Unformatted text preview: PBOBLEM 2-131 The top view of pin A of the pliers is shown. The cross- sectional area of the pin is 4.5 mmz. What is the average
shear stress in the pin when 150~N forces are applied to E
the pliers as shown? ‘
Free Body Qiagmm AB , l ’ 1” 150 N SOLUTION:
The easier way in which to ﬁnd the load in member AB is to draw the F80 of the lower handte and sum
moments about point D.
EM; = 0 = (150 N)(O.13 m) — FAB (sin 23.2“)(033 m)
PM = 1650 N (C) The shear load which must be supported by the pin atjoint A is the load in member AB. The average shear
stress is: raw = (Pas) I (A) z (1650 N) i (4.5 x10‘6m2)
ANS 3mg=367 16 = 57M ...
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- Spring '06