Why do we use multiple reinforcing bars instead of one bar with a big diameter in reinforced concrete?

Question

Why do we use multiple reinforcing bars instead of one with a big diameter in reinforced concrete?

Does increasing the number of reinforcing bars means that the reinforced concrete will have a higher tensile strength?

Answer 1

There are a few reasons.

I'm firstly going to assume you're talking about replacing a bunch of small rebars by a single reasonably-sized one: i.e. instead of $15\phi8$ (7.54 cm²), using $1\phi32$ (8.04 cm²).

One reason is to improve ease of constructibility. Reinforced concrete beams also have transversal reinforcement, and it's very common to place rebars at the four corners of the transversal reinforcement to tie everything together into a self-supporting cage (this also inhibits crushing of concrete at the corners of the transversal reinforcement). But that just means you should place a lower boundary of 2 rebar in each face (upper and lower). But why not adopt $4\phi16$ (8.04 cm²) instead of $15\phi8$?

Multiple small rebar improve the behavior of the concrete-steel interface. The total surface area to transmit the tensile forces from the steel to the concrete at the anchorages is greater, reducing the tensile stress and therefore shortening the anchorage and lap splice lengths. This also reduces the cracks that will develop in the concrete.

Lastly, one can usually also be more efficient with smaller rebar: if what you actually need is 7.5 cm², you'll need either $15\phi8$ (7.54 cm²), $10\phi10$ (7.85 cm²), $7\phi12.5$ (8.59 cm²), $4\phi16$ (8.04 cm²), $3\phi20$ (9.42 cm²), $2\phi25$ (9.82 cm²), or $1\phi32$ (8.04 cm²). $\phi8$ gives the most efficient use of steel, putting as little extra as possible. Obviously, however, this isn't always the case. It just often is.

Now I'm going to assume you're talking about using one ridiculously large steel bar: i.e. instead of $20\phi20$ (68.2 cm²), using $1\phi94$. Some of the points raised here are still valid even if you're talking about being reasonable, but their significance is reduced.

One reason is cost. This concept implies in making custom-diameter rebar for each beam (otherwise, all you're saying is that current rebar is too small), which would impede use of scale to reduce costs. A factory can reduce the unit-cost of a $\phi10$ bar by making millions of them. If each beam in each job site will require a custom diameter, such cost-savings aren't possible.

Also, a single $\phi94$ rebar would weigh almost 54 kg/m, meaning you'd need to use a crane to put it into position, while $20\phi20$ can be easily placed by hand by a single worker, one at a time.

As well, at the extremities of your beam the steel will probably need to be bent. The larger the diameter of the rebar, the larger its bending radius.

And then there's this:

Since the center of gravity of your steel will be farther from the face of the concrete, it will also be less efficient: you'll need more steel to achieve the same strength. This also applies for the first part of this answer, where we were talking about reasonable sizes, but then the difference is obviously much smaller.

But why not use a sheet of steel then, instead of one large circular rebar? Hell, since it would no longer have the spaces between the rebar, you could choose a sheet which is thinner than the multiple rebar's diameters, and which is therefore more efficient as well! But then your surface area would be reduced, increasing the transfer stresses and therefore the anchorage and lap splice lengths. Also, a sheet is a two-dimensional element, so there may be other transversal behaviors which may lead to problems. Also, how would you pour the concrete for the bottom face of the beam?

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At the end of the day, the best rule of thumb (or, well, the one I try to use) is to do your best to fit everything into as few layers of rebar as possible, but filling those layers as much as possible (while still leaving comfortable space for proper concrete pours, including the necessary space for the vibrator). The only exception is the last layer (farthest from the relevant face of the beam), which can be left emptier (but preferably with a number of bars that's allows it to follow a symmetric pattern of the previous layers). Well, balance that with being efficient with the steel: if this requires using large rebar and results in an adopted area of steel much larger than if another layer were added with smaller rebar, then maybe another layer is best.

Answer 2

The main purpose of rebar is to improve the tensile strength of concrete and in practice most of these loads come from bending rather than pure tension.

When a beam in subject to bending forces the greatest stress are at the edges and faces of the beam so just having one big bar running down the centre wouldn't do very much as this part of the structure sees very little load until it starts to fail.

Having smaller diameter rebar distributed throughout the structure also distributes the load from the concrete to the steel more effectively as there is a greater area of contact for adhesion between the two.

In practice the size and placement of the steel reinforcement will be determined by the expected loading on the structure and is a compromise between strength, weight, cost and the practicalities of assembling the steelwork during construction.

Answer 3

From a purely conceptual standpoint, one large bar of the same area as multiple smaller bars provides the same moment capacity for a concrete beam. This is assuming that the center of the bars are all at the same depth.

The distribution of the bars (multiple smaller bars) help to limit cracks by spreading the tension force through a greater width of concrete.

Multiple smaller bars also help when looking at the interaction at the interface of the concrete and reinforcing steel. A single large bar has less surface area than multiple smaller bars. This means that for a given load, the stress between the surface of the bar and the concrete is greater in the single bar case. This has applications to the development length of the reinforcing.

Answer 4

In addition to Wasabi's answer - which is great - large diameter bars need insane development length and lap length.

A rule of thumb estimate for development and lap length is 40 diameters. For 20 mm bars that is a large but reasonable 80 cm, but it becames 128 cm for 32 mm bars and 2 meters for 50 mm bars.

Lap length problem can be avoided by using mechanical connectors, but 2 meters development length would waste a lot of steel and need a large space, very often unavailable.

Answer 5

Concrete and Steel behave in drastically different ways: concrete is assumed not to be able to bear any tensile stress as soon as first cracks appear while steel is ideal material for tensile stress. Optimally it would be ideal to have thin tiny strands of steel as opposed to few large ones, kind of similar to how fiberglass works homogenously.
Specially since we know our design assumptions are highly simplified to make the job of engineering practical. Things like climate and deferential sun exposure, humidity, permeating of corrosive agents through cracks in a member, all impact the behavior of a member. Also as mentioned above the area and contact surface of steel bar are related in a disadvantage ratio of power of two for area of the bar that transfers tension through skin friction to concrete.
On the other hand practical requirements during the course of construction such as bars not easily be deformed under the foot traffic of workers or accommodation space to allow passage of transient pipes or conduits should be permitted. that's were engineering codes come into play to make a uniform and predictable game plan. so the mechanical engineer or hardware designer know what they specify or expect will work.

Answer 6

small diameter of bar center of gravity is near to the concrete surface. larger diameter of bar center of gravity is far from the concrete surface. small diameter of bar easy to handla then the larger diameter of bar. The larger the diameter of the rebar, the larger its bending radius.

Answer 7

I think that 4 bars D1 will better than 1 bar D2 (same tensile area, but more bonding surface).

In addition to that, bigger bars are more brittle while smaller bars have dominant surface yield (more elastic and plastic than brittle). It adds safety when failure happens.