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_sf_y and 0.85f_c′ab provides the couple with lever arm (d – a/2).
• M_n1 = A_sf_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_sf_y (d – a/2) + A_s′f_s′ (d – d′)

Factored flexural capacity is given as,

ΦM_n = ΦA_sf_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_sf_y (d – a/2) + ΦA_s′f_s′ (d – d′) (f)

Where, ΦA_sf_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_stf_y

For A_max = β₁c = 0.85 × 0.375d; A_st will become A_stmax

0.85f_c′β10.375db + As′fs′ = A_stmaxfy

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.

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 in² 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 in²

Total amount of tension reinforcement (A_st) is,

A_st = A_smax (singly) + A_s′= 4.87 + 2.46 = 7.33 in²

Using #8 bar, with bar area Ab = 0.79 in²

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 in² ; A_s′ = 4 × 0.79 = 3.16 in²

A_stmax= 4.87 + 3.16 = 8.036 in² A_st = 7.9 in²

Therefore A_st = 7.9 in² < 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.