Lecture 12

6,864
-1

Published on

Published in: Business, Lifestyle
0 Comments
5 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
6,864
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
0
Comments
0
Likes
5
Embeds 0
No embeds

No notes for slide

Lecture 12

  1. 1. Today’s objectives-Composites II <ul><li>Be able to calculate the upper and lower bounds for the Young’s modulus of a particle composite. </li></ul><ul><li>Be able to calculate the critical length, and the ideal length, of a fiber for a fiber composite. </li></ul><ul><li>Similarly, be able to calculate the strength of the matrix-composite bond (or the matrix shear strength, whichever is weakest) for a fiber composite. </li></ul><ul><li>Be able to calculate the longitudinal and transverse stresses, strains, and elastic moduli (Young’s) for fiber composites. </li></ul>
  2. 2. Types of composites
  3. 3. Review from last class-Particulate Composites <ul><li>For particle-reinforced composites, the elastic modulus (Young’s modulus, E) will fall between an upper (best case) and lower (worse case) limit: </li></ul>
  4. 4. Fiber Reinforced <ul><li>Most common composite type. </li></ul><ul><li>Generally applied for improved strength and stiffness with respect to weight </li></ul><ul><ul><li>Aerospace applications </li></ul></ul><ul><ul><li>High value sporting goods </li></ul></ul><ul><ul><li>Remember the reason for using a composite is that the load is transferred from the matrix to the composite element (here a fiber). </li></ul></ul><ul><li>Since the load cannot be transferred beyond the end of the fiber, a critical fiber length (L c ) likely exists for effective strengthening and stiffening of the overall part </li></ul><ul><ul><ul><li>too small and it doesn’t transfer the load </li></ul></ul></ul><ul><ul><ul><li>too long and it just costs a lot without any added benefit </li></ul></ul></ul>d L c
  5. 5. Fiber composites <ul><li>Why are we using fibers? </li></ul><ul><ul><li>Especially for ceramics, due to Weibull statistics the fracture strength of a small part is usually greater than that of a large component (smaller volume=fewer flaws=fewer big flaws). </li></ul></ul><ul><li>Fibers come in three forms: </li></ul><ul><ul><li>Whiskers (graphite, SiC, Si 3 N 4 , Al 2 O 3 ) </li></ul></ul><ul><ul><ul><li>Single crystals </li></ul></ul></ul><ul><ul><ul><li>Huge length/diameter </li></ul></ul></ul><ul><ul><ul><li>Small, so nearly flaw free </li></ul></ul></ul><ul><ul><ul><li>Strongest known materials </li></ul></ul></ul><ul><ul><ul><li>expensive </li></ul></ul></ul><ul><ul><li>Fibers (aramids, glass, carbon, boron, Si 3 N 4 , Al 2 O 3 ) </li></ul></ul><ul><ul><ul><li>Polycrystalline or amorphous </li></ul></ul></ul><ul><ul><ul><li>Small diameter </li></ul></ul></ul><ul><ul><li>Wires (usually metals) </li></ul></ul><ul><ul><ul><li>Large diameter </li></ul></ul></ul>
  6. 6. Matrix phase <ul><li>Usually a metal or polymer since some ductility is desirable </li></ul><ul><li>Serves several functions for fiber composites </li></ul><ul><ul><li>Bonds with the fibers. </li></ul></ul><ul><ul><li>Protects fibers from surface damage due to abrasion or corrosion (ie avoid cracks on surfaces of fibers). </li></ul></ul><ul><ul><li>Separates the fibers. </li></ul></ul><ul><ul><li>Prevents propagation of brittle cracks between fibers (ie matrix is usually ductile). </li></ul></ul>
  7. 7. Fiber Reinforced <ul><li>Critical fiber length (L c ) depends on: </li></ul><ul><ul><li>σ f * , the fiber ultimate tensile strength (in Pascals) </li></ul></ul><ul><ul><li>d, the fiber diameter </li></ul></ul><ul><ul><li>tau c , either the matrix/fiber bond strength or the matrix shear yield strength ( whichever is smaller ), ( in Pascals ) . </li></ul></ul>L c is approximately 1 mm for glass fiber / Carbon matrix composites (20 to 150 * diameter). <ul><li>Therefore, the bond between matrix and fiber often dictates whether the fiber will improve the properties of the composite by transferring an applied load to the fiber. </li></ul>d L c
  8. 8. Stress along a fiber <ul><li>For L=L c , the optimal fiber load is achieved at the center of the fiber length. </li></ul><ul><li>For L>L c , the optimal fiber load is carried by most of the fiber. These are considered to be “Continuous” fibers and are ideal . </li></ul><ul><li>For L<L c , the optimal fiber load is never reached, so that a weaker, cheaper fiber or even particles could have been used instead. </li></ul>
  9. 9. Optimal fiber length <ul><li>Technically we only need L>L c , but practically optimal fiber lengths are usually about 30*L c . </li></ul>Poorer fiber efficiency Better fiber efficiency • Ex: L c is 1mm for glass fiber in the C matrix, so the optimal fiber length is about 30mm. <ul><li>Thinner fibers help too: as d decreases, the critical length decreases as well (saving money). </li></ul>fiber diameter shear strength of fiber-matrix interface or the matrix itself (whichever is smaller) fiber strength in tension
  10. 10. Continuous and aligned fibers <ul><li>For longitudinal loading, assuming a ductile matrix (e.g. metal) and a brittle fiber (ceramic with no ductility): </li></ul><ul><ul><li>Stage I includes the fiber and matrix both deforming elastically E= σ / ε </li></ul></ul><ul><ul><li>Stage II involves plastic deformation of the matrix and elastic deformation of the fiber. The load actually on the fiber is increased. </li></ul></ul><ul><li>Fibers begin to fail at the failure strain. </li></ul><ul><li>The process is not catastrophic since: </li></ul><ul><ul><li>the fibers are not all the same size and strength. </li></ul></ul><ul><ul><li>the matrix and matrix/fiber bond are still intact except at the fracture points. </li></ul></ul>F
  11. 11. Longitudinal loading <ul><li>The load sustained by the composite is shared by the matrix and the fibers. </li></ul><ul><li>The stress on each component is easily calculated and depends on the relative areas of each component (or volume since V is proportional to A in this orientation). </li></ul><ul><li>If the fiber/matrix bond is good then the strain on fiber and matrix is the same ( isostrain) . </li></ul><ul><li>If deformations are all elastic (stage I), Young’s modulus is proportional to stress/strain. </li></ul><ul><li>Modulus of elasticity is thus easily calculated. </li></ul><ul><li>This is the upper bound of fiber composite properties. </li></ul><ul><li>The force on the fiber vs. the matrix can be determined as well. </li></ul>Remember: E= modulus (GPa), sigma = stress (MPa), epsilon = strain (ratio) F
  12. 12. Transverse loading <ul><li>Now the load is applied at 90 ° to the fiber axis. </li></ul><ul><li>No longer isostrain (same strain on matrix and fiber) </li></ul><ul><li>Instead, isostress applies (stress applied to composite is same as that applied to each component ). </li></ul><ul><li>Stain on each component sums to give the overall strain. </li></ul><ul><li>Again, if deformations are all elastic (stage I), Young’s modulus is proportional to stress/strain. </li></ul><ul><li>This is the lower bound of particulate composite properties. </li></ul>F
  13. 13. Tensile strength review: <ul><li>Longitudinal </li></ul><ul><ul><li>Maximum stress on stress/strain curve </li></ul></ul><ul><ul><li>Usually when the first fibers begin to break </li></ul></ul><ul><ul><li>Once fibers begin to break, the load is transferred to the matrix. </li></ul></ul><ul><ul><li>Longitudinal tensile strength can be calculated. </li></ul></ul><ul><li>Transverse </li></ul><ul><ul><li>The transverse tensile strength is usually at least one order of magnitude less than the longitudinal strength. </li></ul></ul><ul><ul><li>The matrix properties will dominate, along with the fiber/matrix bond strength. </li></ul></ul>
  14. 14. Particle and Fiber variables <ul><li>Once the matrix and disperse phase materials are decided, there are still many options that may effect properties. </li></ul>Volume increases Thinner and longer is better. Longitudinal or transverse equations apply
  15. 15. SUMMARY <ul><li>Be able to calculate the strength of the matrix-composite bond (or the matrix shear strength, whichever is weakest). </li></ul><ul><li>Be able to calculate the longitudinal and transverse stresses, strains, and elastic moduli (young’s) for fiber composites. </li></ul>Reading for next class Short Fibers, Structural Composites Chapter sections: 16.8, 16.10-15

×