Stress and Strain in Plastics Machinery's Handbook, 31st Edition
567
F
F
W
L
Fig. 7a. Adhesive Lap Joint
F
F
Rivet
t
D
L
Fig. 7b. Riveted Lap Joint In Fig. 7b , a similar lap joint is joined with a rivet of diameter D . In this case, the shear area in the rivet is p D 2 ⁄4, a much smaller area. But what if the rivet were aluminum and much stronger than the plastic being joined? A second, very different shear area comes into play, shown as a free body in the lower portion of Fig. 7b , the small bar-end volume adjacent to the rivet. The area shown shaded is equal to L 3 t and is 1 ⁄ 2 of the shear area, so the total shear area equals 2 Lt . Thus, the shear stress t exerted by the rivet and trying to push the plug out of each bar end is F /(2 Lt ). As the force F is increased, the mode of failure in the rivet or the bar end will be decided by which of these shear stresses first exceeds the shear strength of the material on which it is acting. Geometries like those in Fig. 7a and Fig. 7b are not useful for determining shear modulus because there is very little recordable shear deformation prior to failure at the shear strength. However, the very simple shear geometry shown in Fig. 8 is used in ASTM D732 test to measure the shear strengths of plastics. The washer-shaped test sample has an outer diameter of 2 inches (50.8 mm) and a central hole of about 9 ⁄ 16 inch (14.3 mm) to allow the half-inch (12.7 mm) bolt to pass freely through the specimen. The thickness may be any measured value between 0.05 and 0.5 inch (1.27 and 12.7 mm). In this test, the shear area is given by p 3 punch diameter 3 disk thickness. The results of such tests are often described in manufacturers’ marketing data sheets as the shear strength of the material. The shear strength reported from such a test is not a pure shear strength because a considerable part of the load is transferred by bending or compressing, or both, rather than by pure shear, and results can be affected by the susceptibility of the material to the sharpness of the load faces in the test apparatus. Thus, the test cannot be used to develop shear stress-strain curves or to determine the shear modulus. Shear strengths reported from this test have coefficients of variation of about 4 percent. When analyzing plastics in a pure shear situation or when the maximum shear stress is calculated in a complex stress environment, designers often use a shear strength value of about half the tensile strength, or the direct shear strength obtained from the test referred to above, whichever is least.
Copyright 2020, Industrial Press, Inc.
ebooks.industrialpress.com
Made with FlippingBook - Share PDF online