q’ q pt
P(BC) Q 2 R’ p' p O
9. From point P (BC) measure angle 2 in anticlockwise direction and draw a line CR. From R, draw RR’ on OX. Measure OR’ as pn and RR’ as pt.

39. Proof:-
QQ’ = OQ - OQ’ = p - p’
CQ = CQ’ = (p - p’)/2
OC = OQ’ + Q’C = p’ + (p - p’)/2
= (p + p’)/2
Radius of Mohr’s Circle;
CP =  CQ2 + QP2 =  {(p - p’)/2}2 + q2
OA = OC+CA = (p + p’)/2 +  {(p - p’)/2}2 + q2 = p1
OB = OC - CB = (p + p’)/2 -  {(p - p’)/2}2 + q2 = p2
since p1 & p2 are on X-axis the tangential stress is zero. Hence they represent the “Principal Planes”. Here p1 is major principal plane and p2 is minor principal plane.

40. p2 A C B P’ T O p1 pn 2 Q’ p' q pt
p - p’ P Q p T’

41. tan2 = QP/CQ = q / {(p - p’)/2} = 2q / (p - p’)
let QCP = 2 then
tan2 = QP/CQ = q / {(p - p’)/2} = 2q / (p - p’)
The max. tangential stress = max Y-ordinate in the circle
which is equal to the radius of the circle (CT = CT’)
pt max = CT = CP =  {(p - p’)/2}2 + q2= (p1 - p2)/2
the position of the plane CT is at 90 from CA, hence the plane of max. shear is inclined at 45 with principal planes.

42. Example:-10 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other (both tensile) and a shear stress of across the planes are shown in figure. Locate the principal Planes. Calculate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 80 with vertical as shown in figure.
800N/mm2
400N/mm2
350N/mm2 80

43. R P’ pt pt 160 Q O pn Q’ R’ C pn 400 P Example:-10 400 350N/mm280
800N/mm2
400N/mm2 R P’ pt pt
400
160 Q O pn Q’ R’ C pn
400
350N/mm2 P
800N/mm2

44. T P’ pt Q B A p2 O pn Q’ C 400 P p1 Example:-10  400N/mm2 2 350N/mm2

45. Solution:--
Measure:
OR’ = pn = Normal stress = N/mm2
RR’ = pt = Tangential Stress = N/mm2
OR = pr = Resultant Stress =
OA = p1 = Major Principal Stress = N/mm2
OB = p2 = Minor Principal Stress = N/mm2 Angle PCA = 2 = ___ therefore, 1 = 30.32& 2=120.32
CT = pt max=Maximum Tangential Stress= N/mm2
Example:-10

46. Example:-11 A point in a strained material, the Normal stress across two planes at right angles to each other are 500N/mm2 (tensile) and 100N/mm2 (Comp.) and a shear stress of 200N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress(ptmax) and its location. Find Normal (pn),Tangential (pt) stresses on a plane inclined at 30 with vertical as shown in figure.
100N/mm2
200N/mm2 30
500N/mm2
200N/mm2

47. pt P’ 200 R pt pn Q’ - pn O Q R’ C pn 200 P 500 Example:-11 60 100 10030
Example:-11 pt P’
200 R pt pn Q’
- pn O 60 Q C R’
100 pn
200 P
500

48. pt T P’ 200 pn Q B Q’ A - pn O C 200 P 500 p2 p1 Example:-11  2 100

49. Solution:--
Measure:
OR’ = pn = Normal stress = _____
RR’ = pt = Tangential Stress = ______
OR = pr = Resultant Stress = _______
OA = p1 = Major Principal Stress = _______
OB = p2 = Minor Principal Stress = _______
Angle PCA = 2 = ___ therefore, 1 = ___& 2=___
CT = pt max = Maximum Tangential Stress = _______
Example:-11

50. Example:-12 A point in a strained material, the intensities of Normal stress across two planes at right angles to each other are 650N/mm2 (Compressive) and 150N/mm2 (Compressive) and a shear stress of 320N/mm2 across the planes as shown in figure. Locate the principal Planes and evaluate the Principal Stresses (p1 & p2). Also find the planes of maximum shear stress (pt max) & its location. Find Normal(pn),Tangential(pt) stresses on a plane inclined at 60 with vertical as shown in figure.
650N/mm2
320N/mm2 60
150N/mm2
320N/mm2

51. R pt P’ 320 pt Q pn R’ O Q’ C pn P 650 Example:-12 320 650 320 150 32060
150 R pt
320 P’
320 pt Q pn R’ O Q’ C
120 pn
320 P
150
650

52. T pt P’ 320 Q C A pn B O Q’ p2 650 P p1 Example:-12  2 320 650 320
150 T pt
320 P’
320 Q C A pn B O Q’ 2 p2
650
320 P
150 p1

53. Solution:--
Measure:
OR’ = pn = Normal stress = _____
RR’ = pt = Tangential Stress = ______
OR = pr = Resultant Stress = _______
OA = p1 = Major Principal Stress = _______
OB = p2 = Minor Principal Stress = _______
Angle PCA = 2 = ___ therefore, 1 = ___& 2=___
CT = pt max = Maximum Tangential Stress = _______
Example:-12

54. Case:- 2 Element Subjected to direct stress
Stresses only and no shear stress:--
Consider an element subjected to only Direct Stresses p & p’ (both tensile) as shown in figure. Here the shear stress intensity “q” is zero. p' C B pt  p pn p A p'

55. 1. Element Subjected to Direct & Shear Stresses
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)
pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress
Location of Principal Planes; tan2 = 2q / (p - p’)
pt max = (p1- p2)/2 =  {(p-p’)/2}2 + q2
Case-2 Element Subjected to Direct Stresses only:--
If in any case, an element is subjected to only direct stresses & no tangential stresses as shown in figure. In that case the pn, pt, p1, p2 & pt max can be found out by substituting q=0 in equations of CASE- 1.

56. pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)
pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress
Location of Principal Planes; tan2 = 2q / (p - p’)
pt max = (p1- p2)/2 =  {(p -p’)/2}2 + q2
For CASE-2 Substituting q = 0 in above equations;
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 ………….…….….(I)
pt = {(p - p’)/2} * sin2 …………. ……………….……(II)
p1 = p/2 +  p2/4 = p …Major Principal Stress
p2 = p’/2 +  p’2/4 = p’ …Minor Principal Stress
pt max = {(p-p’)/2}2 = (p - p’)/2 = (p1 - p2)/2

57. Case:- 2 Element Subjected to Direct Stress only and no shear stress:--p' C B pt  p pn p A
pn = (p + p’)/2 + {(p - p’)/2} cos2 p'
pt = {(p - p’)/2} sin2
pr = pn2+ pt2
pt max = (p - p’)/2

58. Also find Maximum Tangential stress (pt max)
Example:-13 A point in a strained material is subjected to the stresses as shown in figure. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 50 with vertical as shown in figure.
Also find Maximum Tangential stress (pt max) 50
800 N/mm2
450 N/mm2

59. Solution:- q = 0 p = + 800 N/mm2 (Tensile) p' = + 450 N/mm2 (Tensile)
 = 50
pn = (p + p’)/2 + {(p - p’)/2} cos2 = N/mm2
pt = {(p - p’)/2} sin2 = N/mm2
pr= pn2+ pt2 = N/mm2
pt max = (p - p’)/2 = 175 N/mm2

60. Also find Maximum Tangential stress (pt max)
Example:-14 A point in a strained material is subjected to the stresses as shown in figure. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 30 with vertical as shown in figure.
Also find Maximum Tangential stress (pt max) 30
500 N/mm2
200 N/mm2

61. Solution:- q = 0 p = - 500 N/mm2 (Comp.) p' = + 200 N/mm2 (Tensile)
 = 30 30
500
200
pn = (p + p’)/2+ {(p - p’)/2} cos2 = N/mm2
pt = {(p - p’)/2} sin2 = N/mm2
pr= pn2+ pt2 = N/mm2
pt max = (p - p)/2 = N/mm2

62. Also find Maximum Tangential stress (pt max)
Example:-15 A point in a strained material is subjected to the stresses as shown in figure. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 40 with vertical as shown in figure.
Also find Maximum Tangential stress (pt max) 40
600 N/mm2
300 N/mm2

63. Solution:- q = 0 p = - 600 N/mm2 (Comp.) p' = - 300 N/mm2 (Comp.)
 = 40 40
600
300
pn = (p + p’)/2 + {(p - p’)/2} cos2 = N/mm2
pt = {(p - p’)/2} sin2 = N/mm2
pr= pn2+ pt2 = N/mm2
pt max = (p - p’)/2 = N/mm2

64. p' C B pt  p pn p A p' Mohr’s Stress Circle Method:--
Case:-II If the tangential stresses acting on element is zero, then the normal and tangential stresses across a plane inclined at angle  to the plane carrying p are,
pn = (p + p’)/2 + {(p - p’)/2} cos2
pt = {(p - p’)/2} sin2 p' C B pt  p pn p A p'

65. pt B A O p' C pn p Steps to Draw Mohr’s Circle:-
1. Set off the axis OX on this set off OA = p & OB = p’
2. Bisect BA at C.
3. With C as centre and CA or CB as radius, draw a circle. This circle is known as Mohr’s circle.

66. P pt pr pt 2 B A O p' Q C pn pn p
4. Draw line CP at angle “2” with OX in anticlockwise so that the point P is on the circle.
5. From P, drop perpendicular PQ on X axis. Measure OQ as pn & measure PQ as pt.
6. Join and measure OP which gives resultant stress pr.

67. T P pt pr pt B A O p' C Q pn pn p 2
7. Join & measure CT which represents the plane of maximum shear stress. It inclined at 90 & 270 in Mohr’s circle with principal plane, hence in actual case they are inclined at 45 & 135 with principal planes.

68. pt T P pr pt 2 B A O Q C pn p'
p - p’ pn p

69. Also find Maximum Tangential stress (pt max)
Example:-16 A point in a strained material is subjected to following stresses. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 50 with vertical as shown in figure. Use Mohr’s Circle Method.
Also find Maximum Tangential stress (pt max)
450 N/mm2 50
800 N/mm2
800 N/mm2
450 N/mm2

70. pt Example:-16 800 N/mm2 T P 450 N/mm2 pr pt 100 B A O p' = 450 N/mm2Q C pn pn
p = 800 N/mm2

71. Solution:--
Measure:
OQ = pn = Normal stress = _____
PQ = pt = Tangential Stress = ______
OR = pr = Resultant Stress = _______
CT = pt max = Maximum Tangential Stress = _______
Example:-16

72. Also find Maximum Tangential stress (pt max)
Example:-17 A point in a strained material is subjected to following stresses. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 30 with vertical as shown in figure. Use Mohr’s Circle Method.
Also find Maximum Tangential stress (pt max)
350 N/mm2 30
150 N/mm2
150 N/mm2
350 N/mm2

73. 150 N/mm2 350 N/mm2 Example:-17 pt T P pt pr B C A - pn O p' = 350 Q30
150 N/mm2
350 N/mm2
Example:-17 pt T P pt pr B C A 60
- pn O
p' = 350 Q pn pn
p = 150

74. Solution:--
Measure:
OQ = pn = Normal stress = _____
PQ = pt = Tangential Stress = ______
OR = pr = Resultant Stress = _______
CT = pt max = Maximum Tangential Stress = _______
Example:-17

75. Also find Maximum Tangential stress (pt max)
Example:-18 A point in a strained material is subjected to following stresses. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 50 with vertical as shown in figure. Use Mohr’s Circle Method.
Also find Maximum Tangential stress (pt max)
800 N/mm2 40
600 N/mm2
600 N/mm2
800 N/mm2

76. Example:-18 600 N/mm2 pt T P 800 N/mm2 pr pt A B C O pn Q40
800 N/mm2
600 N/mm2 pt T P pr pt A B C 80 O pn Q
p =- 600 N/mm2 pn
p' = N/mm2

77. Solution:--
Measure:
OQ = pn = Normal stress = _____
PQ = pt = Tangential Stress = ______
OR = pr = Resultant Stress = _______
CT = pt max = Maximum Tangential Stress = _______
Example:-18

78. Case :-3 Stress in One Direction Only:--
Consider an element subjected to stress p in only one direction as shown.
Here, the p’ = 0 & q = 0. 2 1  p p 1 2

79. 1. Element Subjected to Direct & Shear Stresses
pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)
pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress
Location of Principal Planes; tan2 = 2q / (p - p’)
pt max = (p1- p2)/2 =  {(p-p’)/2}2 + q2
Case-3 Element Subjected to Direct Stresses only:--
If in any case, an element is subjected to direct stress in only direction & no tangential stresses as shown in figure. Then the pn, pt, p1, p2 & pt max can be found out by substituting p’ & q = 0 in eq. of CASE- 1.

80. pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I)
pt = {(p - p’)/2} * sin2 - q cos2 ……………….……(II)
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress
Location of Principal Planes; tan2 = 2q / (p - p’)
pt max = (p1- p2)/2 =  {(p -p’)/2}2 + q2
For CASE-3 Substituting p’ & q = 0 in above equations;
pn = p /2 + p /2* cos2 = p cos2 ………….…….….(I)
pt = p /2 * sin2 ………………..……………….……(II)
p1 = p/2 +  p2/4 = p …Major Principal Stress
p2 = p/2 -  p2/4 = 0 …Minor Principal Stress
pt max = {(p-p’)/2}2 = p /2

81. Case :-3 Stresses in One Direction only :--
pn = p cos2
pt = p/2 sin2
pr =  pn2 + pt2
(Inclined at 45 with Principal Planes)
pt max = p/2 2 1  p p 1 2

82. Also find Maximum Tangential stress (pt max)
Example:-19 A point in a strained material is subjected to the stress as shown in figure. Find Normal(pn) , Tangential(pt) and Resultant (pr)stresses on a plane inclined at 50 with vertical as shown in figure.
Also find Maximum Tangential stress (pt max) 50
800 N/mm2
800 N/mm2

83. Solution:--
pn = p cos2 = 800 * cos2(50) = N/mm2
pt = p/2 sin2 = (800/2) * sin(100) = N/mm2
pr =  pn2 + pt2 = N/mm2
(Inclined at 45 with Principal Planes)
pt max = p/2
= 800/2
= 400 N/mm2 50
800 N/mm2
800 N/mm2

84. Example:-20 Find p1, p2 &  if the stresses on two planes are as shown in figure.
100 N/mm2
50 N/mm2
30 N/mm2
60 N/mm2 C A B

85.  C A B D p2 2 p1 100 50 30 60 P’(CD) 30 C Q B Q’ A C’ 50 60 N/mm2
P(AB)
100 N/mm2 p1

86. STEPS:-- 1. Select origin O and draw axis OX.
2. On face AB, Draw pn = OQ = 100 N/mm2, the shear stress = 50 is anticlockwise, Draw QP = 50 (downwards)
3. On face CD, Draw pn = OQ’ = 60 N/mm2, the shear stress = 30 is Clockwise, Draw Q’P’ = 30 (upwards)
4. Join the points P & P’ . Now find the centre of line PP’ as point C’. From point C’ draw perpendicular bisector C’C , which will intersect OX at C.
5. Point C is the Centre of Mohr’s Circle. Join CP & CP’. With C as centre and radius equal to CP or CP’, draw a circle, which will intersect OX at points A & B.
6. Measure OA which gives maximum direct stress and no Shear stress & it is Major principal stress (p1) OA=p1
Measure OB as Minor principal stress (p2) OB = p2.
7. From CP, measure the angle PCP’ in anticlockwise direction as 2. Half of this angle will give .

87. Example:-21 Find p1, p2 &  if the stresses on two planes are as shown in figure.C B
900 N/mm2  30 A 30
1200 N/mm2 D

88. Principal Strain:--
If p1, p2 & p3 are principal stresses, then the principal strains e1, e2 & e3 are given by;
e1 = p1/E -  p2/E -  p3/E (in direction of p1)
e2 = p2/E -  p1/E -  p3/E (in direction of p2)
e3 = p3/E -  p1/E -  p2/E (in direction of p3) p2 p1 p3

89. Example:-22 A rectangular block if steel is subjected to stresses of 1100 Mpa (tensile), 600 Mpa (Comp.), & 450 Mpa (tensile) across three pairs of faces. If  = 0.3, and E = 2*10 5 N/mm2, Calculate strain in each direction.
Solution:-- p1 = 1100 N/mm2 (Ten)
p2 = N/mm2 (Comp.)
p3 = 450 N/mm2 (Ten)
 = 0.3, and E = 2*10 5 N/mm2
e1 = p1/E -  p2/E -  p3/E (in direction of p1) = ______
e2 = p2/E -  p1/E -  p3/E (in direction of p2) = ______
e3 = p3/E -  p1/E -  p2/E (in direction of p3) = ______

90. Example:-23. A circle of 250mm diameter is drawn on a mild steel plate
Example:-23 A circle of 250mm diameter is drawn on a mild steel plate. Then it is subjected to the stresses as shown in figure. Find the lengths of major & minor axis of the ellipse formed due to the deformation of the circle. Take E = 2 * 10 5 N/mm2;  = 0.25
450 MPa
300 MPa
600 MPa
250
600 MPa
300 MPa
450 MPa

91. Solution:-- E = 2 * 10 5 N/mm2;  = 0.25
p = 600 Mpa , p’ = 450 MPa, q = 300 Mpa
p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress
= = MPa
p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress
= = Mpa
e1 = p1/E -  p2/E (in direction of p1)
= 1/E { (215.76)}
= (in direction of p1)
Increase in diameter along Major axis;
d1= e1 . d
= * 250 = mm
Major axis of ellipse , d1 = = mm

92. e2 = p2/E -  p1/E (in direction of p2)
Increase in diameter along Major axis;
d2 = e2 . d
= * 250 = mm
Minor axis of ellipse, d2 = = mm
Major axis of ellipse d1= = mm
Minor axis of ellipse d2= = mm
250

93. 1 = 38 (Anticlockwise with vertical)
tan2 = 2q / (p - p’) = 600 / ( ) = 4
1 = 38 (Anticlockwise with vertical)
2 = 90 + 38 = 128(with vertical)
d1= mm ; d2 = mm
450 MPa
300 MPa
38 P1
600 MPa
600 MPa d2 d1 P2
300 MPa
450 MPa

94. Example:-24 Solve following 4 example using Mohr’s circle method.
400 N/mm2
800 N/mm2
200 N/mm2
800 N/mm2
200 N/mm2

95. 400 N/mm2
200 N/mm2
200 N/mm2

96. Shear stress due twisting moment q = Tr/J = 16T/πd3
SOLID SHAFTS SUBJECTED TO BENDING MOMENT AND TWISTING MOMENT:
(WITHOUT ANY AXIAL FORCE)
Shear stress due twisting moment q = Tr/J = 16T/πd3
Bending stresses are due to bending moment
= p = pbt or pbc = M/Z = 32M/ πd3( pbt and pbc = bending tensile and compressive stresses resp.)
Principal stresses, p1 and p2 = p/2 ± [(p/2)2 + q2]1/2
Using values of p and q
p1 and p2 = 16[ M ±( M2 + T2)1/2] / πd3
Maximum shear stress
τ max = (p1 +p2)/2 = 16 [( M2 + T2)1/2] / πd3
For position of principal planes tan2 = 2q / (p - p’) gives tan2 = 2q / p as p’ is zero

97. For position of principal planes
tan2 = 2q / (p - p’) gives
tan2 = 2q / p as p’ is zero
= T/M
1 =  and  2 =  deg.

98. HOLLOW SHAFTS SUBJECTED TO BENDING MOMENT AND TWISTING MOMENT:
(WITHOUT ANY AXIAL FORCE)
q = Tr/J = 32T(D/2)/π[D4 – d4] …..(1)
p = pbt or pbc = M(D/2)/I = 32M(D)/ π[D4 – d4]……(2)
Principal stresses are given by equations
p1 and p2 = p/2 ± [(p/2)2 + q2]1/2
Inserting values of p and q
p1 and p2 = 16D[ M ±( M2 + T2)1/2]/ π[D4 – d4]
Maximum shear stress τ max = (p1 +p2)/2
= 16 D[( M2 + T2)1/2] /π[D4 – d4]
(Where D and d are outer and inner diameters of the shaft)

99. Q.1 A solid shaft 100mm diameter is subjected to a bending moment of 10kNm and a twisting moment of 12 kNm. Calculate principal stresses and position of plane on which they act.
Ans: M = 10kNm and T = 12kNm
Principal stresses are given by
p1 and p2 = 16[ M ±( M2 + T2)1/2] / πd3
p1 and p2 = 16*106[10 ± ( ] ½ / π100 3
= MPa and MPa
Maximum shear stress
τ max = (p1 +p2)/2 = 16 [( M2 + T2)1/2] / πd3
=(p1 +p2)/2 = MPa
tan2 = T /M = 1.2, So 1 = 25.1 and 2 =115.1deg.

100. Q. 2 The maximum permissible shear stress in hollow shaft is 60MPa
Q. 2 The maximum permissible shear stress in hollow shaft is 60MPa. For shaft the outer diameter shaft is twice the inner diameter. The shaft is subjected to a twisting moment of 10kNm and a bending moment of 12kNm. Calculate diameter of the shaft.
Ans. Given T = 10kNm and M = 12kNm
Maximum shear stress
τ max = (p1 + p2)/2
= 16D[( M2 + T2)1/2] /π[D4 –d4]
60 = 16*D[ ] ½ / π [ D4 - (D/2)4 ]
Outer diameter D = M say 114 mm
Inner diameter d = 57mm

101. Torsion accompanied with bending and axial force: (either tensile or compressive axial force )
The following forces/moments may act of shafts:
*Axial force
* Twisting moment and
* Bending moment
*Direct stresses ’p’ due to axial force on shaft
p = axial force/ cross sectional area of the shaft
For solid shafts, p = P/(πR2), R = radius of shaft
For hollow shafts, p = P/[π(R2 – r2) , Where R is outer radius of the hollow shaft and r is inner radius of the shaft

102. Shear stresses due to twisting moment on shaft:
T/J = q/r = cθ/L (equation for twisting of shaft)
For solid shaft:
Shear stress q = Tr/J = 16T/πd3
For hollow shaft:
Shear stress q = T(D/2)/[π(D4 – d4)/32]
= 16T/ [π(D4 – d4)]
Bending stresses due to bending moment:
Bending tensile or compressive stresses
For solid shafts:
pbt or pbc = M/Z = 32M/ πD3

103. = 32M/ πD3
For hollow shafts:
pbt or pbc = My/I = M(D/2)/ [π(D4 – d4)/64],
where D and d are outer and inner diameter of the shaft and M is bending moment.
Stress at any point on the shaft (either hollow or solid) depends on the application of axial force, bending moment and twisting moment on the shaft.
Principal stresses on shaft =
p1 = (p+p’)/2 + [ {(p-p’)/2}2+q2]1/2 and
p2 = (p+p’)/2 – [{(p-p’)/2}2+q2] 1/2
Where p = p + pbt (or pbc) i.e. algebraic sum of direct and bending stresses and q is shear stresses due to twisting moment

104. = 32M/ πD3
Principal stresses on shaft =
p1 = (p+p’)/2 + [ {(p-p’)/2}2+q2]1/2 and
p2 = (p+p’)/2 – [{(p-p’)/2}2+q2] 1/2
Where p = p + pbt (or pbc) i.e. algebraic sum of direct and bending stresses and q is shear stresses due to twisting moment. The p’ is zero in y direction.
p1 and p2 = (p)/2 ± [ {(p)/2}2+q2]1/2

105. Q. 3 A propeller shaft 200mm diameter transmits 2000kW at 300rpm
Q. 3 A propeller shaft 200mm diameter transmits 2000kW at 300rpm. The propeller is over hanging by 500mm from support and has mass of 5000kg. The propeller thrust is 200kN. Calculate maximum direct stresses induced in the shaft. Also locate position of section where the maximum direct stresses are acting and point on cross section where the maximum direct stresses are act.
Ans.
1. Direct stress due to thrust = force / c. s. area
p = ± 200*1000/ π*1002 = 6.366MPa

106. 2. Bending moment
M = 5000*9.81*500 Nmm (cantilever beam action )
Bending stress = M/Z = 5000*9.81*500/[ π 2003 / 32]
= MPa
3. Power, P = 2 πNT/ 60 or T = 60P/2 πN
T = 60*2000*106/2 π*300 = 63.66*106Nmm
Shear stress due to twisting moment
q = 16T/ πd3 = 16*63.66*106 / π*2003
= MPa.

107. Q. 3A propeller shaft 200mm diameter transmits 2000kW at 300rpm
Q. 3A propeller shaft 200mm diameter transmits 2000kW at 300rpm. The propeller is over hanging by 500mm from support and has mass of 5000kg. The propeller thrust is 200kN. Calculate maximum direct stresses induced in the shaft. Also locate position of section where the maximum direct stresses are acting and point on cross section where the maximum direct stresses are act.
Ans.
1. Direct stress due to thrust = force / c. s. area
σ = ± 200*1000/ π*1002 = 6.366MPa

108. 2. Bending moment
M = 5000*9.81*500 Nmm (cantilever beam action )
Bending stress = M/Z = 5000*9.81*500/[ π 2003 / 32]
= MPa
3. Power, P = 2 πNT/ 60 or T = 60P/2 πN
T = 60*2000*106/2 π*300 = 63.66*106Nmm
Shear stress due to twisting moment
q = 16T/ πd3 = 16*63.66*106 / π*2003
= MPa.

109. A D C
shaft B
Principal stresses on shaft =
p1 and p2 = (p)/2 ± [ {(p)/2}2+q2]1/2
p = σ + pbc and p = σ - pbt
p = =37.592MPa at point B (comp.)
p = = MPa at point A (tensile)
p = MPa (bending stress = zero at points C & D shear stress q = MPa at circumference

110. A D C
shaft B
Principal stresses on shaft =
At Point A p1 = MPa and p2 = 29,96MPa
τ max = (p1 – p2)/ 2 = 42.39MPa
At Point B p1 = MPa and p2 = MPa
τ max = (p1 – p2)/ 2 = 46.37MPa
At Point C & D
p1 = MPa and p2 = MPa
τ max = (p1 – p2)/ 2 = MPa