Plane Stress Transformations



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6 Plane Stress Transformations ASEN 311 - Structures ASEN 311 Lecture 6 Slide 1

Plane Stress State ASEN 311 - Structures Recall that in a bod in plane stress, the general 3D stress state with 9 components (6 independent) reduces to 4 components (3 independent): z z z z zz plane stress 0 0 0 0 0 with = Plane stress occurs in thin plates and shells (e.g. aircraft & rocket skins, parachutes, balloon walls, boat sails,...) as well as thin wall structural members in torsion. In this Lecture we will focus on thin flat plates and associated two-dimensional stress transformations ASEN 311 Lecture 6 Slide

ASEN 311 - Structures Flat Plate in Plane Stress Thickness dimension or transverse dimension z Top surface Inplane dimensions: in, plane ASEN 311 Lecture 6 Slide 3

ASEN 311 - Structures Mathematical Idealization as a Two Dimensional Problem Midplane Plate ASEN 311 Lecture 6 Slide 4

Internal Forces, Stresses, Strains ASEN 311 - Structures Thin plate in plane stress z In-plane internal forces h d d p p p d d + sign conventions for internal forces, stresses and strains d d h In-plane stresses d d = In-plane bod forces h b d d b In-plane strains d d h ε ε γ = γ In-plane displacements h u d d u ASEN 311 Lecture 6 Slide 5

ASEN 311 - Structures Stress Transformation in D (a) P (b) tt tn P nt nn t n z Global aes, sta fied z θ Local aes n,t rotate b θ with respect to, ASEN 311 Lecture 6 Slide 6

ASEN 311 - Structures Problem Statement tt tn nt nn Plane stress transformation problem: given,, and angle θ epress, and in terms of the data nn tt nt t z θ n P This transformation has two major uses: Find stresses along a given skew direction Here angle θ is given as data Find ma/min normal stresses, ma in-plane shear and overall ma shear Here finding angle θ is part of the problem ASEN 311 Lecture 6 Slide 7

Analtical Solution This is also called method of equations in Mechanics of Materials books. A derivation using the wedge method gives ASEN 311 - Structures nn tt = ( ) sin θ cos θ + (cos θ sin θ) nt = cos θ + sin θ + sin θ cos θ = sin θ + cos θ sin θ cos θ For quick checks when θ is 0 or 90, see Notes. The sum of the two transformed normal stresses ο ο nn + = + tt is independent of the angle θ: it is called a stress invariant (mathematicall, this is the trace of the stress tensor). A geometric interpretation using the Mohr's circle is immediate. ASEN 311 Lecture 6 Slide 8

ASEN 311 - Structures Double Angle Version Using double-angle trig relations such as cos θ = cos θ - sin θ and sin θ = sin θ cos θ, the transformation equations ma be rewritten as + nn = + cos θ + sin θ nt = sin θ + cos θ Here is omitted since it ma be easil recovered as + nn tt ASEN 311 Lecture 6 Slide 9

ASEN 311 - Structures Principal Stresses: Terminolog The ma and min values taken b the in-plane normal stress nn when viewed as a function of the angle θ are called principal stresses (more precisel, principal in-plane normal stresses, but qualifiers "in-plane" and "normal" are often omitted). The planes on which those stresses act are the principal planes. The normals to the principal planes are contained in the, plane. The are called the principal directions. The θ angles formed b the principal directions and the ais are called the principal angles. ASEN 311 Lecture 6 Slide 10

Principal Angles ASEN 311 - Structures To find the principal angles, set the derivative of nn with respect to θ to zero. Using the double-angle version, d nn = ( ) sin θ + cos θ = 0 d θ This is satisfied for θ = θ if p tan θ = p (*) It can be shown that (*) provides two principal double angles, o θ p1 and θ p, within the range of interest, which is [0, 360 ] or [ 180 o o,180 ] (range conventions var between tetbooks). o The two values differ b 180. On dividing b we get the principal o angles θ p1 and θ p that differ b 90. Consequentl the two principal directions are orthogonal. ASEN 311 Lecture 6 Slide 11

ASEN 311 - Structures Principal Stress Values Replacing the principal angles given b (*) of the previous slide into the epression for and using trig identities, we get nn + 1, = + in which 1, denote the principal normal stresses. Subscripts 1 and correspond to taking the + and signs, respectivel, of the square root. A staged procedure to compute these values is described in the net slide. ASEN 311 Lecture 6 Slide 1

ASEN 311 - Structures Staged Procedure To Get Principal Stresses 1. Compute + av =, R = + + Meaning: av is the average normal stress (recall that + is an invariant and so is av), whereas R is the radius of Mohr's circle described later. This R also represents the maimum in-plane shear value, as discussed in the Lecture notes.. The principal stresses are = + R, = R 1 av av 3. The above procedure bpasses the computation of principal angles. Should these be required to find principal directions, use equation (*) of the Principal Angles slide. ASEN 311 Lecture 6 Slide 13

Additional Properties ASEN 311 - Structures 1. The in-plane shear stresses on the principal planes vanish. The maimum and minimum in-plane shears are +R and R, respectivel 3. The ma/min in-plane shears act on planes located at +45 and -45 from the principal planes. These are the principal shear planes 4. A principal stress element (used in some tetbooks) is obtained b drawing a triangle with two sides parallel to the principal planes and one side parallel to a principal shear plane For further details, see Lecture notes. Some of these properties can be visualized more easil using the Mohr's circle, which provides a graphical solution to the plane stress transformation problem ASEN 311 Lecture 6 Slide 14

ASEN 311 - Structures Numeric Eample (a) P = 0 psi = =30 psi =100 psi t =10 psi (b) n θ P principal directions θ = 108.44 =110 psi 1 θ 1=18.44 principal planes (d) plane of ma inplane shear (c) 45 = R=50 psi ma P principal stress element P 45 18.44 +45 = 63.44 principal planes principal planes For computation details see Lecture notes ASEN 311 Lecture 6 Slide 15

ASEN 311 - Structures Graphical Solution of Eample Using Mohr's Circle (a) Point in plane stress (a) P = 0 psi = =30 psi =100 psi = 10 50 40 30 0 = shear stress H 10 0 0 40 60 80 100 0 10 C 0 30 40 50 θ = 36.88 +180 = 16.88 ma = 50 Radius R = 50 V = normal stress = 110 1 θ 1 = 36.88 min = 50 (b) Mohr's circle coordinates of blue points are H: (0,30), V:(100,-30), C:(60,0) ASEN 311 Lecture 6 Slide 16

ASEN 311 - Structures What Happens in 3D? This topic be briefl covered in class if time allows, using the following slides. If not enough time, ask students to read Lecture notes (Sec 7.3), with particular emphasis on the computation of the overall maimum shear ASEN 311 Lecture 6 Slide 17

ASEN 311 - Structures General 3D Stress State z z z z zz There are three (3) principal stresses, identified as,, 1 3 ASEN 311 Lecture 6 Slide 18

Principal Stresses in 3D () ASEN 311 - Structures The i turn out to be the eigenvalues of the stress matri. The are the roots of a cubic polnomial (the so-called characteristic polnomial) C() = det 3 z z z 1 = + I I + I = 0 The principal directions are given b the eigenvectors of the stress matri. Both eigenvalues and eigenvectors can be numericall computed b the Matlab function eig(.) z zz 3 ASEN 311 Lecture 6 Slide 19

3D Mohr Circles (Yes, There Is More Than One) = shear stress Overall + ma shear Inner Mohr's circles ASEN 311 - Structures All possible stress states at the material point lie on the gre shaded area between the outer and inner circles = normal stress Principal stress 3 Principal stress 1 Outer Mohr's circle The overall maimum shear, which is the radius of the outer Mohr's circle, is important for assessing strength safet of ductile materials ASEN 311 Lecture 6 Slide 0

ASEN 311 - Structures The Overall Maimum Shear is the Radius of the Outer Mohr's Circle If the principal stresses are algebraicall ordered as 1 3 then overall 1 3 ma = R outer = Note that the intermediate principal stress does not appear. If the are not ordered it is necessar to use the ma function in a more complicated formula that picks up the largest of the three radii: overall 1 3 ma = ma,, 3 1 ASEN 311 Lecture 6 Slide 1

ASEN 311 - Structures Plane Stress in 3D: The 3rd Principal Stress Consider plane stress but now account for the third dimension. One of the principal stresses, call it for the moment 3, is zero:,, =0 1 where 1 and are the inplane principal stresses obtained as described earlier in Lecture 6. The zero principal stress 3 is aligned with the z ais (the thickness direction) while 1 and act in the, plane: z 3 Thin plate 3 1 ASEN 311 Lecture 6 Slide

Plane Stress in 3D: Overall Ma Shear ASEN 311 - Structures Let us now (re)order the principal stresses b algebraic value as 1 3 To compute the overall maimum shear cases are considered: (A) Inplane principal stresses have opposite signs. Then the zero stress is the intermediate one:, and overall inplane 1 3 ma = ma = (B) Inplane principal stresses have the same sign. Then overall 1 If 1 0 and 3 = 0, ma = overall If 3 3 0 and 1 = 0, ma = ASEN 311 Lecture 6 Slide 3

Plane Stress in 3D: Eample ASEN 311 - Structures Plane stress eample treated earlier: P = 0 psi = =30 psi =100 psi = 0 3 = 10 = shear stress 50 40 30 0 overall ma = 55 inplane = 50 ma 10 0 0 0 40 60 80 100 10 0 30 40 50 = normal stress = 110 1 Yellow-filled circle is the in-plane Mohr's circle ASEN 311 Lecture 6 Slide 4