Summary of Main Points

This section has explained the basic concept and the latest formulation of the general microplane model for concrete—a constitutive model in which the nonlinear triaxial behavior is characterized by relations between the stress and strain components on a microplane of any orientation under the constraint that the strains on the microplane are the projections of the macroscopic stress tensor. The microplane model simplifies constitutive modeling because the stress-strain relation on the microplane level involves only a few stress and strain components that have a clear physical meaning. The passage from elastic response to
softening damage defined in terms of different variables is effectively handled by the concept of boundaries in the stress-strain space. The advantage of this recently proposed concept is that various boundaries and the elastic behavior can be defined as a function of different variables (strain components). While the stress-strain boundaries for compression are defined separately for volumetric and deviatoric components, the boundary for tension is defined in terms of the total normal strains. This is necessary to achieve a realistic triaxial response at large tensile strains. A smooth transition from the elastic behavior to the boundary curve has also been formulated. The formulation is fully explicit, that is, the stress can be explicitly calculated from given strains.

Exercises

14.1 Based on symmetry properties, show that the rectangular cartesian components of the tensor A in (14.1.20) satisfy the following properties-, (a) the off-diagonal components arc zero; (b) the diagonal elements are equal, (c) Demonstrate that A =■- /11. where A is a scalar, and (d) compute A. (Hint; compute tr A = ЗА ^ (3/2tt) / df}—Why?)

14.2 Show that A in (14.1.20) can be written as A — (3/4тг) |n гфп ей? where the integral is now extended to the surface of the whole unit sphere and r is the position vector relative to the center of this sphere. Apply the divergence theorem to this surface integral and show that Л — (3/47г) {,, grad rdV where V is the region defined by the unit sphere. Use this expression to determine A.

14.3 Let Віуі, і be the rectangular cartesian components of the fourth-order tensor В in (14.1.20). Show that they satisfy the relations (a) BtjkiSki ~ A, j and B. t]ki6jk = A, i. A basic property of linear elasticity is that the most general cartesian form of an isotropic fourth-order tensor of elastic moduli, say B, such that <r — Be is isotropic, is Bijki ~ BaSijSkt + /А A/Ty, where Bn and B are constants, (b) Use the results in (a) to show that for jB in (14.1.20) Bn 0 and З Вi — tr A. (c) Use the result of the previous exercise to determine Вi.

14.4

Подпись: or Подпись: (14.1.45)
Summary of Main Points Summary of Main Points

Show that В in (14.1.20) can be written as В — (3/47г) JQ r ® r ® n dil where now the integral is extended to the surface of the whole unit spltere and r і s the position vector relative to tiie center of this sphere. Apply the divergence theorem to this surface integral and show that В = (3/4тт) Jv grad (r ® r ® r) dV where V is the region defined by the unit sphere. Show that the component form of this integral can be reduced to Bijki — bnJjk SjiJik t – SkiJij where J is the Huler tensor of inertia products for a sphere of unit radius and unit density with respect to its center.

Use the well-known result that the inertia moment of a homogeneous sphere relative to any diameter is 2mIt2/5, withm — mass and R = radius of the sphere, to prove that J – (1/5)1. Final 1 y, determine the general expression for the components B, jki-