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Chapter 8 · 6 hours

Limit States of Serviceability: Deflection and Cracking

IOE past exam questions

Past questions and answers

5 questions set from this chapter, 1 of them more than once; 1 is most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 3 of 25 exams
  • Asked 3 times
  • 2079 Baisakh · 4 marks
  • 2065 Shrawan (old course) · 7 marks
  • 2065 Kartik (old course) · 6 marks

Explain how the deflection of an RC flexural member is controlled by the empirical (sufficient stiffness) method as per IS 456.

Answer

Idea of the method

The actual calculation of deflection is long, so IS 456 cl. 23.2.1 allows deflection to be controlled indirectly. If the ratio of span to effective depth of a beam, slab or cantilever is not more than a permitted value, the deflection will be within the limits of cl. 23.2 (final deflection including creep, shrinkage and temperature ≤span/250\le \text{span}/250, and the part occurring after finishes ≤span/350\le \text{span}/350 or 20 mm). This is the empirical, or sufficient stiffness (span/depth), method.

Procedure

  1. Take the basic span/effective depth ratio:
MemberBasic l/dl/d
Cantilever7
Simply supported20
Continuous26
  1. Modify it by factors:
    • ktk_t: modification factor for tension steel from Fig. 4 of IS 456, which depends on ptp_t and the service stress fs=0.58fyAst,reqAst,provf_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}}. More steel and lower stress give a larger factor (up to 2.0).
    • kck_c: modification factor for compression steel from Fig. 5, based on pc=100Asc/bdp_c = 100A_{sc}/bd (maximum 1.5).
    • kfk_f: flanged beams factor from Fig. 6, based on bw/bfb_w/b_f and ptp_t.
  2. For spans above 10 m (other than cantilevers) the basic value is multiplied by 10/span (m)10/\text{span (m)}.
  3. The permissible ratio is
(ld)perm=(ld)basic×kt×kc×kf\left(\frac{l}{d}\right)_{perm} = \left(\frac{l}{d}\right)_{basic} \times k_t \times k_c \times k_f
  1. Compare with the actual l/dl/d. If actual ≤\le permissible, deflection is satisfied; if not, increase dd (or the steel) and check again.

Notes

  • For two-way slabs the check is made on the shorter span.
  • This method is a conservative approximation of Annex C (the detailed method, which adds short-term, creep and shrinkage deflections).
  • 2076 Chaitra · 2+2 marks

Describe the method of controlling deflection and cracking in RCC structure.

Answer

Controlling deflection

  • Span/depth ratio method (IS 456 cl. 23.2.1): keep l/dl/d within the basic value (7, 20 or 26 for cantilever, simply supported or continuous member) multiplied by the modification factors for tension steel, compression steel and flanged section.
  • Increase the depth of the member or provide more tension steel (this reduces service stress) and add compression steel to cut creep deflection.
  • Use better concrete (higher modulus), proper curing to reduce shrinkage, and keep a low water-cement ratio.
  • Limit the deflection to span/250\text{span}/250 (total) and span/350\text{span}/350 or 20 mm (after finishes).
  • Camber or pre-camber the member, and delay the fixing of partitions and finishes.

Controlling cracking

  • Limit the bar spacing to the maximum allowed (IS 456 cl. 26.3.3, 43 and Table 15) and use many small bars rather than a few large ones.
  • Provide sufficient cover (not more than 50 mm to the tension bars) and side-face steel (0.1% of web area) in deep beams (D>750D > 750 mm).
  • Limit the steel stress and the crack width to 0.3 mm (IS 456 cl. 35.3.2) - checked by the calculation of Annex F when necessary.
  • Control shrinkage and thermal cracks by proper curing, minimum steel and contraction joints.
  • 2076 Asoj · 3+2 marks

Describe methods of controlling deflection and crack width in RCC structures. Describe the empirical formula for calculating the design surface crack width.

Answer

Methods of controlling deflection

  • Keep the span/effective-depth ratio within the permitted value =(l/d)basic×kt×kc×kf= (l/d)_{basic} \times k_t \times k_c \times k_f (basic 7, 20, 26 for cantilever, simply supported, continuous; IS 456 cl. 23.2.1).
  • Increase the depth or the amount of tension steel, add compression steel, and use good-quality concrete with proper curing.
  • Limit the total deflection to span/250 and the part after finishes to span/350 or 20 mm.

Methods of controlling crack width

  • Limit the spacing of bars (maximum clear spacing from Table 15 and cl. 26.3.3: 300 mm in slabs, 3d3d for main bars) and use small-diameter bars well distributed over the tension zone.
  • Provide adequate cover and side-face reinforcement of 0.1% of the web area in beams deeper than 750 mm.
  • Keep the steel stress low; provide the minimum steel; use proper curing.
  • The surface crack width should not exceed 0.3 mm (0.2 mm in aggressive environments), cl. 35.3.2.

Design surface crack width (IS 456 Annex F)

wcr=3 acr εm1+2 (acr−cmin)h−xw_{cr} = \frac{3\, a_{cr}\, \varepsilon_m}{1 + \dfrac{2\,(a_{cr} - c_{min})}{h - x}}

where

  • acra_{cr} = distance from the point considered (on the surface) to the surface of the nearest longitudinal bar,
  • cminc_{min} = minimum clear cover to the longitudinal bar,
  • hh = overall depth, xx = depth of the neutral axis,
  • εm\varepsilon_m = average strain at the level considered:
εm=ε1−bt (h−x)(a′−x)3 EsAs (d−x)\varepsilon_m = \varepsilon_1 - \frac{b_t\,(h - x)(a' - x)}{3\,E_s A_s\,(d - x)}

ε1\varepsilon_1 is the strain at the level considered (calculated ignoring the tensile stiffening of concrete), btb_t is the width of the section at the centroid of tension steel, a′a' is the distance from the compression face to the point considered, EsE_s and AsA_s are the modulus and area of tension steel.

  • 2073 Shrawan · 5 marks

Discuss the methods of crack control as per IS 456-2000 in RC structures.

Answer

Cracking is a serviceability limit state: cracks spoil the appearance and let moisture reach the steel, causing corrosion. IS 456-2000 does not require a crack width calculation for every member; instead it controls cracking by detailing rules and, where needed, by a check on crack width.

Methods of crack control

  1. Limit the bar spacing (cl. 26.3.3(b), cl. 43 and Table 15).
    • In slabs, the horizontal distance between parallel main bars should not exceed 3d3d or 300 mm, whichever is less; for distribution bars 5d5d or 450 mm.
    • In beams, the clear distance between tension bars is limited by Table 15 (it depends on the service stress, i.e. on fyf_y and the redistribution of moments), e.g. a maximum clear spacing of about 180 to 300 mm.
  2. Use small-diameter bars well distributed around the tension zone instead of a few large bars.
  3. Side-face reinforcement: in beams with overall depth above 750 mm, provide 0.1% of the web area on each face, spaced not more than 300 mm or the web thickness, to control cracks on the sides (cl. 26.5.1.3).
  4. Minimum steel: provide at least the minimum tension steel (e.g. 0.85bd/fy0.85bd/f_y for beams, 0.12-0.15% for slabs) so that cracking does not occur suddenly.
  5. Limit the cover to the tension reinforcement to 50 mm; larger cover needs extra mesh to control cracks. Adequate cover is also needed for durability.
  6. Limit the crack width: the surface width of a crack should not exceed 0.3 mm in members where cracking is not harmful (0.2 mm in aggressive environment, 0.1 mm in severe) - cl. 35.3.2. It is calculated by the formula of Annex F when required.
  7. Control shrinkage and temperature cracks: proper curing, low water-cement ratio, contraction/expansion joints and distribution steel.
  • 2082 Baisakh · 4 marks

Define the limit state of serviceability. Differentiate between short-term and long-term deflection.

Answer

Limit state of serviceability is the state beyond which the structure, though safe against collapse, can no longer be used satisfactorily under service (working) loads because of excessive deflection, cracking or vibration, or because its durability is affected. IS 456 cl. 35 covers deflection, cracking, durability and vibration; deflection and cracking are checked in design with partial safety factor γf=1.0\gamma_f = 1.0 on working loads.

Short-term and long-term deflection

BasisShort-term (immediate) deflectionLong-term deflection
CauseLoad applied soon after casting; elastic behaviour of concreteSustained load causes creep, plus shrinkage of concrete
ModulusEc=5000fckE_c = 5000\sqrt{f_{ck}} MPaEffective modulus Ece=Ec1+θE_{ce} = \dfrac{E_c}{1+\theta} (θ\theta = creep coefficient: 1.6 at 28 days)
OccursImmediately and is recoverableSlowly over months or years and is not recoverable
ValueSmallerLarger (typically 2 to 3 times the short-term)
Calculationa=5wl4384EcIeffa = \dfrac{5wl^4}{384E_cI_{eff}} with IeffI_{eff} from Annex Catotal=ai,perm+acc+asha_{total} = a_{i,perm} + a_{cc} + a_{sh} (creep + shrinkage; shrinkage strain 0.0003)

The total deflection due to all loads, including creep and shrinkage, should not exceed span/250, and the part occurring after construction of partitions and finishes should not exceed span/350 or 20 mm, whichever is less (IS 456 cl. 23.2).

Questions from Old Question Collection (CE 702) (IOE exam papers from 2065 to 2082 (CE 702, BCE IV/I)) and Old Question Collection (CE 702) (4 IOE papers 2075 to 2079 (only 2079 Baisakh not already in the other file)). Answers are written for this site; check them against your class notes.

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