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Sections with stress distribution, moment expressions

Short answer

Balanced: x_u = x_u,max, M_u,lim = 0.36 fck b x_u,max (d - 0.42 x_u,max) = 0.138 fck b d² (Fe415). Under-reinforced: x_u < x_u,max, steel yields, M_u = 0.87 fy A_st (d - 0.42 x_u) = 0.87 fy A_st d (1 - (A_st fy)/(b d fck)). Over-reinforced: x_u > x_u,max, concrete crushes first, moment limited to M_u,lim.

What are balanced, under-reinforced and over-reinforced sections? Explain with neat sketches of the stress distributions and expressions for moment resistance of each section.

Let xux_u be the depth of the neutral axis and xu,maxx_{u,max} its limiting value (0.48d0.48d for Fe415). In the stress block, the compressive force is C=0.36fckb xuC=0.36f_{ck}b\,x_u at lever arm z=d−0.42xuz=d-0.42x_u.

  Strain             Stress block
   0.0035          0.446fck  C=0.36fck.b.xu
   |\               |=====|
   | \ xu           |     | <- 0.42xu from top
 --+--- NA        --+-----+
   |  \             |
   |   \ es         T=0.87fy.Ast

1. Balanced section (xu=xu,maxx_u=x_{u,max})

Concrete reaches 0.0035 and steel reaches yield strain εy=0.87fy/Es+0.002\varepsilon_y=0.87f_y/E_s+0.002 together.

Mu,lim=0.36fckb xu,max(d−0.42xu,max)=0.138fckbd2 (Fe415)M_{u,lim}=0.36f_{ck}b\,x_{u,max}(d-0.42x_{u,max})=0.138f_{ck}bd^2\ (\text{Fe415})

2. Under-reinforced section (xu<xu,maxx_u<x_{u,max})

Steel yields first; xu=0.87fyAst0.36fckbx_u=\dfrac{0.87f_yA_{st}}{0.36f_{ck}b}.

Mu=0.87fyAst(d−0.42xu)=0.87fyAstd(1−Astfybdfck)M_u=0.87f_yA_{st}(d-0.42x_u)=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)

Failure is ductile.

3. Over-reinforced section (xu>xu,maxx_u>x_{u,max})

Concrete crushes first; steel stress is below 0.87fy0.87f_y. As per IS 456 the moment is limited to

Mu=Mu,lim=0.36fckb xu,max(d−0.42xu,max)M_u=M_{u,lim}=0.36f_{ck}b\,x_{u,max}(d-0.42x_{u,max})

Failure is brittle and sudden, so over-reinforced sections are not allowed.

✓ Answer checked by Claude Opus 5.5 (high effort)

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