CHAPTER I. Definitions and General Introduction. STRESSES. Place a brick on end and rest on top of the brick a weight of 200 pounds. (Fig. 1) This weight causes a downward pressure against the brick. In the same time there is developed in the brick an internal resistance of 200 lbs. acting against the weight and keeping the same in equilibrium; otherwise the weight would crush the brick and move downward. The weight represents an external force acting upon the brick. The internal resistance developed in the brick by this weight is called a "stress." We can therefore say, that: "A stress is an internal resistance which balances an external force.” An exterior force acting on a body tends to produce a deformation or change in the shape of the body. We call "Strain or deformation" the change in shape or the distortion caused in a body by an external force acting upon it. Three kinds of simple or direct stresses may be produced by external forces; these are: Tension, caused by forces stretching or tending to pull a body apart. Compression, caused by forces tending to push together or shorten a body. Shear, caused by forces tending to cut across. In all cases unit stress is the stress per unit area. For instance: If a brick 2x4x84 in, stands on one end and a weight of 240 pounds is rested on it, the unit compressive stress equals 30 pounds per sq. in. In the same way, if a weight of 2,000 pounds is hung from a rope having a cross-section of 1⁄2 sq. in., the unit tensile stress equals 2,000÷2=4,000 pounds per sq. in. Consider two plates riveted together, as in Fig. 2. When the plates are in tension, they tend to shear the rivet or cut it. across the area between the two plates. Let the tension in each plate be 2,209 lbs. The cross-sectional area of a 3/4 in. rivet is 3.14Xdiam.=3.14X34=.4418 square ins. The total shear tending to cut the rivet across is 2,209 pounds. This shear divided by the cross-section resisting it is the unit shear. We therefore have: Unit shear-2209.4418 5,000 pounds per square inch. Elastic Limit and Ultimate Strength. When a gradually increasing force is applied to a bar, the deformation of the VIMU bar increases in proportion to the force within certain limits. For instance: If a steel bar one sq. in. in cross-section and 100 in. long is subjected to a gradually increasing tensile force, it elongates as shown in this table: with a tension of 6000 lbs. the bar elongates .02 in. Notice that the elongation was so far proportional with the stress; when the stress was doubled, the deformation also W=200 lbs. 200 200 200 lbs. Fig. 1. doubled. However, with a tension of 42,000 fbs. the bar elongates .15 in. or .16 in., or in other words, the deformation increases faster than the force applied. Elastic Limit is that unit stress at which the deformation. begins to increase in a faster ratio than the stress. When a body is stressed below the elastic limit, upon. removing the force, the body will spring back to its original shape and length. When a body is stressed above the elastic limit, upon removing the force the body does not acquire its original shape and length, but it remains permanently de formed or it acquires a permanent Set. This shows that stressing a body beyond its elastic limit is injurious to the elasticity and strength of the body and should never be allowed in practice. Returning to the above illustration, if the tensile force is Fig. 2-Single Shear. Fig. 3-Double Shear. increased beyond 42,000 lbs., the bar will elongate more and more until finally rupture of the bar takes place. Ultimate strength is the unit stress which occurs just before rupture, and it is the highest unit stress that a bar can bear. The ultimate strength varies with different materials, and is from two to four times the elastic limit. For some materials the ultimate strength is higher in compression than in tension. Unit strain or unit deformation is the deformation per unit length. For instance, in the above table, with a stress of 30,000 lbs. the total elongation in 100 in. was .10 in. and the unit strain or elongation per inch of length of the bar was .10÷100.001 in. Since up to the elastic limit the deformation is nearly uniform throughout the bar, the elongation in any portion of the bar, i. e., in 8 in. will equal 8X.001=.008 in. Modulus of Elasticity is a number which results by dividing unit stress by unit strain and is a constant number for stresses below the elastic limit. For instance, in the last table with 6,000 lbs., per sq. in. of bar area or with a unit |
