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

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.

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

Table XXVII Safe Uniformly Distributed Loads For R 252