Idea Transcript
P4 Stress and Strain
Dr. A.B. Zavatsky HT08
Lecture 8 Plane Strain and Measurement of Strain Plane stress versus plane strain. Transformation equations. Principal strains and maximum shear strains. Mohr’s circle for plane strain. Measurement of strain and strain rosettes. 1
Plane stress versus plane strain Plane Stress
Plane Strain
Stresses
σz=0, τxz=0, τyz=0 σx, σy, τxy may be non-zero.
τxz=0, τyz=0 σx, σy, σz, τxy may be non-zero.
Strains
γxz=0, γyz=0 εx, εy, εz, γxy may be non-zero.
εz=0, γxz=0, γyz=0 εx, εy, γxy may be non-zero.
Plane stress and plane strain do not ordinarily occur simultaneously. One exception is when σz = 0 and σx = -σy, since Hooke’s Law gives εz = 0. 2
Transformation Equations for Plane Strain We want to derive equations for the normal strains εx1 and εy1 and the shear strain γx1y1 associated with the x1y1 axes, which are rotated counterclockwise through an angle θ from the xy axes. Consider the change in length and orientation of the diagonal of a rectangular element in the xy plane after strains εx, εy, and γxy are applied.
y x1
εxdx cos θ
Diagonal increases in length in the x1 direction by εxdx cosθ. Diagonal rotates clockwise by α1.
y1 dy
ds θ
α1ds = ε x dx sin θ
α1
dx
εxdx
x
α1 = ε x
dx sin θ ds
3
y εydy sin θ
y1
Diagonal increases in length in the x1 direction by εydy sinθ.
x1
εydy dy
Diagonal rotates counterclockwise by α2.
α2
α 2 ds = ε y dy cos θ
ds
α2 = ε y
θ
y
x
dx
x1 γxydy cos θ
y1
dy γxy
dy cos θ ds
Diagonal increases in length in the x1 direction by γxydx cosθ. Diagonal rotates clockwise by α3.
ds θ
α 3ds = γ xy dy sin θ α3
dx
γxydy
x
α 3 = γ xy
dy sin θ ds 4
The total increase in the length of the diagonal is: Δ(ds ) = ε x dx cos θ + ε y dy sin θ + γ xy dy cos θ
The normal strain εx1 is the change in length over the original length:
Δ (ds ) dx dy dy ε x1 = = εx cos θ + ε y sin θ + γ xy cos θ ds ds ds ds ds dy
θ
dx = cos θ ds
dy = sin θ ds
dx So, the normal strain εx1 is:
ε x1 = ε x cos 2 θ + ε y sin 2 θ + γ xy sin θ cosθ The normal strain εy1 can be found by substituting θ+90° into the equation for εx1. 5
To find the shear strain γx1y1, we must find the decrease in angle of lines in the material that were initially along the x1y1 axes.
y y1 β
α
θ
x1
γ x1y1 = α + β x
To find α, we just sum α1, α2, and α3, taking the direction of the rotation into account.
α = −α1 + α 2 − α 3 dx dy dy sin θ + ε y cos θ − γ xy sin θ ds ds ds α = −ε x sin θ cosθ + ε y sin θ cos θ − γ xy sin 2 θ
α = −ε x
α = − (ε x − ε y ) sin θ cos θ − γ xy sin 2 θ 6
To find the angle β, we can substitute θ+90 into the equation for α, but we must insert a negative sign, since α is counterclockwise and β is clockwise.
β = (ε x − ε y ) sin(θ + 90) cos(θ + 90) + γ xy sin 2 (θ + 90) β = − (ε x − ε y ) sin θ cos θ + γ xy cos 2 θ So, the shear strain γx1y1 is:
γ x1 y 1 = α + β γ x1 y1 = − (ε x − ε y ) sin θ cosθ − γ xy sin 2 θ − (ε x − ε y ) sin θ cosθ − γ xy cos 2 θ γ x1 y1 = − 2 (ε x − ε y ) sin θ cosθ − γ xy (sin 2 θ − cos 2 θ ) γ x1 y 1 2
= − (ε x − ε y ) sin θ cosθ −
γ xy 2
(sin 2 θ − cos 2 θ )
Using trigonometric identities for sinθ cosθ, sin2θ, and cos2θ gives the strain transformation equations … 7
εx +εy
ε x1 = γ x1 y1 2
2
=
( ε −
x
+
εx −εy
−εy) 2
2
cos 2θ +
sin 2θ +
γ xy 2
γ xy 2
sin 2θ
cos 2θ
Now, compare the strain transformation equations to the stress transformation equations:
σ x1 = τ x1 y1 =
σ x +σ y 2
( σ −
x
+
−σ y ) 2
σ x −σ y 2
cos 2θ + τ xy sin 2θ
sin 2θ + τ xy cos 2θ
The equations have the same form, but with different variables:
ε x1 ⇔ σ x1 εx ⇔ σx εy ⇔σy
γ x1 y1 2
γ xy 2
⇔ τ x1 y1 ⇔ τ xy 8
So, the all the equations that we derived based on the stress transformation equations can be converted to equations for strains if we make the appropriate substitutions. Principal Strains and Principal Angles
ε 1, 2 =
εx +εy 2
±
⎛εx −εy ⎜⎜ ⎝ 2
⎞ ⎛ γ xy ⎞ ⎟⎟ + ⎜⎜ ⎟⎟ ⎠ ⎝ 2 ⎠ 2
2
γ xy tan 2θ p = εx −εy
Remember that εz = 0 (plane strain). Shear strains are zero on the principal planes. Principal stresses and principal strains occur in the same directions.
Maximum Shear Strains
γ max 2
=
⎛εx −εy ⎞ ⎜⎜ ⎝
2
2
⎛ γ xy ⎞
⎟⎟ + ⎜⎜ ⎟⎟ ⎠ ⎝ 2 ⎠
2
⎛εx −εy tan 2θ s = − ⎜ ⎜ γ xy ⎝
⎞ ⎟ ⎟ ⎠
The maximum shear strains are associated with axes at 45° to the directions of the principal strains. 9
Mohr’s Circle for Plane Strain Plot εx1 instead of σx1. Plot (γx1y1/2) instead of τx1y1.
ε2
ε1
c R
γmax/2
εs
εx1
Principal strains ε1, ε2
Maximum shear strain γmax with associated normal strain εs
(γx1y1/2) 10
Example An element of material in plane strain has εx = 340 x 10-6, εy = 110 x 10-6, γxy = 180 x10-6 Find the principal strains, the (in-plane) maximum shear strains, and the strains on an element oriented at an angle θ=30°.
y
Plane strain means that εz = 0.
Equations give ε1 = 371 x 10–6 ε2 = 79 x 10–6 θp = 19.0° and 109.0° γmax= 290 x 10–6 θs = -26.0° and 64.0°
γxy
The transformation equations with θ=30° give εx1 = 360 x 10-6 γx1y1 = -110 x 10-6
εy
x εx (based on Gere & Timoshenko, p 439)
Using εx + εy = εx1 + εy1 gives εy1 = 90 x 10-6
11
Mohr’s Circle c = ε avg = c=
Units on axes are strain x 10-6
εx +ε y
B (θ=90) εy = 110 x10-6 -(γxy/2) = -90 x10-6
2
340 + 110 = 225 2
B (θ=90)
R = (340 − 225) 2 + (180 / 2) 2 R = 1152 + 90 2 = 146
2θp2
ε2
ε
R
ε1,2 = c ± R = 225 ± 146 ε1,2 = 371, 79
A (θ=0)
90 340 − 225 2θ p1 = 38.05° tan 2θ p1 =
2θ p 2 = 2θ p1 + 180°
ε1
c
Principal Strains
θ p1 = 19.0°
2θp1
A (θ=0) εx = 340 x10-6 (γxy/2) = 90 x10-6
γ/2
θ p 2 = 109.0° 12
Units on axes are strain x 10-6 Maximum Shear
B (θ=90) εy = 110 x10-6 -(γxy/2) = -90 x10-6
(γ max / 2) = R = 146
γ max = 292 2θ s1 = −(90 − 2θ p1 ) = −(90 − 38.05)
B (θ=90) 2θs2
2θ s1 = −51.95
2θp1
θ s1 = −26.0°
ε
c
2θ s 2 = 2θ p1 + 90°
R
θ s 2 = 64.0°
2θs1
ε s = c = 225
(γmax/2)
A (θ=0) A (θ=0) εx = 340 x10-6 (γxy/2) = 90 x10-6
γ/2 13
Strains when θ = 30° θ = 30°
Units on axes are strain x 10-6 B (θ=90) εy = 110 x10-6 -(γxy/2) = -90 x10-6
2θ = 60°
ε x1 = c + R cos(2θ − 2θ p1 )
B (θ=90)
ε x1 = 225 + 146 cos(60 − 38.05) ε x1 = 360
C (θ=30) 2θ
ε c
(γ x1 y1 / 2) = − R sin( 2θ − 2θ p1 )
D (θ=30+90) R
(γ x1 y1 / 2) = −146 sin(60 − 38.05)
A (θ=0)
(γ x1 y1 / 2) = −55
γ x1 y1 = −110
A (θ=0) εx = 340 x10-6 (γxy/2) = 90 x10-6
ε y1 = c − R cos(2θ − 2θ p1 ) ε y1 = 225 − 146 cos(60 − 38.05) ε y1 = 90
2θp1
γ/2 14
Principal Strains y
y1
Maximum Shear Strain
No shear strains
y εs
ε2
x1 θp2 θp1
ε1
ε1 = 371 x 10–6 ε2 = 79 x 10 –6 θp1 = 19.0° θp2 = 109.0°
y1
γmax
x
x εs γmax= 290 x 10 –6 θsmax = -26.0° εs= 225 x 10 –6
θsmax
x1
15
Strains when θ = 30° y
y1
x1
εy1
γx1y1
θ
εx1
x
εx1 = 360 x 10-6 εy1 = 90 x 10-6 γx1y1 = -110 x 10-6
16
Measurement of Strain • It is very difficult to measure normal and shear stresses in a body, particularly stresses at a point. • It is relatively easy to measure the strains on the surface of a body (normal strains, that is, not shear strains). • From three independent measurements of normal strain at a point, it is possible to find principal strains and their directions. • If the material obeys Hooke’s Law, the principal strains can be used to find the principal stresses. • Strain measurement can be direct (using electrical-type gauges based on resistive, capacitive, inductive, or photoelectric principles) or indirect (using optical methods, such as photoelasticity, the Moiré technique, or holographic interferometry). 17
Resistance Strain Gauges • Based on the idea that the resistance of a metal wire changes when the wire is subjected to mechanical strain (Lord Kelvin, 1856). When a wire is stretched, a longer length of smaller sectioned conductor results. • The earliest strain gauges were of the “unbonded” type and used pillars, separated by the gauge length, with wires stretched between them.
Lo • Later gauges were “bonded”, with the resistance element applied directly to the surface of the strained member. backing wire grid expanded backing view bonded to surface 18
During the 1950s, foil-type gauges began to replace the wire-type. The foil-type gauges typically consist of a metal film element on a thin epoxy support and are made using printed-circuit techniques. Foil-type gauges can be made in a number of configurations (examples from www.vishay.com): planar single element three-element rosette alignment (0°- 45°- 90°) marks solder tabs for wires Gauge length is typically around 1 mm. Performance of bonded metallic strain gauges depends on: grid material and configuration, backing material, bonding material and method, gauge protection, and associated electrical circuitry. 19
It is possible to derive an equation relating strain ε and the change in resistance of the gauge ΔR:
1 ΔR ε= F R
F = gauge factor (related to Poisson’s ratio and resistivity) R = resistance of the gauge
A typical strain gauge might have F = 2.0 and R = 120 Ω and be used to measure microstrain (10-6).
ΔR = ε F R = (10 −6 )( 2.0)(120) = 0.00024 Ω This is a resistance change of 0.0002%, meaning that something more sensitive than an ohmmeter is required to measure the resistance change. Some form of bridge arrangement (such as a Wheatstone bridge) is most widely used.
cantilever R1 tension R2 compression (Perry & Lissner)20
Strain Rosettes and Principal Strains and Stresses A “0°-60°-120°” strain gauge rosette is bonded to the surface of a thin steel plate. Under one loading condition, the strain measurements are εA = 60 με, εB = 135 με, εC = 264 με. Find the principal strains, their orientations, and the principal stresses.
C
120
We can use more than one approach to find the principal stresses: transformation equations alone, Mohr’s circle alone, or a combination.
B
o
60o A
x (Based on Hibbeler, ex. 15.20 & 15.21) 21
Transformation equations From the measured strains, find εx, εy, and γxy.
εA = 60 με, θA = 0° εB = 135 με, θB = 60° εC = 264 με, θC = 120° ε A = 60 = ε x cos 2 0° + ε y sin 2 0° + γ xy sin 0° cos 0° ε A = 60 = ε x ε B = 135 = ε x cos 2 60° + ε y sin 2 60° + γ xy sin 60° cos 60° ε B = 135 = 0.25 ε x + 0.75 ε y + 0.433 γ xy ε C = 264 = ε x cos 2 120° + ε y sin 2 120° + γ xy sin 120° cos120° ε C = 264 = 0.25 ε x + 0.75 ε y − 0.433 γ xy
3 equations, 3 unknowns Solve to find εx = 60 με, εy = 246 με, γxy = -149 με 22
Use εx, εy, and γxy in the equations for principal strains to find ε1 = 272 με, θp1 = -70.6°, ε2 = 34 με, θp2 = 19.4°. Alternatively, use εx, εy, and γxy to construct the Mohr’s circle for (in-plane) strains and find principal strains and angles.
c = (60+246)/2 = 153 με 2θp1
A
ε2 2θp2
ε1 c R D
A: εx = 60 με (γxy/2)= -74.5 με
ε
D: εy = 246 με (γxy/2)= +74.5 με R = 119 με
γ/2 23
To find the principal stresses, use Hooke’s Law for plane stress (σz = 0)
σx = σy =
E 1 −ν E
2
1 −ν 2
(ε x + νε y )
εx = ε1 = 272 x 10-6 εy = ε2 = 34 x 10-6
(ε y + νε x )
E = 210 GPa ν = 0.3
So, the principal stresses are: σx = σ1 = 65 MPa σy = σ2 = 26 MPa
C
B ε2 19.4o -70.6
o
A
x
ε1 24
εA = 60 με, εB = 135 με, εC = 264 με.
Mohr’s Circle C
B
R? 120
o
2θ
60o A
x
εA = 60 = c + R cos 2θ εB = 135 = c + R cos 2(θ+60) εC = 264 = c + R cos 2(θ+120) 3 equations, 3 unknowns
εA 120o
εB
c?
ε
εC
240o
γ/2
Solve the equations to get c = 153, R = 119, and 2θ = 141.3° When you solve for 2θ, you may get –38.7°. But we have drawn the diagram above such that 2θ is positive, so you should take 2θ = -38.7° + 180° = 141.3°. Next, draw the Mohr’s circle and find principal strains as before. Finally, find principal stresses using Hooke’s Law. 25