Mechanics and Design Chapter 2. Stresses and Strains
Byeng D. Youn System Health & Risk Management Laboratory Department of Mechanical & Aerospace Engineering Seoul National University
Seoul National University CONTENTS
1 Traction or Stress Vector
2 Coordinate Transformation of Stress Tensors 3 Principal Axis 4 Example
2019/1/4 Seoul National University - 2 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Traction Vector Consider a surface element, ∆ S , of either the bounding surface of the body or the fictitious internal surface of the body as shown in Fig. 2.1. Assume that ∆ S contains the point. The traction vector, t, is defined by Δf t = lim (2-1) ∆→S0∆S
Fig. 2.1 Definition of surface traction
2019/1/4 Seoul National University - 3 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Traction Vector (Continued) It is assumed that Δ f and ∆ S approach zero but the fraction, in general, approaches a finite limit. An even stronger hypothesis is made about the limit approached at Q by the surface force per unit area. First, consider several different surfaces passing through Q all having the same normal n at Q as shown in Fig. 2.2.
Fig. 2.2 Traction vector t and vectors at Q
Then the tractions on S , S ′ and S ′′ are the same. That is, the traction is independent of the surface chosen so long as they all have the same normal.
2019/1/4 Seoul National University - 4 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Stress vectors on three coordinate plane Let the traction vectors on planes perpendicular to the coordinate axes be t(1), t(2), and t(3) as shown in Fig. 2.3. Then the stress vector at that point on any other plane inclined arbitrarily to the coordinate axes can be expressed in terms of t(1), t(2), and t(3). (1) Note that the vector t acts on the positive x1-side of the element. The stress vector on the negative side will be denoted by - t(1).
Fig. 2.3 Traction vectors on three planes perpendicular to coordinate axes
2019/1/4 Seoul National University - 5 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Stress components
The traction vectors on planes perpendicular to the coordinate axes, x 1 , x 2 and (1) (2) (3) x3 are t , t , and t . The three vectors can be decomposed into the directions of coordinate axes as (1) t=++TTT11 i 12 jk 13 (2) (2-2) t=++TTT21 i 22 jk 23 (3) t=++TTT31 i 32 jk 33 The nine rectangular components T ij are called the stress components. Here, the first subscript represents the “plane” and the second subscript represents the “direction”.
Tij Direction
Plane
Fig. 2.4 Stress components
2019/1/4 Seoul National University - 6 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Sign convention A stress component is positive when it acts in the positive direction of the coordinate axes and on a plane whose outer normal points in one of the positive coordinate directions.
Fig. 2.5 Sign convention of stress components
Sign convention The stress state at a point Q is uniquely determined by the tensor T which is represented by TTT11 12 13 = T TTT21 22 23 (2-3)
TTT31 32 33
2019/1/4 Seoul National University - 7 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Traction vector on an arbitrary plane: The Cauchy tetrahedron When the stress at a point O is given, then the traction on a surface passing the point Q is uniquely determined. Consider a tetrahedron as shown in Fig. 2.6. The orientation of the oblique plane ABC is arbitrary. Let the surface normal of ∆ ABC be n and the line ON is perpendicular to ∆ ABC . The components of the unit normal vector n are the direction cosine as
n1 =cos( ∠ AON )
n2 =cos( ∠ BON ) (2-4)
n3 =cos( ∠ CON )
Fig. 2.6 Geometry of tetrahedron
2019/1/4 Seoul National University - 8 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Traction vector on an arbitrary plane: The Cauchy tetrahedron (Conti.) If we let ON = h , then
h = OA⋅ n1 = OB ⋅ n2 = OC ⋅ n3 (2-5)
Let the area of ∆ ABC , ∆ OBC , ∆ OCA & ∆ OAB be ∆ S , ∆ S 1, ∆ S 2 & ∆ S 3 respectively. Then the volume of the tetrahedron, ∆ V , can be obtained by 1 1 1 1 ∆V = h ⋅ ∆S = OA⋅ ∆S = OB ⋅ ∆S = OC ⋅ ∆S (2-6) 3 3 1 3 2 3 3
From this we get, h ∆S =∆⋅ S =∆⋅ Sn 11OA h ∆S =∆⋅ S =∆⋅ Sn (2-7) 22OB h ∆S =∆⋅ S =∆⋅ Sn 33OC
2019/1/4 Seoul National University - 9 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Traction vector on an arbitrary plane: The Cauchy tetrahedron (Conti.) Now consider the balance of the force on OABC as shown in Fig. 2.7. The equation expressing the equilibrium for the tetrahedron becomes
(n )* * * (1)* (2)* (3)* t∆+∆−∆−∆−∆=Sρ bt VS123 t S t S0 (2-8) Here the subscript * indicates the average quantity.
Substituting for ∆ V , ∆ S 1 , ∆ S 2 and ∆ S 3 , and dividing through by ∆ S , we get
1 ∗ t(n )*+hρ * bt =++ (1)* nnn t (2)* t (3)* (2-9) 3 123
Fig. 2.7 Forces on tetrahedron
2019/1/4 Seoul National University - 10 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Traction vector on an arbitrary plane: The Cauchy tetrahedron (Conti.) Now let h approaches zero, then the term containing the body force approaches zero, while the vectors in the other terms approach the vectors at the point O. The result is in the limit (n ) (1) (2) (3) (k) tt=++=nnn123 t t t nk (2-10) The important equation permits us to determine the traction t(n) at a point acting on an arbitrary plane through the point, when we know the tractions on only three mutually perpendicular planes through the point. The equation (2-10) is a vector equation, and it can be rewritten by (n) (1) (2) (3) t1 = t1 n1 + t1 n2 + t1 n3 (n) (1) (2) (3) t2 = t2 n1 + t2 n2 + t2 n3 (2-11) (n) (1) (2) (3) t3 = t3 n1 + t3 n2 + t3 n3 Comparing these with eq. (2-2), we get (n) t1 = T11n1 + T21n2 + T31n3 = Tk1nk ()n =++= t2 Tn 12 1 Tn 22 2 Tn 32 3 Tnkk 2 (2-12) (n) t3 = T13n1 + T23n2 + T33n3 = Tk 3nk
2019/1/4 Seoul National University - 11 - Chapter 2 : Stresses and Strains Traction or Stress Vector; Stress Components Cauchy stress tensor Or for simplicity, we put (n) • in indicial notation ti = Tjin j • in matrix notation t(n)= Tn T (2-13) • in dyadic notation t()n =⋅= nT TT ⋅ n
From the derivation of this section, it can be shown that the relation (2-13) also holds for fluid mechanics.
Tij : Cauchy stress tensor.
This stress tensor is the linear vector function which associates with n the traction vector t(n)
2019/1/4 Seoul National University - 12 - Chapter 2 : Stresses and Strains Coordinate Transformation of Stress Tensors Transformation matrix As we discussed in the previous chapter, stress tensor follows the tensor coordinate transformation rule. That is, let x and x be the two coordinate systems and A be a transformation matrix as v= Av or v= AvT
Then the stress tensor T transforms to T as T = AT TA We may consider the stress tensor transformation in two dimensional case. Let the angle between x axis and x axis is θ . Then the transformation matrix A becomes
aa11cosθθ− sin =12 = A 22 aa12sinθθ cos
2019/1/4 Seoul National University - 13 - Chapter 2 : Stresses and Strains Coordinate Transformation of Stress Tensors Transformation matrix (Continued) The stress T transforms to T according to the following
T cosθθ sin TT11 12 cos θ− sin θ T= A TA = −sinθθ cos TT21 22 sin θ cos θ
Evaluating the equation, we get
2 TT11=++ 11 cosθ 2 T21 sin θ cos θ T22 cos θθ sin
22 T12=−+−( TT 22 11 )sincosθθ T12 (cos θ sin θ )
22 TT22=+− 11 sinθ T22 cos θ 2 T12 sincos θθ By using double angle trigonometry, we can get (TT+− )( TT ) TT, =±±11 22 11 22 cos2θθT sin 2 11 22 22 12 ()TT− TT=−+11 22 sin 2θθ cos2 12 2 12
2019/1/4 Seoul National University - 14 - Chapter 2 : Stresses and Strains Coordinate Transformation of Stress Tensors Transformation matrix (Continued) Now we recognize the last two equations are the same as the ones we derived for Mohr circle, (eq 4-25) of Crandall’s book. For the two dimensional stress state, Mohr circle may be convenient because we recognize the stress transformation more intuitively. However, for 3D stress state and computation, it is customary to use the tensor equation directly to calculate the stress components in transformed coordinate system.
Fig. 2.8 Mohr’s circle
2019/1/4 Seoul National University - 15 - Chapter 2 : Stresses and Strains Principal Axes of Stress, Principal Stress, etc. Characteristics of the principal stress
(1) When we consider the stress tensor T as a transformation, then there exist a line which is transformed onto itself by T.
(2) There are three planes where the traction of the plane is in the direction of the normal vector, i.e. tn()n / / or tn()n = λ (2-14)
Definitions • Principal axes • Principal plane • Principal stress
2019/1/4 Seoul National University - 16 - Chapter 2 : Stresses and Strains Principal Axes of Stress, Principal Stress, etc. Determination of the principal stress Let T be the stress at a point in some Cartesian coordinate system, n be a unit vector in one of the unknown directions and λ represent the principal component on the plane whose normal is n. Then tn()n = λ ; that is, nT⋅=λ n In indicial notation, we have
Trsnr = λns = λδ rsnr Rearranging, we have
(Trs − λδ rs )nr = 0 (2-15) The three direction cosines cannot be all zero, since 2 2 2 nr nr = n1 + n2 + n3 =1 A system of linear homogeneous equations, such as eq. (2-15), has solutions which are not all zero if and only if the determinant
Trs − λδ rs = 0 (2-16) The equation (2-16) represents third order polynomial equation w.r.t. λ and it has 3 real roots for λ since T represents a real symmetric matrix.
2019/1/4 Seoul National University - 17 - Chapter 2 : Stresses and Strains Principal Axes of Stress, Principal Stress, etc. Determination of the principal direction
When we get λ = λ1, λ2, λ3 we can substitute the λi into the system of three algebraic equations
T11 − λ T12 T13 n1 − λ = (2-17) T21 T22 T23 n2 0 T31 T32 T33 − λn3
From these, we get the ratio of n1 : n2 : n3.
Since n = n i n i = 1 , we can determine ( n 1 n 2 n 3 ) uniquely. There can be three different cases; (1) 3 distinctive roots (2) Two of the roots are the same (cylindrical) (3) All three roots are the same (spherical)
When the two of the principal stresses, say σ1 and σ2 are not equal, the corresponding principal directions n(1)and n(2)are perpendicular.
2019/1/4 Seoul National University - 18 - Chapter 2 : Stresses and Strains Principal Axes of Stress, Principal Stress, etc. Determination of the principal direction (Continued) Proof> Recall eq. (2-15)
(Trs − λδ rs )nr = 0 (2-15) (i) By substituting λi and n , we get (1) (Trs − λ1δ rs )nr = 0 (2-18) (2) (Trs − λ2δ rs )nr = 0 (1) Note that ( T rs − λ 1 δ rs ) n r represent a vector. From eq. (2-18), we can have (1) (2) (Trs − λ1δ rs )nr ns = 0 (a) (2) (1) (Trs − λ2δ rs )nr ns = 0 (b) By subtracting (b) from (a), we get
(1) (2) (2) (1) (2) (1) (1) (2) Trsnr ns −Trsnr ns + λ2nr nr − λ1nr nr = 0 (c)
2019/1/4 Seoul National University - 19 - Chapter 2 : Stresses and Strains Principal Axes of Stress, Principal Stress, etc. Determination of the principal direction (Continued) The first two terms of eq. (c) become
(1) (2) (2) (1) (1) (2) (1) (2) Trsnr ns −Trsnr ns = Trsnr ns −Tsr nr ns (1) (2) = (Trs −Tsr )nr ns = 0 because T is symmetric. The remaining terms of eq. (c) become
(1) (2) (λ2 − λ1)nr nr = 0 (d)
(1) (2) Since λ 1 ≠ λ 2 and nn = = 1 , eq. (d) implies
(1) (2) (1) (2) nnrr=⋅=nn 0 Therefore, n (1) and n (2) are perpendicular to each other.
2019/1/4 Seoul National University - 20 - Chapter 2 : Stresses and Strains Principal Axes of Stress, Principal Stress, etc. Invariants The principal stresses are physical quantities. Their value does not depend on the choice of the coordinate system. Therefore, the principal stresses are invariants of the stress state. That is, they are invariant w.r.t. the rotation of the coordinate axes. The determinant in the characteristic equation becomes
T11 − λ T12 T13
T21 T22 − λ T23 = 0
T31 T32 T33 − λ Evaluating the determinant, we get 32 λλ−−−=ITTT II λ III 0 (2-20)
2019/1/4 Seoul National University - 21 - Chapter 2 : Stresses and Strains Principal Axes of Stress, Principal Stress, etc. Invariants (Continued) where
ITT =++==T11 T 22 T 33 Tkk tr()
222 IIT =−+()TT11 22 TT 22 33 + TT 33 11 +++ T 23 T 31 T 12 1 = (T T −T T ) 2 ij ij ii jj
TTT11 12 13
IIIT =det ( T ) = TTT21 22 23
TTT31 32 33 1 = e e T T T 6 ijk pqr ip jq kr Since the roots of the cubic equation are invariants, the coefficients should be invariants.
2019/1/4 Seoul National University - 22 - Chapter 2 : Stresses and Strains Example 1. Determine the normal and shear stress at the interface
σ ns σ nn
σ xx = σ 11 = −500 psi
Other stress components are zero. Stress distribution is uniform. σ yy = σ zz =100 psi From the figure we can determine
n=sinαα e12+ cos e We use eq (2-10), TT ()n = n to find out surface traction. That is,
−−500 0 sinαα 500sin ()n = αα = T 0 100 cos 100cos 100 0 0
2019/1/4 Seoul National University - 23 - Chapter 2 : Stresses and Strains Example 1. Determine the normal and shear stress at the interface Therefore, the traction at the surface whose normal is n given by ()n T =−+500sinααee12 100cos Therefore, the surface traction in n direction is given by ()n TTnn = ⋅=−n( 500sinα e1 + 100cos α e 2 ) ⋅ (sin αα ee 12 + cos ) =−+ ( 500sin22αα 100cos ) There may be different ways to obtain shear component of the traction. One way may be the vector subtraction. Then the shear stress at the interface becomes ()n 2 2 TTns = −−( 500sinα + 100cos αα )(sinee12+ cos α ) 22 =− 500sinαe1 + 100cos α e 2 −− ( 500sin α + 100cos αα )(sinee12+ cos α ) 32 =−+− ( 500sinα 500sin α 100cos αα sin )e1 22 ++ (100cosα 500sin αα cos + 500sin αα cos )e2 22 = − 600sinα cos αee12+ 600cos αα sin
The magnitude of T ns becomes Tns = 600sinαα cos
2019/1/4 Seoul National University - 24 - Chapter 2 : Stresses and Strains Example 2 3000− 1000 0 = − Let, T 1000 2000 2000 0 2000 2000 Then determine the normal traction on the surface whose normal is
ne=0.623 + 0.8 e Again we use TT()n = ⋅n 3000− 1000 0 0 ()n =⋅− TTn= 1000 2000 2000 0.6 0 2000 2000 0.8 −600 = =−+ + 2800 600eee123 2800 2800 2800 =()n ⋅= TTnn n 2800 and se= 1 ()n Again we use TTns = ⋅s=( −600 e123 + 2800 e + 2800 e ) ⋅− s= 600
2019/1/4 Seoul National University - 25 - Chapter 2 : Stresses and Strains Example 2 The generalized Hooke’s law
ijkl σij = Cekl
ijkl ijlk ijkl jikl In as much as σij= σ ji and eeij= ji , CC= and CC= According to these symmetry properties, the maximum number of the independent elastic constants is 36.
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