Few subjects are mired in more confusion than the function of steel reinforcement—especially with regard to its effects on cracking. To see what does happen when a steel rod is solidly embedded in concrete that is stretched, imagine a 15-foot-long 18-inch-wide block of 6-inch slab that can be uniformly tensioned lengthwise. Let the block be crack free and made of mature 3500-psi, 150-pcf concrete. How much tension can be applied to the block before it pulls apart, and how much will the block have stretched at that moment?

ACI 318 says 3500-psi concrete weighing 150 pcf will have a tensile strength of 443.7 psi and an elastic modulus of 3,586,616 psi. Because this block has a cross-sectional area of 18 x 6 = 108 square inches, it should break in two when the tensioning force exceeds 443.7 x 108 = 47,919.6 pounds, or equivalently, when its length has increased more than 180 x 443.7 / 3,586,616 = 0.022 inch.

Now let the same block be continuously reinforced with a centered mid-depth layer of welded wire reinforcement (WWR). Just how much will the presence of the WWR change the block’s ability to be stretched without cracking?

**Rule No. 14a: Reinforcement percentage (R _{%}) equals the cross-sectional area of steel (A_{S}) divided by the cross-sectional area of concrete (A_{C}) multiplied by 100: R_{%} = 100 A_{S} / A_{C}. In slabs reinforced with a single mat, or identical multiple mats: R_{%} = 100 m A_{R} / (S_{R} t), where**

*m*

**equals the number of mats, A**

_{R}equals the cross-sectional area of a single rod, S_{R}equals the nominal rod spacing, and*t*

**equals the slab’s depth.**

Adding a layer of 6x6-10/10 WWR to the block thus creates an R_{%} of 0.039; a layer of 6x6-6/6 creates an R_{%} of 0.081; and a layer of 4x4-4/4 creates an R_{%} of 0.167. Note that a #10 wire (10 gage) is equivalent to W1.4 and has a cross section of 0.014 square inches; a #6 wire has a cross section of 0.029 square inches (W2.9); and a #4 has cross section of 0.040 square inches (W4.0).

**Rule No. 14b: In a normal weight slab, the percentage increase (+T _{%}) in precrack tensile capacity obtained through the use of continuous steel reinforcement will equal approximately 500 times the R% divided by the square root of the concrete’s compressive strength: +T_{%} = 500 R_{%}/(f´_{C})^{½}.**

Using **Rule No. 14b**, the percentage increases in the block’s precrack tensile capacity obtained by including the three welded wire sheets indicated above will be 0.33%, 0.68%, and 1.14%, respectively. In other words, the inclusion of the WWR does nothing to increase the block’s ability to resist cracking.

Well, maybe the problem is that not enough steel is being added. So instead of using WWR, let’s put a rebar down the center and see what happens. The R%´ values created by embedding various standard bars in the block will vary from 0.10% for a #3 up to 0.73% for a #8. Again per **Rule No. 14b**, the resulting percentage increases in the block’s precrack tensile capacity will range between 0.86% for the #3 and 6.14% for the #8. As may be seen from the minimal improvement obtained even by using a #8 bar, no matter whether the steel takes the form of WWR or rebar:

**Rule No. 14c: It is not possible for a slab’s precrack tensile capacity to be increased significantly through the use of continuous reinforcement.**

The ACI documents emphasize this fact at every opportunity, but the myth that reinforcement will keep concrete from cracking remains very alive and well—even amongst many design professionals.

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