2.002 MECHANICS AND MATERIALS II Spring, Creep and Creep Fracture: Part III Creep Fracture c L. Anand

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MA MECHANICS AND MATERIALS II Spring, 2004 Creep and Creep Fracture: Part III Creep Fracture c L. Anand

2 Mechanisms of Creep Fracture Figure 1: Creep strain-time curve for constant stress at constant temperature

3 Image removed due to copyright considerations. Figure 2: The upper row refers to low temperatures ( 0.3 T M ) where plastic flow does not depend strongly on temperature or time; the lower row refers to the temperature range ( 0.3 T M )in which materials creep.

4 Figure 3: Schematic of creep fracture mechanisms

5 Figure 4: Schematic of uniaxial stress versus time to fracture data

6 Image removed due to copyright considerations. Figure 5: Map of isothermal fracture data for Nimonic-80A

7 Image removed due to copyright considerations. Figure 6: Map of isothermal fracture data for 304 stainless steel

8 Image removed due to copyright considerations. Figure 7: Map of isothermal fracture data for 316 stainless steel

9 Image removed due to copyright considerations. Figure 8: Micrographs of copper plates illustrating the continuous distribution of creep damage in plates containing notches and subjected to far-field uniaxial tension. Note that it is predominantly the grain boundaries perpendicular to the applied stress that are preferentially damaged.

10 1. Creep Rupture Diagram Creep Fracture

11 logσ = log C (1/ν)logt f C t 1/ν C f = tf = σ σ ν Times to failure are normally presented as creep rupture diagrams. Their application is obvious. If you know the stress and temperature you can read off the life; for a given design life at a certain temperature, you can read off the stress.

12 2. Monkman-Grant t f = σ = C σ ν s { ɛ c } 1/n ss ɛ 0 ν σ ν = s ss ν ɛ 0 ν/n ɛ ν/n t f ( ɛ c ) ss = C (Monkman and Grant) ν 1 and C 0.1 the creep strain to Typically, fracture 10 %

13 3. Time-Temperature Equivalence

14 Data is given in terms of σ in psi, t f in hours and T in degrees Rankine ( F ).

15 Example Problem on Creep A support beam made of 18Cr 8Ni stainless steel is to be used in a chemical reaction chamber operating at C. The beam geometry and loading are idealized as shown below. The performance requirements are that 1. The beam is to carry a constant load F = 600 N.

16 2. No macro-crack formation due to creep fracture in 25 years. 3. Tip-deflection not to exceed 4 cm in 25 years. Determine if the beam meets the performance specifications. If either of the failure criteria are not met, then what is the maximum value of F that the beam can carry and not fail?

17 Data for 304 Stainless Steel ɛ c = Bσ n ; B = ,n =4.5 ss { } n σ ɛ = 1 ɛ 0 ; ɛ 0 = sec, s = 98MP a, n =4.5 s

18 Image removed due to copyright considerations. "Master Rupture Curve for 18-8 Stainless Steel."

19 Solution: M (x) = F (L x) 0 x L M (x) σ(x, y) = y 1/n sgn(y) (1) I n I n = A y 1+1/n da

20 For a rectangular beam { } n I n = bh 2 h 1/n (2) 2+4n 2 n 2 { } v M = ɛ 0 sgn(m ) x 2 si n v x BCs: v = 0 at x = 0, and =0 at x = 0 { } n [ F 1 (L x) n+2 + L n+1 L n+2 ] v = ɛ 0 x si n n +1 n +2 n +2 { } n L n+2 F δ = v (x = L) = ɛ 0 (3) si n n +2

21 Check for Macro-crack Formation From (1) and (2) 2y 1/n M (x) σ(x, y) = { } sgn(y) n bh 2 h 2+4n Since M (x) = F (L x), maximum moment is at x = 0, i.e., M max = FL, and maximum tensile stress occurs at y = h/2, we get σ max = { n FL 600 N 1 m } = { } bh n m

22 σ max =83.33 MP a, and since 1 MP a = 145 psi, we get σ max =12, 083 psi. The temperature is C = F = R From Larson-Miller master curve for 18-8 stainless steel, logσ = logp qθ p and q are constants

23 Solve for q: log(σ a /σ b ) q = ; logp = logσ a + qθ a θ b θ a For σ a =10, 000 psi, θ a =41, 000; for σ b =2, 000 psi, θ b =50, 000 Therefore, q = log(5) 9000 = Solve for p: logp = log10, ( ) 41, 000 = p = Now, with θ LM = T (20 + log(t f )), logσ = logp qt (20 + logt f ) or ( ) 1/(qT ) p log = log(10 20 t f ) σ t f =10 20 (p/σ) 1/(qT )

24 where t f is therupturetimein hours, T is the temperature in degrees Rankine and σ is the stress in psi. t f { 20 =10 20 { } , } =10 = hours Since 1 year = = 8760 hours, t f =28.69 years Therefore, a macro-crack will form on the tensile side of the beam after approximately 28 years. The beam is safe for 25 years. Check for Deflection

25 { } n L n+2 F δ = ɛ 0 si n n +2 n I n = bh 2 2+4n 2 { h } 1/n For b = 0.02 m, h = 0.04 m, n = 4.5, I n = m [3+(1/4.5)], F = 600 N, ɛ 0 =10 9 sec 1, and s = N/m 2 δ = 4.5 (1m) [2+4.5] N sec ( N 2 )( m [3+(1/4.5)]) 6.5 m δ = m/s δ = ( m/s) t t = 25 years = 25y 365da/y 24h/da 3600sec/h = sec

26 Therefore, δ = ( )m/s ( )s = m Too much deflection! A deflection of 4 cm would occur after only 0.04m t = = sec = 3000 hours m/s The load has to be decreased substantially. For a total deflectionof m in 25 years m δ = = m/s 25y 365da/y 24h/da 3600s/h

27 From 1/n ( δ/ɛ 0) F = ( ) si n = 231 N =52 lbf L n+2 n+2 Alternatively, the ratio of the deflections is the ratio of the deflection rates, which in turn are proportional to the load ratio, raised to the power n = 4.5; ( ) 4.5 F.04m/25y = 600 N 2.951m/25y ( ) F = 600 N = 600 N (.01355).2222 = 231 N