Serviceability: deflection and crack control

A beam can be perfectly safe and still be a failure. If a floor sags enough to crack the partitions it carries, or if its cracks are wide enough to stain the soffit and let water reach the bars, the owner will rightly complain. Serviceability limit states are checked under specified loads, not factored loads, and they often govern the depth of floors in long spans. This lesson covers the two main checks: deflection and crack control.

Two ways to control deflection

A23.3 offers two routes. The simple route is to make the member at least as thick as a span-based minimum, in which case deflections need not be calculated, provided the member does not support or attach to partitions or other elements likely to be damaged by large deflections. The second route is to calculate the deflections and compare them with permissible limits.

For the office beam, simply supported over a clear span of about 7.0 m, the minimum depth would be \(7000/16 = 438\) mm. The 600 mm beam passes easily. However, the beam supports gypsum partitions, so the table does not apply strictly and Owen checks the deflection by calculation.

Cracked section properties

Deflection calculations need the stiffness of the member, and that depends on how badly it is cracked. Before cracking, the gross moment of inertia \(I_g\) applies. After cracking, at a section where the moment exceeds \(M_{cr}\), the concrete below the neutral axis is ignored and the steel is transformed into an equivalent area \(nA_s\) of concrete. The neutral axis depth \(kd\) of the cracked transformed section comes from equating first moments of area about the axis:

and the cracked moment of inertia is

For a typical beam \(I_{cr}\) is 30 to 50 per cent of \(I_g\). A real beam is cracked only where the moment exceeds \(M_{cr}\); between cracks the concrete still carries some tension (tension stiffening), so the effective stiffness lies between the cracked and gross values.

Effective moment of inertia

A23.3 uses the Branson equation, which blends \(I_g\) and \(I_{cr}\) with a cubic weighting on the ratio of the cracking moment to the applied moment:

Here \(M_a\) is the maximum service moment in the span at the stage being considered. For continuous members A23.3 permits a weighted average of \(I_e\) at midspan and at the supports. The Branson equation was calibrated on beams with steel ratios around 1 per cent and is known to overestimate stiffness for lightly reinforced slabs, which is one reason careful designers check slabs with more conservative assumptions.

Immediate and long-term deflection

With \(I_e\) known, immediate deflections follow from elastic beam formulas. For a simply supported beam under uniform load the midspan deflection is \(5wL^4/(384E_cI_e)\). Because \(I_e\) depends on the moment, the dead load deflection and the deflection under dead plus live load must be computed separately with their own \(I_e\); the live load deflection is the difference. Superposing a live load deflection computed with \(I_e\) at the full load on a dead load deflection with its own \(I_e\) is the correct way to account for the nonlinearity.

Under sustained load, creep and shrinkage increase the deflection. A23.3 multiplies the immediate deflection due to sustained load by

where \(\rho' = A'_s/bd\) is the compression steel ratio at midspan, and \(s\) is a time-dependent factor: 1.0 at 3 months, 1.2 at 6 months, 1.4 at 12 months and 2.0 at 5 years or more. The sustained load is the dead load plus any live load that is present for long periods, such as storage.

The deflection compared with the l/480 or l/240 limit is the part that occurs after the partitions are attached: the long-term creep deflection under sustained load plus the immediate deflection due to live load. Some designers subtract the part of the dead load deflection that occurs before partitions are built; others conservatively include it all, as below.

The result is not a reason to redesign yet. The simple span model is very conservative for this beam, which is continuous with the columns at one end and with the next span at the other. In CIV-321 the corridor frame analysis gave end moments that reduce the midspan moment substantially; a fixed-pinned beam deflects only about 40 per cent as much as a simply supported one. With the continuous moments and a weighted average \(I_e\), the long-term plus live deflection drops to about 12 mm, inside the 15.6 mm limit. Owen records both calculations, and the team adds 2-15M compression bars at midspan as cheap insurance, which reduces \(\zeta_s\) to about 1.8.

import math
# Office floor beam serviceability, CSA A23.3: Branson effective moment of inertia
fc, fy, Es = 30.0, 400.0, 200000.0
b, h, d, As = 350.0, 600.0, 540.0, 2000.0      # 4-25M
L = 7500.0
wD, wL = 20.0, 12.0                             # specified loads, N/mm (= kN/m)
Ec = 4500 * math.sqrt(fc); n = Es / Ec
fr = 0.6 * math.sqrt(fc)
Ig = b * h**3 / 12; yt = h / 2
Mcr = fr * Ig / yt
# cracked transformed section: b*kd^2/2 = n*As*(d - kd)
kd = (-n * As + math.sqrt((n * As) ** 2 + 2 * b * n * As * d)) / b
Icr = b * kd**3 / 3 + n * As * (d - kd) ** 2
print(f"n = {n:.2f}, Mcr = {Mcr/1e6:.1f} kN.m, kd = {kd:.0f} mm")
print(f"Ig = {Ig/1e6:.0f}e6 mm4, Icr = {Icr/1e6:.0f}e6 mm4")

def Ie(Ma):
    if Ma <= Mcr: return Ig
    return min(Ig, Icr + (Ig - Icr) * (Mcr / Ma) ** 3)

def defl(w, I):                                 # simple span, uniform load
    return 5 * w * L**4 / (384 * Ec * I)

MD = wD * L**2 / 8; MDL = (wD + wL) * L**2 / 8
dD = defl(wD, Ie(MD)); dDL = defl(wD + wL, Ie(MDL))
dL = dDL - dD
zeta = 2.0 / (1 + 50 * 0.0)                    # no compression steel, 5 years or more
dlong = zeta * dD + dL                          # sustained dead load creeps, plus live load
print(f"Ie(D) = {Ie(MD)/1e6:.0f}e6, Ie(D+L) = {Ie(MDL)/1e6:.0f}e6 mm4")
print(f"immediate: dead {dD:.1f} mm, dead+live {dDL:.1f} mm, live {dL:.1f} mm")
print(f"live limit L/360 = {L/360:.1f} mm -> {'OK' if dL <= L/360 else 'exceeds'}")
print(f"long-term + live = {dlong:.1f} mm, limit L/480 = {L/480:.1f} mm -> {'OK' if dlong <= L/480 else 'exceeds'}")

# Crack control parameter z = fs*(dc*A)^(1/3)
fs = 0.6 * fy
dc = 30 + 11.3 + 25.2 / 2
A = 2 * dc * b / 4
z = fs * (dc * A) ** (1 / 3)
print(f"crack control: dc = {dc:.0f} mm, A = {A:.0f} mm2, z = {z:.0f} N/mm (interior limit 30000)")

n = 8.11, Mcr = 69.0 kN.m, kd = 182 mm
Ig = 6300e6 mm4, Icr = 2783e6 mm4
Ie(D) = 3199e6, Ie(D+L) = 2885e6 mm4
immediate: dead 10.5 mm, dead+live 18.5 mm, live 8.1 mm
live limit L/360 = 20.8 mm -> OK
long-term + live = 29.0 mm, limit L/480 = 15.6 mm -> exceeds
crack control: dc = 54 mm, A = 9432 mm2, z = 19155 N/mm (interior limit 30000)

Why cracks need control

Flexural cracks in reinforced concrete are inevitable, because the steel cannot work until the concrete around it has cracked. What matters is their width. Narrow, well-distributed cracks are invisible to the casual eye and close up when load is removed; a few wide cracks look alarming, let water and chlorides reach the steel, and in parking structures accelerate corrosion. Crack width grows with the steel stress and with the distance from the bar to the concrete surface, and it falls when the same steel area is provided as more, smaller bars at closer spacing.

A23.3 controls cracking with the Gergely-Lutz parameter

where \(f_s\) is the steel stress at specified load (which may be taken as \(0.6f_y\)), \(d_c\) is the distance from the tension face to the centre of the nearest bar, and \(A\) is the effective tension area of concrete around the bars divided by the number of bars, the area with the same centroid as the steel. The interior limit corresponds to a crack width of about 0.4 mm and the exterior limit to about 0.33 mm. For members exposed to aggressive environments, designers aim lower.

Two further rules complete crack control. Deep beams, deeper than 750 mm, need skin reinforcement on the side faces near the tension zone, because wide cracks can form in the web between the main bars and the neutral axis. And the parking level of the garage, in a chloride exposure, is detailed to the stricter exterior limit, with the top surface protected by a membrane as the parking structure standard requires.

Other serviceability concerns

• Vibration. Long, light floors can bounce under footfall. Concrete floors are heavy and rarely a problem in offices, but floors with spans over about 10 m or with rhythmic activities deserve a frequency check.
• Shrinkage cracking in slabs. Restraint by walls and columns causes cracks unrelated to load; control joints, pour strips and shrinkage reinforcement manage them.
• Camber and construction loads. Shoring removal and early loading at low concrete strength increase long-term deflection; specifying a minimum strength before stripping helps.
• Durability. Cover, concrete quality and crack control together govern the service life of exposed concrete.