Support settlement on a continuous beam: how much top steel can really be removed
A two-span continuous RCC beam ABC supports a corridor in an institutional building under construction. Each span is 8.0 m, the section is 300 mm × 600 mm, M25 concrete with Fe500 steel, and the factored UDL from the slab, including self-weight, is 20 kN/m. Using the uncracked stiffness, EI = 1.35 × 10^5 kN·m². Supports A and C rest on isolated footings in a filled-up layer; support B is on a footing that, after the beam was cast, settled 15 mm more than the ends, as measured by level survey. The beam was originally designed for the standard continuous-beam envelope: hogging 160 kN·m over B with 4-T20 at top, and sagging 90 kN·m at midspan with 4-T16 at bottom (effective depth 545 mm). The site engineer now argues that since the middle support has dropped, the hogging moment has vanished, and proposes to chip off the top bars over B to save time on a pending concrete pour.
1. Compute the moment at B and the maximum sagging moment with the 15 mm settlement, and compare with the original design envelope. State your assumptions.
Read the worked answer
Method: superposition on the two-span beam, treating B as a redundant reaction.
Step 1 - original (no settlement) values
- MB = −wL²/8 = −20 × 64/8 = −160 kN·m
- RA = 3wL/8 = 60 kN; maximum sagging at x = RA/w = 3.0 m: M = 60 × 3 − 10 × 9 = 90 kN·m ✓ (matches the design).
Step 2 - settlement term. A downward settlement δ at B adds a hogging relief of 3EIδ/L²: MB = −wL²/8 + 3EIδ/L² = −160 + (3 × 1.35 × 10^5 × 0.015)/64 = −160 + 94.9 = −65.1 kN·m
Step 3 - reactions and sagging. RA = 3wL/8 + 3EIδ/L³ = 60 + 11.9 = 71.9 kN; RB = 2wL − 2RA = 176.2 kN. Maximum sagging at x = 71.9/20 = 3.59 m: M = 71.9 × 3.59 − 10 × 3.59² = 258.2 − 128.9 = 129.3 kN·m
Comparison
- Hogging at B: 160 → 65 kN·m (down about 59%).
- Sagging: 90 → 129 kN·m (up about 44%).
- The reversal means the bottom steel now governs.
Assumptions: linear elastic uncracked EI, no creep/shrinkage, no pattern loading, settlement fully and immediately applied, no support rotation restraint. Cracking would reduce EI and hence reduce the settlement-induced moment, so the 65/129 kN·m values are an upper-bound elastic estimate - they must not be used to justify removing steel.
2. Evaluate the site engineer's proposal against the actual capacities, and state what should be done with the reinforcement.
Read the worked answer
Capacity check (b = 300, d = 545, fck = 25, fy = 500) Mu = 0.87 fy Ast d [1 − Ast fy/(b d fck)]
- 4-T16 bottom, Ast = 804 mm²: ratio = 402,000/4,087,500 = 0.098, Mu ≈ 172 kN·m.
- 4-T20 top, Ast = 1257 mm²: ratio = 0.154, Mu ≈ 252 kN·m.
- 2-T20 top, Ast = 628 mm²: ratio = 0.077, Mu ≈ 137 kN·m.
Verdict on the proposal
- The bottom steel already carries 172 kN·m, which is more than the 129 kN·m demand. The beam is not deficient in sagging - the engineer's first premise is wrong.
- The hogging demand of 65 kN·m could technically be met by 2-T20 (137 kN·m), but the top steel at a support must satisfy much more than one load case: the full design envelope (pattern loading, future change of use), minimum and ductility requirements, the requirement to continue a proportion of the support steel for crack control and robustness, and the possibility of moment reversal if the settlement is not stable.
- Chipping hardened concrete at a support damages the cover, the stirrups and the fire resistance, and produces debris that cannot be properly reinstated around congested bars.
What to do: retain all 4-T20 over B and all 4-T16 at the bottom, do not chip anything. The real actions are: (i) re-run the analysis with the settlement included and re-check the serviceability limit state - 15 mm over 8 m is L/533, close to the L/500 limit for brittle finishes, and the increased sagging deflection must be checked against IS 456 limits; (ii) investigate and stop the settlement of footing B; (iii) keep the surface levelling of the floor and partitions flexible until the settlement stabilises.
3. Monitoring shows the settlement is still increasing at about 3 mm/month. What happens at 30 mm, and what design action do you take?
Read the worked answer
Re-evaluate at δ = 30 mm
- MB = −160 + 3 × 1.35 × 10^5 × 0.030/64 = −160 + 189.8 = +29.8 kN·m (sagging) - the support moment has fully reversed. The top steel is no longer stressed in tension by gravity load, but the bottom bars running continuously through B now must carry 29.8 kN·m, which 4-T16 does comfortably.
- RA = 60 + 3 × 1.35 × 10^5 × 0.030/512 = 60 + 23.7 = 83.7 kN.
- Maximum sagging at x = 83.7/20 = 4.19 m: M = RA²/(2w) = 83.7²/40 = 175.1 kN·m.
The governing check: the bottom capacity is about 172 kN·m, so the demand of 175 kN·m now exceeds it. The beam becomes deficient in sagging at about 30 mm settlement, even though the hogging has disappeared. This is exactly the trap in the site engineer's logic.
Actions
- Arrest the movement first. Investigate the fill beneath footing B (boreholes, SPT), and underpin or compaction-grout it; a structural fix is useless while 3 mm/month continues.
- Strengthen for sagging. Add 2-T16 (or an equivalent near-surface/bonded plate) in the bottom over the sagging zone, checked for development length and shear; alternatively prop the beam at B during the repair so that the added bars participate.
- Retain the top steel - if the settlement is later arrested or reversed by grouting, hogging reappears and the top steel is needed again. Do not rely on a single measured value.
- Monitor with levelling pins on B and the midspans, plus crack gauges, and specify a trigger value (say a further 5 mm) at which use is suspended.
- Serviceability: check the increased deflection and vibration, and record the settlement in the as-built file. The design basis of the structure has changed; that must be documented, not settled informally on site.