## Wooden Beams

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:

### Rule

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.

## Table XXVII. Safe Uniformly Distributed Loads For Rectangular Wooden Beams, 1 Inch Thick

 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

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.

### Example

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?

### Solution

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:

### Example

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?

### Solution

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