OUTLINE

1 2

Stress Stress Notation

Stress (or force intensity) at a point δA n depends on P ¾ Position t δF ¾Orientation Stress tensor (a second-order tensor) Stress vector (or traction) is defined as τzz Which has components in normal and tangential directions as τ τzx zy τyz τ xz τyy z

τxy τyx τxx y x [Reference # 2: Page 9 – 27] 3 4

Stress tensor (a second-order tensor) Note : Stress tensor is symmetric (see proof in Reference # 5 section 2.4) τxy = τyx •First index identifies the plan (normal direction). •Second index identifies the direction of the stress. i.e. τij = τji Or τyz = τzy τxz = τzx

e.g. τxy is a stress acting on x-plane, and pointing in y-direction. Sign convention +ive: The component acts τ τ τ are called “normal stresses” xx, yy, zz on +i face in +j direction or on –i face in –j direction τxy, τxz, τyx, …… are called “shearing stresses” -ive: The component acts on +i face in –j direction or on –i face in +j dircetion

5 6 Stress Transformation

Example : ΔS = area of ABC n = unit normal vector

τ τ11 τ 12 ΔS1, ΔS2, ΔS3 = area of 21 OBC,OCA,OAB τ22 τ 13 h = distance between O τ31 τ23 τ 32 and N τ 33 ΔS = ΔS n Normal stresses, Positive (+) Tension i i Negative (−) Compression Homework # 5: Reference # 2 Problem 2.50 7 [Reference # 2: Page 567 – 582] 8

From Newton’s law (ΣFi = mai ) Then, normal stress

mass shear stress Example : Find normal stress on plane whose unit normal vector n = (0,0.60,0.80) As h 0 kPa Called “Cauchy’s formula”

9 10

From stresses τ in x –system, find stresses τ ’ in x’ –system τnn = τji nj ni ij i ij i = τ11 n1 n1 + τ12 n1 n2 +τ13 n1 n3 From x’= β x + τ21 n2 n1 + τ22 n2 n2 +τ23 n2 n3 i ij j Where, β is cosine of angle + τ31 n3 n1 + τ32 n3 n2 +τ33 n3 n3 ij between xi’ and xj

= (2000)(0.6)(0.6) + (2000)(0.6)(0.8) x 2 Unit Vector = ek’

Where nj is unit vector to any = 2,640 kPa plane

11 12 If n is parallel to xk’ axis, then Hence,

nj =ek’ej = βkj y y’ The component in xm’ direction which is stress (τkm’)on xk’ face in xm’ direction is x’ x

z z’ τij τkm’

13 Homework # 6: Reference # 2 Problem 15.7 14

Principal stresses & principal planes Similar as eigenvalue From Cauchy’s formula (for a state of stress at a point) In matrix form :

nj is unit normal vector to any plane, there exist three mutually perpendicular axes (called principal axes) where there is zero or shear stress Ti=τni

Solve for principal stresses (τ) by using eigenvalue problem [Reference # 2: Page 583 – 598] 15 16

Solve for τ Invariants (not change with coordinate system)

Then,

Find 3 roots for τ : τ1, τ2, τ3

17 18 Mohr’s Circle illustrates principal stresses and stress transformations via a There is an orientation at graphical format σ3 which only normal stress σ acts. Mohr's Circle was the leading tool used to visualize relationships 2 between normal and shear stresses and to estimate the maximum σ1 , σ2 , σ3 principal stresses. stresses y >>Plane stress (2D)>> In matrix form τ yy τ = τ = τ = 0 τyx zz zx zy τ In matrix form xy x σ1 τ z xx Homework # 7: Reference # 2 Problem 15.14 19 [Reference # 2: Page 223 – 237] 20

τ σy τyx = τxy τ (σ ,τ ) max So, from the Mohr’s Circle τyx y xy L R τxy σ σavg σ σ K 2 1 σx (σx,-τxy) Where, Sign Convention ƒ Normal stress + tension − compression ƒ Shear stress + clockwise − counterclockwise 21 22

τ τmax Example : Solve : σ (σy,τxy) 2 2θ σ1 R p Max. shear stress L θp K σ2 σavg σ1 σ

y (σx,-τxy) x The angle between the current axes (X and Y) and the principal axes is defined as θp and is equal to 2θp in mohr’s circle

23 24 >>General state of stress(3D)>> From mohr’s circle, determine : Let x, y, z coincide with principal axes z τ3

Principal stresses : y

τ2 x τ1 Maximum stresses : Consider x-z plane get τ1 , τ3 Consider y-z plane get τ2 , τ3 Consider x-y plane get τ1 , τ2 25 26

Equilibrium Equation

Principal stresses : Lastly, stress was considered at one point. But in this section, stress is considered in a small element of deformable body which is in equilibrium condition. Largest shear stress :

Homework # 8: Reference # 2 Problem 7.34 27 [Reference # 2: Page 28 – 45] 28

Canceling terms and dividing by dxdydz, X axis

Similarly for y and z directions, so

29 30 Note : Bx = Body force As equilibrium equations (in differential form) These equations can be written as index notation form Solve :

Example : Find resultant force.

31 32