Acta Technica Corviniensis

ACTA TECHNICA CORVINIENSIS – BULLETIN of ENGINEERINGT 2 = T 0 e -b1y1 sin(ωx - ωv 2 t - a 2 t) (9)As a result:where a 2 and b 2 correspond to the (4) and (5)equations with the correct changes for indices.Thus:q 2 = - K 2 (δT 2 / δy 2 ) y2->0 = K 2 T 0 [a 2 cos(ωx - ωv 2 t)– b 2 sin(ωx + ωv 2 t)] (10)If the wave is stationary and plate is movingrelative to it:andq 2 = K 2 T 0 (b 2 sin ωx + a 2 cos ωx) (11)q = q 1 + q 2 =T 0 [(K 1 b 1 + K 2 b 2 ) sin ωx+(K 2 a 2 - K 1 a 1 ) cos ωx] (12)STATE OF THERMO ELASTIC STRESS IN A PLATESUBJECTED TO A WAVE OF TEMPERATURE THATMOVES UNIFORMLYThe thermo elastic equation of a platedepending on the potential of displacement Ψis:δ 2 Ψ / δx 2 + δ 2 Ψ / δy 2 =(1+ ν 0 )αT 0 e -by sin (ωx + ay- ωvt) (13)where:α - coefficient of thermal expansionν 0 – Poisson’s coefficientThe speed v of the surface on the direction of yis zero (v y->0 ) and Ψ→0 when y→∞, δ Ψ / δy ≡v,resulting in:Ψ 1 = (Ae - ωy)(C cosωx+D sinωx) +(k 1 /vω)(1+v 1 ) α 1 T 0 e -b1y cos(ωx+ a 1 y) (14)Coefficients C and D are evaluated to meet thecondition on the limit. Results that the surfacepressure p 1’will be:p 1’ = E 1 α 1 T 0 k 1 [-(ω-b 1 ) cosωx+ a 1 inωx]/v 1 (15)A similar equation can be written for the bodynumbered 2. At the moment of contact betweenthe two bodies each surface will suffer adisplacement equal and contrary till theequalization of tensions:p 1” = Eωδ/2with δ= δ 0 sinωx ← thermal layer thicknessp= - E 1 ωδ/2 = p 1 ’+ p ” (16)p= E 1 ωδ/2 = p 2 ’+ p ” (17)Given the fact that p must be identical for thetwo bodies (according to the law of balance) δcan be eliminatedp=E 1 E 2 T 0 {[α 2 k 2 (ω-b 2 )/v 2 - α 1 k 1 (ω-b 1 )/v 1 ] cosωx+ [α 2 k 2 a 2 /v 2 +α 1 k 1 a 1 /v 1 ] sinωx}/(E 1 +E 2 ) (18)According to the principles of equilibrium, theheat generated by friction must be equal to theheat from the interface if:mp (v 1 + v 2 )= q (19)(K 1 b 1 +K 2 b 2 )sinωx +(K 2 b 2 -K 1 b 1 ) cosωx =(v 1 + v 2 )μE 1 E 2 {[α 2 k 2 (ω-b 2 )/v 2 - α 1 k 1 (ω-b 1 )/v 1 ] cosωx+[α 2 k 2 a 2 /v 2 +α 1 k 1 a 1 /v 1 ] sinωx }/(E 1 +E 2 ) (20)To satisfy the equation (2):K 1 b 1 +K 2 b 2 =(v 1 +v 2 )μE 1 E 2 [α 2 k 2 a 2 /v 2 +α 1 k 1 a 1 /v 1 ]/(E 1 +E 2 ) (21)K 2 b 2 -K 1 b 1 =(v 1 + v 2 )μE 1 E 2 [α 2 k 2 a 2 (ω-b 2 )/v 2- α 1 k 1 a 1 (ω-b 1 )/v 1 ]/(E 1 +E 2 ) (22)So for bodies of the same material: v 1 =v 2 =v/2,and equation (21) reduces to:μEαka/bK =1= μEαka{[1+[1+(v/kω) 2 ] 1/2 ]/[1+[1+(v/kω) 2 ] 1/2 ] } 1/2 /K (23)and for two bodies of which, one is good conduitfor heat and other heat isolated:k 1 →0, K→0, v 2 →0, and v 1 →v. If v>1,a 1 →ω(v 1 /2k 1 ω) 1/2a 2 →v 2 2k 2 ; b1→ω(v 1 /2k 1 ω) 1/2 ; b 2 →ω[1+(c 2 /k 2 ω) 2 /8]and equation (21) reduces to:CONCLUSIONv 1 =v=2K 2 ω(E 1 +E 2 )/ μE 1 E 2 α 2 (24)The equations from above serve to provide theterms depending on which the pressuredisturbance in a frontal sealing interfaceincreases. In this case load concentrations occur2008/ACTA TECHNICA CORVINIENSIS/Tome I 67