stress of 6,000 lbs. per sq. in. the unit elongation was .0002 in. and the modulus of elasticity for this bar is equal to 6,000+ .0002=30,000,000.

With a unit stress of 12,000 lbs. per sq. in. the unit elongation = .0004 in. and this gives a modulus of elasticity equal to 12000÷.0004-30 mil. This shows that the modulus of elasticity is constant below the elastic limit. The modulus of elasticity is used in figuring out deformations when the stresses are given and vice-versa.

Factor of safety is the number obtained by dividing the ultimate unit strength by the actual unit stress. For instance: A one inch square steel bar supports in tension a load of 5,000 Ibs.. If the ultimate strength or breaking load is 65,000 fbs. per sq. in. then the factor of safety =65000÷5000=13.

The factors of safety must be greater for varying loads than for steady loads. In table on page 5 usual factors of safety are given for steady loads such as in buildings; for varying stresses as in bridges and for shock as in machinery.

Working Unit Stress is the ultimate strength divided by the factor of safety. For instance: Let four be the factor of safety for steel fixed by law or by specifications. If the ultimate strength in tension is 65,000, the working or allowable unit stress equals 65000÷4=16250 lbs. per sq. in. A flat bar 2 in. X I in. section for instance, will not be loaded in tension with more than 2X16250=32500 fbs.

The working unit stresses are higher for dead loads than for variable or moving loads or for impact. The working stress must always be considerably lower than the elastic limit to prevent injury to the material through over-straining.

Coefficient of linear expansion is the increase per unit length of a bar when the temperature of the bar is increased by one degree Fahrenheit. For instance, a steel bar a foot long at 40° F. becomes 1.0000065 ft. at 41° F. The increase per degree of a unit length is therefore .0000065 for steel. This number is the coefficient of linear expansion for steel.

The definitions given in this chapter are of great importance and the reader is advised to thoroughly master them before proceeding any further.

CHAPTER II.

Practical Problems on Stresses and Strains.

To further illustrate the meaning of the various terms defined so far and the relations existing between them, several practical problems are here given. In all problems that follow the values given in the following table shall be used:

I. A square steel bar 2 in. on each side is subject to a tension of 80,000 lbs. To find the unit tensile strength and the factor of safety.

Area of cross-section of bar =2X2=4 sq. in. Unit tensile stress =80000÷4=20000 lbs. per sq. in. The factor of safety equals ultimate strength per sq. in. given in the table as 65000, divided by the actual unit stress; hence: Factor of safety 65000 20000 3.25 or a little over 3.

2. A round cast iron bar 2 in. diam. carries in tension 18850 lbs. Find the unit tensile strength and the factor of safety. Area of cross section of bar =.7854× diam.Xdiam.= .7854=2X2=3.14 sq. in. Unit tensile stress 18850÷3.14 6000 lbs. per sq. in. approximately. Factor of safety =20000 ÷6000=3 1/3. As the factor of safety for cast iron is generally not less than 6, the bar is unsafely loaded.

3. A cast iron block 12X12X2 rests flat on a concrete pier. A column erected on top of this block carries 28800 lbs. Assuming that the block distributes the column load uniformly upon the concrete pier, find the pressure in lbs. per sq. in. on top of the pier. Area of bottom of block =12X12 144 sq. in. Unit pressure on concrete 28800÷144=200 lbs. per sq. in.

4. A wrought iron square bar is to carry in tension 40,000 lbs. with a factor of safety of five. Find its cross section. Ultimate strength for wrought iron in tension 50000 lbs. per sq. in. Since the factor of safety is five, the allowable working load =50000÷5=10000 lbs. per sq. in.

Total required area = 40000 lbs. 10000 lbs. = 4 sq. in. The bar will therefore be a 2 in. square wrought iron bar. 5. A square steel block is to carry a column load of 115,200 lbs. The block rests on a concrete pier. Find the area of the block so that the pressure upon the concrete pier shall not exceed 200 lbs. per sq. in. Ans.: The block must be not less than 2 ft. - o in. X 2 ft. o in. on bottom.

6. Find the diam. of a round steel bar to carry in tension 240,000 lbs., with a factor of safety of four. Ans.: 4.3 in., Use a bar 4 5/16 in. diam.

7. A cast iron square block is to carry 210,000 lbs. in compression. The allowable working stress is 15,000 fbs. Find the area of the block and the factor of safety. Ans.: Area 14 sq. in.; factor

=

6.

8. A square steel block is to carry 280,000 lbs. in compression. What should be its size in order that the unit stress may be one third of the elastic limit? Ans.: 24 sq. ins., or a square about 5 in. X 5 in.

9. A bar of cast iron 3 in. diameter ruptures under a tension of 141,372 pounds. What is its ultimate strength in pounds per sq. inch? Ans.: 20,000 lbs. per sq. in.

IO. A load of 250,000 lbs. is to be carried in tension by means of a round bar. If the factor of safety is five; design a cast iron round bar to support the above load. Design also a wrought iron and a steel round bar, using the same factor of safety. Ans.: Diameters are 8.91 ins. for cast iron; 5.64 ins. for wrought iron; 4.95 ins. for steel.

II. What force will be required to rupture in tension a 3/4 in. round steel bar? Ans.: 28,717 pounds or about 14 tons.

12. A cast iron bar one square inch in cross-section weighs 3.1 lbs. per foot. Find the length of a vertical bar

which ruptures under its own weight when hung on its upper end? Ans.: 6450 ft.

13. In a shearing machine a flat steel bar 2 in. × 1⁄2 in. was sheared exactly at right angles to its length. Find the shearing force? Ans.: 50,000 pounds or about 25 tons.

14. Find the ultimate strength in shear for a 3/4 in. diam. steel rivet, also 5%, 2 and 3? Ans.: 22,090 lbs.; 15,340 lbs.; 9820 lbs.; 5525 lbs.

15. The allowable shearing stress on steel rivets is 10,000 lbs. per sq. in. What is the factor of safety for shear? Ans. 5.

16. How much load could four 3/4 in. steel rivets carry in direct shear with a factor of safety of five? Ans.: 17,672 pounds or 81⁄2 tons.

Fig. 4-Cast Iron Column resting on two
Steel Beams.

17. A cast iron column rests in the middle of two 12 in. steel beams, 6 feet long (Fig. 4). If the beams get an equal share of the load and the column carries 18 tons how many 3/4 in. bolts are required in each end connection of the two 12 in. beams? Ans.: As shown in the figure, there will be a load of 41⁄2 tons or 9,000 pounds at the end of each beam. A 3/4 in. bolt will carry in shear 70% of the amount carried on a 34 in. shop rivet, or 70% of 4418 3000 pounds approximately. It will, therefore, be necessary to provide not less than three 3/4 in. bolts in the ends of each beam.

=

18. A derrick rests at the middle of two 15 in. beams. There are two temporary 3/4 in. bolts in each end of each beam. Should the derrick be used to hoist 16 tons of steel at once? Explain fully? Ans.: The cross-sectional area of a 3/4 in. bolt is .4418 sq. in. The ultimate shearing strength for steel bolts 50,000 lbs. Using a factor of safety 6 on account of varying loads used with derricks, we get: unit working stress=50,000÷6=8333 lbs. per sq. in. and 8333X .4418=3680 lbs. per 3/4 in. rivet The load for 3/4 in. bolts = 70% of 3680 = 2576 lbs. per 3/4 in. bolt For two 3/4 in. bolts

Fig. 5.

you get 2X2576=5152 lbs. The load at each end of beams equals four tons or 8000 lbs. The derrick, therefore, should not be used for 16 tons at a time. To allow for shock and overloading, all the bolt holes in ends of beams should be filled in with good temporary bolts in all connections under and near the derrick.