For 1:3:5 concrete, using as before r = 12, and with a working value for c = 500, and s = 16,000, we find from Equation 28 that the economical percentage of steel equals: p = 1/2 500 500 X 12

16,000 (500 X 12 + 16,000) = .0043

From Table XV we find by interpolation that, for r = 12, and p = .0043, k = .273. Then (see Equation 26): x = 1/3 kd = .091d; and (d-x) = .909d.

Substituting these values in either formula of Equation 29, we have:

M = 62 bd2.

The percentage of steel computed from Equation 28 has been called the most economical percentage, because it is the percentage which will develop the maximum allowed stress in the concrete and the steel at the same time, or by the loading of the beam to some . definite maximum loading. The real meaning of this is best illustrated by a numerical example using another percentage. Assume that the percentage of steel is exactly doubled, or that p = 2 X .0043 = .0086. From Table XV, for r = 12, and p = .0086, we find k=.362; x=.12kZ; and (d - x) = .879d. Substituting these values in both forms of Equation 29, we have:

Mc = 80 bd2; and, Ms = 121 bd2.

The interpretation of these two equations, and also of the equation found above (M = 62 bd2), is as follows: Assume a beam of definite dimensions b and d, and made of concrete whose modulus of elasticity is 1/12 that of the modulus of elasticity of the reinforcing steel; assume that it is reinforced with steel having a cross-sectional area = .00 13 bd. Then, when it is loaded with a load which will develop a moment of 62 bd2, the tension in the steel will equal 16,000 pounds per square inch, and the compression in the concrete will equal 500 pounds per square inch at the outer fibre. Assume that the area of the steel is exactly doubled. One effect of this is to lower the neutral axis (k is increased from .273 to .362), and more of the concrete is available for compression. The load may be increased about 29 per cent, or until the moment equals 80 bd2, before the compression in the concrete reaches 500 pounds per square inch. Under these conditions the steel has a tension of about 10,600 pounds per square inch, and its full strength is not utilized. If the load were increased until the moment was 121 bd2, then the steel would be stressed to 16,000 pounds per square inch, but the concrete would be compressed to over 750 pounds, which would of course be unsafe with such a grade of concrete. If the compression in the concrete is to be limited to 500 pounds per square inch, then the load must be limited to that which will give a moment of 80 bd2. Even for this the steel is doubled in order to increase the load 29 per cent. Whether this is justifiable, depends on several circumstances - the relative cost of steel and concrete, the possible necessity for keeping the dimensions of the beam within certain limits, etc. Usually a much larger ratio of steel than 0.43 per cent is used; 1.0 per cent is far more common; but when such is used, it means that the strength of the steel cannot be fully utilized unless the concrete can stand high compression. A larger value of r will indicate higher values of k, which will indicate higher moments; but r cannot be selected at pleasure. It depends on the character of the concrete used; and, with Es constant, a large value of r means a small value for Ec, which also means a small value for c, the permissible compression stress. Whenever the percentage of steel is greater than the economical percentage, as is usual, then the upper of the two formul of Equation 29 should be used. When in doubt, both should be tested, and that one giving the lower moment should be used.

Table XVI.. Ultimate Load On Slabs Of "Average" Concrete (1:3:5) In Pounds Per Square Foot Weight Of Slab Included

Effective Thickness of

Slab d

Area

OF

Steel in 12-In. Width

Spacing of Bars

SPAN IN FEET (L)

3/8-IN. Sq.

1/2-In. Sq.

4

5

6

7

8

9

10

11

12

13

14

15

2.5

.252

6 3/4

in.

12

in.

1,241

794

551

405

310

245

198

.

.

.

..

..

3 0

.302

5 1/2

,,

10

,,

1,786

1,143

794

583

446

353

286

236

198

.

.

.

3.5

.353

4 3/4

,,

8 1/2

,,

2,432

1,556

1,080

793

608

480

389

322

270

230

198

4.0

.403

4 1/4

,,

7 1/2

,,

3,176

2,033

1,411

1,037

794

627

508

420

353

300

259

226

4.5

.454

3 3/4

,,

6 3/4

,,

4,020

2,573

1,786

1,312

1,005

794

643

531

446

380

328

286

5.0

.504

3 3/8

,,

6

,,

4,962

3,176

2,206

1,620

1,241

980

794

656

551

470

405

353

5.5

.554

3

,,

5 1/2

,,

6,005

3,843

2,669

1,960

1,501

1,186

960

794

667

569

490

427

6 0

.605

.

.

5

,,

..

4,573

3,176

2,334

1,787

1,412

1,142

945

793

677

583

508

7.0

.706

.

.

4 1/4

,,

..

.

4,323

3,176

2,432

1,921

1,556

1,286

1,080

921

794

692

8.0

.806

.

.

3 3/4

,,

.

.

.

4,148

3,176

2,509

2,033

1,680

1,410

1,203

1,037

904

Using p = .0075, and r = 12, we have k = .343; x = .114d; and (d - x) - .886 d. Then, since p is greater than the economical value, we use the upper formula of Equation 29, and have:

M - 76 bd2..(30)

272. Example 1

"What is the working moment for a slab with 5-inch thickness to the steel, the concrete having the properties described above?

Answer. Calling b = 12 inches, M = 76 X 12 X 25 = 22,800 inch-pounds, the permissible moment on a section 12 inches wide.

Example 2. A slab having a span of 8 feet is to support a load of 150 pounds per square foot. The concrete is to be as described above, and the percentage of steel is to be 0.75. What is the required thickness d of the steel?

Answer. A strip 12 inches wide has an area of 8 square feet, and carries a load of 1,200 pounds. The moment = 1/8 Wl = 1/8 X 1,200 X 96 = 14,400 inch-pounds. For a strip 12 inches wide, b = 12 inches, and M = 76 X 12 X d2 = 912d2 = 14,400; d2 = 15.79; d = 3.97 inches - or, say, 4 inches.