ARCH 461/561 Lecture 20 Example 3

Lecture 20
Example Problem

Trapezoidal Loads

Given: the trapezoidal distributed load indicated below.
Determine: the resultant load.

Lecture 20

Solution: As indicated earlier, it is critical to recognize how a problem can be broken down into smaller problems that are more easily solved. This is a very simple case, and one in which the final result is easily determined in parts, but almost impossible if attempted as a whole.

Find the resultants of the two loads. First, the rectangular load;

(200#/ft)(6ft) = 1200# acting at midspan
Next, find the resultant of the triangular load

1/2 (6ft)(300#/ft) = 900#
This force acts at 2 feet from the right end of the beam.

They can now be combined and the reactions found. The total load is equal to the sum of the two smaller resultants or 1200# + 900# = 1200#. Find its location using sum MA:

     2100#(x) + B(6ft) = 1200#(3ft) + 900#(4ft) + B(6ft)
              2100#(x) = 3600#ft +3600#ft
                 2100x = 7200#ft
                     x = 3.43 ft from A

So, the resultant load equals 2100# acting 3.43 ft from A.

Lecture 20 Example Problem

Lecture 20
Example Problem