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The doubly reinforced concrete beam design may be required when a beam’s cross-section is limited because of architectural or other considerations. As a result, the concrete cannot develop the compression force required to resist the given bending moment. In that case, steel bars are added to the beam's compression zone to improve it at compression.

Therefore, a beam reinforced with tension steel and compression steel is called a doubly reinforced concrete beam. The moment of resistance of a doubly reinforced concrete beam is greater than that of a singly reinforced concrete beam for the same cross-section, steel grade, and concrete.

The use of compression reinforcement has decreased considerably due to the use of the strength method of design, which accounts for the full strength-potential of concrete in the compression zone. Nonetheless, compression reinforcement can be used for reasons other than strength like decreasing long term deflection of beam, account for minimum-moment loading, and holding stirrups at their positions.

Contents:

**Why is Compression Reinforcement Used in Beam?**

- To increase the strength of the concrete beam.
- To reduce long-term deflections of members.
- For minimum moment loading.
- For positioning stirrups (by tying them to the compression bars) and keeping them in place during concrete placement and vibration.

When compression reinforcement is added for purposes other than strength, the presence of the compression bars is neglected in the flexural calculations.

## Design Guidelines

### 1. Geometry of Beam

The designer may not have much control over the beam's dimension due to architectural or any other considerations that restrict the geometry of the beam.

The dimensions of the beam are set by architectural engineers, so the structural engineer knows the dimension of the beam and the only unknown would be the area of reinforcement.

### 2. **Steel reinforcement**

ACI 318-19 specifies the maximum tensile reinforcement ratio (*p*_{max}) that can be put into a singly reinforced concrete beam. The (*p*_{max}) can be calculated using the following expression:

If the calculated reinforcement ratio (*p*) of the beam under consideration is greater than (*p*_{max}), it should be designed as a doubly reinforced concrete beam.

The ACI 318-19 has also determined the maximum reinforcement ratio (*p*_{max}) that can be put into a doubly reinforced concrete beam to ensure steel bars in the beam's tension zone yield. Therefore, the (*p*_{max}) ensures that the beam fails by tensile yielding of steel instead of crushing of concrete which is sudden and undesirable.

Where:

B_{1}: is equal to 0.85 if (fc'=<28 MPa) otherwise use equation 3 to compute it.

fc': concrete compressive strength, MPa

fy: yield stress of steel bar, MPa

Epsilon, cu: ultimate concrete compressive strain which is 0.003, as per ACI 318-19

*p*': compression reinforcement ratio, which is computed according to equation 4

### 3. **Steel Bar Sizes**

Usually, it is advised to avoid the use of large bar sizes for beams. This is because such bars cause flexural cracking and require greater length to develop their strength. However, the cost of placement of large bar size is lower than the installation cost of a large number of small bar sizes.

Moreover, common bar sizes for beams range from No.10 to No.36 (SI unit) or No.3 to No.10 (US customary unit), and the two larger diameter bars No.43 (NO.14) and No.57 (No.18) are used for columns.

Furthermore, it is possible to mix different bar diameters to meet the steel area requirements more closely. Finally, the maximum number of bars installed in a beam of given width is controlled by the bar diameter, minimum spacing, maximum aggregate size, stirrup diameter, and concrete cover requirements.

### 4. **Spacing between Bars**

ACI 318-19 specifies minimum spacing between bars equal to bar diameter or 25mm. This minimum spacing shall be maintained to guarantee proper placement of concrete around steel bars, and prevent air pockets below reinforcements that consequently ensure a good bond between steel bras and concrete. If two layers of steel bars are placed in a beam, then the distance between them shall not be less than 25 mm.

### 5. **Concrete Protection for Reinforcement**

The designer must maintain minimum thickness or concrete cover outside of the outermost steel to provide the steel with adequate concrete protection against fire and corrosion.

According to ACI Code, concrete cover of 40 mm for cast in place beams, not exposed directly to the ground or weather and at least 50 mm cover if the concrete surface is to be exposed to the weather or in contact is recommended.

To simplify construction and reduce costs, the overall dimensions of beams, b, and h are almost rounded up to the nearest 25 mm.

**Design of Doubly Reinforced Concrete Beams**

Commonly, beams are designed as singly reinforced concrete beam, but the failure of concrete to develop adequate compression force would necessitate the addition of steel bars to the compression zone of concrete.

The distribution of stress and strains in the doubly reinforced concrete beam is shown in Figure-1. When reinforcement is added to the beam's compression zone, the same quantity is added to the tension zone of the beam to encounter the force of the compression steel bar, as shown in Figure-1, stress-strain (B).

There are two cases for the design of doubly reinforced concrete beam based on modes of failure of the beam:

**Case 1: Both Tension and Compression Steel at Yield Stress**

Assume that both compression steel (A's) and tensile steel (As) are stressed to failure (yield stress (fy)):

**fs=fs ^{'}=fy**

The total resisting moment of the beam can be thought of in two parts. The first part (Mn_{1}) is provided by the couple consisting of the force in compression steel (A′s) and the force in an equal tension steel area; see Figure-1, stress diagram (B):

The second part (Mn_{2}) is the contribution of the remaining tension-steel (As-A′s) acting with compression concrete; see Figure-1, stress diagram (A):

Where:

A's: compression steel, mm^{2}

fy: yield stress of steel bar, MPa

d: effective depth measured from the compression face to the center of steel bar in tension zone, mm

d': distance from compression face to center of compression steel bars, mm

As: total steel area in tension zone of the beam, mm^{2}

a: depth of rectangular stress block, mm , and it can be computed using equation 3

Where:

fc^{'}: concrete compressive strength, MPa

b: width of concrete beam, mm

The total resisting moment of doubly reinforced concrete beam can be computed using the following formula:

**Case 2: Compression Steel below Yield Stress**

The compression reinforcement may not reach yield stress if:

- The beam is wide or shallow.
- It has uncommon concrete cover over compression reinforcement.
- The tensile reinforcement ratio of the beam is low.
- Or the beam is reinforced with high yield strength steel.

When the compression steels are below yield stress at beam failure, the assumption of (**fs=fs ^{'}**) is no longer valid, hence new equations should be developed to account for the compressive steel stress which is below the yield stress.

Use minimum tensile reinforcement ratio (*p*^{-}_{cy}) to determine whether compression steel has yielded or not:

If the tensile reinforcement ratio (*p*) is less than the minimum tensile reinforcement ratio (*p*^{-}_{cy}), the neutral axis is sufficiently high that the compression steel is less than the yield stress.

So, new equations for **compression steel stress (f' _{s}) **and flexural strength would be as follows:

Where:

E_{s}: modulus of elasticity of steel which is 200000 MPa

c: neutral axis depth, mm, see Figure-1.

### Summary of Procedure for Design of Doubly Reinforced Concrete Beam

**Step 1:** Calculate the maximum moment or resisting moment (M_{n}) that can be resisted by the under reinforced section with *ρ* =*ρ _{max}*. The corresponding tensile steel area As=

*ρ*bd,

_{max}Compute (a) from equation 7, and calculate (d) using the following equation:

d=beam height-cover- stirrup diameter-0.5*longitudinal bar diameter **Equation 14**

**Step 2:** Find the excess moment (M_{1}), if any, that must be resisted and set M_{2}=M_{n} as calculated in Step 1:

Where M_{u} is applied or ultimate moment calculated from imposed loads.

**Step 3:** Define (As), which is calculated from Step 1, as As2, i.e., that part of the tension steel area in the doubly reinforced beam that works with the compression force in the concrete, in (As-A's= As_{2}).

**Step 4:** Tentatively assume fs′= fy , then compute compression steel area (A's) using Equation 5.

**Step 5:** Add an additional amount of tensile steel (As_{1}= A′s), then the total tensile steel area (As) in the tension zone of the beam is (As_{2}) from Step 2 plus (As_{1}).

**Step 6:** Analyze the doubly reinforced concrete beam to see if fs′= fy, i.e, check the tensile reinforcement ratio (*p*) against ρ^{-}_{cy}. Calculate (*p*) by using Equation 4 and use (As) from (**Step 5**).

**Step 7:** If ρ >ρ^{-}_{cy}, the compression steel stress is equal to (fy), and the design is ended. However, ρ <ρ^{-}_{cy}, the compression steel stress is less than (fy) and the compression steel area must be increased to provide the needed force.

**Step 8:** Calculate neutral axis depth which is equal to rectangular stress block divided by B_{1} i.e, either 0.85 or the value from Equation 3.

**Step 9:** Calculate compression steel stress using Equation 10, and finally estimate revised compression reinforcement area using Equation 5 in which plug value of (fs') instead of (fy).

**Step 10:** Determine number of bars and draw detail of the design.

## Example:

Design a rectangular beam that must carry a service live load of 45 KN/m and a calculated dead load of 20 KN/m (including self-weight) on a 5.5 m span is limited in cross-section for architectural reasons to 250 mm width and 500 mm total depth. If fy= 420 Mpa and fc′=28 MPa.

## Solution:

1- Initially, compute the ultimate distributed load (w_{u}) using a suitable load combination provided by ACI 318-19:

** w _{u}=1.2DL+1.6LL= 1.2(15.5)+1.6(38)= 79.4 KN/m**

2- Calculate applied or ultimate moment using the equation of maximum moment for simply supported beam:

**M _{u}=(w_{u}l^{2})/8= (79.4*5.5^{2})/8= 300.23 KN.m**

Use Equation 13 to estimate the maximum resistant moment (M_{n}) that a singly reinforced concrete beam can provide:

*Determine unknown parameters of equation 13:*

Use Equation 14 to determine the effective depth (d), assume bar diameter 29 mm, two layers of bars, and bar size of 10 mm for stirrups.

**d= 500-40-10-29-(25/2)= 408.5 mm**

**Note:**

If Equation 1 is used to compute (*p _{max})*, then the value of strength reduction factor for beam, which is 0.9, should be checked. That is why ACI 318-19 recommends to use a tensile strain of 0.005 in Equation 1 instead of 0.004 to avoid this check and ensure the strength reduction factor is 0.9:

*p _{max}*= 0.85*0.85*(28/420)*(0.003/(0.003+0.005))= 0.01806

As= *p*bd= 0.01806*250*412.5= 1862.695 mm^{2}, this value would be defined as As2 (discussed in Step 3 of design procedure) if the beam is designed as a doubly reinforced concrete beam

a= (1862.695*420)/(0.85*28*250)= 131.484 mm

**M _{n}= 1862.695*420*(408.5-(131.484/2))= 268150517.4 N.mm= 268.150 KN.m**

**M _{d}= strength reduction factor*M_{n}= 0.9*268.150= 241.335 KN.m**

Since **M _{d}**<

**M**, the beam should be designed as a doubly reinforced concrete beam.

_{u}3- Calculate excess moment (M_{1}) using Equation 15:

**M _{1}=(300.23/0.9)-241.335= 92.253**

**KN.m**

4- Assume compression steel stress reaches yield stress fs′= fy, then compute compression steel area (A's) using equation 5. To compute (d'), assume a compression bar diameter of 22 mm.

d'=cover-diameter of stirrup bar+0.5*diameter of longitudinal bar

d'=40+10+11= 61 mm

**A's= (****92.253***10^{6})/(420*(**408.5**-61))= 632.086 mm^{2}= As1

5- Compute total tensile reinforcement area check if (fs′= fy):

**As= As1+As2= 1862.695+ 632.086= 2494.78 mm ^{2}**

*p*=As/bd= 2494.78/(250*408.5)= 0.02442

*p*'=A's/ bd= 632.086/(250*408.5)= 6.189*10^{-3}

*p*^{-}_{cy}=0.85*0.85*(28/420)*(61/408.5)*(0.003/(0.003-0.0021))+6.189*10^{-3}= 0.03016

Since *p*^{-}_{cy}>*p*, the compression steel is not yielded and the compression steel area should be revised.

6- Compute neutral axis depth (c):

**c=a/B _{1}= 131.484/0.85= 154.687 mm**

7- Use Equation 7 to calculate stress in compression steel:

**fs ^{'}=0.003*200000*((154.687-61)/154.687)= 363.393 MPa **

**As'= (92.253*10 ^{6})/(363.393*(408.5-61))= 730.548 mm^{2}**

8- Determine the number of bars for both compression and tension reinforcement area:

No. of bars for tension zone= Total reinforcement area/ area of single bar

**No. of bar=2494.78/(PI/4*29 ^{2})= 3.7= 4**

No. of bars for compression zone= Total reinforcement area/ area of single bar

**No. of bar= 730.548/ (PI/4*22 ^{2})= 1.92= 2**

Therefore, there are four bars in two layers in tension side of the beam, and two bars in one layer in compression side of the beam.

It should be known that the number of bars has been rounded, hence the area of reinforcement in both compression and tension zone increases. The new steel areas are as follows:

As= 4*(PI/4*29^{2})= 2642.079 mm^{2}

As'= 2*(PI/4*22^{2})= 760.265 mm^{2}

Since the reinforcement area increases, we should check whether the compression steel reaches **363.393** **MPa** or not:

As-As'=2642.079-632.086= 2009.993 mm^{2}

a= (2009.993*420)/(0.85*28*250)= 141.881 mm

c= 141.881/0.85= 166.919 mm

**fs ^{'}=0.003*200000*((166.919-61)/166.919)= 380.733**

**MPa> 363.393 MPa**

For more details regarding checking and estimating the strength reduction factor, please click here.

dt=500-40-10-(29/2)= 435.5mm

c/dt= 166.919/435.5=0.383>0.375, therefore the strength reduction factor is 0.9.

Finally, use equation estimate moment resistant (M_{d}) of the beam, and it should be greater than the applied moment (**M _{u}= 300.23**):

Finally, use equation 12 to estimate design moment resistant (M_{d}) of the beam, and it should be greater than the applied moment (**M _{u}= 300.23**):

**M _{n}=2009.993*420*(408.5-(141.881/2))+380.733*760.265(408.5-61)= 385553383.5 N.mm= 385.553 KN.m**

**M _{d}= 0.9*385.553= 346.998 KN.m**>

**300.23**, therefore the design is safe.

## FAQs

**What is doubly reinforced concrete beam?**

A reinforced concrete beam with steel reinforcement both in tension and compression zone is called a doubly reinforced beam.

**Why is compression reinforcement provided in concrete beams?**

The compression reinforcement is provided to increase the load-carrying capacity of beams. It is possible to increase the moment-carrying capacity of the beam by increasing its dimension.

However, it is not always possible to change dimensions due to architectural considerations. Consequently, a doubly reinforced concrete beam would be the only option to consider.

**What is the difference between doubly reinforced concrete beam and singly reinforced concrete beam?**

A beam that is reinforced in the tension zone only is called a singly reinforced concrete beam, whereas a doubly reinforced concrete beam is reinforced on both tension and compression faces.

**What are the advantages of compression reinforcement other than strength improvement?**

1- Decreased long term deflection of beam.

2- Holds stirrups in their position during the concrete pouring and compaction process.

3. For minimum loading moment

**What are the situations in which compression reinforcement may not yield?**

1- The beam is shallow or wide.

2- Tensile reinforcement ratio of the beam is low.

3. Grade of reinforcement is high.

4- Unusual concrete cover is used over compression reinforcement.

**Read More**

Design of Rectangular Reinforced Concrete Beam

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