1. True / False questions (mark the correct answer) : (1 point each) [10]

Transcription

1 NE 409 / 509 1of 4 +1 NUCLEAR MATERIALS - Test # NAME 4/15/1999 Total Points : 50 Wite down the appopiate equations and intemediate steps neatly so that you may qualify fo patial cedit even if the final answe is wong. Use the backside if moe oom is needed. - RELEVANT INFORMATION AND DATA GIVEN ON THE LAST PAGE tea it fo convenient usage - Units should be placed at appopiate places and definitely at the final answe. 1. Tue / False questions (mak the coect answe) : (1 point each) [10] a. An edge dislocation slips pependicula to itself (its line). Tue o false b. A scew dislocation slips along its Buge's vecto. Tue o false c. A sessile dislocation can move easily unde an applied stess. Tue o false d. A dislocation pinned between points emains to be semicicula (adius of cuvatue is half the pinning distance) unde no foce. Tue o false e. Like scew dislocations epel each othe. Tue o false f. A pismatic loop has Buge s vecto pependicula to the plane of the loop. Tue o false g. Kinks and jogs on dislocations make the cystals softe. Tue o false h. Some jogs on scew dislocations have scew chaacte. Tue o false i. A scew dislocation may change its glide plane by coss slip. Tue o false j. An edge dislocation may change its glide plane by coss slip. Tue o false. Evaluate the foce on a scew dislocation with Buges vecto along x axis due to an extenal stess σyy [3]

2 NE 409 / 509 of a. What is the mean distance between dislocations if the dislocation density is ρ? [] b. What will be the stengthening due to dislocations of density ρ (just wite down the fomula and define the tems) [] 4. A cylindical fcc single cystal with diamete of 0.1 mm is pulled in tension with a stess of 1000 MPa. [see fig.]. The loading diection is along [101] while the slip system is (111)[110]. (G = 00 GPa and a = Å) [101] a. What is the esolved shea stess along the slip diection in the slip plane? [4] b. If CRSS fo the slip system fo the mateial is 550 MPa, would the cystal defom? [1] An edge dislocation (AB) is situated pependicula to the slip diection (see Fig.) of the slip system and pinned at two points (A and B) sepaated by 1000 Å. The Buge s vecto of the dislocation is along the slip diection. c. Detemine the adius of cuvatue of the dislocation due to the applied stess? [4]

3 NE 409 / 509 3of 4 +1 d. What is the citical stess fo a Fank-Read souce to opeate (wite down the equation and define the tems)? [] e. Compute the minimum applied load (nomal to the specimen coss-section) in kn equied fo the pinned dislocation (AB) to opeate as a Fank-Read souce [5] f. If the dislocation line in the figue is a scew dislocation, what will be its Buges vecto? [3] g. What is the unit line vecto ( ˆ t ) of the dislocation (AB)? []

4 NE 409 / 509 4of Conside the following two dislocations in a bcc metal : a b 1 = [1 11] fig.) a and b = [111] on {110} planes (see C A (101) B (10 1 ) a. On the figue, identify the #1 and # dislocations (i.e., A and B) : [1] A = B = The two dislocations (#s 1 and ) glide on the two planes to inteact with each othe at point C to become one dislocation (b 3 ) : b 1 + b b 3 b. Wite down the equation and find the Buges vecto of #3 dislocation (at C) : [] c. Show that the eaction is valid (satisfy the two conditions) [3] c. What is the plane on which dislocation #3 lies? [4] d. Is this a glissile dislocation o sessile dislocation (explain)? []

5 NE 409 / 509 5of 4 +1 Relevant Equations etc σ = E ε = G γ G = σ = K ε n U el = σ E Ε (1+ν) κ = E 3(1-ν) J = K εn+1 u n + 1 G πw = σ cosφ cosλ P N = exp( ) ε& = α ρ b v 1 ν b ΓoT = αgb Gb h = 8π(1 ν) E el Gb 8π = R = αgb L = ngb σ = α Gb π l ε& = A σ n e -Q/RT v = o m F = G x t = (σ.b)xt l = 1 ρ l = 1 3 c f v = D l σ = αgb ρ σ = σ i + k y D σ = αgb 3 c (o αgb 3 αgb f v N) σ = D σ xz = σ zx = - Gb π σ yz = σ zy = Gb π Scew Dislocation b = bkˆ; tˆ = kˆ y x + y = - Gb sinθ π x x + y = Gb cosθ π STRESSES AROUND DISLOCATIONS Edge Dislocation b = bî; tˆ = kˆ Gb 3x σ xx = - y +y Gb x σ yy = y - y Gb x σ xy = σ yx = x - y σ zz = ν (σ xx + σ yy ) Cu : (fcc) a = 3Å, G = 110 GPa, ν = 0.33, Q v = 8kCal/mole, Q D = 55kCal/mole, σ el = 3b, E d = 4 ev Fe & Steels : (bcc) a =.8Å, G = 110GPa, Q v =35kCal/mole, Q D = 60kCal/mole, σ el = 3b, E d = 40eV CONSTANTS AND CONVERSIONS R = Cal/mole-K k = 13.8x10-4 J/K = 8.31J/mole-K N A = 6.0x10 3 1N = 0.4 lbf = 0.10 kgf, 1J = 0.39Cal/mole = 6.4x10 18 ev 1Pa = 1N/m = 0.145x10-3 psi 1psi = 6.895x10 3 Pa = 6.895x10 4 dy/cm = 7.03x10 - kgf/cm 10 6 J/m = MPa-m ν D = pe sec 1 ev/atom = 3.05 kcal/mole 1 amu = 1.66x10-4 g