MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MA 02139 2.002 MECHANICS AND MATERIALS II Spring, 2004 Creep and Creep Fracture: Part I c�L. AnandRATE-DEPENDENCE AND RATE-INDEPENDENCE OF PLASTIC RESPONSE• Plastic deformation in metals is thermally-activated and inherently rate-dependent.• However, the plastic stress-strain response of most single and polycrystalline materials at absolute tem-peratures T < 0.35 Tm, where Tm is the melting tem-perature of the material in degrees absolute, is only slightly rate-sensitive, and in this temperature regime it is often be modeled as rate-independent. We shall first consider a rate-independent theory. Material Melting Temp, C Tm, K 0.35 Tm, K ≡ C Ti 1668 Fe 1536 Cu 1083 Al 660 Pb 327 1941 679 406 1809 633 360 1356 452 201 933 327 54 660 231 -42Consequences of Viscoplastic Deformation at High Homologous Temperature For isothermal deformation (T=const)Creep Test • A typical creep test consists of instantaneously load-ing a cylindrical test specimen of a material to a con-stant stress, which is maintained at a constant tem-perature. The resulting strain is measured as a func-tion of time.Idealization of Creep Curve • For deformation analysis at constant temperature, the strain-time response may be idealized as �c� = �e + ˙ tss�c�˙ = �˙e + ˙ss σ˙�e˙ = EStress Dependence of �˙ssc at Constant Temperature ˙�c ss = ˙�0 �σ�n sTemperature-Dependence of �˙c at Constant σ -ss Preliminaries • Avogadro’s number: NA = 6.022×1023 atoms/molecules per mole • Boltzmann’s constant k = 1.381 × 10−23 J/K • Universal gas constant R = NAk = 8.314 J(mol)−1K−1 • Mole: 1 mole of any substance is that mass of the substance containing NA atoms/molecules; e.g., the mass of 1 mole of C12 atoms is 12 grams.� � � � � � Temperature Dependence of �˙c at Constant Stress ss • The slope of the curve is m/1 = −Q/R, where Q is called the activation energy for creep, with units J(mol)−1. Then for a constant C(σ), the above curve can be mathematically represented as 1 Qln�˙c = lnC − Q ⇐⇒ �˙c = Cexp −ss R T ss RT� � �� An Important Observation �c• Let us evaluate the increase in ˙ for a material with ss Q = 270 kJ/mol when the temperature is increased from T1 = 8000 C = 1073 K to T2 = 8200 C = 1093 K. �c Q 1 1˙ss1 = exp − = 0.5746 �c R T2 T1˙ss2 • Therefore, with a temperature increase of only 200C, the creep rate almost doubled!! • Caution: the temperature T must be expressed in kelvinsCombined Stress and Temperature Dependence of �c˙ ss � � ��� �nQ σ �˙c = Aexp − RT s A pre-exponential factor (s−1) Q creep activation energy (J/mol) n creep exponent s reference stress which produces a strain rate ˙�0 ˙�0 = Aexp � − Q � RTSummary of One-Dimensional Creep Equation �˙ = �˙e + ˙�c (1) σ˙�˙e = ; E = E(T ) (2) E � � ��� �nQ σ �˙c = Aexp − (3) RT s • Note that equation (3) states that the rate of creep (or ‘viscoplastic’) strain increases exponentially with temperature, so that the time required for a given amount of creep strain decreases exponentially with temperature.Example Problem: Stress Relaxation • Consider a bolt with pre-tension σ = σi at time t = 0. Given that the bolt is maintained at constant tem-perature, determine the pre-tension at some time t. The isothermal constitutive equation for steady state creep is given by ˙�c = Bσn with n �= 1Example Problem: Stress Relaxation (cont.) �c�˙ = �˙e + ˙�c ⇒ 0 = �˙e + ˙ since � = const in the bolt σ˙ 1 dσ ⇒ 0 = + Bσn ⇒ = −Bσn E E dt � t� σ(t) ⇒ σ−ndσ = −EB dt ⇒ σ−ndσ = −EB dt σi=σ(0) ⇒ σ(t)−(n−1) − σ−(n−1) = (n − 1)EBt i σi ⇒ σ(t) = [1 + (n − 1) t Bσin (E/σi)]1/(n−1) • Defining the characteristic relaxation time tr such that σ(tr) = σi/2, we get 02(n−1) − 1 2(n−1) − 1 tr = = (n − 1)EBσi (n−1) (n −