Reinforced concrete beam design
Beam stresses under loads
Moment and shear diagram of a beam under dead and live
loads are shown below.
Failure modes and reinforcements
Ultimate Strength design of flexural reinforcements
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 fc’. 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 = 0.85fc’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. 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 = Asfy = C = 0.85f’c ab
where fy is the yield strength of reinforcing steel and As is the area of steel. Therefore,
The depth of stress block,
a = Asfy/(0.85f’c b), or a = Asfyd/(0.85f’c bd),
Let the reinforcement ratio, r = As/bd, then
a = rfyd/0.85f’c
Let m = fy/0.85f’c , then, a = rdm..The nominal moment strength of the section,
Mn = C (d-a/2) = 0.85f’c ab(d-a/2)
Then, The nominal moment strength of the section,
Mn = Asfy (d-a/2) = Asfy (d-rdm/2) = Asfy d- Asfy drm/2
ACI code requires that the factored moment,
Mu £ f Mn
Where, f = 0.9, is the strength reduction factor for beam design. Let Mu = f Mn , We have Mu = f (Asfy d- Asfy drm/2)
Divide both side by bd2, we have Mu/fbd = (As/bd)fy -(As/bd) fy rm/2) = rfy - fy r2m/2)
Let Rn = Mu/fbd2, and we can rewrite the equation as
r2(m/2) - r - Rn/fy = 0
Solving the equation, the reinforcement ratio,
r = (1/m)[1-(1-2mRn/fy)1/2]
The area of reinforcement is As = rbd
Ductile and brittle failures, Balance condition, Maximum and minimum reinforcement ratio
There are two situations when a reinforced concrete beam
fails due to bending. One is when
the reinforcing steel reaches its yield stress, fy.
The other is when the concrete reach 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
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 balanced condition. The reinforcement ratio based on ACI code is
rb = (0.85f’c/fy) b1 [87000/(87000+fy)] [f’c and fy are in psi (lb/in2)]
rb = (0.85f’c/fy) b1 [600/(600+fy)] [f’c and fy are in MPa (MN/m2)]
= 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 beam, ACI code limits the maximum reinforcement ratio to 0.75rb. On the other hand, when the amount of steel is too small, the beam will fail when concrete reach 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 rmin = 200/fy (psi).
A simply supported reinforced concrete beam is supporting uniform dead and live loads
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
Requirement: Design flexural reinforcement for bending
1. Calculate factored moment:
Weight of beam: WB = 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)(202)/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: Rn = (201000)(12)/[(0.9)(16)(21.52)]=362.4 psi,
m = 60000/[(0.85)(4000)]=17.65
r = (1/m)(1-2mRn/fy)1/2)=0.0064
Minimum reinforcemnet ratio: rmin = 200/fy=0.0033
Maximum reinforcement ratio; rmin = (0.75)(0.85f’c/fy) b1 [87000/(87000+fy)]=0.021
Required reinforcement, As = rbd = 2.2 in2.
Use 4#8 bar area of reinforcement is 0.79 in2x4 = 2.37 in2.
Shear strength of concrete
The direct shear shrength according to ACI is
fvc =0.85[1.9Öfc’+2500rw(Vud/Mu)] £ 0.85(3.5Öfc’)
where rw (» 0.002) is reinforcement ratio, Vu is factored shear stress, Mu is factored moment at the critical section. Or
ACI code requirements for shear reinforcement
Av = 50 bw s /fy
Where s is spacing of web reinforcement, fy is yield strength of steel, Av is cross section area of web reinforcement, bw is width of beam web.
The shear force that is resisted by shear reinforcements is Vs = (Vu - fVc). Normally, stirrup is spaced vertically at a spacing, s, for shear reinforcement. Within an effective depth d, the shear strength provided by Avfyd/s, where Av is area of stirrup, fy is yield strength of reinforcing steel. The shear strength multiply by a reduction factor, f, needs to be larger than Vs. Therefore, Vs = f (Avfyd/s). The spacing of stirrup is calculated as
s = (fAvfyd)/Vs
ACI code requirements for placing stirrup are as follows.
Design shear reinforcement for the beam in the previous example
Support column size: 12”x12”
Calculate factored shear:
distance between support, Ln = 19 ft
shear Vu = WuLn/2 = 38.2 kips
strength of concrete:
fVc = 0.85(2Ö4000) d b = 37 kips
= 18.5 kips
length that required no shear reinforcement is
= (Ln /2)(18.5/38.2) = 4.6 ft
from center of beam that required minimum reinforcment is
= (Ln/2)( fVc
/Vu) = 9.2 ft close to Ln/2
= 9.5 ft
#3 stirrup the area of stirrup, area of steel: Av = 2(0.11 in2)
= 0.22 in2.
spacing, s = (0.22 in2)(60000 psi) /[(50 psi)(16 in)] = 16.5 in
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 Ls = (5)(10.75 in)+5 in = 4.9 ft O.K.
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