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Incipient failure of a circular cylinder under gravity
International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
Incipient failure of a circular cylinder under gravity J.A. Chamberlaina; ∗ , J.E. Sadera , K.A. Landmana , D.J. Horrobina , L.R. Whiteb a
b
Department of Mathematics and Statistics, The University of Melbourne, Victoria 3010, Australia Department of Chemical Engineering, Carnegie Mellon University, Pittsburgh, PA 15213-3890, USA Received 23 October 2000; accepted 15 April 2002
Abstract We consider the incipient failure of a circular cylinder under gravitational loading. Axisymmetric slip-line 4eld theory is used to solve for the stress distribution in the plastic region of the cylinder at incipient failure. This extends similar work on the incipient failure of a rectangular block under gravity (Int. J. Mech. Sci. 43(3) (2000) 793), to the axisymmetric geometry. Our calculations show that the height of incipient failure depends not only on the yield stress and density of the material, but also on the radius of the cylinder and the coe8cient of friction at the base. In addition, we 4nd that there exists a critical coe8cient of friction above which the height of incipient failure is una9ected. These results suggest a method of measuring the yield stress by observing the height of incipient failure. ? 2002 Published by Elsevier Science Ltd. Keywords: Slip-lines; Slump test; Yield stress; Cylinder; Axisymmetric; Gravity
1. Introduction The yield stress of a slurry is an important parameter used in industrial processes. In mining engineering, the yield stress is a control parameter governing the pumping and disposal of mine tailings [1], whereas in the construction industry, the yield stress of fresh concrete a9ects its mixing, transportation, placing and consolidation behaviour [2,3]. Laboratory instruments such as the vane rheometer [4] can be used to measure the yield stress of these materials. However, in the 4eld the “slump test” presents a simpler approach [1]. In this test a frustum of a cylinder or a cone is placed on a Aat base and 4lled with slurry. Lifting the frustum causes the slurry to slump down to a lower height, which can then be related to the yield stress. Rather than considering this Aow problem, we examine the height of incipient failure, i.e., the height of material required to just initiate Aow. This height is of fundamental importance to the understanding and operation of the slump test, since it ∗
Corresponding author. Fax: +61-3-8344-4599. E-mail address: [email protected] (J.A. Chamberlain).
0020-7403/02/$ - see front matter ? 2002 Published by Elsevier Science Ltd. PII: S 0 0 2 0 - 7 4 0 3 ( 0 2 ) 0 0 0 4 4 - 9
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Nomenclature H R0 y y g R Z
ij ij ˆ Z ∇ n Fbase ˆ R h r0 ij I r z hp hpc c p pA ; pB ; : : : A ; B ; : : : rA ; rB ; : : : zA ; zB ; : : : rC(1) (2) C
rC(0)
unscaled height of incipient failure unscaled radius of cylinder uniaxial yield stress of the material shear yield stress of the material density of the material acceleration due to gravity unscaled radial coordinate in the cylindrical coordinate system unscaled vertical coordinate in the cylindrical coordinate system angular coordinate in the cylindrical coordinate system the ij component of the unscaled stress tensor unscaled stress tensor the ij component of the strain tensor unit vector in the upward direction, Z increasing di9erential del operator unit normal to a surface stress applied by the material acting on the base unit vector in the radial direction, R increasing coe8cient of friction on the base scaled height of incipient failure scaled radius of the cylinder scaled stress tensor the ij component of the scaled stress tensor unit tensor scaled radial coordinate in the cylindrical coordinate system scaled vertical coordinate in the cylindrical coordinate system scaled height of plastic material on the vertical free surface scaled height of plastic material on the vertical free surface at the onset of the stress discontinuity critical coe8cient of friction. Increasing friction above this value has no e9ect on the solution equal to −(rr + zz )=2, one of the two variables used to specify the stress state angle between the r-axis and the -direction measured in the anti-clockwise direction. Used along with p to specify the stress state value of p at point A, point B; : : : value of at point A, point B; : : : value of r at point A, point B; : : : value of z at point A, point B; : : : 4nal estimate of r at point C, Fig. 2, using Chitkara and Butt’s method 4nal estimate of at point C, Fig. 2, using Chitkara and Butt’s method initial estimate of r at point C, Fig. 2, using Chitkara and Butt’s method
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
zC(0)
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initial estimate of z at point C, Fig. 2, using Chitkara and Butt’s method
(0) C (1) C (1) zC F
G pC(0) p+ p− + −
initial estimate of at point C, Fig. 2, using Chitkara and Butt’s method intermediate estimate of at point C, Fig. 2, using Chitkara and Butt’s method 4nal estimate of z at point C, Fig. 2, using Chitkara and Butt’s method equals cot A + cot (1) C , used in Chitkara and Butt’s method
equals cot B + cot (1) C , used in Chitkara and Butt’s method 4nal estimate of p at point C, Fig. 2, using Chitkara and Butt’s method value of p at the stress discontinuity, on the side closest to the vertical free surface value of p at the stress discontinuity, on the side closest to the centreline value of at the stress discontinuity, on the side closest to the vertical free surface value of at the stress discontinuity, on the side closest to the mid-line
must be exceeded for Aow to occur. Also, calculation of the point of incipient failure can be used to develop a method for determining the yield stress, as we shall discuss. Since cylindrical geometry is used typically in practice, we extend previous work on a plane-strain rectangular block [5] to the case of an axisymmetric cylinder. A simplistic approach commonly used to analyse the slump test is to assume that all stresses are vertical, and uniform across any horizontal plane [1,3,6]. This model can also be used directly to calculate the height of incipient failure H , and gives H = y =(g);
(1)
where y is the uniaxial yield stress, is the density and g is the acceleration due to gravity. This shall henceforth be referred to as the “uniform stress model”. It predicts that the radius of the cylinder and friction conditions at the base do not a9ect the height of incipient failure, and that the material yields only along the base. In this paper, we rigorously analyse the height of incipient failure of a cylinder. We assume that the slurry is a perfectly rigid-plastic material. A 4nite-di9erencing implementation of the slip-line 4eld method is used [7–9], incorporating the Haar–Karman hypothesis. We allow for frictional e9ects at the base, which are modelled as Coulombic in nature. In so doing, we show that there exists a critical coe8cient of friction, where increasing the coe8cient of friction above this critical value does not change the height of incipient failure. The predictions of the uniform stress model are assessed using these results. Finally, we compare the slip-line 4eld cylinder solutions with analogous results for the plane-strain problem. In the next section, we describe the governing equations and the underlying assumptions in the model. This will be followed in Section 3 by a discussion of how to construct slip-line 4elds for the axisymmetric circular cylinder, with the details relegated to the appendices. Example slip-line 4eld diagrams are given for several cases of base friction and radius. In Section 4, the results of the slip-line 4eld calculations are discussed with reference to (i) predicting the height of incipient failure,
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(ii) measuring the yield stress, (iii) comparing the results to those for a rectangular block, and (iv) discussing the implications of the Haar–Karman hypothesis. Conclusions are given in Section 5.
2. Model A schematic illustration of the circular cylinder and the coordinate system used is given in Fig. 1. The angular coordinate in this cylindrical coordinate system is denoted by . We assume that the stress 4eld and the incipient strain 4eld are symmetric about the vertical axis of the cylinder, with no incipient displacement in the direction. The strain components r and z are therefore zero. We also assume that the corresponding stress components r and z are zero, although these conditions on r and z can also be derived for particular yield conditions using an associated Aow rule (see Ref. [10, p. 15] or [11, p. 119]). The non-zero stress components are rr , zz , rz and
. The force balance equation is ˆ = 0; ∇· − gZ
(2)
where is the stress tensor, with components ij , is the density of the material, g is the magnitude ˆ is the unit vector in the upward direction. We use the of the acceleration due to gravity, and Z convention that a positive stress component implies tension. The component of this force balance equation is identically zero, and the cylindrical nature of the problem is reAected in the presence of the
stress tensor component in the R component of this force balance equation. The force balance is three dimensional, ensuring that force equilibrium is maintained throughout the cylinder.
Z
H
R R0
Fig. 1. Problem geometry.
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The cylinder is assumed to be a perfectly rigid-plastic body, with the yield behaviour described by a yield criterion that does not depend on the hydrostatic stress. Also, the yield stress is assumed to have the same magnitude in tension and in compression, i.e., there is no Bauschinger e9ect (see Ref. [12, p. 8] or Ref. [11, p. 76]). Examples of yield criteria that satisfy these requirements are those due to Tresca and von Mises [11, p. 127], but the analysis presented here is appropriate for any perfectly plastic material. Note that in the plane-strain solution calculated in Ref. [5], the stress state in the region covered by the slip-line 4eld is a pure shear stress state superimposed on a hydrostatic component, i.e., the intermediate principal stress is half-way between the major and minor principal stresses. This plane-strain analysis is also valid for any perfectly plastic material in a state of pure shear superimposed on a hydrostatic component, with no hydrostatic dependence in the yield criterion. However, in the plane-strain case the yielding is characterised by a shear yield stress whereas in the axisymmetric case the yielding is characterised by a uniaxial yield stress, under the Haar–Karman assumption (see below). The problem is formulated by combining the yield condition with the force balance equation, giving di9erential equations satis4ed in the plastically deforming region of the cylinder. Under plane-strain conditions, the stress problem formulated in this way is well determined. However, with the addition of the hoop stress
, the equivalent axisymmetric stress problem is under-determined. We invoke the Haar–Karman hypothesis [9] to close the problem and ensure that it is hyperbolic, and as a consequence the need to distinguish between the di9erent yield criteria disappears. The Haar–Karman hypothesis states that the intermediate principal stress
is equal to one or other of the principal stresses in the RZ plane: 2 : (3)
= ( RR + ZZ )=2 ± ( ZZ − RR )2 =4 + RZ This implies that every element of material in the plastic region is in a state of uniaxial stress superimposed on a hydrostatic component. The algebraically larger choice of
is used because we expect the hoop stress to be non-compressive on the vertical free surface. Under the Haar–Karman hypothesis, the yield condition can be written as 2 = y2 ; ( RR − ZZ )2 + 4 RZ
(4)
where y is the yield stress in a state of uniaxial tension or compression. The uniaxial yield stress is related to the shear yield stress y in a way which depends on the yield criterion. For example, √ a Tresca material satis4es y = 2y , while a von Mises material satis4es y = 3y [11, p. 127]; [13, p. 128]. The boundary condition at the top surface and sides of the cylinder is the usual free surface condition n· = 0;
(5)
where n is the outward pointing normal to the surface. The boundary conditions at the base of the cylinder are examined next. The stress applied by the material acting on the base is given by ˆ + ZZ Z; ˆ Fbase = RZ R
(6)
ˆ is the unit vector in the radial direction. We expect that the material applies a downward where R force on the base so that ZZ is negative there. Since the base resists the outward Aow of the material,
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the shear component RZ is positive. The base frictional conditions are modelled as Coulombic in nature: RZ = min(− ZZ ; y =2);
(7)
where is the coe8cient of friction. Note that Coulomb’s law is modi4ed so that the magnitude of RZ does not exceed half the uniaxial yield stress y . As well, the line R = 0 is assumed to be a line of symmetry so that RZ = 0 on the Z-axis. All boundary conditions are compatible with the Haar–Karman hypothesis. Stresses are scaled by y and lengths are scaled by y =(g). The scaled height and scaled radius are denoted by h and r0 , respectively. To solve the problem, we 4rst remove the inhomogeneous term in Eq. (2) using the substitution = ( − gZI)=y ;
(8)
where is the scaled stress. This substitution allows us to apply standard axisymmetric slip-line 4eld theory [7–9] to the new force balance equation ∇· = 0;
(9)
where the gradient operator ∇ is relative to scaled coordinates r = Rg=y and z = Zg=y . The yield condition (4) becomes 2 (rr − zz )2 + 4rz =1
and the Haar–Karman hypothesis becomes 2:
= (rr + zz )=2 + (zz − rr )2 =4 + rz
(10) (11)
The boundary condition on the vertical surface of the cylinder is now rr = −z
and
rz = 0;
(12)
whereas the boundary condition on the base is rz = min(−zz ; 12 ):
(13)
By symmetry we require that rz = 0 at r = 0, i.e., along the centreline of the cylinder. 3. Construction of plastic eld The axisymmetric slip-line 4eld solutions are constructed using an approach similar to that used for the analogous plane-strain problem [5]. The slip-line 4eld method applied here is based on propagating information along two sets of characteristics known as - and "-lines. Fig. 2 shows an -line CA, a "-line CB and the slip-line angle between the r-axis and the -line. Further details on how the stress 4eld at C is calculated from the stress 4eld at A and B are given in Appendix A. This calculation forms the basis of the solution method. The di9erences between the axisymmetric and plane-strain rectangular block calculations arise from the presence of 1=r terms in the 4nite di9erence equations for the axisymmetric case
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
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B
β z C
φ
α A
r Fig. 2. Schematic for calculating the stress 4eld at point C from points A and B.
(see Appendix A). In the plane-strain solution we build increments on a stress 4eld for small base-width to give a series of solutions for larger values of the base-width [5], moving the centreline with each increment. However, in the cylinder case the centreline is 4xed at r = 0, and due to the 1=r terms in the slip-line 4eld equations, a 4eld for small r0 does not form part of a 4eld for a larger value of r0 . A completely new solution is therefore required for every given value of r0 . In the case of the circular cylinder, the 4rst step in constructing a solution is to choose a value for the radius r0 , which is 4xed for the rest of the calculation. Below we will outline the iterative technique for constructing the slip-line 4eld, the details of which are relegated to Appendix B. We observe in experiments of incipient failure that the material yields near the base, so that there is a region of plastic material on the vertical free surface from the base up to a certain height. In the theoretical analysis considered here, this height will be determined iteratively. We choose an initial height hp of plastic material on the vertical free surface. We then calculate the stress completely on the free surface up to height hp . This stress information is propagated into the cylinder to form a slip-line 4eld in the shape of a distorted triangle (e.g., CAD in Fig. 3), and then continued towards the base, taking into account the base friction boundary condition. To satisfy the symmetry condition at r = 0, as well as the boundary conditions at the base and on the vertical free surface, we adjust hp by iteration. Only one value of hp gives a stress 4eld which satis4es all the boundary conditions for a particular radius r0 and coe8cient of friction . The slip-line 4elds constructed in this way are statically determined, because they are uniquely speci4ed by the stress boundary conditions on the free surface, base and centreline, without reference to any kinematic boundary conditions. As hp increases, the slip-lines propagating upward and towards the centre from the bottom of the vertical free surface coincide and a stress discontinuity emerges. This feature is solved by adding slices across the top of the 4eld and building the stress discontinuity step by step, in a similar way to that used for the plane-strain case (see Appendix C and Ref. [5]). Examples of the slip-line 4elds, produced using the above procedure, are given below for three possible cases of base friction; perfect slip, large friction and intermediate friction. We discuss the
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centre-line
0.6 β
0.4
C
α
z 0.2 0.0
T
A
base
0.0
0.2
0.4
0.6
0.8
1.0
r
Fig. 3. Numerical slip-line 4eld for a cylinder of radius r0 = 1:0 with zero friction on the base. The plastic height hp is equal to the length of AD.
1.6 1.5 1.4 hpc 1.3 axisymmetric plane strain
1.2 1.1 1.0 1.5
2.0
2.5 r0
3.0
3.5
Fig. 4. Graph of the plastic height at the onset of the stress discontinuity, hpc = $=2 − arctan(1=2r0 ), as a function of r0 (Appendix D). The analogous plane-strain case hpc = $=2 is also shown.
range of values of r0 , from small values where no stress discontinuity exists, through the onset and growth of a stress discontinuity for large values of r0 . 3.1. Perfect slip The case of perfect slip ( = 0) is now considered. In Fig. 3, an example 4eld is presented for the case of r0 = 1:0, where no stress discontinuity exists. Note that the slip-lines are curved rather than straight, and the families of - and "-lines are non-parallel. This is due to the inAuence of gravity on the problem; in zero gravity the lines would be straight and parallel. As in the plane-strain problem [5], the plastic material extends all the way along the base. As r0 increases, the stress discontinuity appears 4rst at r0 = 3:49 with hp = 1:44. This feature is caused by convergence of the -lines at point C (see Fig. 3) for su8ciently large hp . The plastic height giving the onset of a stress discontinuity will henceforth be denoted hpc . The height hpc depends only on stress information propagated from the vertical free surface into 4eld CAD. Since the stress 4eld CAD depends only on r0 and hp and not (see above), we 4nd that hpc is purely a function of r0 . Therefore, hpc can be calculated using r0 only (Appendix D). The locus of possible values for hpc as a function of r0 is shown in Fig. 4, together with the value of hpc (= $2 ) for the analogous plane-strain problem [5].
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
β
1.5 z
stress discontinuity
D
1787
E
centre-line α
1.0 0.5 T
0.0
T0
0
A
base
1
2 r
3
4
Fig. 5. Numerical slip-line 4eld for the case of zero friction on the base and r0 = 4:0. The distance AE is equal to hpc , the plastic height required for the onset of the stress discontinuity. The stress discontinuity starts from where two -lines coincide. This point is where the -line emanating from point A crosses the "-line T0 E. 2.0 1.5 µ increasing
hp 1.0
µ ≥ µc µ = 0.3 µ = 0.2 µ = 0.1 µ = 0.05 µ=0
0.5 0.0 0
1
2
3
4
5
6
7
r0
Fig. 6. Plastic height hp as a function of scaled radius r0 for six cases of the coe8cient of friction at the base. The case = 0:3 is virtually indistinguishable from the case ¿ c .
As r0 is increased further, a stress discontinuity propagates. An example 4eld for the case of r0 = 4:0 is given in Fig. 5, where the stress discontinuity is present. Again, note that there is plastic material all the way along the base, and the -lines emanating from the bottom of the vertical free surface converge towards the stress discontinuity. A plot of hp as a function of r0 is given in Fig. 6, with the lowest curve corresponding to perfect slip. Note that the plastic height hp increases with increasing r0 . 3.2. Large friction In discussing the large friction case, we need to introduce the concept of a critical coe8cient of friction, c . This critical value is the coe8cient of friction above which there is no change in the solution if base friction increases. The general form of the slip-line 4eld depends not only on whether the friction coe8cient is above or below the critical value, but also on whether the radius r0 is large or small. We shall begin by describing the situation for small radii. In Fig. 7 an example 4eld for r0 = 1:0 and ¿ c is given. This illustrates how the critical coe8cient of friction arises from the slip-line 4eld. There is a fan emanating from a singularity at point A A because of the incompatibility of the boundary conditions along the base and the vertical free surface. The angular size of the fan is limited by two factors: (1) the base friction condition at A and (2) the symmetry condition = 7$=4 at the centre. The friction condition provides a
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D 1.0
centre-line β
0.8
C
α
0.6 z
B 0.4 0.2
rigid
A
0.0
A′
base
0.0
0.2
0.4
0.6
0.8
1.0
r
Fig. 7. Numerical slip-line 4eld for r0 = 1:0 and large friction, ¿ c . Line BCD is a "-line and CA is an -line. The plastic height hp is equal to the length of AD.
maximum permissible fan angle which cannot be exceeded. As the coe8cient of friction increases, the maximum angle increases from zero for perfect slip ( = 0) to $=4 for full stick (rz = y =2 on the base). At point C the value of is always less than 7$=4. As the fan angle increases from zero, the value of at point B also increases from its initial value at C. In Fig. 7, point B satis4es the symmetry boundary condition at a fan angle smaller than the limit imposed by the friction condition at A . Consequently, the angular size of the fan in this case is determined by the centre symmetry condition = 7$=4 rather than the limit due to friction. Therefore, the material is rigid on the base, except at point A A. For a given radius r0 , the coe8cient of friction is equal to the critical coe8cient of friction c when the fan angle due to the symmetry condition and the limiting angle due to friction are the same. When ¿ c , the slip-line 4eld is not a9ected by the coe8cient of friction and the plastic 4eld does not see the base boundary condition except at A A. For coe8cients of friction smaller than the critical value, the fan angle is at the limit set by friction and a plastic region (which varies with friction) is required below the fan and along part of the base to satisfy the symmetry condition (see Fig. 10). The value of c increases with increasing r0 until it reaches a maximum value of 2=(2+$)(=0:389) [5], see Fig. 8. 1 The analysis above applies to cylinders with radius r0 such that c ¡ 2=(2 + $). For large r0 , the rigid (and static) material does not extend across the whole base for friction coe8cients either above or below the critical value. Where there is sliding along the base when the friction coe8cient is above the critical value, the associated boundary condition is a perfect stick condition (rz = y =2), as we shall discuss below.
1
This is the same result as for the plane-strain case because the maximum value of the critical coe8cient of friction depends only on the stress singularity at point A A. At this point dr and d z are zero so that the 1=r terms are not present in 4nite di9erence equations linking the vertical free surface to the base.
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
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0.4 0.3
µc 0.2
axisymmetric plane strain
0.1 0.0 0.0
0.5
1.0
1.5
2.0
r0 Fig. 8. Critical coe8cient of base friction c , calculated from numerical slip-line 4elds, as a function of scaled radius r0 . The critical coe8cient of friction for the equivalent plane-strain problem [5] is also shown.
2.5
stress discontinuity
centre-line β
2.0
D
α
1.5 z 1.0
E T
0.5
rigid
0.0
base
0
1
T0 2 r
S
A A′
3
4
Fig. 9. Numerical slip-line 4eld for r0 = 4:0 and large friction, ¿ c . The plastic height required for the onset of the stress discontinuity, hpc , is the length of AE. The stress discontinuity starts where two -lines coincide. This point can be seen as the intersection of the -line propagating from A (on the right-hand side of the fan) with the "-line T0 E.
The analogous curve for critical coe8cient of friction as a function of base half-width for the plane-strain case is also shown in Fig. 8 [5]. 2 Interestingly, this curve is close to the axisymmetric results, with slight di9erences due to the 1=r terms. For the case ¿ c , the stress discontinuity occurs 4rst at hpc = 1:23 and r0 = 1:4. These values of hpc and r0 are smaller than the corresponding values for the perfect slip ( = 0) case, see above. For ¿ c and r0 ¿ 1:4, a stress discontinuity exists. An example slip-line 4eld for r0 = 4:0 is given in Fig. 9. Note that there is a large fan emanating from point A A, and on part of the base, SA , the -lines meet the base tangentially indicating that the shear stress is equal to half the uniaxial yield stress. This full stick condition (see Eq. (13)) is independent of any increase in the coe8cient of friction above the critical value c = 2=(2 + $). The relationship between the plastic height hp and radius r0 for the case of ¿ c is shown as the solid curve in Fig. 6.
2
In the plane-strain calculation a slightly di9erent scaling for lengths is employed: the scaled half-width is equal to the true half-width divided by 2y =g where y is the material’s shear yield stress. In plotting Figs. 4, 8 and 13, we had to assume a relationship between y and y . We chose y = 2y , which corresponds to Tresca’s yield criterion.
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1.0
D
centre-line
0.8 C
β
0.6
B
α
z 0.4
T
0.2 rigid
0.0
0.0
0.2
A A′
base
S
0.4
0.6
0.8
1.0
r
Fig. 10. Numerical slip-line 4eld for r0 = 1:0 and intermediate friction, = 0:2. Line TBCD is a "-line and CA is an -line. The plastic height hp is equal to the length of AD.
2.5
centre-line
stress discontinuity
β
2.0
D
α
z
1.5 1.0
E T
0.5
rigid
0.0
base
0
1
T0 2 r
A A′
S 3
4
Fig. 11. Numerical slip-line 4eld for r0 = 4:0 and intermediate friction, = 0:2. The length of line AE is the plastic height at the onset of the stress discontinuity, hpc . The stress discontinuity starts where two -lines coincide. This point can be seen as the intersection of the -line propagating from A (on the right-hand side of the fan) with "-line T0 E.
3.3. Intermediate friction In the case of intermediate friction, 0 ¡ ¡ c , the slip-line 4elds exhibit hybrid features from both of the extreme cases above. We will discuss the particular case of = 0:2, which is in the intermediate range of friction if r0 ¿ 0:3 (Fig. 8). An example slip-line 4eld for r0 = 1:0 where no stress discontinuity exists is given in Fig. 10. The onset of the stress discontinuity occurs at hpc = 1:29 and r0 = 1:69. A slip-line 4eld for r0 = 4:0 which contains a stress discontinuity is given in Fig. 11. In both these solutions (r0 = 1:0 and 4:0) there is a fan emanating from A A with smaller fan angle than in the case of large friction, and part of the base is plastic, i.e., the line SA shown in Figs. 10 and 11. In the graph of hp versus r0 for several values of coe8cient of friction (Fig. 6), the curves for intermediate friction are bounded by the curves for =0 and ¿ c , with hp increasing as increases.
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
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2.4 2.0 h 1.6
µ ≥ µc µ = 0.3 µ = 0.2 µ = 0.1 µ = 0.05 µ=0
µ increasing
1.2
uniform stress model
0.8 0
2
4
6
8
r0 Fig. 12. Scaled height of incipient failure h = (g=y )H versus scaled radius r0 = (g=y )R0 for six values of coe8cient of friction at the base. The case = 0:3 is virtually indistinguishable from the case ¿ c . The prediction of the uniform stress model is also shown.
4. Results and discussion The height of incipient failure for a circular cylinder loaded by gravity can be calculated from the stress 4elds constructed in the previous section, using a method based on that used for the plane-strain analysis [5]. The slip-line 4eld calculation gives the stress tensor along a line from the top of the plastic part of the vertical free surface at height hp to the centre of the cylinder (e.g., TBCD in Fig. 10). This line de4nes a surface of revolution in the axisymmetric geometry. The upward force exerted by the material below this surface of revolution must equal the weight of the material above, when the cylinder is at the point of incipient failure [5]. This force balance gives a scaled height of incipient failure h for each stress 4eld. Note that the scaled height of incipient failure depends only on the radius and the coe8cient of friction on the base, whereas the actual height of incipient failure H = (y =(g))h depends on the yield stress, gravity and the density as well. Performing the force balance calculation for a series of stress 4elds gives a graph of h as a function of r0 for several values of (Fig. 12). Also shown in Fig. 12 is the prediction of the uniform stress model (h = 1) which is independent of the coe8cient of friction and the radius r0 . In the slip-line 4eld results, as r0 approaches zero, the scaled height of incipient failure approaches unity for all . As r0 increases from zero, h increases for all friction cases. The e9ect of friction is to increase the height of incipient failure. This continues until the critical value of coe8cient of friction c is reached (see Fig. 8). At this point the curves for di9erent friction cases coincide, so that h remains constant as is increased above c . From Fig. 12, it is clear that the predictions of the uniform stress model can be greatly in error for r0 ¿ 0. A comparison of these three-dimensional axisymmetric results with analogous results for a planestrain rectangular block [5] (also see footnote 2) reveals that the curves coincide at r0 = 0, and qualitatively similar behaviour is observed for r0 ¿ 0 (see Fig. 13). The di9erence between the two sets of data increases as the base half-width and radius increase, with the cylinder incipient height data always lower than that for the rectangular block except at r0 = 0. Nonetheless, the di9erence is smaller than 12% in all cases. In the axisymmetric case a plot of h versus hp (Fig. 14) shows that the ratio of hp to h is virtually independent of the coe8cient of friction . This phenomenon should
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Fig. 13. Graph comparing the cylinder and plane-strain results for height of incipient failure h versus radius or half-width. The horizontal shading corresponds to the region covered by the range of friction cases in the plane-strain case, and the vertical shading corresponds to the range of friction cases in the cylinder case.
2.5 2.0 h
µ ≥ µc µ = 0.3 µ = 0.2 µ = 0.1 µ = 0.05 µ=0
1.5 1.0 0.0
0.5
1.0
1.5
2.0
2.5
hp
Fig. 14. Plot of scaled height of incipient failure, h, versus the height of plastic material on the vertical free surface, hp .
be observable experimentally. The corresponding plane-strain result in Ref. [5] is very similar to this result, with the curves for the axisymmetric case within 5% of the plane-strain results (also see footnote 2). The height of incipient failure data can also be presented as a plot of uniaxial yield stress y multiplied by a factor 1=(Hg) versus aspect ratio r0 =h = R0 =H (Fig. 15). This suggests a method for determining the yield stress of a slurry. Measuring the height of incipient failure H and the radius R0 gives the aspect ratio at incipient failure on the horizontal axis of the graph in Fig. 15. After choosing which friction curve is appropriate, the yield stress can be read o9 the vertical axis of the graph using the known factor 1=(Hg). The experimental implementation of this method will be detailed elsewhere. The Haar–Karman hypothesis restricts the stress 4eld in the yielded region to have the hoop stress
equal to the algebraically greater principal stress in the r–z plane. Assuming that the stress 4eld can be continued into the non-deforming region while satisfying the equilibrium equations and stress boundary conditions, the slip-line solutions give a lower bound on the height of incipient failure [10, p. 19]. How close this lower bound is to an upper bound solution or an exact solution and how well the Haar–Karman hypothesis is satis4ed in the exact solution could be investigated using models which do not invoke the Haar–Karman hypothesis. For example, models which involve constructing kinematically admissable upper bound solutions or models which simultaneously solve the stress and
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1.1 (1/(Hρg)) σy
1.0
uniform stress model
0.9 0.8
µ increasing
0.7
µ=0 µ = 0.05 µ = 0.1 µ = 0.2 µ = 0.3 µ ≥ µc
0.6 0.5 0.4 0.0
0.5
1.0
1.5 2.0 R0 /H
2.5
3.0
Fig. 15. Plot of 1=h = (1=(Hg))y versus aspect ratio r0 =h = R0 =H where h is the scaled height of incipient failure for six values of the coe8cient of friction . The friction case = 0:3 is virtually indistinguishable from the case ¿ c . We also show the prediction of the uniform stress model.
kinematic equations could be used for this purpose. These investigations are outside the scope of the present study.
5. Conclusions We have investigated the incipient failure of a cylinder under gravitational loading using slip-line 4eld theory. The cylinder was assumed to behave as a perfectly rigid-plastic material and satisfy the Haar–Karman hypothesis. The friction at the base was modelled using a modi4ed Coulomb law. The stress 4eld in the deformable plastic region of the cylinder at the point of incipient failure has been calculated. This gives a prediction of the height of incipient failure of a cylinder as a function of the uniaxial yield stress and density of the material, acceleration due to gravity, cylinder radius and base friction. The results for the height of incipient failure of a cylinder suggest a method of determining the yield stress of a concentrated suspension such as a mineral tailings slurry. Details of an experimental study of this method will be reported elsewhere. We found that the results for the cylinder case are very similar to the results for the plane-strain rectangular block [5] (also see footnote 2) with the height of incipient failure increasing as the radius of the cylinder increases. However, the height of incipient failure is lower for the cylinder than for the rectangular block with the same frictional conditions on the base and the radius of the cylinder equal to the half-width of the block. We have examined the e9ect of base friction on the height of incipient failure. In the limit as the radius r0 → 0, the height of incipient failure is independent of friction and equal to y =(g), which is the prediction of the uniform stress model. As r0 increases from zero, the e9ect of friction becomes greater with the height of incipient failure increasing with increasing friction, until the critical coe8cient of friction is reached. At this point raising the coe8cient of friction has no e9ect on the height of incipient failure. The results given here show that the uniform stress model is inaccurate for all values of radius except when r0 1. This agrees with the similar 4nding in Ref. [5] for the rectangular block.
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Acknowledgements We acknowledge the support of the Particulate Fluids Processing Centre (PFPC), an Australian Research Council Special Research Centre. Appendix A. Slip-line eld theory Eqs. (9) and (10) with the Haar–Karman hypothesis (11) form a hyperbolic system with characteristics running along the lines of maximum shear stress. The two sets of characteristics (slip-lines) are labelled as - and "-lines according to the following rule: the - and "-lines must form a right-handed coordinate system, and the algebraically greater of the two principal stresses must act in the 4rst and third quadrants of the –" coordinate system. We write the equations using the variables p, where p = −(rr + zz )=2 and is the angle between the r-axis and the -direction measured in the anti-clockwise direction. The stresses can be written in terms of p and : rr = −p − 12 sin 2; rz = 12 cos 2;
zz = −p + 12 sin 2;
= −p + 12 :
(A.1)
The hoop stress
is chosen to be equal to the algebraically greater principal stress in the r–z plane, according to the Haar–Karman hypothesis as discussed in Section 2 (also see Ref. [9]). Information is propagated along the slip-lines using the following equations based on Morrison and Richmond [8] and Shield [9], modi4ed by our scaling and choice in the Haar–Karman hypothesis. For an -line we have dp + d +
dr + d z = 0; 2r
tan =
dz dr
(A.2)
and for a "-line we have dp − d +
dr − d z = 0; 2r
cot = −
dz : dr
(A.3)
The slip-line 4eld is constructed using 4nite di9erence versions of these equations. For any two adjacent points A and C along an -line (see Fig. 2) we have r C − rA + z C − z A p C − pA + C − A + = 0; (A.4) rC + r A A + C − (zC − zA ) = 0: (A.5) (rC − rA ) tan 2 For any two adjacent points B and C along a "-line we have pC − pB − (C − B ) +
B + C (rC − rB ) cot 2
rC − rB − (zC − zB ) = 0; rC + r B
(A.6)
+ (zC − zB ) = 0:
(A.7)
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In constructing the slip-line 4elds illustrated in this paper, the following sub-problem occurs frequently: if we know p and at points A and B and these points are connected to point C by an -line AC and a "-line BC then we wish to 4nd p, , r and z at point C. Once we have formulated a procedure to solve this problem then this is used repeatedly to build up a slip-line 4eld. We formulate the problem in terms of the four equations (A.4) – (A.7) in four unknowns, pC , C , rC and zC . These equations are reduced to two by eliminating pC from Eqs. (A.4) and (A.6), and substituting for zC using Eq. (A.5): −pA + pB + 2C − A − B C + A rC − rA 1 + tan + rA + r C 2 C + A 1 − zB = 0; rB − rC + zA + (rC − rA ) tan + rB + r C 2 B + C (rC − rB ) cot 2 C + A − zB = 0: + zA + (rC − rA ) tan 2
(A.8)
(A.9)
We employ a root 4nding procedure to solve for rC and C , and then calculate zC and pC using Eqs. (A.4) and (A.5). The 4rst guess for the root 4nding procedure, (rC(1) ; (2) C ), is found using the method of Chitkara and Butt [7]. For completeness, the method of Chitkara and Butt is given below with our notation and scaling. An approximation for the position of point C is given by extending the tangents to the - and "-lines at points A and B, respectively, and 4nding the point of intersection of these two tangents: zC(0) = (−rA + rB + zB tan B + zA cot A )=(cot A + tan B );
(A.10)
rC(0) = rA + (zC(0) − zA ) cot A :
(A.11)
Putting rC(0) and zC(0) into Eqs. (A.6) and (A.4), and taking the di9erence between these two equations, eliminates pC and gives an estimate of at point C: 2(0) C = p A − pB + A + B − +
1 rB + rC(0)
1 rA + rC(0)
(rC(0) − rA + zC(0) − zA )
(rC(0) − rB − zC(0) + zB ):
(A.12)
Based on the fact that the intersection of the tangents at A and B is not exactly point C, a second approximation to C is found using (1) C =
(0) C + A : 2
(A.13)
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Second estimates for rC and zC are calculated by solving the following two equations for rC(1) and zC(1) , where average slopes of AC and BC have been used to determine the position of point C. rC(1) − rA = 12 (zC(1) − zA )(cot A + cot (1) C );
(A.14)
zB − zC(1) = 12 (rC(1) − rB )(cot B + cot (1) C ):
(A.15)
This gives zC(1) =
FGzA + 4zB + 2G(rB − rA ) ; FG + 4
(A.16)
rC(1) = rA + 12 F(zC(1) − zA );
(A.17)
(1) where F = cot A + cot (1) C and G = cot B + cot C . The 4nal estimates of C and pC in Chitkara and Butt’s method are given by putting rC(1) and zC(1) into the following formulae:
2(2) C = p A − pB + A + B + +
1 rB +
rC(1)
1 rA +
rC(1)
(rA − rC(1) + zA − zC(1) )
(rC(1) − rB − zC(1) + zB );
pC(0) = pA − (2) C + A −
1 rA +
rC(1)
(rC(1) − rA + zC(1) − zA );
(A.18) (A.19)
where Eq. (A.18) is obtained in the same way as Eq. (A.12), and Eq. (A.19) is from Eq. (A.4). (0) Chitkara and Butt’s method calculates point C as the point which satis4es C = (2) C , pC = pC , (1) (1) rC = rC , zC = zC . This estimate is normally close to the solution of the 4nite di9erence equations (A.4) – (A.7), and gives fast convergence in the root 4nding procedure discussed above.
Appendix B. Details of construction of plastic elds Assuming that the material is plastic on the vertical free surface up to height hp implies that the stress along this free surface is given by p = 1=2 + z and = 7$=4, using an argument similar to that in Ref. [5]. The initial choice of hp is chosen to be an underestimate of the actual value of hp for the given radius and base friction. After propagating the stress 4eld into the body in the shape of a distorted triangle (CAD in Fig. 16), the "-line along the top has ¡ 7$=4 at point C. However, the centre symmetry condition demands that = 7$=4. Depending on the friction conditions, we add 4elds to the initial triangle so that the closest point to the centre satis4es = 7$=4. Fig. 16 shows the additional 4elds BA AC and TSA B. In some cases, not all of the pieces of the 4eld in Fig. 16 are needed. If there is no friction at the base then no fan (BA AC) is needed because the boundary conditions at the vertical free surface are compatible with the base boundary condition at point A. If the base coe8cient of friction is su8ciently large and r0 is su8ciently small then the material in
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z D rigid
β
C hp
B T
fan
π/4
α
rigid S
A A'
r
r0 Fig. 16. Schematic of a slip-line 4eld solution for an initial choice of hp and non-zero friction on the base.
the 4eld TSA B is not needed and the size of the fan is chosen so that = 7$=4 at B. Otherwise, the 4eld TSA B is constructed so that = 7$=4 at point T . Usually, the initial choice of hp is not large enough for the 4eld to reach the centre and satisfy the symmetry condition = 7$=4. After completing the 4eld for the initial choice of hp , the plastic height hp is incremented, and a new slip-line 4eld constructed by adding to the 4eld discussed above. A narrow slice is added above the "-line TBCD, followed by a slice below -line TS, chosen to maintain the symmetry condition = 7$=4 at the closest point to the centreline. This process is repeated until the 4eld reaches the centre. However, normally the 4eld overshoots the centre and the last increment of hp needs to be iterated to obtain a 4eld which just touches the centre. Appendix C. Construction of the stress discontinuity For su8ciently large hp , the -lines coincide at a point. This point can be seen as the intersection of the -line from point A (on the right-hand side of the fan) with the "-line T0 E in Fig. 11, or as point C in Fig. 17. The stress discontinuity forms a line emanating upward from this point. The tangential component of stress is discontinuous and the normal and shear components of stress are continuous across the stress discontinuity. This leads to the following two conditions used in the construction of the stress discontinuity: (i) for p and we have [5] p+ − p− = |sin(+ − − )|;
(C.1)
where the superscript + refers to the side of the stress discontinuity closest to the vertical free surface and the superscript − refers to the opposite side; and (ii) the line of the stress discontinuity bisects any two -lines emanating from a single point on the stress discontinuity, where one of the -lines is on the left-hand side of the discontinuity and the other -line is on the right-hand side. We construct the stress discontinuity in a series of steps during the process of incrementally increasing hp described in Appendix B. At each increment of hp , the slice above the "-line T0 E (Fig. 11) is constructed from the free surface down to the point where the stress discontinuity occurs. The 4eld to the right of the increment of the stress discontinuity is known, and we wish to
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z
stress discontinuity
D β
C hpc
α A r r0 Fig. 17. Schematic for calculating the plastic height at the onset of the stress discontinuity, hpc .
4nd the 4eld in the form of the next point on the stress discontinuity and a slice to the left of the stress discontinuity. This 4eld is required to be compatible with the existing 4eld ("-line T0 E and below, Fig. 11). Formulating suitable 4nite di9erence relations based on Eqs. (A.4) – (A.7) applied to points near the stress discontinuity and the two conditions above, gives a well-determined problem. Solving these equations in an iterative fashion yields the new point on the stress discontinuity. This point is uniquely determined by conditions (i) and (ii), and the existing slip-line 4eld below. The 4eld is extended to form a slice above "-line T0 E (Fig. 11) or TBCD (Fig. 16). A 4eld is added below the -line TS (Fig. 16) to maintain the condition = 7$=4 at the closest point to the centre. This procedure is repeated, building up the stress discontinuity as shown in Figs. 5, 9 and 11.
Appendix D. Onset of stress discontinuity In the equivalent plane-strain incipient failure problem the height of plastic material on the vertical free surface which gives the onset of the stress discontinuity is $=2 [5]. In the axisymmetric case, this height, denoted by hpc , is less than $=2 and depends on r0 . It can be calculated as follows. We consider a triangle ACD with AD along the vertical free surface and D at the top of the plastic region (Fig. 17). The stress 4eld within this triangle is entirely determined by the boundary conditions on AD. These conditions (constant and constant, unit, pressure gradient along the boundary) imply that must be a function of r only in ACD and p must equal z plus a function of r only in ACD. Under these circumstances, the axial force balance equation can be integrated exactly to give 2 r cos 2 = −r 02 − 1 : (D.1) r
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
The length AD (which we have been calling hp ) can then be calculated as 3$ dr hp = tan(2$ − ) dr + tan − 2 CA CD r0 2 = dr 1 − cos2 2 rc cos 2 4r02 + cos2 2 − sin2 2 1 + arctan = − arctan 2r0 sin 2(cos 2 + 4r02 + cos2 2)
:
1799
(D.2)
=C
All that remains is to determine C when C lies at the bottom of the stress discontinuity. From Eq. (D.1), as r decreases from r0 , cos 2 decreases from zero until at some r value it reaches −1. This forms a natural boundary of the stress 4eld de4ned by the boundary conditions on AD, where the - and "-lines become vertical and horizontal, respectively. It is when C reaches this boundary that a stress discontinuity becomes necessary, as the -lines that intersect the vertical free surface and those that intersect the base would otherwise start to cross above C. Therefore, C = 6$=4, and since
cos 2 4r02 + cos2 2 − sin2 2 $ (D.3) = lim + arctan 2 6$ sin 2(cos 2 + 4r02 + cos2 2) → 4
we have
1 $ hpc = − arctan : 2 2r0
This result coe8cients larger, hpc footnote 2,
(D.4)
is illustrated in Fig. 4 for the range of values for r0 compatible with the possible of friction, given that the solution is at the onset stress discontinuity. As r0 becomes for the axisymmetric geometry approaches the plane-strain result [5], see Fig. 4 and as expected.
References [1] Pashias N, Boger DV, Summers J, Glenister DJ. A 4fty cent rheometer for yield stress measurement. Journal of Rheology 1996;40(6):1179–89. [2] Tanigawa Y, Mori H. Analytical study on deformation of fresh concrete. Journal of Engineering Mechanics 1989;115(3):493–508. [3] Christensen G. Modelling the Aow of fresh concrete: the slump test. Ph.D thesis, Department of Chemical Engineering, Princeton University, 1991. [4] Nguyen QD, Boger DV. Measuring the Aow properties of yield stress Auids. Annual Review of Fluid Mechanics 1992;24:47–88. [5] Chamberlain JA, Sader JE, Landman KA, White LR. Incipient plane-strain failure of a rectangular block under gravity. International Journal of Mechanical Sciences 2001;43(3):793–815. [6] Murata J. Flow and deformation of fresh concrete. MatUeriaux et Constructions 1984;17(98):117–29.
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[7] Chitkara NR, Butt MA. A general numerical method of construction of axisymmetric slip-line 4elds. International Journal of Mechanical Sciences 1992;34(11):833–48. [8] Morrison HL, Richmond O. Large deformation of notched perfectly plastic tensile bars. Journal of Applied Mechanics 1972;39:971–7. [9] Shield RT. On the plastic Aow of metals under conditions of axial symmetry. Proceedings of the Royal Society, London Series A 1955;233:267–87. [10] Chakrabarty J. Applied plasticity. New York: Springer-Verlag, 1999. [11] Lubliner J. Plasticity theory. New York: Macmillan, 1990. [12] Hill R. The mathematical theory of plasticity. Oxford: Clarendon Press, 1950. [13] Desai CS, Siriwardane HJ. Constitutive laws for engineering materials with emphasis on geologic materials. Englewood Cli9s, NJ: Prentice-Hall, 1984.
Incipient failure of a circular cylinder under gravity J.A. Chamberlaina; ∗ , J.E. Sadera , K.A. Landmana , D.J. Horrobina , L.R. Whiteb a
b
Department of Mathematics and Statistics, The University of Melbourne, Victoria 3010, Australia Department of Chemical Engineering, Carnegie Mellon University, Pittsburgh, PA 15213-3890, USA Received 23 October 2000; accepted 15 April 2002
Abstract We consider the incipient failure of a circular cylinder under gravitational loading. Axisymmetric slip-line 4eld theory is used to solve for the stress distribution in the plastic region of the cylinder at incipient failure. This extends similar work on the incipient failure of a rectangular block under gravity (Int. J. Mech. Sci. 43(3) (2000) 793), to the axisymmetric geometry. Our calculations show that the height of incipient failure depends not only on the yield stress and density of the material, but also on the radius of the cylinder and the coe8cient of friction at the base. In addition, we 4nd that there exists a critical coe8cient of friction above which the height of incipient failure is una9ected. These results suggest a method of measuring the yield stress by observing the height of incipient failure. ? 2002 Published by Elsevier Science Ltd. Keywords: Slip-lines; Slump test; Yield stress; Cylinder; Axisymmetric; Gravity
1. Introduction The yield stress of a slurry is an important parameter used in industrial processes. In mining engineering, the yield stress is a control parameter governing the pumping and disposal of mine tailings [1], whereas in the construction industry, the yield stress of fresh concrete a9ects its mixing, transportation, placing and consolidation behaviour [2,3]. Laboratory instruments such as the vane rheometer [4] can be used to measure the yield stress of these materials. However, in the 4eld the “slump test” presents a simpler approach [1]. In this test a frustum of a cylinder or a cone is placed on a Aat base and 4lled with slurry. Lifting the frustum causes the slurry to slump down to a lower height, which can then be related to the yield stress. Rather than considering this Aow problem, we examine the height of incipient failure, i.e., the height of material required to just initiate Aow. This height is of fundamental importance to the understanding and operation of the slump test, since it ∗
Corresponding author. Fax: +61-3-8344-4599. E-mail address: [email protected] (J.A. Chamberlain).
0020-7403/02/$ - see front matter ? 2002 Published by Elsevier Science Ltd. PII: S 0 0 2 0 - 7 4 0 3 ( 0 2 ) 0 0 0 4 4 - 9
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Nomenclature H R0 y y g R Z
ij ij ˆ Z ∇ n Fbase ˆ R h r0 ij I r z hp hpc c p pA ; pB ; : : : A ; B ; : : : rA ; rB ; : : : zA ; zB ; : : : rC(1) (2) C
rC(0)
unscaled height of incipient failure unscaled radius of cylinder uniaxial yield stress of the material shear yield stress of the material density of the material acceleration due to gravity unscaled radial coordinate in the cylindrical coordinate system unscaled vertical coordinate in the cylindrical coordinate system angular coordinate in the cylindrical coordinate system the ij component of the unscaled stress tensor unscaled stress tensor the ij component of the strain tensor unit vector in the upward direction, Z increasing di9erential del operator unit normal to a surface stress applied by the material acting on the base unit vector in the radial direction, R increasing coe8cient of friction on the base scaled height of incipient failure scaled radius of the cylinder scaled stress tensor the ij component of the scaled stress tensor unit tensor scaled radial coordinate in the cylindrical coordinate system scaled vertical coordinate in the cylindrical coordinate system scaled height of plastic material on the vertical free surface scaled height of plastic material on the vertical free surface at the onset of the stress discontinuity critical coe8cient of friction. Increasing friction above this value has no e9ect on the solution equal to −(rr + zz )=2, one of the two variables used to specify the stress state angle between the r-axis and the -direction measured in the anti-clockwise direction. Used along with p to specify the stress state value of p at point A, point B; : : : value of at point A, point B; : : : value of r at point A, point B; : : : value of z at point A, point B; : : : 4nal estimate of r at point C, Fig. 2, using Chitkara and Butt’s method 4nal estimate of at point C, Fig. 2, using Chitkara and Butt’s method initial estimate of r at point C, Fig. 2, using Chitkara and Butt’s method
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
zC(0)
1781
initial estimate of z at point C, Fig. 2, using Chitkara and Butt’s method
(0) C (1) C (1) zC F
G pC(0) p+ p− + −
initial estimate of at point C, Fig. 2, using Chitkara and Butt’s method intermediate estimate of at point C, Fig. 2, using Chitkara and Butt’s method 4nal estimate of z at point C, Fig. 2, using Chitkara and Butt’s method equals cot A + cot (1) C , used in Chitkara and Butt’s method
equals cot B + cot (1) C , used in Chitkara and Butt’s method 4nal estimate of p at point C, Fig. 2, using Chitkara and Butt’s method value of p at the stress discontinuity, on the side closest to the vertical free surface value of p at the stress discontinuity, on the side closest to the centreline value of at the stress discontinuity, on the side closest to the vertical free surface value of at the stress discontinuity, on the side closest to the mid-line
must be exceeded for Aow to occur. Also, calculation of the point of incipient failure can be used to develop a method for determining the yield stress, as we shall discuss. Since cylindrical geometry is used typically in practice, we extend previous work on a plane-strain rectangular block [5] to the case of an axisymmetric cylinder. A simplistic approach commonly used to analyse the slump test is to assume that all stresses are vertical, and uniform across any horizontal plane [1,3,6]. This model can also be used directly to calculate the height of incipient failure H , and gives H = y =(g);
(1)
where y is the uniaxial yield stress, is the density and g is the acceleration due to gravity. This shall henceforth be referred to as the “uniform stress model”. It predicts that the radius of the cylinder and friction conditions at the base do not a9ect the height of incipient failure, and that the material yields only along the base. In this paper, we rigorously analyse the height of incipient failure of a cylinder. We assume that the slurry is a perfectly rigid-plastic material. A 4nite-di9erencing implementation of the slip-line 4eld method is used [7–9], incorporating the Haar–Karman hypothesis. We allow for frictional e9ects at the base, which are modelled as Coulombic in nature. In so doing, we show that there exists a critical coe8cient of friction, where increasing the coe8cient of friction above this critical value does not change the height of incipient failure. The predictions of the uniform stress model are assessed using these results. Finally, we compare the slip-line 4eld cylinder solutions with analogous results for the plane-strain problem. In the next section, we describe the governing equations and the underlying assumptions in the model. This will be followed in Section 3 by a discussion of how to construct slip-line 4elds for the axisymmetric circular cylinder, with the details relegated to the appendices. Example slip-line 4eld diagrams are given for several cases of base friction and radius. In Section 4, the results of the slip-line 4eld calculations are discussed with reference to (i) predicting the height of incipient failure,
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(ii) measuring the yield stress, (iii) comparing the results to those for a rectangular block, and (iv) discussing the implications of the Haar–Karman hypothesis. Conclusions are given in Section 5.
2. Model A schematic illustration of the circular cylinder and the coordinate system used is given in Fig. 1. The angular coordinate in this cylindrical coordinate system is denoted by . We assume that the stress 4eld and the incipient strain 4eld are symmetric about the vertical axis of the cylinder, with no incipient displacement in the direction. The strain components r and z are therefore zero. We also assume that the corresponding stress components r and z are zero, although these conditions on r and z can also be derived for particular yield conditions using an associated Aow rule (see Ref. [10, p. 15] or [11, p. 119]). The non-zero stress components are rr , zz , rz and
. The force balance equation is ˆ = 0; ∇· − gZ
(2)
where is the stress tensor, with components ij , is the density of the material, g is the magnitude ˆ is the unit vector in the upward direction. We use the of the acceleration due to gravity, and Z convention that a positive stress component implies tension. The component of this force balance equation is identically zero, and the cylindrical nature of the problem is reAected in the presence of the
stress tensor component in the R component of this force balance equation. The force balance is three dimensional, ensuring that force equilibrium is maintained throughout the cylinder.
Z
H
R R0
Fig. 1. Problem geometry.
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
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The cylinder is assumed to be a perfectly rigid-plastic body, with the yield behaviour described by a yield criterion that does not depend on the hydrostatic stress. Also, the yield stress is assumed to have the same magnitude in tension and in compression, i.e., there is no Bauschinger e9ect (see Ref. [12, p. 8] or Ref. [11, p. 76]). Examples of yield criteria that satisfy these requirements are those due to Tresca and von Mises [11, p. 127], but the analysis presented here is appropriate for any perfectly plastic material. Note that in the plane-strain solution calculated in Ref. [5], the stress state in the region covered by the slip-line 4eld is a pure shear stress state superimposed on a hydrostatic component, i.e., the intermediate principal stress is half-way between the major and minor principal stresses. This plane-strain analysis is also valid for any perfectly plastic material in a state of pure shear superimposed on a hydrostatic component, with no hydrostatic dependence in the yield criterion. However, in the plane-strain case the yielding is characterised by a shear yield stress whereas in the axisymmetric case the yielding is characterised by a uniaxial yield stress, under the Haar–Karman assumption (see below). The problem is formulated by combining the yield condition with the force balance equation, giving di9erential equations satis4ed in the plastically deforming region of the cylinder. Under plane-strain conditions, the stress problem formulated in this way is well determined. However, with the addition of the hoop stress
, the equivalent axisymmetric stress problem is under-determined. We invoke the Haar–Karman hypothesis [9] to close the problem and ensure that it is hyperbolic, and as a consequence the need to distinguish between the di9erent yield criteria disappears. The Haar–Karman hypothesis states that the intermediate principal stress
is equal to one or other of the principal stresses in the RZ plane: 2 : (3)
= ( RR + ZZ )=2 ± ( ZZ − RR )2 =4 + RZ This implies that every element of material in the plastic region is in a state of uniaxial stress superimposed on a hydrostatic component. The algebraically larger choice of
is used because we expect the hoop stress to be non-compressive on the vertical free surface. Under the Haar–Karman hypothesis, the yield condition can be written as 2 = y2 ; ( RR − ZZ )2 + 4 RZ
(4)
where y is the yield stress in a state of uniaxial tension or compression. The uniaxial yield stress is related to the shear yield stress y in a way which depends on the yield criterion. For example, √ a Tresca material satis4es y = 2y , while a von Mises material satis4es y = 3y [11, p. 127]; [13, p. 128]. The boundary condition at the top surface and sides of the cylinder is the usual free surface condition n· = 0;
(5)
where n is the outward pointing normal to the surface. The boundary conditions at the base of the cylinder are examined next. The stress applied by the material acting on the base is given by ˆ + ZZ Z; ˆ Fbase = RZ R
(6)
ˆ is the unit vector in the radial direction. We expect that the material applies a downward where R force on the base so that ZZ is negative there. Since the base resists the outward Aow of the material,
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the shear component RZ is positive. The base frictional conditions are modelled as Coulombic in nature: RZ = min(− ZZ ; y =2);
(7)
where is the coe8cient of friction. Note that Coulomb’s law is modi4ed so that the magnitude of RZ does not exceed half the uniaxial yield stress y . As well, the line R = 0 is assumed to be a line of symmetry so that RZ = 0 on the Z-axis. All boundary conditions are compatible with the Haar–Karman hypothesis. Stresses are scaled by y and lengths are scaled by y =(g). The scaled height and scaled radius are denoted by h and r0 , respectively. To solve the problem, we 4rst remove the inhomogeneous term in Eq. (2) using the substitution = ( − gZI)=y ;
(8)
where is the scaled stress. This substitution allows us to apply standard axisymmetric slip-line 4eld theory [7–9] to the new force balance equation ∇· = 0;
(9)
where the gradient operator ∇ is relative to scaled coordinates r = Rg=y and z = Zg=y . The yield condition (4) becomes 2 (rr − zz )2 + 4rz =1
and the Haar–Karman hypothesis becomes 2:
= (rr + zz )=2 + (zz − rr )2 =4 + rz
(10) (11)
The boundary condition on the vertical surface of the cylinder is now rr = −z
and
rz = 0;
(12)
whereas the boundary condition on the base is rz = min(−zz ; 12 ):
(13)
By symmetry we require that rz = 0 at r = 0, i.e., along the centreline of the cylinder. 3. Construction of plastic eld The axisymmetric slip-line 4eld solutions are constructed using an approach similar to that used for the analogous plane-strain problem [5]. The slip-line 4eld method applied here is based on propagating information along two sets of characteristics known as - and "-lines. Fig. 2 shows an -line CA, a "-line CB and the slip-line angle between the r-axis and the -line. Further details on how the stress 4eld at C is calculated from the stress 4eld at A and B are given in Appendix A. This calculation forms the basis of the solution method. The di9erences between the axisymmetric and plane-strain rectangular block calculations arise from the presence of 1=r terms in the 4nite di9erence equations for the axisymmetric case
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
1785
B
β z C
φ
α A
r Fig. 2. Schematic for calculating the stress 4eld at point C from points A and B.
(see Appendix A). In the plane-strain solution we build increments on a stress 4eld for small base-width to give a series of solutions for larger values of the base-width [5], moving the centreline with each increment. However, in the cylinder case the centreline is 4xed at r = 0, and due to the 1=r terms in the slip-line 4eld equations, a 4eld for small r0 does not form part of a 4eld for a larger value of r0 . A completely new solution is therefore required for every given value of r0 . In the case of the circular cylinder, the 4rst step in constructing a solution is to choose a value for the radius r0 , which is 4xed for the rest of the calculation. Below we will outline the iterative technique for constructing the slip-line 4eld, the details of which are relegated to Appendix B. We observe in experiments of incipient failure that the material yields near the base, so that there is a region of plastic material on the vertical free surface from the base up to a certain height. In the theoretical analysis considered here, this height will be determined iteratively. We choose an initial height hp of plastic material on the vertical free surface. We then calculate the stress completely on the free surface up to height hp . This stress information is propagated into the cylinder to form a slip-line 4eld in the shape of a distorted triangle (e.g., CAD in Fig. 3), and then continued towards the base, taking into account the base friction boundary condition. To satisfy the symmetry condition at r = 0, as well as the boundary conditions at the base and on the vertical free surface, we adjust hp by iteration. Only one value of hp gives a stress 4eld which satis4es all the boundary conditions for a particular radius r0 and coe8cient of friction . The slip-line 4elds constructed in this way are statically determined, because they are uniquely speci4ed by the stress boundary conditions on the free surface, base and centreline, without reference to any kinematic boundary conditions. As hp increases, the slip-lines propagating upward and towards the centre from the bottom of the vertical free surface coincide and a stress discontinuity emerges. This feature is solved by adding slices across the top of the 4eld and building the stress discontinuity step by step, in a similar way to that used for the plane-strain case (see Appendix C and Ref. [5]). Examples of the slip-line 4elds, produced using the above procedure, are given below for three possible cases of base friction; perfect slip, large friction and intermediate friction. We discuss the
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J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800 D
centre-line
0.6 β
0.4
C
α
z 0.2 0.0
T
A
base
0.0
0.2
0.4
0.6
0.8
1.0
r
Fig. 3. Numerical slip-line 4eld for a cylinder of radius r0 = 1:0 with zero friction on the base. The plastic height hp is equal to the length of AD.
1.6 1.5 1.4 hpc 1.3 axisymmetric plane strain
1.2 1.1 1.0 1.5
2.0
2.5 r0
3.0
3.5
Fig. 4. Graph of the plastic height at the onset of the stress discontinuity, hpc = $=2 − arctan(1=2r0 ), as a function of r0 (Appendix D). The analogous plane-strain case hpc = $=2 is also shown.
range of values of r0 , from small values where no stress discontinuity exists, through the onset and growth of a stress discontinuity for large values of r0 . 3.1. Perfect slip The case of perfect slip ( = 0) is now considered. In Fig. 3, an example 4eld is presented for the case of r0 = 1:0, where no stress discontinuity exists. Note that the slip-lines are curved rather than straight, and the families of - and "-lines are non-parallel. This is due to the inAuence of gravity on the problem; in zero gravity the lines would be straight and parallel. As in the plane-strain problem [5], the plastic material extends all the way along the base. As r0 increases, the stress discontinuity appears 4rst at r0 = 3:49 with hp = 1:44. This feature is caused by convergence of the -lines at point C (see Fig. 3) for su8ciently large hp . The plastic height giving the onset of a stress discontinuity will henceforth be denoted hpc . The height hpc depends only on stress information propagated from the vertical free surface into 4eld CAD. Since the stress 4eld CAD depends only on r0 and hp and not (see above), we 4nd that hpc is purely a function of r0 . Therefore, hpc can be calculated using r0 only (Appendix D). The locus of possible values for hpc as a function of r0 is shown in Fig. 4, together with the value of hpc (= $2 ) for the analogous plane-strain problem [5].
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
β
1.5 z
stress discontinuity
D
1787
E
centre-line α
1.0 0.5 T
0.0
T0
0
A
base
1
2 r
3
4
Fig. 5. Numerical slip-line 4eld for the case of zero friction on the base and r0 = 4:0. The distance AE is equal to hpc , the plastic height required for the onset of the stress discontinuity. The stress discontinuity starts from where two -lines coincide. This point is where the -line emanating from point A crosses the "-line T0 E. 2.0 1.5 µ increasing
hp 1.0
µ ≥ µc µ = 0.3 µ = 0.2 µ = 0.1 µ = 0.05 µ=0
0.5 0.0 0
1
2
3
4
5
6
7
r0
Fig. 6. Plastic height hp as a function of scaled radius r0 for six cases of the coe8cient of friction at the base. The case = 0:3 is virtually indistinguishable from the case ¿ c .
As r0 is increased further, a stress discontinuity propagates. An example 4eld for the case of r0 = 4:0 is given in Fig. 5, where the stress discontinuity is present. Again, note that there is plastic material all the way along the base, and the -lines emanating from the bottom of the vertical free surface converge towards the stress discontinuity. A plot of hp as a function of r0 is given in Fig. 6, with the lowest curve corresponding to perfect slip. Note that the plastic height hp increases with increasing r0 . 3.2. Large friction In discussing the large friction case, we need to introduce the concept of a critical coe8cient of friction, c . This critical value is the coe8cient of friction above which there is no change in the solution if base friction increases. The general form of the slip-line 4eld depends not only on whether the friction coe8cient is above or below the critical value, but also on whether the radius r0 is large or small. We shall begin by describing the situation for small radii. In Fig. 7 an example 4eld for r0 = 1:0 and ¿ c is given. This illustrates how the critical coe8cient of friction arises from the slip-line 4eld. There is a fan emanating from a singularity at point A A because of the incompatibility of the boundary conditions along the base and the vertical free surface. The angular size of the fan is limited by two factors: (1) the base friction condition at A and (2) the symmetry condition = 7$=4 at the centre. The friction condition provides a
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J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
D 1.0
centre-line β
0.8
C
α
0.6 z
B 0.4 0.2
rigid
A
0.0
A′
base
0.0
0.2
0.4
0.6
0.8
1.0
r
Fig. 7. Numerical slip-line 4eld for r0 = 1:0 and large friction, ¿ c . Line BCD is a "-line and CA is an -line. The plastic height hp is equal to the length of AD.
maximum permissible fan angle which cannot be exceeded. As the coe8cient of friction increases, the maximum angle increases from zero for perfect slip ( = 0) to $=4 for full stick (rz = y =2 on the base). At point C the value of is always less than 7$=4. As the fan angle increases from zero, the value of at point B also increases from its initial value at C. In Fig. 7, point B satis4es the symmetry boundary condition at a fan angle smaller than the limit imposed by the friction condition at A . Consequently, the angular size of the fan in this case is determined by the centre symmetry condition = 7$=4 rather than the limit due to friction. Therefore, the material is rigid on the base, except at point A A. For a given radius r0 , the coe8cient of friction is equal to the critical coe8cient of friction c when the fan angle due to the symmetry condition and the limiting angle due to friction are the same. When ¿ c , the slip-line 4eld is not a9ected by the coe8cient of friction and the plastic 4eld does not see the base boundary condition except at A A. For coe8cients of friction smaller than the critical value, the fan angle is at the limit set by friction and a plastic region (which varies with friction) is required below the fan and along part of the base to satisfy the symmetry condition (see Fig. 10). The value of c increases with increasing r0 until it reaches a maximum value of 2=(2+$)(=0:389) [5], see Fig. 8. 1 The analysis above applies to cylinders with radius r0 such that c ¡ 2=(2 + $). For large r0 , the rigid (and static) material does not extend across the whole base for friction coe8cients either above or below the critical value. Where there is sliding along the base when the friction coe8cient is above the critical value, the associated boundary condition is a perfect stick condition (rz = y =2), as we shall discuss below.
1
This is the same result as for the plane-strain case because the maximum value of the critical coe8cient of friction depends only on the stress singularity at point A A. At this point dr and d z are zero so that the 1=r terms are not present in 4nite di9erence equations linking the vertical free surface to the base.
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
1789
0.4 0.3
µc 0.2
axisymmetric plane strain
0.1 0.0 0.0
0.5
1.0
1.5
2.0
r0 Fig. 8. Critical coe8cient of base friction c , calculated from numerical slip-line 4elds, as a function of scaled radius r0 . The critical coe8cient of friction for the equivalent plane-strain problem [5] is also shown.
2.5
stress discontinuity
centre-line β
2.0
D
α
1.5 z 1.0
E T
0.5
rigid
0.0
base
0
1
T0 2 r
S
A A′
3
4
Fig. 9. Numerical slip-line 4eld for r0 = 4:0 and large friction, ¿ c . The plastic height required for the onset of the stress discontinuity, hpc , is the length of AE. The stress discontinuity starts where two -lines coincide. This point can be seen as the intersection of the -line propagating from A (on the right-hand side of the fan) with the "-line T0 E.
The analogous curve for critical coe8cient of friction as a function of base half-width for the plane-strain case is also shown in Fig. 8 [5]. 2 Interestingly, this curve is close to the axisymmetric results, with slight di9erences due to the 1=r terms. For the case ¿ c , the stress discontinuity occurs 4rst at hpc = 1:23 and r0 = 1:4. These values of hpc and r0 are smaller than the corresponding values for the perfect slip ( = 0) case, see above. For ¿ c and r0 ¿ 1:4, a stress discontinuity exists. An example slip-line 4eld for r0 = 4:0 is given in Fig. 9. Note that there is a large fan emanating from point A A, and on part of the base, SA , the -lines meet the base tangentially indicating that the shear stress is equal to half the uniaxial yield stress. This full stick condition (see Eq. (13)) is independent of any increase in the coe8cient of friction above the critical value c = 2=(2 + $). The relationship between the plastic height hp and radius r0 for the case of ¿ c is shown as the solid curve in Fig. 6.
2
In the plane-strain calculation a slightly di9erent scaling for lengths is employed: the scaled half-width is equal to the true half-width divided by 2y =g where y is the material’s shear yield stress. In plotting Figs. 4, 8 and 13, we had to assume a relationship between y and y . We chose y = 2y , which corresponds to Tresca’s yield criterion.
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1.0
D
centre-line
0.8 C
β
0.6
B
α
z 0.4
T
0.2 rigid
0.0
0.0
0.2
A A′
base
S
0.4
0.6
0.8
1.0
r
Fig. 10. Numerical slip-line 4eld for r0 = 1:0 and intermediate friction, = 0:2. Line TBCD is a "-line and CA is an -line. The plastic height hp is equal to the length of AD.
2.5
centre-line
stress discontinuity
β
2.0
D
α
z
1.5 1.0
E T
0.5
rigid
0.0
base
0
1
T0 2 r
A A′
S 3
4
Fig. 11. Numerical slip-line 4eld for r0 = 4:0 and intermediate friction, = 0:2. The length of line AE is the plastic height at the onset of the stress discontinuity, hpc . The stress discontinuity starts where two -lines coincide. This point can be seen as the intersection of the -line propagating from A (on the right-hand side of the fan) with "-line T0 E.
3.3. Intermediate friction In the case of intermediate friction, 0 ¡ ¡ c , the slip-line 4elds exhibit hybrid features from both of the extreme cases above. We will discuss the particular case of = 0:2, which is in the intermediate range of friction if r0 ¿ 0:3 (Fig. 8). An example slip-line 4eld for r0 = 1:0 where no stress discontinuity exists is given in Fig. 10. The onset of the stress discontinuity occurs at hpc = 1:29 and r0 = 1:69. A slip-line 4eld for r0 = 4:0 which contains a stress discontinuity is given in Fig. 11. In both these solutions (r0 = 1:0 and 4:0) there is a fan emanating from A A with smaller fan angle than in the case of large friction, and part of the base is plastic, i.e., the line SA shown in Figs. 10 and 11. In the graph of hp versus r0 for several values of coe8cient of friction (Fig. 6), the curves for intermediate friction are bounded by the curves for =0 and ¿ c , with hp increasing as increases.
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
1791
2.4 2.0 h 1.6
µ ≥ µc µ = 0.3 µ = 0.2 µ = 0.1 µ = 0.05 µ=0
µ increasing
1.2
uniform stress model
0.8 0
2
4
6
8
r0 Fig. 12. Scaled height of incipient failure h = (g=y )H versus scaled radius r0 = (g=y )R0 for six values of coe8cient of friction at the base. The case = 0:3 is virtually indistinguishable from the case ¿ c . The prediction of the uniform stress model is also shown.
4. Results and discussion The height of incipient failure for a circular cylinder loaded by gravity can be calculated from the stress 4elds constructed in the previous section, using a method based on that used for the plane-strain analysis [5]. The slip-line 4eld calculation gives the stress tensor along a line from the top of the plastic part of the vertical free surface at height hp to the centre of the cylinder (e.g., TBCD in Fig. 10). This line de4nes a surface of revolution in the axisymmetric geometry. The upward force exerted by the material below this surface of revolution must equal the weight of the material above, when the cylinder is at the point of incipient failure [5]. This force balance gives a scaled height of incipient failure h for each stress 4eld. Note that the scaled height of incipient failure depends only on the radius and the coe8cient of friction on the base, whereas the actual height of incipient failure H = (y =(g))h depends on the yield stress, gravity and the density as well. Performing the force balance calculation for a series of stress 4elds gives a graph of h as a function of r0 for several values of (Fig. 12). Also shown in Fig. 12 is the prediction of the uniform stress model (h = 1) which is independent of the coe8cient of friction and the radius r0 . In the slip-line 4eld results, as r0 approaches zero, the scaled height of incipient failure approaches unity for all . As r0 increases from zero, h increases for all friction cases. The e9ect of friction is to increase the height of incipient failure. This continues until the critical value of coe8cient of friction c is reached (see Fig. 8). At this point the curves for di9erent friction cases coincide, so that h remains constant as is increased above c . From Fig. 12, it is clear that the predictions of the uniform stress model can be greatly in error for r0 ¿ 0. A comparison of these three-dimensional axisymmetric results with analogous results for a planestrain rectangular block [5] (also see footnote 2) reveals that the curves coincide at r0 = 0, and qualitatively similar behaviour is observed for r0 ¿ 0 (see Fig. 13). The di9erence between the two sets of data increases as the base half-width and radius increase, with the cylinder incipient height data always lower than that for the rectangular block except at r0 = 0. Nonetheless, the di9erence is smaller than 12% in all cases. In the axisymmetric case a plot of h versus hp (Fig. 14) shows that the ratio of hp to h is virtually independent of the coe8cient of friction . This phenomenon should
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J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
Fig. 13. Graph comparing the cylinder and plane-strain results for height of incipient failure h versus radius or half-width. The horizontal shading corresponds to the region covered by the range of friction cases in the plane-strain case, and the vertical shading corresponds to the range of friction cases in the cylinder case.
2.5 2.0 h
µ ≥ µc µ = 0.3 µ = 0.2 µ = 0.1 µ = 0.05 µ=0
1.5 1.0 0.0
0.5
1.0
1.5
2.0
2.5
hp
Fig. 14. Plot of scaled height of incipient failure, h, versus the height of plastic material on the vertical free surface, hp .
be observable experimentally. The corresponding plane-strain result in Ref. [5] is very similar to this result, with the curves for the axisymmetric case within 5% of the plane-strain results (also see footnote 2). The height of incipient failure data can also be presented as a plot of uniaxial yield stress y multiplied by a factor 1=(Hg) versus aspect ratio r0 =h = R0 =H (Fig. 15). This suggests a method for determining the yield stress of a slurry. Measuring the height of incipient failure H and the radius R0 gives the aspect ratio at incipient failure on the horizontal axis of the graph in Fig. 15. After choosing which friction curve is appropriate, the yield stress can be read o9 the vertical axis of the graph using the known factor 1=(Hg). The experimental implementation of this method will be detailed elsewhere. The Haar–Karman hypothesis restricts the stress 4eld in the yielded region to have the hoop stress
equal to the algebraically greater principal stress in the r–z plane. Assuming that the stress 4eld can be continued into the non-deforming region while satisfying the equilibrium equations and stress boundary conditions, the slip-line solutions give a lower bound on the height of incipient failure [10, p. 19]. How close this lower bound is to an upper bound solution or an exact solution and how well the Haar–Karman hypothesis is satis4ed in the exact solution could be investigated using models which do not invoke the Haar–Karman hypothesis. For example, models which involve constructing kinematically admissable upper bound solutions or models which simultaneously solve the stress and
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
1793
1.1 (1/(Hρg)) σy
1.0
uniform stress model
0.9 0.8
µ increasing
0.7
µ=0 µ = 0.05 µ = 0.1 µ = 0.2 µ = 0.3 µ ≥ µc
0.6 0.5 0.4 0.0
0.5
1.0
1.5 2.0 R0 /H
2.5
3.0
Fig. 15. Plot of 1=h = (1=(Hg))y versus aspect ratio r0 =h = R0 =H where h is the scaled height of incipient failure for six values of the coe8cient of friction . The friction case = 0:3 is virtually indistinguishable from the case ¿ c . We also show the prediction of the uniform stress model.
kinematic equations could be used for this purpose. These investigations are outside the scope of the present study.
5. Conclusions We have investigated the incipient failure of a cylinder under gravitational loading using slip-line 4eld theory. The cylinder was assumed to behave as a perfectly rigid-plastic material and satisfy the Haar–Karman hypothesis. The friction at the base was modelled using a modi4ed Coulomb law. The stress 4eld in the deformable plastic region of the cylinder at the point of incipient failure has been calculated. This gives a prediction of the height of incipient failure of a cylinder as a function of the uniaxial yield stress and density of the material, acceleration due to gravity, cylinder radius and base friction. The results for the height of incipient failure of a cylinder suggest a method of determining the yield stress of a concentrated suspension such as a mineral tailings slurry. Details of an experimental study of this method will be reported elsewhere. We found that the results for the cylinder case are very similar to the results for the plane-strain rectangular block [5] (also see footnote 2) with the height of incipient failure increasing as the radius of the cylinder increases. However, the height of incipient failure is lower for the cylinder than for the rectangular block with the same frictional conditions on the base and the radius of the cylinder equal to the half-width of the block. We have examined the e9ect of base friction on the height of incipient failure. In the limit as the radius r0 → 0, the height of incipient failure is independent of friction and equal to y =(g), which is the prediction of the uniform stress model. As r0 increases from zero, the e9ect of friction becomes greater with the height of incipient failure increasing with increasing friction, until the critical coe8cient of friction is reached. At this point raising the coe8cient of friction has no e9ect on the height of incipient failure. The results given here show that the uniform stress model is inaccurate for all values of radius except when r0 1. This agrees with the similar 4nding in Ref. [5] for the rectangular block.
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Acknowledgements We acknowledge the support of the Particulate Fluids Processing Centre (PFPC), an Australian Research Council Special Research Centre. Appendix A. Slip-line eld theory Eqs. (9) and (10) with the Haar–Karman hypothesis (11) form a hyperbolic system with characteristics running along the lines of maximum shear stress. The two sets of characteristics (slip-lines) are labelled as - and "-lines according to the following rule: the - and "-lines must form a right-handed coordinate system, and the algebraically greater of the two principal stresses must act in the 4rst and third quadrants of the –" coordinate system. We write the equations using the variables p, where p = −(rr + zz )=2 and is the angle between the r-axis and the -direction measured in the anti-clockwise direction. The stresses can be written in terms of p and : rr = −p − 12 sin 2; rz = 12 cos 2;
zz = −p + 12 sin 2;
= −p + 12 :
(A.1)
The hoop stress
is chosen to be equal to the algebraically greater principal stress in the r–z plane, according to the Haar–Karman hypothesis as discussed in Section 2 (also see Ref. [9]). Information is propagated along the slip-lines using the following equations based on Morrison and Richmond [8] and Shield [9], modi4ed by our scaling and choice in the Haar–Karman hypothesis. For an -line we have dp + d +
dr + d z = 0; 2r
tan =
dz dr
(A.2)
and for a "-line we have dp − d +
dr − d z = 0; 2r
cot = −
dz : dr
(A.3)
The slip-line 4eld is constructed using 4nite di9erence versions of these equations. For any two adjacent points A and C along an -line (see Fig. 2) we have r C − rA + z C − z A p C − pA + C − A + = 0; (A.4) rC + r A A + C − (zC − zA ) = 0: (A.5) (rC − rA ) tan 2 For any two adjacent points B and C along a "-line we have pC − pB − (C − B ) +
B + C (rC − rB ) cot 2
rC − rB − (zC − zB ) = 0; rC + r B
(A.6)
+ (zC − zB ) = 0:
(A.7)
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
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In constructing the slip-line 4elds illustrated in this paper, the following sub-problem occurs frequently: if we know p and at points A and B and these points are connected to point C by an -line AC and a "-line BC then we wish to 4nd p, , r and z at point C. Once we have formulated a procedure to solve this problem then this is used repeatedly to build up a slip-line 4eld. We formulate the problem in terms of the four equations (A.4) – (A.7) in four unknowns, pC , C , rC and zC . These equations are reduced to two by eliminating pC from Eqs. (A.4) and (A.6), and substituting for zC using Eq. (A.5): −pA + pB + 2C − A − B C + A rC − rA 1 + tan + rA + r C 2 C + A 1 − zB = 0; rB − rC + zA + (rC − rA ) tan + rB + r C 2 B + C (rC − rB ) cot 2 C + A − zB = 0: + zA + (rC − rA ) tan 2
(A.8)
(A.9)
We employ a root 4nding procedure to solve for rC and C , and then calculate zC and pC using Eqs. (A.4) and (A.5). The 4rst guess for the root 4nding procedure, (rC(1) ; (2) C ), is found using the method of Chitkara and Butt [7]. For completeness, the method of Chitkara and Butt is given below with our notation and scaling. An approximation for the position of point C is given by extending the tangents to the - and "-lines at points A and B, respectively, and 4nding the point of intersection of these two tangents: zC(0) = (−rA + rB + zB tan B + zA cot A )=(cot A + tan B );
(A.10)
rC(0) = rA + (zC(0) − zA ) cot A :
(A.11)
Putting rC(0) and zC(0) into Eqs. (A.6) and (A.4), and taking the di9erence between these two equations, eliminates pC and gives an estimate of at point C: 2(0) C = p A − pB + A + B − +
1 rB + rC(0)
1 rA + rC(0)
(rC(0) − rA + zC(0) − zA )
(rC(0) − rB − zC(0) + zB ):
(A.12)
Based on the fact that the intersection of the tangents at A and B is not exactly point C, a second approximation to C is found using (1) C =
(0) C + A : 2
(A.13)
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Second estimates for rC and zC are calculated by solving the following two equations for rC(1) and zC(1) , where average slopes of AC and BC have been used to determine the position of point C. rC(1) − rA = 12 (zC(1) − zA )(cot A + cot (1) C );
(A.14)
zB − zC(1) = 12 (rC(1) − rB )(cot B + cot (1) C ):
(A.15)
This gives zC(1) =
FGzA + 4zB + 2G(rB − rA ) ; FG + 4
(A.16)
rC(1) = rA + 12 F(zC(1) − zA );
(A.17)
(1) where F = cot A + cot (1) C and G = cot B + cot C . The 4nal estimates of C and pC in Chitkara and Butt’s method are given by putting rC(1) and zC(1) into the following formulae:
2(2) C = p A − pB + A + B + +
1 rB +
rC(1)
1 rA +
rC(1)
(rA − rC(1) + zA − zC(1) )
(rC(1) − rB − zC(1) + zB );
pC(0) = pA − (2) C + A −
1 rA +
rC(1)
(rC(1) − rA + zC(1) − zA );
(A.18) (A.19)
where Eq. (A.18) is obtained in the same way as Eq. (A.12), and Eq. (A.19) is from Eq. (A.4). (0) Chitkara and Butt’s method calculates point C as the point which satis4es C = (2) C , pC = pC , (1) (1) rC = rC , zC = zC . This estimate is normally close to the solution of the 4nite di9erence equations (A.4) – (A.7), and gives fast convergence in the root 4nding procedure discussed above.
Appendix B. Details of construction of plastic elds Assuming that the material is plastic on the vertical free surface up to height hp implies that the stress along this free surface is given by p = 1=2 + z and = 7$=4, using an argument similar to that in Ref. [5]. The initial choice of hp is chosen to be an underestimate of the actual value of hp for the given radius and base friction. After propagating the stress 4eld into the body in the shape of a distorted triangle (CAD in Fig. 16), the "-line along the top has ¡ 7$=4 at point C. However, the centre symmetry condition demands that = 7$=4. Depending on the friction conditions, we add 4elds to the initial triangle so that the closest point to the centre satis4es = 7$=4. Fig. 16 shows the additional 4elds BA AC and TSA B. In some cases, not all of the pieces of the 4eld in Fig. 16 are needed. If there is no friction at the base then no fan (BA AC) is needed because the boundary conditions at the vertical free surface are compatible with the base boundary condition at point A. If the base coe8cient of friction is su8ciently large and r0 is su8ciently small then the material in
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
1797
z D rigid
β
C hp
B T
fan
π/4
α
rigid S
A A'
r
r0 Fig. 16. Schematic of a slip-line 4eld solution for an initial choice of hp and non-zero friction on the base.
the 4eld TSA B is not needed and the size of the fan is chosen so that = 7$=4 at B. Otherwise, the 4eld TSA B is constructed so that = 7$=4 at point T . Usually, the initial choice of hp is not large enough for the 4eld to reach the centre and satisfy the symmetry condition = 7$=4. After completing the 4eld for the initial choice of hp , the plastic height hp is incremented, and a new slip-line 4eld constructed by adding to the 4eld discussed above. A narrow slice is added above the "-line TBCD, followed by a slice below -line TS, chosen to maintain the symmetry condition = 7$=4 at the closest point to the centreline. This process is repeated until the 4eld reaches the centre. However, normally the 4eld overshoots the centre and the last increment of hp needs to be iterated to obtain a 4eld which just touches the centre. Appendix C. Construction of the stress discontinuity For su8ciently large hp , the -lines coincide at a point. This point can be seen as the intersection of the -line from point A (on the right-hand side of the fan) with the "-line T0 E in Fig. 11, or as point C in Fig. 17. The stress discontinuity forms a line emanating upward from this point. The tangential component of stress is discontinuous and the normal and shear components of stress are continuous across the stress discontinuity. This leads to the following two conditions used in the construction of the stress discontinuity: (i) for p and we have [5] p+ − p− = |sin(+ − − )|;
(C.1)
where the superscript + refers to the side of the stress discontinuity closest to the vertical free surface and the superscript − refers to the opposite side; and (ii) the line of the stress discontinuity bisects any two -lines emanating from a single point on the stress discontinuity, where one of the -lines is on the left-hand side of the discontinuity and the other -line is on the right-hand side. We construct the stress discontinuity in a series of steps during the process of incrementally increasing hp described in Appendix B. At each increment of hp , the slice above the "-line T0 E (Fig. 11) is constructed from the free surface down to the point where the stress discontinuity occurs. The 4eld to the right of the increment of the stress discontinuity is known, and we wish to
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z
stress discontinuity
D β
C hpc
α A r r0 Fig. 17. Schematic for calculating the plastic height at the onset of the stress discontinuity, hpc .
4nd the 4eld in the form of the next point on the stress discontinuity and a slice to the left of the stress discontinuity. This 4eld is required to be compatible with the existing 4eld ("-line T0 E and below, Fig. 11). Formulating suitable 4nite di9erence relations based on Eqs. (A.4) – (A.7) applied to points near the stress discontinuity and the two conditions above, gives a well-determined problem. Solving these equations in an iterative fashion yields the new point on the stress discontinuity. This point is uniquely determined by conditions (i) and (ii), and the existing slip-line 4eld below. The 4eld is extended to form a slice above "-line T0 E (Fig. 11) or TBCD (Fig. 16). A 4eld is added below the -line TS (Fig. 16) to maintain the condition = 7$=4 at the closest point to the centre. This procedure is repeated, building up the stress discontinuity as shown in Figs. 5, 9 and 11.
Appendix D. Onset of stress discontinuity In the equivalent plane-strain incipient failure problem the height of plastic material on the vertical free surface which gives the onset of the stress discontinuity is $=2 [5]. In the axisymmetric case, this height, denoted by hpc , is less than $=2 and depends on r0 . It can be calculated as follows. We consider a triangle ACD with AD along the vertical free surface and D at the top of the plastic region (Fig. 17). The stress 4eld within this triangle is entirely determined by the boundary conditions on AD. These conditions (constant and constant, unit, pressure gradient along the boundary) imply that must be a function of r only in ACD and p must equal z plus a function of r only in ACD. Under these circumstances, the axial force balance equation can be integrated exactly to give 2 r cos 2 = −r 02 − 1 : (D.1) r
J.A. Chamberlain et al. / International Journal of Mechanical Sciences 44 (2002) 1779 – 1800
The length AD (which we have been calling hp ) can then be calculated as 3$ dr hp = tan(2$ − ) dr + tan − 2 CA CD r0 2 = dr 1 − cos2 2 rc cos 2 4r02 + cos2 2 − sin2 2 1 + arctan = − arctan 2r0 sin 2(cos 2 + 4r02 + cos2 2)
:
1799
(D.2)
=C
All that remains is to determine C when C lies at the bottom of the stress discontinuity. From Eq. (D.1), as r decreases from r0 , cos 2 decreases from zero until at some r value it reaches −1. This forms a natural boundary of the stress 4eld de4ned by the boundary conditions on AD, where the - and "-lines become vertical and horizontal, respectively. It is when C reaches this boundary that a stress discontinuity becomes necessary, as the -lines that intersect the vertical free surface and those that intersect the base would otherwise start to cross above C. Therefore, C = 6$=4, and since
cos 2 4r02 + cos2 2 − sin2 2 $ (D.3) = lim + arctan 2 6$ sin 2(cos 2 + 4r02 + cos2 2) → 4
we have
1 $ hpc = − arctan : 2 2r0
This result coe8cients larger, hpc footnote 2,
(D.4)
is illustrated in Fig. 4 for the range of values for r0 compatible with the possible of friction, given that the solution is at the onset stress discontinuity. As r0 becomes for the axisymmetric geometry approaches the plane-strain result [5], see Fig. 4 and as expected.
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