Concrete and Steel Structures: Limit State, Bond, Prestress and Plastic Analysis

Two materials, one design philosophy, and the philosophy is where the marks are. Working stress design kept the material in its elastic range and divided the failure stress by a single factor of safety, which is safe but says nothing about how close to collapse the structure really is. Limit state design asks the question directly: it factors the loads up and the material strengths down, separately, and then checks that the factored structure can still carry the factored load. The separation is the whole point, because the uncertainty in a load has nothing to do with the uncertainty in a material — so IS 456 multiplies loads by 1.5 and divides concrete by 1.5 and steel by 1.15, giving design stresses of 0.446 f_ck and 0.87 f_y; IS 800 divides steel by 1.10 for yielding and 1.25 for rupture at a bolt hole. Steel gets the smaller material factor because it is manufactured and tested, concrete the larger because it is mixed and cured on site. From there, reinforced concrete flexure is one question repeated: where is the neutral axis? Put in more steel and it descends, until at x_u,max the concrete crushes at the same moment the steel yields, and beyond that point adding steel buys a brittle failure rather than a stronger beam — which is why the code caps x_u/d and why M_u,lim = 0.138 f_ck b d² for Fe415 is a ceiling and not a target. Bond decides whether the steel can be relied on at all: a bar develops its strength only over a length, L_d ≈ 47φ for Fe415 in M20, and a bar cut short is a bar not there. Prestress inverts the problem by pre-compressing the concrete so that the service moment brings it to a uniform stress rather than cracking its tension face. And on the steel side, plastic analysis abandons the elastic diagram once the first hinge forms and asks only what mechanism finally collapses.

Two Partial Factors, and Why They Are Two

Working stress design took a single factor of safety, applied it to the material strength, and kept the structure elastic. That is safe, but it hides the answer to the question an engineer actually needs: how much more load before it falls down? Limit state design answers it by separating the two sources of uncertainty. The load factor covers the chance that the loading exceeds what was assumed — IS 456 uses 1.5 on the combination of dead and imposed load. The material factor covers the chance that the material is weaker than specified, and it is applied to each material independently: 1.5 for concrete and 1.15 for steel, so the design stresses become 0.67 f_ck/1.5 = 0.446 f_ck and f_y/1.15 = 0.87 f_y. The asymmetry is deliberate and worth being able to explain: steel is manufactured under control and tested by the coil; concrete is mixed, placed, compacted and cured on the site by people in a hurry. The larger factor is bought by that difference, not by anything intrinsic to the material.

Flexure in Reinforced Concrete: Where Is the Neutral Axis?

Every reinforced concrete flexure calculation is the same calculation. Concrete takes the compression, steel takes the tension, and the neutral axis sits where the two are in equilibrium: 0.36 f_ck b x_u = 0.87 f_y A_st. Add steel and x_u descends; add concrete and it rises. The moment of resistance is then that force couple times its lever arm, M_u = 0.87 f_y A_st (d − 0.42 x_u), and the 0.36 and 0.42 are the area and centroid of the assumed parabolic-rectangular stress block, not physical constants. What turns this into a design rule is a limit on how far the neutral axis is allowed to descend, and the reason for the limit is the failure mode rather than the strength.

🎯 Under-reinforced, over-reinforced, and the reason the code forbids the stronger one
In an under-reinforced beam the steel reaches yield first. It then stretches a great deal at almost constant force, the neutral axis rises, cracks open and widen visibly, the deflection grows — and only after all that does the concrete crush. The beam gives a long, loud, visible warning before it fails. In an over-reinforced beam the concrete reaches its crushing strain first, while the steel is still elastic. There is no yield plateau, no large deflection, no crack pattern to see: the compression face simply goes, suddenly and without notice. The over-reinforced beam is stronger. It is also the one the code will not let you build, and that trade — strength surrendered for warning — is the single most examinable idea in the subject. The boundary between the two is the balanced section, and the code fixes it by capping x_u,max/d at 0.53 for Fe250, 0.48 for Fe415 and 0.46 for Fe500 — the higher-strength steel needing a shallower neutral axis because it must strain further to yield. Substituting the cap into the moment expression gives the ceiling M_u,lim = 0.36(x_u,max/d)(1 − 0.42 x_u,max/d) f_ck b d², which evaluates to 0.148, 0.138 and 0.134 f_ck b d² for those three grades. Read it as a ceiling on what a singly reinforced section of given size can be allowed to carry: past it, you do not add more tension steel, you deepen the beam or add compression steel and design it as doubly reinforced.
⚠️ The beam from chapter 1, in concrete, does not fit
Follow the same beam through all four chapters and watch the answer change. Chapter 1 found a working bending moment of 165 kN·m. Chapter 2 put it on a 300 × 450 section and read an elastic extreme-fibre stress of 16.3 MPa. Now design it in M20 concrete with Fe415. First factor the load: M_u = 1.5 × 165 = 247.5 kN·m. Then find what the section can take: M_u,lim = 0.138 × 20 × 300 × 450² = 167.7 kN·m. The demand is 48 per cent larger than the ceiling, so this section cannot carry that moment as a singly reinforced beam — and no amount of extra tension steel will change that, because the ceiling is set by the concrete and the neutral axis limit, not by the steel provided. The options are the real ones an engineer has: deepen the beam to d = √(247.5 × 10⁶/(0.138 × 20 × 300)) = 547 mm; keep the depth and add compression steel, designing it doubly reinforced; or raise the concrete grade, since M_u,lim is proportional to f_ck. Notice what the elastic calculation of chapter 2 did not tell you. 16.3 MPa looked like a modest number beside steel’s 165 MPa allowable, and for a steel beam it would have been — but plain concrete of grade M20 has a permissible compressive stress in bending of about 7 MPa, so that same figure was already a warning. The stress that is comfortable in one material is a failure in another, and this is why reinforced concrete is designed by limit state and not by carrying an elastic stress across from a steel calculation.

Bond, Development Length, and Prestressed Concrete

Reinforcement works only if the concrete can transfer force into it, and that transfer happens over a length. Equating the force a bar can carry to the bond available round its surface, (π φ²/4)(0.87 f_y) = (π φ L_d) τ_bd, gives the development length L_d = φ(0.87 f_y)/(4 τ_bd). With τ_bd = 1.2 MPa for M20 raised by the code’s factor of 1.6 for deformed bars, that is 47φ for Fe415 — so a 20 mm bar needs about 940 mm of embedment before it can be counted on at full strength. The consequences are all practical: a bar must extend past the point of maximum moment, a lap splice must be long enough to hand the force over, a bar that stops short of a support has simply not delivered its steel, and higher-grade steel needs a longer anchorage, not a shorter one — 56.6φ for Fe500 against 28.3φ for Fe250 — because bond strength depends on the concrete while the force to be transferred depends on the steel.

🧠 A prestressed beam whose service load makes it uniformly compressed
Concrete is strong in compression and unreliable in tension. Prestress exploits that by pre-compressing the section so hard that the service load only brings it back to zero rather than into tension. The analysis is the combined-stress formula of the previous chapter with an extra term: σ = P/A ± Pe/Z ∓ M/Z, checked at two load stages, because they are different problems. Take a rectangular beam 300 × 600 with a prestressing force of 900 kN at an eccentricity of 150 mm below the centroid. A = 180 000 mm² and Z = 18 × 10⁶ mm³, so P/A = 5.0 MPa and Pe/Z = 7.5 MPa. At transfer, before the beam carries anything, the top fibre reads 5.0 − 7.5 = −2.5 MPa — tension, at the top, in a beam that has not yet been loaded, which is exactly the state that cracks a beam during handling and is the reason transfer is checked at all. The bottom reads +12.5 MPa. At service, add the moment. Under 90 kN·m the fibres read 2.5 and 7.5. Under 135 kN·m the applied M/Z = 7.5 MPa exactly cancels the prestress moment and both fibres read 5.0 MPa — the section is in uniform compression, using its material as evenly as it is possible to use it, which is the condition the design was aiming at. Under 180 kN·m they read 7.5 and 2.5, the reverse of the transfer state. That progression is the whole idea: a prestressed beam is designed for two load cases at opposite ends of its life, and a section that is comfortable at service can be cracking at transfer.

Steel Members, Connections, and Plastic Analysis

A steel tension member looks like the easiest thing in the paper and carries two independent checks, which is the whole of it. Gross section yielding takes the full area at the yield stress with the smaller partial factor: T_dg = A_g f_y/1.10. Net section rupture takes the area left after the bolt holes at the ultimate stress with a larger factor and a further 0.9: T_dn = 0.9 A_n f_u/1.25. The design capacity is the lesser, and neither can be skipped because which one governs depends on the hole pattern and the grade. Run a plate 100 × 10 of Fe410 with one 22 mm hole: T_dg = 1000 × 250/1.10 = 227.3 kN and T_dn = 0.9 × 780 × 410/1.25 = 230.3 kN. Gross yielding governs — by 1.3 per cent. That margin is the argument for doing both calculations every time: a slightly wider hole, a second bolt in the same line, or a higher-grade steel flips which one wins, and the two answers are close enough that intuition cannot be trusted to pick.

A compression member is the other half of the pair and behaves nothing like it, because a tension member cannot buckle and a compression member usually does. Its capacity is governed by slenderness λ = KL/r, and the design compressive stress is read from a buckling curve that blends the two limiting behaviours: crushing at f_y for a stocky member and Euler buckling at π²E/λ² for a slender one. The two are equal at λ = π√(E/f_y), which for E = 200 GPa and f_y = 250 MPa is λ = 88.9. Below that the Euler stress exceeds the yield stress and yielding governs — at λ = 40 the Euler value is 1234 MPa, which the steel will never reach — and above it buckling governs, with the Euler stress down to 137 MPa at λ = 120 and 61 MPa at λ = 180. Real curves sit below both idealisations, because an actual column is neither perfectly straight nor free of residual stress from rolling and welding, and the codes carry several curves selected by section shape and axis for exactly that reason. A beam-column carries both axial load and bending, and is checked against an interaction expression rather than either limit alone, since the axial force amplifies the bending through the P–δ effect and the bending reduces the axial capacity.

Plastic analysis takes the ductility of steel seriously. Once the extreme fibre of a section yields, further moment does not break the beam: yielding spreads inwards until the whole section is at f_y, at which point the section is a plastic hinge and carries a constant M_p = f_y Z_p, where Z_p is the plastic modulus and takes the full area about the equal-area axis. The ratio Z_p/Z_e is the shape factor, and it is a pure property of the shape: exactly 1.5 for a rectangle, 16/3π = 1.698 for a solid circle, 2.0 for a diamond, and only about 1.12–1.14 for a rolled I-section — because an I already has most of its material at the extreme fibres, so there is little left to recruit. That last number is the reason plastic design is worth much less on an I-beam than the rectangle suggests. Collapse then comes not when the first hinge forms but when enough hinges form to make a mechanism: for a UDL, 8M_p/L² on a simply supported span, 11.66M_p/L² propped, and 16M_p/L² fixed at both ends — exactly twice the simply supported collapse load, because the fixed beam must form three hinges instead of one.

Shape factors, and what plastic design is actually worth
SectionShape factor Z_p/Z_eReserve past first yield
Rectangle1.500 exactly50%
Solid circle16/3π = 1.69870%
Diamond2.000100%
Rolled I-sectionabout 1.12–1.14only 12–14%

Key takeaways

  • Limit state factors loads up and materials down, separately: 1.5 on load, 1.5 on concrete, 1.15 on steel, giving 0.446 f_ck and 0.87 f_y. Concrete gets the larger factor because it is made on site.
  • An over-reinforced beam is stronger and forbidden: it crushes without warning. The code caps x_u,max/d at 0.53 / 0.48 / 0.46 for Fe250 / Fe415 / Fe500, giving M_u,lim = 0.148 / 0.138 / 0.134 f_ck b d².
  • L_d = φ(0.87 f_y)/(4 τ_bd) — about 47φ for Fe415 in M20. Stronger steel needs a longer anchorage, because the force grows with the steel while the bond depends on the concrete.
  • A prestressed section is checked at two stages. The 300 × 600 beam with 900 kN at e = 150 has −2.5 MPa tension on top at transfer, and at a service moment of 135 kN·m both fibres read exactly 5.0 MPa — uniform compression.
  • A steel tension member needs both checks — A_g f_y/1.10 and 0.9 A_n f_u/1.25. On the worked plate they differ by 1.3%, which is why neither can be guessed.
  • Shape factor is Z_p/Z_e: exactly 1.5 for a rectangle but only 1.12–1.14 for a rolled I, because an I already has its metal at the extreme fibres. A fixed-ended beam collapses at 16M_p/L², twice the simply supported value.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In limit state design of reinforced concrete, an over-reinforced beam is not permitted mainly because

    1. it has a lower moment of resistance than a balanced section
    2. it fails by sudden crushing of concrete with no warning
    3. the steel corrodes faster when it is not yielding
    4. the neutral axis rises above the compression face
    Show answer

    Answer: B — it fails by sudden crushing of concrete with no warning

    It fails by sudden crushing, with no warning. In an over-reinforced beam the concrete reaches its crushing strain while the steel is still elastic, so there is no yield plateau, no large deflection and no visible crack pattern before the compression face goes. In an under-reinforced beam the steel yields first and then stretches a long way at nearly constant force — the neutral axis rises, cracks open and widen, the deflection grows, and the beam gives a slow, visible warning before the concrete eventually crushes. Option A is the trap and it is the reverse of the truth: an over-reinforced section is stronger, not weaker, and that is what makes the rule interesting. The code deliberately gives up strength to buy warning, and it enforces the trade by capping x_u,max/d at 0.53, 0.48 and 0.46 for Fe250, Fe415 and Fe500. Option D describes nothing that happens — adding tension steel drives the neutral axis down, towards the tension face, which is precisely why the cap is expressed as an upper limit on x_u. Option C is invented. The design consequence is the one to remember: once the demand exceeds M_u,lim for the section, more tension steel is not the answer. Deepen the beam, raise the concrete grade, or add compression steel and design it as doubly reinforced.

  2. For M20 concrete and Fe415 steel, the limiting moment of resistance of a singly reinforced section 300 mm wide with an effective depth of 450 mm, in kN·m, is ______ (round to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 167.67

    167.67 kN·m. For Fe415 the code caps x_u,max/d at 0.48, and substituting into M_u,lim = 0.36(x_u,max/d)(1 − 0.42 x_u,max/d) f_ck b d² gives the coefficient 0.36(0.48)(1 − 0.2016) = 0.138. Then M_u,lim = 0.138 × 20 × 300 × 450² = 0.138 × 20 × 300 × 202 500 = 167.67 × 10⁶ N·mm = 167.67 kN·m. The coefficient is grade-specific and mixing them up is the commonest error here: it is 0.148 for Fe250 and 0.134 for Fe500, the higher-strength steel giving the smaller value because it needs a shallower neutral axis to reach its larger yield strain. What makes this number worth computing is what it means. M_u,lim is a ceiling on the section, not a property of the reinforcement — provide more tension steel than the balanced amount and the section does not carry more moment in any way the code will accept, it merely becomes over-reinforced and brittle. And it exposes a design failure in the beam this course has carried through four chapters: that beam has a working moment of 165 kN·m, which factors to 1.5 × 165 = 247.5 kN·m, some 48 per cent above this ceiling. The section needs d = √(247.5 × 10⁶/(0.138 × 20 × 300)) = 547 mm, or compression steel, or a stronger concrete — M_u,lim scales directly with f_ck, so M25 would raise it to 209.6 kN·m and still not be enough.

  3. For the same concrete grade, the development length required for a deformed bar of a given diameter is

    1. shorter for Fe500 than for Fe250, because higher-grade steel bonds better
    2. longer for Fe500 than for Fe250, because the force to be transferred is larger
    3. the same for all grades of steel
    4. independent of the bar diameter
    Show answer

    Answer: B — longer for Fe500 than for Fe250, because the force to be transferred is larger

    Longer for Fe500, because the force to be transferred is larger. Equate the bar force to the bond available over its surface: (πφ²/4)(0.87 f_y) = (πφ L_d) τ_bd, so L_d = φ(0.87 f_y)/(4 τ_bd). The bond stress τ_bd is a property of the concrete — 1.2 MPa for M20, raised by the code factor 1.6 for deformed bars to 1.92 — while the force to be developed is a property of the steel. Raise f_y and the numerator grows with nothing in the denominator to match, so the anchorage gets longer: 28.3φ for Fe250, 47.0φ for Fe415, 56.6φ for Fe500. Option A inverts the physics and also invents a mechanism; bond depends on the concrete and on the bar surface, and the grade of the steel inside is irrelevant to how well the concrete grips it. Option C ignores the f_y in the numerator. Option D is contradicted by the φ sitting in front — although the interesting part is how it depends on diameter: L_d is proportional to φ, so a 20 mm Fe415 bar in M20 needs about 940 mm, while a 12 mm bar needs 564 mm. The practical reading is that high-strength steel is not a free saving. It reduces the area of steel needed but lengthens every anchorage and every lap, and in a shallow member or a narrow column there may simply not be room for the anchorage the bar requires — which is why detailers sometimes choose more bars of a smaller diameter rather than fewer large ones.

  4. A rectangular prestressed beam 300 mm × 600 mm carries a prestressing force of 900 kN at an eccentricity of 150 mm below the centroid. At transfer, with no external load, the stress at the top fibre in MPa is ______ (take tension as negative)

    Numerical answer — type the value.

    Show answer

    Answer: -2.5

    −2.5 MPa, that is 2.5 MPa of tension. A = 300 × 600 = 180 000 mm² and Z = 300 × 600²/6 = 18 × 10⁶ mm³. The direct term is P/A = 900 000/180 000 = 5.0 MPa of compression everywhere. The eccentricity is below the centroid, so its moment puts tension at the top: Pe/Z = 900 000 × 150/(18 × 10⁶) = 7.5 MPa. The top fibre therefore reads 5.0 − 7.5 = −2.5 MPa and the bottom 5.0 + 7.5 = +12.5 MPa. Tension at the top of an unloaded prestressed beam surprises candidates every year, and it is exactly why the code requires a separate check at transfer: the beam is at its most eccentrically stressed before it carries anything, and beams have cracked on the casting yard for this reason. Follow the same section into service and the picture reverses. A service moment of 90 kN·m adds M/Z = 5.0 MPa at the top and subtracts it at the bottom, giving 2.5 and 7.5. At 135 kN·m the applied M/Z = 7.5 MPa exactly cancels the prestress moment, and both fibres read 5.0 MPa — uniform compression, the most even use of the material the section can achieve, and the state the eccentricity was chosen to produce. At 180 kN·m they read 7.5 and 2.5, the mirror of transfer. Two load cases at opposite ends of the beam’s life, and a section satisfactory at one can be cracking at the other.

  5. A plate 100 mm × 10 mm of Fe410 steel (f_y = 250 MPa, f_u = 410 MPa) has one bolt hole of 22 mm diameter. Taking γ_m0 = 1.10 and γ_m1 = 1.25, the design tensile strength in kN is ______ (round to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 227.27

    227.27 kN, governed by gross section yielding. Both checks must be run and the smaller taken. Gross section yielding: T_dg = A_g f_y/γ_m0 = (100 × 10)(250)/1.10 = 227 273 N = 227.27 kN. Net section rupture: the hole removes 22 × 10 = 220 mm², leaving A_n = 780 mm², and T_dn = 0.9 A_n f_u/γ_m1 = 0.9 × 780 × 410/1.25 = 230 256 N = 230.26 kN. The design strength is the lesser, 227.27 kN. The reason this is a good question is the margin: the two answers differ by 1.31 per cent, so there is no rule of thumb that tells you which governs, and both must be computed every time. Change the geometry slightly and the winner changes — widen the hole, put a second bolt in the same line, or specify a higher-grade steel with a larger f_y/f_u ratio, and rupture takes over. The structure of the two expressions is worth understanding rather than memorising. Gross yielding uses the yield stress with the small factor 1.10, because a member yielding over its whole length is a serviceability disaster but not a collapse. Net rupture uses the ultimate stress with the larger factor 1.25 and an extra 0.9, because fracture at a hole is sudden and local, and the stress concentration round the hole means the average net-section stress understates what the material near the hole actually sees. Two failure modes with different consequences get different factors, which is the whole logic of limit state design applied within a single member.

  6. The shape factor of a rectangular cross-section is ______

    Numerical answer — type the value.

    Show answer

    Answer: 1.5

    1.5 exactly. The shape factor is Z_p/Z_e. The elastic modulus is Z_e = bd²/6. The plastic modulus takes the whole section at yield, so for a rectangle the tension and compression blocks are each bd/2 acting at d/4 from the equal-area axis, giving Z_p = 2(bd/2)(d/4) = bd²/4. The ratio is (bd²/4)/(bd²/6) = 1.5, so a rectangular section has 50 per cent of reserve moment capacity beyond first yield at the extreme fibre. The number that matters more in practice is the one for a rolled I-section: about 1.12 to 1.14, and the reason is worth seeing. The shape factor measures how much material is available to recruit as yielding spreads inwards from the extreme fibres — and an I-section has already put nearly all of its metal in the flanges, at the extreme fibres, precisely to maximise Z_e. There is very little left in the web to bring in, so the plastic reserve is small. A rectangle, which wastes material near its neutral axis where the elastic stress is low, has plenty to recruit; a diamond, which concentrates material at the neutral axis, has the most, at a shape factor of 2.0, and a solid circle sits at 16/3π = 1.698. The practical conclusion is the one candidates most often get backwards: plastic design offers the least additional capacity on exactly the sections steelwork actually uses. Its real value in a frame is not the section reserve but the redistribution — a fixed-ended beam collapses at 16M_p/L² against 8M_p/L² simply supported, twice the load, because three hinges must form instead of one.

  7. In limit state design to IS 456, the partial safety factor for concrete is 1.5 and for steel 1.15. The difference exists because

    1. concrete is stronger than steel and can afford a larger factor
    2. concrete is made and cured on site, so its in-place strength is more variable
    3. steel is more expensive, so a smaller factor keeps the cost down
    4. the factors are arbitrary and carry no physical meaning
    Show answer

    Answer: B — concrete is made and cured on site, so its in-place strength is more variable

    Concrete is made and cured on site, so its in-place strength is far more variable. A partial material factor is a statement about scatter, not about strength. Reinforcing steel is rolled in a mill under process control, tested coil by coil, and arrives on site as a known product; its yield strength has a tight distribution, so a factor of 1.15 covers the tail. Concrete is batched, transported, placed, compacted and cured on the site, and every one of those steps can go wrong in ways that reduce the strength actually present in the member — extra water at the mixer, poor compaction round congested reinforcement, curing stopped early in hot weather. The distribution is wide, so the factor is larger. Option A gets it backwards twice: concrete is far weaker than steel per unit area, and in any case it is the variability, not the mean, that sets the factor. Option C confuses a safety criterion with an economic one; the factors are calibrated on reliability, and if cost drove them the cheaper material would be given the slack. Option D is the answer worth arguing against most firmly — the factors are chosen, not arbitrary, and the reasoning behind them is examinable. The same logic explains the split within steel design in IS 800: γ_m0 = 1.10 for yielding, which is ductile and gradual, against γ_m1 = 1.25 for rupture at a bolt hole, which is sudden and local. The factor tracks the consequence and the uncertainty of the failure mode, not the material’s reputation.

  8. A prismatic steel beam of plastic moment capacity M_p carries a uniformly distributed load. Which of the following collapse loads are correct? (One or more)

    1. Simply supported: w_u = 8M_p/L²
    2. Fixed at both ends: w_u = 16M_p/L²
    3. The fixed-ended beam collapses at twice the simply supported load
    4. Collapse occurs as soon as the first plastic hinge forms
    Show answer

    Answer: A — Simply supported: `w_u = 8M_p/L²`; B — Fixed at both ends: `w_u = 16M_p/L²`; C — The fixed-ended beam collapses at twice the simply supported load

    A, B and C. For a simply supported span the maximum moment is w L²/8, so collapse arrives when that reaches M_p, giving w_u = 8M_p/L² — A is right, and note that here one hinge is enough, because a simply supported beam is determinate and a single hinge turns it into a mechanism. Fixing both ends makes the beam twice indeterminate: hinges form first at the two supports, and the beam keeps carrying load with those hinges rotating at constant M_p until a third hinge forms at midspan. Equating internal and external work for that three-hinge mechanism gives w_u = 16M_p/L² — B is right, and C follows immediately as the ratio 16/8 = 2. D is false, and it is the whole point of plastic analysis. In a determinate structure the first hinge does end it; in an indeterminate one it does not, because the structure simply redistributes — the hinge caps the moment at that section, and further load is carried by the parts that have not yet yielded. That redistribution is the real gain of plastic design in frames, and it is much larger than the section-level gain: the shape factor of a rolled I-section is only about 1.12, so the reserve within the section is 12 per cent, while the reserve from redistribution in a fixed-ended beam is 100 per cent. It also carries a condition worth stating: redistribution needs ductility, so the argument holds for structural steel and fails for a brittle material or for a steel section slender enough to buckle locally before the hinge can form. That is why plastic design is confined to sections classified as plastic or compact.