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

1. Concrete is assumed to resist compression only, tension shall be resisted by reinforcements.  Reinforcements shall be placed at the side of the beam that has tension.  For a simply supported beam, tension is at the bottom of beam.  For a cantilever end, tension is at the top of the beam. 
2. Shear is at its maximum at edge of supports.  Diagonal shear cracks is normally developed close to the support.  Stirrup for shear reinforcement is normally placed vertically to intercept the crack.  They are normally closer spaced near the support and gradually spread out toward center of the beam.

Ultimate Strength design of flexural reinforcements

Reinforcement calculation

Design assumption:

1. Strain distribute linearly across the section.
2. Concrete resists only compressive stresses.

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.  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_sf_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_sf_y/(0.85f’_c b), or a = A_sf_yd/(0.85f’_c bd),

Let the reinforcement ratio, r = As/bd, then

a = rf_yd/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_sf_y (d-a/2) = A_sf_y (d-rdm/2) = A_sf_y d- A_sf_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_sf_y d- A_sf_y drm/2)

Divide both side by bd², we have M_u/fbd =  (A_s/bd)f_y -(A_s/bd) f_y rm/2) = rf_y - f_y r²m/2)

Let R_n = M_u/fbd², and we can rewrite the equation as

r²(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

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, f_y.  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 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 balanced condition.  The reinforcement ratio based on ACI code is

r_b = (0.85f’_c/f_y) b₁ [87000/(87000+f_y)]            [f’_c and f_y are in psi (lb/in²)]

r_b = (0.85f’_c/f_y) b₁ [600/(600+f_y)]                    [f’_c and f_y are in MPa (MN/m²)]

Where b₁ = 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 0.75r_b.  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 r_min = 200/f_y (psi).

Design examples

Situation:

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

Requirement:  Design flexural reinforcement for bending

Solution:

1. 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)(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: R_n = (201000)(12)/[(0.9)(16)(21.5²)]=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) b₁ [87000/(87000+f_y)]=0.021

Required reinforcement, As = rbd = 2.2 in².

Use 4#8 bar area of reinforcement is 0.79 in²x4 = 2.37 in².

Shear Reinforcement

Shear strength of concrete

The direct shear shrength according to ACI is

fv_c =0.85[1.9Öf_c’+2500r_w(V_ud/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

1. When shear stress, v_u £ ½ fv_c ,no shear reinforcement is required.
2. When ½ fv_c < v_u £ fv_c, use minimum reinforcement

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.

3. When fv_c < v_u , use v_u £ f(v_c + v_s), where v_s is shear strength provided by shear reinforcement.

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_vf_yd/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_vf_yd/s).  The spacing of stirrup is calculated as

s = (fA_vf_yd)/V_s

ACI code requirements for placing stirrup are as follows.

1. When ½ fv_c < v_u £ fv_c, max s = d/2 £ 24 in.
2. When fv_c < v_s £ 4Öf_c, max s = d/2 £ 24 in.
3. When fv_c < v_s £ 8Öf_c, max s = d/4 £ 12 in.

Design example

Situation:

Design shear reinforcement for the beam in the previous example

Support column size: 12”x12”

Solution:

1. Calculate factored shear:

Clear distance between support, L_n = 19 ft

Factor shear V_u = W_uL_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

L₁ = (L_n /2)(18.5/38.2) = 4.6 ft

Distance from center of beam that required minimum reinforcment is

L₂ = (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 in²) = 0.22 in².

Maximum spacing, s = (0.22 in²)(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.