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# Introduction to materials

Chapter 4 polymer composites
4.1 overview
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4.2 mixing principle
The mixture we discussed must satisfy the following conditions:
The fiber must be tightly bonded to the substrate. We generally regard the combination of fiber and matrix as chemical bond. But in fact, this kind of bonding is only physical, that is to say, the matrix material infiltrates into the irregular surface of the fiber to form% 26 × 8220; Interlock% 26 ×
The fibers must be continuous or lapped in the length direction. If the short fibers are not overlapped in the length direction, a section without fibers will appear. When the stress exceeds the ultimate strength of the matrix, it will fracture in this section first. Compounding with fibers is meaningless. Conversely, if the fiber is lapped, the stress will be transferred from the matrix to the adjacent fiber, and the fiber will bear the stress, which greatly improves the strength.
There is a critical fiber volume fraction V f crit, above which fiber reinforcement can occur.
There is a critical fiber length above which reinforcement can occur. The critical length depends on the fiber diameter DF.
Let's make a simple mathematical analysis of fiber composites
4.2.1 the continuous fiber is parallel to the external force
When the fiber length is long enough to overlap in any section, we consider the fiber to be continuous. When the stress acts on the length direction of the fiber arrangement [Fig. 4-3 (a)], the stress conforms to the mixing law: SC = vfsf + VM SM (4-1), where V represents the volume fraction, s represents the stress, F and M represent the fiber and matrix respectively. Substituting VM = (1-vf) into SC = vfsf + (1-vf) Sm (4-2) in most composites, vfsf is the controlling factor. It is assumed that the strains on the two components are equal. If only elastic deformation occurs, SF = EF, SM = em, SC = EC, where e is the strain. So we have: EC = VF EF + VM EM em. Since all strains should be equal, the modulus of the composite is: EC = EF VF + Em VM = EF VF + (1-vf) em (4-3) Fig. 4-3 prediction of the modulus of the composite with different fiber orientations (a) e1 = EC（ b) When E2 = EC, the stress acting along the fiber length direction is called parallel action, and similar action occurs when the material is under heating load or electric load. Because the load on any section of the composite is evenly distributed, the strain on each fiber and matrix is the same. The contribution of EF and VF to the composite modulus EC is the main one. If the modulus of the fiber is two orders of magnitude higher than that of the matrix, the contribution of the matrix can be ignored. The above equation is derived from continuous fibers, but it is also applicable if the fibers are long enough to cause enough laps. This is because the stress at the end of the fiber is transferred to the adjacent fiber through the matrix. This problem can be seen more clearly in Figure 4-4. The distribution of normal stress and shear stress along the fiber length is shown in the figure. At the end of the fiber, the normal stress is zero and the shear stress is very high, which leads to the stress transfer to the adjacent fiber. Due to the coupling effect between fiber and matrix, the fracture of single fiber will not lead to the failure of the whole composite. When the two fiber ends are far away from each other (Fig. 4-4b), in order to resist this movement, a plastic flow parallel to the stress occurs in the matrix, resulting in shear stress in the matrix. Of course, unbroken fibers have to bear more stress Figure 4-4 (a) normal and shear stresses acting along the fiber axis（
4.2.2 continuous fiber perpendicular to external force
When the stress is perpendicular to the fiber axis, an equal stress state occurs [Fig. 4-3 (b)]. In this state, the cross-sectional area of the fiber is small enough, and it is impossible for the fiber to make a significant contribution to the strength of the composite unless the volume content of the fiber is very high; 30%）。
4.2.3 plastic deformation of matrix
In most cases, when the stress exceeds the yield strength of the matrix, the fiber will not undergo plastic deformation. This is especially true when the matrix is plastic. The ultimate strength of the composite can be expressed as SCU = SFU VF + S% 26 × 8217; M (1-vf) (4-7), where SFU is the ultimate tensile strength of the fiber, S% 26 × 8217; M is the flow stress of the strain hardening matrix. The ultimate strength of the composite is higher than that of the matrix (SCU% 26 # 179; SCU% 26 # 179); smu）。

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