Machinery's Handbook, 31st Edition
590 DESIGNING PLASTIC PARTS When the plastics selection for the part has been finalized, the suppliers of the plastic resin can provide recommendations of grades and wall thicknesses appropriate to those grades. Table 8 shows typical nominal wall thicknesses for various thermoplastics. Table 8. Typical Nominal Wall Thicknesses for Various Classes of Thermoplastics Thermoplastics Group Typical Working Range Thermoplastics Group Typical Working Range inch mm inch mm Acrylonitrile-butadiene-styrene (ABS) 0.045–0.140 1.14–3.56 Polyester elastomer 0.025–0.125 0.64–3.18 Acetal 0.030–0.120 0.76–3.05 Polyethylene 0.030–0.200 0.76–5.08 Acrylic 0.025–0.150 0.64–3.81 Polyphenylene sulfide 0.020–0.180 0.51–4.57 Liquid-crystal polymer 0.008–0.120 0.20–3.05 Polypropylene 0.025–0.150 0.64–3.81 Long-fiber-reinforced plastics 0.075–1.000 1.91–25.4 Polystyrene 0.035–0.150 0.89–3.81 Modified polyphenylene ether 0.045–0.140 1.14–3.56 Polysulfone 0.050–0.150 1.27–3.81 Nylon 0.010–0.115 0.25–2.92 Polyurethane 0.080–0.750 2.03–19.05 Polyarylate 0.045–0.150 1.14–3.81 Polyvinyl chloride (PVC) 0.040–0.150 1.02–3.81 Polycarbonate 0.040–0.150 1.02–3.81 Styrene-acrylonitrile (SAN) 0.035–0.150 0.89–3.81 Polyester 0.025–0.125 0.64–3.18 … … … If the plastics part is to carry loads, load-bearing areas should be analyzed for stress and deflection. When stress or deflection is too high, solutions are to use ribs or contours to increase section modulus; to use a higher-strength, higher-modulus (fiber-reinforced) material; or to increase the wall thickness if it is not already too thick. Where space allows, adding or thickening ribs can increase structural integrity without thickening walls. Equations (15), (16), and (17) on pages 569–570 can be related to formulas using the section modulus and moment of inertia on page 257, where for Case 2, stress at the beam center is given by s = − Wl ⁄4 Z . On page 256, note that Z = I 4 distance from neutral axis to extreme fiber ( h 4 2 in the plastics example). The rectangular beam section diagrammed on page 242 gives the equivalent of I = bh 3 ⁄12 for the rectangular section in the plastics example. Therefore, Z I bh h bh 12 2 6 3 2 # = = = h ⁄ 2 In s = − Wl ⁄4 Z , the ( − ) sign indicates that the beam is supported at the ends, so that the upper fibers are in compression and the lower fibers are in tension. Also, W = F and l = L in the respective equations, so that stress, s = FL ⁄4( bh 2 )⁄6, and s = 3 FL ⁄2 bh 2 . To calculate the maximum deflection y at load, equate Equation (16) from page 569 and Y = Wl 3 ⁄48 EI from page 257, where W = F , l = L , E = E , and I = bh 3 ⁄12. Therefore, y E ( bh FL Ebh FL 48 12) 4 3 3 3 3 = = ⁄ Example: Assume that a beam (see Fig. 17) as described in connection with Equations (15), (16), and (17) on pages 569–570 is 0.75 inch wide with a constant wall thickness of 0.08 inch, so that the cross-sectional area is 0.06 in 2 , and there is a central load W of 5 lb. Based on a bending or flexural modulus of 300,000 lb/in 2 , the maximum stress is calcu lated at 6250 lb/in 2 and the maximum deflection at 0.694 inch. Both the stress and the deflection are too high, so a decision is made to add a rib measuring 0.04 inch thick by 0.4 inch deep, with a small draft of 1 ⁄ 2 degree per side to reinforce the structure. The equations in the BEAMS section starting on page 256, the drawing on page 248 representing the ribbed section (neglecting fillets), and the accompanying formulas permit calculation of the maximum stress and deflection for the ribbed section. With the new cross-sectional area only slightly larger at 0.0746 in 2 , the calculated stress is reduced to 2270 lb/in 2 , and the deflection goes down to 0.026 inch, which is acceptable for both the material and the application. To achieve the same result from a heavier beam would require a thickness of 0.239 inch, tripling the weight of the beam and increasing molding difficulties. The rib adds only 24.3 percent to the total section weight.
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