ビンガム物体に対する平行板プラストメーターの理論

抄録

The purpose of this paper is to show how to determine yield velue f and plastic, viscosity η of a Bingham body by means of a parallel plate plastometer. The material under test in the form of a cylinder is placed between parallel circular plates. A constant force F is applied perpendicularly to one plate, the other plate being fixed. From the meaurement of the displacement of the other plate, the rheological behavior of the specimen may be determined.
We have treated the problem under the following assumptions: i) the material is incompressible; ii) no body force acts on the material; iii) the motion is very slow; iv) there is no relative motion between the plates and the material in immediate contact with the plates; v) the distance h between the plates is so small compared to the radius R of the sample that the velocity component in the perpendicular direction is negligible.
Then it is found that the pressure p is a function of time t or h and the distance r from the axis of the sample and satisfies the following equation,
6ηr(p'r+p)³h=p'{h(p'r+p)-fr}{h(p'r+p)+2fr}². (1)
Since p may be regarded as a function of r, h and f, we developed p with regard to h and f and neglected second order terms. Then the pressure is given by
p=p₀-3ηh/h³(R²-r²)+(R-r)f/h, (2)
where p₀ is the atmospheric pressure.
Since the applied force F together with πR²p₀ is in equilibrium with the pressure in the material, We get
-3πR⁴η/2h³h+πR³f/3h=F. (3)
We then treat the following two arrangements: i) the specimen is larger than the plates; thus the area under compression is constant. ii) the plates are larger than the specimen; thus the volume V of the specimen is constant.
i) We may take R=a, and eq. (3) can be written as
D=(τ-f)/η (4)
with D=-9ah/2h² and τ=3hF/πa³. Thus if D is plotted against τ, a straight line of slope tan^-1 (1/η) is obtained. From the intercept on the abscissa, the yield value f can be obtained.
ii) If we express eq. (3) in terms of V, we have the same relation as (4) with D=-3V1/2h/ 2π^1/2h^5/2 and τ=3π^1/2h^5/2F/V^3/2.
Eq. (3) may be integrated in the special case πa³f/3Fh<<1. Thus we have
3πa⁴/4(1/h²-1/h₀²)+πa⁷f/6F(1/h³-1/h₀³)=F/ηt. (5)
Similarly we get
3V²/8π(1/h⁴-1/h₀⁴)+fV^7/2/13π^3/2F(1/h^13/2-1/h₀^13/2)=F/ηt (6)
in the special case where V^3/2f/3π^1/2h^5/2F<<1. If we put f=0 in eqs. (5) and (6), these reduce to the well-known formulate for Newtonian liquids.