Mechanics.Of.Materials.Saouma Episode 4 pps

...  J+J c ) (4. 73) or: E 11 = ∂u 1 ∂X 1 (4. 74- a) E 12 = 1 2  ∂u 1 ∂X 2 + ∂u 2 ∂X 1  (4. 74- b) ··· = ··· (4. 74- c) Note the similarity with Eq. 4. 7. Victor Saouma Mechanics of Materials II Draft 4. 3 Strain ... T 0 n 0 (4. 149 ) and for which df = t 0 dA 0 = tdA ⇒ t 0 = dA ddA 0 t (4. 150) using Eq. 4. 147 -b and 4. 149 the preceding equation becomes T 0 n 0 = dA dA 0 Tn =...

Ngày tải lên: 11/08/2014, 15:21

... B 12 C 11 D 21 + B 12 C 12 D 22 A 12 = B 11 C 11 D 11 + B 11 C 12 D 12 + B 12 C 11 D 21 + B 12 C 12 D 22 A 21 = B 21 C 11 D 11 + B 21 C 12 D 12 + B 22 C 11 D 21 + B 22 C 12 D 22 A 22 = B 21 C 21 D 11 + ... Mechanics 1 11. 2 Stress Intensity Factors 2 11 .3 Fracture Properties of Materials 10 11 .4 Examples 11 11 .4 .1 Example...

Ngày tải lên: 13/08/2014, 17:20

... σ (2) and n 2 i be the direction cosines of this axis,    2n 2 1 + n 2 2 + n 2 3 =0 n 2 1 − n 2 2 +2n 2 3 =0 n 2 1 +2n 2 2 − n 2 3 =0 ⇒ n 2 1 = 1 √ 3 ; n 2 2 = − 1 √ 3 ; n 2 3 = − 1 √ 3 (2. 44) Finally, ... (2. 27) II σ = −(σ 11 σ 22 + σ 22 σ 33 + σ 33 σ 11 )+σ 2 23 + σ 2 31 + σ 2 12 (2. 28) = 1 2 (σ ij σ ij − σ ii σ jj )= 1 2 σ ij σ ij − 1 2 I...

Ngày tải lên: 13/08/2014, 17:20

... e 2 ]+x 2 1 x 2 [e 1 ⊗ e 3 ] +x 2 2 x 3 [e 2 ⊗ e 1 ]+2x 1 x 2 x 3 [e 2 ⊗ e 2 ]+x 1 x 2 2 [e 2 ⊗ e 3 ] (3. 39-a) +x 2 x 2 3 [e 3 ⊗ e 1 ]+x 1 x 2 3 [e 3 ⊗ e 2 ]+2x 1 x 2 x 3 [e 3 ⊗ e 3 ] = x 1 x 2 x 3   2 x 1 /x 2 x 1 /x 3 x 2 /x 1 2 ... (3. 27-a) =8i − j − 10k at (1, −2, −1) (3. 27-b) n = 2i − j − 2k  (2) 2 +(−1) 2 +(−2) 2 = 2 3 i − 1 3 j − 2 3 k (3. 27-c) ∇φ...

Ngày tải lên: 13/08/2014, 17:20

... X 2 plane. Solution: [F]=   λ 1 00 00−λ 3 0 λ 2 0   (4. 43-a) det F = λ 1 λ 2 λ 3 (4. 43-b) ∆V = λ 1 λ 2 λ 3 (4. 43-c) ∆A 0 = 1 (4. 43-d) n 0 = −e 3 (4. 43-e) ∆An = (1)(det F)(F −1 ) T (4. 43-f) = λ 1 λ 2 λ 3   1 λ 1 00 00− 1 λ 3 0 1 λ 2 0      0 0 −1    =    0 λ 1 λ 2 0    (4. 43-g) ∆An ... 4. 101 du =(u∇ X ) P dX in the direction of −X 2 or {du} =   41 0 012 0...

Ngày tải lên: 13/08/2014, 17:20

... P 2 (5. 5-a)  C A·dr = 0 along a closed contour line (5. 5-b) 5. 3 Integration by Parts 5 The integration by part formula is  b a u(x)v  (x)dx = u(x)v(x)| b a −  b a v(x)u  (x)dx (5. 6) 5. 4 Gauss; ... deformation to the corresponding spatial vector dx at x. Thus, if we let d ˜ f = ˜ tdA 0 (4. 155 -a) and df = Fd ˜ f (4. 155 -b) where d ˜ f is the pseudo differential force which tra...

Ngày tải lên: 13/08/2014, 17:20

... =  V T: ˙ εdV +  V v·(∇·T)dV (6. 35-b) 41 We now evaluate P ext in Eq. 6. 31 P ext =  S t i v i dS +  V ρv i b i dV (6. 36- a) =  V v·(ρb + ∇·T)dV +  V T: ˙ εdV (6. 36- b) Using Eq. 6. 17 (T ij,j + ρb i = ... publish his law when first discovered it in 166 0. Instead he published it in the form of an anagram “ceiinosssttuu” in 167 6 and the solution was given in 167 8. Ut tensio sic vi...

Ngày tải lên: 13/08/2014, 17:20

... any deformation W>0thus λ + 2 3 G ≡ K>0 (7. 47- a) G>0 (7. 47- b) ruling out K = G = 0, we are left with E>0; −1 <ν< 1 2 (7. 48) 37 The isotropic strain energy function can be alternatively ... substituting into Eq. 7. 37 we obtain:    σ xx σ yy τ xy    = 1 1 − ν 2   1 ν 0 ν 10 00 1−ν 2      ε xx ε yy γ xy    (7. 55-a) ε zz = − 1 1 − ν ν(ε xx + ε yy ) (...

Ngày tải lên: 13/08/2014, 17:20

... − K III (2πr) 1 2 sin θ 2 (10. 38- a) τ yz = K III (2πr) 1 2 cos θ 2 (10. 38- b) σ xx = σ y = σ z = τ xy = 0 (10. 38- c) w = K III à 2r 1 2 sin 2 (10. 38- d) u = v = 0 (10. 38- e) where =34 for plane strain, ... − a 2 r 2  +  1+3 a 4 r 4 − 4a 2 r 2  1 2 σ 0 cos 2θ (9. 78- a) σ θθ = σ 0 2  1+ a 2 r 2  −  1+ 3a 4 r 4  1 2 σ 0 cos 2θ (9. 78- b) σ rθ = −  1 − 3a 4 r 4 + 2a 2 r 2  1 2...

Ngày tải lên: 13/08/2014, 17:20

... .90 29 . 797 1 1.60 1.1 899 1.0571 .92 42 .92 42 .8264 1.80 1.1476 1.0 495 .95 13 .95 13 .8677 2.00 1.11 49 1.04 09 .96 70 .96 70 . 895 7 2.20 1. 090 4 1.0336 .97 68 .97 68 .91 54 2.50 1.06 49 1.0252 .98 55 .98 55 .93 58 3.00 ... .6 898 .6 898 .5688 1.20 1.2208 .98 51 .7 494 .7 494 .6262 1.25 1.2405 1.0168 . 792 9 . 792 9 .6701 1.30 1.2457 1.0358 .82 59 .82 59 .7053 1.40 1.2350 1.0536 ....

Ngày tải lên: 13/08/2014, 17:20

... from E √ 10 to a critical value of magnitude E 100 √ 10 , or a factor of 100 . 20 Also σ theor max =2σ act cr  a a o a =10 6 m =1àm a o =1 A = =10 10 m theor max =2 act cr 10 6 10 10 = 200 act cr (12.29) 21 ... 000a 0 σ act cr =  Eγ 4a γ = Ea 0 10  σ act cr =  E 2 40 a o a a a 0 =2, 500  σ act cr =  E 2 100 ,000 = E 100 √ 10 (12.28) 19 Thus if we set a flaw si...

Ngày tải lên: 13/08/2014, 17:20

... and σ min = 50 MPa. The aircraft is made up of aluminum which has R =15 kJ m 2 ,E=70GPa, C =5ì 10 11 m cycle ,andn = 3. The smallest detectable aw is 4mm. How long would it be before the crack will ... can be rewritten as : ∆N = ∆a C [∆K(a)] n (15.3) or N =  dN =  a f a i da C [∆K(a)] n (15.4) 11 Thus it is apparent that a small error in the SIF calculations would be magnified greatly as...

Ngày tải lên: 13/08/2014, 17:20

... c 2233 E 22 E 33 +2c 2223 E 22 E 23 +2c 2231 E 22 E 31 +2c 2 212 E 22 E 12 + 1 2 c 3333 E 2 33 +2c 3323 E 33 E 23 +2c 3331 E 33 E 31 +2c 3 312 E 33 E 12 +2c 2323 E 2 23 +4c 2331 E 23 E 31 +4c 2 312 E 23 E 12 +2c 3131 E 2 31 +4c 3 112 E 31 E 12 +2c 121 2 E 2 12 (18.31) we ... W and obtain for instance T 12 = ∂W ∂E 12 =2c 6 + c 1 112 E 11 + c 2 212 E 22 + c 3 312 E 33 + c...

Ngày tải lên: 13/08/2014, 17:20

... (hardening) uniformly (see figure 19 .13) . s 1 s 2 s 3 n+1 n+1 (Hardening) F = 0 for p n F = 0 for (Perfectly Plastic) H = 0 (Softening) F = 0 for H < 0 p p H > 0 Figure 19 .13: Isotropic Hardening/Softening 1. ... φ (19.63) Victor Saouma Mechanics of Materials II Draft 19.4 Plastic Yield Conditions (Classical Models) 13 c Cot π Plane 3 2 1 σ σ ξ ρ σ Φ σ 1 ’ ρ ty ρ cy σ t σ t σ 1...

Ngày tải lên: 13/08/2014, 17:20

... elastic stiffness E o to the secant stiffness E s due to material damage is defined by E s =(1−D)E o (20 .14) where the damage parameter D provides a measure of the reduction of the stiffness due to the

Ngày tải lên: 13/08/2014, 17:20