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LONG-TERM LOADING Generally, long-term loading is
either a constant applied load or a constant induced strain. Plastic parts subjected to a constant load, such as pressure vessels or structures supporting weight, tend to creep and show increased deformation over time. Other design elements, such as a press-fit boss or spring finger, undergo continuous, fixed deformation or strain. These features stress relax over time and show a loss in retention force. See the Long-Term Mechanical Properties section in this chapter for an explanation of creep and stress relaxation.
Creep data, such as isochronous stress-strain curves, provide a means for predicting a material’s behavior. Figure -7 shows a typical set of time-dependent curves at 40°C for 0% Glass- Fiber reinforced polyamide 6 resin. Each curve represents the material behavior for different loading durations. To predict creep, substitute an apparent modulus for the instantaneous elastic or Young’s modulus in structural calculations. Many people confuse actual modulus and creep modulus. Except for environmental effects the material’s elasticity does not decrease over time; nor does its strength. Because of visco- elasticity, deformation occurs over time in response to a constant load. While the instantaneous tensile modulus of the material remains constant, the apparent modulus decreases over time (see figure -8). We use this hypothetical, time-dependent creep modulus to predict the amount of sag or deformation that occurs over time.
Isochronous Stress-Strain Figure 3-27
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Stress relaxation is the decrease in stress that occurs in a material that is subjected to constant, prolonged strain at a constant temperature. Measuring stress relaxation involves varying the load over a period of time to maintain a constant strain rate. This test is more difficult than the test for creep that measures the change in deflection over time in a specimen under constant stress. For this reason, creep curves are often used to calculate stress relaxation, generally resulting in a ±0% margin of error.
To find the apparent modulus from isochronous strain-strain data, divide the calculated stress by the corresponding strain on the curve for the selected load duration. For example, if a flat part made of 0% GF Nylon 6 at 90°C (see figure -7), has a tensile stress of ,000 psi ( MPa) and a load duration of ,000 hours, you can calculate an apparent modulus of 00,000 psi from the isochronous stress-strain curve. Significantly lower than the instantaneous value of 600,000 psi, this lower apparent modulus will account for the added deflection that occurs because of creep when it is substituted into deflection calculations.
Creep Modulus vs. Time Figure 3-28
Creep modulus for Durethan BKV30, 30% Glass-Fiber reinforced Polyamide 6 at 90°C. Creep (apparent) modulus decreases over time, but the actual modulus remains constant.
For a given strain, read vertically through the isochronous stress-strain curves to predict the effects of stress relaxation. Again, using the curves in figure -7, you can see that for an applied strain of %, the tensile stress drops from an instantaneous value of 8,400 psi (58 MPa) to approximately 5,00 psi (5 MPa) after 0,000 hours.
Page 7 of 68: This document contains important information and must be read in its entirety. Example 3-6:
Plate Deflection Considering Creep Find the deflection in the circular plate of example 3-3 after 10,000 hours at 90°C. The geometry and loading are shown in figure 3-15.
As in the short-term case, the first step is to calculate the stress. Because the stress calculation does not depend on modulus, the result is the same as in example 3-3:
σmax = 4,90 psi
To find the appropriate modulus value requires a set of isochronous stress- strain curves at 90°C as shown in figure 3-27. On the 10,000 hour curve, a stress of 5,100 psi corresponds to roughly 2% strain. Calculate the apparent (creep) modulus by dividing stress by strain. Use the result of 255,000 psi to calculate the actual deflection after 10,000 hours.
δmax= pr4(5-4ν-ν) 6Ecreept
= (75)(0.75)4[5-4(0.8)-(0.8)] 6(55,000)(0.0) = 0.067 inches
The deflection at 10,000 hours is more than double the instantaneous value of 0.0113 inches!
Table 3-7
Example 3-7: Stress Relaxation
A permanently deflected ABS cantilever snap arm is used to hold a metal part in position. The arm is 1 inch long, 0.080 inch thick and 0.25 inch wide. The deflection of the arm is 0.1 inch. What is the instantaneous retention force of the arm? After four days (~102 hours)? After six weeks (~103 hours)?
First, find the strain level in the arm from the formula shown below. This can be derived from y = PL3 / 3EI (Table 3-5) and letting E = σb / ε.
ε = yh = (0.)(0.08) = .% strain L (.0)
Now, using figure 3-10, find the stress corresponding to 1.2% strain on the desired time curves. Then calculate the relaxation modulus (Er = σ / 0.012) and find the retention force using P = 3ErIy / L3. The results are shown in Table 3-7.
Note that the drop off in retention force is proportional to the drop in stress. For a given strain, the stress drops a similar amount during each logarithmic increase in time period. The initial retention force drops by 38% in the four days, then by an additional 20% over the next six weeks. For this reason, designs that rely on such retention forces are not recommended in thermoplastics.
STRUCTURAL DESIGN
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