Shaft Deflection
Single Point Shaft Deflection In applications where a support rail is not used, shaft deflection can become critical in the function of the bearing. If deflection is greater than the misalignment capabilities of a standard pillow block, binding can occur. One solution would be to increase shaft and bearing size (to lessen the amount of deflection) or to use an open bearing configuration with a support rail. Follow the formulas below to check shaft deflection and sag.
Total Deflection Eample ø1 in. Shaft 24 in. Length (L) 250 lb. load (W) F • L 3 48E I = 250 lb. • (24 in.) 3 48(1.45 • 10 6 ) in. 2 lb. = 3,456,000 in. 3 lb. 69,600,000 in. 2 lb.
Deflection =
Deflection at Center
1/2 L
F
Displacement Angle
Deflection plus Sag
Deflection =
0.0497 in.
Formula for Inch and Metric Shafting Deflection Total shaft deflection in horizontal applications: Total Deflection ( δ ) = Def + Sag = FL 3 + 5wL 4 48E I 384E I Deflection = (FL 3 ) / (48E I) Sag = (5wL 4 ) / (384E I) Deflection = Deflection due to load at center of shaft* Sag = Deflection of shaft due to its own weight* L = Shaft unsupported length* F = load being applied at center of shaft** a = Displacement angle
5wL 4 384E I
SAG
=
= 5(0.222)(24 4 ) lb. in 3 384(1.45 • 10 6 ) lb. in 2
368,271 lb. in 3 556,800,000 lb in 2
SAG
=
SAG 0.000661 in. Total Deflection ( δ ) = Deflection + SAG =
E I = (Modulus of Elasticity, E ) • (Second Moment of Area, I ) w = Unit weight of shaft. (lb/in for inch shafts and N/mm for metric shafts)
= 0.0497 in. + 0.000661 in.
* inch for inch shafts, mm for metric shafts ** lb for inch shafts, N for metric shafts Notes: I = π • diam 4 / 64 sw = π • diam 2 / 4 • density
Total Deflection ( δ ) = 0.0504 in.
E = Modulus of Elasticity (Young’s modulus)
pbclinear.com • Engineering Your Linear Motion Solutions 145
Made with FlippingBook - professional solution for displaying marketing and sales documents online