798 



APPLIED MECHANICS. 



[COHESIVE STRENGTH. 



fig. go. 



fiult. Multiply the width by 

 the thickness, or generally, I'uid 

 the sectional area in square 

 inches, and multiply by the 

 uumlr in column 2 of the 

 table. 



Example 2. Required the 

 cohesive strength of a bar of 

 wrought copper of the form and 

 dimensions given in Fig. 50. 



The upper part of the figure 

 is an oblong or rectangle, of which the width is 4 inches 

 and the thickness is 1 inch ; its area is therefore 4x1, 

 or 4 square inches. 



The lower part of the figure is a trapezoid, of which the 

 thickness at one edge is 2 inches, and at the other 1 ; the 

 mean thickness is therefore li inch, and the width being 2 

 inches, its area is 1J x 2, or 3 square inches. The total 

 area of the section is therefore 7 square inches : and 7 

 multiplied by 5 tons, the strength of wrought copper in 

 column 2 of table, gives the strength of the bar 35 tons. 



3. When the bar is cylindrical. 



Rule. Multiply the diameter in inches by itself, and 

 by the number in column 3. 



Example 3. Required the cohesive strength of a fir 

 pole 4 inches in diameter. 4 x 4 x 1"26 give 20 '16, say 

 20 tons. 



4. When the bar is of elliptical section. 



Rule. Multiply the greater diameter by the less, and 

 by the number in column 3. 



Example 4. Required the cohesive strength of an 

 elliptical lead bar, of which the greater diameter is 2 

 inches and the less 1} inch : 2 x If X 600 give 1500 Ibs. 



5. When the bar is cylindrical and hollow like a pipe. 

 Rule. Multiply the outer diameter by itself, and also 



the inner by itself ; subtract the one product from the 

 other, and multiply the remainder by the number in 

 column 3. 



Example 5. Required the cohesive strength of a brass 

 tube 2 inches diameter outside, and 1J diameter inside. 



2 



X2=4 

 X lj - 2j 



X 2-12 = 3-7 tons. 



6. As a general rule, whatever be the form of section, 

 find the area either in square inches or circular inches, 

 and use as a multiplier the number in column 2 in the 

 one case, and that in column 3 in the other. Or, find the 

 strengths due to the several parts of the section, and add 

 them together for the total strength. 



imple 5. Required the cohesive strength of a cast- 

 iron bar of the form and dimensions given in Fig. 57. 



Taking first the hollow semicircular part, 

 Outer dia.5 in. X 5= 25 

 Innerdia.3in.x3= 9 



Take J for semicir. ) 1 



8 circ. in. X 2 -12 tns.(col. 3) - 16-96 tns. 

 Two upright sides each 



2 in. by 1 in. have an 



area2xlx2 . - 4sq.ins. 

 Two lower flanges each 



2 1 in. by 1 in . have an 



area2x2Jxl . - 5 



Square inches 9x27 tons (col 2) =24 3 tns. 



Total cohesive strength 41 -26 tns. 

 say 41 tons. 



II. The converse operations are for calculating the 

 dimensions of a bar required to carry a given load. 



7. \Vhcn the bar is square or oblong. 



Rule. Multiply the number in column 4, by the given 

 number of tons, and the product gives the number of 

 square inches of sectional area. 



Example 7- Required the area of a bar of wrought- 

 iron capable of sustaining 10 tons. 



Number from coL 4 

 Multiply by ... 



0-120 

 10 



1 2 square inch. 



This sectional area may be attained by making the 

 bar square about 1 -fa inch of a side ; or making it 



Fl. ST.; 



*-- 2t~ -X 



2k- --.-> 



oblong, such as 2 inches by | inch thick, or any dimen- 

 sions such that their product is at least 1'2 square 

 inch. 



8. When the bar is cylindrical 



Rule. Multiply the number in col. 5 by the weight 

 in tons, and take the square root of the product ; it will 

 be the diameter in inches. 



Example 8. Required the diameter of a cylindrical 

 bar of wrought copper to carry 35 tons. 



Number from col. 5 ... 0-255 

 Multiply by 35 



9" nearly. 



The square root of 9 is 3 inches, the diameter of the 

 bar. 



9. For bars having sections of various forms and 

 proportions, the calculations must be trials to a certain 

 extent. The number of square inches required is found 

 as in Example 7, and these may be disposed in any suit- 

 able form that may be required. 



Before we leave the subject of tensive strength, we 

 will discuss a case which frequently occurs in practice, 

 and where the cohesive force of materials is employed in 

 a vessel to resist a bursting or exploding strain arising 

 from the pressure of a fluid within it. Were a boiler, 

 pipe, or other vessel, intended to contain a fluid exerting 

 considerable pressure, made of any form except circular 

 in section, the internal pressure would change the form 

 of section. Thus, suppose the section were of a square 

 form, as in Fig. 68, the internal pressure acting equally 



i 



\ 



T 7 



T T 



on every part of the casing, as marked by the arrows, 

 would bulge the flat sides, as indicated by the dotted 

 lines ; and, finally, were the material sufficiently pliable, 

 would extend it into a circular form. After having 

 attained a circular section, no farther change of form 

 would be effected, and the pressure of the internal fluid 



