Week 10 – Lecture 9
Introduction
σ_x, σ_y, σ_z  normal stress
ͳ_xy, ͳ_yz, ͳ_zx
ͳ_xy = ͳ_yx, ͳ_yz = ͳ_zy, ͳ_zx = ͳ_xz
Complimentary shear stress: a shear stress on a given
plane is accompanied by another of equal magnitude
on a plane perpendicular to the first.
Introduction
ͳ_xy  the shear stress acting in the direction of y-axis on the
face ABEF.
ͳ_yx  the shear stress acting in the x-direction of x-axis on
face ADEH
Note: for shear stress, the first subscript denotes the plane on which
the shear stress acts and the second subscript denotes the direction
of the shear stress.
Introduction
Plane stress: state of stress in which two faces of the
cubic element are free of stresses.
Transformation of Plane Stress
• Figure 1
• Figure 3
• Figure 2
• Figure 2b
5
Transformation of Plane Stress
• Figure 8
• Figure 8
6
Transformation of Plane Stress
• Figure 9
• Figure 9
7
ΣF_x’ = 0
σ_x’ ΔA – σ _x (ΔA cosθ)cos θ - ͳ_xy (ΔA cosθ)sin θ - σ _y (ΔA sinθ)sin θ - ͳ_xy (ΔA
sinθ)cos θ = 0
ΣF_y’ = 0
ͳ_x’y’ (ΔA) + σ _x (ΔA cosθ)sin θ - ͳ_xy (ΔA cosθ)cos θ - σ _y (ΔA sinθ)cos θ + ͳ_xy (ΔA
sinθ)sin θ = 0
σ_x’ = [(σ_x + σ_y)/2) + [(σ_x - σ_y)/2] cos2θ + ͳ_xy sin2 θ ----(1)
σ_y’ = [(σ_x + σ_y)/2] - [(σ_x - σ_y)/2] cos2θ - ͳ_xy sin2 θ ------(2)
ͳ_x’y’ = -[(σ_x - σ_y)/2] sin2 θ + ͳ_xy cos2 θ ------(3)
sin2 θ = 2sin θcos θ
cos2 θ = (cos θ)^2 – (sin θ)^2
σ_x’ = [(σ_x + σ_y)/2) + [(σ_x - σ_y)/2] cos2θ + ͳ_xy sin2 θ ----(1)
σ_x’ - [(σ_x + σ_y)/2) = + [(σ_x - σ_y)/2] cos2θ + ͳ_xy sin2 θ
Taking square on both sides
[σ_x’ - [(σ_x + σ_y)/2)]^2 = [[(σ_x - σ_y)/2] cos2θ + ͳ_xy sin2 θ]^2
(a+b)^2 = (a+b)(a+b) = a^2 + b^2 +2ab
(a-b)^2 = (a-b)(a-b) = a^2 + b^2 -2ab
[σ_x’ - [(σ_x + σ_y)/2)]^2 = [[(σ_x - σ_y)/2] cos2θ + ͳ_xy sin2 θ]^2
= [(σ_x - σ_y)/2]^2 (cos2θ)^2 + ͳ_xy^2 (sin2 θ)^2 + 2[(σ_x - σ_y)/2] cos2θ * ͳ_xy sin2 θ ----(4)
ͳ_x’y’ = -[(σ_x - σ_y)/2] sin2 θ + ͳ_xy cos2 θ ----- (3)
Squaring equation 3
ͳ_x’y’^2 = ͳ_xy^2 (cos2 θ)^2 + [(σ_x - σ_y)/2]^2 (sin2 θ)^2 - 2 (ͳ_xy cos2 θ ) * [(σ_x - σ_y)/2] sin2 θ ----(5)
Adding 4 and 5:
[σ_x’ - [(σ_x + σ_y)/2)]^2 + ͳ_x’y’^2 = [(σ_x - σ_y)/2]^2 (cos2θ)^2 + ͳ_xy^2 (sin2 θ)^2 + 2[(σ_x - σ_y)/2] cos2θ *
ͳ_xy sin2 θ + ͳ_xy^2 (cos2 θ)^2 + [(σ_x - σ_y)/2]^2 (sin2 θ)^2 - 2 (ͳ_xy cos2 θ ) * [(σ_x - σ_y)/2] sin2 θ
[σ_x’ - [(σ_x + σ_y)/2)]^2 + ͳ_x’y’^2 = [(σ_x - σ_y)/2]^2 (cos2θ)^2 + ͳ_xy^2 (sin2 θ)^2 + ͳ_xy^2 (cos2 θ)^2 + [(σ_x
- σ_y)/2]^2 (sin2 θ)^2
[σ_x’ - [(σ_x + σ_y)/2)]^2 + ͳ_x’y’^2 = [(σ_x - σ_y)/2]^2 (cos2θ)^2 + ͳ_xy^2 (sin2 θ)^2 + ͳ_xy^2 (cos2 θ)^2 + [(σ_x
- σ_y)/2]^2 (sin2 θ)^2
(σ_x + σ_y)/2 = σ_ave
R = sqrt [{(σ_x + σ_y)/2}^2 + ͳ_xy^2]
(σ_x’ - σ_ave)^2 + ͳ_x’y’^2 = R^2
σ_max = σ_ave + R = (σ_x + σ_y)/2 + sqrt [{(σ_x +
σ_y)/2}^2 + ͳ_xy^2]
σ_min = σ_ave – R = (σ_x + σ_y)/2 - sqrt [{(σ_x +
σ_y)/2}^2 + ͳ_xy^2]
ͳ_max = R = sqrt [{(σ_x + σ_y)/2}^2 + ͳ_xy^2]
Tan2θ_P = 2 ͳ_xy/ (σ_x - σ_y)
Tan2θ_s = -(σ_x - σ_y)/ 2 ͳ_xy
Principal Stresses
• The previous equations are combined to yield
parametric equations for a circle,
𝜎𝑥 ′ − 𝜎ave
where
2
+ 𝜏𝑥2′ 𝑦′ = 𝑅 2
𝜎𝑥 + 𝜎𝑦
𝜎ave =
2
• Figure 10
2
2
+ 𝜏𝑥𝑦
• Principal stresses occur on the principal
planes of stress with zero shearing stresses.
 max,min 
tan 2 p 
• Figure 11
𝑅=
𝜎𝑥 − 𝜎𝑦
2
 x  y
2
  y 
2
  x
  xy

 2 
2
2 xy
 x  y
Note: defines two angles separated by 90o
11
Principal Stresses
• Figure 10
• Figure 11
Maximum Shearing Stress
• Maximum shearing
stress occurs for
 x   ave
 x  y 
2
 max  R  



xy
2


 x  y
tan 2s  
2 xy
2
•
Figure 7.7 Circular relationship of
transformed stresses.
Note: defines two angles separated by 90o and
offset from  p by 45o
    ave 
•
x  y
2
Figure 7.10 Maximum shearing stress.
13
σ_x = 50 Mpa
σ_y = -10 Mpa
τ_xy = 40 MPa
 max,min 
tan 2 p 
 x  y
2
  y 
2
  x


xy

 2 
2
2 xy
 x  y
Note: defines two angles separated by 90o
σ_max = (50 – 10)/2 + sqrt [[(50 + 10)/2]^2 + 40^2]
= 20 + 50 = 70MPa
σ_min = (50 – 10)/2 - sqrt [[(50 + 10)/2]^2 + 40^2]
= 20 - 50 = -30MPa
Tan 2θ_P = 2 x 40 / (50 + 10)
Tan 2θ_P = 80/60 = 8/6 = 4/3
2θ_P = Tan^-1 (4/3)
2θ_P = 53.1
θ_P = 26.6o, 26.6 + 90 = 116.6
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σ_x = 50 Mpa
σ_y = -10 Mpa
τ_xy = 40 MPa
𝑅=
𝜎𝑥 − 𝜎𝑦
2
2
2
+ 𝜏𝑥𝑦
Maximum shear stress = +-50
𝜎𝑥 − 𝜎𝑦
tan 2 𝜃s = −
2𝜏𝑥𝑦
Tan 2θ_s = - (50 +10)/2x40
= - 60/80 = - ¾
θ_s = -18.4, -18.4 + 90 = 71.6
(c) σ_n = (50 -10) /2 = 20 MPa
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σ_x = σ_l = Pr/2xt = (4.2 x 10^6 x 40 )/(2x 4) = 21 Mpa
σ_y = σ_c = Pr/t = 42MPa
d_ o – d_i = 2t  d_o = 2t + d_i = 88mm
τ_xy = Tr/J = (1.6x10^3 x 44 x 10^-3)/ [Pi/32 (0.088^4 –0.08^4)
τ_xy = 37.7 MPa
2
 x  y
 x  y 
2
 max,min 
 
  xy

2
 2 
tan 2 p 
2 xy
 x  y
Note: defines two angles separated by 90o
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σ_x = σ_l = Pr/2xt = (4.2 x 10^6 x 40 )/(2x 4) = 21 Mpa
σ_y = σ_c = Pr/t = 42MPa
d_ o – d_i = 2t  d_o = 2t + d_i = 88mm
τ_xy = Tr/J = (1.6x10^3 x 44 x 10^-3)/ [Pi/32 (0.088^4 –0.08^4)
τ_xy = 37.7 MPa
 max,min 
tan 2 p 
 x  y
2
 x  y 
2
 
  xy

 2 
2
2 xy
 x  y
Note: defines two angles separated by 90o
σ_max/min = (21 + 42)/2 +- sqrt [[(21 - 42)/2]^2 + 37.7^2]
= 31.5 +- 39.1
σ_max = 31.5 + 39.1 = 70.65 MPa
σ_min = 31.5 – 39.1 = -7.65
τ_max/min = +- 39.1
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 max,min 
tan 2 p 
 x  y
2
 x  y 
2
 
  xy

 2 
2
2 xy
 x  y
Note: defines two angles separated by 90o
Tan 2θ_P = 2 x 37.7/(21-42)
θ_P = -37.2, -37.2 + 90 = 52.8
𝜎𝑥 − 𝜎𝑦
tan 2 𝜃s = −
2𝜏𝑥𝑦
Tan 2θ_S = - (21-42)/2x37.7 =
θ_S = 7.8, 7.8 + 90 = 97.8
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P = ωT = 2PiN/60 x T
T = P/ (2PiN/60 ) = 200 x 1063/ (2x Pi x 800/60) = 2387.3 N-m
τ_xy = Tr/J
= 2387.3 x (d/2)/(Pi d^4/32) = 16 x 2387.3/ Pi x d^3 = 12158 / d^3
σ_x = My/I = 4000 x (d/2)/ (Pi x d^4/64) = 40744/d^3
τ_ max = 𝑅 =
𝜎𝑥
2
2
2
+ 𝜏𝑥𝑦
20
P = ωT = 2PiN/60 x T
T = P/ (2PiN/60 ) = 200 x 1063/ (2x Pi x 800/60) = 2387.3 N-m
τ_xy = Tr/J
= 2387.3 x (d/2)/(Pi d^4/32) = 16 x 2387.3/ Pi x d^3 = 12158 / d^3
σ_x = My/I = 4000 x (d/2)/ (Pi x d^4/64) = 40744/d^3
τ_ max = 𝑅 =
𝜎𝑥
2
2
2
+ 𝜏𝑥𝑦
sqrt[(40744/d^3 /2)^2 + (12158 / d^3)^2] <= 120 MPa
d >=58.3mm
σ_x = 40744/0.0583^3 = 206 MPa
τ_xy = 12158 / 0.0583^3 = 61 Mpa
σ_y = 0
 max,min 
tan 2 p 
 x  y
2
  y 
2
  x


xy

 2 
2
2 xy
 x  y
Note: defines two angles separated by 90o
21