MIT - 16.20 Fall, 2002 Unit 8 Solution Procedures Readings : R Ch. 4 T & G 17, Ch. 3 (18-26) Ch. 4 (27-46) Ch. 6 (54-73) Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Summarizing what we ’ ve looked at in elasticity, we have: 15 equations in 15 unknowns - 3 equilibrium - 6 strains - 6 strain-displacement - 3 displacements - 6 stress-strain - 6 stresses These must be solved for a generic body under some generic loading subject to the prescribed boundary conditions (B.C. ’s) There are two types of boundary conditions: 1. Normal (stress prescribed) 2. Geometric (displacement prescribed) --> you must have one or the other To solve this system of equations subject to such constraints over the continuum of a generic body is, in general, quite a challenge. There are basically two solution procedures: 1. Exact -- satisfy all the equations and the B.C. ’ s 2. Numerical -- come as “close as possible ” (energy methods, etc.) Paul A. Lagace ? 2001 Unit 8 - p . 2 MIT - 16.20 Fall, 2002 Le t ’s consider “exact” techniques. A common, and classic, one is: Stress Functions ? Relate six stresses to (fewer) functions defined in such a manner that they identically satisfy the equilibrium conditon ? Can be done for 3-D case ? Can be done for anisotropic (most often orthotropic ) case --> See: Lekhnitskii , Anisotropic Plates , Gordan & Breach, 1968. --> Let’ s consider ? plane stress ? (eventually) isotropic 8 equations in 8 unknowns - 2 equilibrium - 3 strains - 3 strain-displacement - 2 displacements - 3 stress-strain - 3 stresses Paul A. Lagace ? 2001 Unit 8 - p . 3 MIT - 16.20 Fall, 2002 Define the “ Airy ” Stress Function = φ (x, y) English a scalar mathematician ? 2 φ σ xx = 2 + V (8 - 1 ) ? y ? 2 φ σ yy = ? x 2 + V (8 - 2 ) 2 φ σ xy = ? ? (8 - 3 ) xy ?? where: V = potential function for body forces f x and f y f x = ? ? V f y = ? ? V ? x ? y V exists if ? x f = 0 ( curl ) that is: ? f x = ? f y ? y ? x Paul A. Lagace ? 2001 Unit 8 - p . 4 MIT - 16.20 Fall, 2002 Recall that curl f = 0 ? “ conservative ” field - gravity forces - spring forces - etc. What does that compare to in fluids? Irrotational flow Look at how φ has been defined and what happens if we place these equations (8-1 - 8-3) into the plane stress equilibrium equations: ? σ xx + ? σ xy + f x = 0 (E1) ? x ? y ? σ ? σ xy yy ? x + ? y + f y = 0 (E2) we then get: Paul A. Lagace ? 2001 Unit 8 - p . 5 ? ?? ?? MIT - 16.20 Fall, 2002 2 ? ? ? 2 φ E 1 () : ? 2 + V ? ? + ? ? ? ? ? φ ?? ? ? V = 0 ? x ? x ? ? y ? ? y ? ?? y ? ? x ? 3 φ + ? V ? ? 3 φ ? ? V = 0 ( yes ) ? ?? xy xy 2 ? x ?? 2 ? x 2 ? ? ? ? ? φ ? ? ? ? 2 φ ( E 2 ) : ? x ? ?? y ?? + ? y ? ? ? x 2 + V ??? ? ? V = 0 ? x ? y ? 3 φ + ? 3 φ + ? V ? ? V = 0 ( yes ) ? 2 2 xy xy ? y ? y ? Equilibrium automatically satisfied using Airy stress function! Does that mean that any function we pick for φ (x, y) will be valid? No , it will satisfy equilibrium, but we still have the strain-displacement and stress-strain equations. If we use these, we can get to the governing equation: Paul A. Lagace ? 2001 Unit 8 - p . 6 (E MIT - 16.20 Fall, 2002 Step 1 : Introduce φ into the stress-strain equations (compliance form): 1 E 3 ε xx = E ( σ xx ? ν σ yy ) () 1 ε yy = ( ? ν σ xx + σ yy ) (E4) E ε xy = 2 ( 1 E + ν ) σ xy (E5) So: ε xx = 1 ? ? 2 φ ? ν ? 2 φ ? + ( 1 ? ν ) V ( E 3 ′ ) E ? ? ? y 2 ? x 2 ?? E ε yy = 1 ? ? 2 φ ? ν ? 2 φ ? + ( 1 ? ν ) V (E 4 ′ ) E ? ? ? x 2 ? y 2 ? ? E ε xy = ? 2 ( 1 + ν ) ? 2 φ (E 5 ′ ) xy E ?? Paul A. Lagace ? 2001 Unit 8 - p . 7 MIT - 16.20 Fall, 2002 Step 2 : Use these in the plane stress compatibility equation: ? 2 ε xx ? 2 ε yy ? 2 ε xy + = xy ? y 2 ? x 2 ?? ( E 6 ) ? we get quite a mess! After some rearranging and manipulation, this results in: V V ?? + ? ?? + ?? = ? ? ( ) ? + ? ( ) ? ?? ? ? ? ? ? ( ) ?? + ?? ?? ? ?? ? 4 4 4 2 2 4 2 2 2 2 2 2 2 2 1 φ φ α x y y E T x T y y 2 ? ? 2 φ ν x x ? (*) temperature term we haven ’ t yet considered α = coefficient of thermal expansion ? T = temperature differential This is the basic equation for isotropic plane stress in Stress Function form Recall : φ is a scalar Paul A. Lagace ? 2001 Unit 8 - p . 8 MIT - 16.20 Fall, 2002 If we recall a little mathematics, the Laplace Operator in 2-D is: ? 2 ? 2 ? 2 = + ? x 2 ? y 2 ? 4 ? 4 ? 4 ?? 2 ? 2 = ? 4 = ? x 4 + 2 2 2 + ? y 4 x ?? y This is the biharmonic operator (also used in fluids) So the (*) equation can be written: ? 4 φ = ? E α ? 2 ( ? T ) ? ( 1 ? ν ) ? 2 V (*) Finally, in the absence of temperature effects and body forces this becomes: ? 4 φ = 0 homogeneous form What happened to E, ν ?? Paul A. Lagace ? 2001 Unit 8 - p . 9 MIT - 16.20 Fall, 2002 ? this function, and accompanying governing equation, could be defined in any curvilinear system (we ’ ll see one such example later) and in plane strain as well. But …what ’ s this all useful for??? This may all seem like “ magi c ” . Why were the σ ’s assumed as they were? This is not a direct solution to a posed problem, per se , but is known as… The Inverse Method In general, for cases of plane stress without body force or temp ( ? 4 φ = 0 ): 1. A stress function φ (x, y) is assumed that satisfies the biharmonic equation 2. The stresses are determined from the stress function as defined in equations (8-1) - (8-3) 3. Satisfy the boundary conditions (of applied tractions) 4. Find the ( structural ) problem that this satisfies Paul A. Lagace ? 2001 Unit 8 - p . 1 0 MIT - 16.20 Fall, 2002 ? Mathematicians actually did this and created many solutions. So there are many stress functions that have been found to solve specific structural problems (see, for example, Rivell o , pp, 72-73 also T & G) ? These are linear solutions and thus the “ Principle of Superposition ” applies such that these can be combined to solve any particular problem ? The inverse method yields an exact solution . In real life, an exact solution generally cannot be obtained. We often “ notch it down one ” and resort to the … The Semi-Inverse Method This is basically the same as the Inverse Method except that the solution is not exact in that we … ? Make simplifying assumptions to get solvable equations. These can be with regard to: – stress components – displacement components Paul A. Lagace ? 2001 Unit 8 - p . 1 1 MIT - 16.20 Fall, 2002 ? Assumptions are based on physical intuition, experimental evidence, prior experience (sheer need?) ? Assumptions may be due to … – boundary conditions not properly met – etc. ( Note : plane stress is generally an assumption) ? There is an important concept that allows us to make such assumptions: St. Venant ’ s Principle “If the loading distribution on the small section of the surface of an elastic body is replaced by another loading which has the same resultant force and moment as the original loading, then no appreciable changes will occur in the stresses in the body except in the region near the surface where the loading is altered”. What does this really say? ? “Far” from the specifics of load introduction / boundary conditions, the specifics of such are unimportant ? This is a ramification of the issue of “scale” as discussed earlier in this course and in Unified. Paul A. Lagace ? 2001 Unit 8 - p . 1 2 MIT - 16.20 Fall, 2002 ?( “ Rule of Thumb ” ) Generally, the area where “specifics ” are important extends into the body for a distance equal to about the greatest linear dimension of the portion of the surface on which the loading / B.C. occurs. ? This allows us to get solutions for most parts of a structure via such a method. But failure often originates/occurs in a region of load introduction/boundary condition ( example : where do nailed/screwed boards break?) Examples (for stress functions) Example 1 (assume ? T = 0, V = 0 [no body forces]) Pick φ = C 1 y 2 this satisfies ? 4 φ = 0 C 1 is a constant …will be determined by satisfying the B.C.’ s Paul A. Lagace ? 2001 Unit 8 - p . 1 3 MIT - 16.20 Fall, 2002 Using the definitions: ? 2 φ σ xx = 2 = 2 C 1 ? y ? 2 φ σ = = 0 yy ? x 2 σ = 0 xy gives the state of stress What problem does this solve??? Uniaxial loading Figure 8.1 Representation of Uniaxial Plane Stress Loading applied stress (uniform stress on end) Paul A. Lagace ? 2001 Unit 8 - p . 1 4 MIT - 16.20 Fall, 2002 Check the B.C. ’ s : @ x = l , 0 σ σ xx = σ o = 2 C 1 ? C 1 = o 2 σ = 0 xy @ y = ± w 2 σ = 0 yy σ = 0 xy Thus: φ = σ o y 2 2 Paul A. Lagace ? 2001 Unit 8 - p . 1 5 MIT - 16.20 Fall, 2002 ? The strains can then be evaluated using the stress-strain equations (compliance form). ? The displacements can then be determined by using the strain-displacement relations and integrating and applying the displacement B.C. ’ s . In this case, you get: σ o u = x E v = ? ν σ o y E ( Note : elastic constants now come in) --> Let’ s look at the “real ” case and see where/why we have to apply St. Venant ’ s Principle and the “Semi - Inverse Method ” . Consider a test coupon in mechanical grips (end of specimen is fixed): Paul A. Lagace ? 2001 Unit 8 - p . 1 6 MIT - 16.20 Fall, 2002 Figure 8.2 Representation of uniaxial test specimen and resultant stress state x y These grips allow no displacement in the y-direction --> v = 0 both @ x = 0, l The solution gives: ? νσ v = y ≠ 0 i n general @ x = 0 , l E But , “far ” from the effects of load introduction, the solution φ holds. Near the grips, biaxial stresses arise. Often failure occurs here. ( Note : not in a pure uniaxial field. This is a common problem with test specimens.) Paul A. Lagace ? 2001 Unit 8 - p . 1 7 MIT - 16.20 Fall, 2002 Example 2 Let ’s now get a bit more involved and consider the … Stress Distribution Around a Hole Figure 8.3 Configuration of uniaxially loaded plate with a hole Large ( “ infinite ”) plate subjected to uniform far - field tension Since the “local specific ” of interest is a circle, it makes sense to use polar coordinates. By using the transformations to polar coordinates of Unit #7, we find: Paul A. Lagace ? 2001 Unit 8 - p . 1 8 MIT - 16.20 Fall, 2002 1 ? φ 1 ? 2 φ σ rr = r ? r + r 2 ? θ 2 + V ? 2 φ + V σ θθ = ? r 2 1 ? 2 φ 1 ? φ + r σ r θ = ? r ?? θ r 2 ? θ These are, again, defined such that equilibrium equations are automatically satisfied. where: f r = ? ? V f θ = ? 1 ? V ? r r ? θ ? f θ + f θ 1 ? f r and V exists if: = ? r r r ? θ The governing equation is again: ? 4 φ = ? E α ? 2 ( ? T ) ? ( 1 ? ν ) ? 2 V for plane stress, isotropic Paul A. Lagace ? 2001 Unit 8 - p . 1 9 ? ? MIT - 16.20 Fall, 2002 but the Laplace operator in polar coordinates is: ? 2 = ? 2 + 1 ? + 1 ? 2 ? r 2 r ? r r 2 ? θ 2 Let ’s go through the steps … Step 1 : Assume a φ (r, θ ) “We know ” the correct function is: φ = [ A 0 + B 0 ln r + C 0 r 2 + D 0 r 2 ln r ] + ?? A 2 r 2 + B 2 2 + Cr 4 + D 2 ?? cos 2 θ 2 r Does this satisfy equilibrium? It must . One can show it satisfies ? 4 φ However, it can also be shown that non-zero values of D 0 result in mulitvalued displacements in the y-direction (v), so we must get D 0 = 0 [don’ t worry about this, just “accept ” it …I do!] Paul A. Lagace ? 2001 Unit 8 - p . 2 0 ? ? ? ? ? ? MIT - 16.20 Fall, 2002 Step 2 : Determine stresses Performing the derivatives from the φ - σ relations in polar coordinates results in: σ rr = B 2 0 + 2 C 0 + ?? ? 2 A 2 ? 6 B 4 2 ? 4D 2 ? cos 2 θ r r r 2 ? σ θθ = ? B 2 0 + 2 C 0 + ?? 2 A 2 + 6 B 4 2 + 12C 2 r 2 ?? cos 2 θ r r 2 in σ r θ = ? ? 2 A 2 ? 6 B 4 2 + 6C 2 r 2 ? 2D 2 ?? s2 θ r r Note that we have a term involving r 2 . As r gets larger, the stresses would become infinite. This is not possible. Thus, the coefficient C 2 must be zero: C 2 = 0 So we have five constants remaining: A 2 , B 0 , B 2 , C 0 , D 2 we find these by … Paul A. Lagace ? 2001 Unit 8 - p . 2 1 MIT - 16.20 Fall, 2002 Step 3 : Satisfy the boundary conditions What are the boundary conditions here? ? at the edge of the hole there are no stresses (stress-free edge) ? @ r = a: σ rr = 0, σ θ r = 0 ? at the y-edge, there are no stresses ? @ y = ± ∞ : σ yy = 0, σ yx = 0 ? at the x-edges, the stress is equal to the applied stress and there is no strain ? @ x = ± ∞ : σ xx = σ 0 , σ xy = 0 Paul A. Lagace ? 2001 Unit 8 - p . 2 2 MIT - 16.20 Fall, 2002 Since we are dealing with polar coordinates, we need to change the last two sets of B. C . ’ s to polar coordinates. We use: σ ? αρ = l 2 ? σ l ? βγ l σγ and look at r = ∞ σ ? xx = σ rr = cos 2 θ σ o σ ? yy = σ θθ = s in 2 θ σ o σ ? xy = σ r θ = ? sin θ cos θ σ o Figure 8.4 Representation of local rotation of stresses from polar to Cartesian system We use the double angle trigonometric identities to put this in a more convenient form: Paul A. Lagace ? 2001 Unit 8 - p . 2 3 MIT - 16.20 Fall, 2002 cos 2 θ =� 1 ( 1 + cos 2 θ ) 2 1 sin θ cos θ = sin 2 θ 2 sin 2 θ =� 1 ( 1 ? cos 2 θ ) 2 Thus at r = ∞ σ σ o σ = o + cos 2 θ rr 2 2 σ σ r θ = ? o s in 2 θ 2 Why don ’t we include σ θθ ? At the boundary in polar coordinates, this stress does not act on the edge/boundary. Paul A. Lagace ? 2001 Unit 8 - p . 2 4 MIT - 16.20 Fall, 2002 Figure 8.5 Stress condition at boundary of hole So we (summarizing) appear to have 4 B. C. ’s: σ σ σ rr = o + o cos 2 θ @ r = ∞ 2 2 σ σ r θ = ? o s in 2 θ @ r = ∞ 2 σ rr = 0 @ r = a σ r θ = 0 @ r = a Paul A. Lagace ? 2001 Unit 8 - p . 2 5 MIT - 16.20 Fall, 2002 And we have 5 constants to determine. What happened? The condition of σ rr at r = ∞ is really two B. C. ’s ? a constant part ? a part multiplying cos 2 θ Going through (and skipping) the math, we end up with: σ σ θ σ σ θ σ σ θθ θ rr o 2 2 o 2 2 4 4 o 2 2 o 4 4 o 2 2 1 a r 2 1 4 a r 3 a r 2 1 a r 2 1 3 a r 2 1 2 a r = ? ? ? ? ? ? ? + ? + ? ? ? ? ? ? = ? ? ? ? ? ? ? + ? ? ? ? ? ? = ? + cos cos 2 2 r 22 4 4 3 a r in ? ? ? ? ? ? ? s2 θ σ σ + for: … Paul A. Lagace ? 2001 Unit 8 - p . 2 6 MIT - 16.20 Fall, 2002 Figure 8.6 Polar coordinate configuration for uniaxially loaded plate with center hole So we have the solution to find the stress field around a hole. Let’ s consider one important point. Wher e ’s the largest stress? At the edge of the hole. Think of “flo w ” around the hole: Figure 8.7 Representation of stress “flow ” around a hole Paul A. Lagace ? 2001 Unit 8 - p . 2 7 MIT - 16.20 Fall, 2002 @ θ = 90 °, r = a σ ? a 2 ? σ ? a 4 ? o σ θθ = σ xx = 2 o ? ? 1 + a 2 ?? ? 2 ?? 1 + 3 a 4 ? ? ( ? 1 ) = 3 σ o Define the: Stress Concentration Facto r (SCF) = local stress far - field stress SCF = 3 at hole in isotropic plate The SCF is a more general concept. Generally the “ sharper” the discontinuity, the higher the SCF. The SCF will also depend on the material. For orthotropic materials, it depends on E x and E y getting higher as E x / E y increases. In a un i -directional composite, can have SCF = 7. Can do stress functions for orthotropic materials, but need to go to complex variable mapping --> (See Lekhnitskii as noted earlier) Paul A. Lagace ? 2001 Unit 8 - p . 2 8 MIT - 16.20 Fall, 2002 Example 3 -- Beam in Bending (we’ll save for a problem set or a recitation) There are many other cases (use earlier references) ? circular disks ? rotary disks . . “ c lassic ” cases But …what ’ s really the point? Are these still used? Yes! This is a very powerful technique which is especially well-suited for preliminary design and exploratory development ? parametric study ? know assumptions and resulting limitations and then interpret results accordingly ? linear solutions --> can use principle of superposition ? can find many solutions in books (See attached page) Paul A. Lagace ? 2001 Unit 8 - p . 2 9 MIT - 16.20 1. 2. 3. Fall, 2002 Roark, Raymond J., “Roark’ s Formulas for Stresses and Strain, 6th Ed., ” New York, McGraw-Hill, 1989. Peterson, Rudolph E., “ Stress Concentration Factors: Charts and Relations Useful in Making Strength Calculations for Machine Parts and Structural Elements, ” New York, Wiley, 1974. Pilkey , Walter D., “Formulas for Stress, Strain, and Structural Matrices, ” New York, John Wiley, 1994. Paul A. Lagace ? 2001 Unit 8 - p . 3 0