ARCH 461/561 Lecture 20 Example 1

Lecture 20
Example Problem

Distributed Loads

Given:
the uniformly distributed loads indicated below.
Determine:
the resultant load for each of the uniformly distributed loads.
Lecture 20

Solution:
Both systems are uniformly distributed loads. To find the resultant of a uniformly distributed load simply multiply the magnitude of the load in #/ft by the length of the load. The line of action of the resultant of a uniformly distributed load will alw ays act through the midpoint of the original load.

I. The resultant of the loads of the system at the left are found by taking the load of 100 #/ft and multiplying it by the total length of the beam, 6 feet. This yields a total of 600 pounds of total load and acting down through the midpoint of the lengt h or:

100#/ft (6ft) = 600# acting 3 ft from A

II. The problem on the right can be solved by considering each distributed load separately. The resultants for each of the loads is found first, and then they can combined to determine the resultant of the system. First,

600#/ft (2ft) = 1200 #

This force acts 1 ft to the right of A.
Second,

400#/ft (4ft) = 1600#

This force acts at 1ft to the left of B or 6 ft from A.
These forces can be combined in ways discussed in earlier examples to find the resultant for the system.

Lecture 20 Example Problem

Lecture 20
Example Problem