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Normal Stresses in Shear Flow Revised A.J AN007 Normal stresses in shear flow revised A.J. Franck, TA Instruments Keywords: normal force, ARES, elasticity, secondary flows INTRODUCTION gasoline (to make flame thrower liquids) exhibited very unusual flow behavior. When such Historically, normal stresses in shear flow were solutions were mixed they clirnbed up the stirrer the first evidence of elasticity in liquids. However shaft (see Figure 1) and even out of the container! today, due to the wide availability of more Karl Weissenberg /3/ (1947) reported this and convenient measurements like the elastic showed that rod climbing was caused by normal modulus, G’, via sinusoidal oscillations, few stresses generated by the shearing. normal stress measurements are made. Yet there are many applicafions where shear normal These same shear normal stresses cause a stresses can be very important. Whenever shear cross-linked rubber rod to get longer when it is strains are large, measuring normal stresses is twisted. Figure 2 illustrates a rod twisted by a both appropriate to the problem and sensitive to torque M. Each element of the rod is acted upon by a shear stress, T or T , and also experiences the microstructure generated by the large strain. zq 21 normal stresses T and T . Because the rod is /1/ This review describes shear normal stresses 22 33 curved, T is a hoop stress, which squeezes and what causes them. More details on actually 11 making normal stress measurements can be around the rod circumference and forces the found in the Product brief: ”Measuring Normal material to deform along the z axis. To prevent Forces /2/. the rod from elongating in z direction a force F must be exerted on the ends WHAT ARE NORMAL STRESSES? During World War II it was observed that high F polymers like natural rubber when dissolved in q T11 T22 T21 T33 T11 T11 T21 T22 T11 Figure 2. Torque and normal forces are generated when an elastic rod is twisted. The direction of the deformation is q (or x1). Thus the stress component Newtonian fluid Viscoelastic fluid in the direction of twisting acting on the element Inertia causes the fluid to Elastic forces cause the flow away from the center fluid to climb up the rod surface that was perpendicular to that direction is designated Tqq or T11;. Similarly, Tzz or T22 acts Figure 1. A rod is rotated in a beaker of motor oil normal to the horizontal surfaces and Trr or T33 acts on the left and in a solution of 1% polyisobutylene normal to the surfaces perpendicular to the in oil on the right. cylinder’s radius 1 AN007 80 T22 70 60 Normal Stress [kPa] 60 & T21 50 Shear Stress [kPa] & T11 & T11 40 & T & 21 T33 30 40 && 20 & ) & )) 10 20 )) && )) 0 ) )) )) && )) -10 )) & ) )))) & ))) x2 -20 0 )))))))) , normal stress N1 [kPa] & Stress [kPa] & s -30 && T -40 -20 && 22 -50 && s -60 && T21 -70 -40 & T11 T21 shear stress & -0.4 &-0.2 0.0 0.2 0.4 & T33 && -60 strain s/x2 -0.4 -0.2 0 0.2 0.4 Strain s/x2 Figure 3. a) Shear and normal stresses versus shear strain for a silicone rubber sample subject to simple shear shown schematically. The data points indicate the normal stress difference, necessary to keep the block at constant thickness X2, while the squares are the shear stress data(from Macosko, 1994). The lines are for an ideal rubber with G = 163 kPa If the same rubber is twisted between a small predicts the normal stress: angle cone and a disk in a cone/plate rheometer, the shear and normal stresses can be simply and T - T = N = Gg 2 directly related to the torque and normal forces: 11 22 1 (5) In figure 3 these relations fit the data well. Note M T = the quadratic shape of T -T and that when the 21 3 (1) 11 22 pR strain is reversed it stays positive. Since there 2F T11 - T22 = N1 = (2) pR 2 Figure 3 shows shear stress and normal 5000 shear stress at 1.4 s-1 [kPa] shear stress at 1 s-1 [kPa] stress data for silicone rubber vs. shear strain. Normal stress at 1.4 s-1 [kPa] 4000 Strain is directly related to the twist angle: Normal stress at 1.0 s-1 [kPa] Second Normal stress at 1 s-1 [kPa] Rq Rq 3000 g = @ for small a (3) R sin a Ra 2000 where a is the angle of the cone. Note that a 1000 normal stress difference generates the force F. 0 This is due to the curvature of the sample as in shear stress, normal stress N1 [kPa] twisting the rubber rod. -1 0 1 2 3 4 5 6 7 8 strain g Ideal rubber obeys Hooke’s law in shear, i.e. Figure 4: Shear stress and normal stress the shear stress is proportional to the strain differences for a polystyrene melt (at 220oC) vs. shear strain after the start up of steady shearing). T21 = Gg (4) The dotted lines are the predictions for an ideal rubber, eq. 4-6. Data taken on ARES with force where G is the shear modulus. Figure 3 shows rebalance transducer. T12 and N1 were measured that shear stress is linear with strain over a wide with cone and plate, eq. 1 and 2. N2 was range and G = 163 kPa. If Hooke’s law is determined by subtracting cone and plate normal extended to large strains (neo-Hookean) it stress data from parallel plate results on the same material. 2AN007 are up to three components of the normal stress normal stress, N , is not zero but N =- 0.25 N . there can be another independent normal stress 2 2 1 This is typical for polymeric liquids. difference: This same behavior illustrated in Figure 4 qualitatively occurs for dilute polymer solutions T - T = N (6) 22 33 2 because even a single polymer coil will deform For an ideal rubber N =0. For real rubbers and due to the flow (See Figure 6a). Brownian motion 2 tries to randomize the chain bringing it back to polymeric liquids N2 is typically significantly an un-deformed coil which creates tension, T11, smaller than N2 and of opposite sign. This is shown in Figure 4. along the streamlines. Tiny rod-like particles will also orient in shear flows. Brownian motion tries Polymeric liquids behave much like rubber at to randomize this orientation. As with the random short times. Figure 4 illustrates shear and normal coil this generates tension along stream-lines, stresses for a polymer melt. At small strains the which can generate shear normal stresses which response looks like that for rubber. The shear will increase with the degree of orientation. stress can be fit to eq. 4 and N1 to eq. 5. At high Emulsion droplets deform due to flow and strains the stresses decrease and eventually reach interfacial tension tries to restore them to a a steady value. These steady state values increase spherical shape also generating normal stresses. with the shear rate as shown in Figure 4. Of course, stresses for these dilute systems are typically much smaller than those for entangled At low shear rates the shear stress increases polymers linearly with the shear rate and the normal stresses stress with the shear rate squared. In For polymer melts normal stresses are very order to study the shear rate effects the data are sensitive to molecular weight and molecular reduced with the shear rate. Thus, the shear stress weight distribution. For entangled melts the low coefficient or viscosity and the first and second shear rate shear viscosity depends strongly on normal stress coefficients are defined as: the average molecular weight N N T21 1 2 = Y = Y2 unentangled (10) = h 2 1 2 ho ~Mw Mw £ Mentangle g& g& g& 3 4 (7) (8) (9) ' entangled ho ~Mw Mw > Mentangle WHAT CAUSES NORMAL STRESSES? N 1 2 Figure 4 shows that typical polymer melts (11) lim 2 = Y1,0 » Mw Mz behave like cross-linked rubber at short times g&®0 g& and high strain rates. This is because the chains in typical polymer melts are highly entangled and Thus, we might expect normal stresses to be the entanglements form a temporary network. At very useful in polymer characterization. large strain the entanglement network comes However, at small shear rates the first normal apart. At higher shear rates the stresses can go stress is just twice the elastic or storage modulus through a maximum as the chains disentangle. measured in sinusoidal shearing. Eventually equilibrium between entangling (due to Brownian motion) and disentangling (due to the shearing) is reached, generating steady stresses (see Figure 4). At lower shear rates the G/R steady state stresses decrease because there is R more time for entanglement As the shear rate goes to zero, so do the stresses as expected for a Figure 6:. Normal stresses arise in shear flow of a) true liquid. Note also in Figure 4 that the second polymer coils and b) rod-like particles due to Brownian motion and in c) liquid drops due to interfacial tension, G 3 AN007 lim N1 = lim 2G' g&®0 w®0 (12) ] 2 104 In Figure 5 the dynamic data are compared to the steady shear viscosity and normal stress. The correlation holds over a wide range of shear rate 3 10 and frequency. At higher rates or frequencies, the from G'' [Pas] from torque [Pas] dynamic and steady results deviate. from G' [Pas2] from N1 [Pas2] stress coefficint [Pas], [Pas from N2 [Pas2] Measuring the dynamic moduli only perturbs 102 polymer chains around their equilibrium 0.1 1 10 configuration.
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