Midterm Examination Wed Oct Please initial the statement below to show that you have read it

Transcription

1 EN10: Continuum Mechnics Midterm Exmintion Wed Oct School of Engineering Brown University NAME: Generl Instructions No collbortion of ny kind is permitted on this exmintion. You my bring double sided pges of reference notes. No other mteril my be consulted Write ll your solutions in the spce provided. No sheets should be dded to the exm. Mke digrms nd sketches s cler s possible, nd show ll your derivtions clerly. Incomplete solutions will receive only prtil credit, even if the nswer is correct. If you find you re unble to complete prt of question, proceed to the next prt. Plese initil the sttement below to show tht you hve red it `By ffixing my nme to this pper, I ffirm tht I hve executed the exmintion in ccordnce with the Acdemic Honor Code of Brown University. PLEASE WITE YOU NAME ABOVE ALSO! 1 (5 points). (5 points) 3. (5 points) 4. (5 points) 5. (8 points) TOTAL (8 points)

2 T 1. Let be proper orthogonl tensor stisfying = = I, det()=1, nd let,b be vectors. Using index nottion, show tht ( ) ( b) = ( b ). You will need the following result for determinnt: ( ) ( b) lmn det( A ) = ijk Ail Ajm Akn ijk jn n kmbm = qjk diqjnkmb n m = qjk ql il jnkmb n m T ( ) = il qjk ql jnkm = il det( ) lnm b n m = il lnm b n m = b (5 POINTS). Let, b be two unit vectors. Find expressions for the eigenvlues nd eigenvectors of S= b+ b, in terms of b, nd b Since Su must lwys lie in the,b plne, one eigenvector is b nd the corresponding eigenvlue is zero. We cn construct the other eigenvectors s + ββ where β is to be found. The definition of n eigenvector/vlue pir yields S( + ββ) = β+ β ( + ββ) = ( β) + β + β 1 + β( β) = γ( + ββ ) ( ) [ ] [ ] We cn see by inspection tht choosing 1 β = [ ] [ ] [ ] ( b) b 1 + ( b) = 1 + ( b) ( + b ) so γ = 1+ b is n eigenvlue nd ( + b ) is n eigenvector. The other eigenvector must be perpendiculr to this one so ± ( b ) is the other eigenvector You cn lso work step-by-step through the lgebr: β+ β + ( β) 1 + β( β) = γ 1 + β( β) [ ] [ ] [ ] ( β) [ β+ β] + [ 1 + β( β) ] = γ[ β+ β] [ β β] β [ β β ][ β β] β [ β β][ β β ] [ β β ] + + ( ) 1 + ( ) + ( ) ( ) 1 + ( ) = 0 ( β ) β 1 ( ) + ( β) 1= 0 β =± 1 Thus ( ) [ ] ( ) [ ] The eigenvlues re therefore [ 1 + ( b )], [ 1 ( b )] b+ b ( + b) = ( b) + b+ + b( b) = 1 + ( b) ( + b) b+ b ( b) = ( b) + b b( b) = 1 ( b) ( b) (5 POINTS)

3 3. The figure shows n incompressible sphericl shell with inner rdius nd outer rdius b. The interior is being inflted t constnt rte d / dt =. 3.1 Assume tht the velocity field is rdil v= v () r e. Clculte n expression for the velocity grdient in terms of v, nd hence show tht v = /. e 3 θ e e e θ φ φ e e 1 ( ) v v L= yv= e e + e θ e θ + e φ e φ Incompressibility requires tr(l)=0, which yields dv = v log( v) = log( ) v = d [3 POINTS] 3. Suppose tht fly wlks round the outer surfce of the shell with constnt speed w reltive to the surfce. Find formul for the ccelertion of the fly. The ccelertion is v = + ( yv) v= e + e 3 e + 3( eθ eθ + eφ eφ) e ( 1 + w βeθ ± β eθ t y 4 = e ( 1 ) w βe 3 θ ± β eθ < β < [ POINTS] 3

4 4. The figure shows the reference nd deformed configurtions for solid. The corners A nd B mp to nd b fter deformtion. Mteril fibers perpendiculr to the plne retin their length nd orienttion during deformtion. The stress-strin lw for the mteril reltes the mteril stress Σ to the Lgrnge strin E by Σ= 4E Clculte the trction cting on the plne c-d shown in the figure. L/ e A L B L e e 1 e 1 c b 45 0 d 4L Undeformed Deformed ecll F=V 1 JF σf = ( F F I) σ = FF FF FF = V V V V = J J J V V ( ) ( ) ( ) 1 T T T T T T T 4 4 1/ 0 1/ 0 3/8 0 σ = = The trction follows s 1 3/8 n σ = 4 Alterntively, the deformtion grdient cn be constructed s sequence of stretch nd rottion / F = 1 0 = 0 1/ 0 The Lgrnge strin follows s 1 T / E= ( F F I ) = = 1/ /4 (Cn lso get this s ( U I )/ The mteril stress follows s T 6 0 Σ= ( F F I ) = 0 3/ The Cuchy stress is then T 0 1/ / 0 1 3/8 0 σ = JFΣF = 0 = = 0 3/ 1/ 0 0 3/ [5 POINTS] 4

5 5. The figure shows n infinitely long cylindricl tube with verticl xis. A fluid with mss density ρ nd viscosity µ flows down the tube with stedy velocity v= vr () e z under the ction of n xil grvittionl force b= ρ ge z. The Cuchy stress in the fluid is σ= µ D where D is the stretch rte. e z e z r e r 5.1 Find n expression for the velocity grdient for the flow using the cylindriclpolr coordinte system shown in the figure. 1 v L= v= vr () e z e r + eθ + e z = e z e r r θ z e θ r e r [ POINTS] 5. Using the principle of virtul work, show tht the velocity distribution must stisfy 1 v r g r = r µ Integrte by prts: Substitute bck into PVW v d v d P = µ D: dd b d v dv = µ πrdr + rgdvπrdr = 0 V V 0 0 v d v v v µ π rdr = µ rdv dv r πdr r r r r 0 0 v v v 1 v = πµ bd v( b) d v( ) µd v + π rdr r 0 v v v 1 v r g πµ bd v( b) d v( ) µd v + π rdr = 0 r µ The integrnd must vnish which yields the nswer. b [4 POINTS] 5

6 5.3. Clculte v(r) 1 v rg v 1 rg v 1 rg C r = r = r + C = r + r µ µ µ r 1 r g v= r + Clog( r) + D 4 µ r g v= r 4µ The velocity must be bounded t r=0 so C=0, nd v=0 t r= whence ( ) [ POINTS] 6