Issue 39

A. Risitano et alii, Frattura ed Integrità Strutturale, 39 (2017) 202-215; DOI: 10.3221/IGF-ESIS.39.20 206 dissipation to zero   r int 0   . Consequently, the volumetric density of the heat sources of a mechanical nature  sm can be written as the following: e sm e 0 :           (2) The only term that describes the dissipative process is the thermoelastic coupling. This equation is easily resolvable when considering an isotropic and homogeneous material for which e e , , 0         . Indeed, assuming the previous positions, the Helmholtz free energy becomes:           K Tr Tr 2 2 0 , ,                 (3) where:   0 (  ) is the initial free energy when the material is unstrained;   is the temperature variation in comparison with the reference temperature  0 ;   and  are the Lamé coefficients;  K is the bulk modulus;   is the thermal dilatation coefficient. Using Eq. (3) and Schwarz’s theorem, it is possible to write the Duhamel-Neumann equation calculating the derivate of     with respect to e    :       K Tr I , 2                   (4) where by, the derivative with respect to  can be written       K I K I , ,                   (5) Assuming the application of this equation to the transformation for which the elastic property and the thermal dilatation coefficient are temperature independent, Eq. (2) can be written as the following:     e e e e e sm e K I K Tr 0 0 0 : : : ,                                   (6) Analysing Eq. (6), it is possible to note that during a tensile test (strain is positive) all the coefficients are positive; therefore the equation will yield a negative result. This result implies that the material absorbs heat from the outside for elastic deformation and previous hypothesis. In this phase, the behaviour of the material is perfectly thermo-elastic and the temperature of the specimen decreases. Performing the test at a constant strain rate, the quantity  sm will be constant (the coefficients are considered to be independent of the stress state). When local plasticisation occurs, the positive heat sources are activated   r int 0   , causing a consequent increasing in temperature. The next step is to solve the previous expression. Assuming the following geometry of the specimen as the integration domain {L, a, e} , it is possible to write Eq. (1) using average values: c i T C T T s , ,                   (7)