This section is from the book "The Building Trades Pocketbook", by International Correspondence Schools. Also available from Amazon: Building Trades Pocketbook: a Handy Manual of reference on Building Construction.

These are principally used to support a uniformly distributed load, consequently a convenient formula for determining directly the safe strength of rectangular beams uniformly loaded is useful, and may be deduced from the foregoing principles as follows: The bending moment M in foot-pounds is WL/8, or in inch-pounds, 12WL/8. Placing this equal to the resisting moment, M1, or Q S, there results

12WL/8 = QS, or l2WL = 8QS; whence W =8QS/12L =2/3 QS/L.. For a rectangular beam, the section modulus Q = bd2/6, and the above formula may be written W = 2/3 S/L bd2/6= Sbd2/9L ; b and d are in inches and L in feet. This formula expressed in words is as follows:

The safe uniformly distributed load in pounds for a rectangular beam is equal to the safe unit fiber stress multiplied by the breadth in inches and by the square of the depth in inches, and the product divided by nine times the span in feet.

Span in Feet. | Depth of Beam. | |||||||

6" | 7" | 8" | 9" | 10" | 12" | 14" | 16" | |

5 | 800 | 1,090 | 1,420 | 1,800 | 2,220 | 3,200 | 4,380 | 5,690 |

6 | 665 | 910 | 1,190 | 1,500 | 1,850 | 2,670 | 3,650 | 1.740 |

7 | 570 | 780 | 1,020 | 1,290 | 1,590 | 2,290 | 3,130 | 4.060 |

8 | 500 | 680 | 890 | 1,130 | 1,390 | 2,000 | 2,740 | 3,560 |

9 | 445 | 610 | 790 | 1,000 | 1,230 | 1,780 | 2,430 | 3,160 |

10 | 400 | 540 | 710 | 900 | 1,110 | 1,600 | 2,190 | 2,840 |

11 | 365 | 495 | 650 | 820 | 1,010 | 1,450 | 1,990 | 2590 |

12 | 335 | 450 | 590 | 750 | 930 | 1,330 | 1,820 | 2,370 |

13 | 310 | 420 | 550 | 690 | 860 | 1,230 | 1,690 | 2,200 |

14 | 285 | 390 | 510 | 640 | 800 | 1,150 | l.570 | 2.040 |

15 | 265 | 360 | 480 | 600 | 740 | 1,070 | 1,460 | 1,900 |

16 | 250 | 340 | 450 | 560 | 700 | 1.000 | 1,370 | 1,780 |

17 | 235 | 320 | 420 | 530 | 650 | 940 | 1,290 | 1,680 |

18 | 220 | 300 | 400 | 500 | 620 | 890 | 1,220. | 1,590 |

19 | 210 | 290 | 380 | 480 | 590 | 840 | 1.150 | 1,500 |

20 | 200 | 272 | 360 | 450 | 560 | 800 | 1.090 | 1.420 |

21 | 190 | 260 | 340 | 430 | 530 | 760 | 1,040 | 1360 |

22 | 180 | 248 | 325 | 410 | 510 | 730 | 1,000 | 1.300 |

23 | 175 | 237 | 310 | 390 | 480 | 700 | 950 | 1.240 |

24 | 165 | 228 | 297 | 380 | 460 | 670 | 910 | 1.190 |

25 | 160 | 218 | 285 | 360 | 450 | 640 | 880 | 1.140 |

26 | 155 | 210 | 275 | 350 | 430 | 620 | 840 | 1.100 |

27 | 149 | 202 | 265 | 330 | 410 | 590 | 810 | 1,060 |

28 | 143 | 195 | 255 | 315 | 400 | 570 | 780 | 1,020 |

29 | 138 | 188 | 248 | 307 | 380 | 550 | 750 | 980 |

30 | 134 | 182 | 237 | 297 | 370 | 530 | 730 | 950 |

Safe load for any thickness = safe load for 1 in. x thick-ness in inches. Thickness for any load = load / safe load for 1 in.

The values given are safe loads for spruce or white-pine beams, with a factor of safety of 4. For oak, or Northern yellow pine, multiply tabular values by 1 1/2, and for Georgia yellow pine, multiply tabular values by 1 3/4.

Table XXVII is calculated for a maximum fiber stress of 1,000 lb. per sq. in., but may be used with any fiber stress by dividing that stress by 1,000 and multiplying by the tabular value; thus, for a stress of 800 lb., the safe load is .8 the tabular value.

Loads given below the heavy zigzag line produce deflections likely to crack plastered ceilings.

What uniformly distributed load will a hemlock joist 3 in. wide X 14 in. deep safely support, the span being 20 ft., and the factor of safety, 5?

In the formula W =Sbd2 / 9L, S=3,500 lb., the modulus of rupture of hemlock (see Table IX), / 5, the factor of safety, = 700 lb. per sq. in.; 5 = 3 in.; d2 = 14 in. X 14 in. = 196; and L = 20 ft. Then the safe uniformly distributed load W=(700x3x196)/(9x20)=2,287 lb

Otherwise, from Table XXVII, the safe load with a fiber stress of 1,000 lb. per sq. in. for this beam = the load for 1 in. width X the width, or 1,090 X 3 = 3,270 lb.; but for a safe stress of 7001b. per sq. in., the safe load is 700 / 1000 =.7 of 3,270 lb. = 2,289 lb. The following example shows how to calculate the size of a rectangular beam necessary to support a given load:

What size Georgia yellow-pine joist would be required to support two concentrated loads placed as shown in Fig. 19, using a factor of safety of 4?

The reaction R2 = [(3000x3)+(2000x15)] / 20

= 1,950 lb., and R1 = (3,000 + 2,000) -1,950 = 3,0501b. The greatest bending moment M is found, by trial, to be under the 2,000 lb. load, and is equal to 1,950 lb. X 5 ft. = 9,750 ft.-lb., or 117,000 in.-lb. Such a section must be obtained that will have a resisting moment equal to this bending moment of

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