Determine handle diameter , lever arm cross-secion , and journal diameter if :
effort = 400 N
bending sterss = 50 N/mm^2
bear stress = 40 N/mm^2
Let D = Diameter of the handle
Bending moment, M = P × L = 400 × 450 = 180 × 10³ N-mm
The twisting moment depends upon the point of application of the effort.
T = 400 × 100 = 40 × 10³ N-mm
We know that equivalent twisting moment,
"T_e = \\sqrt{(M\u00b2+T\u00b2)} = 184.4 \u00d7 10\u00b3 N-mm"
We also know that equivalent twisting moment (Te
),
"184.4 \u00d7 10\u00b3 = \\dfrac{\u03c0}{16} \u00d7 Torque \u00d7 D\u00b3"
"D\u00b3 = \\dfrac{184.4\u00d7 10\u00b3}{10.8} = 17.1 \u00d7 10\u00b3"
D = 25.7 mm
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