Principal Stresses & Strains

 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

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

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

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

 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

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)

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

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

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.

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

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

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

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

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)

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

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 = 160000/(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

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.

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 = 23.094kN. Ans.

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 = (120+100)/2 + [(120 – 100)/2 ]cos 2*30 = 115MPa and σn = [(120 -100)/2]sin 2*30 = 8.66MPa. . . .Ans

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 = (120- 60)/2 +[ (120 + 60)/2 ]cos 2*30 = 75 MPa σn = [(120 + 60)/2]sin 2*30 = 77.94 MPa. . . .Ans

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

Solution: Given, p = 800 N/mm2, p’= 350 N/mm2, q = 400 N/mm2  = 50 (i) pn & pt pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I) = 575 - 39.07 + 393.93 = 929.85 N/mm2 pt = {(p - p’)/2} * sin2 - q cos2 ………………..……(II) = 221.58 - ( - 69.46) = 291.04 N/mm2

(ii) Principal Planes:-- p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress = 575 + 458.94 = 1033.94 N/mm2 p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress = 575 - 458.94 = 116.06 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 = 90 + 30.32 =120.32  (with vertical.) (iv) pt max = (p1- p2)/2 =  {(p -p’)/2}2 + q2 = 458.94 N/mm2 (at 45 with principal planes.)

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

Solution: Given, p = 500 N/mm2, p’= -100 N/mm2, q = 200 N/mm2  = 30 (i) pn & pt pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I) = 200 + 150 + 173.20 = 523.20 N/mm2 pt = {(p - p’)/2} * sin2 - q cos2 ………………..……(II) = 259.80 - (100) = 159.80 N/mm2

(ii) Principal Planes:-- p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress = 200 + 360.55 = 560.55 N/mm2 p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress = 200 - 360.55 = -160.55 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 = 90 + 16.85 =106.85 (with vertical.) (iv) pt max = (p1- p2)/2 =  {(p -p’)/2}2 + q2 = 360.55 N/mm2 (at 45 with principal planes.)

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

Solution: Given, p = - 650 N/mm2, p’= - 150 N/mm2, q = 320 N/mm2  = 60 (i) pn & pt pn = {(p + p’)/2} + {(p - p’)/2}* cos2 + q sin2…….….(I) = - 400 + 125 + 277.13 = 2.13 N/mm2 pt = {(p - p’)/2} * sin2 - q cos2 ………………..……(II) = - 216.50 - ( - 160) = - 56.50 N/mm2

(ii) Principal Planes:-- p1 = (p+p’)/2 +  {(p-p’)/2}2+q2 …Major Principal Stress = - 400 + 406.08 = 6.08 N/mm2 p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress = - 400 - 406.08 = - 806.08 N/mm2 (iii) Location of Principal Planes:-- tan2 = 2q / (p - p’)  = [ tan-1{ 2q / (p - p’) } ]/2 1 = [ tan-1{ 2q / (p - p’) } ]/2 = - 26.00 (with vertical.) 2 = 90 + 1 = 90 + 16.85 =+ 64.00 (with vertical.) (iv) pt max = (p1- p2)/2 =  {(p -p’)/2}2 + q2 = 406.08 N/mm2 (at 45 with principal planes.)

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

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)

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

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.

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.

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.

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.

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.

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.

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

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.

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

R P’ pt pt 160 Q O pn Q’ R’ C pn 400 P Example:-10 400 350N/mm2 80 800N/mm2 400N/mm2 R P’ pt pt 400 160 Q O pn Q’ R’ C pn 400 350N/mm2 P 800N/mm2

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

Solution:-- Measure: OR’ = pn = Normal stress = 929.85 N/mm2 RR’ = pt = Tangential Stress = 291.04 N/mm2 OR = pr = Resultant Stress = OA = p1 = Major Principal Stress = 1033.94 N/mm2 OB = p2 = Minor Principal Stress = 116.06 N/mm2 Angle PCA = 2 = ___ therefore, 1 = 30.32& 2=120.32 CT = pt max=Maximum Tangential Stress=929.85 N/mm2 Example:-10

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

pt P’ 200 R pt pn Q’ - pn O Q R’ C pn 200 P 500 Example:-11 60 100 100 30 Example:-11 pt P’ 200 R pt pn Q’ - pn O 60 Q C R’ 100 pn 200 P 500

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

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

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

R pt P’ 320 pt Q pn R’ O Q’ C pn P 650 Example:-12 320 650 320 150 320 60 150 R pt 320 P’ 320 pt Q pn R’ O Q’ C 120 pn 320 P 150 650

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

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

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'

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.

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

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

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

Solution:- q = 0 p = + 800 N/mm2 (Tensile) p' = + 450 N/mm2 (Tensile)  = 50 pn = (p + p’)/2 + {(p - p’)/2} cos2 = 594.61 N/mm2 pt = {(p - p’)/2} sin2 = 172.34 N/mm2 pr= pn2+ pt2 = 619.08 N/mm2 pt max = (p - p’)/2 = 175 N/mm2

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

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 = - 325 N/mm2 pt = {(p - p’)/2} sin2 = - 303.11 N/mm2 pr= pn2+ pt2 = 444.41 N/mm2 pt max = (p - p)/2 = - 350 N/mm2

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

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 = - 476.04 N/mm2 pt = {(p - p’)/2} sin2 = - 443.16 N/mm2 pr= pn2+ pt2 = 650.39 N/mm2 pt max = (p - p’)/2 = - 450 N/mm2

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'

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.

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.

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.

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

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

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

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

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

150 N/mm2 350 N/mm2 Example:-17 pt T P pt pr B C A - pn O p' = 350 Q 30 150 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

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

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

Example:-18 600 N/mm2 pt T P 800 N/mm2 pr pt A B C O pn Q 40 800 N/mm2 600 N/mm2 pt T P pr pt A B C 80 O pn Q p =- 600 N/mm2 pn p' = - 800 N/mm2

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

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

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.

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

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

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

Solution:-- pn = p cos2 = 800 * cos2(50) = 330.54 N/mm2 pt = p/2 sin2 = (800/2) * sin(100) = 393.93 N/mm2 pr =  pn2 + pt2 = 514.23 N/mm2 (Inclined at 45 with Principal Planes) pt max = p/2 = 800/2 = 400 N/mm2 50 800 N/mm2 800 N/mm2

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

 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

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 .

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

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

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 = - 600 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) = ______

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

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 = 525 + 309.23 =834.24 MPa p2 = (p+p’)/2 -  {(p-p’)/2}2+q2 …Minor Principal Stress = 525 - 309.23 = 215.76 Mpa e1 = p1/E -  p2/E (in direction of p1) = 1/E { 834.24 - 0.25(215.76)} = 0.0039 (in direction of p1) Increase in diameter along Major axis; d1= e1 . d = 0.0039 * 250 = 0.975 mm Major axis of ellipse , d1 = 250 + 0.975 = 250.975 mm

e2 = p2/E -  p1/E (in direction of p2) Increase in diameter along Major axis; d2 = e2 . d = 0.000036 * 250 = 0.009 mm Minor axis of ellipse, d2 = 250 + 0.009 = 250.009 mm Major axis of ellipse d1= 250 + 0.975 = 250.975 mm Minor axis of ellipse d2= 250 + 0.009 = 250.009 mm 250 250.975 250.009

1 = 38 (Anticlockwise with vertical) tan2 = 2q / (p - p’) = 600 / (600 - 450) = 4 1 = 38 (Anticlockwise with vertical) 2 = 90 + 38 = 128(with vertical) d1= 250.975 mm ; d2 = 250.009 mm 450 MPa 300 MPa 38 P1 600 MPa 600 MPa d2 d1 P2 300 MPa 450 MPa

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

400 N/mm2 200 N/mm2 200 N/mm2

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

For position of principal planes tan2 = 2q / (p - p’) gives tan2 = 2q / p as p’ is zero = T/M 1 =  and  2 =  + 90 deg.

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)

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 ± ( 10 2 + 12 2 ] ½ / π100 3 = 130.48 MPa and 28.625MPa Maximum shear stress τ max = (p1 +p2)/2 = 16 [( M2 + T2)1/2] / πd3 =(p1 +p2)/2 = 50.928MPa tan2 = T /M = 1.2, So 1 = 25.1 and 2 =115.1deg.

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[ 122 + 102 ] ½ / π [ D4 - (D/2)4 ] Outer diameter D = 112.25M say 114 mm Inner diameter d = 57mm

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

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

= 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

= 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

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

2. Bending moment M = 5000*9.81*500 Nmm (cantilever beam action ) Bending stress = M/Z = 5000*9.81*500/[ π 2003 / 32] = 31.226MPa 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 = 40.527MPa.

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

2. Bending moment M = 5000*9.81*500 Nmm (cantilever beam action ) Bending stress = M/Z = 5000*9.81*500/[ π 2003 / 32] = 31.226MPa 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 = 40.527MPa.

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 = 6.366 + 31.226 =37.592MPa at point B (comp.) p = 6.366 - 31.226 = - 24.86MPa at point A (tensile) p = 6.366 MPa (bending stress = zero at points C & D shear stress q = 40.527MPa at circumference

A D C shaft B Principal stresses on shaft = At Point A p1 = -54.82MPa and p2 = 29,96MPa τ max = (p1 – p2)/ 2 = 42.39MPa At Point B p1 = 65.166MPa and p2 = - 27.574MPa τ max = (p1 – p2)/ 2 = 46.37MPa At Point C & D p1 = 43.835MPa and p2 = - 37.469MPa τ max = (p1 – p2)/ 2 = 40.652MPa