Load Distribution Example Problem

Lecture 19
Example Problem

Load Distribution

Given:
the floor framing diagram below with a total design load of 50 psf (10 psf Dead Load + 40 psf Live Load); the floor joist spacing is 2 feet on center; the joists are 18 feet long; the beams are 16 feet long
Determine:
the total load on each of the columns.

Solution:
The tributary area of one joist is the area supported by that joist; the area equals half the distance to the adjacent joists to either side multiplied by the full length of the joist. The joists rest on beams and provide the load on the beams. In this example, each beam supports one end of each joist (usually beams support joists on each side, rather than on only one side as shown in this example.) The area contributing to the beam load is half the area supported by the joists. The beams are supported in turn by columns; for this problem, the tributary area of a column is half the area supported by a beam.

joist loading

Each foot along the length of a joist supports an area spanning two feet. This is the center to center spacing of the joists.

The load on a one foot length of a joist is 50 psf x 2 sqf/ft = 100 plf (pounds per linear foot). This is the design load on the floor multiplied by the tributary area for the joist per linear foot. Therefore, the total load supported by one joist equals the linear load multiplied by the length of the joist or 100 plf x 18 feet = 1800 pounds.

Lecture 19

Since the joists are supported by the beams, the sum of the reactions at the ends of the joists is the total load on the beams. Since the joists are symmetrically loaded, they distribute their load equally to the supporting beams at either end. The concentrated load on a beam from one joist is 900 pounds every two feet. Loads from the joists on their supporting beams are usually considered to be uniformly distributed if there are five or more equally loaded and spaced joists.

Lecture 19

Assume that the load in this example is uniformly distributed.
What is the load/foot on the beam? The load on the beam is equivilent to the total of each point load contributed by the joists "spread" out over the length of the beam. That is, 900 pounds every 2 feet for 16 feet which is a total load of 900 pounds x 8 = 7200 pounds. If this was distributed over the entire beam, it would be a linear load of 450 pounds/ft.

What is the load on a column?
The load on the column has the same value as the reaction at the beam end. The reaction from the element above becomes the action on the element below. Thus, the load on the column is 3600 pounds; 7200 pounds divided by two.

Lecture 19 Example Problem

Lecture 19
Example Problem