Email address: [e-mail]

Clarence McChristian

9711 Parkfield Dr

Austin, Texas 78758

United States

Email address: [e-mail]

Telephone no: 01 512 837 2572

Mobile no: 01 512 554 3969

Date Of Birth: 24/07/1953 Clarence E. McChristian B.S.

The school year is over for NEISD in San Antonio. I am seeking an AutoCAD

job in your area. My work experience includes Aerospace, Avionics, and Architectural;

residential and commercial, Civil/Survey one year, Mechanical and sheet metal.

I have 15 years working in high tech manufacturing, and as an instructor, I have 9 years AutoCAD teaching and course development experience at the adult vocation level.

For the last six months I have been substitute teaching Algebra, Trig, Reading Music and Art courses for San Antonio Texas in the North East Independent School District.

APPLICATIONS OF GRAPHICAL INTEGRATION

Integration used to study the strength of materials to determine shear, moments and deflections of beams. A truck exerts a total force of 36000 lb on a beam that is used to span a portion of a bridge.

Step 1.) Determine the resultants supporting each of the beams. Draw a force diagram using Bow’s notational system, laying out the vectors in sequence. Locate Pole point “O”, draw lines from the end of the vectors to “O”.

Step 2.) Draw ray “OA” in the associated interval of the funicular diagram. Continue for each ray. Ray “OE” is transferred to the vector diagram by drawing a parallel through point “O” to intersect the vector diagram.

Note: The vector “DE” is the right-end resultant of 20.1 kips (one kip equals 1000 lb).

Vector “EA” is the left-hand resultant of 15.9 kips.

Step 3.) The origin of the resultant forces is found by extending “AO” and “OD” in the funicular diagram to their point of intersection. In this case 36 kips (the sum of “AO” and “OD”).

Step 4.) Using the load diagram use integration to find the shear diagram. This indicates points in the beam where failure due to crushing is most critical. These concentrated loads will yield a shear diagram composed of straight-line segments.

Step 5.) Draw the left-end resultant of 15.9 kips to scale from the axis. The first load of 4 kips, acting in a downward direction, is subtracted from 15.9. The second load of 16 kips exerting a downward force is subtracted from 11.9 kips (15.0-4). The third load of 16 kips exerting a downward force is also subtracted. The right-end resultant brings the shear diagram back to the X-axis.

Note: The beam must be designed to withstand maximum shear at each support and minimum shear at the center.

Step 6.) The moment diagram is used to evaluate the bending characteristics of applied loads in foot-pounds at any interval along the beam. Any X-value in the moment diagram represents the cumulative foot-pounds in the shear diagram as measured from either end of the beam.

Step 7.) Using the Pole point a rectangular area of 200 ft-kips is found in the shear diagram. The estimated total area to be less than 600 ft-kips; use a convenient scale unit of 600 for the moment diagram.

Step 8.) Locate the area of 200 ft-kips in the moment diagram by projecting the 20-foot mark on the X-axis until it intersects with a 200-unit projection from the Y-axis, locating point “K”. The diagonal, “OK”, is transferred to the shear diagram, drawn from the ordinate of the given rectangle to point “P” on the extension of X-axis.

Note: Rays “AP”, “BP”, “CP” and “DP” are found in the shear diagram by projecting horizontally from the various values of shear. In the moment diagram, draw these rays in their in respective intervals to form a straight-line curve that represents the cumulative area of the shear diagram in units of ft-kips.

Conclusions: Maximum bending will occur at the center of the beam, where shear is zero. The bending is scaled to be about 560 ft-kips. The beam selected for this span must be capable of withstanding a shear of 20.1 kips and a bending moment of 506 ft-kips.

Дата добавления: 2015-11-05