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Chapter 7 · 4 hours

Reinforcement detailing: Codal Provisions

IOE past exam questions

Past questions and answers

5 questions set from this chapter, 1 of them more than once. Most repeated first.

  • Asked 2 times
  • 2079 Baisakh · 4 marks
  • 2067 Asar (old course) · 8 marks

What do you understand by curtailment of tension steel in simple beams? Discuss the actual and theoretical point of curtailment of re-bar, showing by illustrating a neat sketch.

Answer

Meaning of curtailment

Curtailment means stopping (cutting off) some of the tension bars of a beam where the bending moment has fallen so that fewer bars are needed. The main bars are designed for the maximum bending moment at mid-span, but near the supports the moment is small, so the same amount of steel is not needed there. Curtailing the extra bars saves steel and reduces congestion at the supports, but it must be done so that the remaining bars can still carry the moment and the shear and the bar has enough bond length.

Theoretical and actual cut-off points

  • Theoretical cut-off point: the section where the bending moment diagram shows that the bar is no longer required, i.e. the moment of resistance of the remaining bars is equal to the applied moment MuM_u.
  • Actual cut-off point: the bar cannot be stopped exactly at the theoretical point, because the stress in the bar is still developing there and the real moment diagram may differ from the assumed one. IS 456 cl. 26.2.3.2 therefore requires every bar to be extended beyond the theoretical cut-off point by a distance not less than the effective depth dd or 12ϕ12\phi, whichever is greater.
 BM diagram of a simply supported beam (sagging)

        ______ Mu = Mu1 (all bars)
      /          \
     /            \___ Mu2 (remaining bars)
    /                  \
   A----|------|-------|------------->
        P     Q   extension
        P = theoretical cut-off point
        Q = actual cut-off point
        PQ >= max(d, 12 phi)

At the theoretical point P the moment of resistance of the remaining bars is just equal to the applied moment. The bar is carried on to Q, so that it is fully effective up to P.

Conditions to be satisfied (IS 456 cl. 26.2.3)

  1. The bar must also have a development length LdL_d measured from the point of maximum stress, i.e. the length of bar beyond the section of maximum moment must be at least LdL_d.
  2. Flexural bars must not be cut in a tension zone unless one of the following is satisfied at the cut-off point: (a) the shear there does not exceed two-thirds of the permitted shear; or (b) extra stirrups are provided over a distance of 3d/43d/4 from the cut-off; or (c) for bars of 36 mm or less, the continuing bars have double the area needed for flexure and the shear does not exceed three-quarters of the permitted value.
  3. At least one-third of the positive moment steel (one-fourth in continuous members) must be carried into the support.
  4. At a support, at least one-third of the negative moment steel must extend beyond the point of inflection by at least dd, 12ϕ12\phi or l/16l/16, whichever is greater.
  5. Bars are curtailed symmetrically and in stages, never all at one section.
  • 2076 Chaitra · 2 marks

What is splicing and why is it required in RCC structures?

Answer

Splicing is the joining of two reinforcing bars end to end so that the force in one bar is transferred to the other. The joint may be a lap, a weld or a mechanical coupler.

It is required because:

  • Bars are supplied in standard lengths (usually 12 m), while beams, columns and walls are often longer, or are built in stages (e.g. column bars lapped at every floor).
  • It gives continuity of the steel across construction joints.
  • It lets short or left-over lengths be used and makes bars easy to transport and handle.

The splice must be located where stress is low and must be long enough to transfer the full bar force through bond, as per IS 456 cl. 26.2.5.

  • 2071 Chaitra · 2 marks

Enlist and make sketch of three kinds of mechanical splices.

Answer

Mechanical splices join bars end to end with a coupling device, without overlapping the bars. Three common kinds are:

  1. Threaded (screwed) coupler - the bar ends are threaded (parallel or taper) and screwed into a long internally threaded sleeve.
  2. Swaged (cold-pressed) sleeve - a steel sleeve is slipped over the two bar ends and squeezed with a hydraulic press so that it grips the ribs of both bars.
  3. Grouted sleeve (or wedge-lock sleeve) - the bar ends are inserted into a steel sleeve that is filled with high-strength non-shrink grout (or locked by wedges).
1. Threaded coupler
   ====[ thread |  thread ]====
       <- internal thread ->

2. Swaged sleeve
   ======(##########)======
         pressed zone

3. Grouted sleeve
   =====[ grout filled   ]=====
        <- sleeve + grout ->

Mechanical splices are used for large bars (above 36 mm) and where lapping would cause congestion.

  • 2068 Baisakh (old course) · 5 marks

What do you understand by splicing of bars? Write down the primary conditions for the application of splicing in reinforced concrete structures.

Answer

Splicing of bars is the process of joining two bars end to end so that stress is transferred from one to the other. The usual methods are (i) lap splice, where the bars overlap and force passes through bond with the surrounding concrete, (ii) welded splice, and (iii) mechanical splice with couplers.

Primary conditions for splicing (IS 456 cl. 26.2.5)

  1. Location: splices should be away from sections of maximum tensile stress. They are best placed where the bending moment is small, e.g. column bars just above floor level, beam bottom bars near the supports.
  2. Staggering: splices should be staggered, and not more than about half of the bars should be spliced at one section. Bars in a lap are placed in contact, or with a clear gap not more than 4ϕ4\phi so that stress is transferred properly.
  3. Lap length (including anchorage value of hooks):
    • flexural tension: not less than LdL_d or 30ϕ30\phi, whichever is greater;
    • direct tension: not less than 2Ld2L_d or 30ϕ30\phi;
    • compression: not less than LdL_d in compression or 24ϕ24\phi.
  4. Bar size: lap splices are not used for bars larger than 36 mm. Such bars are welded or joined with mechanical couplers (the strength of the joint must be at least that of the bar).
  5. Transverse steel: closely spaced stirrups/ties are provided around the lap zone to confine the concrete and prevent splitting.
  6. In members subjected to repeated or reversing load, welding and mechanical splices are preferred to laps, and in seismic zones lap splices are placed at the middle half of the column height.
  • 2079 Bhadra · 3+3 marks

Discuss about requirements for good detailing. Also describe bar bending schedule.

Answer

Requirements for good detailing

Reinforcement detailing means the exact arrangement of bars shown on the drawings. Good detailing should:

  1. Follow the code (IS 456, SP 34, IS 13920 in seismic regions) for cover, spacing, anchorage, lap and curtailment.
  2. Provide proper anchorage and development length to every bar, with hooks or bends where space is limited.
  3. Keep reinforcement simple: use few bar diameters, standard shapes and regular spacing so that it is easy to fabricate and place.
  4. Avoid congestion, especially at beam-column joints, so that concrete can flow and be compacted; keep the minimum clear spacing between bars.
  5. Give the specified cover (spacers, cover blocks) for durability and fire resistance.
  6. Place bars where the stresses are: tension steel on the tension face, curtailed at correct points, with stirrups closely spaced near supports.
  7. Show clear drawings: plan, elevation and sections with bar marks, diameters, spacing, lap positions and notes.
  8. Allow for construction: bars must be capable of being placed in the sequence of erection, and lap positions and construction joints must be shown.

Bar bending schedule (BBS)

A bar bending schedule is a table prepared from the structural drawing that lists every bar of a member with its mark, diameter, shape, number, length of each bar, total length and weight. It is used for ordering, cutting, bending, fixing and checking the steel, and for estimating the quantity (and cost).

Steps:

  1. Number each different bar (bar mark) on the drawing.
  2. Find the cutting length of each bar = straight length + hooks + bends - bend deductions (a 90° bend deducts 1ϕ1\phi to 2ϕ2\phi, a 45° bend 0.5ϕ0.5\phi, a hook adds 9ϕ9\phi for a standard U hook in IS 2502 practice).
  3. Total length = number × length; weight = total length × unit weight, where unit weight =ϕ2162= \dfrac{\phi^2}{162} kg/m (ϕ\phi in mm).
Bar | Dia | Shape  | No. | Length | Total | Weight
mark| mm  |        |     | (m)    | L (m) | (kg)
----+-----+--------+-----+--------+-------+-------
 A  | 16  | straight|  3 | 4.20   | 12.60 | 19.9
 B  | 12  | cranked |  2 | 4.40   |  8.80 |  7.8
 C  |  8  | stirrup | 20 | 1.10   | 22.00 |  8.7

The schedule is checked against the drawing before the steel is cut.

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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