Shear
Flow in Beams
Example: * (continued) |
Note: Q is the first moment of
the area between where the shear stress is to be calculated |
Therefore
Q = A’ ybar
= (25)(130)(130 + 12.5) =
4.63 x 106 mm3
V
= FI/Qs where F =
1200 N, I
= (1/12)(260)(310)3
– (1/12)(160)(260)3 = 411 x 106 mm4 s
= 100 mm V
= Vmax =
(1200)(411 x 106) / (4.63 x 106 )(100) =
10,650 N (result) Click here for another
example. |
Return to Notes on Solid Mechanics |
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