Deflection of RC BEAM Using Calculation Method – Worked Example

This article presents a worked example on how to calculate the deflection of a reinforced concrete beam using the rigorous calculation method of cl 7.4.3 in EN 1992-1-1. The beam which deflection is to be checked has already being designed here as a singly reinforced concrete beam. In that article, the beam was designed for flexural and shear strength, and then deflection check was carried out using the deemed-to-satisfy span-effective-depth ratio method. In this article, we shall carry out the deflection check using the calculation method.

 

Beam Design Details

The dimension and the reinforcement details from strength design are given thus:

Beam Dimensions

Beam breadth: 225mm

Beam Depth: 450mm

Beam Length: 5000mm

Reinforcement Details

Beam type: Singly reinforced

Provided Area of tensile reinforcement: 599.8mm2

Required Area of tensile reinforcement: 440.9mm2

 

Loading

For deflection check, serviceability quasi-permanent load should be use. The moment due to quasi-permanent load is derived below

Characteristic Permanent Load (gk): 10.42KN/m

Characteristic Variable Load (qk): 6.25KN/m

Quasi-Permanent Factor (ψ2) = 0.3 (Category A/B Buildings)

Design Quasi-Permanent Line Load: 10.42 + (0.3 x 6.25) = 12.30kN/m

\text { Quasi-Permanent Moment }=\frac{12.3 \times 5^2}{8}=38.42 \mathrm{KNm}Quasi-Permanent Moment =  = 38.42KNm

 

Cement type and exposure conditions

 Cement Class: Class N

Exposure condition: Two short sides & one long side exposed

Relative humidity: 80%

 

Deflection Check

The calculations of deflection check reproduced below was carried out using fppSuite software. You can download the pdf report which is directly from fppSuite here. To also use fppSuite for your structural designs and calculations, click here

Below are the detailed deflection check using calculation method:

 

1. EXPOSED PERIMETER, U 

u= TWO SHORT SIDES + ONE LONG SIDE EXPOSED

u = 2b + h

u = 900.0 mm

 

2.  CROSS-SECTIONAL AREA, Ac

Ac = b × h = 225.0 × 450.0

Ac  = 101250 mm²

 

     3. NOTIONAL SIZE,

h₀ = 2 . Ac / u = 2 × 101250 / 900.0

h₀ = 225.0 mm

 

4. Shrinkage Strain

MEAN CONCRETE STRENGTH,  f_cm

f_cm = f_ck + 8 = 30.0 + 8 = 38N/mm²

 

BASIC DRYING SHRINKAGE COEFFICIENT, εcdo ANNEX B,

For CEMENT CLASS N →  α_ds1 = 4,    α_ds2 = 0.12

ε_cdo = 0.85 × [(220 + 110α_ds1 ) × exp(−α_ds2 × 0.1 × f_cm)] × 1.55 × [1 − (0.01RH)³] × 10−6    (Cl. 3.1.4, Annex B )

ε_cdo = 0.85 × [(220 + 110 × 4) × exp(−0.12 × 0.1 × 38.0)] × 1.55 × [1 − (0.01 × 80.0)³] × 10−6

ε_cdo = 0.00026895

 

NOTIONAL SIZE COEFFICIENT,  TABLE 3.3

h₀ = 225.0 mm → 200 <  h₀ ≤ 300 mm →  h₀ = 0.75

h₀ = 0.75

DRYING SHRINKAGE STRAIN, ε_cd

ε_cd = ε_cdo × k_h = 0.00026895 × 0.75

ε_cd = 0.00020171

 

AUTOGENOUS SHRINKAGE STRAIN,  ε_ca(∞)

ε_ca = 2.5 × (fck − 10) × 10−6

ε_ca = 2.5 × (30.0 − 10) × 10−6

ε_ca = 5e-05

 

TOTAL SHRINKAGE STRAIN, ε_cs

ε_cs = ε_cd + ε_ca = 0.00020171 + 5e-05

ε_cs = 0.00025171

 

    5. Creep Coefficient

f_cm = 38.0 N/mm² > 35 N/mm² → high-strength correction factors   α₁ , α₂ apply.

HIGH-STRENGTH CORRECTION FACTORS

α₁ = (35/f_cm)0.7 = (35/38.0)^0.7 = 0.9441

α₂ = (35/f_cm)0.2 = (35/38.0)^0.2 = 0.9837

 

RELATIVE HUMIDITY FACTOR, Φ_RH

Φ_RH = [1 + (1 − 0.01RH)/ (0.1 × h₀ x 0.333) × α₁] × α₂

Φ_RH = [1 + (1 − 0.01 × 80.0)/ (0.1 × 225.00.333) × 0.9441] × 0.9837

Φ_RH = 1.2896

 

ADJUSTED AGE AT LOADING,

T0 CEMENT CLASS EXPONENT

α  = 0

t₀ = t_0,T × [9/(2 + t_0,T x 1.2) + 1]^ α

t₀  = 28 × [9/(2 + 28 x 1.2) + 1] ^ 0

t₀ = 28.0 days

 

Deflection assessed at t = ∞ → βc (t,t 0 ) = 1.00 (full long-term creep development).

 

CREEP COEFFICIENT,  φ_(t,t0)

φ_(t,t0) = Φ_RH × [16.8 / √fcm] × [1 / (0.1 + t0 0.2)] × βc(t,t0)

φ_(t,t0) = 1.2896 × [16.8 / √38.0] × [1 / (0.1 + 28.00.2)] × 1

φ_(t,t0) = 1.7167

 

 

          6. Effective Modulus of Elasticity

MEAN SECANT MODULUS AT 28 DAYS, E_cm,28    (Annex B.1 Cl. 3.1.3, Table 3.1 & Cl. 7.4.3(5))

E_cm,28 = 22 × [(f_ck + 8)/10]0.3 = 22 × [(30.0 + 8)/10]0.3

E_cm,28 = 32.84 GPa (kN/mm²)

 

Age at loading t_0,T = 28 days → no strength-gain adjustment required,  E_cm = E_cm,28

 

EFFECTIVE MODULUS OF ELASTICITY,  E_c,eff   EQ. 7.20

E_c,eff =  E_cm/ (1 + φ(t,t0)) = 32.84 / (1 + 1.7167)

E_c,eff = 12.09 GPa (kN/mm²)

 

        7.    Modular Ratio

SHORT-TERM (ELASTIC) MODULAR RATIO,   α_e

α_e  = Es / Ecm = 200 / 32.84

α_e = 6.091

LONG-TERM (EFFECTIVE) MODULAR RATIO, α_e,LT

α_e,LT = Es / Ec,eff = 200 / 12.09

α_e,LT = 16.547

 

 

     8. Cracking Moment & Distribution Coefficient

MEAN TENSILE STRENGTH,  f_ctm

f_ctm= 0.3 ×  f_ck 2/3

f_ctm = 0.3 × 30.0  x 2/3

f_ctm = 2.896 N/mm²

 

CRACKING MOMENT, M_cr

M_cr = f_ctm × (b × h²) / 6 × 10−6

M_cr = 2.896 × (225.0 × 450.0²) / 6 × 10−6

M_cr = 21.995 kNm

 

M = 38.42 kNm > Mcr = 21.995 kNm → SECTION CRACKED UNDER SERVICE MOMENT

 

DISTRIBUTION COEFFICIENT (SHORT-TERM), ζ

Β = 1.0

ζ = max[0, 1 − β × ( M_cr /M)²] = max[0, 1 − 1 × (21.995/38.42)²]

ζ = 0.6723

 

DISTRIBUTION COEFFICIENT (LONG-TERM),

Β = 0.5 (SUSTAINED/CYCLIC LOADING)

ζ_LT  = max[0, 1 − β × (Mcr/M)²] = max[0, 1 − 0.5 × (21.995/38.42)²]

ζ_LT  = 0.8361

 

 

          9. Second Moment of Area

 

Cracked Neutral Axis depth (SHORT-TERM)

x = [−α_e  As + √((α_e As)² + 2 b α_e As d)] / b

x = [−6.091 × 599.8 + √((6.091 × 599.8)² + 2 × 225.0 × 6.091 × 599.8 × 407.0)] / 225.0

x = 99.87 mm

 

CRACKED SECOND MOMENT OF AREA,  Icr (SHORT-TERM)

Icr = (b × x³)/3 + α_e × As × (d − x) ²

Icr = (225.0 × 99.87³)/3 + 6.091 × 599.8 × (407.0 − 99.87)²

Icr = 419314151 mm⁴

 

CRACKED NEUTRAL AXIS DEPTH, x_LT (LONG-TERM)

x_LT = [−α_e,LT As + √((α_e,LT  As)² + 2 b α_e,LT As  d)] / b

x_LT = 150.44 mm

 

CRACKED SECOND MOMENT OF AREA, ICR,LT (LONG-TERM)

Icr,_LT = (b × xLT³)/3 + α_e,LT × As × (d − x_LT)²

Icr,_LT   =   908637835mm⁴

 

UNCRACKED (GROSS) SECOND MOMENT OF AREA, I_UNCR

I_UNCR = b × h³ / 12 = 225.0 × 450.0³ / 12

I_UNCR = 1708593750 mm⁴

 

 

       10.   Curvature Due to Loading

 

CRACKED CURVATURE, (1/r)_cr  — SHORT-TERM & LONG-TERM

(1/r)_cr = M × 10⁶ / (Ecm × 10³ × Icr)

(1/r)_cr = 38.42 × 10⁶ / (32.84 × 10³ × 419314151)

(1/r)_cr = 2.79e-06 mm−1

 

(1/r)_cr,LT = M × 106 / (Ec,eff × 10³ × I_cr,LT)

(1/r)_cr,LT = 38.42 × 106 / (12.09 × 10³ × 908637835)

(1/r)_cr,LT = 3.498e-06 mm−1

 

UNCRACKED CURVATURE, (1/r)_uncr — SHORT-TERM & LONG-TERM

(1/r)_uncr = M × 106 / (Ecm × 10³ × I_,uncr)

(1/r)_uncr = 38.42 × 106 / (32.84 × 10³ × 1708593750)

(1/r)_uncr = 6.85e-07 mm−1

 

(1/r)_uncr,LT = M × 10⁶ / (E_c,eff × 10³ × I_uncr)

(1/r)_uncr,LT  = 38.42 × 10⁶ / (12.09 × 10³ × 1708593750)

(1/r)_uncr,LT = 1.86e-06 mm−1

 

INTERPOLATED (LOADING) CURVATURE,  (1/r)_m

(1/r)_m  = ξ (1/r)_cr + (1 − ξ)   (1/r)_uncr

(1/r)_m = 0.6723 × 2.79e-06 + (1 − 0.6723) × 6.85e-07

 

(1/r)_m,LT  =    ζ_LT (1/r)_cr,LT + (1 − ζ_LT)   (1/r)_uncr

(1/r)_m,LT = 0.8361 × 3.498e-06 + (1 − 0.8361) × 1.86e-06

(1/r)_m,LT = 3.23e-06 mm−1

 

 

             11.  Curvature due to Shrinkage

FIRST MOMENT OF REINFORCEMENT AREA ABOUT CENTROID,  Scr

 

Instantaneous cracked section

Scr = As × (d − x)

Scr = 599.8 × (407.0 − 99.87)

Scr = 184218 mm³

 

Long-term cracked section

S_cr,LT = As × (d −  x_LT)

S_cr,LT = 599.8 × (407.0 − 150.44)

S_cr,LT = 153882 mm³

 

Uncracked Section

S_uncr = As × (d − h/2)

S_uncr = 599.8 × (407.0 − 450.0/2)

S_uncr = 109164 mm³

 

CRACKED SHRINKAGE CURVATURE, (1/RCS)CR

(1/r_cs)cr = S_cr × α_e ×  S_cr/ Icr

(1/r_cs)cr   = 0.00025171 × 6.091 × 184218 / 419314151

(1/r_cs)cr  = 6.74e-07 mm−1

 

(1/r_cs)cr,LT = S_cr,LT × α_e,LT × Scr,LT / Icr,LT

(1/r_cs)cr,LT = 0.00025171 × 16.547 × 153882 / 908637835

(1/r_cs)cr,LT = 7.05e-07 mm−1

 

UNCRACKED SHRINKAGE CURVATURE, (1/RCS)UNCR

(1/r_cs)uncr = ε_cs × α_e × S_uncr /  I_uncr

(1/r_cs)uncr = 0.00025171 × 6.091 × 109164 / 1708593750

(1/r_cs)uncr = 9.8e-08 mm−1

 

(1/r_cs)uncr,LT = ε_cs × α_eLT × S_uncr / I_uncr

(1/r_cs)uncr,LT   = 0.00025171 × 16.547 × 109164 / 1708593750

(1/r_cs)uncr,LT = 2.66e-07 mm−1

 

INTERPOLATED SHRINKAGE CURVATURE, (1/RCS)

(1/r_cs) = ζ × (1/rcs)cr + (1 − ζ) × (1/rcs)uncr (1/rcs)

(1/r_cs)  = 0.6723 × 6.74e-07 + (1 − 0.6723) × 9.8e-08

(1/r_cs) = 4.85e-07 mm−1

 

(1/r_cs,LT) = 0.8361 × 7.05e-07 + (1 − 0.8361) × 2.66e-07

(1/r_cs,LT) = 6.33e-07 mm−1

 

Total Curvature (Loading & Shrinkage)

 

(1/r)total = (1/r)M + (1/rcs)

(1/r)total = 2.1e-06 + 4.85e-07

(1/r)total = 2.585e-06 mm−1

 

(1/r)total,LT = (1/r)M,LT + (1/rcs,LT)

(1/r)total,LT  = 3.23e-06 + 6.33e-07

(1/r)total,LT = 3.863e-06 mm−1

 

 

             12.  DEFLECTION COEFFICIENT, K

STRUCTURAL SYSTEM: SIMPLY SUPPORTED, k = 0.104

INSTANTANEOUS (SHORT-TERM) DEFLECTION, a

a = k × L² × (1/r)total

a = 0.104 × 5000.0² × 2.585e-06

a = 6.72 mm

 

LONG-TERM DEFLECTION,   a_LT

a_LT = k × L² × (1/r)total,LT

a_LT = 0.104 × 5000.0² × 3.863e-06

a_LT = 10.04 mm

 

Deflection Check

LIMITING DEFLECTION,  a_lim

a_lim = L/250 = 5000.0 / 250

a_lim = 20.0 mm

 

Instantaneous deflection a = 6.72 mm vs a_lim = 20.0 mm, deflection passed

Long-term deflection a_LT = 10.04 mm vs a_lim = 20.0 mm, deflection passed