Materials for Biomedical Engineering. Mohamed N. RahamanЧитать онлайн книгу.
the strain can also be defined in two ways. For a specimen of length l that is deformed by an infinitesimally small value dl, the true strain ε is defined as the integral of all the infinitesimally small strains, that is
(4.18)
where, lo is the initial length of the specimen. The relation between the true strain and the engineering (nominal) strain εn is
(4.19)
The true stress σ, that is, the applied load divided by the instantaneous area, can also be found from the engineering or nominal stress σn and εn according to the equation
(4.20)
For small strains, such as strains in the elastic region of most solids but not in rubbers, a distinction between the true value of the stress and strain and their engineering (nominal) value is not relevant because their true value is almost equal to their engineering value. On the other hand, at large deformations in the ductile region, it is important to distinguish between true and engineering values for stress and strain.
Example 4.1
Tensile testing of a metal alloy using a cylindrical dog‐bone specimen of gage length 60.00 mm and diameter 12.8 mm yielded the following data:
| Force (N) | Length (mm) |
|---|---|
| 0 | 60.00 |
| 15 442 | 60.06 |
| 30 883 | 60.12 |
| 46 325 | 60.18 |
| 61 766 | 60.24 |
| 77 208 | 60.30 |
| 92 649 | 60.36 |
| 105 517 | 60.42 |
| 107 834 | 60.48 |
| 109 506 | 60.54 |
| 110 150 | 60.60 |
| 110 664 | 60.66 |
| 108 992 | 60.72 |
| 106 547 | 60.78 |
Determine (a) the Young’s modulus, (b) the elastic limit, and (c) the yield strength of the metal alloy.
Solution:
As the elongation at the highest force is small, only 1.3% of the original length, the engineering values of the stress and strain are almost identical to their true values and, thus, we can use the engineering values. While the data can be converted to stress and strain, it is not necessary in solving this problem as the area remains nearly constant during the deformation. It is often preferable to determine the requisite properties from a plot of force versus elongation but due to the simple nature of the data in this problem, we will determine them directly from an examination of the data.
1 The Young’s modulus is the slope of the linear part of the stress versus strain curve. Examining the data we find that the force increases linearly with elongation up to a force of 92 649 N. The stress σ at this point is given by where d = 12.8 × 10−3 m is the diameter of the specimen. Substituting for d gives σ = 720 × 106 Pa or 720 MPa. The strain ε at this point is the elongation divided by the original length, equal to 0.36 ÷ 60.00 = 0.006 (or 0.6%). The Young’s modulus is 720 × 106 Pa ÷ 0.006 = 120 × 109 Pa or 120 GPa.
2 The stress–strain curve becomes nonlinear above a force of 92 649 N. Although the changeover from elastic to plastic deformation cannot be accurately determined, to a high approximation, the elastic limit is the stress or strain found in part (a), that is σ = 720 MPa or ε = 0.006.
3 Based on the slowly varying force above the elastic limit, this material does not show a clearly defined yield point. Thus, the yield stress σy will be taken as the stress at an offset strain of 0.002 or an offset elongation of 0.12 mm. Starting at this elongation, a line parallel to the linear portion of the force versus elongation data will intersect the force data at approximately 109 506 N. Dividing this force by the cross‐sectional area of the specimen gives σy = 851 MPa.
It is left as an exercise to confirm these mechanical properties from a plot of force versus extension.
Viscoelasticity
When metals and ceramics are subjected to a constant stress below their elastic limit at temperatures relevant to biomedical applications and the majority of engineering applications, the resulting strain remains constant with time. In comparison, if a constant tensile stress is applied to a polymer specimen, the resulting strain does not remain constant but will increase slowly with time (Figure 4.4a). This slow extension with time is called creep. It is due to rearrangement of the mobile sections of the polymer chain in response to the stress. If the stress is removed, the polymer chains return slowly to their original conformation and the strain decreases slowly to zero. This strain recovery in polymers has its origins in the overall entropy of the polymer chains. Rearrangement of the polymer chains during deformation leads to a decrease in their entropy. Consequently, upon removal of the stress, there is a driving force for the chains to return to their higher entropy conformation. If a polymer specimen is extended to a certain constant strain, the tensile stress required to maintain this strain is not constant but decreases slowly with time (Figure 4.4b). This effect is called stress relaxation.
Figure 4.4 Linear viscoelastic behavior of polymers. (a) In creep, a constant stress σ applied at t = 0 leads to time‐dependent strain ε (t); at a given time, the strain increases linearly with the applied stress. (b) In stress relaxation, a constant strain ε applied at t = 0 leads to a time‐dependent stress σ (t); at a given time, the stress increases with increasing strain.
Creep and stress relaxation are manifestations of a general property of polymers called viscoelasticity. The overall response is a combination of two responses:
An elastic response because, given sufficient time, the solid recovers completely upon removal of the stress
A viscous response that results in creep or stress relaxation, in much the same way that a very thick syrup of high viscosity will deform slowly