2006
DOI: 10.1201/9780203968253
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Design Analysis in Rock Mechanics

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Cited by 63 publications
(53 citation statements)
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“…Values Y = (10-100) × 10 9 N m −2 are typical for rock (Pariseau 2006), while the thermal expansivities of common rocks are within a factor of a few of α = 10 −5 K −1 (Lauriello 1974;Richter & Simmons 1974), which we take as our best-guess estimate of the expansivity of Phaethon. Substituting δT = 500 K (from the previous section) in Equation (2) we obtain thermal stress S = (5-50) × 10 7 N m −2 (500-5000 bars) corresponding to the temperature changes experienced on Phaethon as it moves around its orbit.…”
Section: Dust Production: Thermal Fracturementioning
confidence: 99%
“…Values Y = (10-100) × 10 9 N m −2 are typical for rock (Pariseau 2006), while the thermal expansivities of common rocks are within a factor of a few of α = 10 −5 K −1 (Lauriello 1974;Richter & Simmons 1974), which we take as our best-guess estimate of the expansivity of Phaethon. Substituting δT = 500 K (from the previous section) in Equation (2) we obtain thermal stress S = (5-50) × 10 7 N m −2 (500-5000 bars) corresponding to the temperature changes experienced on Phaethon as it moves around its orbit.…”
Section: Dust Production: Thermal Fracturementioning
confidence: 99%
“…4 and Eq. (3) is for a thrust-fault earthquake specifically; that is, at the initial state at t 0 , the vertical stress is larger than the horizontal stress because the rock's lateral earth coefficient is smaller than 1.0 (Pariseau, 2007). However, near the failure state, the vertical stress will be smaller than the horizontal stress because of the increasing external force in the horizontal direction from tectonic compression.…”
Section: The Basics and Algorithms Of The New Non-stationary Modelmentioning
confidence: 99%
“…4) can be expressed as a function of the minor principal stress σ 3 , and two strength parameters of the shearing plane, i.e., cohesion c and friction angle ϕ (Pariseau, 2007):…”
Section: The Basics and Algorithms Of The New Non-stationary Modelmentioning
confidence: 99%
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“…Shear strain g can be explained as a change in the position of one surface of a solid, right rectangular prism relative to a parallel surface Dl caused by shear stress t applied to the first surface, without an accompanying volume change, divided by the thickness of the prism h o (Pariseau 2012).…”
Section: Definitionmentioning
confidence: 99%