Machinery's Handbook, 31st Edition
Deflection in Beam Design 273 Deflection as a Limiting Factor in Beam Design.— For some applications, a beam must be stronger than required by the maximum load it is to support in order to prevent exces sive deflection. Maximum allowable deflections vary widely for different classes of ser vice, so a general formula for determining them cannot be given. When exceptionally stiff girders are required, one rule is to limit the deflection to 1 inch per 100 feet of span; hence, if l = length of span in inches, deflection = l ÷ 1200. According to another formula, deflection limit = l ÷ 360 where beams are adjacent to materials like plaster, which would be broken by excessive beam deflection. Some machine parts of the beam type must be very rigid to maintain alignment under load. For example, the deflection of a punch press column may be limited to 0.010 inch or less. These examples merely illustrate variations in practice. It is impracticable to give general formulas for determining the allowable deflection in any specific application because the allowable amount depends on the condi- tions governing each class of work. Procedure in Designing for Deflection: Assume that a deflection equal to l ÷ 1200 is to be the limiting factor in selecting a wide-flange (W-shape) beam having a span length of 144 inches. Supports are at both ends and load at center is 15,000 pounds. Deflection y is to be limited to 144 ÷ 1200 = 0.12 inch. According to the formula on page 257 (Case 2), in which W = load on beam in pounds, l = length of span in inches, E = modulus of elasticity of material in psi, and I = moment of inertia of cross section in inches 4 : , . , , , . y EI Wl I yE Wl 48 48 48 0 12 29 000 000 15 000 144 2681 Deflection hence 3 3 3 # # # = = = = A structural wide-flange beam, see Steel Wide-Flange Sections on page 2694 , having a depth of 12 inches and weighing 35 pounds per foot has a moment of inertia I of 285 and a section modulus ( Z or S ) of 45.6. Checking now for maximum stress s (Case 2, page 257 ): . , , s Z Wl 4 4 460 15 000 144 11 842 lbs/in 2 # # = = = Although deflection is the limiting factor in this case, the maximum stress is checked to make sure that it is within the allowable limit. As the limiting deflection is decreased, for a given load and length of span, the beam strength and rigidity must be increased, and, consequently, the maximum stress is decreased. Thus, in the preceding example, if the maximum deflection is 0.08 inch instead of 0.12 inch, then the calculated value for the moment of inertia I will be 402; hence a W 12 3 53 beam having an I value of 426 could be used (nearest value above 402). The maximum stress then would be reduced to 7640 lb/in 2 and the calculated deflection is 0.076 inch. A similar example using metric SI units is as follows. Assume that a deflection equal to l ÷ 1000 mm is to be the limiting factor in selecting a W-beam having a span length of 5 meters. Supports are at both ends and the load at the center is 30 kN. Deflection y is to be limited to 5000 ÷ 1000 = 5 mm. The formula on page 257 (Case 2) is applied, and W = load on beam in N; l = length of span in mm; E = modulus of elasticity (assume 200,000 N/mm 2 in this example); and I = moment of inertia of cross section in mm 4 . Thus,
3
Wl
Deflection y
=
48
EI
hence
, 48 5 200 000 30 000 5000 78 125 000 mm , , , 3 # # #
3
Wl
I
4
yE = Although deflection is the limiting factor in this case, the maximum stress is checked to make sure that it is within the allowable limit using the formula from page 257 (Case 2): 48 = =
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