Lec.22.pptx - 11/6/11 22. Stresses Around a Hole...

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Unformatted text preview: 11/6/11 22. Stresses Around a Hole (II) I Main Topics A General solu=on for a plane strain case B Boundary condi=ons C Solu=on that honors boundary condi=ons D Significance of solu=on E Superposi=on F Stress concentra=ons 11/6/11 GG303 1 21. Stresses Around a Hole (I) Hydraulic Fracture Ship Rock, New Mexico hKp://jencarta.com/images/aerial/ShipRock.jpg 11/6/11 GG303 From Wu et al., 2007 2 1 11/6/11 22. Stresses Around a Hole (II) II General solu=on for a plane strain case Start with the governing equa=on 0= d 2ur 1 dur ur + − dr 2 r dr r 2 Consider a power series solu=on for ur and its deriva=ves ur = … C−3r −3 + C−2 r −2 + C−1r −1 + C0 r 0 + C1r1 + C2 r 2 + C3r 3 + ... dur = … − 3C−3r −4 − 2C−2 r −3 − 1C−1r −2 + 0C0 r −1 + 1C1r 0 + 2C1r1 + 3C1r 2 + ... dr d 2 ur = … 12C−3r −5 + 6C−2 r −4 + 2C−1r −3 + 0C0 r −2 + 0C1r −1 + 2C1r 0 + 6C1r1 ... dr 2 11/6/11 GG303 3 22. Stresses Around a Hole (II) II General Solu=on for a plane strain case Now subs=tute the series solu=ons into the governing equa=on 0 = d u + 1 du − 1 u 2 r dr 2 r r dr r2 r 0 = … 12C−3r −5 + 6C−2 r −4 + 2C−1r −3 + 0C0 r −2 + 0C1r −1 + 2C1r 0 + 6C1r1 ... 1 + … − 3C−3r −4 − 2C−2 r −3 − 1C−1r −2 + 0C0 r −1 + 1C1r 0 + 2C1r1 + 3C1r 2 + ... r 1 − 2 …C−3r −3 + C−2 r −2 + C−1r −1 + C0 r 0 + C1r1 +C2 r 2 + C3r 3 + ... r ( ) ( ) 0 = … 12C−3r −5 + 6C−2 r −4 + 2C−1r −3 + 0C0 r −2 + 0C1r −1 + 2C0 r 0 + 6C1r1 + ... ( − (… C ) + ...) + … − 3C−3r −5 − 2C−2 r −4 − 1C−1r −3 + 0C0 r −2 + 1C1r −1 + 2C0 r 0 + 3C1r1 + ... 11/6/11 −3 r −5 + C−2 r −4 + C−1r −3 + C0 r −2 + C1r −1 + C2 r 0 + C3r1 GG303 4 2 11/6/11 22. Stresses Around a Hole (II) II General Solu=on for a plane strain case 0 = … 12C−3r −5 + 6C−2 r −4 + 2C−1r −3 + 0C0 r −2 + 0C1r −1 + 2C0 r 0 + 6C1r1 + ... ( + (… − 1C ) − ...) + … − 3C−3r −5 − 2C−2 r −4 − 1C−1r −3 + 0C0 r −2 + 1C1r −1 + 2C0 r 0 + 3C1r1 + ... −3 r −5 − 1C−2 r −4 − 1C−1r −3 − 1C0 r −2 − 1C1r −1 − 1C2 r 0 −1C3r1 Now collect terms of the same powers 0 = … 8C−3r −5 + 3C−2 r −4 + 0C−1r −3 − 1C0 r −2 + 0C1r −1 + 3C0 r 0 + 8C1r1 + ... 11/6/11 GG303 5 22. Stresses Around a Hole (II) II General Solu=on for a plane strain case 0 = … 8C−3r −5 + 3C−2 r −4 + 0C−1r −3 − 1C0 r −2 + 0C1r −1 + 3C0 r 0 + 8C1r1 + ... For this to hold for all values of r, the product of each leading coefficient and constant must equal 0 because the powers of r are linearly independent. All coefficients except C ­1 and C1 thus must be zero. ur = …C−3r −3 + C−2 r −2 + C−1r −1 + C0 r 0 + C1r1 + C2 r 2 + C3r 3 + ... ur = C−1r −1 + C1r1 11/6/11 • General solu=on for radial displacements • Solve for constants via boundary condi=ons GG303 6 3 11/6/11 22. Stresses Around a Hole (II) III Boundary condi=ons A Two boundary condi=ons must be specified to solve our problem because our general solu=on has two unknown coefficients: ur = C−1r −1 + C1r1 11/6/11 GG303 7 22. Stresses Around a Hole (II) III Boundary condi=ons: Radial displacements B ur = u0 at the wall of the hole: ur|r=a = u0 C ur = 0 at infinity: ur|r=b=∞ = 0 11/6/11 GG303 8 4 11/6/11 22. Stresses Around a Hole (II) IV Solu=on that honors boundary condi=ons Gov. eq.: ur = C ­1r  ­1 + C1r1 BC 1: ur|r=a = u0 BC 2: ur|r=b=∞ = 0 As r∞, ur C1r BC 2 requires C1 = 0, so ur = C ­1r  ­1 11/6/11 GG303 9 22. Stresses Around a Hole (II) IV Solu=on that honors boundary condi=ons Gov. eq.: ur = C ­1r ­1 + C1r1 BC 1: ur|r=a = u0 BC 2: ur|r=b=∞ = 0 As r ∞, ur C1r BC 2 requires C1 = 0, so ur = C ­1r  ­1 By BC 1, ur|r=a = u0 = C ­1a  ­1 So, C ­1 =a u0 ur = (a/r) u0 11/6/11 Solu=on sa=sfying boundary condi=ons GG303 10 5 11/6/11 22. Stresses Around a Hole (II) IV Solu=on that honors boundary condi=ons (cont.) ur = (a/r) u0 • The hole radius a provides a scale • The displacements decay with distance r from the hole (as suspected) • The displacements scale with u0 • Problems with different boundary condi=ons have different solu=ons 11/6/11 GG303 11 22. Stresses Around a Hole (II) IV Solu=on that honors boundary condi=ons (cont.) Strains −1 ∂ ( r −1 ) ∂ ur ∂ ( u0 ar ) ε rr = = = u0 a ∂r ∂r ∂r − u0 a = 2 = −u0 ar −2 r εθθ = 11/6/11 ur u0 ar −1 u0 a = = 2 = u0 ar −2 r r r GG303 ε rθ = 0 12 6 11/6/11 22. Stresses Around a Hole (II) IV Solu=on that honors boundary condi=ons (cont.) Stresses (in terms of u0) Hooke’s Law σ rr = E⎡ ν ε rr + (ε rr + εθθ )⎤ ⎥ (1 + ν ) ⎢ (1 − 2ν ) ⎣ ⎦ = E ⎡ − u0 a ν ⎛ − u0 a u0 a ⎞ ⎤ + + 2 ⎟⎥ ⎜ (1 + ν ) ⎢ r 2 (1 − 2ν ) ⎝ r 2 r ⎠⎦ ⎣ = E ⎡ − u0 a ⎤ (1 + ν ) ⎢ r 2 ⎥ ⎣ ⎦ 11/6/11 Strains GG303 13 22. Stresses Around a Hole (II) IV Solu=on that honors boundary condi=ons (cont.) Stresses (in terms of u0) Hooke’s Law σ rr = E⎡ ν ε rr + (ε rr + εθθ )⎤ ⎥ (1 + ν ) ⎢ (1 − 2ν ) ⎣ ⎦ = E ⎡ − u0 a ν ⎛ − u0 a u0 a ⎞ ⎤ + + 2 ⎟⎥ ⎜ (1 + ν ) ⎢ r 2 (1 − 2ν ) ⎝ r 2 r ⎠⎦ ⎣ = E ⎡ − u0 a ⎤ (1 + ν ) ⎢ r 2 ⎥ ⎣ ⎦ Strains σ θθ = E⎡ ν εθθ + (εθθ + ε rr )⎤ ⎥ (1 + ν ) ⎢ (1 − 2ν ) ⎣ ⎦ = E ⎡ u0 a ν ⎛ u0 a − u0 a ⎞ ⎤ + + 2 ⎟⎥ ⎜ (1 + ν ) ⎢ r 2 (1 − 2ν ) ⎝ r 2 r ⎠⎦ ⎣ = E ⎡ u0 a ⎤ (1 + ν ) ⎢ r 2 ⎥ ⎣ ⎦ 11/6/11 Same absolute magnitude, opposite sign; σrr =  ­σθθ σ rθ = 2Gε rθ = 0 GG303 14 7 11/6/11 22. Stresses Around a Hole (II) IV Solu=on that honors boundary condi=ons (cont.) Stresses (in terms of trac=ons) First evaluate the radial stress on the wall of the hole (r=a), which equals trac=on T, and from that solve for u0 E ⎡ − u0 a ⎤ E ⎡ − u0 ⎤ = =T (1 + ν ) ⎢ a 2 ⎥ (1 + ν ) ⎢ a ⎥ ⎣ ⎦ ⎣ ⎦ E u0 = − aT (1 + ν ) σ rr r=a = Now subs=tute for u0 in the general expression for σrr on previous page ⎡⎛ E⎞⎤ aT a 2 ⎝ ⎠⎥ E ⎢⎜ (1 + ν ) ⎟ ⎥ ⎛ a⎞ ⎢ σ rr = =T⎜ ⎟ ⎝ r⎠ ⎥ (1 + ν ) ⎢ r2 ⎢ ⎥ ⎣ ⎦ 11/6/11 ⎛ a⎞ σ θθ = −σ rr = −T ⎜ ⎟ ⎝ r⎠ 2 GG303 15 22. Stresses Around a Hole (II) V Significance For a pressure in a hole with no remote load at r = ∞: A The radial normal stress σrr is a principal stress because σrθ = 0; σrr is the most compressive stress. B The circumferen=al normal stress σθθ is a principal stress because σθr = 0; σθθ is the most tensile stress. C A high pressure could cause radial cracking (e.g., radial dikes around a magma chamber). 11/6/11 GG303 ⎛ a⎞ σ θθ = −T ⎜ ⎟ ⎝ r⎠ ⎛ a⎞ σ rr = T ⎜ ⎟ ⎝ r⎠ 2 2 16 8 11/6/11 Stresses Around a Hole (II) VI Superposi=on 11/6/11 GG303 17 22. Stresses Around a Hole (II) VII Stress concentra=ons A The hole causes a doubling of the normal stress far from the hole (i.e.,a stress concentra=on) B The magnitude of the circumferen=al stress at the hole is independent of the size of the hole, so a =ny cylindrical hole causes the same stress concentra=on as a large one. C A =ny hole near the wall of a larger hole might be expected to have an even larger stress concentra=on (why?) 11/6/11 GG303 18 9 11/6/11 22. Stresses Around a Hole (II) VI Stress concentra=ons D Stress concentra=ons explain myriad phenomena, such as 1  Why paper doesn’t explode when pulled upon hard 2  Why paper tears along “the doKed line” 3  Why cracks in riveted steel plates start from the rivet holes 4  Why cracks in drying mudflats originate from the where grass stems have poked through the mud etc. 5  Why “strong” rocks can fail under “low” stresses 6  Why dikes can propagate through the Earth’s crust 7  Why seismic ruptures grow so large 11/6/11 GG303 19 10 ...
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This note was uploaded on 12/05/2011 for the course GEOLOGY 300 taught by Professor Stephenmartel during the Fall '11 term at University of Hawaii, Manoa.

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