This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

The dimensions of iron beams, whether wrought or cast, are to be ascertained by the rules already given, when the beams are of rectangular form in their cross-section; these rules are applicable alike to wood and iron (Art. 93), and may be used for any material, provided the constant appropriate to the given material be used. But when the form of cross-section is such as that which is usual for rolled-iron beams (Fig. 45), the rules need modifying. Without' attempting to explain these modifications (referring for this to Transverse Strains, Art. 457 and following article), it may be remarked that the elements of resistance to flexure in a beam constitute what is termed the Moment of Inertia. This, in a beam of rectangular cross-section, is equal to 1/12 of the breadth into the cube of the depth; or -

I=1/12 bd3. (66.)

Fig. 45.

This would be appropriate to rolled-iron beams if the hollow on each side were filled with metal, so as to complete the form of cross-section into a rectangle. The proper expression for them may be obtained by taking first the moment for the beam as if it were a solid rectangle, and from this deducting the moment for the part which on each side is wanting, or for the rectangles of the hollows. In accordance with this view of the case, we have -

I=1/12(b,d3-b,d3) (67.)

in which b is the breadth of the beam or width of the flanges; b, is the breadth of the two hollows, or is equal to b less the thickness of the web or stem; d is the depth including top and bottom flanges; and d, is the depth in the clear between the top and bottom flanges.

Now, if equation (32.) be divided by 12, we shall have -

1/12=bd3=Wl3/12 Fδ;

and since 1/12 b d3 represents the moment of inertia, we have -

I = Wl3/12 Fδ (68.)

This gives the value of I for a beam of any form in cross. section loaded at the middle. By this equation the values of I have been computed for rolled-iron beams of many sizes, and the results recorded in Table XVII., Transverse Strains. A few of these are included in Table IV., as follows:

Name. | Depth. | Weight per yard. | / = |

4 | 30 | 7.84 | |

Paterson.... | 5 | 30 | 12.082 |

Phoenix... . | 5 | 36 | 14.317 |

Trenton.... | 6 | 40 | 23.761 |

Phoenix .... | 7 | 55 | 42.43 |

Trenton .... | 7 | 60 | 46.012 |

Buffalo...... | 3 | 65 | 64.526 |

Paterson.... | 8 | So | 34.735 |

Phoenix..... | 9 | 70 | 92.207 |

Phoenix .... | 9 | 84 | 107.793 |

Name. | Depth. | Weight per yard. | I = |

Buffalo...... | 9 | 90 | 109.117 |

Phoenix..... | 9 | 150 | 190.63 |

Buffalo..... | 10 1/2 | 90 | 151.436 |

Buffalo...... | 10 1/2 | 105 | 175.645 |

Trenton..... | 10 1/2 | 135 | 241.478 |

Buffalo..... | 12 1/4 | 125 | 286.019 |

Paterson. .. . | 12 1/4 | 125 | 292.05 |

Paterson. ... | 12 1/4 | 170 | 398.936 |

Buffalo...... | 12 1/4 | 180 | 418.945 |

15 3/16 | 150 | 528.223 |

169. - Rolled . Iron Beams: Dimension*; Weight at

Middle - If, in equation (68.), there be substituted for F its value for wrought iron, as in Table III., we shall have -

I=W13/12x62000δ

or -

I=Wl3/744000δ. (69.)

This is a rule by which to ascertain the size of a rolled-iron beam to sustain a given weight at middle with a given deflection, and, in words at length, is as follows:

Rule XLVIII - Multiply the weight in pounds by the cube of the length in feet; divide the product by 744000 times the deflection in inches, and the quotient will be the moment of inertia of the required beam, and may be found, or the next nearest number, in Table IV. in column headed I. Opposite to the number thus found, to the left, will be found the name, depth, and weight per yard of the required beam.

Example. - Which of the beams of Table IV. would be proper to carry 10,000 pounds at the middle with a deflection of one inch, the length between bearings being 20 feet? Here we have, substituting for the symbols their values -

1 =Wl3/744000δ=10000x203/744000x1=80000000/744000=107.527;

or, the moment of inertia of the required beam is 107.527, the nearest to which, in the table, is 107.793, pertaining to the Phoenix 9-inch, 84-pound beam. This, then, is the required beam.

170. - Rolled-Iron Beams: Deflection when Weight is at Middle. - By a transposition of symbols in equation (69.), we have -

δ=Wl3/744000 I' (70)

or a rule for the deflection of rolled-iron beams when the weight is at the middle. This, in words, is -

Rule XLIX. - Multiply the weight in pounds by the cube of the length in feet; divide the product by 744000 times the value of / for the given beam, and the quotient will be the required deflection in inches.

Example. - What will be the deflection of a Phoenix 9-inch, 70-pound beam 20 feet long, loaded at the middle with 7500 pounds? The value of I for this beam, in Table IV., is 92.207; therefore, substituting for the symbols their values, and proceeding by the rule, we have -

δ= W13/744000 I= 7500x203/744000 x 92.207=0.87461;

or, the deflection will be, say, 7/8 of an inch.

171. - Rolled-Iron Beams: Weight when at Middle. -

A transposition of factors in equation (70.) gives -

W=744000 Iδ/l3 (71.)

This is a rule for the weight at middle, and, in words, is - Rule L. - Multiply 744000 times the value of I by the deflection in inches; divide the product by the cube of the length, and the quotient will be the required weight in pounds.

Example. - What weight at the middle of a Buffalo 9-inch, 90-pound beam will deflect it one inch, the length between bearings being 20 feet? The value of I for this beam, in

Table IV., is 109.117; therefore -

W=744000 Iδ/l3=744000x109.117x1/203=10147.88;

or, the required weight is, say, 10,148 pounds.

172. - Rolled-Iron Beams: Weight at any Point. - The equation for a load at any point is (Transverse Strains, Art. 485)-

W=186000 Iδ/lmn;

in which m and n represent the two parts in feet into which the point where the load rests divides the length. This, in words, is as follows:

Rule LI. - Multiply 186000 times the value of I by the deflection in inches; divide the product by the product of the length into the rectangle formed by the two parts into which the point where the load rests divides the length; the quotient will be the required weight in pounds.

Example. - What weight is required, located at 10 feet from one end, to deflect 1 1/2 inches a Paterson 12 1/4-inch, 125-pound beam 25 feet long between bearings? The value of I for this beam, in Table IV., is 292.05; m = 10, and n = l - m = 25 - 10 = 15; therefore -

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