# M.

Vable

Intermediate Mechanics of Materials: Chapter 1

Stress and Strain
The learning objectives in this chapter are:

• Understanding the concept of stress and the use of double subscripts in
determining the direction of stress components on a surface.

• Understanding the concept of strain and the use of small strain and finite
difference approximation.

• Understanding the stress and strain transformation in three dimension.

1-1

M. Vable

Intermediate Mechanics of Materials: Chapter 1

Internally Distributed Force System

• The intensity of internal distributed forces on an imaginary cut surface of a
body is called the stress on a surface. • The intensity of internal distributed force that is normal to the surface of an imaginary cut is called the normal stress on a surface. • The intensity of internal distributed force that is parallel to the surface of an imaginary cut surface is called the shear stress on the surface. • Relating stresses to external forces and moments is a two step process.
Stresses Static Equivalency Internal Forces and Moments Equilibrium External Forces and Moments

Static equivalency
Uniform Normal Stress σavg Uniform Shear Stress τavg Normal stress x linear in y y z Normal stress linear in z x y z T Uniform shear stress in tangential direction.

x N = σ avg A V = τ avg A y z Mz y z My

x

Axial

Bending

Bending

Torsion

1-2

(2) ΔFj and ΔAi are both negative. direction of the internal force component. Vable Intermediate Mechanics of Materials: Chapter 1 Stress at a Point Outward normal i ΔAi ΔFj Internal Force ⎛ ΔF j⎞ σ ij = lim ⎜ -------. σ xx • Stress Matrix in 3-D: τ yx τ zx τ xy σ yy τ zy τ xz τ yz σ zz Table 1. • A stress component is positive if numerator and denominator have the same sign. Thus σij is positive if: (1) ΔFj and ΔAi are both positive. • ΔAi will be considered positive if the outward normal to the surface is in the positive i direction.M.⎟ ΔA i⎠ ΔA i → 0⎝ direction of outward normal to the imaginary cut surface. Comparison of number of components Quantity Scaler Vector Stress 1-D 1=10 1=11 1=12 2-D 1=20 2=21 4=22 3-D 1=30 3=31 9=32 1-3 .1.

1-4 . σ xx τ yx 0 τ xy σ yy 0 0 0 0 Symmetric Shear Stresses: τ xy = τ yx τ yz = τ zy τ zx = τ xz • A pair of symmetric shear stress points towards the corner or away from the corner. Stress cube showing all positive stress components σ xx τ yx τ zx τ xy σ yy τ zy τ xz τ yz σ zz Plane Stress: All stress components on a plane are zero. Vable Intermediate Mechanics of Materials: Chapter 1 Stress Element • Stress element is an imaginary object that helps us visualize stress at a point by constructing surfaces that have outward normal in the coordinate directions.M.

B. and C faces of the cube shown in Figure P1.M.4.4 1-5 . P1.3 Show the non-zero stress components on the A. σ xx = 0 τ yx = – 15 ksi τ zx = 0 τ xy = – 15 ksi σ yy = 10ksi ( C ) τ zy = 25ksi τ xz = 0 τ yz = 25ksi σ zz = 20ksi ( T ) Fig.3 1. Vable Intermediate Mechanics of Materials: Chapter 1 1.3 and Figure P1.

(c) y t n n θ Horizontal Plane Matrix Notation n x = cos θ ⎧ ⎫ ⎪ nx ⎪ {n} = ⎨ ⎬ ⎪ ny ⎪ ⎩ ⎭ n y = sin θ ⎧ ⎪ {t} = ⎨ ⎪ ⎩ t x = cos λ ty = nx [σ] = T True only in 2D: λ = 90 + θ t x = – n y The symmetry of shear stresses T dP z lan e x θ λ x σ nn = σ xx cos θ + σ yy sin θ + 2τ xy sin θ cos θ 2 τ nt = – σ xx cos θ sin θ + σ yy sin θ cos θ + τ xy ⎛ cos 2θ – sin θ⎞ ⎝ ⎠ 2 2 σ tt = σ xx sin θ + σ yy cos θ – 2τ xy cos θ sin θ 2 2 t y = sin λ ⎫ tx ⎪ ⎬ ty ⎪ ⎭ [σ] σ xx τ xy τ yx σ yy = [σ] σ nn = { n } [ σ ] { n } τ nt = { t } [ σ ] { n } σ tt = { t } [ σ ] { t } T T 1-6 .M. Vable Intermediate Mechanics of Materials: Chapter 1 Stress transformation in two dimension (a) t In ne cli Vertical Plane y Outward normal to the inclined plane.

• stress is a second order tensor. Vable Intermediate Mechanics of Materials: Chapter 1 Traction or Stress vector Mathematically the stress vector {S} is defined as: {S} = [σ ]{n} S x = σ xx n x + τ xy n y S y = τ yx n x + σ yy n y • pressure is a scaler quantity. • traction is a vector quantity. τnt (dA) σnn(dA) σxx ( nx dA) σxx( nx dA) Sy (dA) Sx (dA) τxy (nx dA) τyx (ny dA) σyy (ny dA) τxy (nx dA) τyx (n dA) y σyy (n dA) y { S } = σ nn { n } + τ nt { t } Stress vector in different coordinate systems. Statically equivalent force wedge. y Sy τnt{t} {S} σnn{n} Sx x 1-7 ..M.

1-8 . 2 = xy ⎝ ⎠ ⎝ ⎠ 2 2 • The eigenvalues of the stress matrix are the principal stresses.M. 2 = 2 2 ( σ xx + σ yy ) ± ( σ xx + σ yy ) – 4 ( σ xx σ yy – τ xy ) ⁄ 2 OR 2 2 σ xx + σ yy σ xx – σ yy 2 ⎛ -----------------------⎞ ± ⎛ ---------------------.⎞ + τ 2 σ 1. Vable Intermediate Mechanics of Materials: Chapter 1 Principal Stresses and Directions { S } = [ σ ]{ p } = σp { p } OR ⎧ σ xx τ xy ⎪ p x {S} = ⎨ τ yx σ yy ⎪ p y ⎩ OR ⎫ ⎧ ⎫ σp 0 ⎪ px ⎪ ⎪ ⎬ = ⎨ ⎬ 0 σp ⎪ py ⎪ ⎪ ⎭ ⎩ ⎭ ( σ xx – σ p ) τ yx Characteristic equation ⎧ ⎫ ⎪ px ⎪ ⎨ ⎬ = 0 ( σ yy – σ p ) ⎪ p y ⎪ ⎩ ⎭ τ xy σ p – σ p ( σ xx + σ yy ) + ( σ xx σ yy – τ xy ) = 0 Roots: σ 1. • The eigenvectors of the stress matrix are the principal directions.

M. Vable Intermediate Mechanics of Materials: Chapter 1 Stress Transformation in 3-D ⎧ ⎫ ⎪ nx ⎪ ⎪ ⎪ { n } = ⎨ ny ⎬ ⎪ ⎪ ⎪ nz ⎪ ⎩ ⎭ ⎧ ⎫ Sx ⎪ ⎪ ⎪ ⎪ { S } = ⎨ Sy ⎬ ⎪ ⎪ ⎪ Sz ⎪ ⎩ ⎭ T σ xx [σ] = τ yx τ zx τ xy σ yy τ zy τ xz τ yz σ zz σ nn = { n } [ σ ] { n } τ nt = { t } [ σ ] { n } σ tt = { t } [ σ ] { t } Equilibrium condition: T T {S} = [σ ]{n} { S } = σ nn { n } + τ nt { t E } 1-9 .

θ y. • The eigenvectors of the stress matrix are the principal directions. px + py + pz = 1 Principal stress convention Ordered principal stresses in 3-D: Ordered principal stresses in 2-D: Principal Angles 2 2 2 σ1 > σ2 > σ3 σ1 > σ2 0 ≤ θ x. Vable Intermediate Mechanics of Materials: Chapter 1 Principal Stresses and Directions { S } = [ σ ]{ p } = σp { p } OR σ xx τ yx τ zx τ xy σ yy τ zy ⎧ ⎫ ⎧ ⎫ σp 0 0 ⎪ px ⎪ τ xz ⎪ p x ⎪ ⎪ ⎪ ⎪ ⎪ τ yz ⎨ p y ⎬ = 0 σ p 0 ⎨ p y ⎬ ⎪ ⎪ ⎪ ⎪ σ zz ⎪ p z ⎪ 0 0 σp ⎪ pz ⎪ ⎩ ⎭ ⎩ ⎭ OR ( σ xx – σ p ) τ yx τ zx τ xy ( σ yy – σ p ) τ zy ⎧ ⎫ ⎪ px ⎪ ⎪ ⎪ τ yz ⎨ py ⎬ = 0 ⎪ ⎪ ( σ zz – σ p ) ⎪ p z ⎪ ⎩ ⎭ τ xz • The eigenvalues of the stress matrix are the principal stresses. θ z ≤ 180 o o 1-10 .M.

Vable Intermediate Mechanics of Materials: Chapter 1 Characteristic equation 3 2 σp – I1 σp + I2 σp –I3 = 0 Stress Invariants I 1 = σ xx + σ yy + σ zz I2 = σ xx τ xy σ yy τ yz σ xx τ xz + + τ yx σ yy τ zy σ zz τ zx σ zz σ xx I3 = 3 2 τ xy σ yy τ zy τ xz τ yz σ zz τ yx τ zx x –I1 x + I2 x – I3 = 0 Roots: x 1 = 2A cos α + I 1 ⁄ 3 A = ( I1 ⁄ 3 ) – I2 ⁄ 3 3 2 x 2.M. 3 = – 2 A cos ( α ± 60 ) + I 1 ⁄ 3 o cos 3α = [ 2 ( I 1 ⁄ 3 ) – ( I 1 ⁄ 3 )I 2 + I 3 ] ⁄ ( 2A ) σ1 Principal Stress Matrix [σ] = σ2 σ3 I1 = σ1 + σ2 + σ3 I2 = σ1 σ2 + σ2 σ3 + σ3 σ1 I3 = σ1 σ2 σ3 1-11 3 .

------------------⎞ τ max = max ⎝ 2 2 2 ⎠ 1-12 . ------------------. Vable Intermediate Mechanics of Materials: Chapter 1 Maximum Shear Stress σ1 – σ2 σ2 – σ3 σ3 – σ1 ⎛ ------------------.M.

( σ 1 – σ 2 ) + ( σ 2 – σ 3 ) + ( σ 3 – σ 1 ) 3 1-13 .M. 2 2 2 σ nn = σ 1 n 1 + σ 2 n 2 + σ 3 n 3 τ nt = 2 2 2 2 2 2 ( σ 1 n 1 + σ 2 n 2 + σ 3 n 3 ) – σ nn 2 n1 = n2 = n3 = 1 ⁄ 3 σ oct = ( σ 1 + σ 2 + σ 3 ) ⁄ 3 2 2 2 1 τ oct = -. Vable Intermediate Mechanics of Materials: Chapter 1 Octahedral stresses • A plane that makes equal angles with the principal planes is called an octahedral plane.

(e) maximum shear stress 18 12 9 12 12 – 6 ksi 9 –6 6 1-14 . Vable Intermediate Mechanics of Materials: Chapter 1 1. 120o.35 The stress at a point is given by the stress matrix shown. to x.M. (b) the principal stresses (c) the second principal direction and (d) the magnitude of the octahedral shear stress. and z direction respectively. and 70. Determine: (a) the normal and shear stress on a plane that has an outward normal at 37o.43o. y.

• Eulerian strain is computed from deformation by using the final deformed geometry as the reference geometry. • Relating strains to displacements is a problem in geometry. Contractions (Lf < Lo) result in negative normal strains. • Lagrangian strain is computed from deformation by using the original undeformed geometry as the reference geometry.= ----Lo Lo • Elongations (Lf > Lo) result in positive normal strains. 1-15 . • The relative movement of a point with respect to another point on the body is called deformation. Vable Intermediate Mechanics of Materials: Chapter 1 Strain • The total movement of a point with respect to a fixed reference coordinates is called displacement. Kinematics Displacements Strains Average normal strain Lf – Lo δ ε av = ---------------.M.

– α 2 Units of average strain • To differentiate average strain from strain at a point. or cm/cm.001 in / 1-16 . Increase in the angle (α > π / 2) result in negative shear strain π γ av = -. • in/in.M.5% is equal to a strain of 0. • prefix: μ = 10-6. 0. or m/m (for normal strains) • rads (for shear strains) • percentage.005 1000 μ in / in is equal to a strain 0. Vable Intermediate Mechanics of Materials: Chapter 1 Average shear strain Undeformed grid Wooden Bar with Masking Tape A Deformed grid Wooden Bar with Masking Tape A A1 γ B α C B π/2 C Wooden Bar with Masking Tape Wooden Bar with Masking Tape • Decreases in the angle (α < π / 2) result in positive shear strain.

M.41 Determine the deformation in bars AP and BP in Fig. 1. Vable Intermediate Mechanics of Materials: Chapter 1 Small Strain Approximation • Small-strain approximation may be used for strains less than 0. P1. • In small shear strain (γ) calculations the following approximation may be used for the trigonometric functions: tan γ ≈ γ sin γ ≈ γ cos γ ≈ 1 • Small-strain calculations result in linear deformation analysis.41 1-17 .41 using small strain approximation. P1. Fig.01 • Small normal strains are calculated by using the deformation component in the original direction of the line element regardless of the orientation of the deformed line element. • Drawing approximate deformed shape is very important in analysis of small strains.

Vable Intermediate Mechanics of Materials: Chapter 1 Strain at a point Engineering Strain Δu ε xx = lim ⎛ ------⎞ ⎝ Δx⎠ Δx → 0 = ∂u ∂x Δv ε yy = lim ⎛ ----.+ ----.⎞ = + ⎝ Δy Δx⎠ ∂y ∂x Δx → 0 Δy → 0 Δv Δw ∂v ∂w γ yz = γ zy = lim ⎛ ----.⎞ ⎝ Δy⎠ Δy → 0 = ∂v ∂y Δw ε zz = lim ⎛ -------⎞ ⎝ Δz-⎠ Δz → 0 = ∂w ∂z Δu Δv ∂u ∂v γ xy = γ yx = lim ⎛ -----. ε xx = du (x) dx 1-18 . then the partial derivative with respect to that coordinate will be same as ordinary derivative.+ -------⎞ = + ⎝ Δz Δy ⎠ ∂z ∂y Δy → 0 Δz → 0 Δw Δu ∂w ∂u γ zx = γ xz = lim ⎛ ------.M.+ ------⎞ = + Δz ⎠ ∂x ∂z Δx → 0⎝ Δx Δz → 0 • The partial derivative with respect to a coordinate implies that during the process of differentiation the other coordinates are held constant. • If a displacement is only a function of one coordinate.

Vable Intermediate Mechanics of Materials: Chapter 1 Finite Difference Approximation • Forward difference approximates the slope of the tangent using the point ahead of point i as: ui + 1 – ui ( ε xx ) = ---------------------i xi + 1 – xi • Backward difference approximates the slope of the tangent using the point behind i as: ui – ui – 1 ( ε xx ) = ---------------------i xi – xi – 1 • Central difference takes the average value using the point ahead and behind as: 1 ui + 1 – ui ui – ui – 1 ( ε xx ) = -.---------------------.+ ---------------------i 2 xi + 1 – xi xi – xi – 1 1-19 .M.

58.624 0.160 -0. 1-20 .480 0. ε yy .336 0.288 v (μmm) 0.384 0. and γ xy at points 1 and 4 shown in Figure P1.112 -0.224 Point 9 10 11 12 13 14 15 16 u (μmm) 0.128 -0.240 0.272 -0.128 -0.096 Determine the strains ε xx .080 -0.000 -0.58 The displacements u and v in the x and y directions respectively were measured by Moire' interferometry.048 0.256 0.128 -0.256 0.112 -0.304 0. Point 1 2 3 4 5 6 7 8 u (μmm) 0.144 0. Displacements of 16 points on the body and are as given below.112 0.M.032 -0. Vable Intermediate Mechanics of Materials: Chapter 1 1.224 0.336 0.048 -0.144 0.032 v (μmm) 0.048 -0.000 0.176 0.

Vable Intermediate Mechanics of Materials: Chapter 1 Strain Transformation Strain transformation equations in 2-D ε nn = ε xx cos θ + ε yy sin θ + γ xy sin θ cos θ ε tt = ε xx sin θ + ε yy cos θ – γ xy sin θ cos θ γ nt = – 2ε xx sin θ cos θ + 2ε yy sin θ cos θ + γ xy ( cos θ – sin θ ) Stress transformation equations in 2-D σ nn = σ xx cos θ + σ yy sin θ + 2τ xy sin θ cos θ σ tt = σ xx sin θ + σ yy cos θ – 2τ xy cos θ sin θ 2 τ nt = – σ xx cos θ sin θ + σ yy sin θ cos θ + τ xy ⎛ cos 2θ – sin θ⎞ ⎝ ⎠ 2 2 2 2 2 2 2 2 2 2 • tensor normal strains = engineering normal strains • tensor shear strains = (engineering shear strains)/ 2 Tensor strain matrix from engineering strains ε xx [ ε ] = ε yx = γ yx ⁄ 2 ε zx = γ zx ⁄ 2 ε xy = γ xy ⁄ 2 ε yy ε zy = γ zy ⁄ 2 T ε xz = γ xz ⁄ 2 ε yz = γ yz ⁄ 2 ε zz ε nn = { n } [ ε ] { n } ε nt = { t } [ ε ] { n } T T γ nt = 2ε nt ε tt = { t } [ ε ] { t } 1-21 .M.

= max ⎝ 2 2 2 2 ⎠ ε xy ε yy ε zy ε xz ε zz = ε1 ε2 + ε2 ε3 + ε3 ε1 ε xz ε yz ε zz = ε1 ε2 ε3 1-22 . Vable Intermediate Mechanics of Materials: Chapter 1 Characteristic equation 3 2 εp – I1 εp + I2 εp –I3 = 0 Strain invariants I 1 = ε xx + ε yy + ε zz = ε 1 + ε 2 + ε 3 I2 = ε xx ε xy ε ε ε + yy yz + xx ε yx ε yy ε zy ε zz ε zx ε xx I3 = ε yx ε zx Maximum shear strain ε1 – ε2 ε2 – ε3 ε3 – ε1 γ max ⎛ ----------------. ----------------. ----------------⎞ ----------.M.
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