Mathematical Background in Aircraft Structural Mechanics
CHAPTER 1. Linear Elasticity
SangJoon Shin School of Mechanical and Aerospace Engineering Seoul National University
Active Aeroelasticityand Rotorcraft Lab. Basic equation of Linear Elasticity
Structural analysis … evaluation of deformations and stresses arising within a solid object under the action of applied loads
- if time is not explicitly considered as an independent variable → the analysis is said to be static → otherwise, structural dynamic analysis or structural dynamics
Small deformation Under the assumption of { Linearly elastic material behavior - Three dimensional formulation → a set of 15 linear 1st order PDE involving displacement field (3 components) { stress field (6 components) strain field (6 components) plane stress problem → simpler, 2-D formulations { plane strain problem For most situations, not possible to develop analytical solutions
→ analysis of structural components … bars, beams, plates, shells
1-2 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.1 The concept of Stress
1.1.1 The state of stress at a point
State of stress in a solid body… measure of intensity of forces acting within the solid - distribution of forces and moments appearing on the surface of the cut … equipollent force F , and couple M - Newton’s 3rd law → a force and couple of equal magnitudes and opposite directions acting on the two forces created by the cut
Fig. 1.1 A solid body cut by a plane to isolate a free body
1-3 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.1 The concept of Stress
A small surface of area A n located at point P on the surface generated by the cut → equipollent force F n , couple M n
- limiting process of area → concept of “stress vector”
Fn τn = lim (1.1) dA →0 n dAn existence of limit : “fundamental assumption of continuum mechanics”
- Couple M n → 0 as dA n → 0 … couple is the product of a differential element of force by a differential element of moment arm → negligible, second order differential quantity
- Total force acting on a differential element of area, dAn
Fn= dA nnτ (1.2) Unit : force per unit area, Nm / 2 or Pa
1-4 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.1 The concept of Stress
Surface orientation, as defined by the normal to the surface, is kept constant during the limiting process
- Fig. 1.2 … Three different cut and the resulting stress vector
first… solid is cut at point P by a plane normal to axis i1 : τ differential element of surface with an area dA 1 , stress vector 1 → No reason that those three stress vectors should be identical.
Fig. 1.2 A rigid body cut at point P by three planes orthogonal to the Cartesian axes.
1-5 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.1 The concept of Stress
Component of each stress vectors acting on the three forces
τσ1=++ 1iii 1 τ 12 2 τ 13 3 (1.3a)
direct shearing → both act on the force normal to stress Stress {normal {shear axis in the direction of and i1 i2 i3 → “engineering stress components” unit : force/area, Pa “positive” force … the outward normal to the face, i.e., the normal pointing away from the body, is in the same direction as the axis → sign convention (Fig.1.3) 9 components of stress components → fully characterize the state of stress at P Force … vector quantity, Fig. 1.3 sign conventions 3 components of the force vector (1st order tensor) Stress … 9 quantities (2nd order tensor) Strain tensor Bending stiffness of a beam Mass moments of inertia
1-6 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.1 The concept of Stress
1.1.2 Volume equilibrium eqn.
Stress varies throughout a solid body
- Fig. 1.4 axial stress component at the negative face : σ 2
axial stress component at the positive face at coordinate x22222++ dx:(σ x dx )
if σ 22 () x is an analytic function, using a Taylor series expansion
∂σ 2 σσ22()()x+= dx 2 22 x + dx2 +... h . o . terms in dx2 ∂x2 ∂σ x2 σσ()x+=+ dx 2 dx 22 2 2∂x 2 - body forces b … gravity, inertial, electric, magnetic origin 2 → 3 b=++ bi11 bi 2 2 bi 33 unit : force/volume, Nm/
Fig. 1.4 Stress components acting on a differential element of volume.
1-7 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.1 The concept of Stress
i) Force equilibrium
direction of axis i1
∂∂στ1 21 ∂τ 31 + + +=b1 0 ∂∂xx123 ∂ x (1.4a) . . . . . . must be satisfied at all points inside the body - equilibrium should be enforced on the DEFORMED configuration (strictly) Unknown, unfortunately “linear theory of elasticity”… assumption that the displacements of the body under the applied loads are very small, and hence the difference between deformed and undeformed is very small.
ii) Moment equilibrium
about axis i1 → ττ23−= 32 0 → “principle of reciprocity of shear stress” (Fig. 1.5) - only 6 independent components in 9 stress components Fig. 1.5 Reciprocity of the → symmetry of the stress tensor (1.6) shearing stresses acting on two orthogonal faces
1-8 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.1 The concept of Stress
1.1.3 Surface equilibrium eqn.
At the outer face of the body. equilibrium stress acting Externally applied t { inside the body} { Surface tractions }
t=++ ti11 ti 2 2 ti 33 Fig. 1.6 … free body in the form of a differential tetrahedron bounded by iii,, { 3 negative faces cut through the body in directions normal to axes 123 A fourth face, ABC, of area dAn
Fig. 1.6 A tetrahedron with one face along the outer surface of the body.
1-9 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.1 The concept of Stress
Force equilibrium along i 1 , and dividing by dAn
tnnn1=++στ 1 1 21 2 τ 31 3 (1.9a) body force term vanishes since it is a h.o. differential term. A body is said to be in equilibrium if eqn (1.4) is satisfied at all points inside the body and eqn (1.9) is satisfied at all points of its external surface.
1-10 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.2 Analysis of the state of stress at a point
It is fully defined once the stress components acting on three mutually orthogonal faces at a point are known.
1.2.1 Stress components acting on an arbitrary face
Fig 1.7… “Cauchy’s tetrahedron” with a fourth face normal to unit vector n of arbitrary orientation
Fig. 1.6 Differential tetrahedron element with one face, ABC, normal to unit vector n and the other three faces normal to axes and , respectively. ii12, i3
1-11 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.2 Analysis of the state of stress at a point
- force equilibrium
ττ11dA++=+ 2 dA 2 τ 33 dA τnn dA bdV
dividing by dA n and neglecting the body force term (since it is multiplied by a h.o. term)
ττn =++11nnn τ 2 2 τ 33 - Expanding the 3 stress vectors,
τστn =(1i 1 ++ 12 i 2 τ 13 in 3 )( 1 + τ 21 i 1 ++ σ 2 i 2 τ 23 in 3 )( 2 + τ 31 i 1 ++ τ 32 i 2 σ 3 in 3 ) 3 (1.10)
- To determine the direct stress σ n , project this vector eqn in the direction of n
n⋅=τσn (1 n 1 + τ 12 n 2 + τ 13 nn 3)( 1 + τ 21 n 1 + σ 2 n 2 + τ 23 nn 3)( 2 + τ 31 n 1 + τ 32 n 2 + σ 3 nn 3) 3 222 σσn =+++1n 1 σ 2 n 2 σ 3 n 32 τ 23 nn 2 3 + 22 τ 13 nn 1 3 + τ 12 nn 1 2 (1.11)
- stress component acting in the plane of face ABC : τ ns … by projecting Eq. (1.10) along vector s
τσns =1ns 1 1 + σ 2 ns 2 2 + σ 3 ns 3 3 + τ 12( ns 2 1 ++ ns 1 2 )( τ 13 ns 1 3 ++ ns 3 1 )( τ 23 ns 2 3 + ns 3 2 )(1.12)
1-12 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.2 Analysis of the state of stress at a point
Eqns (1.11), (1.12)… Once the stress components acting on 3 mutually orthogonal faces are known, the stress components on a face of arbitrary orientation can be readily computed.
How much information is required to fully determine the state of stress at a point P of a solid? Compute definition of the state of stress at a point only requires knowledge of the stress vectors, or equivalently of the stress tensor components, acting on three mutually orthogonal faces.
1.2.2 Principal stresses
Is there a face orientation for which the stress vector is exactly normal to the face ? Does a particular orientation, n exist for which the stress vector acting on σ this face consists solely of τσ np = n , where p is the yet unknown? … projecting Eq.(1.10) along axes iii 123 ,, → 3 scalar eqns σσ− τ τ n 1 p 12 13 1 τ σσ−= τ 21 2 p 23n 2 0 (1.13) τ τ σσ− 31 32 3 p n3
1-13 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.2 Analysis of the state of stress at a point - Determinant of the system = 0, non-trivial sol. exists. 32 σσσpp−III123 + p −=0 (1.14)
“stress invariants” (1.15) - Solution of Eq (1.14) … “principal stress” σσσ,, 3 solutions pp 123 p → non-trivial sol. for the direction cosines “principal stress direction” Homogeneous eqns → arbitrary constant → enforcing the normality condition
222 nnn1++= 23 I
1.2.3 Rotation of stresses
* * ** *** Arbitrary basis : I= (, i1 ii 23 )→ σσσ1,, 23 { *** τττ23,, 13 12 - orientation of basis J * relative to J → matrix of direction cosine, or rotation matrix R (A.36)
*** cos(ii11 , ) cos( ii 21 , ) cos( ii 31 , ) lmn 1 11 *** (A.36) = cos(ii , ) cos( ii , ) cos( ii , ) = l n m R 12 22 32 2 2 2 = ***lmn cos(ii13 , ) cos( ii 23 , ) cos( ii 33 , ) 3 33
1-14 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.2 Analysis of the state of stress at a point
* Eq(1.11) → σ 1 in terms of those resolved in axis J
n⋅=τσn (1 n 1 + τ 12 n 2 + τ 13 nn 3)( 1 + τ 21 n 1 + σ 2 n 2 + τ 23 nn 3)( 2 + τ 31 n 1 + τ 32 n 2 + σ 3 nn 3) 3 222 σσn =+++1n 1 σ 2 n 2 σ 3 n 32 τ 23 nn 2 3 + 22 τ 13 nn 1 3 + τ 12 nn 1 2 (1.11)
*222 σσ1=+++ 1l 1 σ 2 l 2 σ 3 l 32 τ 23 ll 2 3 + 22 τ 13 ll 1 3 + τ 12 ll 1 2 (1.18)
* lll123,, : direction cosines of unit vector i1 * Similar eqns for σ 2⋅⋅⋅ mmm 123,, * σ 3⋅⋅⋅ nnn 123,, Shear component : Eq. (1.12) →
* τσ12= 1lm 1 1 + σ 2 lm 2 2 + σ 3 lm 3 3 + τ 12( lmlm 2 1 ++ 1 2 )( τ 13 lmlm 3 1 ++ 1 3 )( τ 23 lmlm 2 3 + 3 2 )(1.19) Compact matrix eqn.
σττ** * σττ 1 12 13 1 12 13 τ* στ **= T τ στ 21 2 23 RR21 2 23 (1.20) ττσ** * ττσ 31 32 3 31 32 3
1-15 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.2 Analysis of the state of stress at a point
“Stress invariant”. … invariant w.r.t. a change of coordinate system. (1.21)
*** I1=++=++σσσσσσ 1 2 3 1 2 3, (1.21a) * * * * * * *2 *2 *2 I2=σσ 1 2 + σσ 2 3 + σσ 3 1 −−− τ 12 τ 13 τ 23 222 =σσ1 2 + σσ 2 3 + σσ 3 1 −−− τ 12 τ 13 τ 23 , (1.21b) *** * *2 * *2 * *2 *2 *2 *2 I3=σσσ 1 23 − σ1τ 23−−+ στ 2 13 στ 3 122 τ 12 τ 13 τ 23 222 =σσσ1 2 3 −−−+ στ 1 23 στ 2 13 στ 3 122 τ 12 τ 13 τ 23 (1.21c)
1-16 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
All stress components acting along the direction of axis i 3 are assumed to vanish
or to be negligible. Only non-vanishing components :σστ 1,, 2 12
Independent of x3 Vary thin plate or sheet subject to loads applied in its own plane (Fig. 1.11)
1-17 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
1.3.1 Equilibrium eqns
Considerably simplified from the general, 3-D case → 2 remaining eqns
∂∂στ1 21 ∂ τ12 ∂ σ 2 + +=bb120, + +=0 (1.26) ∂∂xx12 ∂∂ xx12 Surface tractions
tn1=+=+ 1στ 1 n 2 21, tn 2 1 τ 121 n 2 σ 2 (1.27) Very thin plate or sheet subject to loads applied in its own plane Fig. 1.11
Fig. 1.11 Plane stress problem in thin sheet with in-plane tractions
1-18 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
Fig. 1.11 … outer normal unit vector n= ni11 + ni 2 2,
nnn1=cosθθ , 23 = sin , = 0
tangent unit vector s= si11 + si 2 2
s1=−=sinθθ , ss 23 cos , = 0 22 Eq.(1.11) → tn =++cosθσ1 sin θσ2 2sin θ cos θτ12 (1.28) → 22 Eq.(1.12) ts =sinθ cos( θσ2 −+ σ 1 ) (cos θ − sin θτ ) 12 (1.29)
1.3.2 Stress acting on an arbitrary face within the sheet
Fig.1.12 : 2-D version of Cauchy’s tetrahedron (Fig. 1.7)
Fig. 1.12 Differential element with a face at an angle θ
1-19 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
- Equilibrium of forces 1 τττdx+=+ dx ds bdx dx 21 12 n 122 dividing by ds 1 ττ=+−n τ n bdx dx n 11 2 2 1 22ds Neglected since multiplied by h.o. term
τn =+( σ1ii 1 τ 12 2 )cos θτ ++( 21 ii 1 σ 2 2 )sin θ (1.30)
- Projecting in the dir. of unit vector n → σ n
22 σn =++ σ1 cos θσ2 sin θ 2 τ12 cossin θ θ (1.31)
- Projecting in the dir. of normal to n →τ ns
22 τns =−+ σ1 cos θθσθθτ sin2 cos sin +−12 (cos θ sin θ ) (1.32)
→ Knowledge of σστ 1 ,, 2 12 or 2 orthogonal faces allows computation of the stress components acting on a face with an arbitrary orientation
1-20 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
1.3.3 Principal stress
Simply write Eqn. (1.13)-(1.15) with στ3= 13 = τ 23 = 0
or, using Eq. (1.31)… particular orientation, θ p , that maximizes (or minimizes) σ n dσ 2τ sin 2θ →=→=n 0 tan 2θ 12 = p dθ p σσ− cos 2 θ (1.33) 12 p
2 sol.s θ p and θπ p + /2 corresponding to 2 mutually orthogonal principal stress directions. τ() σσ− θθ= 12 = 1 2 where ∆ is determined by (1.34) sin 2pp , cos 2 , ∆∆2 22 1/2 sin 2θθpp+= cos 2 1 −σσ12 2 ∆= +(τ12 ) (1.35) 2
Unique solution for θ p - Max./Min. axial stress : “principal stress” by introducing Eq.(1.34) into (1.31)
σσ12++σσ12 σσ= +∆; = −∆ (1.36) pp1222 Where the shear stress vanishes
1-21 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
dτ ns Max. shear stress → θ s → = 0 using Eq.(1.32) (1.37) dθ σσ− 1 →=−=θ 12 tan 2 s (1.38) 2τθ12 tan 2 p
2 sol.s θs and θπ s + /2 corresponding to 2 mutually orthogonal faces directions. σσ− - Max. shear stress τ =∆= pp12 (1.40) max 2 π θθ= − (1.41) sp4 Max. shear stress occurs at a face inclined at a 45° angle w.r.t. the principal stress directions σσ+ σσ+ σσ= =12 = pp12 12ss 22 (1.42)
1-22 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
1.3.4 Rotation of Stresses
*2 2 - Eq (1.31) σ1=++ σ 1 cos θσ2 sin θ 2 τ12 sin θ cos θ (1.45)
* 22 (1.46) (1.32) τ12=−+ σθθσθθτ 1 sin cos2 sin cos +−12 ( cos θ sin θ)
- Compact matrix form σ*2cos θ sin 2 θ 2cossin θθ σ 11 σ*2= θ 2 θ− θ θσ 22sin cos 2cossin (1.47) τ*−− θθ θθ22 θ θτ 12 sin cos sin cos cos sin 12
can be easily inverted by simply replacing θ by −θ σ - Knowledge of the stress components σ 1, 2 , τ 12 on 2 orthogonal faces allows computation of those acting on a face with an arbitrary orientation
1-23 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
1.3.5 Special state of stresses
¡) Hydrostatic stress state …… σσ= = P “hydrostatic pressure” pp12
τ12 = 0 With any arbitrary orientation
ii) Pure shear state
σσ= − (Fig. 1.13) pp21 At the face inclined at a 45° angle w.r.t. the principal stress direction τσ* = − ** 12 p1 σσ12= = 0 (1.51)
iii) Stress state in thin-walled pressure vessels
Fig.14 … cylindrical pressure vessel subjected to internal
pressure Pi
1-24 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
σ a (axial direction) 2 in-plane stress components σ h (circumferential or “hoop” direction)
Possibly a shear stress,τ ah 2 Left/right - Axial force equilibrium σπaiRt= p π R /2 Area of cylinder cross-section in the direction of axial 2 σ ai= pR/2 t
- tangential (hoop) direction 22σ hibt= p Rb Internal area of cylinder projected σ hi= pR/ t to the tangential (hoop) direction
τ ah = 0
1.3.6 Mohr’s circle for plane stress
σ σ ◦ p1 , p2 … Principal stresses at a point
◦ Eq.(1.49) -> stresses acting on a face oriented at an angle w.r.t. the
principal stress direction 𝜃𝜃 * σσ=a + R cos 2 θ = sin2 (1.52) 1-25 Active∗ Aeroelasticity and Rotorcraft Lab., Seoul National University 𝜏𝜏 −𝑅𝑅 𝜃𝜃 1.3 The state of plane stress
◦ where , σa=( σσ pp + ) /2 R =(σσpp − ) /2 12 12
*22 *2 (σσ−+a ) ( τ) =R (1.53)
◦ equation of a circle “Mohr’s circle”
σ * :horizontal axis, τ * :vertical axis (“inverted”)
center at a coordinate σ a on the horizontal axis R :radius, ……each point on Mohr’s circle represents the state of stress acting at a face at a specific orientation
◦ Observations
P σσ* = * 1) At point, 1 , p 1 , τ = 0 …..principal stress direction P2 , second principal stress direction
π * 2) At point E , θ = , τ max = R = ( σσ pp − ) /2 -> Max. shear stress orientation 1 4 12 E2
3) At point, A 1, A 2 two faces oriented 90° apart, the shear stresses are equal in magnitude and of opposite sign -> principle of reciprocity
1-26 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
◦ Construction procedure στ 1) First point A 1 at ( 1, 12 )
2) Second point A 2 , ( στ 2 , − 12 ) at a 90° angle counterclockwise w.r.t. the first point
3) Straight line joining A 1 and A2
4) Stress component at β an angle
…….a new diameter B 12 OB rotated 2 β degree from the ref. diameter AOA12
◦ Important features
σσ, 1) Principal stress pp12 -> points P 1 and P 2 , direct stress Max/Min. shear stress=0
τ= σσ − /2 2) Max. shear stress…….. Vertical line E 1=radius, max ( PP12)
direction…..45° since POE11=90 ° 3) Stress components acting on 2 mutually orthogonal faces ……..2 diametrically opposite points on Mohr’s circle 4) All the point on Mohr’s circle represent the same state of stress at one point of the solid
1-27 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.3 The state of plane stress
1.3.7 Lame’s ellipse
Eq. (1.30) When selecting the principal stress direction
** τσnp=1cos θii 12 + σp sin θ 2
** ( xx1, 2 ) : Tip of the stress v e c t o r, τn =xi11 + xi 2 2 = σθ x11= σθp cos x22p sin Eliminating 22 xx 𝜃𝜃 12 += 1 σσ (1.54) 12pp σ σ Equation of ellipse with semi-axis equal to p 1 and p2 (Fig 1.17) ◦ Pure shear…..ellipse circle (Fig.1.18)
1-28 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.4 The concept of strain
◦ State of strain……..characterization of the deformation in the neighborhood of a material point in a solid
at a given point P, located by a position vector r=++ xi11 xi 2 2 xi 33 (Fig.1.22)
small rectangular parallelepiped PQRST of differential size “ reference configuration,” undeformed state
“deformed configuration” PQRST
◦ displacement vector …..measure of how much a material point moves. (1.56) ¯ Rigid body motion…..translation, rotation -> does not produce strain two 𝑢𝑢 parts Deformation or straining -> strain-displacement relation
1-29 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.4 The concept of strain
1.4.1 The state of strain at a point
◦ Material line PR in the ref. conf. Material line PR in the ref. conf. in the deformed configuration ◦ 2 factors in the measure of state of strain assumed to be still straight, but a parallelogram Stretching of a material line …… εεε123,,
Angular distortion between 2 material lines……γγγ23,, 13 12 ¡) Relative elongations or extensional strain . . . : magnitude PRdef − PR ref ε = (1.57) 1 …. Non‖-dimensional‖ quantity PR ref
PRref = dx11 i = dx 1
PRdef = dx11 i + u( x 1 +− dx 1) u() x 1 ∂∂uu PRdef = dx11 i ++ u() x 1 + dx1 − u () x 1 = dx 11 i + dx1 ∂∂xx11
∂∂uu12∂u3 =+++i11 dx i 1 i 2 i 3 dx 1 (1.59) ∂∂xx11 ∂ x 1 1-30 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.4 The concept of strain
222 ∂∂uu11 ∂ u 2 ∂u3 =+++12 + dx1 ∂∂xx11 ∂ x 1 ∂ x 1
Then
222 ∂∂uu ∂ u ∂u ε =+++11 2 +3 − 1112 dx 1 (1.60) ∂∂xx11 ∂ x 1 ∂ x 1
◦ fundamental assumption of linear elasticity….all displacement components remain very small so that all 2nd order terms can be neglected. And, using the binomial expansion
∂∂uu11 ε1 ≅+11 −= (1.62) ∂∂xx11 “direct strains” or “axial strains” ∂u ∂u ε = 2 ε = 3 (1.63) 2 , 3 ∂ ∂x2 x3
1-31 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.4 The concept of strain
¡¡) Angular distortions or shear strains
γ 23 ….between two material lines PT and PS, defined as the change of the initially right angle π γ =−=−TPS TPS TPS (1.64) 23 ref def 2 def Non-dimensional quantities <……> : angle between segment π sinγ 23 =−= sin TPScos TPS 2 def def (1.65) by law of cosine
=22 +− (1.66) TSdef PTdef PSdef 2cos TPSdef PT def PS def
2sinγ 23 2+− 22 PTdef PSdef TS def (1.67) γ 23 = arcsin 2 PTdef PS def ∂u ∂u PTdef =+= i33 dx A, PSdef =+= i22 dx B ∂x3 ∂x2 PSdef=−=− PS def PTdef B A 1-32 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.4 The concept of strain
Numerator
∂u2∂∂uu33 ∂∂ uu11 ∂∂ uu 2 2 ∂ u 2 N =2 ++ + + dx23 dx ∂∂∂∂∂∂∂∂xxxxxxxx32232323
Denominator
D=⋅⋅2 AA BB
-with the help of small displacement assumption
∂u2 ∂u3 ∂u2 ∂u3 N ≅+2dx23 dx D ≅++21dx23 dx ∂∂xx32 ∂∂xx23
∂u ∂u γ ≅+2 3 23 (1.70) ∂∂xx32 “shearing strain”
∂u1 ∂u3 ∂∂uu12 “shear strain” γ13 ≅+ γ12 ≅+ (1.71) ∂∂xx31 ∂∂xx21
-Strain-displacement relationship, Eqs. (1.63), (1,71) ….. Under the small displacement assumption
Large displacement Eqns. (1.60),(1.67) should be used
1-33 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.4 The concept of strain
¡¡¡) Rigid body rotation
∂ 1 u3 ∂u2 (1.73a) ω1 = − 2 ∂∂xx23
T Rotation vectorω = { ωωω 123 ,, } ……the rotation of the solid about axes iii123 ,, respectively
1.4.2 The volumetric strain
◦ After deformation
ν≈+(111 ε1)( + ε 2)( + ε 3)dx 123 dx dx ≈++( 1εεε 1 2 + 3)dx 123 dx dx (1.74)
where high order strain quantities are neglected
◦ relative change in volume
e =++εεε123 “volumetric strain” (1.75)
1-34 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.5 Analysis of the state of strain at a point
* *** - Arbitrary reference frame J= ( iii1,, 23)
-> Strain-displacement relationship in J * (1.76), (1.77)
1.5.1 Rotation of strains
◦ chain rule ∂u** ∂∂ ux ∂∂ ux * ∂ u *∂x ∂ u * ∂ u * ∂ u * ε * ==1 11 + 12 + 13 =++ 1 1 1 1* ***lll123 (1.78) ∂x1 ∂∂ xx 11 ∂∂ xx 21 ∂∂ xx 31 ∂ x 1 ∂ x 2 ∂ x 3
Where Eq. (A.39) is used
* - Next, u 1 in terms of the components in J
* ∂∂∂ ε1=l 1( lu 11 ++ lu 2 2 lu 33) + l 2( lu 11 ++ lu 2 2 lu 33) + l 3( lu 11 ++ lu 2 2 lu 33) (1.79) ∂∂∂xxx123
Using Eq. (1.63) and (1.71)
*222 εε1=+++ 1l 1 ε 2 l 2 ε 3 l 3 γ 12 ll 1 2 + γ 13 ll 1 3 + γ 23 ll 2 3
1-35 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.5 Analysis of the state of strain at a point
Similar eqns. (1.80), (1.81)
γ 23 γ γ ε = 13 12 Engineering shear strain comp. (1.82) 23 ε13 = ε = 2 2 12 2 Tensor shear strain component pp* pp* 1111 * * T (A.39) p22= Rp p22= Rp * * pp33 pp33 - Compact matrix form εεε** * εεε 1 12 13 1 12 13 ε* εε **= T ε εε 12 2 23 RR13 2 23 (1.83) εε** ε * εε ε 13 23 3 13 23 3
1.5.2 Principal strains
◦ Is there a coordinate system J * for which the shear strains vanish?
ε* 00 εε ε 1 1 12 13 ε* = T ε εε 002 RR13 2 23 ε* εε ε 00 3 13 23 3
1-36 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.5 Analysis of the state of strain at a point
- Pre-multiplying and reversing the equality ¯ ε 00 εε1 12 ε 13 p1 𝑅𝑅 ε εεRR= 00 ε 13 2 23 p2 εε13 23 ε 3 00ε p3 where the orthogonality of , Eq. (A.37), is used ¯ 𝑅𝑅 l1 l 2 l 31 lmn 11 100 T (A.37) RRmmmlmn= 1 2 32 2 2=010 = I = = = nn1 2 n 33 lmn 33 001
εεε,, - pp123 p: sol. of 3 system of 3 eqns εε ε nn 1 12 13 1 1 ε εε= ε 13 2 23nn 2 p 2 εε13 23 ε 3 nn 3 3
1-37 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.5 Analysis of the state of strain at a point
Determinant of the system vanishes-> non-trivial solution.
32 Cubic eqn. εεεpp−III123 + p −=0 (1.85)
“Strain (1.86) invariant” 3 sol:εεε ,, -> corresponding “principal strain direction” pp123 p
-> homogeneous eqn.-> arbitrary constant ->normality condition
◦ displacement component along i3 is assumed to vanish, or to be negligible
Example: a very long buried pipe aligned with i 3 dir.
1-38 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.6 The state of plane strain
1.6.1 Strain-displacement relations for plane strain
∂u ∂u 1 2 ∂∂uu12 (1.87) ε1 = ε 2 = γ = + ∂x ∂x 12 1 2 ∂∂xx21
1.6.2 Rotation of strains
◦ chain rule ∂u** ∂∂ ux ∂∂ ux * ∂ u * ∂ u * ε* ==+=1 11 12 1 θθ +1 1 * **cos sin ∂x1 ∂∂ xx 11 ∂∂ xx 21 ∂ x 1 ∂ x2
Eq (A.43)
1-39 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.6 The state of plane strain
- in terms of those in J ∗ 𝑢𝑢1 ∂∂ * (1.88) εθ1=cos(uu 12 cos θθθ ++ sin) sin( uu12 cos θθ + sin ) ∂∂xx12 -Then,
*2 2 (1.89) ε1 =++cos θε1 sin θε2 sin θ cos θγ12 -Matrix form
ε*2 cos θ sin 2 θ 2sincos θθ ε 11 ε*2= θ 2 θ− θθε (1.91) 22sin cos 2sincos ε*−− θθ θθ22 θ θε 12 sin cos sin cos cos sin 12
can be readily inverted by replacing by -
𝜃𝜃 𝜃𝜃
1-40 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.6 The state of plane strain
1.6.3 Principal strains
◦ θ p , in which the max (or min) elongation occurs
dε* εε− γ 1 ==0−1 2 2sin2θθ +=12 2cos2 0 (1.95) dθ 22pp
γ12 /2 tan 2θ p = (1.96) (εε12− ) /2 π 2 sols. --- θ , θθ= + ….. 2 mutually orthogonal principal strain directions p1 pp212 εε+ εε+ ε =12 +∆ ε =12 −∆ (1.99) p1 2 p2 2
where shear strain vanishes Principal stresses - The orientation of the are not necessarily identical Principal strains
1-41 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.6 The state of plane strain
1.6.4 Mohr’s circle for plane strain
- Strains along a direction defined by angle w.r.t. the principal strain direction
* * 𝜃𝜃γ εε= + R cos 2 θ = −Rsin 2θ (1.100) a 2 Where, εε+ εε− ( pp) ( pp12) ε = 12 R = a 2 2
* 2 γ *2 (1.101) => (εε−+a ) =R Mohr’s circle 2
Fig.1.23, positive angle …. counterclockwise dir. shear strain …….positive downward 𝜃𝜃 γ Vertical axis ……strain tensor, 12 2
1-42 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.7 Measurement of strains
No practical experimental device for direct measurement of STRESS …….indirect measurement of strain first constitutive laws ¡) strain gauges ◦ measurement of extensional strains on the body’s external surfaces - Very thin electric wire, or an etched foil pattern - extension …. wire’s cross-section reduced by Poisson’s effect. Slightly increasing its electrical resistance compression….increasing its electrical reduced resistance - Wheatstone bridge …..accurate measurement
“micro-strains” … / =10 / ¡¡) Chevron strain gauges −6 𝜇𝜇 𝑚𝑚 𝑚𝑚 𝑚𝑚 𝑚𝑚 Fig 1.24….. e+45 and e-45, experimentally measured relative elongations Using Eq (1.94a),
εε1+ 2 γ 12 εε1+ 2 γ 12 e = + e− = − +45 22 45 22 ……Not sufficient to determine the strain state at the point
εεγ1,, 2 12 3 measurements would be required 2 principal strains & dir.
However, can uniquely determine γ12=ee+− 45 − 45 (1.102)
1-43 Active Aeroelasticity and Rotorcraft Lab., Seoul National University 1.7 Measurement of strains
¡¡¡) Strain gauge rosette
Fig.1.25……3 independent measurements, “delta rosette” 2 e 2 1 Eq. (1.94a) ε = e ε 2=ee 23 +− γ12 =(ee2 − 3 ) (1.103) 11 32 3
Fig.1.26 …..various arrangement of strain gauges
1-44 Active Aeroelasticity and Rotorcraft Lab., Seoul National University