THE TRENCH FLEXURE PROBLEM and

Transcription

1 THE TRENCH FLEXURE PROBLEM and THE YIELD STRENGTH OF THE OUTER RISE Presented by: Paul Evans, Christine Wittich, and Qian Yao For: SIO234 Geodynamics, Fall 2011

2 BRIEF OVERVIEW Discussion of the trench flexure problem and history Derivation of the analytical deflection equation for a trench (elastic) Static analysis of the trench Bathymetric profile comparison Observations Yield strength envelope of the outer rise 2

3 THE TRENCH FLEXURE PROBLEM 3

4 THE TRENCH FLEXURE PROBLEM 4

5 λ 2 = S + α = (D/K) 1/4 S2 4DK 2D x >,ω = 0 C 1 = C 2 = 0 x = 0,ω = 0 C 3 = C 4 Dλ 4 + Sλ 2 + K = 0 ε = S 2(DK) 1/ 2 λ 1 = [( 1 ε 2 )1/ 2 + i( 1 +ε 2 )1/ 2 ]α 1 λ 2 = [( 1 ε 2 )1/ 2 i( 1 +ε 2 )1/ 2 ]α 1 λ 3 = [( 1 ε 2 )1/ 2 + i( 1 +ε 2 )1/ 2 ]α 1 λ 4 = [( 1 ε 2 )1/ 2 i( 1 +ε 2 )1/ 2 ]α 1 ω = 2iC 3 e α 1 ( 1 ε 2 )1/2 x sin( α 1 ( 1+ε 2 )1/2 x ) = Ae α 1 ( 1 ε 2 )1/2 x sin( α 1 ( 1+ε 2 )1/2 x ) A = 2iC 3 D d4 ω dx 4 + S d2 ω dx 2 + Kω = 0 ω = C 1 e λ 1x + C 2 e λ 2x + C 3 e λ 3x + C 4 e λ 4 x DERIVATION = C 1 e [(1 ε 2 )1/2 +i( 1+ε 2 )1/2 ]α 1 x + C2 e [(1 ε 2 )1/2 i( 1+ε 2 )1/2 ]α 1 x + C3 e [(1 ε 2 )1/2 +i( 1+ε 2 )1/2 ]α 1 x + C4 e [(1 ε 2 )1/2 i( 1+ε 2 )1/2 ]α 1 x λ 2 = S ± S 2 4DK 2D = 2(DK)1/ 2 ε ± 4DK(ε 2 1) 2D = [ ε ± ε 2 1](2(DK) 1/ 2 ) 2D = (ε ± i 1 ε 2 )(K /D) 1/ 2 = [( 1 ε 2 )1/ 2 ± i( 1 +ε 2 )1/ 2 ] 2 α 2

6 DERIVATION One dimensional bending of a thin plate D d4 ω dx 4 + S d2 ω dx 2 + Kω = 0 Boundary condition: x >,ω = 0 x = 0,ω = 0

7 DERIVATION D d4 ω dx 4 + S d2 ω dx 2 + Kω = 0 Dλ 4 + Sλ 2 + K = 0 ω = C 1 e λ 1x + C 2 e λ 2x + C 3 e λ 3x + C 4 e λ 4 x λ 1,λ 2,λ 3,λ 4 C 1,C 2,C 3,C 4 are the nonrepeated roots of Dλ 4 + Sλ 2 + K = 0 are constants

8 DERIVATION λ 2 = S ± S 2 4DK 2D = 2(DK)1/ 2 ε ± 4DK(ε 2 1) 2D = [ ε ± ε 2 1](2(DK) 1/ 2 ) 2D = (ε ± i 1 ε 2 )(K /D) 1/ 2 = [( 1 ε 2 )1/ 2 ± i( 1 +ε 2 )1/ 2 ] 2 α 2 ε = S 2(DK) 1/ 2 α = (D/K) 1/4

9 DERIVATION λ 1 = [( 1 ε 2 )1/ 2 + i( 1 +ε 2 )1/ 2 ]α 1 λ 2 = [( 1 ε 2 )1/ 2 i( 1 +ε 2 )1/ 2 ]α 1 λ 3 = [( 1 ε 2 )1/ 2 + i( 1 +ε 2 )1/ 2 ]α 1 λ 4 = [( 1 ε 2 )1/ 2 i( 1 +ε 2 )1/ 2 ]α 1 ε if not equal 1 or -1 ω = C 1 e λ 1x + C 2 e λ 2x + C 3 e λ 3x + C 4 e λ 4 x = C 1 e [(1 ε 2 )1/2 +i( 1+ε 2 )1/2 ]α 1 x + C2 e [(1 ε 2 )1/2 i( 1+ε 2 )1/2 ]α 1 x + C3 e [(1 ε 2 )1/2 +i( 1+ε 2 )1/2 ]α 1 x + C4 e [(1 ε 2 )1/2 i( 1+ε 2 )1/2 ]α 1 x

10 DERIVATION ω = C 1 e λ 1x + C 2 e λ 2x + C 3 e λ 3x + C 4 e λ 4 x = C 1 e [(1 ε 2 )1/2 +i( 1+ε 2 )1/2 ]α 1 x + C2 e [(1 ε 2 )1/2 i( 1+ε 2 )1/2 ]α 1 x + C3 e [(1 ε 2 )1/2 +i( 1+ε 2 )1/2 ]α 1 x + C4 e [(1 ε 2 )1/2 i( 1+ε 2 )1/2 ]α 1 x x >,ω = 0 C 1 = C 2 = 0 x = 0,ω = 0 C 3 = C 4 ω = 2iC 3 e α 1 ( 1 ε 2 )1/2 x sin(e α 1 ( 1+ε 2 )1/2 x ) = Ae α 1 ( 1 ε 2 )1/ 2 x sin(e α 1 ( 1+ε 2 )1/2 x ) A = 2iC 3

11 if ε equals 1 or -1 ω = Ae α 1 ( 1 ε 2 )1/2 x sin(e α 1 ( 1+ε 2 )1/2 x ) is also our solution Therefore our solution is ω = Ae α 1 ( 1 ε 2 )1/2 x sin(e α 1 ( 1+ε 2 )1/2 x ) ε = S 2(DK) 1/ 2 α = (D/K) 1/4 11

12 TRENCH FLEXURE: CRITICAL PARAMETERS, characteristic bending length Can be estimated easily by understanding material, dimensionless parameter related to buckling load How can we estimate this value? What is the horizontal load on the plate? Estimate and continue If, and typical 12

13 TRENCH FLEXURE: CRITICAL PARAMETERS!=" sin [ %/' 1+* ], %/' 1 * Maxima cannot be explicitly determined Take corrected bathymetric profiles Assume small * (Caldwell % / = '0/4 [1+( 4/0 1)*] and! / = "/ 2, 0 4 (1+ 0/4 *) 1976) If *=0 or *=0.3, Error for determining ' from xb and A from wb less than 5% and less than 25% Horizontal load is not significant! S cannot be determined from bathymetry anyway. 13

14 TRENCH FLEXURE: STATIC ANALYSIS Axial load is small How about shears and moments? (Caldwell 1976) 14

15 TRENCH FLEXURE: STATIC ANALYSIS Simplify: normalize by the maximum Normalized Deflection at a Trench Outer Rise Normalized Deflection, w/w max Normalized Lateral Distance, x/x max 15

16 TRENCH FLEXURE: STATIC ANALYSIS 4=5 6 2!/6 % 2 8 Normalized Moment at a Trench 6 4 Normalized Moment Normalized Lateral Distance, x/x max 16

17 TRENCH FLEXURE: STATIC ANALYSIS 7=5 6 3!/6 % 3 8 Normalized Shear at a Trench 6 Normalized Shear Normalized Lateral Distance, x/x max 17

18 TRENCH FLEXURE: BATHYMETRIC COMPARISON (Caldwell 1976) 18

19 YIELD STRENGTH OF THE OUTER RISE Previous information about trenches considered only purely elastic flexure Assumed that yield stress, σ y was infinitely large Therefore, the moment induced in the trench lithosphere is where Note this moment equation applies only to elastic bending 19

20 YIELD STRENGTH OF THE OUTER RISE If a yield stress, σ y, is prescribed, there is the possibility for elasto-plastic behavior to occur During elasto-plastic behavior, the plate will behave elastically until a moment is applied that will cause the yield stress to be reached Under elastic behavior, the curvature of the plate will be related linearly to the moment Once the yield stress is reached, the curvature will continue to increase without any increase in stress and result in a non-linear relationship between curvature and moment 20

21 ELASTO-PLASTIC BEHAVIOR Bending stress diagrams σ B σ y Neutral Axis h h/2 ELASTIC ELASTO- PLASTIC In elasto-plastic behavior, the stresses are constrained by the yield stress, σ y 21

22 ELASTO-PLASTIC BEHAVIOR ELASTIC ELASTO- PLASTIC PLASTIC (Plastic Hinge) 22

23 GENERAL MOMENT EQUATION The moment is defined as the vertical integral of the fiber stresses weighted by the distance from the neutral plane [McNutt] Solving this integral for the elastic case resulted in the previously mentioned equation Solving for the elasto-plastic case is not straightforward, as the momentcurvature relationship is non-linear How can we determine the moment for the elasto-plastic behavior? 23

24 DETERMINING THE MOMENT From the general moment equation, it can be seen that the moment is determined by summing the stress times the distance from the neutral axis (z n ) along the height of the lithosphere By creating a force couple based on the shape of the stress diagram, the moment can be easily calculated 24

25 DETERMINING THE MOMENT Break the diagram into simple shapes, determine the resultant force of each shape, and assume the force acts at the centroid of the shape Neutral Axis h h/2 σ B C (2/3)*h/2 C = ½ (σ B *h/2) Moment arm = 2/3 * h/2 Moment = 2 x C x Moment arm M = h 2 / 6 σ B σ B = 6M / h 2 Note This allows us to solve for the bending stress 25

26 DETERMINING THE MOMENT Do the same procedure for the elasto-plastic case σ y h h/2 h/2 h /2 C2 C1 C1 = (σ y *h ) C2 = ½ [σ y *(h/2- h )] C1 Moment arm = h/2 h /2 C2 Moment arm = 2/3 * (h/2-h /2) h Moment = 2 x [(C1 x MA1) + (C2 x MA2)] 26

27 WHY IS THIS IMPORTANT? Using elastic theory at a location in which the stresses are high enough that plastic behavior is occurring can result in a large underestimate of plate thickness, as will be shown Note At features such as seamounts, the curvature is minimal, and therefore the stresses in the plate are typically not high enough to cause yielding At trenches, curvature is extreme enough that it is very likely that the yield stress of the plate will be exceeded 27

28 EXAMPLE OF UNDERESTIMATING PLATE THICKNESS Assume h = T e = 40 km, z n = 20 km, K = curvature = -5e-7 /m, σ y = 500 MPa, E = 8e10 N/m 2, ν =.25 σ B = MPa Based on purely elastic bending, M E = 2.275x10 17 N-m - from the previous diagram, σ B = MPa For elasto-plastic behavior, the bending stress cannot exceed σ y 28

29 EXAMPLE OF UNDERESTIMATING PLATE THICKNESS Assume h = T e = 40 km, z n = 20 km, K = curvature = -5e-7 /m, σ y = 500 MPa, E = 8e10 N/m 2, ν =.25 σ B = MPa Based on purely elastic bending, M E = 2.275x10 17 N-m - from the previous diagram, σ B = MPa For elasto-plastic behavior, the bending stress cannot exceed σ y σ y = 500 MPa 29

30 EXAMPLE OF UNDERESTIMATING PLATE THICKNESS Assume h = T e = 40 km, z n = 20 km, K = curvature = -5e-7 /m, σ y = 500 MPa, E = 8e10 N/m 2, ν =.25 σ B = MPa Based on purely elastic bending, M E = 2.275x10 17 N-m - from the previous diagram, σ B = MPa For elasto-plastic behavior, the bending stress cannot exceed σ y σ y = 500 MPa 30

31 EXAMPLE OF UNDERESTIMATING PLATE THICKNESS Now, using the method explained previously to find the elasto-plastic moment, M EP = 1.77x10 17 N-m < M E M EP < M E makes sense, as a weaker material would not be able to resist a larger moment Knowing the actual moment in the plate but using elastic theory to solve for plate thickness, H would be H = 36.7 km < Actual H = 40 km Note this error will increase as curvature increases 31

32 YIELD STRENGTH ENVELOPE Based on 2 different failure modes of the plate 1. Brittle Failure At near surface, relatively cool depths, the plate will fail by movement along localized fractures (new faults or reactivated abyssal hills) and the strength varies under tension and compression 2. Plastic Flow At higher temperatures and pressures at lower depths, the plate will fail due to ductile or plastic flow, which is independent of load direction If there is not moment saturation, there will be an elastic zone between the two locations in the upper and lower plate where yield stress has been reached Combining these two failure modes, a strength envelope can be developed 32

33 YIELD STRENGTH ENVELOPE Example of a Yield Envelope [Kirby, 1980] 33

34 YIELD STRENGTH ENVELOPE Simplified yield strength envelope used by McNutt You can see that for the shaded stress diagram, there is extensive plate yielding (plate yields when stress diagram crosses the envelope) 34

35 TONGA TRENCH New faults created due to brittle failure of the plate surface (stars represent earthquakes) 35

36 IN SUMMARY It is important to consider elasto-plastic conditions at trenches, as curvature at these locations is large, and therefore yielding is likely The example presented was simplified Constant yield strength - As can clearly be seen from the yield envelopes, the yield strength can vary greatly over a plate, and different values must be used for compression and tension (positive and negative bending) Axial load = 0 If an axial load is present, it will result in a shift in the neutral axis from the center of the bending zone Note that this issue relates to seismicity, as outer rise earthquakes have been attributed to the brittle tension failure at the surface due to plates failing in bending 36

37 REFERENCES In addition to the images cited in the presentation, Caldwell, J.G., Haxby, W.F., Karig, D.E., Turcotte, D.L. (1976) On the Applicability of a Universal Elastic Trench Profile. Earth and Planetary Science Letters, 31, Mcnutt, M.K., Menard, H.W. (1982) Constraints on Yield Strength in the Oceanic Lithosphere Derived from Observations on Flexure. Geophysical Journal of Royal Astronomical Society, 71, Mofjeld, H.O., et.al.(2004) Tsunami Scattering and Earthquake Faults in the Deep Pacific Ocean. Oceanography, 17, Mueller, S.,Phillips,R.J.(1995) On the Reliability of Lithospheric Constraints Derived from Models of Outer-rise Flexure. Geophysics Journal Int., 123,

38 QUESTIONS Thank You.