lecture 5.4 - uniaxial strain
TRANSCRIPT
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Geodynamics www.helsinki.fi/yliopisto
Geodynamics
Basics of elasticity Lecture 5.4 - Uniaxial strain
Lecturer: David Whipp [email protected]
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Goals of this lecture
• Present an example of uniaxial strain in the context of sedimentation
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Stresses as a result of burial
• How does elastic stress change in sedimentary rocks as a result of burial?
• What stress/strain conditions are appropriate for this scenario?
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Uniaxial strain
• Uniaxial strain occurs when only one component of the principal strains is nonzero (𝜀1 in this example)
• In this case, if we consider 𝜀2 = 𝜀3 = 0, the equations for linear elasticity reduce to
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�2 = �3 =⌫
(1� ⌫)�1
�1 =(1� ⌫)E"1
(1 + ⌫)(1� 2⌫)
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Uniaxial strain
• Let’s consider a parcel of rock initially at the surface that now has been buried by sediments of density 𝜌 to a depth ℎ
• In this case, we can assume 𝜎1 is vertical and equal to the weight of the overburden, 𝜎1 = 𝜌𝑔ℎ
• From the equations on the previous slide, we find
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�2 = �3 =⌫
(1� ⌫)⇢gh
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p =1
3(�1 + �2 + �3) =
(1 + ⌫)
3(1� ⌫)⇢gh
Uniaxial strain
• Let’s now consider the effects on the deviatoric stress, the principal stresses minus pressure 𝑝
• Which results in deviatoric stresses
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�01 = �1 � p =
2(1� 2⌫)
3(1� ⌫)⇢gh
�02 = �2 � p = �0
3 = �3 � p = � (1� 2⌫)
3(1� ⌫)⇢gh
![Page 7: Lecture 5.4 - Uniaxial strain](https://reader030.vdocuments.site/reader030/viewer/2022012809/61bf07015c1b1f02747dd2a8/html5/thumbnails/7.jpg)
Uniaxial strain
• Let’s now consider the effects on the deviatoric stress, the principal stresses minus pressure 𝑝
• Which results in deviatoric stresses
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p =1
3(�1 + �2 + �3) =
(1� ⌫)
3(1� ⌫)⇢gh
�01 = �1 � p =
2(1� 2⌫)
3(1� ⌫)⇢gh
�02 = �2 � p = �0
3 = �3 � p = � (1� 2⌫)
3(1� ⌫)⇢ghUnder tension!
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Let’s see what you’ve learned…
• If you’re watching this lecture in Moodle, you will now be automatically directed to the quiz!
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