ARCH 461/561 Lecture 20 Example 2

Lecture 20
Example Problem

Triangular Loads

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

Solution:
The total load for the problem at the left is determined by the area under the loading envelope. Since this load is an iscosoles triangle simply take 1/2 the base and multiply it by the maximum magnitude of the loading which is equal to the height;

1/2 (6 ft)(1200 #/ft) = 3600# down
Since the loading is symmetrical, the resultant will act along the centerline of the triagular loading, or 3 ft from A.

The hydrostatic problem on the right is a bit more complex. First note that water weighs 62.5 pcf and that deeper beneath the surface of water, the greater the pressure and loading on the indicated wall. The greatest magnitude of the load is taken as th e weight of water multiplied by the depth; 62.5 pcf x 5 feet = 375 psf at the base of the wall. The total load is then found in a manner similar to above. Multiply 1/2 (base)(greatest magnitude of the force).

1/2 (5ft)(375 #/ft^2) = 937.5#/ft of wall
Note that this equation gives the force acting on a piece of wall 5 ft high by 1 ft wide simply because of the nature of the weight of water being a volumetric measurement.

The resultant of the load will pass throught the cetroid of the area represented by the load. This is at a point 1/3 the distance from the greater side of the triangle and perpendicular to the member that is being loaded. In this case 5ft / 3 or 1.66 ft from the bottom.

Another way to determine this point is to draw a line from each point of the triangle to the midpoint of the opposite side. The point at which these lines intersect is also the centroid of that area.

Lecture 20 Example Problem

Lecture 20
Example Problem