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Deformation in Al-z.sbnd;SiC composites due to thermal stresses
Materials Science and Engineering, 75 (1985) 151-167
151
Deformation in AI-SiC Composites Due to Thermal Stresses Y. FLOM and R. J. ARSENAULT Metallurgical Materials Laboratory, University of Maryland, College Park, MD 20742 (U.S.A.)
(Received January 2, 1985; in revised form March 13, 1985)
ABSTRACT
Plastic strains and the e x t e n t o f the plastic zone due to differential thermal expansion were experimentally determined in an A l - S i C composite. The combined plastic shear strains %pss at the A I - S i C interface for furnace-cooled, air-cooled and quenched samples were found to be 1.32%, 1.23% and O.99% respectively. Profiles o f %pss were plotted versus distance from the interface and compared with the theoretical distribution o f effective strain ~. The theoretical e x t e n t o f the plastic zone measured from the interface was found to be 1.3 multiplied by the particle radius. This value was slightly less than the observed value. The plastic deformation on the heating half o f the thermocycle was found to be at least equal to the deformation on the cooling half. A theoretical treatment o f the local plastic deformation in a short composite cylinder was suggested, from which the effective plastic strain ~ and the e x t e n t of the plastic zone were determined. 1. INTRODUCTION When a composite material is subjected to a temperature change, local plastic deformation can occur. The plastic deformation is due to a stress created by the difference between the coefficients of thermal expansion of the component phases, and this stress is at the matrixreinforcement interfaces. The magnitude of the stress is equal to A s AT where A s is the difference between the coefficients of thermal expansion of the phases under consideration and AT is the temperature change. The relatively large (10 to 1) difference between the coefficient of thermal expansion of SiC and t h a t of aluminum should result in the creation of a substantial misfit strain at the A1-SiC interface during cooling from the fabricating or 0025-5416/85/$3.30
annealing temperature. The mechanical properties of the composite should be affected by the magnitude and extent of the plastic deformation that takes place in the soft matrix around a hard particle as a result of the misfit relaxation in the interface region. Recent investigations [1-4] have indeed shown the important role of the interfacial regions in the composite strengthening. Thus, knowledge of the m a ~ i t u d e of plastic strains and the size of the plastic zone as well as other characteristics of the interfacial region (the bond strength, the microstructure etc.) should contribute to the understanding of the mechanism of composite strengthening. The experimental determinations of plastic strains and plastic zone radii about a particle in the matrix due to the difference between the coefficients of thermal expansion of the phases, to our knowledge, have not been reported in the literature. Several theoretical investigations have been undertaken to predict the magnitude of the plastic strain in the plastic zone around a particle. The relaxation of the misfit caused by the introduction of an oversized spherical particle into a spherical hole in the matrix has been analytically described b y Lee et al. [5]. Using the misfitting sphere model, they calculated strains in the plastic zone that surrounds a hard sphere and also the plastic zone radius. Hoffman [6] calculated the overall total strains in the tungsten-fiberreinforced 80Ni-20Cr matrix, using a thickwall long~ylinder approach and assuming that a hydrostatic stress state exists within each constituent. Garmong [ 7 ], assuming uniformity of the stresses and strains in the matrix, calculated deformation parameters for a hypothetical eutectic composite and reported values of matrix plastic strains that were of the order of 0.4%. © Elsevier Sequoia/Printed in The Netherlands
152 Dvorak e t al. [8] developed a new axisymmetric plasticity theory of fiberous composites involving large thermal changes. The long~omposite-cylinder model was adopted as a composite unit cell, and the microstress distribution as well as the yielding surfaces were obtained for AI-W composites. Mehan [9] calculated the residual strains in an AI-~-A1203 composite due to cooling from the fabricating temperature. He considered an idealized composite consisting of a long sapphire cylinder surrounded by an aluminum matrix. This is equivalent to the long-composite-cylinder model used by Dvorak e t al. T h i s model implies that the thermal strain along the cylinder axis is a constant, i.e. that dl ez = - - = constant l where l is the length of the cylinder. This simplifies the procedure of obtaining the radial er and the tangential e0 strains which was done using incompressibility and boundary conditions. The effective strain ~ at the Al-a-A1203 interface was f o u n d to be 1.6%. However, the above-mentioned composite models do n o t give an accurate description of the plastic strain state in the short-composite-cylinder model. A short-cylinder model nearly duplicates the situation in the whisker and platelet A1-SiC composites at present produced. The purpose of this work was to determine experimentally the magnitude of the local plastic strain produced in the aluminum matrix around a short SiC cylinder during a thermocycle and also to estimate the extent of the plastic zone around the cylinder. Also an effort was made to develop a theoretical
TABLE
model of the plastic zone around the short cylinder.
2. EXPERIMENTAL PROCEDURE The small particle interspacing (several micrometers) in a commercially available A1-SiC composite makes it impossible to measure the local plastic strain at the A1-SiC interface directly. Thus a composite model consisting of an SiC cylinder embedded in an aluminum matrix was fabricated so t h a t direct strain measurements could be attempted. Aluminum of 99.99% purity (to eliminate the influence of the alloying elements) and commercial carborundum were used to produce the composite model. Selected properties of these materials are given in Table 1. Platelets of SiC were separated from carborundum conglomerates that are used in the production of abrasives. These platelets were spark planed on an electric discharge machine to plates approximately 1 mm thick. These flat plates were cut into rectangular rods approximately 1 m m × 1 ram. After this, each rod was spark machined to cylinders of about 1 mm diameter. Pure aluminum rods, 12.5 mm in diameter in the as-received condition, were cut into studs 37 mm long. Two aluminum studs and one SiC rod were assembled together and put in a specially built compaction die, where they were hot pressed to produce one compact. Compaction was done on the Instron testing machine. During the entire compaction cycle (Table 2), a vacuum of about 10 -a Tort was maintained using a mechanical vacuum pump.
1
Selected properties of the aluminum and SiC used in this paper
Material
A1 SiC
Yield strength
Elastic modulus
Bulk modulus
(MPa)
(MPa)
(MPa)
11.73 62 X103 34.5 + (tension) 483 × 103 1380 + (compression)
57.5 X103 96.6 × 103
Poisson 's Thermal expansion ratio coefficient
(m m -1 K -1) in the temperature range 293-773 K
0.31 0.19
28 × 1 0 - 6 3 x 10 -6
Melting point (K)
Reference
933 3373
[10, 11] [12,13]
153 TABLE 2 Flow chart of the compaction of an AI-SiC composite cylinder
%pss = K N S
Hea ring Heating time Temperature
~ 1 h ~ 843 K
Compression Cross-head s p e e d T o t a l travel
0.1 c m m i n -1 10 m m
Holding at constant load Time Pressure
2 h 2.28 MPa
Incremental cooling under pressure Time Pressure
sic,.~]
(a)
1
duced where %pas is equivalent to the product of slip band density N and the a m o u n t of slip S, i.e.
12 h 2.28 MPa
i
._~
(b)
(c)
(d)
Fig. 1. A s c h e m a t i c d i a g r a m of t h e f a b r i c a t i o n s e q u e n c e o f t h e A1-SiC c o m p o s i t e m o d e l .
The central portion of the compacted sample was sliced in the transverse direction into disks 1 m m thick using the electric discharge machine, set at a low power, and each disk contained an SiC cylinder very close to the perfect center (Fig. 1). All A1-SiC disks were metallographically and then electrolytically polished to remove the thin cold-worked surface layer of aluminum. The m e t h o d adopted for the evaluation of plastic deformation was based on the direct observation of slip bands on the polished surface of the sample around the SiC particles. The a m o u n t of slip is a characteristic of the a m o u n t of plastic deformation (when deformation occurs by slip) in a crystalline solid. The plastic strain can be evaluated if the number of slip bands and the displacement on each band are known. The concept of combined plastic shear strain %ps~ has been intro-
(1)
where K is a coefficient which takes into consideration different crystallographic situations. A detailed treatment of %pss is given in Appendix A. Thus, the m e t h o d reduces the data collection to the measurements of slip band densities and their heights in the area of interest. The electropolished AI-SiC disks were separated into three groups: A, B and C. Each group was heated to about 823 K and then cooled as follows: group A, furnace cooled; group B, air cooled; group C, quenched in alcohol. Since the surface of each disk had a high quality polish, slip bands could be observed around the SiC in an optical microscope. Slip band density and height measurements were obtained using a Zeiss interference microscope. Areas containing slip bands were photographed in white light and in green monochromatic light. Pictures taken in white light gave the actual image of the slip bands. Pictures of the same areas taken in monochromatic light gave interference fringe patterns (Figs. 2-4). Thus, the correlation between slip bands and interference fringes was established. When a furrow is present in the plane surface o f the object, the straight interference bands are deflected by the furrow and the furrow depth t can be determined from the deflection: t -
dk b2
(2)
where d is the deflection of the interference band, b is the band interval, taken as the distance from the middle of one band to the middle of the next band and k/2 = 0.27 #m for the thallium light. Band deflection can be estimated to be one-tenth of the band interval. Thus the height measurements can be as accurate as -+0.1 × 0.27 = -+0.027 pm. The slip band density was determined by using the " m e s h " m e t h o d which consisted of the following. Pictures of slip band images and interference patterns were enlarged to 10 cm × 12.5 cm on high contrast photographic paper. After that, a transparent plastic
154
~(a)
SiC
AI
N (b) Fig. 2. (a) Slip bands and (b) interference fringe patterns of different areas around the SiC. (Magnification, 32 × .)
SiC
AI
(b) Fig. 3. (a) Slip bands and (b) interference fringe patterns of different areas around the SiC. (Magnification, 32 × .)
155
(a)
AI
SiC
Z
Y
(b) Fig. 4. (a) Slip bands and (b) interference fringe patterns of different areas around the SiC. (Magnification, 32 x.)
film with a square mesh of a specific size (1.0 cm X 1.0 cm square for pictures with a magnification o f 80X and 2.5 cm X 2.5 cm square f o r pictures with a magnification of 200X) was overlaid on the p h o t o g r a p h and t h e total length o f the slip bands within each square along th e radial direction was measured. If M is th e magnification of the picture, L (m) the total length o f slip bands in one square and A (mm 2) the area o f the square, t hen t he slip band density N is L N = M--
X 10 -3 pm -1
(3)
a Leitz optical microscope. A reduct i on in size was necessary in order to fit the holder into the h o t stage. The entire t h e r m o c y c l e was recorded on videotape using an RCA industrial television camera. The time was recorded by means of a digital generator interfaced to the recording unit, and t he t e m p e r a t u r e was recorded on the sound track of t he tape at 20 K intervals. The total length of the t h e r m o c y c l e was approxi m at el y 8 min. The m a x i m u m temperatures were around 873 K. The heating and cooling rates were around 100 K min -1.
A The accuracy o f slip band density measurements is mu ch higher than t ha t of height measurements. Therefore, the main source of error in combined plastic shear strain ~'cpss evaluation is the height of the slip band (provided that, of course, the assumptions and results of discussion given in Appendix A are reasonably correct).
2.1. Hot-stage experiment Two A1-SiC disks were reduced in diameter to 7 mm and t h e r m o c y c l e d in the h o t stage o f
3. EXPERIMENTAL RESULTS
3.1. Combined plastic shear strain The experimentally det erm i ned values of com bi ned plastic shear strain %pss are presented in Tables 3 - 5 . These com bi ned plastic shear strains %pss are pl ot t ed versus distance in t he form of a histogram in Fig. 5. In the same figure we have a plot of effective strain which was det erm i ned theoretically (see Appendix B). The histogram represents the
156 TABLE 3 C o m b i n e d plastic shear s t r a i n 7cpss for g r o u p A samples ( f u r n a c e c o o l e d )
Sample
Path a
Combined plastic shear strain b 7cpss (%) for the following distances f r o m the A l - S i C interface c 6.25×10-2mm
18.75×10-2mm
31.25X10-2mm
43.75×10-2mm
56.25×10-2mm
0.09 0.06 0.08 0.16 0.13 0.08
0.026 0.07
1 1 1 1 1 1
1 2 3 4 5 6
0.86 0.61 0.41 0.26 0.42 0.51
0.35 0.40 0.140 0.21 0.32 0.41
0.18 0.22 0.24 0.19 0.32 0.30
2 2
1 2
0.30 0.45
0.21 0.28
0.06
3 3 3 3 3
1 2 3 4 5
0.43 0.26 0.39 0.46 0.45
0.24 0.27 0.22 0.18 0.26
0.22 0.22 0.26 0.23
4
1
0.43
0.24
0.04
0.44
0.24
0.21
Average
0.03
0.22 0.11 0.16
0.22
0.16
0.05
a T h e p a t h is t h e r o u t e in t h e radial d i r e c t i o n f r o m t h e A I - S i C i n t e r f a c e i n t o t h e m a t r i x , w h i c h was selected for slip b a n d d e n s i t y a n d h e i g h t m e a s u r e m e n t s . S e l e c t i o n was b a s e d o n t h e a m o u n t o f slip t h a t o c c u r r e d , a n d t h o s e r o u t e s were selected in w h i c h t h e a m o u n t o f slip b a n d s a p p e a r e d t o be t h e greatest. b T h e given s t r a i n is a c t u a l l y t h e r e s u l t o f t h e p r o d u c t NS. It does n o t include t h e c o e f f i c i e n t K w h i c h was f o u n d t o b e e q u a l t o 3 (see A p p e n d i x A). Values o f Tcpss t h a t i n c o r p o r a t e K are given in Table 6. CThe d i s t a n c e given here is t h e d i s t a n c e f r o m t h e A I - S i C i n t e r f a c e t o t h e c e n t e r s o f t h e first, s e c o n d etc. squares. T h e r e f o r e , values o f 7cpss r e p r e s e n t t h e s t r a i n over t h e e n t i r e square. TABLE 4 C o m b i n e d plastic s h e a r s t r a i n 7cpss in g r o u p B s a m p l e s (air c o o l e d )
Sample
Path a
Combined plastic shear strain b ~/cpss (%) for the following distances from the A l - S i C interface c 6.25 x l O-2 m m
18.75 × l O-2 m m
31.25 x l O-2 m m
43.75 × l O-2 m m
56.25 x l O-2 m m
68.75 X l O-2 m m
0.35 0.17 0.19
0.13 0.19 0.11
0.13
0.11
0.014
0.08
0.13
0.11
1 1 1 1 1
1 2 3 4 5
0.34 0.25 0.56 0.47 0.25
0.32 0.25 0.31 0.55 0.07
2
1
0.23
0.17
4 4 4 4 4
1 2 3 4 5
0.65 0.47 0.54 0.31 0.48
0.41 0.22 0.29 0.21 0.25
0.41
0.27
Average
0.1 0.25 0.18
0.12
a T h e p a t h is t h e r o u t e in t h e radial d i r e c t i o n f r o m t h e A I - S i C i n t e r f a c e i n t o t h e m a t r i x , w h i c h was selected for slip b a n d d e n s i t y a n d h e i g h t m e a s u r e m e n t s . S e l e c t i o n was based o n t h e a m o u n t of slip t h a t o c c u r r e d , a n d t h o s e r o u t e s were selected in w h i c h t h e a m o u n t o f slip b a n d s a p p e a r e d t o b e t h e greatest. b T h e given s t r a i n is a c t u a l l y t h e r e s u l t o f t h e p r o d u c t NS. It d o e s n o t include t h e c o e f f i c i e n t K w h i c h was f o u n d t o b e e q u a l t o 3 (see A p p e n d i x A). Values o f 7cpss t h a t i n c o r p o r a t e K are given in Table 6. CThe d i s t a n c e given here is t h e d i s t a n c e f r o m t h e A I - S i C i n t e r f a c e t o t h e c e n t e r s o f t h e first, s e c o n d etc. squares. T h e r e f o r e , values o f ~/cpss r e p r e s e n t t h e s t r a i n over t h e e n t i r e square.
157 TABLE
5
Combined plastic shear strain "~cpssin group C samples (quenched) Sample
C o m b i n e d plastic shear strain b 7cpss (%) f o r the f o l l o w i n g distances f r o m the A l - S i C interface c
Path a
6.25
x
10 - 2
mm
1 8 . 7 5 x 10 - 2
mm
3 1 . 2 5 x 10 - 2
1 1 1
1 2 3
0.40 0.19 0.69
0.16 0.27
2 2 2 2
1 2 3 4
0.37 0.31 0.22 0.55
0.29 0.11 0.24 0.23
0.16 0.05 0.12 0.14
3 3
1 2
0.31 0.44
0.18 0.24
0.23
4 4 4
1 2 3
0.25 0.25 0.18
0.21 0.10 0.13
0.08 0.04 0.05
0.33
0.20
0.1
Average
mm
aThe path is the route in the radial direction from the AI-SiC interface into the matrix, which was selected for slip band density and height measurements. Selection was based on the amount of slip that occurred, and those routes were selected in which the amount of slip bands appeared to be the greatest. b The given strain is actually the result of the product N S . It does not include the coefficient K which was found to be equal to 3 (see Appendix A). Values of 7cpss that incorporate K axe given in Table 6. CThe distance given here is the distance from the AI-SiC interface to the centers of the first, second etc. squares. Therefore, values of 7cpss represent the strain over the entire square.
50
tv
2.5
5 20' c~
I~1 k
~
i2mm
1.5
THEORETICAL PLASTIC
I ° o,F _z O
OF 05
zo,
....
I 06
I 0.7
I 0.8
I
I; 0.9
, IX:)
I~ ' I.I
DISTANCE FROM THE CENTER OF THE DISK
1.2
I
I I.:3
1.4
r (ram)
Fig. 5. T h e o r e t i c a l e vs. r and e x p e r i m e n t a l 7cpss v s . p r o f i l e s a r o u n d the SiC in A I - S i C c o m p o s i t e disks t h e r m o c y c l e d b e t w e e n 298 and 823 K (p is the plastic zone radius): - - - , g r o u p A ; , g r o u p B; -----, g r o u p C.
r
actual discrete c h a r a c t e r o f t h e m e a s u r e m e n t s o f %pss. Each h o r i z o n t a l p o r t i o n o f t h e h i s t o g r a m c o r r e s p o n d s t o the average value o f
%pss o b t a i n e d f r o m the i n c r e m e n t o f t h e area of the specimen. T h e largest c o m b i n e d plastic shear strain %pss o f 1.3% was observed at t h e A1-SiC interface in g r o u p A samples; g r o u p C s h o w e d t h e smallest a m o u n t o f strain, 0.99% (Table 6). In a d d i t i o n , t h e e x t e n t o f t h e plastic z o n e (i.e. t h e largest distance f r o m the interface at w h i c h slip bands can still be m e a s u r e d ) was smaller in samples f r o m g r o u p C t h a n in samples f r o m g r o u p s A a n d B. This result c o r r e s p o n d s t o t h e e f f e c t o f strain rate (which is p r o p o r t i o n a l t o t h e cooling rate) o n t h e relative a m o u n t o f plastic and elastic strains d u r i n g d e f o r m a t i o n . Generally, higher h e a t i n g a n d c o o l i n g rates will cause t h e elastic stresses t o be larger, a n d l o w e r heating a n d c o o l i n g rates will allow greater plastic relaxat i o n [ 7 ]. T h e h e i g h t o f t h e slip b a n d s was m e a s u r e d t o + 0.1 o f a b a n d interval. This gives an error o f + 0.1 X 0 . 2 7 / ~ m = + 0 . 0 2 7 # m in the height values. As was m e n t i o n e d b e f o r e , t h e slip b a n d d e n s i t y m e a s u r e m e n t s i n t r o d u c e m u c h less error. T h e r e f o r e , t h e error range f o r
158 TABLE 6 Values of 7cpss incorporating K for group A, B and C samples Group
A B C
Combined plastic shear strain ?cpss (%) for the following distances from the AI-SiC interface 6.25 × lO-2mm
18.75 × lO-2mm
31.25 × lO-2mm
43.75 X lO-2mm
56.25 × lO-2mm
68.75 x lO-2mm
1.32 1.23 0.99
0.72 0.81 0.60
0.63 0.54 0.3
0.48 0.36
0.15 0.39
0.33
%pss determination can be evaluated as + (0.027/0.135) × 100 ~ 20% where 0.135 pm is taken as the average slip band height. This is obviously a large error range. 3.2. Hot-stage observations 3.2.1. H e a t i n g
Slip bands began to appear even before the temperature reached 373 K and were forming as widely spaced deep lines. As the temperature increased, the density of the slip bands increased also, reaching an apparent maxim u m at around 573 K. At around 533 K a new group of broken lines began to form with no relation to the previously formed slip bands. The development of these lines became more intensive at higher temperatures. When the temperature approached about the 653 K mark, the separation between the matrix and the SiC became fairly visible. At approximately 693 K, slip bands began to fade and disappeared almost totally by the time the temperature reached 8 5 3 - 8 7 3 K. At the m a x i m u m temperatures, the surface did n o t appear as flat as it did at the beginning. 3.2.2. Cooling
On cooling, previously formed slip band patterns began to show up again but they did not reach the size and extent of the former slip band patterns. The development of the slip bands in new areas was not observed. The broken line which encircled the entire SiC particle began to develop radial spokes that connected this line to the interface. When the sample cooled to room temperature, the surface of the sample remained rippled. Subsequent thermocycles did not show any changes in slip band morphology. A broken line around the SiC cylinder became much more clearly defined with repeated thermocycles. The matrix appeared to be separated from the particle all the time.
4. DISCUSSION Composite models discussed in the literature for the theoretical determination of stresses and strains in the matrix (spherical particle in a matrix, long composite cylinder etc.) cannot be applied to a short composite disk. Therefore, an a t t e m p t was made to evaluate the radius of the plastic zone and the plastic strains around the SiC particle using the approach described in Appendix B. The resultant profile of the effective plastic strain ~ around the SiC particle and the extent of the plastic zone are shown in Fig. 5. As can be seen, p ~ 1.15 mm, and g = 2.5% at the interface. The theoretical plastic strain is higher than the observed plastic strain, and the theoretical plastic zone radius is smaller than the observed plastic zone radius. The theoretical plastic zone size is estimated on the assumption t h a t deformation is homogeneous, i.e. uniform around the SiC particle. In the real case we have "bursts" of plastic flow in accordance with a particular crystallographic situation. Thus, the extent of the burst of plastic deformation can be larger than that of homogeneous deformation since the same a m o u n t of metal flow must be accommodated. The explanation of the difference between observed and calculated strains can be given as follows. On heating, because of the differential expansion between the aluminum and the SiC, the aluminum matrix tends to pull away from the SiC. If the bonding between the aluminum and the SiC is sufficient to resist the pulling action, the matrix undergoes deformation proportional to AT of the cycle. If, however, the bonding is weak, the matrix breaks away from the SiC at a certain temperature T~, and subsequent temperature increase has no effect on matrix deformation since it is now free to expand.
159 Hot-stage observations showed t h a t slip band formation started at a temperature of less than 373 K. Accurate evaluation of the temperature at which slip became visible was not possible because the thermocouple in the h o t stage was n o t sensitive enough in the temperature range between 298 and 373 K. It was noticed that more slip bands were formed on the heating half of the thermocycle than on the cooling half. The slip band arrangement on the cooling half repeated the arrangement developed during the heating half. When the temperature approached about 773 K, slip bands began to disappear. This can be explained from the surface tension point of view. Another possible explanation of the stoppage of slip band development is that the aluminum matrix breaks away from the SiC and continues to expand freely without any restraint from the SiC. It is also possible t h a t very fine slip bands still continued to form (high temperature creep, for example), but we did n o t see this because of the limitation in resolution of our optical system. It should be mentioned, however, that the disappearance of the slip bands corresponds to a similar observation made by Lammers e t al. [14] in their in situ transmission electron microscopy investigation of &l-SiC composites. They observed "slip lines" in thermally cycled transmission electron microscopy foils, and these slip lines disappeared at high temperatures. The fact that slip band patterns formed on cooling repeat themselves shows that the same slip systems are engaged in the "reversed" deformation, proving at least partially t h a t there is a certain reversibility of the plastic deformation t h a t occurs when the load is reversed. The broken lines that form around SiC appeared to be the boundaries of recrystallized grains. If this is indeed the case, recrystallization t o o k place at the interface and occurred very rapidly (2-3 min).
5. CONCLUSION From the experimental data and the theoretical model the following conclusions were obtained.
(1) A new experimental technique for measuring local plastic strains was developed and utilized in the &l-SiC system. (2) The theoretical treatment of the short composite cylinder gave the distribution and extent of the plastic strains, which are in fair agreement with the experimental results. There is a disagreement, however, between the experimental and theoretical values of plastic strains immediately at the &l-SiC interface. (3) An A1-SiC bond is a very important factor influencing the plastic deformation around SiC particles. REFERENCES 1 R.J. Arsenault and R. M. Fisher, Scr. MetaU., 1 7 (1983) 67. 2 R. J. Arsenault and R. M. Fisher, in J. Carl~on and M. Ohlson (eds.), Proc. 4th Int. Conf. on the Mechanical Behavior o f Materials, Stockholm,
1983, Vol. 1, Pergamon, Oxford, 1983, p. 451. 3 R. J. Arsenault, Mater. Sci. Eng., 64 (1984) 171. 4 R. J. Arsenault and C. S. Pande, Scr. Metall., 18 (1984) 1131. 5 J. K. Lee, Y. Y. Earmme, H. I. Aaronson and K. C. Russell, Metall. Trans. A, 11 (1980) 1837. 6 C.A. Hoffman, J. Eng. Mater. Technol., 95(1973) 55. 7 G. Garmong, Metall. Trans., 5 (1974) 2183. 8 G. J. Dvorak, M. S. M. Rao and J. Q. Tam, J. Compos. Mater., 7 (1973) 194. 9 R. L. Mehan, Metal Matrix Composites, ASTM Spec. Tech. Publ. 438, 1968, p. 29. 10 G. E. Dieter, Mechanical Metallurgy, McGrawHill, New York, 2nd edn., 1976, p. 51. 11 Metals Handbook, Vol. 1, American Society for Metals, Metals Park, OH, 1961. 12 A. Taylor and R. M. Jones, in J. R. O'Connor and J. Smiltens (eds.), Silicon Carbide, Pergamon, Oxford, 1960, p. 147. 13 Handbook of Material Science, Vol. II, CRC Press, West Palm Beach, FL, 1975. 14 M. E. Lammers, R. J. Arsenault and R. M. Fisher, to be published. APPENDIX A C o m b i n e d p l a s t i c shear strain 7cpss The relation between the slip band density N, the a m o u n t S of slip or displacement on the band and the plastic strain 7 has been discussed in a number of publications [A1A3]. In their classical works, Yamaguchi e t al. [A1] and Brown [A2] found direct proportionality between the strain and a m o u n t of slip during plastic deformation. Cottrell A.1.
160
[A3] summarized their results and arrived at the expression 7 = N S for the case when there is only one slip system in operation. For a multiple~slip~ystem case, however, this expression must be modified to take into account the different crystallographic orientations of the slip systems. The a t t e m p t to carry o u t this modification is offered in the present work and consists of the following: the highlights of the one-slipsystem case; a description of the rigorous approach for multislip cases; a simplified approximation with the introduction of the combined plastic shear strain % , ~ concept. Let us assume first that slip occurs in one slip system only. Let us consider an imaginary block (Fig. A1) that is cut out of material in such a way that slip planes are perpendicular to the facets of the block. This block is oriented in such a way that its height is measured along the z axis. Let us suppose that a shear stress ry~ is applied to the block which causes the block to slip as shown in Fig. A1. By definition, the shear strain in this case is expressed as 7~z-
where S i is the slip or the displacement in the ith band*. (Here Si is analogous to the furrow depth t in eqn. (2).) If n is the total number of slip bands involved in the slip and Sy is the average value of slip per one band, then
~=~ n~
As can be seen from Fig. A1, n is the number of intersects that the z axis makes with the slip bands and is equal to the total number of slip bands along the z direction. If the number N of slip bands per unit length along z is known (i.e. the slip band density), then n=gz
where N (cm -1) is the slip band density along the z direction. Now the total strain 7yz can be expressed [A3] as ~kS~ ')'yz =
Z
w n~
z _
Z
Si i=1
(AI)
NzS~ Z
where AS~ is the total displacement in the y direction and Z is the height of the block. The total displacement ASy is equal to the sum of the displacements on each slip plane, i.e.
,~Z~=~ Si
N can be evaluated as the total slip band length divided by the total area. To do this we can take eqn. (At) and multiply the numerator and the denominator by the width L of the block:
L~S~
i=I
'Yyz--
z
_
TYZ
z
n
I
-I
.
t=ASy
Ty z
l
Si
n~
L
z
L
(nL )Sy zL y.,
// x
Fig. A1. Hypothetical block with one slip system. S i is the displacement on the slip plane; ~Sy is the total displacement along the y axis.
*Generally, the slip direction does not have to be perpendicular to the y plane as shown in Fig. A1. However, despite this and other rather crude assumptions that are made later, the purpose is to show the complexity of the rigorous treatment.
~'Vz
z2l Zl
z2
I
22
161 (3) Combine the terms determined in step (3) with the terms from step (1) having the same indices. Now we have obtained the total strain tensor eij in the x l , Yl, zl coordinate system due to all slip systems:
eu = •
I
0
7yz
%y
0
Y12
Tyz
(4) Determine the principal components of the strain tensor e u solving the following equation:
Xl x2
Fig. A2. Hypothetical block with two operational slip systems 1 and 2.
-x
= 0
(A3)
%y Here, n L is the total length of the slip bands on the side of the block, z L is the total area of this side and NA (cm -1) = n L / z L is the slip band density. Thus,
21/2 =
7yz = NSy In the case when more than one slip system is involved in plastic deformation, slip bands are n o t parallel but instead each slip system has its own orientation* (Fig. A2). In this situation the equation for shear strain has to be modified to take into account the different orientations of the slip bands. The rigorous way to do this would be as follows. (1) Determine the plastic strain %,~, using the same approach described previously for slip system 1. (2) Rotate the coordinate system x l , Yl, zl to align it with system xz, Y2, z2 where z z is the direction perpendicular to slip bands in slip system 2. In the coordinate system x2, Y2, zz, determine %2z2 using the relation (A2)
7 Y : z 2 = a y 2 iaz~ j "fij
where ay2iaz~ j is the rotation matrix and 7ij is the strain tensor in the x l , y~, zx coordinate system and is given by
7ii =
for ~. The three roots k l , ~2 and k 3 are the three principal strains et, e 2 and e 3 respectively. (5) Determine the effective strain
ii0 0j 0
7y,z,
%,y,
0
J
*Here, the hypothetical assumption is made that slip planes from the second slip system are also perpendicular to the facets of the block.
3
{(cl -
e 2 ) 2 + (e2 -
e 3 ) 2 + (e3 - e l ) 2 )
v2
As a result of all these steps we finally get one data point on the plot ~ = f(r). This procedure is simple in principle but complicated to implement, since the crystallographic orientation has to be determined for each grain. As an alternative way of estimating the a m o u n t of plastic deformation that occurred around the SiC particle, the following approximation is offered. First, the concept of c o m b i n e d plastic shear strain 7cpss is introduced. We shall define ~¢cpss as %pss
=
KNS
where S is the average step height. In general, the slip plane is not perpendicular to the surface and therefore the coefficient K is incorporated. K is a coefficient which will be evaluated later. N is the slip band density, which is obtained from the total slip band length of all slip systems divided by the area where the total slip band length is the length of all the slip bands enclosed in a selected area (e.g. a square mesh of some net). The selection of the size of the area is based on considerations of the scale. It is realized that the combined plastic shear strain %pss has no direct correlation with the effective strain ~, the octahedral shear strain %ct or any other
162
specific strain. However, the relative simplicity in the determination of %p~ and the fact that % p , takes into account direct proportionality between the amount of plastic deformation and the slip band density makes it very attractive for use as a criterion for plastic deformation in our case and in other similar cases in general.
Now we shall assume that the amounts of strain N1SM1, N2SM2 and NsSM3 are approximately equal because of the s y m m e t r y of the specimen and provided that no voids or cracks are created. Furthermore,
A.2. Determination o f the coefficient K The coefficient K takes into account the different crystallographic orientations of slip systems. Let us consider the case when we have three slip systems operating, i.e. we see three slip band groups. The individual slip systems contribute the strain N1S1, N2S2 and NsS3 to the total %p,, i.e.
or
%pss
1 K __ - - + sin 01
sin 01 SM2 s i n 02
and _
sin 03
1
+ ~1 sin 02 sin 03
Let us evaluate the range within which K can change by considering a unit triangle (Fig. A3). Let the directions shown on the triangle correspond to the poles of the surface of the sample. Then the values of K would be as shown in Table A1. The average value of K is a b o u t 4.5 when three slip systems are operating. Thus, the average correction factor per one slip system is 4.5/3 = 1.5. In general, slip band densities were measured along directions going through areas where two slip systems,.i.e, two sets of slip bands, were present. Thus for, two slip systems, K = 2 × 1.5 = 3.
&l
S3
1)
where
where N1, N2 andN8 are the slip band densities and $1, $2 and $3 are the average displacements on the band within each slip system, given by
_
1
÷--+ sin 01 sin 02
'~Cpss ~- KNSM
%p. = N1 1 + N25 + N35
$2
(1
NSM
SM3 sin 0a
where S i n , SM2and SM3 are the average heights o f the slip bands in the three slip systems, measured perpendicularly to the surface, and 01, 02 and 03 are the angles between the slip planes and the surface plane. Thus, 1 7cpss = N I ~ I
300~311 321 / 310 320
1
0----sin 1 ÷ N2SM2 sin 0 2 +
1
Fig. A3. Some orientations of the sample for which values of K were determined.
÷ N3SM3 - -
sin 83
TABLE A1 Values of K corresponding to the orientations shown in Fig. A3
Surface plane pole K for (111) slip plane
(111) 3.2
(110) 3.7
(100) 3.7
(210) 3.66
(211) 5.1
(221) 5.9
(310) 3.6
(311) 4.2
(320) 3.1
(321) 4.9
163 SiC and aluminum. If we did n o t have SiC in the center of the aluminum disk, the aluminum ring would shrink w i t h o u t any restraint and the size o f the bore would be reduced by
References for Appendix A A1 K. Yamaguchi, Sci. Paper Inst. Phys. Chem. Res. Tokyo, 8 (1928) 289; 11 (1929) 151. A2 A. F. Brown, J. Inst. Met., 80 (1951) 115. A3 A. H. Cottretl, Dislocations and Plastic Flow in Crystals, Clarendon, Oxford, 1965, p. 157.
Ua = Aa = a A a AT APPENDIX B
where a is the radius of the SiC, A s ( ~ 25 × 10 -6 K -1) is the difference between the coefficient of thermal expansion o f aluminum and t hat o f SiC, AT ( ~ 500 K) is the temperature interval and Ua is the displacement at t he interface. The reduct i on in the radius o f t he bore does n o t occur when SiC is present in the center o f the aluminum disk. The a m o u n t o f aluminum t h a t is n o t allowed to go towards the center of the bore would cause plastic flow o f the adjacent matrix in all directions away f r o m the SiC. Thus, some of the matrix will spill o u t (as shown in Fig. B1). As we go f u r t h e r away from the interface, the plastic flow is restricted by the matrix t hat surrounds t he central port i on. Two regions can be considered: t he plastic region t hat we assume is adjacent to the SiC and the elastic region t hat encloses t he plastic region. Let us make the following assumptions. (1) In the plastic zone the matrix is a perfectly plastic material, i.e. no w ork hardening 0 ccurs. (2) At the starting t e m p e r a t u r e T of 773 K, SiC and aluminum axe just in c o n t a c t with one another. (3) The profile o f t he aluminum which has spilt o u t is a straight line.
Among the several composite models described in th e literature, there is one t ha t appears to be close to our case; this is k n o w n as the long-composite-cylinder model. It consists of long c ont i nuous fiber surrounded b y the metal matrix. This m odel implies t hat plastic strain in the direction of the fiber (the z direction) resulted f r o m unequal expansion or c o n t r a c t i o n o f fiber and t h a t the matrix is a constant, i.e. ez = dl/l = constant where l is the length o f the fiber. Therefore, the total plastic strain state becomes a plane strain case which enables t wo o t h e r plastic strains e r and e 0 to be f o u n d w i t h o u t difficulty. Although short c om pos i te cylinders have the same s y m m e t r y as long composite cylinders, the strain in the z direction is n o t a constant but is in general a f u n c t i o n o f two variables: (1) t he distance f r o m the interface and (2) the position along the z axis. Since ez =/= constant, the longcylinder approach c a n n o t be used in o u r case. We shall discuss the case shown in Fig. B1. Let us consider an Al-SiC disk t h a t cools f r o m a t e m p e r a t u r e T. As a result o f cooling, a certain a m o u n t of aluminum will be pushed back because o f t he differential shrinkage of
F" ~ ~ - -
A,,,
I
"H"
"
PLANE
", --T "
I .
i
--i "o-
~'
!
/
I
l
Fig. B1. Schematic representation of the plastic-elastic shell around the SiC.
" PLANE
164
(4) T h e plastic-elastic f r o n t is also a straight line. N o w let us consider t h e radius p o f t h e plastic z o n e . It is a f u n c t i o n o f the vertical p o s i t i o n o f the transverse plane in which p is considered. According t o o u r a s s u m p t i o n , P = Po - - q z
(B2) with a general expression f o r p (eqn. (B1)) w h e n z = ~1 H Uz = A h
P0 -- 2(Po
r
--PH) 1 --
P0--2(P0--
p
1
g)~--a
1
Ah = ~H Ao~ A T and
(the equation of the straight line) where Po is t h e plastic z o n e radius in t h e " 0 " plane, which is t h e plane o f s y m m e t r y b e t w e e n t h e u p p e r and l o w e r halves o f t h e c o m p o s i t e disk. T o d e t e r m i n e q we let z = ~z H , i.e. t h e t o p o f t h e " H " plane is considered. When 1 z =- ~ H , p -- P H . T h e r e f o r e PH = PO - - q~ H
F r o m this 1 q = 2(P0 -- PH) H
1 _
1
Uz - EH A a A T
1 Aa
P = P0 -- 2(P0 -- PH) - ~
(B1)
L e t us consider n o w t h e p o r t i o n o f t h e m a t r i x t h a t spills o u t . F r o m the triangle s h o w n in Fig. B2 we can write
Po -- 2(Po
--
A T Po - - 2 ( P 0
Uz-~ z
--PH)~
z
-- r
P 0 _ 2 ( P 0
--PH)
p
_
1 -- r
1
(B2)
PH --a
where U~ is t h e vertical displacement. Generally speaking, the vertical d i s p l a c e m e n t U~ is a f u n c t i o n o f z a n d r, i.e. Uz = ¢ ( f ( z ) , g(r)} E q u a t i o n (B2) gives o n l y an expression f o r U= w h e n t h e " H " plane is considered. The p r o b l e m t h u s is t o find a general e x p r e s s i o n f o r U=. T o do this, let us replace pn in eqn.
for z = 1~ H or if we n o r m a l i z e z = z / H -- - 5z" L e t us c h e c k the values o f Uz at d i f f e r e n t z values. First, let z = 1 ( " H " plane), and so p 1 po--2(po-H)~--r U ~ = ~1 A ~ A T p 1 Po--2(Po-H){--a 1 U~ = ~ A~ A T = Ah
and, w h e n r
= PH ,
U~ = O
L e t z = 0 ( " 0 " plane), and so Uz = 0, since t h e " 0 " plane is the plane o f vertical symm e t r y . A t this p o i n t we can see a certain logic in eqn. (B4), w h i c h enables us t o suggest t h e general expression f o r Uz in the following form: U~ = z' A~ A T P0 - - 2 ( P o - - P H ) Z ' - - r Po -- 2(Po - - P H ) Z ' - - a
i °;'r
PH
(B4)
H)2 - a
T h e n , w h e n r = a,
PH - - r
U= = Ah - -
~
(B3)
PH) ~ - - a
If we simplify eqn. (B3) it reduces to eqn. (B2), b u t let us leave eqn. (B3) in its f o r m and e x a m i n e it in m o r e detail. First let us substit u d e H = 1 m m in eqn. (B3), since this is a c o n v e n i e n t way o f simplifying this expression and H = 1 m m h a p p e n s t o be t h e actual height o f o u r c o m p o s i t e disks. N o w we can rewrite eqn. (B3) as _
and n o w
Po -- 2(P0
_
"H" P L A N E
J
z = I/: ) H
Fig. B2. " S p i l l - o u t " p o r t i o n o f the a l u m i n u m m a t r i x .
(B5)
(since H = 1). It is realized t h a t eqn. (B5) lacks rigorous p r o o f . H o w e v e r , since it inc o r p o r a t e s o u r b o u n d a r y c o n d i t i o n s and since we d o n o t have a n y additional i n f o r m a t i o n , it is r e a s o n a b l e t o a d o p t eqn. (B5) and t o see w h a t results we can o b t a i n and t h e n t o judge t h e validity o f this expression. L e t us consider t h e cylindrical c o o r d i n a t e s y s t e m in which t h e z axis coincides with t h e z axis o f o u r sample. By d e f i n i t i o n
165
6r --
T o find t h e c o n s t a n t C we use t h e b o u n d a r y c o n d i t i o n U~ =a = Ua = a A a AT:
dr
Vr 60 = - -
Ua = a A o ~ A T
6 z --
C N a +--+--a 2 a 3
r
M
6z --
M
....
C = aU a 4---
dz
a 2--
N
2
2
a3
3
N o w eqn. (B10) b e c o m e s dz
= A~ A T
Po - - 2 ( P o - - P H ) Z
M Ur------r+
--r
+
Uaa
2
(B5)*
Ma 2 4-
N a 3 ~_Nr2_
(BII)
- -
r
2r
3r
3
Po--2(Po--PH)Z--a
2(P0
+z A~AT
--
PH) (a
--
N o w we can find the strains er, e0 and e~ f o r t h e " 0 " plane:
r)
[P0 -- 2(Po - - p ~ ) z - - a]
dU r _
I n c o m p r e s s i b i l i t y requires t h a t
6r
er + eo + e~ = O
Ur eo
or
dUr 4- Vr 4-
-dr
r
6 z ----0
(B6)
L e t us c o n s i d e r t h e " 0 " plane, w h e r e z = O; then PO - - r
e~ = A~ A T - Po - - a
(B7)
-
dUr+
Ur +
dr
-0
= 0
2
Ma 2
Uaa +
7
Na 2
2
2r 2 + --3r2 + --3 N r -t
Ma 2
Na s
-
- -
-
2r 2
3r 2
N + - - r
3
(B12)
L e t us consider t h e tip o f t h e plastic-elastic i n t e r f a c e , i.e. r = Po. T h e n (ez = 0 f r o m eqn. (B7)) 6r =--6 0
Upo
(BI3)
P0 (B8)
P0 - - a
M--Nr
r2
ez = M - - N r
_
Po - - r
Uaa
2 M -
r
E q u a t i o n (B6) t h e n b e c o m e s d U r + Ur + A a A T dr r
M
dr
(B9)
r
w h e r e Up0 is t h e d i s p l a c e m e n t o f t h e tip o f t h e plastic-elastic f r o n t :
Vr=p0 = Up0 Each p o i n t o f t h e plastic-elastic f r o n t is in t h e state o f i n c i p i e n t yielding, i.e.
where M = Aa
6-~- 60
Do
AT -
-
w h e r e g is t h e effective strain at t h e i n t e r f a c e and e 0 is t h e yielding strain t a k e n f r o m t h e yielding c o n d i t i o n o f t h e tensile sample in uniaxial loading:
Po - - a and 1 N = A~ AT - PO - - a
Oy = E 6 0
T h e s o l u t i o n o f eqn. (B9) is M
C
N
Ur=----r4---4---r
2
r
2
3
Therefore
(B10)
* F r o m here o n we o m i t t h e superscript prime in z ' and use z i n s t e a d t o p r e v e n t c o n f u s i o n , b u t it s h o u l d be b o r n e in m i n d t h a t t h e value z is n o r m a l i z e d w i t h r e s p e c t to H.
O'y 60 - ~ - - -
E
ii.73 62 × l 0 s 2 X 10 -4
166
where Oy = 11.73 MPa and E = 62 X 10 a MPa for pure aluminum. Thus we have a system o f equations:
up0
er (r =po ) --
Po
_Uo
eO(r=po)
-
-
(B14)
-
P0
ez(r=po )
0
=
metrical a b o u t the " 0 " plane, it goes through its local m a x i m u m or m i n i m u m point when it crosses the " 0 " plane and t herefore dUz/dz = 0 when z = 0. Thus we f o u n d that, within the " 0 " plane, all the shear c o m p o n e n t s of the strain tensor are equal to zero, which means t hat e r = el , eo = e2 and ez = ea for the " 0 " plane. Thus, if we go back to the yon Mises condition o f yield, the effective strain g can now be expressed as 21/2
Applying the yon Mises yielding condition,
g= 3
2~2 {~ : - - { ( e
I - - e 2 ) 2 -~- (e I - - e 3 ) 2 + (e 2 - - e 3 ) 2 } 1]2
3 where el, ez and e a are the principal strains. Generally speaking, the strains er, e0 and ez do n o t coincide with the principal strains el , e2 and ea, because o f the existence o f the shear c o m p o n e n t in the total strain tensor, which is given by
d~ 7zr =
dr
dUr dz
e=
3 (\Po]
\ Po /
\Po/
J
= 1.155 Up-~° Po However,
= 2X I0-4
The c o m p o n e n t s 70~ and %0 axe equal to zero because o f the s y m m e t r y around SiC. dU=
Aa A T z
dr
Po --2(P0 --PH)Z --a
/I,,'
= 1 . 1 5 5 Up0 P0 Therefore
for the " 0 " plane; when z = O, dUz/dr = 0. To evaluate dU~/dz, let us consider the region of the matrix adjacent t o the " 0 " plane as shown in Fig. B3. Point A is displaced to poi nt A', and p o in t B to point B'. (B is a mirror image o f A.) The displacements UAA'and Usa' are equal because o f the s y m m e t r y and f r o m the assumption th at no voids are created. The displacement Ur varies as we go along the z axis and presents a continuous f u n c t i o n of z. Since the f u n c t i o n is continuous and sym-
UP° Po
(BI5)
1.7 × 1 0 -4
From eqn. (BI1) we get M
UP°
=
2
U~a P o nu
Maz 't-
P0
Na3 --
2P0
3Po
N ~ --Po
3
2
and, substituting for M and N, we obtain
Upo__Aa_~TIOo2+aPo+a2 Po t 3
(a+Po)Pot_aZ } 2 (BI6)
A'
Equations (B15) and (BI6) can be solved as a system of two equations with two unknowns,
'
Up, and P0. Solving t hem , we obtain P0 = 1.15 mm. In order to find PH we can follow the same procedure. However, for t he " H " plane case t he strains er, eo and ez are n o t equal to el, e 2 and e a. Therefore the procedure has to be modified. The new procedure would be to find an expression for
,
AI
SiC
"0" PLANE
Ur
B~--~B
and, from eqn. (B14),
~=C 0
+ --
r
((er-- eo)2 + (eo - - e z ) 2 + (er--ez) 2 }1/2
Fig. B3. Displacements AA p and BB' of the matrix symmetrical a b o u t the " 0 " plane.
dUz + dUr dr
dz
167 1 for z = ~ a n d t h e n t o solve t h e e q u a t i o n
L%r
0
%.]
e0 --k
0
0
e z -- k
= 0
This is a cubic equation. Three roots give the values for the principal strains: ~'I = el k2
=
62
and •3
an idea o f t h e e x t e n t o f t h e plastic z o n e o n t h e surface o f t h e s a m p l e . A d d i t i o n a l l y , if t h e m a g n i t u d e o f %r is small, t h e d e v i a t i o n o f er, eo a n d ez f r o m e l , e2 a n d e 8 will also be small. 1 N o w , considering z = ~ a n d using t h e yielding c o n d i t i o n at t h e p l a s t i c - e l a s t i c f r o n t , we can f o l l o w steps similar t o t h o s e t h a t we used f o r t h e " 0 " plane. As a result PH = 1.11 m m . It is useful t o find a general e x p r e s s i o n f o r g in t h e " 0 " plane. A f t e r s u b s t i t u t i o n o f t h e n u m e r i c a l values f o r a a n d p, eqn. (B12) b e c o m e s , f o r z = 0,
= 63
This is s i m p l e in p r i n c i p l e b u t v e r y c o m p l e x to implement for the following reason. The individual c o m p o n e n t s o f this t e n s o r are o f t h e t y p e e x p r e s s e d b y eqns. ( B l l ) a n d (B15). If the coefficients of the cubic equation were n u m e r i c a l , t h e n w e c o u l d solve it using t h e t r i a l - a n d - e r r o r m e t h o d . In o u r case, all t h e c o e f f i c i e n t s h a v e a general e x p r e s s i o n . T h e general s o l u t i o n o f t h e c u b i c e q u a t i o n has t h e form
ez -
1.15 - - r 0.65
Aa AT
0. 513r 8 + 0.407 -- 0.885r 2 60 =
e~ =
r2
0.667r 8 -- 0.575r 2 -- 0.265 0.65r2 A~ A T
= 9 . 0 7 7 X 10 -8 X X (4.668r 2 -- 10.35r + 5.952 \
kl =
--2 +
+
+
Aa &T
0.531 r
0.42~ ~2
VJ
Each q and p would be a combination of the e x p r e s s i o n s f o r er, eo, e~ a n d %r- Because o f t h e d i f f i c u l t y o f d e t e r m i n i n g PH in a r i g o r o u s w a y w e can use an a p p r o x i m a t e s o l u t i o n b y t r e a t i n g strains e , eo a n d e~ as t h e principal strains e l , 62 a n d e 8. This implies t h a t we ignore t h e i n f l u e n c e o f t h e shear strain %~. It is realized t h a t t h e resulting value f o r PH will n o t be a t r u e value b u t it will at least give
This e f f e c t i v e strain is p l o t t e d as a f u n c t i o n o f r in Fig. 5. When ~ = eo = 2 X 10 -4 is substit u t e d in t h e e x p r e s s i o n a b o v e , t h e t h e o r e t i c a l plastic z o n e radius can b e o b t a i n e d f r o m r = p = 1.15 m m . T h e radius o f t h e f i b e r is 0.5 m m . T h u s p = 1 . 1 5 / 0 . 5 = 2.3 X f i b e r radius. This radius includes t h e f i b e r size. T h u s t h e e x t e n t o f t h e plastic z o n e m e a s u r e d f r o m t h e i n t e r f a c e is (2.3 - - 1) X f i b e r radius = 1.3 X f i b e r radius.
151
Deformation in AI-SiC Composites Due to Thermal Stresses Y. FLOM and R. J. ARSENAULT Metallurgical Materials Laboratory, University of Maryland, College Park, MD 20742 (U.S.A.)
(Received January 2, 1985; in revised form March 13, 1985)
ABSTRACT
Plastic strains and the e x t e n t o f the plastic zone due to differential thermal expansion were experimentally determined in an A l - S i C composite. The combined plastic shear strains %pss at the A I - S i C interface for furnace-cooled, air-cooled and quenched samples were found to be 1.32%, 1.23% and O.99% respectively. Profiles o f %pss were plotted versus distance from the interface and compared with the theoretical distribution o f effective strain ~. The theoretical e x t e n t o f the plastic zone measured from the interface was found to be 1.3 multiplied by the particle radius. This value was slightly less than the observed value. The plastic deformation on the heating half o f the thermocycle was found to be at least equal to the deformation on the cooling half. A theoretical treatment o f the local plastic deformation in a short composite cylinder was suggested, from which the effective plastic strain ~ and the e x t e n t of the plastic zone were determined. 1. INTRODUCTION When a composite material is subjected to a temperature change, local plastic deformation can occur. The plastic deformation is due to a stress created by the difference between the coefficients of thermal expansion of the component phases, and this stress is at the matrixreinforcement interfaces. The magnitude of the stress is equal to A s AT where A s is the difference between the coefficients of thermal expansion of the phases under consideration and AT is the temperature change. The relatively large (10 to 1) difference between the coefficient of thermal expansion of SiC and t h a t of aluminum should result in the creation of a substantial misfit strain at the A1-SiC interface during cooling from the fabricating or 0025-5416/85/$3.30
annealing temperature. The mechanical properties of the composite should be affected by the magnitude and extent of the plastic deformation that takes place in the soft matrix around a hard particle as a result of the misfit relaxation in the interface region. Recent investigations [1-4] have indeed shown the important role of the interfacial regions in the composite strengthening. Thus, knowledge of the m a ~ i t u d e of plastic strains and the size of the plastic zone as well as other characteristics of the interfacial region (the bond strength, the microstructure etc.) should contribute to the understanding of the mechanism of composite strengthening. The experimental determinations of plastic strains and plastic zone radii about a particle in the matrix due to the difference between the coefficients of thermal expansion of the phases, to our knowledge, have not been reported in the literature. Several theoretical investigations have been undertaken to predict the magnitude of the plastic strain in the plastic zone around a particle. The relaxation of the misfit caused by the introduction of an oversized spherical particle into a spherical hole in the matrix has been analytically described b y Lee et al. [5]. Using the misfitting sphere model, they calculated strains in the plastic zone that surrounds a hard sphere and also the plastic zone radius. Hoffman [6] calculated the overall total strains in the tungsten-fiberreinforced 80Ni-20Cr matrix, using a thickwall long~ylinder approach and assuming that a hydrostatic stress state exists within each constituent. Garmong [ 7 ], assuming uniformity of the stresses and strains in the matrix, calculated deformation parameters for a hypothetical eutectic composite and reported values of matrix plastic strains that were of the order of 0.4%. © Elsevier Sequoia/Printed in The Netherlands
152 Dvorak e t al. [8] developed a new axisymmetric plasticity theory of fiberous composites involving large thermal changes. The long~omposite-cylinder model was adopted as a composite unit cell, and the microstress distribution as well as the yielding surfaces were obtained for AI-W composites. Mehan [9] calculated the residual strains in an AI-~-A1203 composite due to cooling from the fabricating temperature. He considered an idealized composite consisting of a long sapphire cylinder surrounded by an aluminum matrix. This is equivalent to the long-composite-cylinder model used by Dvorak e t al. T h i s model implies that the thermal strain along the cylinder axis is a constant, i.e. that dl ez = - - = constant l where l is the length of the cylinder. This simplifies the procedure of obtaining the radial er and the tangential e0 strains which was done using incompressibility and boundary conditions. The effective strain ~ at the Al-a-A1203 interface was f o u n d to be 1.6%. However, the above-mentioned composite models do n o t give an accurate description of the plastic strain state in the short-composite-cylinder model. A short-cylinder model nearly duplicates the situation in the whisker and platelet A1-SiC composites at present produced. The purpose of this work was to determine experimentally the magnitude of the local plastic strain produced in the aluminum matrix around a short SiC cylinder during a thermocycle and also to estimate the extent of the plastic zone around the cylinder. Also an effort was made to develop a theoretical
TABLE
model of the plastic zone around the short cylinder.
2. EXPERIMENTAL PROCEDURE The small particle interspacing (several micrometers) in a commercially available A1-SiC composite makes it impossible to measure the local plastic strain at the A1-SiC interface directly. Thus a composite model consisting of an SiC cylinder embedded in an aluminum matrix was fabricated so t h a t direct strain measurements could be attempted. Aluminum of 99.99% purity (to eliminate the influence of the alloying elements) and commercial carborundum were used to produce the composite model. Selected properties of these materials are given in Table 1. Platelets of SiC were separated from carborundum conglomerates that are used in the production of abrasives. These platelets were spark planed on an electric discharge machine to plates approximately 1 mm thick. These flat plates were cut into rectangular rods approximately 1 m m × 1 ram. After this, each rod was spark machined to cylinders of about 1 mm diameter. Pure aluminum rods, 12.5 mm in diameter in the as-received condition, were cut into studs 37 mm long. Two aluminum studs and one SiC rod were assembled together and put in a specially built compaction die, where they were hot pressed to produce one compact. Compaction was done on the Instron testing machine. During the entire compaction cycle (Table 2), a vacuum of about 10 -a Tort was maintained using a mechanical vacuum pump.
1
Selected properties of the aluminum and SiC used in this paper
Material
A1 SiC
Yield strength
Elastic modulus
Bulk modulus
(MPa)
(MPa)
(MPa)
11.73 62 X103 34.5 + (tension) 483 × 103 1380 + (compression)
57.5 X103 96.6 × 103
Poisson 's Thermal expansion ratio coefficient
(m m -1 K -1) in the temperature range 293-773 K
0.31 0.19
28 × 1 0 - 6 3 x 10 -6
Melting point (K)
Reference
933 3373
[10, 11] [12,13]
153 TABLE 2 Flow chart of the compaction of an AI-SiC composite cylinder
%pss = K N S
Hea ring Heating time Temperature
~ 1 h ~ 843 K
Compression Cross-head s p e e d T o t a l travel
0.1 c m m i n -1 10 m m
Holding at constant load Time Pressure
2 h 2.28 MPa
Incremental cooling under pressure Time Pressure
sic,.~]
(a)
1
duced where %pas is equivalent to the product of slip band density N and the a m o u n t of slip S, i.e.
12 h 2.28 MPa
i
._~
(b)
(c)
(d)
Fig. 1. A s c h e m a t i c d i a g r a m of t h e f a b r i c a t i o n s e q u e n c e o f t h e A1-SiC c o m p o s i t e m o d e l .
The central portion of the compacted sample was sliced in the transverse direction into disks 1 m m thick using the electric discharge machine, set at a low power, and each disk contained an SiC cylinder very close to the perfect center (Fig. 1). All A1-SiC disks were metallographically and then electrolytically polished to remove the thin cold-worked surface layer of aluminum. The m e t h o d adopted for the evaluation of plastic deformation was based on the direct observation of slip bands on the polished surface of the sample around the SiC particles. The a m o u n t of slip is a characteristic of the a m o u n t of plastic deformation (when deformation occurs by slip) in a crystalline solid. The plastic strain can be evaluated if the number of slip bands and the displacement on each band are known. The concept of combined plastic shear strain %ps~ has been intro-
(1)
where K is a coefficient which takes into consideration different crystallographic situations. A detailed treatment of %pss is given in Appendix A. Thus, the m e t h o d reduces the data collection to the measurements of slip band densities and their heights in the area of interest. The electropolished AI-SiC disks were separated into three groups: A, B and C. Each group was heated to about 823 K and then cooled as follows: group A, furnace cooled; group B, air cooled; group C, quenched in alcohol. Since the surface of each disk had a high quality polish, slip bands could be observed around the SiC in an optical microscope. Slip band density and height measurements were obtained using a Zeiss interference microscope. Areas containing slip bands were photographed in white light and in green monochromatic light. Pictures taken in white light gave the actual image of the slip bands. Pictures of the same areas taken in monochromatic light gave interference fringe patterns (Figs. 2-4). Thus, the correlation between slip bands and interference fringes was established. When a furrow is present in the plane surface o f the object, the straight interference bands are deflected by the furrow and the furrow depth t can be determined from the deflection: t -
dk b2
(2)
where d is the deflection of the interference band, b is the band interval, taken as the distance from the middle of one band to the middle of the next band and k/2 = 0.27 #m for the thallium light. Band deflection can be estimated to be one-tenth of the band interval. Thus the height measurements can be as accurate as -+0.1 × 0.27 = -+0.027 pm. The slip band density was determined by using the " m e s h " m e t h o d which consisted of the following. Pictures of slip band images and interference patterns were enlarged to 10 cm × 12.5 cm on high contrast photographic paper. After that, a transparent plastic
154
~(a)
SiC
AI
N (b) Fig. 2. (a) Slip bands and (b) interference fringe patterns of different areas around the SiC. (Magnification, 32 × .)
SiC
AI
(b) Fig. 3. (a) Slip bands and (b) interference fringe patterns of different areas around the SiC. (Magnification, 32 × .)
155
(a)
AI
SiC
Z
Y
(b) Fig. 4. (a) Slip bands and (b) interference fringe patterns of different areas around the SiC. (Magnification, 32 x.)
film with a square mesh of a specific size (1.0 cm X 1.0 cm square for pictures with a magnification o f 80X and 2.5 cm X 2.5 cm square f o r pictures with a magnification of 200X) was overlaid on the p h o t o g r a p h and t h e total length o f the slip bands within each square along th e radial direction was measured. If M is th e magnification of the picture, L (m) the total length o f slip bands in one square and A (mm 2) the area o f the square, t hen t he slip band density N is L N = M--
X 10 -3 pm -1
(3)
a Leitz optical microscope. A reduct i on in size was necessary in order to fit the holder into the h o t stage. The entire t h e r m o c y c l e was recorded on videotape using an RCA industrial television camera. The time was recorded by means of a digital generator interfaced to the recording unit, and t he t e m p e r a t u r e was recorded on the sound track of t he tape at 20 K intervals. The total length of the t h e r m o c y c l e was approxi m at el y 8 min. The m a x i m u m temperatures were around 873 K. The heating and cooling rates were around 100 K min -1.
A The accuracy o f slip band density measurements is mu ch higher than t ha t of height measurements. Therefore, the main source of error in combined plastic shear strain ~'cpss evaluation is the height of the slip band (provided that, of course, the assumptions and results of discussion given in Appendix A are reasonably correct).
2.1. Hot-stage experiment Two A1-SiC disks were reduced in diameter to 7 mm and t h e r m o c y c l e d in the h o t stage o f
3. EXPERIMENTAL RESULTS
3.1. Combined plastic shear strain The experimentally det erm i ned values of com bi ned plastic shear strain %pss are presented in Tables 3 - 5 . These com bi ned plastic shear strains %pss are pl ot t ed versus distance in t he form of a histogram in Fig. 5. In the same figure we have a plot of effective strain which was det erm i ned theoretically (see Appendix B). The histogram represents the
156 TABLE 3 C o m b i n e d plastic shear s t r a i n 7cpss for g r o u p A samples ( f u r n a c e c o o l e d )
Sample
Path a
Combined plastic shear strain b 7cpss (%) for the following distances f r o m the A l - S i C interface c 6.25×10-2mm
18.75×10-2mm
31.25X10-2mm
43.75×10-2mm
56.25×10-2mm
0.09 0.06 0.08 0.16 0.13 0.08
0.026 0.07
1 1 1 1 1 1
1 2 3 4 5 6
0.86 0.61 0.41 0.26 0.42 0.51
0.35 0.40 0.140 0.21 0.32 0.41
0.18 0.22 0.24 0.19 0.32 0.30
2 2
1 2
0.30 0.45
0.21 0.28
0.06
3 3 3 3 3
1 2 3 4 5
0.43 0.26 0.39 0.46 0.45
0.24 0.27 0.22 0.18 0.26
0.22 0.22 0.26 0.23
4
1
0.43
0.24
0.04
0.44
0.24
0.21
Average
0.03
0.22 0.11 0.16
0.22
0.16
0.05
a T h e p a t h is t h e r o u t e in t h e radial d i r e c t i o n f r o m t h e A I - S i C i n t e r f a c e i n t o t h e m a t r i x , w h i c h was selected for slip b a n d d e n s i t y a n d h e i g h t m e a s u r e m e n t s . S e l e c t i o n was b a s e d o n t h e a m o u n t o f slip t h a t o c c u r r e d , a n d t h o s e r o u t e s were selected in w h i c h t h e a m o u n t o f slip b a n d s a p p e a r e d t o be t h e greatest. b T h e given s t r a i n is a c t u a l l y t h e r e s u l t o f t h e p r o d u c t NS. It does n o t include t h e c o e f f i c i e n t K w h i c h was f o u n d t o b e e q u a l t o 3 (see A p p e n d i x A). Values o f Tcpss t h a t i n c o r p o r a t e K are given in Table 6. CThe d i s t a n c e given here is t h e d i s t a n c e f r o m t h e A I - S i C i n t e r f a c e t o t h e c e n t e r s o f t h e first, s e c o n d etc. squares. T h e r e f o r e , values o f 7cpss r e p r e s e n t t h e s t r a i n over t h e e n t i r e square. TABLE 4 C o m b i n e d plastic s h e a r s t r a i n 7cpss in g r o u p B s a m p l e s (air c o o l e d )
Sample
Path a
Combined plastic shear strain b ~/cpss (%) for the following distances from the A l - S i C interface c 6.25 x l O-2 m m
18.75 × l O-2 m m
31.25 x l O-2 m m
43.75 × l O-2 m m
56.25 x l O-2 m m
68.75 X l O-2 m m
0.35 0.17 0.19
0.13 0.19 0.11
0.13
0.11
0.014
0.08
0.13
0.11
1 1 1 1 1
1 2 3 4 5
0.34 0.25 0.56 0.47 0.25
0.32 0.25 0.31 0.55 0.07
2
1
0.23
0.17
4 4 4 4 4
1 2 3 4 5
0.65 0.47 0.54 0.31 0.48
0.41 0.22 0.29 0.21 0.25
0.41
0.27
Average
0.1 0.25 0.18
0.12
a T h e p a t h is t h e r o u t e in t h e radial d i r e c t i o n f r o m t h e A I - S i C i n t e r f a c e i n t o t h e m a t r i x , w h i c h was selected for slip b a n d d e n s i t y a n d h e i g h t m e a s u r e m e n t s . S e l e c t i o n was based o n t h e a m o u n t of slip t h a t o c c u r r e d , a n d t h o s e r o u t e s were selected in w h i c h t h e a m o u n t o f slip b a n d s a p p e a r e d t o b e t h e greatest. b T h e given s t r a i n is a c t u a l l y t h e r e s u l t o f t h e p r o d u c t NS. It d o e s n o t include t h e c o e f f i c i e n t K w h i c h was f o u n d t o b e e q u a l t o 3 (see A p p e n d i x A). Values o f 7cpss t h a t i n c o r p o r a t e K are given in Table 6. CThe d i s t a n c e given here is t h e d i s t a n c e f r o m t h e A I - S i C i n t e r f a c e t o t h e c e n t e r s o f t h e first, s e c o n d etc. squares. T h e r e f o r e , values o f ~/cpss r e p r e s e n t t h e s t r a i n over t h e e n t i r e square.
157 TABLE
5
Combined plastic shear strain "~cpssin group C samples (quenched) Sample
C o m b i n e d plastic shear strain b 7cpss (%) f o r the f o l l o w i n g distances f r o m the A l - S i C interface c
Path a
6.25
x
10 - 2
mm
1 8 . 7 5 x 10 - 2
mm
3 1 . 2 5 x 10 - 2
1 1 1
1 2 3
0.40 0.19 0.69
0.16 0.27
2 2 2 2
1 2 3 4
0.37 0.31 0.22 0.55
0.29 0.11 0.24 0.23
0.16 0.05 0.12 0.14
3 3
1 2
0.31 0.44
0.18 0.24
0.23
4 4 4
1 2 3
0.25 0.25 0.18
0.21 0.10 0.13
0.08 0.04 0.05
0.33
0.20
0.1
Average
mm
aThe path is the route in the radial direction from the AI-SiC interface into the matrix, which was selected for slip band density and height measurements. Selection was based on the amount of slip that occurred, and those routes were selected in which the amount of slip bands appeared to be the greatest. b The given strain is actually the result of the product N S . It does not include the coefficient K which was found to be equal to 3 (see Appendix A). Values of 7cpss that incorporate K axe given in Table 6. CThe distance given here is the distance from the AI-SiC interface to the centers of the first, second etc. squares. Therefore, values of 7cpss represent the strain over the entire square.
50
tv
2.5
5 20' c~
I~1 k
~
i2mm
1.5
THEORETICAL PLASTIC
I ° o,F _z O
OF 05
zo,
....
I 06
I 0.7
I 0.8
I
I; 0.9
, IX:)
I~ ' I.I
DISTANCE FROM THE CENTER OF THE DISK
1.2
I
I I.:3
1.4
r (ram)
Fig. 5. T h e o r e t i c a l e vs. r and e x p e r i m e n t a l 7cpss v s . p r o f i l e s a r o u n d the SiC in A I - S i C c o m p o s i t e disks t h e r m o c y c l e d b e t w e e n 298 and 823 K (p is the plastic zone radius): - - - , g r o u p A ; , g r o u p B; -----, g r o u p C.
r
actual discrete c h a r a c t e r o f t h e m e a s u r e m e n t s o f %pss. Each h o r i z o n t a l p o r t i o n o f t h e h i s t o g r a m c o r r e s p o n d s t o the average value o f
%pss o b t a i n e d f r o m the i n c r e m e n t o f t h e area of the specimen. T h e largest c o m b i n e d plastic shear strain %pss o f 1.3% was observed at t h e A1-SiC interface in g r o u p A samples; g r o u p C s h o w e d t h e smallest a m o u n t o f strain, 0.99% (Table 6). In a d d i t i o n , t h e e x t e n t o f t h e plastic z o n e (i.e. t h e largest distance f r o m the interface at w h i c h slip bands can still be m e a s u r e d ) was smaller in samples f r o m g r o u p C t h a n in samples f r o m g r o u p s A a n d B. This result c o r r e s p o n d s t o t h e e f f e c t o f strain rate (which is p r o p o r t i o n a l t o t h e cooling rate) o n t h e relative a m o u n t o f plastic and elastic strains d u r i n g d e f o r m a t i o n . Generally, higher h e a t i n g a n d c o o l i n g rates will cause t h e elastic stresses t o be larger, a n d l o w e r heating a n d c o o l i n g rates will allow greater plastic relaxat i o n [ 7 ]. T h e h e i g h t o f t h e slip b a n d s was m e a s u r e d t o + 0.1 o f a b a n d interval. This gives an error o f + 0.1 X 0 . 2 7 / ~ m = + 0 . 0 2 7 # m in the height values. As was m e n t i o n e d b e f o r e , t h e slip b a n d d e n s i t y m e a s u r e m e n t s i n t r o d u c e m u c h less error. T h e r e f o r e , t h e error range f o r
158 TABLE 6 Values of 7cpss incorporating K for group A, B and C samples Group
A B C
Combined plastic shear strain ?cpss (%) for the following distances from the AI-SiC interface 6.25 × lO-2mm
18.75 × lO-2mm
31.25 × lO-2mm
43.75 X lO-2mm
56.25 × lO-2mm
68.75 x lO-2mm
1.32 1.23 0.99
0.72 0.81 0.60
0.63 0.54 0.3
0.48 0.36
0.15 0.39
0.33
%pss determination can be evaluated as + (0.027/0.135) × 100 ~ 20% where 0.135 pm is taken as the average slip band height. This is obviously a large error range. 3.2. Hot-stage observations 3.2.1. H e a t i n g
Slip bands began to appear even before the temperature reached 373 K and were forming as widely spaced deep lines. As the temperature increased, the density of the slip bands increased also, reaching an apparent maxim u m at around 573 K. At around 533 K a new group of broken lines began to form with no relation to the previously formed slip bands. The development of these lines became more intensive at higher temperatures. When the temperature approached about the 653 K mark, the separation between the matrix and the SiC became fairly visible. At approximately 693 K, slip bands began to fade and disappeared almost totally by the time the temperature reached 8 5 3 - 8 7 3 K. At the m a x i m u m temperatures, the surface did n o t appear as flat as it did at the beginning. 3.2.2. Cooling
On cooling, previously formed slip band patterns began to show up again but they did not reach the size and extent of the former slip band patterns. The development of the slip bands in new areas was not observed. The broken line which encircled the entire SiC particle began to develop radial spokes that connected this line to the interface. When the sample cooled to room temperature, the surface of the sample remained rippled. Subsequent thermocycles did not show any changes in slip band morphology. A broken line around the SiC cylinder became much more clearly defined with repeated thermocycles. The matrix appeared to be separated from the particle all the time.
4. DISCUSSION Composite models discussed in the literature for the theoretical determination of stresses and strains in the matrix (spherical particle in a matrix, long composite cylinder etc.) cannot be applied to a short composite disk. Therefore, an a t t e m p t was made to evaluate the radius of the plastic zone and the plastic strains around the SiC particle using the approach described in Appendix B. The resultant profile of the effective plastic strain ~ around the SiC particle and the extent of the plastic zone are shown in Fig. 5. As can be seen, p ~ 1.15 mm, and g = 2.5% at the interface. The theoretical plastic strain is higher than the observed plastic strain, and the theoretical plastic zone radius is smaller than the observed plastic zone radius. The theoretical plastic zone size is estimated on the assumption t h a t deformation is homogeneous, i.e. uniform around the SiC particle. In the real case we have "bursts" of plastic flow in accordance with a particular crystallographic situation. Thus, the extent of the burst of plastic deformation can be larger than that of homogeneous deformation since the same a m o u n t of metal flow must be accommodated. The explanation of the difference between observed and calculated strains can be given as follows. On heating, because of the differential expansion between the aluminum and the SiC, the aluminum matrix tends to pull away from the SiC. If the bonding between the aluminum and the SiC is sufficient to resist the pulling action, the matrix undergoes deformation proportional to AT of the cycle. If, however, the bonding is weak, the matrix breaks away from the SiC at a certain temperature T~, and subsequent temperature increase has no effect on matrix deformation since it is now free to expand.
159 Hot-stage observations showed t h a t slip band formation started at a temperature of less than 373 K. Accurate evaluation of the temperature at which slip became visible was not possible because the thermocouple in the h o t stage was n o t sensitive enough in the temperature range between 298 and 373 K. It was noticed that more slip bands were formed on the heating half of the thermocycle than on the cooling half. The slip band arrangement on the cooling half repeated the arrangement developed during the heating half. When the temperature approached about 773 K, slip bands began to disappear. This can be explained from the surface tension point of view. Another possible explanation of the stoppage of slip band development is that the aluminum matrix breaks away from the SiC and continues to expand freely without any restraint from the SiC. It is also possible t h a t very fine slip bands still continued to form (high temperature creep, for example), but we did n o t see this because of the limitation in resolution of our optical system. It should be mentioned, however, that the disappearance of the slip bands corresponds to a similar observation made by Lammers e t al. [14] in their in situ transmission electron microscopy investigation of &l-SiC composites. They observed "slip lines" in thermally cycled transmission electron microscopy foils, and these slip lines disappeared at high temperatures. The fact that slip band patterns formed on cooling repeat themselves shows that the same slip systems are engaged in the "reversed" deformation, proving at least partially t h a t there is a certain reversibility of the plastic deformation t h a t occurs when the load is reversed. The broken lines that form around SiC appeared to be the boundaries of recrystallized grains. If this is indeed the case, recrystallization t o o k place at the interface and occurred very rapidly (2-3 min).
5. CONCLUSION From the experimental data and the theoretical model the following conclusions were obtained.
(1) A new experimental technique for measuring local plastic strains was developed and utilized in the &l-SiC system. (2) The theoretical treatment of the short composite cylinder gave the distribution and extent of the plastic strains, which are in fair agreement with the experimental results. There is a disagreement, however, between the experimental and theoretical values of plastic strains immediately at the &l-SiC interface. (3) An A1-SiC bond is a very important factor influencing the plastic deformation around SiC particles. REFERENCES 1 R.J. Arsenault and R. M. Fisher, Scr. MetaU., 1 7 (1983) 67. 2 R. J. Arsenault and R. M. Fisher, in J. Carl~on and M. Ohlson (eds.), Proc. 4th Int. Conf. on the Mechanical Behavior o f Materials, Stockholm,
1983, Vol. 1, Pergamon, Oxford, 1983, p. 451. 3 R. J. Arsenault, Mater. Sci. Eng., 64 (1984) 171. 4 R. J. Arsenault and C. S. Pande, Scr. Metall., 18 (1984) 1131. 5 J. K. Lee, Y. Y. Earmme, H. I. Aaronson and K. C. Russell, Metall. Trans. A, 11 (1980) 1837. 6 C.A. Hoffman, J. Eng. Mater. Technol., 95(1973) 55. 7 G. Garmong, Metall. Trans., 5 (1974) 2183. 8 G. J. Dvorak, M. S. M. Rao and J. Q. Tam, J. Compos. Mater., 7 (1973) 194. 9 R. L. Mehan, Metal Matrix Composites, ASTM Spec. Tech. Publ. 438, 1968, p. 29. 10 G. E. Dieter, Mechanical Metallurgy, McGrawHill, New York, 2nd edn., 1976, p. 51. 11 Metals Handbook, Vol. 1, American Society for Metals, Metals Park, OH, 1961. 12 A. Taylor and R. M. Jones, in J. R. O'Connor and J. Smiltens (eds.), Silicon Carbide, Pergamon, Oxford, 1960, p. 147. 13 Handbook of Material Science, Vol. II, CRC Press, West Palm Beach, FL, 1975. 14 M. E. Lammers, R. J. Arsenault and R. M. Fisher, to be published. APPENDIX A C o m b i n e d p l a s t i c shear strain 7cpss The relation between the slip band density N, the a m o u n t S of slip or displacement on the band and the plastic strain 7 has been discussed in a number of publications [A1A3]. In their classical works, Yamaguchi e t al. [A1] and Brown [A2] found direct proportionality between the strain and a m o u n t of slip during plastic deformation. Cottrell A.1.
160
[A3] summarized their results and arrived at the expression 7 = N S for the case when there is only one slip system in operation. For a multiple~slip~ystem case, however, this expression must be modified to take into account the different crystallographic orientations of the slip systems. The a t t e m p t to carry o u t this modification is offered in the present work and consists of the following: the highlights of the one-slipsystem case; a description of the rigorous approach for multislip cases; a simplified approximation with the introduction of the combined plastic shear strain % , ~ concept. Let us assume first that slip occurs in one slip system only. Let us consider an imaginary block (Fig. A1) that is cut out of material in such a way that slip planes are perpendicular to the facets of the block. This block is oriented in such a way that its height is measured along the z axis. Let us suppose that a shear stress ry~ is applied to the block which causes the block to slip as shown in Fig. A1. By definition, the shear strain in this case is expressed as 7~z-
where S i is the slip or the displacement in the ith band*. (Here Si is analogous to the furrow depth t in eqn. (2).) If n is the total number of slip bands involved in the slip and Sy is the average value of slip per one band, then
~=~ n~
As can be seen from Fig. A1, n is the number of intersects that the z axis makes with the slip bands and is equal to the total number of slip bands along the z direction. If the number N of slip bands per unit length along z is known (i.e. the slip band density), then n=gz
where N (cm -1) is the slip band density along the z direction. Now the total strain 7yz can be expressed [A3] as ~kS~ ')'yz =
Z
w n~
z _
Z
Si i=1
(AI)
NzS~ Z
where AS~ is the total displacement in the y direction and Z is the height of the block. The total displacement ASy is equal to the sum of the displacements on each slip plane, i.e.
,~Z~=~ Si
N can be evaluated as the total slip band length divided by the total area. To do this we can take eqn. (At) and multiply the numerator and the denominator by the width L of the block:
L~S~
i=I
'Yyz--
z
_
TYZ
z
n
I
-I
.
t=ASy
Ty z
l
Si
n~
L
z
L
(nL )Sy zL y.,
// x
Fig. A1. Hypothetical block with one slip system. S i is the displacement on the slip plane; ~Sy is the total displacement along the y axis.
*Generally, the slip direction does not have to be perpendicular to the y plane as shown in Fig. A1. However, despite this and other rather crude assumptions that are made later, the purpose is to show the complexity of the rigorous treatment.
~'Vz
z2l Zl
z2
I
22
161 (3) Combine the terms determined in step (3) with the terms from step (1) having the same indices. Now we have obtained the total strain tensor eij in the x l , Yl, zl coordinate system due to all slip systems:
eu = •
I
0
7yz
%y
0
Y12
Tyz
(4) Determine the principal components of the strain tensor e u solving the following equation:
Xl x2
Fig. A2. Hypothetical block with two operational slip systems 1 and 2.
-x
= 0
(A3)
%y Here, n L is the total length of the slip bands on the side of the block, z L is the total area of this side and NA (cm -1) = n L / z L is the slip band density. Thus,
21/2 =
7yz = NSy In the case when more than one slip system is involved in plastic deformation, slip bands are n o t parallel but instead each slip system has its own orientation* (Fig. A2). In this situation the equation for shear strain has to be modified to take into account the different orientations of the slip bands. The rigorous way to do this would be as follows. (1) Determine the plastic strain %,~, using the same approach described previously for slip system 1. (2) Rotate the coordinate system x l , Yl, zl to align it with system xz, Y2, z2 where z z is the direction perpendicular to slip bands in slip system 2. In the coordinate system x2, Y2, zz, determine %2z2 using the relation (A2)
7 Y : z 2 = a y 2 iaz~ j "fij
where ay2iaz~ j is the rotation matrix and 7ij is the strain tensor in the x l , y~, zx coordinate system and is given by
7ii =
for ~. The three roots k l , ~2 and k 3 are the three principal strains et, e 2 and e 3 respectively. (5) Determine the effective strain
ii0 0j 0
7y,z,
%,y,
0
J
*Here, the hypothetical assumption is made that slip planes from the second slip system are also perpendicular to the facets of the block.
3
{(cl -
e 2 ) 2 + (e2 -
e 3 ) 2 + (e3 - e l ) 2 )
v2
As a result of all these steps we finally get one data point on the plot ~ = f(r). This procedure is simple in principle but complicated to implement, since the crystallographic orientation has to be determined for each grain. As an alternative way of estimating the a m o u n t of plastic deformation that occurred around the SiC particle, the following approximation is offered. First, the concept of c o m b i n e d plastic shear strain 7cpss is introduced. We shall define ~¢cpss as %pss
=
KNS
where S is the average step height. In general, the slip plane is not perpendicular to the surface and therefore the coefficient K is incorporated. K is a coefficient which will be evaluated later. N is the slip band density, which is obtained from the total slip band length of all slip systems divided by the area where the total slip band length is the length of all the slip bands enclosed in a selected area (e.g. a square mesh of some net). The selection of the size of the area is based on considerations of the scale. It is realized that the combined plastic shear strain %pss has no direct correlation with the effective strain ~, the octahedral shear strain %ct or any other
162
specific strain. However, the relative simplicity in the determination of %p~ and the fact that % p , takes into account direct proportionality between the amount of plastic deformation and the slip band density makes it very attractive for use as a criterion for plastic deformation in our case and in other similar cases in general.
Now we shall assume that the amounts of strain N1SM1, N2SM2 and NsSM3 are approximately equal because of the s y m m e t r y of the specimen and provided that no voids or cracks are created. Furthermore,
A.2. Determination o f the coefficient K The coefficient K takes into account the different crystallographic orientations of slip systems. Let us consider the case when we have three slip systems operating, i.e. we see three slip band groups. The individual slip systems contribute the strain N1S1, N2S2 and NsS3 to the total %p,, i.e.
or
%pss
1 K __ - - + sin 01
sin 01 SM2 s i n 02
and _
sin 03
1
+ ~1 sin 02 sin 03
Let us evaluate the range within which K can change by considering a unit triangle (Fig. A3). Let the directions shown on the triangle correspond to the poles of the surface of the sample. Then the values of K would be as shown in Table A1. The average value of K is a b o u t 4.5 when three slip systems are operating. Thus, the average correction factor per one slip system is 4.5/3 = 1.5. In general, slip band densities were measured along directions going through areas where two slip systems,.i.e, two sets of slip bands, were present. Thus for, two slip systems, K = 2 × 1.5 = 3.
&l
S3
1)
where
where N1, N2 andN8 are the slip band densities and $1, $2 and $3 are the average displacements on the band within each slip system, given by
_
1
÷--+ sin 01 sin 02
'~Cpss ~- KNSM
%p. = N1 1 + N25 + N35
$2
(1
NSM
SM3 sin 0a
where S i n , SM2and SM3 are the average heights o f the slip bands in the three slip systems, measured perpendicularly to the surface, and 01, 02 and 03 are the angles between the slip planes and the surface plane. Thus, 1 7cpss = N I ~ I
300~311 321 / 310 320
1
0----sin 1 ÷ N2SM2 sin 0 2 +
1
Fig. A3. Some orientations of the sample for which values of K were determined.
÷ N3SM3 - -
sin 83
TABLE A1 Values of K corresponding to the orientations shown in Fig. A3
Surface plane pole K for (111) slip plane
(111) 3.2
(110) 3.7
(100) 3.7
(210) 3.66
(211) 5.1
(221) 5.9
(310) 3.6
(311) 4.2
(320) 3.1
(321) 4.9
163 SiC and aluminum. If we did n o t have SiC in the center of the aluminum disk, the aluminum ring would shrink w i t h o u t any restraint and the size o f the bore would be reduced by
References for Appendix A A1 K. Yamaguchi, Sci. Paper Inst. Phys. Chem. Res. Tokyo, 8 (1928) 289; 11 (1929) 151. A2 A. F. Brown, J. Inst. Met., 80 (1951) 115. A3 A. H. Cottretl, Dislocations and Plastic Flow in Crystals, Clarendon, Oxford, 1965, p. 157.
Ua = Aa = a A a AT APPENDIX B
where a is the radius of the SiC, A s ( ~ 25 × 10 -6 K -1) is the difference between the coefficient of thermal expansion o f aluminum and t hat o f SiC, AT ( ~ 500 K) is the temperature interval and Ua is the displacement at t he interface. The reduct i on in the radius o f t he bore does n o t occur when SiC is present in the center o f the aluminum disk. The a m o u n t o f aluminum t h a t is n o t allowed to go towards the center of the bore would cause plastic flow o f the adjacent matrix in all directions away f r o m the SiC. Thus, some of the matrix will spill o u t (as shown in Fig. B1). As we go f u r t h e r away from the interface, the plastic flow is restricted by the matrix t hat surrounds t he central port i on. Two regions can be considered: t he plastic region t hat we assume is adjacent to the SiC and the elastic region t hat encloses t he plastic region. Let us make the following assumptions. (1) In the plastic zone the matrix is a perfectly plastic material, i.e. no w ork hardening 0 ccurs. (2) At the starting t e m p e r a t u r e T of 773 K, SiC and aluminum axe just in c o n t a c t with one another. (3) The profile o f t he aluminum which has spilt o u t is a straight line.
Among the several composite models described in th e literature, there is one t ha t appears to be close to our case; this is k n o w n as the long-composite-cylinder model. It consists of long c ont i nuous fiber surrounded b y the metal matrix. This m odel implies t hat plastic strain in the direction of the fiber (the z direction) resulted f r o m unequal expansion or c o n t r a c t i o n o f fiber and t h a t the matrix is a constant, i.e. ez = dl/l = constant where l is the length o f the fiber. Therefore, the total plastic strain state becomes a plane strain case which enables t wo o t h e r plastic strains e r and e 0 to be f o u n d w i t h o u t difficulty. Although short c om pos i te cylinders have the same s y m m e t r y as long composite cylinders, the strain in the z direction is n o t a constant but is in general a f u n c t i o n o f two variables: (1) t he distance f r o m the interface and (2) the position along the z axis. Since ez =/= constant, the longcylinder approach c a n n o t be used in o u r case. We shall discuss the case shown in Fig. B1. Let us consider an Al-SiC disk t h a t cools f r o m a t e m p e r a t u r e T. As a result o f cooling, a certain a m o u n t of aluminum will be pushed back because o f t he differential shrinkage of
F" ~ ~ - -
A,,,
I
"H"
"
PLANE
", --T "
I .
i
--i "o-
~'
!
/
I
l
Fig. B1. Schematic representation of the plastic-elastic shell around the SiC.
" PLANE
164
(4) T h e plastic-elastic f r o n t is also a straight line. N o w let us consider t h e radius p o f t h e plastic z o n e . It is a f u n c t i o n o f the vertical p o s i t i o n o f the transverse plane in which p is considered. According t o o u r a s s u m p t i o n , P = Po - - q z
(B2) with a general expression f o r p (eqn. (B1)) w h e n z = ~1 H Uz = A h
P0 -- 2(Po
r
--PH) 1 --
P0--2(P0--
p
1
g)~--a
1
Ah = ~H Ao~ A T and
(the equation of the straight line) where Po is t h e plastic z o n e radius in t h e " 0 " plane, which is t h e plane o f s y m m e t r y b e t w e e n t h e u p p e r and l o w e r halves o f t h e c o m p o s i t e disk. T o d e t e r m i n e q we let z = ~z H , i.e. t h e t o p o f t h e " H " plane is considered. When 1 z =- ~ H , p -- P H . T h e r e f o r e PH = PO - - q~ H
F r o m this 1 q = 2(P0 -- PH) H
1 _
1
Uz - EH A a A T
1 Aa
P = P0 -- 2(P0 -- PH) - ~
(B1)
L e t us consider n o w t h e p o r t i o n o f t h e m a t r i x t h a t spills o u t . F r o m the triangle s h o w n in Fig. B2 we can write
Po -- 2(Po
--
A T Po - - 2 ( P 0
Uz-~ z
--PH)~
z
-- r
P 0 _ 2 ( P 0
--PH)
p
_
1 -- r
1
(B2)
PH --a
where U~ is t h e vertical displacement. Generally speaking, the vertical d i s p l a c e m e n t U~ is a f u n c t i o n o f z a n d r, i.e. Uz = ¢ ( f ( z ) , g(r)} E q u a t i o n (B2) gives o n l y an expression f o r U= w h e n t h e " H " plane is considered. The p r o b l e m t h u s is t o find a general e x p r e s s i o n f o r U=. T o do this, let us replace pn in eqn.
for z = 1~ H or if we n o r m a l i z e z = z / H -- - 5z" L e t us c h e c k the values o f Uz at d i f f e r e n t z values. First, let z = 1 ( " H " plane), and so p 1 po--2(po-H)~--r U ~ = ~1 A ~ A T p 1 Po--2(Po-H){--a 1 U~ = ~ A~ A T = Ah
and, w h e n r
= PH ,
U~ = O
L e t z = 0 ( " 0 " plane), and so Uz = 0, since t h e " 0 " plane is the plane o f vertical symm e t r y . A t this p o i n t we can see a certain logic in eqn. (B4), w h i c h enables us t o suggest t h e general expression f o r Uz in the following form: U~ = z' A~ A T P0 - - 2 ( P o - - P H ) Z ' - - r Po -- 2(Po - - P H ) Z ' - - a
i °;'r
PH
(B4)
H)2 - a
T h e n , w h e n r = a,
PH - - r
U= = Ah - -
~
(B3)
PH) ~ - - a
If we simplify eqn. (B3) it reduces to eqn. (B2), b u t let us leave eqn. (B3) in its f o r m and e x a m i n e it in m o r e detail. First let us substit u d e H = 1 m m in eqn. (B3), since this is a c o n v e n i e n t way o f simplifying this expression and H = 1 m m h a p p e n s t o be t h e actual height o f o u r c o m p o s i t e disks. N o w we can rewrite eqn. (B3) as _
and n o w
Po -- 2(P0
_
"H" P L A N E
J
z = I/: ) H
Fig. B2. " S p i l l - o u t " p o r t i o n o f the a l u m i n u m m a t r i x .
(B5)
(since H = 1). It is realized t h a t eqn. (B5) lacks rigorous p r o o f . H o w e v e r , since it inc o r p o r a t e s o u r b o u n d a r y c o n d i t i o n s and since we d o n o t have a n y additional i n f o r m a t i o n , it is r e a s o n a b l e t o a d o p t eqn. (B5) and t o see w h a t results we can o b t a i n and t h e n t o judge t h e validity o f this expression. L e t us consider t h e cylindrical c o o r d i n a t e s y s t e m in which t h e z axis coincides with t h e z axis o f o u r sample. By d e f i n i t i o n
165
6r --
T o find t h e c o n s t a n t C we use t h e b o u n d a r y c o n d i t i o n U~ =a = Ua = a A a AT:
dr
Vr 60 = - -
Ua = a A o ~ A T
6 z --
C N a +--+--a 2 a 3
r
M
6z --
M
....
C = aU a 4---
dz
a 2--
N
2
2
a3
3
N o w eqn. (B10) b e c o m e s dz
= A~ A T
Po - - 2 ( P o - - P H ) Z
M Ur------r+
--r
+
Uaa
2
(B5)*
Ma 2 4-
N a 3 ~_Nr2_
(BII)
- -
r
2r
3r
3
Po--2(Po--PH)Z--a
2(P0
+z A~AT
--
PH) (a
--
N o w we can find the strains er, e0 and e~ f o r t h e " 0 " plane:
r)
[P0 -- 2(Po - - p ~ ) z - - a]
dU r _
I n c o m p r e s s i b i l i t y requires t h a t
6r
er + eo + e~ = O
Ur eo
or
dUr 4- Vr 4-
-dr
r
6 z ----0
(B6)
L e t us c o n s i d e r t h e " 0 " plane, w h e r e z = O; then PO - - r
e~ = A~ A T - Po - - a
(B7)
-
dUr+
Ur +
dr
-0
= 0
2
Ma 2
Uaa +
7
Na 2
2
2r 2 + --3r2 + --3 N r -t
Ma 2
Na s
-
- -
-
2r 2
3r 2
N + - - r
3
(B12)
L e t us consider t h e tip o f t h e plastic-elastic i n t e r f a c e , i.e. r = Po. T h e n (ez = 0 f r o m eqn. (B7)) 6r =--6 0
Upo
(BI3)
P0 (B8)
P0 - - a
M--Nr
r2
ez = M - - N r
_
Po - - r
Uaa
2 M -
r
E q u a t i o n (B6) t h e n b e c o m e s d U r + Ur + A a A T dr r
M
dr
(B9)
r
w h e r e Up0 is t h e d i s p l a c e m e n t o f t h e tip o f t h e plastic-elastic f r o n t :
Vr=p0 = Up0 Each p o i n t o f t h e plastic-elastic f r o n t is in t h e state o f i n c i p i e n t yielding, i.e.
where M = Aa
6-~- 60
Do
AT -
-
w h e r e g is t h e effective strain at t h e i n t e r f a c e and e 0 is t h e yielding strain t a k e n f r o m t h e yielding c o n d i t i o n o f t h e tensile sample in uniaxial loading:
Po - - a and 1 N = A~ AT - PO - - a
Oy = E 6 0
T h e s o l u t i o n o f eqn. (B9) is M
C
N
Ur=----r4---4---r
2
r
2
3
Therefore
(B10)
* F r o m here o n we o m i t t h e superscript prime in z ' and use z i n s t e a d t o p r e v e n t c o n f u s i o n , b u t it s h o u l d be b o r n e in m i n d t h a t t h e value z is n o r m a l i z e d w i t h r e s p e c t to H.
O'y 60 - ~ - - -
E
ii.73 62 × l 0 s 2 X 10 -4
166
where Oy = 11.73 MPa and E = 62 X 10 a MPa for pure aluminum. Thus we have a system o f equations:
up0
er (r =po ) --
Po
_Uo
eO(r=po)
-
-
(B14)
-
P0
ez(r=po )
0
=
metrical a b o u t the " 0 " plane, it goes through its local m a x i m u m or m i n i m u m point when it crosses the " 0 " plane and t herefore dUz/dz = 0 when z = 0. Thus we f o u n d that, within the " 0 " plane, all the shear c o m p o n e n t s of the strain tensor are equal to zero, which means t hat e r = el , eo = e2 and ez = ea for the " 0 " plane. Thus, if we go back to the yon Mises condition o f yield, the effective strain g can now be expressed as 21/2
Applying the yon Mises yielding condition,
g= 3
2~2 {~ : - - { ( e
I - - e 2 ) 2 -~- (e I - - e 3 ) 2 + (e 2 - - e 3 ) 2 } 1]2
3 where el, ez and e a are the principal strains. Generally speaking, the strains er, e0 and ez do n o t coincide with the principal strains el , e2 and ea, because o f the existence o f the shear c o m p o n e n t in the total strain tensor, which is given by
d~ 7zr =
dr
dUr dz
e=
3 (\Po]
\ Po /
\Po/
J
= 1.155 Up-~° Po However,
= 2X I0-4
The c o m p o n e n t s 70~ and %0 axe equal to zero because o f the s y m m e t r y around SiC. dU=
Aa A T z
dr
Po --2(P0 --PH)Z --a
/I,,'
= 1 . 1 5 5 Up0 P0 Therefore
for the " 0 " plane; when z = O, dUz/dr = 0. To evaluate dU~/dz, let us consider the region of the matrix adjacent t o the " 0 " plane as shown in Fig. B3. Point A is displaced to poi nt A', and p o in t B to point B'. (B is a mirror image o f A.) The displacements UAA'and Usa' are equal because o f the s y m m e t r y and f r o m the assumption th at no voids are created. The displacement Ur varies as we go along the z axis and presents a continuous f u n c t i o n of z. Since the f u n c t i o n is continuous and sym-
UP° Po
(BI5)
1.7 × 1 0 -4
From eqn. (BI1) we get M
UP°
=
2
U~a P o nu
Maz 't-
P0
Na3 --
2P0
3Po
N ~ --Po
3
2
and, substituting for M and N, we obtain
Upo__Aa_~TIOo2+aPo+a2 Po t 3
(a+Po)Pot_aZ } 2 (BI6)
A'
Equations (B15) and (BI6) can be solved as a system of two equations with two unknowns,
'
Up, and P0. Solving t hem , we obtain P0 = 1.15 mm. In order to find PH we can follow the same procedure. However, for t he " H " plane case t he strains er, eo and ez are n o t equal to el, e 2 and e a. Therefore the procedure has to be modified. The new procedure would be to find an expression for
,
AI
SiC
"0" PLANE
Ur
B~--~B
and, from eqn. (B14),
~=C 0
+ --
r
((er-- eo)2 + (eo - - e z ) 2 + (er--ez) 2 }1/2
Fig. B3. Displacements AA p and BB' of the matrix symmetrical a b o u t the " 0 " plane.
dUz + dUr dr
dz
167 1 for z = ~ a n d t h e n t o solve t h e e q u a t i o n
L%r
0
%.]
e0 --k
0
0
e z -- k
= 0
This is a cubic equation. Three roots give the values for the principal strains: ~'I = el k2
=
62
and •3
an idea o f t h e e x t e n t o f t h e plastic z o n e o n t h e surface o f t h e s a m p l e . A d d i t i o n a l l y , if t h e m a g n i t u d e o f %r is small, t h e d e v i a t i o n o f er, eo a n d ez f r o m e l , e2 a n d e 8 will also be small. 1 N o w , considering z = ~ a n d using t h e yielding c o n d i t i o n at t h e p l a s t i c - e l a s t i c f r o n t , we can f o l l o w steps similar t o t h o s e t h a t we used f o r t h e " 0 " plane. As a result PH = 1.11 m m . It is useful t o find a general e x p r e s s i o n f o r g in t h e " 0 " plane. A f t e r s u b s t i t u t i o n o f t h e n u m e r i c a l values f o r a a n d p, eqn. (B12) b e c o m e s , f o r z = 0,
= 63
This is s i m p l e in p r i n c i p l e b u t v e r y c o m p l e x to implement for the following reason. The individual c o m p o n e n t s o f this t e n s o r are o f t h e t y p e e x p r e s s e d b y eqns. ( B l l ) a n d (B15). If the coefficients of the cubic equation were n u m e r i c a l , t h e n w e c o u l d solve it using t h e t r i a l - a n d - e r r o r m e t h o d . In o u r case, all t h e c o e f f i c i e n t s h a v e a general e x p r e s s i o n . T h e general s o l u t i o n o f t h e c u b i c e q u a t i o n has t h e form
ez -
1.15 - - r 0.65
Aa AT
0. 513r 8 + 0.407 -- 0.885r 2 60 =
e~ =
r2
0.667r 8 -- 0.575r 2 -- 0.265 0.65r2 A~ A T
= 9 . 0 7 7 X 10 -8 X X (4.668r 2 -- 10.35r + 5.952 \
kl =
--2 +
+
+
Aa &T
0.531 r
0.42~ ~2
VJ
Each q and p would be a combination of the e x p r e s s i o n s f o r er, eo, e~ a n d %r- Because o f t h e d i f f i c u l t y o f d e t e r m i n i n g PH in a r i g o r o u s w a y w e can use an a p p r o x i m a t e s o l u t i o n b y t r e a t i n g strains e , eo a n d e~ as t h e principal strains e l , 62 a n d e 8. This implies t h a t we ignore t h e i n f l u e n c e o f t h e shear strain %~. It is realized t h a t t h e resulting value f o r PH will n o t be a t r u e value b u t it will at least give
This e f f e c t i v e strain is p l o t t e d as a f u n c t i o n o f r in Fig. 5. When ~ = eo = 2 X 10 -4 is substit u t e d in t h e e x p r e s s i o n a b o v e , t h e t h e o r e t i c a l plastic z o n e radius can b e o b t a i n e d f r o m r = p = 1.15 m m . T h e radius o f t h e f i b e r is 0.5 m m . T h u s p = 1 . 1 5 / 0 . 5 = 2.3 X f i b e r radius. This radius includes t h e f i b e r size. T h u s t h e e x t e n t o f t h e plastic z o n e m e a s u r e d f r o m t h e i n t e r f a c e is (2.3 - - 1) X f i b e r radius = 1.3 X f i b e r radius.