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Draft 32 KINEMATIC thus the preceding equations can be combined to yield df = F ˜ Tn 0 dA 0 (4.157) we also have from Eq. 4.148 and 4.149 df = t 0 dA 0 = T 0 n 0 ddA 0 (4.158) and comparing the last two equations we note that ˜ T = F −1 T 0 (4.159) which gives the relationship between the first Piola-Kirchoff stress tensor T 0 and the second Piola- Kirchoff stress tensor ˜ T. 105 Finally the relation between the second Piola-Kirchoff stress tensor and the Cauchy stress tensor can be obtained from the preceding equation and Eq. 4.153 ˜ T = (det F) F −1 T F −1 T (4.160) and we note that this second Piola-Kirchoff stress tensor is always symmetric (if the Cauchy stress tensor is symmetric). It can also be shown that it is energitically correct. 106 To determine the corresponding stress vector, we solve for ˜ T first, then for dA 0 and n 0 from dA 0 n 0 = 1 det F F T n (assuming unit area dA), and finally ˜ t = ˜ Tn 0 . Example 4-14: Piola-Kirchoff Stress Tensors 4.7 Hydrostatic and Deviatoric Strain 85 The lagrangian and Eulerian linear strain tensors can each be split into spherical and deviator tensor as was the case for the stresses. Hence, if we define 1 3 e = 1 3 tr E (4.161) then the components of the strain deviator E are given by E ij = E ij − 1 3 eδ ij or E = E − 1 3 e1 (4.162) We note that E measures the change in shape of an element, while the spherical or hydrostatic strain 1 3 e1 represents the volume change. Victor Saouma Mechanics of Materials II Draft 4.7 Hydrostatic and Deviatoric Strain 33 Piola−Kirchoff Stress Tensors The deformed configuration of a body is described by x 1 =X 1 ê 2, x 2 =−X 2 /2, x 3 =4X 3 ; If the Cauchy stress tensor is given by i k j j j j j j j j 100 0 0 0 00 0 0 0 y { z z z z z z z z MPa; What are the corresponding first and second Piola−Kirchoff stress tensors, and calculate the respective stress tensors on the e 3 plane in the deformed state. ‡ F tensor CST = 880, 0, 0<, 80, 0, 0<, 80, 0, 100<< 880, 0, 0<, 80, 0, 0<, 80, 0, 100<< F = 881 ê 2, 0, 0<, 80, 0, − 1 ê 2<, 80, 4, 0<< 99 1 ÄÄÄÄ Ä 2 ,0,0=, 90, 0, - 1 ÄÄÄÄ Ä 2 =, 80, 4, 0<= Finverse = Inverse@FD 982, 0, 0<, 90, 0, 1 ÄÄÄÄ Ä 4 =, 80, - 2, 0<= ‡ First Piola−Kirchoff Stress Tensor Tfirst = Det@FD CST . Transpose@FinverseD 880, 0, 0<, 80, 0, 0<, 80, 25, 0<< MatrixForm@%D i k j j j j j j 000 000 0250 y { z z z z z z ‡ Second Piola−Kirchoff Stress Tensor Tsecond = Inverse@FD . Tfirst 980, 0, 0<, 90, 25 ÄÄÄÄÄÄÄÄ 4 ,0=, 80, 0, 0<= MatrixForm@%D i k j j j j j j j j 000 0 25 ÄÄÄÄÄÄ 4 0 000 y { z z z z z z z z ‡ Cuchy stress vector Can be obtained from t=CST n tcauchy = MatrixForm@CST . 80, 0, 1<D i k j j j j j j 0 0 100 y { z z z z z z ‡ Pseudo−Stress vector associated with the First Piola−Kirchoff stress tensor For a unit area in the deformed state in the e 3 direction, its undeformed area dA 0 n 0 is given by dA 0 n 0 = F T n ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ det F detF = Det@FD 1 n = 80, 0, 1< 80, 0, 1< 2 m−piola.nb MatrixForm@Transpose@FD .nê detFD i k j j j j j j 0 4 0 y { z z z z z z Thus n 0 =e 2 and using t 0 =T 0 n 0 we obtain t01st = MatrixForm@Tfirst . 80, 1, 0<D i k j j j j j j 0 0 25 y { z z z z z z We note that this vector is in the same direction as the Cauchy stress vector, its magnitude is one fourth of that of the Cauchy stress vector, because the undeformed area is 4 times that of the deformed area ‡ Pseudo−Stress vector associated with the Second Piola−Kirchoff stress tensor t0second = MatrixForm@Tsecond . 80, 1, 0<D i k j j j j j j j j 0 25 ÄÄÄÄÄÄ 4 0 y { z z z z z z z z We see that this pseudo stress vector is in a different direction from that of the Cauchy stress vector (and we note that the tensor F transforms e 2 into e 3 ). m−piola.nb 3 Victor Saouma Mechanics of Materials II Draft 34 KINEMATIC ε ε I εε γ 2 III II Figure 4.8: Mohr Circle for Strain 4.8 Principal Strains, Strain Invariants, Mohr Circle 86 Determination of the principal strains (E (3) <E (2) <E (1) , strain invariants and the Mohr circle for strain parallel the one for stresses (Sect. 2.3) and will not be repeated here. λ 3 − I E λ 2 − II E λ − III E =0 (4.163) where the symbols I E , II E and III E denote the following scalar expressions in the strain components: I E = E 11 + E 22 + E 33 = E ii =trE (4.164) II E = −(E 11 E 22 + E 22 E 33 + E 33 E 11 )+E 2 23 + E 2 31 + E 2 12 (4.165) = 1 2 (E ij E ij − E ii E jj )= 1 2 E ij E ij − 1 2 I 2 E (4.166) = 1 2 (E : E − I 2 E ) (4.167) III E = detE = 1 6 e ijk e pqr E ip E jq E kr (4.168) 87 In terms of the principal strains, those invariants can be simplified into I E = E (1) + E (2) + E (3) (4.169) II E = −(E (1) E (2) + E (2) E (3) + E (3) E (1) ) (4.170) III E = E (1) E (2) E (3) (4.171) 88 The Mohr circle uses the Engineering shear strain definition of Eq. 4.91, Fig. 4.8 Example 4-15: Strain Invariants & Principal Strains Victor Saouma Mechanics of Materials II Draft 4.8 Principal Strains, Strain Invariants, Mohr Circle 35 Determine the planes of principal strains for the following strain tensor 1 √ 30 √ 300 001 (4.172) Solution: The strain invariants are given by I E = E ii = 2 (4.173-a) II E = 1 2 (E ij E ij − E ii E jj )=−1 + 3 = +2 (4.173-b) III E = |E ij | = −3 (4.173-c) The principal strains by E ij − λδ ij = 1 − λ √ 30 √ 3 −λ 0 001− λ (4.174-a) =(1− λ) λ − 1+ √ 13 2 λ − 1 − √ 13 2 (4.174-b) E (1) = λ (1) = 1+ √ 13 2 =2.3 (4.174-c) E (2) = λ (2) = 1 (4.174-d) E (3) = λ (3) = 1 − √ 13 2 = −1.3 (4.174-e) The eigenvectors for E (1) = 1+ √ 13 2 give the principal directions n (1) : 1 − 1+ √ 13 2 √ 30 √ 3 − 1+ √ 13 2 0 001− 1+ √ 13 2 n (1) 1 n (1) 2 n (1) 3 = 1 − 1+ √ 13 2 n (1) 1 + √ 3n (1) 2 √ 3n (1) 1 − 1+ √ 13 2 n (1) 2 1 − 1+ √ 13 2 n (1) 3 = 0 0 0 (4.175) which gives n (1) 1 = 1+ √ 13 2 √ 3 n (1) 2 (4.176-a) n (1) 3 = 0 (4.176-b) n (1) ·n (1) = 1+2 √ 13 + 13 12 +1 n (1) 2 2 =1⇒ n 1 2 =0.8; (4.176-c) ⇒ n (1) = 0.80.60 (4.176-d) For the second eigenvector λ (2) =1: 1 − 1 √ 30 √ 3 −10 001− 1 n (2) 1 n (2) 2 n (2) 3 = √ 3n (2) 2 √ 3n (2) 1 − n (2) 2 0 = 0 0 0 (4.177) which gives (with the requirement that n (2) ·n (2) =1) n (2) = 001 (4.178) Victor Saouma Mechanics of Materials II Draft 36 KINEMATIC Finally, the third eigenvector can be obrained by the same manner, but more easily from n (3) = n (1) ×n (2) = det e 1 e 2 e 3 0.80.60 001 =0.6e 1 − 0.8e 2 (4.179) Therefore a j i = n (1) n (2) n (3) = 0.80.60 001 0.6 −0.80 (4.180) and this results can be checked via [a][E][a] T = 0.80.60 001 0.6 −0.80 1 √ 30 √ 300 001 0.80 0.6 0.60−0.8 01 0 = 2.30 0 01 0 00−1.3 (4.181) Example 4-16: Mohr’s Circle Construct the Mohr’s circle for the following plane strain case: 00 0 05 √ 3 0 √ 33 (4.182) Solution: 1 2 1 3456 60 o 2 B D E F ε ε s n 2 3 We note that since E (1) = 0 is a principal value for plane strain, ttwo of the circles are drawn as shown. Victor Saouma Mechanics of Materials II Draft 4.9 Initial or Thermal Strains 37 4.9 Initial or Thermal Strains 89 Initial (or thermal strain) in 2D: ε ij = α∆T 0 0 α∆T Plane Stress =(1+ν) α∆T 0 0 α∆T Plane Strain (4.183) note there is no shear strains caused by thermal expansion. 4.10 † Experimental Measurement of Strain 90 Typically, the transducer to measure strains in a material is the strain gage. The most common type of strain gage used today for stress analysis is the bonded resistance strain gage shown in Figure 4.9. Figure 4.9: Bonded Resistance Strain Gage 91 These gages use a grid of fine wire or a metal foil grid encapsulated in a thin resin backing. The gage is glued to the carefully prepared test specimen by a thin layer of epoxy. The epoxy acts as the carrier matrix to transfer the strain in the specimen to the strain gage. As the gage changes in length, the tiny wires either contract or elongate depending upon a tensile or compressive state of stress in the specimen. The cross sectional area will increase for compression and decrease in tension. Because the wire has an electrical resistance that is proportional to the inverse of the cross sectional area, Rα 1 A , a measure of the change in resistance can be converted to arrive at the strain in the material. 92 Bonded resistance strain gages are produced in a variety of sizes, patterns, and resistance. One type of gage that allows for the complete state of strain at a point in a plane to be determined is a strain gage rosette. It contains three gages aligned radially from a common point at different angles from each other, as shown in Figure 4.10. The strain transformation equations to convert from the three strains a t any angle to the strain at a point in a plane are: a = x cos 2 θ a + y sin 2 θ a + γ xy sin θ a cos θ a (4.184) b = x cos 2 θ b + y sin 2 θ b + γ xy sin θ b cos θ b (4.185) c = x cos 2 θ c + y sin 2 θ c + γ xy sin θ c cos θ c (4.186) 93 When the measured strains a , b ,and c , are measured at their corresponding angles from the reference axis and substituted into the above equations the state of strain at a point may be solved, namely, x , y ,andγ xy . In addition the principal strains may then be computed by Mohr’s circle or the principal strain equations. 94 Due to the wide variety of styles of gages, many factors must be considered in choosing the right gage for a particular application. Operating temperature, state of strain, and stability of installation all influence gage selection. Bonded resistance strain gages are well suited for making accurate and practical strain measurements because of their high sensitivity to strains, low cost, and simple operation. Victor Saouma Mechanics of Materials II Draft 38 KINEMATIC Figure 4.10: Strain Gage Rosette 95 The measure of the change in electrical resistance when the strain gage is strained is known as the gage factor. The gage factor is defined as the fractional change in resistance divided by the fractional change in length along the axis of the gage. GF = ∆R R ∆L L Common gage factors are in the range of 1.5-2 for most resistive strain gages. 96 Common strain gages utilize a grid pattern as opposed to a straight length of wire in order to reduce the gage length. This grid pattern causes the gage to be sensitive to deformations transverse to the gage length. Therefore, corrections for transverse strains should be computed and applied to the strain data. Some gages come with the tranverse correction calculated into the gage factor. The transverse sensitivity factor, K t , is defined as the transverse gage factor divided by the longitudinal gage factor. K t = GF transverse GF longitudinal These sensitivity values are expressed as a percentage and vary from zero to ten percent. 97 A final consideration for maintaining accurate strain measurement is temperature compensation. The resistance of the gage and the gage factor will change due to the variation of resistivity and strain sen- sitivity with temperature. Strain gages are produced with different temperature expansion coefficients. In order to avoid this problem, the expansion coefficient of the strain gage should match that of the specimen. If no large temperature change is expected this may be neglected. 98 The change in resistance of bonded resistance strain gages for most strain measurements is very small. From a simple calculation, for a strain of 1 µ (µ =10 −6 ) with a 120 Ω gage and a gage factor of 2, the change in resistance produced by the gage is ∆R =1× 10 −6 × 120 × 2 = 240 × 10 −6 Ω. Furthermore, it is the fractional change in resistance that is important and the number to be measured will be in the order of a couple of µ ohms. For large strains a simple multi-meter may suffice, but in order to acquire sensitive measurements in the µΩ range a Wheatstone bridge circuit is necessary to amplify this resistance. The Wheatstone bridge is described next. 4.10.1 Wheatstone Bridge Circuits 99 Due to their outstanding sensitivity, Wheatstone bridge circuits are very advantageous for the mea- surement of resistance, inductance, and capacitance. Wheatstone bridges are widely used for strain measurements. A Wheatstone bridge is shown in Figure 4.11. It consists of 4 resistors arranged in a diamond orientation. An input DC voltage, or excitation voltage, is applied between the top and bottom of the diamond and the output voltage is measured across the middle. When the output voltage is zero, the bridge is said to be balanced. One or more of the legs of the bridge may be a resistive transducer, such as a strain gage. The other legs of the bridge are simply completion resistors with resistance equal to that of the strain gage(s). As the resistance of one of the legs changes, by a change in strain from a resistive strain gage for example, the previously balanced bridge is now unbalanced. This unbalance causes a voltage to appear across the middle of the bridge. This induced voltage may be measured with a voltmeter or the resistor in the opposite leg may be adjusted to re-balance the bridge. In either case the change in resistance that caused the induced voltage may be measured and converted to obtain the Victor Saouma Mechanics of Materials II Draft 4.10 † Experimental Measurement of Strain 39 engineering units of strain. Figure 4.11: Quarter Wheatstone Bridge Circuit 4.10.2 Quarter Bridge Circuits 100 If a strain gage is oriented in one leg of the circuit and the other legs contain fixed resistors as shown in Figure 4.11, the circuit is known as a quarter bridge circuit. The circuit is balanced when R 1 R 2 = R gage R 3 . When the circuit is unbalanced V out = V in ( R 1 R 1 +R 2 − R gage R gage +R 3 ). 101 Wheatstone bridges may also be formed with two or four legs of the bridge being composed of resistive transducers and are called a half bridge and full bridge respectively. Depending upon the type of application and desired results, the equations for these circuits will vary as shown in Figure 4.12. Here E 0 is the output voltage in mVolts, E is the excitation voltage in Volts, is strain and ν is Poisson’s ratio. 102 In order to illustrate how to compute a calibration factor for a particular experiment, suppose a single active gage in uniaxial compression is used. This will correspond to the upper Wheatstone bridge configuration of Figure 4.12. The formula then is Victor Saouma Mechanics of Materials II Draft 40 KINEMATIC Figure 4.12: Wheatstone Bridge Configurations Victor Saouma Mechanics of Materials II Draft 4.10 † Experimental Measurement of Strain 41 E 0 E = F(10 −3 ) 4+2F(10 −6 ) (4.187) 103 The extra term in the denominator 2F(10 −6 ) is a correction factor for non-linearity. Because this term is quite small compared to the other term in the denominator it will be ignored. For most measurements a gain is necessary to increase the output voltage from the Wheatstone bridge. The gain relation for the output voltage may be written as V = GE 0 (10 3 ), where V is now in Volts. so Equation 4.187 becomes V EG(10 3 ) = F(10 −3 ) 4 V = 4 FEG (4.188) 104 Here, Equation 4.188 is the calibration factor in units of strain per volt. For common values where F =2.07, G = 1000, E = 5, the calibration factor is simply 4 (2.07)(1000)(5) or 386.47 microstrain per volt. Victor Saouma Mechanics of Materials II [...]... Saouma Mechanics of Materials II Draft 6 6.3.2 FUNDAMENTAL LAWS of CONTINUUM MECHANICS †Moment of Momentum Principle The moment of momentum principle states that the time rate of change of the total moment of momentum of a given set of particles equals the vector sum of the moments of all external forces acting on the particles of the set 25 26 Thus, in the absence of distributed couples (this theory of. .. condition of incompressibility 6.3 6.3.1 Linear Momentum Principle; Equation of Motion Momentum Principle The momentum principle states that the time rate of change of the total momentum of a given set of particles equals the vector sum of all external forces acting on the particles of the set, provided Newton’s Third Law applies The continuum form of this principle is a basic postulate of continuum mechanics. .. integral of the divergence of that function over the volume enclosed by the surface 8 For 2D-1D transformations, we have ∇·qdA = A 9 qT nds (5. 9) s This theorem is sometime refered to as Green’s theorem in space 5. 4.1 †Green-Gauss ΦvT ndΓ − Φ∇·vdΩ = Ω Γ (∇Φ)T vdΩ (5. 10) ∂Φ ΨdΩ ∂x (5. 11) Ω If we select vT = [ Ψ 0 0 ], we obtain Φ Ω Victor Saouma ∂Ψ dΩ = ∂x ΦΨnx dΓ − Γ Ω Mechanics of Materials II Draft 5. 5... Theorem 5. 5 10 3 Stoke’s Theorem Stoke’s theorem states that A·dr = (∇×A)·ndS = C (5. 12) (∇×A)·dS S S where S is an open surface with two faces confined by C Stoke’s theorem says that the line integral of the tangential component of a vector function over some closed path equals the surface integral of the normal component of the curl of that function integrated over any capping surface of the path 5. 5.1... medium of density ρ fills the volume at time t, then the total mass in V is ρ(x, t)dV M= (6.6) V where ρ(x, t) is a continuous function called the mass density We note that this spatial form in terms of x is most common in fluid mechanics 15 The rate of increase of the total mass in the volume is ∂M = ∂t V ∂ρ dV ∂t (6.7) 16 The Law of conservation of mass requires that the mass of a specific portion of the... 17 FUNDAMENTAL LAWS of CONTINUUM MECHANICS The chain rule will in turn give ∂(ρvi ) ∂ρ ∂vi =ρ + vi ∂xi ∂xi ∂xi (6.11) It can be shown that the rate of change of the density in the neighborhood of a particle instantaneously at x by ∂ρ ∂ρ ∂ρ dρ = + v·∇ρ = + vi (6.12) dt ∂t ∂t ∂xi 18 where the first term gives the local rate of change of the density in the neighborhood of the place of x, while the second... coordinates This equation shows that the divergence of the velocity vector field equals (−1/ρ)(dρ/dt) and measures the rate of flow of material away from the particle and is equal to the unit rate of decrease of density ρ in the neighborhood of the particle If the material is incompressible, so that the density in the neighborhood of each material particle remains constant as it moves, then the continuity... = 0 C 5. 3 5 Integration by Parts The integration by part formula is b a 5. 4 6 b b u(x)v (x)dx = u(x)v(x)|a − v(x)u (x)dx (5. 6) a Gauss; Divergence Theorem In the most general case we have δF = Ω F (5. 7) δΩ The divergence theorem (also known as Ostrogradski’s Theorem) comes repeatedly in solid mechanics and can be stated as follows: 7 ∇·vdΩ = Ω v.ndΓ or vi,i dΩ = Γ Ω vi ni dΓ (5. 8) Γ The flux of a vector... 2 ∂x 2 ∂x ∂vx ∆x∆y∆z ∂x (5. 15- a) (5. 15- b) Similarly ∆Vy ∆Vz ∂vy ∆x∆y∆z ∂y ∂vz ∆x∆y∆z ∂z = = (5. 16-a) (5. 16-b) Hence, the total increase per unit volume and unit time will be given by ∂vx ∂x + ∂vy ∂y ∂vz ∂z + ∆x∆y∆z ∆x∆y∆z = div v = ∇·v (5. 17) Furthermore, if we consider the total of fluid crossing dS during ∆t, Fig 5. 1-b, it will be given by (v∆t)·ndS = v·ndS∆t or the volume of fluid crossing dS per unit... crossing dS per unit time is v·ndS Thus for an arbitrary volume, Fig 5. 1-c, the total amount of fluid crossing a closed surface S per unit time is ∇·vdV (Eq 5. 17), thus v·ndS But this is equal to S V ∇·vdV v·ndS = S (5. 18) V which is the divergence theorem Victor Saouma Mechanics of Materials II Draft Chapter 6 FUNDAMENTAL LAWS of CONTINUUM MECHANICS 6.1 Introduction We have thus far studied the stress tensors . Saouma Mechanics of Materials II Draft 6 FUNDAMENTAL LAWS of CONTINUUM MECHANICS 6.3.2 †Moment of Momentum Principle 25 The moment of momentum principle states that the time rate of change of the. − Ω ∂Φ ∂x ΨdΩ (5. 11) Victor Saouma Mechanics of Materials II Draft 5. 5 Stoke’s Theorem 3 5. 5 Stoke’s Theorem 10 Stoke’s theorem states that C A·dr = S (∇×A)·ndS = S (∇×A)·dS (5. 12) where. P 2 (5. 5-a) C A·dr = 0 along a closed contour line (5. 5-b) 5. 3 Integration by Parts 5 The integration by part formula is b a u(x)v (x)dx = u(x)v(x)| b a − b a v(x)u (x)dx (5. 6) 5. 4 Gauss;
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