(a) Bending Stress (really a combination of tension on one side and compression on the other).

(b) Torsional or Twisting Stress (a form of shearing stress).

127. Ultimate Strengths.-By taking specimens of the different materials and loading them until they break, it has been possible to find out just what each kind of material will stand. The load to which each square inch of cross-section must be subjected in order to break it, is called the Ultimate strength of the material. The strength of most materials differs for the different methods. of loading shown in Fig. 83.

The Tensile Strength of a material is the resistance offered by its fibers to being pulled apart.

The Compressive Strength of a material is the resistance offered by its fibers to being crushed.

The Shearing Strength of a material is the resistance offered by its fibers to being cut off.

The following table gives the average values for the most used materials.

128. Safe Working Stresses. Having found how great a stress is required to break one square inch of material, we naturally would not allow anywhere near this stress to come on a piece of material in actual service. The Ultimate Strength is usually divided by some number, known as the Factor of Safety, and the quotient is used as the Safe Working Stress.

For example, if 60,000 lb. per square inch will break a piece of soft steel and we use a factor of safety of 5, this would give:

Safe working stress =

60000
5

= 12000 lb.

per square inch.

The following table gives the Safe Working Stresses of the most used materials.

Instead of writing "safe loads in pounds per square inch" for tension, compression, or shear, the symbols St, Sc, and S, are used. So if Aarea in square inches, then the load W which can be carried safely = Area × safe load per square inch or

Perhaps more often we would want to find the area necessary in order to support a certain weight or load. In this case, we would want a formula which would give A.

If we divide the total load by the safe stress, we will get the necessary area; or

=

This simply says, area of metal necessary total weight to be carried divided by safe load in pounds per square inch. From the area of a bolt or rod, its diameter can be easily found.

129. Strengths of Bolts.-There is a well-known saying that a chain is only as strong as its weakest link." This means, in general, that any mechanism must be so designed that its weakest part will be strong enough to stand the greatest load that may come on it. In figuring the size of a bolt to hold a certain load, we would not calculate the full diameter of the bolt and make the area there just sufficient, but we must see to it that the bolt has a cross-sectional area at the root of the threads large enough to support the load. Then the body of the bolt will have a surplus of strength.

Example:

What size of steel eyebolt will support a weight of 5000 lb.? Take 12,000 lb. as the safe load in tension.

.416 sq. in. is then the necessary area to support the weight. Of course, the example could be completed by saying .7854 D2= .416 sq. in., where D= diameter at the root of the thread. By then solving for D we would get the diameter at the root of the threads. But the Bolt Tables afford an easier method than this. In the following table, .4193 is given as the area of a 3 in. bolt at the root of the thread. Therefore, a in. eyebolt would probably be used.

In figuring the allowable loads for steel bolts, it is best not to allow over 12,000 lb. stress per square inch and 10,000 lb. is perhaps even more usual on account of the sharp root of the threads, which makes a bolt liable to develop cracks at this point.

BOLT TABLE.-U. S. S. THREADS

130. Strength of Hemp Ropes.—It is quite common in calculating the strength of ropes and cables to assume that the section of the rope is a solid circle. Of course, the strands of the rope do not completely fill the circle but, if we find by test the allowable safe strength per square inch on this basis, it will be perfectly safe to make calculations for other sizes of ropes on the same basis. The safe working stress based on the full area of the circle is 1420 lb. per square inch. The Nominal Area (as the area of the full circle by which the rope is designated is called) is A = .7854XD2. The safe stress is 1420 lb. per square inch and, consequently, the weight that can be supported by a rope of diameter D is

W=SXA

= 1420X.7854 XD2

Here we have two constant numbers (1420 and .7854) that would be used every time we were to calculate the safe strength of a rope. If this were to be done often we would not want to multiply these together every time, so we can combine them now, once and for all.

1420.7854-1120, approximately

Hence

Example:

W=1120XD2

Find the safe load on a hemp rope of 1⁄2 in. diameter.

131. Wire Ropes and Cables.-For wire ropes made of crucible steel, a safe working load of 15,000 lb. per square inch of nominal area is allowable. For cables of Swedish iron but half this value should be used.

132. Strength of Chains.-It has been demonstrated by repeated tests that a welded joint cannot be safely loaded as heavily as a solid piece of material. Of course, there are often welds that are practically as strong as the stock, but it is not safe to depend on them. For this reason, the safe working load per

square inch for chain links is often given as 9000 lb., which is just of 12,000 lb.

If D

the diameter of the rod of which the links are made A=2X.7854 XD2

W = S1XA

W=9000X2X.7854X D2

Combining the constant numbers, this can be simplified into W=14,000 X D2

This is used in the same way as the formula for a rope.

133. Columns.-The previous examples were cases of tension. The size of a rod or timber subjected to compression is computed in the same way unless it is long in comparison with its thickness. When a bar under compression has a length greater than ten times its least thickness, it is called a Column and must be considered by the use of complicated formulas which take account of its length. It can be seen by taking a yardstick, or similar piece, that it is much easier to break than a piece of shorter length but otherwise of the same dimensions. A long piece, when compressed, will buckle in the center and break under a light thrust or compression. An example of this can be found in the piston rod on a steam engine, where, on account of the length of the rod, it is necessary to use much lower stresses than those given in the tables. The compressive stress allowed in piston rods varies with the judgment of different designers but is generally about 5000 lb. per square inch, using a pressure on the piston of 125 lb. per square inch.

Example:

Find the size of rod for a 30 in. by 52 in. Corliss engine with 125 lb. steam pressure.

30 in. is the diameter of the cylinder and 52 in. is the stroke, which is not considered in the problem except in that it has reduced the allowable stress in the rod.

.7854 X 302=706.86 sq. in., area of piston. 706.86X125=88357.5 lb., total pressure on piston.

Using 5000 lb. per square inch, allowable stress in the rod.

883585000=17.67 sq. in. sectional area of rod,

From the table of areas of circles, it is seen that this is the area of a circle nearly 4 in. in diameter, so we would use a 4 in. rod.

PROBLEMS

Note. In all examples involving screw threads, to get areas at root of thread, use the table given in this chapter. Give sizes of bolts always as diameters.