Page 226 - Analysis and Design of Energy Geostructures
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Deformation in the context of energy geostructures  199


                    r. For the cubic concrete sample:
                          The thermally induced volumetric stress is proportional to the volu-
                       metric thermal strain:

                                                      σ 5 3Kε   th
                                                        th
                          As the sample is fully restrained, the prevented strain is equal to the
                                                 th
                       volumetric thermal strain ε 5 αΔT.
                                                           26
                                  th
                          Thus: σ 5 3 KαΔT 5 3 20 10 10  10 5 6 MPa
                       For the cubic soil sample:
                          The same calculation is performed with a 30 MPa bulk modulus and
                       an equal coefficient of thermal expansion.
                                  th
                          Thus: σ 5 9 kPa
                    s. i. The principal stresses are the normal stresses acting on the principal
                              planes, along which zero shear stresses are observed. For a stress
                              tensor in the form of a diagonal matrix, the principal stresses are
                              thus the diagonal components σ kk of the tensor, that is:

                                           σ 1 5 10 kPa   acting on the yz plane


                                           σ 2 5 5 kPa    acting on the xz plane


                                           σ 3 5 2 kPa    acting on the xz plane
                       ii. The three stress invariants are defined as

                          8
                                            I 1 5 tr σ ij 5 σ ii 5 σ xx 1 σ yy 1 σ zz
                          >
                          >
                          >
                          <      1                                              2     2    2
                            I 2 5   σ ii σ jj 2 σ ij σ ij 5 σ xx σ yy 1 σ yy σ zz 1 σ zz σ xx 2 σ 2 σ 2 σ
                                 2                                              xy    yz   zx
                          >
                          >
                          >                                             2        2        2
                          :   I 3 5 detσ ij 5 σ xx σ yy σ zz 1 2σ xy σ yz σ zx 2 σ xx σ 2 σ yy σ 2 σ zz σ
                                                                        yz       zx       xy
                          with
                                              2               3   2          3
                                                                    10 0   0
                                                σ xx  σ xy  σ xz
                                         σ ij 5  4  σ yx  σ yy  σ yz  5  5  4  0  5  0  5
                                                                    0   0  2
                                                σ zx  σ zy  σ zz
                          Thus
                                              8
                                                   I 1 5 10 1 5 1 2 5 17
                                              <
                                                I 2 5 10 5 1 5 2 1 2 10 5 80
                                                     I 3 5 10 5 2 5 100
                                              :
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