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Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 57

We’re going to ignore the rounded-off corners at the ends and the inside corner of the angle and consider the cross-section to be composed of two rectangles:

  1. a thin vertical rectangle, 0.75 inches wide and 7.25 inches high; and
  2. a thin horizontal rectangle, 0.75 inches wide and 4 inches long.

We’ll put the origin of our coordinate system at the top left corner of the cross-section with the x axis running to the right and the y axis running down. Combining the two rectangles, we get

A = A_1 + A_2 = 0.75 \cdot 7.25 + 0.75 \cdot 4 = 8.4375\,\rm{in^2}
\bar x = \frac{A_1 \cdot 0.375 + A_2 \cdot 2}{A} = \frac{8.0391}{8.4375} = 0.953\,\rm{in}
\bar y = \frac{A_1 \cdot (0.75 + 7.25/2) + A_2 \cdot 0.375}{A} = \frac{24.914}{8.4375} = 2.95\,\rm{in}

So the center of gravity is 0.953 in to the right of the top left corner, 2.95 in down from the top left corner, and (trivially) 3 ft in from either end.

Problem 58Problem 56

Last modified: January 22, 2009 at 8:32 PM.