Design of Doubly Reinforced Beam in Flexure

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)

Flexural Capacity

 

The total nominal capacity of doubly reinforced beam is therefore,

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

 

Flexural Capacity

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).

Maximum Reinforcement

Maximum Reinforcement

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.

Design a doubly reinforced concrete beam for an ultimate flexural

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.

Drafting

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