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More Steel, Less Design Strength: Net Tensile Strain and φ in NSCP 2015

By 8 min readUpdated

When a beam comes out short, the usual reflex is to add bottom bars. On site, an extra bar is sometimes added just to be sure (para sigurado). More steel, stronger beam. Usually true.

But past a certain point, extra bottom bars can lower the beam's design strength and make it fail with less warning. Here is a sample beam checked to NSCP 2015.

What happens inside a beam in bending

In a simply supported beam, the concrete at the top is squeezed and the bottom bars are stretched. (In a frame, the same applies at the supports with the top bars in tension.)

Concrete is taken to crush at a strain of 0.003 (NSCP 422.2.2.1). How far the lowest layer of bars has stretched at that moment is the net tensile strain εt.

  • Large εt: the steel yields well before the concrete crushes. The beam cracks, sags and warns. It is ductile.
  • Small εt: the concrete crushes when the steel has barely yielded, or not at all. Little warning. It is brittle, over-reinforced in the code's sense.

Add bottom bars, and the bigger steel force needs a bigger concrete compression zone, so the neutral axis moves down. The bars end up closer to it and stretch less. More bottom steel, smaller εt.

How NSCP 2015 handles it

First, NSCP sets the strength reduction factor φ for bending (Table 421.2.2). With fy = 420 MPa and Es = 200,000 MPa, the yield strain is εy = 420 / 200,000 = 0.0021 (some books round it to 0.002).

Net tensile strain εt Section φ
0.005 or moretension-controlled0.90
between 0.0021 and 0.005transition0.65 + 0.25 (εt − 0.0021) / (0.005 − 0.0021)
0.0021 or lesscompression-controlled0.65

Second, for non-prestressed beams (Pu below 0.10 f'c Ag), it sets a floor: εt must be at least 0.004 (409.3.3.1).

So too much bottom steel lowers φ. Mn still rises, but φ can drop faster, and φMn can go down.

Strength reduction factor phi against net tensile strain Graph of phi for bending against net tensile strain. Phi is 0.65 up to the yield strain 0.0021, rises in a straight line to 0.90 at 0.005 and stays at 0.90. The zone below 0.004 is shaded as not allowed for beams. Case A is at 0.0045 and phi 0.858, case B at 0.0026 and 0.696 inside the shaded zone, case C at 0.0048 and 0.882, case D at 0.0062 and 0.90. φ depends on the net tensile strain εt NSCP 2015 Table 421.2.2, fy = 420 MPa · the dots are the four cases of the test beam 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.65 0.70 0.75 0.80 0.85 0.90 φ net tensile strain εt compression-controlled transition tension-controlled εt below 0.004: not allowed for beams εy = 0.0021 A: εt 0.0045 φ 0.858 B: εt 0.0026 φ 0.696 C: εt 0.0048, φ 0.882 D: εt 0.0062 φ 0.90
Figure 1. How φ for bending changes with the net tensile strain εt: 0.65 up to the yield strain of 0.0021, then a straight line up to 0.90 at 0.005. The shaded zone below 0.004 is not allowed for beams. The dots are the four cases worked out below; Case B, at εt = 0.0026 and φ = 0.696, is inside the shaded zone.

The test beam

I used the sample beam in our RC Beam Section Designer (NSCP 2015), B-1 of a two-storey residence:

  • 350 × 500 mm, f'c = 28 MPa, fy = 420 MPa
  • 40 mm clear cover, 10 mm stirrups, 6.0 m clear span, simply supported
  • I set the factored moment to Mu = 350 kN·m and changed only the bars (and, for Case D, the depth).
The four beam sections of the test, drawn to scale Four beam cross-sections, all 350 mm wide. Case A: 350 by 500 with six 25 mm bottom bars in two layers, four plus two, net tensile strain 0.0045, pass. Case B: 350 by 500 with eight 25 mm bottom bars, four plus four, 0.0026, fail. Case C: case B plus four 20 mm top bars, 0.0048, pass. Case D: six 25 mm bottom bars in a 350 by 600 beam, 0.0062, pass. The four sections, drawn to the same scale f'c = 28 MPa · fy = 420 MPa · 40 mm cover · 10 mm stirrups · only the bars in the calculation are shown Case A 6-Ø25 bottom 500 350 εt = 0.0045 · φ = 0.858 PASS Case B 8-Ø25 bottom 500 350 εt = 0.0026 · φ = 0.696 FAIL · εt < 0.004 Case C 8-Ø25 + 4-Ø20 top 500 350 εt = 0.0048 · φ = 0.882 PASS Case D 6-Ø25, h = 600 600 350 εt = 0.0062 · φ = 0.90 PASS
Figure 2. The four sections drawn to the same scale: Case A with 6-Ø25, Case B with 8-Ø25, Case C with 8-Ø25 plus 4-Ø20 at the top, and Case D with 6-Ø25 in a 600 mm deep beam. Only Case B, with εt = 0.0026, fails the 0.004 limit.

Case A and Case B: adding two bars

Case A has 6-Ø25 (six 25 mm bars) at the bottom only. Case B adds two more: 8-Ø25, or 33% more steel. The sheet's four-bars-per-layer limit gives 4 + 2 and 4 + 4.

Excel result Case A: 6-Ø25 Case B: 8-Ø25
Steel area As2,945 mm²3,927 mm²
Neutral axis depth c174.7 mm232.9 mm
Net tensile strain εt0.00450.0026
φ0.8580.696
Nominal strength Mn428.7 kN·m517.1 kN·m
Design strength φMn367.8 kN·m359.9 kN·m
Ductility, εt ≥ 0.004PASSFAIL
OverallSECTION ADEQUATEREVISE SECTION

Mn rose about 21%. But φ fell from 0.858 to 0.696, so the design strength fell from 367.8 to 359.9 kN·m. More steel, lower design strength, less warning.

Hand check of Case B

With β1 = 0.85 for f'c = 28 MPa (Table 422.2.2.4.3):

c = As · fy / (0.85 · f'c · β1 · b) = 3,927 × 420 / (0.85 × 28 × 0.85 × 350) = 232.9 mm

The lowest layer sits at dt = 500 − 40 − 10 − 25/2 = 437.5 mm. Then:

εt = 0.003 × (437.5 − 232.9) / 232.9 = 0.00263, below 0.004, FAIL
φ = 0.65 + 0.25 × (0.00263 − 0.0021) / (0.005 − 0.0021) = 0.696
a = 0.85 × 232.9 = 198.0 mm; d (centroid of both layers) = 412.5 mm
Mn = 3,927 × 420 × (412.5 − 198.0/2) / 10⁶ = 517.1 kN·m
φMn = 0.696 × 517.1 = 359.9 kN·m

Like the sheet, this assumes all eight bars yield. But the upper layer at 387.5 mm reaches only εs = 0.003 × (387.5 − 232.9) / 232.9 = 0.00199, below 0.0021. A layer-by-layer solve gives c = 230.2 mm and φMn = 360.3 kN·m: same conclusion.

When the concrete crushes, the lowest layer is at only 1.25 times the yield strain and the upper layer has not yielded. That is the brittle behaviour the 0.004 rule prevents.

The sheet's flexure line for Case B still says PASS (359.9 > 350); only the ductility line catches it. With φ taken as 0.90 by habit, φMn would be 0.90 × 517.1 = 465.4 kN·m, about 29% too high.

The Design Check Summary of Case B in the RC Beam Section Designer (NSCP 2015): flexure passes at 359.92 against 350.00 kN·m, but ductility fails with εt = 0.00263 against 0.004, and the overall result is REVISE SECTION.
Figure 3. The Design Check Summary of Case B in the RC Beam Section Designer (NSCP 2015): flexure passes at 359.92 against 350.00 kN·m, but ductility fails with εt = 0.00263 against 0.004, and the overall result is REVISE SECTION.

Case C and Case D: two fixes that work

Excel result Case C: 8-Ø25 bottom + 4-Ø20 top Case D: 6-Ø25 bottom, h = 600 mm
εt0.00480.0062
φ0.8820.90
Design strength φMn500.3 kN·m497.2 kN·m

Case C: compression bars at the top. The four 20 mm top bars carry 455.7 kN of compression (net of the concrete they displace), so the concrete gives only 1,193.6 kN instead of 1,649.3 kN. So c = 168.6 mm, and εt = 0.003 × (437.5 − 168.6) / 168.6 = 0.0048.

Case D: a deeper beam. The same six bars, 100 mm deeper. c stays at 174.7 mm, but the bars are farther down: dt = 537.5 mm, so εt = 0.003 × (537.5 − 174.7) / 174.7 = 0.0062. Above 0.005, so φ = 0.90.

Strain diagrams of cases A, B and C when the concrete crushes Three strain diagrams over the 500 mm depth, each with 0.003 compression strain at the top. Case A: neutral axis at 174.7 mm, strain at the lowest bars 0.0045. Case B: neutral axis at 232.9 mm, strain 0.0026, below the 0.004 minimum. Case C: neutral axis at 168.6 mm, strain 0.0048. Strain over the depth when the concrete reaches 0.003 350 × 500 beam, to scale · dot = strain at the lowest bars (dt = 437.5 mm) · dashed line = 0.004 minimum Case A · 6-Ø25 bottom 0.003 neutral axis 0.004 lowest bars c = 174.7 mm εt = 0.0045 PASS Case B · 8-Ø25 bottom 0.003 neutral axis 0.004 lowest bars c = 232.9 mm εt = 0.0026 FAIL · below 0.004 Case C · 8-Ø25 + 4-Ø20 top 0.003 neutral axis 0.004 lowest bars c = 168.6 mm εt = 0.0048 PASS
Figure 4. Strain over the 500 mm depth when the concrete reaches 0.003, drawn to scale. The neutral axis is at c = 174.7 mm in Case A, 232.9 mm in Case B and 168.6 mm in Case C, which gives εt = 0.0045, 0.0026 and 0.0048 at the lowest bars.

The sheet says SECTION ADEQUATE for both. Each gives about 500 kN·m, against 360 for Case B, and a beam that warns before it fails.

Design strength of the four cases against the factored moment Bar chart of design strength phi Mn in kN m against Mu of 350. Case A, six 25 mm bars: 367.8. Case B, eight 25 mm bars: 359.9, lower than case A and failing the ductility check. Case C, eight 25 mm bars plus four 20 mm top bars: 500.3. Case D, six 25 mm bars in a 600 mm deep beam: 497.2. Design strength φMn of the four cases kN·m, from the Excel runs · the dashed line is the factored moment Mu = 350 kN·m 0 100 200 300 400 500 600 kN·m Case A 6-Ø25 bottom 367.8φ 0.858 Case B 8-Ø25 bottom 359.9φ 0.696 · FAIL Case C 8-Ø25 + 4-Ø20 top 500.3φ 0.882 Case D 6-Ø25, h = 600 mm 497.2φ 0.90 Mu = 350
Figure 5. Design strength φMn of the four cases against Mu = 350 kN·m: 367.8 for Case A, 359.9 for Case B, 500.3 for Case C and 497.2 for Case D.

A note for frames. In a special moment frame, NSCP 418.6.3.1 also caps the steel ratio at 0.025. Cases B and C have ρ = 3,927 / (350 × 412.5) = 0.027 and would not pass; Case D, at 2,945 / (350 × 520.8) = 0.016, would. Extra beam bars also raise the earthquake shear on the beam, joint and columns.

The lesson

Adding bottom bars helps only while the concrete at the top can keep up. Past that, help the compression side or give the steel more depth.

This is not about any one trade. Adding a bar feels safe. But an over-reinforced beam looks stronger, and nothing on site shows that it is not.

What to check

  1. On site, do not add or upsize bars without the engineer of record's OK. Any change in the bars changes εt and, in a frame, the earthquake forces.
  2. In design, compute εt and φ for every beam. Check εt ≥ 0.004 (409.3.3.1), and aim for 0.005 or more so that φ = 0.90.
  3. If a beam needs more strength and εt is already low, add top compression bars or make the beam deeper, not more bottom bars.
  4. Treat designed top bars as real reinforcement, not just stirrup hangers, enclosed in the stirrups. Their tie spacing is limited to the least of 16 bar diameters, 48 tie diameters and the beam size (409.7.6.4.3): here 16 × 20 = 320 mm, against 150 mm provided. Frame beams need closer hoops at the ends.
  5. Watch the congestion. Eight 25 mm bars in a 350 mm beam already need two layers. Keep at least 25 mm clear between layers, with the upper bars directly above the lower ones (425.2.2), using short 25 mm bar spacers, and vibrate well.

General information only. Follow the approved structural plans, and refer any change in reinforcement to the engineer of record.