# Strain II.

## Presentation on theme: "Strain II."— Presentation transcript:

Strain II

Mohr Circle for Strain For stress Mohr circle, we plot the normal stress n against the shear stress s, and we use equations which represent a circle We will learn about stress Mohr circle later Because geologists deal with deformed rocks, when using the Mohr circle for strain, we would like to deal with measures that represent the deformed state (not the undeformed state!)

Reciprocal strain ellipse
Undeformed Strain ellipse  Deformed 1 1 + 1 '1 '1 + Deformed + reciprocal strain ellipse  Undeformed

Equations for Mohr circle for strain
We already know the quadratic elongation:  =s2 = (1+e)2 Let’s introduce two new parameters: ´ = 1/ (reciprocal quadratic elongation) ´ represents the abscissa for the Mohr circle coordinates ´ = / = ´ ´ represents the ordinate of the Mohr circle coordinates

Equations for the Mohr circle
The two equations for the Mohr circle are in terms of ´ and ´ (not  and ) because we want to deal with the deformed state ´ = (´1+´3)/2–(´3-´1)/2 cos2´ ´ = (´3-´1)/2 sin2´

Mohr Circle for Strain …
Parametric equations of the Mohr circle: ´ = (´1+´3)/2–(´3-´1)/2 cos2´ Center, c =(´1+´3)/2 (mean strain) Radius, r =(´3-´1)/2 (1/2 differential stress) ´ = c – r cos2´ NOTE: If ´=0; 2´=0, then cos2´=1 and ´ = (´1+´3)/2–(´3-´1)/2 = ´1+´3 -´3+´1/ 2  ´= ´1 If ´=90; 2´=180, cos2´= -1 and ´ = (´1+´3)/2–(´3-´1)/2 *(-1) = ´1+´3 +´3-´1/ 2  ´= ´3 ´ = (´3-´1)/2 sin2´ ´ = r sin2´ NOTE: ´=0 for 2´=0 & 180

Real world r Mohr world c
CW + Real world r Mohr world CW + O c

Mohr Circle for Strain The coordinates of any point on the circle satisfy the above two equations The Mohr circle always plots to the right of the origin because we plot the reciprocal quadratic elongation ’, which is always + for both + extension and – extension (i.e., shortening):  =(1+e)2 and ´=1/

(this is opposite to theconvention with stress!)
Sign Conventions The 2 angle is from the c´1 line to the point on the circle, where c is the center of the Mohr circle NOTE: Points on the circle represent lines in the real world! Since we use the reciprocal quadratic elongations ´1 & ´3 Clockwise (cw) 2 from c´1 is ‘+’, and Counterclockwise (ccw) 2 is ‘-’ (this is opposite to theconvention with stress!) cw  in real world is cw 2 in the Mohr circle, and vice versa! However, ccw , from the O1´ line to any point on the circle, is ‘+’, and cw  is ‘-’, where O is the origin

Real world r Mohr world c
CW + Real world r Mohr world CW + + O c

Simple example: A unit sphere is shortened by 50% and extended by 100%
e1 = 1, and e3 = -0.5 S1 = X = l´/lo = 1+e1= 2 & S3= Z = l´/lo= 1+e3 = 0.5 1 = (1+e1)2 = S12 = 4 &  3 = (1+e3)2 = S32 = 0.25  1´ = 1/1 = 0.25 & 3´ = 1/3 = 4 Note the area remains constant: XZ =  1 3 =  4 0.25 = 1 c = (´1+ ´3)/2 = (0.25+4)/2 = 2.125 r = (´3 - ´1)/2 = (4-0.25)/2 = 1.90 Having ´c´ and ´r´, we can plot the circle!

C = 2.125 r = 1.90 ’ 1 ’1 ’3 ’ c 4 1 2 3 -1

Example: deformed brachiopod
Undeformed Mo = 5.4 Deformed m’ ~ 45o ho = 12.4 m’= 9 1 m’ 1 m’ cw ccw h’ ccw h’ h’= 12.4 m’ Strain ellipse

Median line (m): (’, ’) = (0.36, 0.36)
Relative Length in mm (measured on pc screen at 100% magnification view!) Median line (m): (’, ’) = (0.36, 0.36) mo m’ em m ’m m = tan m ’m =m/m 54 90 0.67 2.78 0.36 1 Hinge line (h): (’, ’) = (1, -1) = lnfe ho h’ eh h ’h h = tan h ’h =h/h 124 1 -1

1 m h = lnfe m’ h’ Mohr World ’ = 1 ’ ’3 ’1 ’ m’ h’ 1 +2m’
Median line (m): (’, ’) = (0.36, 0.36) m’ Hinge line (h): (’, ’) = (1, -1) m’ cw ccw ’ = 1 h’ ’ ccw h’ 1 m bisector +2m’ ’3 +m’ ’1 ’ -h’ c 1 2 3 -2h’ Mohr World ’h’ is ccw -35o ’m’ is cw +13o h’ is cw -45o m’ is ccw +45o -1 h = lnfe

Example: Simple shear of a brachiopod
1 Undeformed mo ccw h’ ’h’ cw ho h’ ’m’ ccw cw m’ 1 m’ Example: Simple shear of a brachiopod ’h’ is cw +45o ’m’ is ccw -20o h’ is ccw +32o m’ is cw -30o

’h’ is cw ~27o ’m’ is ccw ~35o h’ is ccw 32o m’ is cw 30o
Relative Length in mm (measured on pc screen at 100% magnification view!) ’h’ is cw ~27o ’m’ is ccw ~35o h’ is ccw 32o m’ is cw 30o Median line (m): (’, ’) = (1.1, ) mo m’ em m ’m m = tan m ’m =m/m 44 42 -0.045 0.912 1.1 -0.577 -0.633 Hinge line (h): (’, ’) = (0.8, 0.50) ho h’ eh h ’h h = tan h ’h =h/h 82 92 0.122 1.259 0.8 0.625 0.50

h m 1 Physical World ’h’ is cw ~27o ’m’ is ccw ~35o
h’ is ccw 32o m’ is cw 30o ccw h’ ’h’ cw h’ ’= 1 ’m’ ccw cw m’ ’ (0.8, 0.50) m’ h 0.5 2’h’ bisector ’ ’1 ’3 h’ m’ c 0.5 1.5 2 2’m’ Mohr World -0.5 m (1.1, )

Lines of no finite elongation - lnfe
Elongation along the lnfe is zero They don’t change length during deformation, i.e., elnfe=0, and lnfe=1, and therefore ´lnfe=1 Draw the vertical line of ´ =1  the ´ axis! The intersection of this line with the Mohr circle defines the lnfe (there are two of them!). Numerical Solution: tan2 ´ = (1-´1)/(´2-1)

Lnfe c Real world Mohr world 1 m ’ h Lnfe 2 (’, ’) m 1 + 2'm’
’3 ’1 ’ - c 2 2'h’ 1 Mohr world h (’, ’) Lnfe

Finding the angular shear
Because ´=/ and ´=1/ then: ´= ´ which yields  = ´/´ Since shear strain:  = tan , and  = ´/´: tan  = ´/´ i.e.,  is the angle of slope of a straight line through the origin The Mohr circle does not directly provide the shear strain  or the angular shear , it only provides ´ But, since  = ´/´, then ´=  if ´=1

Angular Shear  = ´/´ The above equation means that we can get the angular shear () for any line (i.e., any point on the circle) from the ´/´ of the coordinates of that point Thus,  is the angle between the ´ axis and a line connecting the origin (O) to any point (´, ´) on the circle ccw and cw  in the Mohr circle translate into ccw and cw in the physical world (i.e., same sense), respectively!

Lines of Maximum Shear (lms)
’1 lms ’ ‘lms Real world lms lnfe ‘lms lms 1 + 2‘lms ’3 ’1 - c ’ 2‘lms 1 2 lms Mohr world lnfe

Finding lines of maximum shear (lms) strain (max)
Draw two tangents (±) to the Mohr circle from the origin, and measure the 2´ (±) where the two lines intersect the circle. These are the lms Numerical Solution: Orientation: tan ´lms= (2/1) (Note: these are , not ´) Amount: max = (1-2 )/2 12

Lines of Maximum Shear (lms)
’1 lms ’ ‘lms Real world lms lnfe ‘lms lms 1 + 2‘lms ’3 ’1 - c ’ 2‘lms 1 2 lms Mohr world lnfe

Another Example

rosette c Two brachiopod problem on page 318 Ragan
Note: our sign convention is different! rosette ’ arm b =30 OL arm a arm b 30 arm a 1 arm b k’1 k’3 + 27 c ’ -35 arm a Arbitrarily locate OL along arm b. Put origin of rosette at OL. Line up both arm ‘b’s. Find ‘pa’ where arm ‘a’ of rosette intersects other arm ‘a’. Draw circle

Two trilobites problem on page 316-317 Ragan
rosette ’ m=20 arm m S1 OL m arm m k’3 + 36 2m=40 k’1 c ’ Arbitrarily locate OL along arm m. Put origin of rosette at OL. Line up both arm ‘m’s. Find k’1 where arm m of rosette intersects the x axis. Draw circle