This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

Nearly all the information revealed by such a test can be well represented in a diagram called a stress-deformation diagram. It is made as follows: Lay off the values of the unit-deformation (fourth column) along a horizontal line, according to some convenient scale, from some fixed point in the line. At the points on the horizontal line representing the various unit-elongations, lay off perpendicular distances equal to the corresponding unit-stresses. Then connect by a smooth curve the upper ends of all those distances, last distances laid off. Thus, for instance, the highest unit.

elongation (0.20) laid off from o (Fig. 4) fixes the point a, and a perpendicular distance to represent the highest unit-stress (83,000) fixes the point b. All the points so laid off give the curve ocb. The part oc, within the elastic limit, is straight and nearly vertical while the remainder is curved and more or less horizontal, especially toward the point of rupture b. Fig. 5 is a typical stress-deformation diagram for timber, cast iron, wrought iron, soft and hard steel, in tension and compression.

12. Working Stress and Strength, and Factor of Safety. The greatest unit-stress in any part of a structure when it is sustaining its loads is called the working stress of that part. If it is under tension, compression and shearing stresses, then the corresponding highest unit-stresses in it are called its work-ingr stress in tension, in Com-pression, and in shear respectively; that is, we speak of as many working stresses as it has kinds of stress.

By working strength of a material to be used for a certain purpose is meant the highest unit-stress to which the material ought to be subjected when so used. Each material has a working strength for tension, for compression, and for shear, and they are in general different.

By factor of safety is meant the ratio of the ultimate strength of a material to its working stress or strength. Thus, if Su denotes ultimate strength, Sw denotes working stress or strength, and f denotes factor of safety, then f = Su/Sw; also Sw = Su/f (3)

When a structure which has to stand certain loads is about to be designed, it is necessary to select working strengths or factors of safety for the materials to be used. Often the selection is a matter of great importance, and can be wisely performed only by an experienced engineer, for this is a matter where hard-andfast rules should not govern but rather the judgment of the expert. But there are certain principles to be used as guides in making a selection, chief among which are:

Fig. 4.

1. The working strength should be considerably below the elastic limit. (Then the deformations will be small and not permanent.)

Fig. 5. (After Johnson.)

2. The working strength should be smaller for parts of a structure sustaining varying loads than for those whose loads are steady. (Actual experiments have disclosed the fact that the strength of a specimen depends on the kind of load put upon it, and that in a general way it is less the less steady the load is.)

3. The working strength must be taken low for non-uniform material, where poor workmanship may be expected, when the loads are uncertain, etc. Principles 1 and 2 have been reduced to figures or formulas for many particular cases, but the third must remain a subject for display of judgment, and even good guessing in many cases.

The following is a table of factors of safety* which will be used in the problems:

Factors of Safety.

Materials. | For steady stress. (Buildings.) | For varying stress. (Bridges.) | For shocks. (Machines.) |

Timber | 8 | 10 | 15 |

15 | 25 | 30 | |

Cast iron | 6 | 15 | 20 |

Wrought iron | 4 | 6 | 10 |

Steel | 5 | 7 | 15 |

They must be regarded as average values and are not to be adopted in every case in practice.

Examples. 1. A wrought-iron rod 1 inch in diameter sustains a load of 30,000 pounds. What is its working stress? If its ultimate strength is 50,000 pounds per square inch, what is its factor of safety ?

The area of the cross-section of the rod equals 0.7854 X (diam-eter)2=0.7854 X 12=0.7854 square inches. Since the whole stress on the cross-section is 30,000 pounds, equation 1 gives for the unit working stress

S=30,000/0.7854 = 38,197 pounds per square inch.

Equation 3 gives for factor of safety f=50,000 / 38,197 = 1.3

2. How large a steel bar or rod is needed to sustain a steady pull of 100,000 pounds if the ultimate strength of the material is G5,000 pounds ?

The load being steady, we use a factor of safety of 5 (see table above); hence the working strength to be used (see equation 3) is

S = 65,000/5 = 13,000 pounds per square inch.

The proper area of the cross-section of the rod can now be com-puted from equation 1 thus:

*Taken from Merriman's "Mechanics of Materials."

A =P/S=1000,000/13000=7.692 square inches.

A bar 2X4 inches in cross-section would be a little stronger than necessary. To find the diameter (d) of a round rod of sufficient strength, we write 0.7854 d2 = 7.692, and solve the eqation for d; thus: d 2=7.692/0.7854=9.794, OR D=3.129 inches.

3. How large a steady load can a short timber post safely sustain if it is 10 x 10 inches in cross-section and its ultimate compressive strength is 10,000 pounds per square inch ?

According to the table (page 12) the proper factor of safety is 8, and hence the working strength according to equation 3 is

S=10,000/8=1,250 pounds per square inch

The area of the cross-section is 100 square inches; hence the safe load (see equation 1) is

P = 100 X 1,250 = 125,000 pounds.

4. "When a hole is punched through a plate the shearing strength of the material has to be overcome. If the ultimate shearing strength is 50,000 pounds per square inch, the thickness of the plate ½ inch, and the diameter of the hole ¾ inch, what is the value of the force to be overcome?

The area shorn is that of the cylindrical surface of the hole or the metal punched out; that is

3.1416 X diameter X thickness = 3.1416 X¾ X½ = 1.178 sq. in. Hence, by equation 1, the total shearing strength or resistance to punching is

P = 1.178 X 50,000 = 58,900 pounds.

Continue to: