Strength of Materials. 9/18/07. Topics to Cover. Pressure Stress Strain Scale and Square-Cube Effect Tension Compression Shear. Effect of force on non-rigid objects. Besides the force, what else is important? Examples to consider: Ice on a pond Hammering a nail into wood Pressure!
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8" x 8" post.
Determine:a. the compression stress in the 8" x 8" post b. the bearing stress between the post and the steel plate c. the bearing stress between the steel plate and the concrete.
Solution:a. The compressive force in the post : 6400 pounds/ 64 in2 = 100 psi b. The stress at the point between the post and the steel bearing plate is dictated by the smallest area being loaded. The stress between the two is again 6400 pounds divided by 64 square inches for a total stress of 100 psi. c. The stress between the plate and the concrete floor is also determined by the smallest area. In this case it is the 10" x 10" plate: 6400 pounds / 100 in2. The resulting stress is 64 psi, a clear reduction in the stress.
Solution:a. The internal force in that member is 1000 lb tension and the tension stress (intensity of the force per unit of area) is 1000 pounds divided by the area of 1 square inch. This is a stress of 1000 psi. b. The 1000 lb load is now distributed evenly across an area of 4 square inches; thus, 1000 lb divided by 4 in2 is a stress of 250 psi. This clearly demonstrates the inverse relationship between stress and area.
Lecture for 2/8/07