In this article reinforced concret beam design is described in detail with solved examples. Beam design is described more in detail in these articles: Flexural Design of Reinforced Concrete Beams, Serviceability of Reinforced Concrete Beams, and Shear Design of Reinforced Concrete Beams.
Loads (Dead & Live), bending moment, and shear diagram of a concrete beam are shown below respectively:
Design assumption:
Therefore, the stress distribution across the section of the beam is as shown below:
At an ultimate strain of 0.003, the stress at extreme fiber of the beam reaches ultimate strength of concrete f_{c}’. The distribution of the compressive stresses is a complex curve. For calculation purpose, a stress block of 0.85fc’ spread over a depth, a, is used. Therefore, the total compressive stress in a rectangular beam is
C = 0.85f_{c}’ab
Where b is the width of the beam.
At ultimate stress situation, the concrete at top portion is subjected to compression. The compressive stresses distribute uniformly over a depth a. The resultant of compressive stress, C is located at a distance, a/2, from the top surface. The tensile force is taken by rebars at an effective distance, d, from the top surface. By equilibrium, the tensile force is equal to the compression resultant,
T = A_{s}f_{y} = C = 0.85f’_{c} ab
where f_{y} is the yield strength of reinforcing steel and A_{s} is the area of steel. Therefore,
The depth of stress block,
a = A_{s}f_{y}/(0.85f’_{c} b), or a = A_{s}f_{y}d/(0.85f’_{c} bd),
Let the reinforcement ratio, r = As/bd, then
a = rf_{y}d/0.85f’_{c}
Let m = f_{y}/0.85f’_{c} , then, a = rdm..The nominal moment strength of the section,
M_{n} = C (d-a/2) = 0.85f’_{c} ab(d-a/2)
Then, The nominal moment strength of the section,
M_{n} = A_{s}f_{y} (d-a/2) = A_{s}f_{y} (d-rdm/2) = A_{s}f_{y} d- A_{s}f_{y} drm/2
ACI code requires that the factored moment,
M_{u} £ f M_{n}
Where, f = 0.9, is the strength reduction factor for beam design. Let M_{u} = f M_{n} , We have M_{u} = f (A_{s}f_{y} d- A_{s}f_{y} drm/2)
Divide both side by bd^{2}, we have M_{u}/fbd = (A_{s}/bd)f_{y} -(A_{s}/bd) f_{y} rm/2) = rf_{y} - f_{y} r^{2}m/2)
Let R_{n} = M_{u}/fbd^{2}, and we can rewrite the equation as
r^{2}(m/2) - r - R_{n}/f_{y} = 0
Solving the equation, the reinforcement ratio,
r = (1/m)[1-(1-2mR_{n}/f_{y})^{1/2}]
The area of reinforcement is A_{s} = rbd
There are two situations when a reinforced concrete beam fails due to bending. One is when the reinforcing steel reaches its yield stress, f_{y}. The other is when the concrete reaches it maximum compressive stress, f’_{c}. When a reinforced concrete beam fails in yielding of steel, the failure is ductile because the steel can stretch for a long period of time before it actually breaks. When it fails in concrete, the failure is brittle because concrete breaks when it reach maximum strain.
When concrete reaches its maximum strain at the same time as the steel reach is yielding stress, it is called a balance condition. Using a maximum strain, 0.003 of concrete and assume a linear distribution of strain across beam section, one can determine the reinforcement ratio at the balanced condition. The reinforcement ratio based on ACI code is
r_{b} = (0.85f’_{c}/f_{y}) b_{1} [87000/(87000+f_{y})] [f’_{c} and f_{y} are in psi (lb/in^{2})]
r_{b} = (0.85f’_{c}/f_{y}) b_{1} [600/(600+f_{y})] [f’_{c} and f_{y} are in MPa (MN/m^{2})]
Where b_{1} = 0.85 for 4000 psi (30 Mpa) concrete, and reduce 0.05 for each 1000 psi of f’_{c} in excess of 4000 psi.
To ensure a ductile failure of the beam, ACI code limits the maximum reinforcement ratio to 0.75r_{b}. On the other hand, when the amount of steel is too small, the beam will fail when concrete reaches its tensile strength. It needs to have a minimum amount of steel to ensure a ductile failure mode. The minimum reinforcement ratio in ACI code is r_{min} = 200/f_{y} (psi).
The direct shear strength according to ACI is
fv_{c} =0.85[1.9Öf_{c}’+2500r_{w}(V_{u}d/M_{u})] £ 0.85(3.5Öf_{c}’)
where r_{w} (» 0.002) is reinforcement ratio, V_{u} is factored shear stress, M_{u} is factored moment at the critical section. Or
fv_{c} =0.85(2Öf_{c}’)
ACI code requirements for shear reinforcement:
When shear stress, v_{u} £ ½ fv_{c} ,no shear reinforcement is required.
A_{v} = 50 b_{w }s /f_{y}
Where s is spacing of web reinforcement, f_{y} is yield strength of steel, A_{v} is cross section area of web reinforcement, b_{w} is width of beam web.
Stirrup reinforcements:
The shear force that is resisted by shear reinforcements is Vs = (Vu - fV_{c}). Normally, stirrup is spaced vertically at a spacing, s, for shear reinforcement. Within an effective depth d, the shear strength provided by A_{v}f_{y}d/s, where A_{v} is area of stirrup, f_{y} is yield strength of reinforcing steel. The shear strength multiply by a reduction factor, f, needs to be larger than Vs. Therefore, V_{s} = f(A_{v}f_{y}d/s). The spacing of stirrup is calculated as
s = (fA_{v}f_{y}d)/V_{s}
ACI code requirements for placing stirrups:
When ½ fv_{c} < v_{u} £ fv_{c}, max s = d/2 £ 24 in.
When fv_{c} < v_{s} £ 4Öf_{c}, max s = d/2 £ 24 in.
Given:
A simply supported reinforced concrete beam is supporting uniform dead and live loads
Design data:
Dead load: 1500 lb/ft
Live load: 800 lb/ft
Length of beam: 20 ft
Width of beam: 16 in
Depth of beam: 24 in
Minimum concrete cover: 1.5 in
Diameter of stirrup, 0.5 in
Compressive strength of concrete: 4000 psi
Yeild strength of steel: 60000 psi
Support column size: 12”x12”
Requirement: Design flexural reinforcement for bending
Solution:
1. Design for longitudinal bars:
Calculate factored moment:
Weight of beam: W_{B} = 150 lb/ft x 1.33 ft x 2 ft = 400 lb/ft
Factored load: Wu = 1.4(400+1500)+1.7(800) = 4020 lb/ft
Factored moment: Mu = (4020)(20^{2})/8 = 201000 ft-lb
Assume the main reinforcement bar is 1" in diameter (#8 bar)
Effective depth: d:24-1.5-0.5-0.5 = 21.5 in
Factor: R_{n} = (201000)(12)/[(0.9)(16)(21.5^{2})]=362.4 psi,
m = 60000/[(0.85)(4000)]=17.65
Reinforcement ratio
r = (1/m)(1-2mR_{n}/f_{y})^{1/2})=0.0064
Minimum reinforcemnet ratio: r_{min} = 200/f_{y}=0.0033
Maximum reinforcement ratio; r_{min} = (0.75)(0.85f’_{c}/f_{y}) b_{1} [87000/(87000+f_{y})]=0.021
Required reinforcement, As = rbd = 2.2 in^{2}.
Use 4#8 bar area of reinforcement is 0.79 in^{2}x4 = 2.37 in^{2}.
2. Design for shear bars (stirrups):
Calculate factored shear:
Clear distance between support, L_{n} = 19 ft
Factor shear V_{u} = W_{u}L_{n}/2 = 38.2 kips
Shear strength of concrete:
fV_{c} = 0.85(2Ö4000) d b = 37 kips
1/2fV_{c} = 18.5 kips
The length that required no shear reinforcement is
L_{1} = (L_{n} /2)(18.5/38.2) = 4.6 ft
Distance from center of beam that required minimum reinforcment is
L_{2} = (L_{n}/2)( fV_{c} /Vu) = 9.2 ft close to L_{n}/2 = 9.5 ft
Use #3 stirrup the area of stirrup, area of steel: A_{v} = 2(0.11 in^{2}) = 0.22 in^{2}.
Maximum spacing, s = (0.22 in^{2})(60000 psi) /[(50 psi)(16 in)] = 16.5 in
Maximum spacing d/2 = 10.75 in (Govern)
Use 6 stirrups at 10.75 inch spacing, with first stirrup at 5". Total length cover by stirrups is L_{s} = (5)(10.75 in)+5 in = 4.9 ft O.K.