Design of Doubly Reinforced Beam in Flexure

Background

  • The problem in increasing the capacity of the beam is the restriction that As should not exceed Asmax. This places a restriction on the maximum flexural capacity of the beam.
  • If As exceeds Asmax, 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 As′ on both faces (tension and compression) of the beam. This will increase the range of Asmax
  • 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 Asfy and 0.85fc′ab provides the couple with lever arm (d – a/2).
  • Mn1 = Asfy (d – a/2) (c)
  • The forces As′fy and As′fs′ provide another couple with lever arm (d – d′).
  • Mn2 = As′fs′ (d – d′) (d)

 

The total nominal capacity of doubly reinforced beam is therefore,

Mn = Mn1 + Mn2 = Asfy (d – a/2) + As′fs′ (d – d′)

 

Factored flexural capacity is given as,

ΦMn = ΦAsfy (d – a/2) + ΦAs′fs′ (d – d′) (e)

Avoid failure, ΦMn ≥ Mu. For ΦMn = Mu, we have from equation (e), Mu = ΦAsfy (d – a/2) + ΦAs′fs′ (d – d′) (f)

Where, ΦAsfy (d – a/2) is equal to ΦMnmax (singly) for As = Asmax

Therefore, Mu = ΦMnmax (singly) + ΦAs′fs′ (d – d′)

{Mu – ΦMnmax (singly)} = ΦAs′fs′ (d – d′)

As′ = {Mu – ΦMnmax (singly)}/ {Φfs′ (d – d′)} ……….……. (g) ; where, fs′ ≤ fy

Maximum Reinforcement

Cc + Cs = T [ ∑Fx = 0 ]

0.85fc′ab + As′fs′ = Astfy

For Amax = β1c = 0.85 × 0.375d; Ast will become Astmax

0.85fc′β10.375db + As′fs′ = Astmaxfy

Astmax = β10.31875bdfc′/fy + As′fs′/fy

Astmax = Asmax (singly) + As′fs′/fy

Cc = Compression force due to concrete in compression region,

Cs = Compression force in steel in compression region needed to

balance the tension force in addition to the tension force provided by Asmax (singly).

  • Astmax = Asmax (singly) + As′fs′/fy
  • The total steel area actually provided Ast as tension reinforcement must be less than Astmax in above equation i.e. Ast ≤ Astmax
  • Astmax (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 ΦMnmax (singly)
  • Step No. 02: Moment to be carried by compression steel
  • Step No. 03: Find εs′ and fs′
  • Step No. 04: Calculation of As′ and Ast.
  • 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: Ast ≤ Astmax
  • 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 fc′ = 3 ksi and fy = 40 ksi.

Step No. 01: Calculation of ΦMnmax (singly)

ρmax (singly) = 0.0203

Asmax (singly) = ρmax (singly)bd = 4.87 in2 ΦMnmax (singly) = 2948.88 in-kip

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

Mu (extra) = Mu – ΦMnmax (singly)

= 4500 – 2948.88 = 1551.12 in-kip

Step No. 03: Find εs′ and fs

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 fs′ = fy Alternatively,

εs′ = (0.003 – 0.008d′/d) (i)

εs′ = (0.003 – 0.008 × 2.5/20) = 0.002 > εy = 40/29000 = 0.00137

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

Step No. 04: Calculation of As′ and Ast.

As′ = Mu(extra)/{Φfs′(d – d′)}=1551.12/{0.90×40×(20–2.5)}= 2.46 in2

Total amount of tension reinforcement (Ast) is,

Ast = Asmax (singly) + As′= 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 = Ast/ Ab= 7.33/ 0.79 = 9.28

No. of bars to be provided on compression side=As′/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:

Ast ≤ Astmax

Ast , which is the total steel area actually provided as tension reinforcement must be less than Astmax .

Astmax = Asmax (singly) + As′fs′/fy

Astmax (singly ) is a fixed number for the case under consideration and As′ is steel area actually placed on compression side.

Asmax (singly) = 4.87 in2 ; As′ = 4 × 0.79 = 3.16 in2

Astmax= 4.87 + 3.16 = 8.036 in2 Ast = 7.9 in2

Therefore Ast = 7.9 in2 < Astmax 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.

 

 

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