P4 Stress and Strain Dr. A.B. Zavatsk HT08 Lecture 8 Plane Strain and Measurement of Strain Plane stress versus plane strain. Transformation equations. Principal strains and maimum shear strains. Mohr s circle for plane strain. Measurement of strain and strain rosettes.
Plane stress versus plane strain Stresses Plane Stress σ z 0, τ z 0, τ z 0 σ, σ, τ ma be non-zero. Plane Strain τ z 0, τ z 0 σ, σ, σ z, τ ma be non-zero. Strains z 0, z 0,, z, ma be non-zero. z 0, z 0, z 0,, ma be non-zero. Plane stress and plane strain do not ordinaril occur simultaneousl. One eception is when σ z 0 and σ -σ, since Hooke s Law gives z 0.
Transformation Equations for Plane Strain We want to derive equations for the normal strains and and the shear strain associated with the aes, which are rotated counterclockwise through an angle from the aes. Consider the change in length and orientation of the diagonal of a rectangular element in the plane after strains,, and are applied. d cos Diagonal increases in length in the direction b d cos. Diagonal rotates clockwise b α. d ds d α d α ds d α d ds sin sin 3
d sin Diagonal increases in length in the direction b d sin. d Diagonal rotates counterclockwise b α. d α ds d d cos α ds d d α ds cos cos Diagonal increases in length in the direction b d cos. Diagonal rotates clockwise b α 3. d ds α ds 3 d sin d α 3 d α 3 d ds sin 4
The total increase in the length of the diagonal is: Δ( ds) d cos + d sin d + cos The normal strain is the change in length over the original length: Δ( ds) d d cos + sin + ds ds ds d ds cos ds d d ds d cos sin ds d So, the normal strain is: cos + sin + sin cos The normal strain can be found b substituting +90 into the equation for. 5
To find the shear strain, we must find the decrease in angle of lines in the material that were initiall along the aes. β α α + β α α + α α d d d α sin + cos sin ds ds ds α sin cos + sin cos sin To find α, we just sum α, α, and α 3, taking the direction of the rotation into account. α ( )sin cos 3 sin 6
To find the angle β, we can substitute +90 into the equation for α, but we must insert a negative sign, since α is counterclockwise and β is clockwise. β ( )sin( + 90) cos( + 90) + β ( )sin cos + cos sin ( + 90) So, the shear strain is: α + β ( )sin cos ( )sin cos ( )sin cos sin (sin (sin ( )sin cos cos cos Using trigonometric identities for sin cos, sin, and cos gives the strain transformation equations ) ) cos 7
8 The equations have the same form, but with different variables: ( ) cos sin sin cos + + + + Now, compare the strain transformation equations to the stress transformation equations: ( ) τ σ σ τ τ σ σ σ σ σ cos sin sin cos + + + + σ σ σ τ τ
9 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. Remember that z 0 (plane strain). Shear strains are zero on the principal planes. Principal stresses and principal strains occur in the same directions. Maimum Shear Strains ma + s tan The maimum shear strains are associated with aes at 45 to the directions of the principal strains. Principal Strains and Principal Angles, + ± + p tan
Mohr s Circle for Plane Strain Plot instead of σ. Plot ( /) instead of τ. ma / c R Principal strains, s Maimum shear strain ma with associated normal strain s ( /) 0
Eample An element of material in plane strain has 340 0-6, 0 0-6, 80 0-6 Find the principal strains, the (in-plane) maimum shear strains, and the strains on an element oriented at an angle 30. Plane strain means that z 0. Equations give 37 0 6 79 0 6 p 9.0 and 09.0 ma 90 0 6 s -6.0 and 64.0 The transformation equations with 30 give 360 0-6 -0 0-6 (based on Gere & Timoshenko, p 439) Using + + gives 90 0-6
Mohr s Circle c 340 + 0 c R R,, avg (340 5) 5 + + 90 5 + (80 / ) 46 Principal Strains c ± R 5 ± 46 37, 79 90 tan p 340 5 38.05 p p p p 9.0 p 09.0 + 80 Units on aes are strain 0-6 B (90) / p c R B (90) 0 0-6 -( /) -90 0-6 p A (0) A (0) 340 0-6 ( /) 90 0-6
Maimum Shear ( ma ma s s s s s / ) R 46 9 (90 5.95 6.0 s p 64.0 c 5 p + 90 ) (90 38.05) Units on aes are strain 0-6 B (90) / c s ( ma /) s p R B (90) 0 0-6 -( /) -90 0-6 A (0) A (0) 340 0-6 ( /) 90 0-6 3
Strains when 30 30 60 ( ( ( c + R cos( p 5 + 46 cos(60 38.05) 360 / ) R sin( / ) 46 sin(60 38.05) / ) 55 0 c R cos( p ) p 5 46 cos(60 38.05) 90 ) ) Units on aes are strain 0-6 B (90) / c D (30+90) R B (90) 0 0-6 -( /) -90 0-6 p C (30) A (0) A (0) 340 0-6 ( /) 90 0-6 4
Principal Strains Maimum Shear Strain No shear strains s p ma p 37 0 6 79 0 6 p 9.0 p 09.0 s ma 90 0 6 sma -6.0 s 5 0 6 sma 5
Strains when 30 360 0-6 90 0-6 -0 0-6 6
Measurement of Strain It is ver difficult to measure normal and shear stresses in a bod, particularl stresses at a point. It is relativel eas to measure the strains on the surface of a bod (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 obes Hooke s Law, the principal strains can be used to find the principal stresses. Strain measurement can be direct (using electrical-tpe gauges based on resistive, capacitive, inductive, or photoelectric principles) or indirect (using optical methods, such as photoelasticit, the Moiré technique, or holographic interferometr). 7
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, 856). When a wire is stretched, a longer length of smaller sectioned conductor results. The earliest strain gauges were of the unbonded tpe and used pillars, separated b the gauge length, with wires stretched between them. L o Later gauges were bonded, with the resistance element applied directl to the surface of the strained member. epanded view backing wire grid backing bonded to surface 8
During the 950s, foil-tpe gauges began to replace the wire-tpe. The foil-tpe gauges tpicall consist of a metal film element on a thin epo support and are made using printed-circuit techniques. Foil-tpe gauges can be made in a number of configurations (eamples from www.visha.com): single element alignment marks solder tabs for wires Gauge length is tpicall around mm. planar three-element rosette (0-45 - 90 ) Performance of bonded metallic strain gauges depends on: grid material and configuration, backing material, bonding material and method, gauge protection, and associated electrical circuitr. 9
It is possible to derive an equation relating strain and the change in resistance of the gauge ΔR: F ΔR R F gauge factor (related to Poisson s ratio and resistivit) R resistance of the gauge A tpical strain gauge might have F.0 and R 0 Ω and be used to measure microstrain (0-6 ). ΔR F R (0 This is a resistance change of 0.000%, 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 widel used. cantilever R tension 6 )(.0)(0) 0.0004 Ω R compression (Perr & Lissner) 0
Strain Rosettes and Principal Strains and Stresses A 0-60 -0 strain gauge rosette is bonded to the surface of a thin steel plate. Under one loading condition, the strain measurements are A 60 μ, B 35 μ, C 64 μ. Find the principal strains, their orientations, and the principal stresses. C B We can use more than one approach to find the principal stresses: transformation equations alone, Mohr s circle alone, or a combination. 0 o 60 o A (Based on Hibbeler, e. 5.0 & 5.)
Transformation equations From the measured strains, find,, and. A 60 μ, A 0 B 35 μ, B 60 C 64 μ, C 0 A A B B C C 60 60 35 35 0.5 64 cos cos cos 64 0.5 0 + 60 + + 0.75 + 0.75 0 + sin sin + 0.433 0 + sin 60 + 0.433 0 + sin 0 cos0 sin 60 cos60 sin0 cos0 3 equations, 3 unknowns Solve to find 60 μ, 46 μ, -49 μ
Use,, and in the equations for principal strains to find 7 μ, p -70.6, 34 μ, p 9.4. Alternativel, use,, and to construct the Mohr s circle for (in-plane) strains and find principal strains and angles. A p p c R D c (60+46)/ 53 μ A: 60 μ ( /) -74.5 μ D: 46 μ ( /) +74.5 μ R 9 μ / 3
To find the principal stresses, use Hooke s Law for plane stress (σ z 0) σ σ E ν E ν ( ( + ν + ν ) ) 7 0-6 34 0-6 E 0 GPa ν 0.3 So, the principal stresses are: σ σ 65 MPa σ σ 6 MPa C B 9.4 o -70.6 o A 4
Mohr s Circle A 60 μ, B 35 μ, C 64 μ. C B R? 0 o 60 o A A 60 c + R cos B 35 c + R cos (+60) C 64 c + R cos (+0) A B c? C 0 o 40 o 3 equations, 3 unknowns Solve the equations to get c 53, R 9, and 4.3 When ou solve for, ou ma get 38.7. But we have drawn the diagram above such that is positive, so ou should take -38.7 + 80 4.3. Net, draw the Mohr s circle and find principal strains as before. Finall, find principal stresses using Hooke s Law. / 5