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 ofSteel perinch of Width SPAN IN FEET (L) Twice Dead Load perFoot 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.