Problems for Chapter 2

 

Governing Equations

 

 

 

2.3.  Equations of motion and equilibrium for deformable solids

 

 

2.3.1.      A prismatic concrete column of mass density ρ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHbpGCaaa@3485@  supports its own weight, as shown in the figure.  (Assume that the solid is subjected to a uniform gravitational body force of magnitude g per unit mass).

2.3.1.1.            Show that the stress distribution

σ 22 =ρg(H x 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHdpWCdaWgaaWcbaGaaGOmaiaaik daaeqaaOGaeyypa0JaeyOeI0IaeqyWdiNaam4zaiaacIcacaWGibGa eyOeI0IaamiEamaaBaaaleaacaaIYaaabeaakiaacMcaaaa@3FD7@

satisfies the equations of static equilibrium

σ ij x i +ρ b j =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaWcaaqaaiabgkGi2kabeo8aZnaaBa aaleaacaWGPbGaamOAaaqabaaakeaacqGHciITcaWG4bWaaSbaaSqa aiaadMgaaeqaaaaakiabgUcaRiabeg8aYjaadkgadaWgaaWcbaGaam OAaaqabaGccqGH9aqpcaaIWaaaaa@4206@

and also satisfies the boundary conditions σ ij n i =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHdpWCdaWgaaWcbaGaamyAaiaadQ gaaeqaaOGaamOBamaaBaaaleaacaWGPbaabeaakiabg2da9iaaicda aaa@3A72@

on all free boundaries.

2.3.1.2.            Show that the traction vector acting on a plane with normal n=sinθ e 1 +cosθ e 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWHUbGaeyypa0Jaci4CaiaacMgaca GGUbGaeqiUdeNaaGPaVlaaykW7caWHLbWaaSbaaSqaaiaaigdaaeqa aOGaey4kaSIaci4yaiaac+gacaGGZbGaeqiUdeNaaGPaVlaaykW7ca WHLbWaaSbaaSqaaiaaikdaaeqaaaaa@489C@  at a height x 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWG4bWaaSbaaSqaaiaaikdaaeqaaa aa@34AA@  is given by

T=ρg(H x 2 )cosθ e 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWHubGaeyypa0JaeyOeI0IaeqyWdi Naam4zaiaacIcacaWGibGaeyOeI0IaamiEamaaBaaaleaacaaIYaaa beaakiaacMcaciGGJbGaai4BaiaacohacqaH4oqCcaaMc8UaaCyzam aaBaaaleaacaaIYaaabeaaaaa@452D@

2.3.1.3.            Deduce that the normal component of traction acting on the plane is

T n =ρg(H x 2 ) cos 2 θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWGubWaaSbaaSqaaiaad6gaaeqaaO Gaeyypa0JaeyOeI0IaeqyWdiNaam4zaiaacIcacaWGibGaeyOeI0Ia amiEamaaBaaaleaacaaIYaaabeaakiaacMcaciGGJbGaai4Baiaaco hadaahaaWcbeqaaiaaikdaaaGccqaH4oqCaaa@43E4@

2.3.1.4.            show also that the tangential component of traction acting on the plane is

T t =ρg(H x 2 )sinθcosθ(cosθ e 1 sinθ e 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWHubWaaSbaaSqaaiaadshaaeqaaO Gaeyypa0JaeqyWdiNaam4zaiaacIcacaWGibGaeyOeI0IaamiEamaa BaaaleaacaaIYaaabeaakiaacMcaciGGZbGaaiyAaiaac6gacqaH4o qCcaaMc8Uaci4yaiaac+gacaGGZbGaeqiUdeNaaGPaVlaacIcaciGG JbGaai4BaiaacohacqaH4oqCcaaMc8UaaCyzamaaBaaaleaacaaIXa aabeaakiabgkHiTiGacohacaGGPbGaaiOBaiabeI7aXjaaykW7caWH LbWaaSbaaSqaaiaaikdaaeqaaOGaaiykaaaa@5BE4@

(the easiest way to do this is to note that T= T n n+ T t MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWHubGaeyypa0JaamivamaaBaaale aacaWGUbaabeaakiaah6gacqGHRaWkcaWHubWaaSbaaSqaaiaadsha aeqaaaaa@3A85@  and solve for the tangential traction).

2.3.1.5.            Suppose that the concrete contains a large number of randomly oriented microcracks.  A crack which lies at an angle θ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH4oqCaaa@347B@  to the horizontal will propagate if

| T t |+μ T n > τ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaadaabdaqaaiaahsfadaWgaaWcbaGaam iDaaqabaaakiaawEa7caGLiWoacqGHRaWkcqaH8oqBcaWGubWaaSba aSqaaiaad6gaaeqaaOGaeyOpa4JaeqiXdq3aaSbaaSqaaiaaicdaae qaaaaa@4040@

where μ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH8oqBaaa@347B@  is the friction coefficient between the faces of the crack and τ 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHepaDdaWgaaWcbaGaaGimaaqaba aaaa@3570@  is a critical shear stress that is related to the size of the microcracks and the fracture toughness of the concrete, and is therefore a material property.

2.3.1.6.            Assume that μ=1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH8oqBcqGH9aqpcaaIXaaaaa@363C@ .  Find the orientation of the microcrack that is most likely to propagate.  Hence, find an expression for the maximum possible height of the column.

 

 

 

2.3.2.      Is the stress field given below in static equilibrium?  If not, find the acceleration or body force density required to satisfy linear momentum balance

σ 11 =C x 1 x 2 σ 12 = σ 21 =C( a 2 x 2 2 ) σ 33 = σ 23 = σ 13 =0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rk0le9 v8qqaqFD0xXdHaVhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakqaabeqaaiabeo8aZnaaBaaaleaacaaIXa GaaGymaaqabaGccqGH9aqpcaWGdbGaamiEamaaBaaaleaacaaIXaaa beaakiaadIhadaWgaaWcbaGaaGOmaaqabaGccaaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlabeo8aZnaaBaaaleaacaaIXaGaaGOmaa qabaGccqGH9aqpcqaHdpWCdaWgaaWcbaGaaGOmaiaaigdaaeqaaOGa eyypa0Jaam4qaiaacIcacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaey OeI0IaamiEamaaDaaaleaacaaIYaaabaGaaGOmaaaakiaacMcacaaM c8UaaGPaVdqaaiabeo8aZnaaBaaaleaacaaIZaGaaG4maaqabaGccq GH9aqpcqaHdpWCdaWgaaWcbaGaaGOmaiaaiodaaeqaaOGaeyypa0Ja eq4Wdm3aaSbaaSqaaiaaigdacaaIZaaabeaakiabg2da9iaaicdaaa aa@7337@

 

 

 

2.3.3.      Let ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabew9aMbaa@3801@  be a twice differentiable, scalar function of position.  Derive a plane stress field from ϕ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFKI8=feu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabew9aMbaa@3801@  by setting

σ 11 = 2 ϕ x 2 2 σ 22 = 2 ϕ x 1 2 σ 12 = σ 21 = 2 ϕ x 1 x 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=wi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabeo8aZnaaBaaaleaacaaIXaGaaGymaa qabaGccqGH9aqpdaWcaaqaaiabgkGi2oaaCaaaleqabaGaaGOmaaaa kiabew9aMbqaaiabgkGi2kaadIhadaqhaaWcbaGaaGOmaaqaaiaaik daaaaaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlabeo8aZnaaBaaaleaacaaIYaGa aGOmaaqabaGccqGH9aqpdaWcaaqaaiabgkGi2oaaCaaaleqabaGaaG Omaaaakiabew9aMbqaaiabgkGi2kaadIhadaqhaaWcbaGaaGymaaqa aiaaikdaaaaaaOGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8Uaeq4Wdm3aaSbaaSqaaiaaigdacaaIYaaabeaakiabg2da9i abeo8aZnaaBaaaleaacaaIYaGaaGymaaqabaGccqGH9aqpcqGHsisl daWcaaqaaiabgkGi2oaaCaaaleqabaGaaGOmaaaakiabew9aMbqaai abgkGi2kaadIhadaWgaaWcbaGaaGymaaqabaGccqGHciITcaWG4bWa aSbaaSqaaiaaikdaaeqaaaaakiaaykW7caaMc8UaaGPaVdaa@97B2@

Show that this stress field satisfies the equations of stress equilibrium with zero body force.

 

 

 

 

2.3.4.      The stress field

σ ij = 3 P k x k x i x j 4π R 5 R= x k x k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiabeo8aZnaaBaaaleaacaWGPbGaamOAaa qabaGccqGH9aqpdaWcaaqaaiabgkHiTiaaiodacaWGqbWaaSbaaSqa aiaadUgaaeqaaOGaamiEamaaBaaaleaacaWGRbaabeaakiaadIhada WgaaWcbaGaamyAaaqabaGccaWG4bWaaSbaaSqaaiaadQgaaeqaaaGc baGaaGinaiabec8aWjaadkfadaahaaWcbeqaaiaaiwdaaaaaaOGaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaamOuaiabg2da9maakaaabaGaamiEamaaBaaaleaacaWGRbaabe aakiaadIhadaWgaaWcbaGaam4Aaaqabaaabeaaaaa@57B2@

represents the stress in an infinite, incompressible elastic solid that is subjected to a point force with components P k MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=wi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFH e9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaa caqabeaacmqaamaaaOqaaiaadcfadaWgaaWcbaGaam4Aaaqabaaaaa@3269@  acting at the origin (you can visualize a point force as a very large body force which is concentrated in a very small region around the origin).

2.3.4.1.            Verify that the stress field is in static equilibrium

2.3.4.2.            Consider a spherical region of material centered at the origin.  This region is subjected to (1) the body force acting at the origin; and (2) a force exerted by the stress field on the outer surface of the sphere.   Calculate the resultant force exerted on the outer surface of the sphere by the stress, and show that it is equal in magnitude and opposite in direction to the body force.

 

 

 

2.3.5.      In this problem, we consider the internal forces in the polymer specimen described in Problem 2.1.29 (you will need to solve 2.1.29 before you can attempt this one). Suppose that the specimen is homogeneous, has mass density ρ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l bbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHbpGCaaa@337F@  in the reference configuration, and may be idealized as a viscous fluid, in which the Kirchhoff stress is related to stretch rate by

τ=μD+pI MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l bbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacaWHepGaeyypa0JaeqiVd0MaaCirai abgUcaRiaadchacaWHjbaaaa@3941@

where p is an indeterminate hydrostatic pressure and μ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l bbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaH8oqBaaa@3375@  is the viscosity.

 

2.3.5.1.            Find expressions for the Cauchy stress tensor, expressing your answer as components in { e r , e θ , e z } MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8rkY=vi pgYlH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8ku c9pgc9q8qqaq=dir=f0=yqaiVgFr0xfr=xfr=xb9adbaqaaeGaciGa biaabeqaaiqabaWaaaGcbaGaai4EaiaahwgadaWgaaWcbaGaamOCaa qabaGccaGGSaGaaCyzamaaBaaaleaacqaH4oqCaeqaaOGaaiilaiaa hwgadaWgaaWcbaGaamOEaaqabaGccaGG9baaaa@3B47@

2.3.5.2.            Assume steady, quasi-static deformation (neglect accelerations).  Express the equations of equilibrium in terms of the angle ϕ(r,t) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l bbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHvpGzcaGGOaGaamOCaiaacYcaca WG0bGaaiykaaaa@3780@

2.3.5.3.            Solve the equilibrium equation, together with appropriate boundary conditions, to calculate ϕ(r,t) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l bbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacqaHvpGzcaGGOaGaamOCaiaacYcaca WG0bGaaiykaaaa@3780@  

2.3.5.4.            Find the torque necessary to rotate the external cylinder

2.3.5.5.            Calculate the acceleration of a material particle in the fluid

2.3.5.6.            Estimate the rotation rate α ˙ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaebbnrfifHhDYfgasaacH8srps0l bbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0R Yxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGa caGaaeqabaGadeaadaaakeaacuaHXoqygaGaaaaa@3367@  where inertia begins to play a significant role in determining the state of stress in the fluid