In Table XVII has been computed for convenience the ultimate total load on rectangular beams made of average concrete (1:3:5) and with a width of 1 inch. For other widths, multiply by the width of the beam. Since M0 = 1/8 W0 l; and since by Equation 23, for this grade of concrete, M0 = 397 b d2', and since for a computation of beams 1 inch wide, b = 1, we may write 1/8 W0 l = 397 d2. For l we shall substitute 12 L. Making this substitution and solving for W0, we have W0 = 265 d2 ÷ L. Since b = 1, A, the area of steel per inch of width of the beam = .0084 d.

Table XVII. Ultimate Total Load On Rectangular Beams Of Average Concrete (1:3:5), One Inch Wide

For other widths, multiply by width of beam. Formulas: W0 = 265 d2 ÷ L; A = .0084 d. Ultimate compression in concrete 2,000 pounds per sq. in.; ultimate tension in steel 55,000 pounds per sq. in.

Effective Depth of Beam, d

Area of

Steel perinch of Width

SPAN IN FEET (L)

Twice Dead Load per

Foot of Beam

4

5

6

7

8

9

10

11

13

13

14

15

16

17

18

19

20

4

.0336

1,060

848

707

606

530

471

424

385

353

326

303

283

265

249

236

223

212

10

5

.0420

1,656

1,324

1,104

946

828

736

662

603

553

510

473

441

414

390

368

349

331

12

6

.0504

2,385

1,908

1,590

1,363

1,192

1,060

954

867

795

734

681

636

596

561

530

502

477

15

7

.0588

3,246

2,596

2,164

1,855

1,623

1,443

1,298

1,180

1,083

999

927

865

812

764

721

683

649

17

8

.0673

4,240

3,393

2,827

2,423

2.120

1,884

1,696

1,542

1,413

1,305

1,211

1,131

1,060

998

943

893

848

20

9

.0756

5,366

4,292

3,577

3,066

2,683

2,385

2,146

1,951

1,789

1,651

1,533

1,431

1,341

1,263

1,193

1,130

1,073

22

10

.0840

6,625

5,300

4,417

3,786

3,312

2,944

2,650

2,409

2,208

2,038

1,893

1,767

1,656

1,559

1,473

1,395

1,325

24

11

.0924

8,016

6,412

5,344

4,581

4,008

3,563

3,206

2,915

2,672

2,466

2,290

2,137

2,004

1,886

1,781

1,688

1,603

26

12

.1008

9,540

7,632

6,360

5,451

4,770

4,240

3,816

3,469

3,180

2,935

2,726

2,544

2.385

2,345

2,120

2,008

1,908

28

13

.1092

11,196

8,957

7,464

6,398

5.598

4,976

4,478

4,071

3,732

3,445

3,199

2,986

2,799

2,634

2,488

3,357

2,239

30

14

.1176

12,985

10,388

8,657

7,420

6,492

5,771

5,194

4,723

4,328

3,995

3,710

3,463

3,246

3,055

2,886

3,734

2,597

32

15

.1260

14,906

11,924

9,937

8,518

7,453

6,625

5,962

5,430

4,969

4,586

4,259

3,975

3,726

3,508

3,312

3,138

2,981

34

16

.1344

16,960

13,568

11,307

9,691

8,480

7,538

6,784

6,167

5,653

5,218

4,845

4,523

4,240

3,991

3,769

3,571

3,392

36

17

.1428

19,146

15,317

12,764

10,941

9,573

8,509

7,658

6,963

6,382

5,891

5,470

5,106

4,786

4,505

4,255

4,031

3,929

38

18

1513

21,465

17,173

14,310

12,266

10,733

9,540

8,586

7.805

7,155

6,605

6,133

5,724

5,366

5,051

4,770

4,519

4,293

40

19

.1596

23.916

19,133

15,944

13,666

11,958

10,629

9,566

8,697

7,973'

7,359

6,833

6,378

5,979

5,627

5,315

5,035

4,783

42

20

.1680

26,500

21,200

17,667

15,143

13,350

11,778

10,600

9,636

8,833

8,154

7,571

7,067

6,625

6,235

5,889

5,579

5,300

44

For values in the lower left-hand corner of the table, possible failure by diagonal shear must be very carefully tested and provided for.

Example. What is the ultimate total load on a simple beam having a depth of 16 inches to the reinforcement, 12 inches.wide, and having a span of 20 feet?

Answer. Looking in Table XVII, under L = 20, and opposite d = 16, we find that a beam 1 inch wide will sustain a total load of 3,392 pounds. For a width of 12 inches, the total ultimate load will be 12 X 3,392 = 40,704 pounds. At 144 pounds per cubic foot, the beam will weigh 3,S40 pounds. Using a factor of 2 on this, we shall have 7,680 pounds, which, subtracted from 40,704, gives 33,024. Dividing this by 4, we have 8,256 lbs. as the allowable live load on such a beam.