# Background

- The problem in increasing the capacity of the beam is the restriction that A
_{s}should not exceed A_{smax}. This places a restriction on the maximum flexural capacity of the beam. - If A
_{s}exceeds A_{smax}, the strain in concrete will reach to a value of 0.003 before εs reaches to 0.005, thus violating the ACI code recommenation for ensuring ductile behavior. - However, If either the strength of concrete is increased or some reinforcement is placed on compression side, the load at which strain will reach to a value of 0.003 will be increased, When this happens As on tension side can be increaseed without compromising ductility, which will also increase the flexural capacity of the beam.
- Practically this can achieved simply by placing some amount of additional reinforcement A
_{s′}on both faces (tension and compression) of the beam. This will increase the range of A_{smax} - In this case the beam is called as doubly reinforced beam.

# Flexural Capacity

- Consider figure d and e, the flexural capacity of doubly reinforced beam consists of two couples:
- The forces A
_{s}f_{y}and 0.85f_{c}′ab provides the couple with lever arm (d – a/2). - M
_{n1}= A_{s}f_{y}(d – a/2) (c) - The forces A
_{s}′f_{y}and A_{s}′f_{s}′ provide another couple with lever arm (d – d′). - M
_{n2}= A_{s}′f_{s}′ (d – d′) (d)

The total nominal capacity of doubly reinforced beam is therefore,

M_{n} = M_{n1} + M_{n2} = A_{s}f_{y} (d – a/2) + A_{s}′f_{s}′ (d – d′)

Factored flexural capacity is given as,

ΦM_{n} = ΦA_{s}f_{y} (d – a/2) + ΦA_{s}′f_{s}′ (d – d′) (e)

Avoid failure, ΦM_{n} ≥ M_{u}. For ΦM_{n} = M_{u}, we have from equation (e), M_{u} = ΦA_{s}f_{y} (d – a/2) + ΦA_{s}′f_{s}′ (d – d′) (f)

Where, ΦA_{s}f_{y} (d – a/2) is equal to ΦM_{nmax} (singly) for A_{s} = A_{smax}

Therefore, Mu = ΦM_{nmax} (singly) + ΦA_{s}′f_{s}′ (d – d′)

{Mu – ΦM_{nmax} (singly)} = ΦA_{s}′f_{s}′ (d – d′)

A_{s}′ = {M_{u} – ΦM_{nmax} (singly)}/ {Φf_{s}′ (d – d′)} ……….……. (g) ; where, f_{s}′ ≤ f_{y}

# Maximum Reinforcement

C_{c} + C_{s} = T [ ∑F_{x} = 0 ]

0.85f_{c}′ab + A_{s′}f_{s}′ = A_{st}f_{y}

For A_{max} = β_{1}c = 0.85 × 0.375d; A_{st} will become A_{stmax}

0.85f_{c}′β10.375db + As′fs′ = A_{stmax}fy

A_{stmax} = β10.31875bdf_{c}′/f_{y} + A_{s}′f_{s}′/f_{y}

A_{stmax} = A_{smax} (singly) + A_{s}′f_{s}′/f_{y}

C_{c} = Compression force due to concrete in compression region,

C_{s} = Compression force in steel in compression region needed to

balance the tension force in addition to the tension force provided by A_{smax} (singly).

- A
_{stmax}= A_{smax}(singly) + A_{s}′f_{s}′/f_{y} - The total steel area actually provided A
_{st}as tension reinforcement must be less than A_{stmax}in above equation i.e. A_{st}≤ A_{stmax} - A
_{stmax}(singly ) is a fixed number, whereas As′ is steel area actually placed on compression side. (For more clarification, see example) - Note that Compression steel in the above equation may or may not yield when tension steel yields.

**Conditions at which fs′ = fy when tension steel yields.**

- By similarity of triangle (fig b), compression steel strain (εs′) is,
- εs′ = εu (c – d′)/ c (h)
- For tensile steel strain (εs) = εt = 0.005 (for under reinforced behavior):
- c = 0.375d
- Substituting the value of c in eqn. (h), we get,
- εs′ = εu (0.375d – d′)/ 0.375d = (0.003 – 0.008d′/d) (i)
- Equation (i) gives the value of εs′ for the condition at which reinforcement on tension side is at strain of 0.005 ensuring ductility.

**Conditions at which fs′ = fy when tension steel yields.**

- εs′ = {0.003 – 0.008d′/d} ……..……………….. (i) OR
- d′/d = (0.003 – εs′)/0.008 (j)
- Substituting εs′ = εy,in equation (j).
- d′/d = (0.003 – εy)/0.008 (k)
- Equation (k) gives the value of d′/d that ensures that when tension steel is at a strain of 0.005 (ensuring ductility), the compression steel will also be at yield.
- Therefore for compression to yield, d′/d should be less than the value given by equation (k).

**Conditions at which fs′ = fy when tension steel yields.**

- Table 3 gives the ratios (d′/d) and minimum beam effective depths (d) for compression reinforcement to yield.
- For grade 40 steel, the minimum depth of beam to ensure that compression steel will also yields at failure is 12.3 inch.

Table 3: Minimum beam depths for compression reinforcement to yield |
||

fy, psi | Maximum d’/d | Minimum d for d’ = 2.5″ (in.) |

40000 | 0.2 | 12.3 |

60000 | 0.12 | 21.5 |

# Design Steps

- Step No. 01: Calculation of ΦM
_{nmax}(singly) - Step No. 02: Moment to be carried by compression steel
- Step No. 03: Find εs′ and fs′
- Step No. 04: Calculation of A
_{s}′ and A_{st.} - Step No. 05: Ensure that d′/d < 0.2 (for grade 40) so that selection of bars does not create compressive stresses lower than yield.
- Step No. 06: Ductility requirements: A
_{st}≤ A_{stmax} - Step No. 07: Drafting

# Practical Example

Design a doubly reinforced concrete beam for an ultimate flexural demand of 4500 in-kip. The beam sectional dimensions are restricted. Material strengths are f_{c}′ = 3 ksi and fy = 40 ksi.

**Step No. 01: Calculation of ΦMnmax (singly)**

ρ_{max} (singly) = 0.0203

A_{smax} (singly) = ρ_{max} (singly)bd = 4.87 in^{2} ΦM_{nmax} (singly) = 2948.88 in-kip

## Step No. 02: Moment to be carried by compression steel

M_{u} (extra) = M_{u} – ΦM_{nmax} (singly)

= 4500 – 2948.88 = 1551.12 in-kip

## Step No. 03: Find ε_{s}′ and f_{s}′

From table 2, d = 20″ > 12.3″, and for d′ = 2.5″, d′/d is 0.125 < 0.20 for grade 40 steel. So compression steel will yield.

Stress in compression steel f_{s}′ = f_{y} Alternatively,

ε_{s}′ = (0.003 – 0.008d′/d) (i)

ε_{s}′ = (0.003 – 0.008 × 2.5/20) = 0.002 > ε_{y} = 40/29000 = 0.00137

As εs′ is greater than εy, so the compression steel will yield.

## Step No. 04: Calculation of A_{s}′ and A_{st}.

A_{s}′ = M_{u}(extra)/{Φf_{s}′(d – d′)}=1551.12/{0.90×40×(20–2.5)}= 2.46 in^{2}

Total amount of tension reinforcement (A_{st}) is,

A_{st} = A_{smax} (singly) + A_{s}′= 4.87 + 2.46 = 7.33 in^{2}

Using #8 bar, with bar area Ab = 0.79 in^{2}

No. of bars to be provided on tension side = A_{st}/ Ab= 7.33/ 0.79 = 9.28

No. of bars to be provided on compression side=A_{s}′/Ab = 2.46/ 0.79 = 3.11

**Provide 10 #8 (7.9 in2 in 3 layers) on tension side and**

**4 #8 (3.16 in2 in 1 layer) on compression side.**

**Step No. 05: **

**Ensure that d′/d < 0.2 (for grade 40) so that selection of bars does not create compressive stresses lower than yield.**

With tensile reinforcement of 10 #8 bars in 3 layers and compression reinforcement of 4 #8 bars in single layer, d = 19.625″ and d′ = 2.375

d′/d = 2.375/ 19.625 = 0.12 < 0.2, OK

**Step No. 06: Ductility requirements: **

A_{st} ≤ A_{stmax}

A_{st} , which is the total steel area actually provided as tension reinforcement must be less than A_{stmax} .

A_{stmax} = A_{smax} (singly) + A_{s}′f_{s}′/f_{y}

A_{stmax} (singly ) is a fixed number for the case under consideration and A_{s}′ is steel area actually placed on compression side.

A_{smax} (singly) = 4.87 in^{2} ; A_{s}′ = 4 × 0.79 = 3.16 in^{2}

A_{stmax}= 4.87 + 3.16 = 8.036 in^{2} A_{st} = 7.9 in^{2}

Therefore A_{st} = 7.9 in^{2} < A_{stmax} OK.

#### Step No. 07: Drafting

**Provide 10 #8 (7.9 in2 in 3 layers) on tension side and 4 #8 (3.16 in2 in 1 layer) on compression side.**