Design of beams: Behavior in Flexure

In limit state design, the type of beam section is decided by the depth of neutral axis $x_u$ compared with the limiting depth $x_{u,max}$ (IS 456, cl. 38.1): $0.53d$ (Fe250), $0.48d$ (Fe415), $0.46d$ (Fe500).

Under-reinforced section ($x_u<x_{u,max}$)

• Steel area is less than that for the balanced section; steel yields before concrete reaches its ultimate strain 0.0035.
• Failure mode: ductile. Steel yields, cracks widen and deflection becomes large, giving ample warning before concrete crushes.
• Moment: $M_u=0.87f_yA_{st}d\left(1-\dfrac{A_{st}f_y}{bdf_{ck}}\right)$.
• Preferred in design.

Balanced section ($x_u=x_{u,max}$)

• Steel yields and concrete reaches 0.0035 at the same time.
• Moment: $M_{u,lim}=0.36f_{ck}b\,x_{u,max}(d-0.42x_{u,max})$, i.e. $0.138f_{ck}bd^2$ for Fe415.

Over-reinforced section ($x_u>x_{u,max}$)

• Excess steel; concrete crushes (strain 0.0035) before steel yields.
• Failure mode: brittle and sudden with no warning. Not allowed in design; capacity is limited to $M_{u,lim}$ and compression steel is added if more moment is needed.

  Section     Strain             Stress block
  +----+                         0.446fck
  |    |   0.0035               ||====||
  |    |   |\                    ||    ||
  |  x |   | \ xu        xu      ||    ||
  |    |   |  \                  ||    ||
  +----+  -+---  NA            C
  | As |   |  /
  +----+   | / eps_s >= ey (under), = ey (balanced), < ey (over)

T- and L-beams occur when a slab is cast monolithically with the beam, so the slab acts as the flange. Design steps (IS 456 cl. 23.1.2 and Annex G-2) are as follows.

1. Fix dimensions: assume web width $b_w$ (usually 230-300 mm) and depth $D\approx L/12$ to $L/10$; slab thickness $D_f$ is known from slab design; $d=D-$ cover.
2. Effective flange width $b_f$:
   • T-beam: $b_f=\dfrac{l_0}{6}+b_w+6D_f$
   • L-beam: $b_f=\dfrac{l_0}{12}+b_w+3D_f$
   • limited to the actual centre-to-centre spacing of beams. $l_0$ is the distance between points of zero moment (span for simply supported).
3. Loads: self weight of web and slab, finishes, live load; factor by 1.5.
4. Moment: $M_u=w_uL^2/8$ (or from analysis).
5. Locate neutral axis: find the moment capacity with $x_u=D_f$: $$M_{u,f}=0.36f_{ck}b_fD_f(d-0.42D_f)$$
   • If $M_u\le M_{u,f}$, NA lies in the flange: design as a rectangular beam of width $b_f$.
   • If $M_u>M_{u,f}$, NA lies in the web: go to step 6.
6. NA in the web:
   • If $D_f/d\le0.2$: $M_u=0.36f_{ck}b_wx_u(d-0.416x_u)+0.45f_{ck}(b_f-b_w)D_f(d-D_f/2)$.
   • If $D_f/d>0.2$: replace $D_f$ by $y_f=0.15x_u+0.65D_f$.
   • Solve for $x_u$ and check $x_u\le x_{u,max}$ (if not, increase section or add compression steel).
7. Tension steel: from $0.87f_yA_{st}=0.36f_{ck}b_wx_u+0.45f_{ck}(b_f-b_w)D_f$ (or $y_f$).
8. Checks: minimum steel $0.85b_wd/f_y$, maximum 4% of $b_wD$, shear, deflection ($L/d$ with modification factor and $b_w/b_f$ factor), cracking and detailing. Flange transverse steel is provided in the slab.

Given: $L=4.5$ m, superimposed load 45 kN/m, point load 75 kN at mid-span (service), effective cover 45 mm, M25, Fe415.

Step 1: Section (assumed)

Take $b=300$ mm, $D=600$ mm, so $d=600-45=555$ mm.

Step 2: Loads

$$\begin{aligned} \text{Self weight}&=0.30\times0.60\times25=4.50\ \text{kN/m}\\ w&=45+4.50=49.50\ \text{kN/m},\quad w_u=1.5w=74.25\ \text{kN/m}\\ P_u&=1.5\times75=112.5\ \text{kN} \end{aligned}$$

Step 3: Factored moment

$$M_u=\frac{w_uL^2}{8}+\frac{P_uL}{4}=187.95+126.56=314.51\ \text{kN·m}$$

Step 4: Limiting moment (IS 456 cl. 38.1; Fe415: $x_{u,max}=0.48d$)

$$M_{u,lim}=0.138f_{ck}bd^2=0.138\times25\times300\times555^2=318.72\ \text{kN·m}\ \ge M_u$$

A singly reinforced section is sufficient.

Step 5: Tension steel (IS 456 Annex G-1.1(b))

$$M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)$$

Solving this quadratic for $M_u=314.51$ kN·m gives $A_{st,req}=1948$ mm$^2$.

Provide 4-25 mm bars, $A_{st,prov}=1963$ mm$^2$ (fits in one layer in 300 mm width).

• Minimum steel: $0.85bd/f_y=341$ mm$^2$, OK.
• $x_u=\dfrac{0.87f_yA_{st}}{0.36f_{ck}b}=262.6$ mm $<0.48d=266.4$ mm, OK.

Step 6: Deflection check (IS 456 cl. 23.2, simply supported basic ratio 20)

$$p_t=\frac{100A_{st}}{bd}=1.18\%,\qquad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=238.8\ \text{N/mm}^2$$

Modification factor (IS 456 Fig. 4): $k_t=1.05$.

$$\left(\frac{L}{d}\right)_{allow}=20\times1.05=21.1,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{4500}{555}=8.11$$

Actual $<$ allowable, so the beam is safe in deflection.

Answer: Beam 300 x 600 mm, d = 555 mm, $M_u=314.51$ kN·m, $A_{st}=1948$ mm$^2$ (provide 4-25 mm bars); deflection check satisfied. Add 2-12 mm hanger bars and 8 mm stirrups (shear design to be done separately).

Given: $L=4$ m, $b=250$ mm, $D=475$ mm, effective cover 45 mm, so $d=430$ mm and $d'=45$ mm; superimposed load 45 kN/m and point load 75 kN at mid-span (service), M25. "TMT bars" is taken as Fe500 ($f_y=500$ N/mm$^2$).

Step 1: Loads and moment

$$\begin{aligned} \text{Self weight}&=0.25\times0.475\times25=2.97\ \text{kN/m},\quad w=47.97\ \text{kN/m}\\ w_u&=1.5w=71.95\ \text{kN/m},\quad P_u=1.5\times75=112.5\ \text{kN}\\ M_u&=\frac{w_uL^2}{8}+\frac{P_uL}{4}=143.91+112.50=256.41\ \text{kN·m} \end{aligned}$$

Step 2: Limiting moment (Fe500: $x_{u,max}=0.46d$)

$$M_{u,lim}=0.1336f_{ck}bd^2=0.1336\times25\times250\times430^2=154.40\ \text{kN·m}<M_u$$

The section is too shallow for singly reinforced design, so it is designed as doubly reinforced.

Doubly reinforced design

$$M_{u2}=M_u-M_{u,lim}=256.41-154.40=102.01\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.46d=197.8$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{45}{197.8}\right)=0.00270$$

From the design stress-strain curve of Fe500 (IS 456 Fig. 23): $f_{sc}=410.2$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=11.15$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{102.01\times10^6}{(410.2-11.15)(430-45)}=664\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1023\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=609\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=1632\ \text{mm}^2$$

Provide: tension 6-20 mm dia ($1885$ mm$^2$), compression 4-16 mm dia ($804$ mm$^2$). Provide in two layers for tension steel as needed within 250 mm width, with 8 mm stirrups.

Step 3: Deflection check (IS 456 cl. 23.2)

$$p_t=\frac{100A_{st}}{bd}=1.75\%,\quad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=251.1\ \text{N/mm}^2,\quad k_t=1.13$$

Compression steel $p_c=\dfrac{100A_{sc}}{bd}=0.75\%$, modification factor $k_c=\dfrac{1.6p_c}{p_c+0.275}=1.17$.

$$\left(\frac{L}{d}\right)_{allow}=20\times k_t\times k_c=26.6,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{4000}{430}=9.30$$

Actual $<$ allowable, so the beam is safe in deflection.

Answer: $M_u=256.41$ kN·m; $A_{st}=1632$ mm$^2$ (6-20 mm), $A_{sc}=664$ mm$^2$ (4-16 mm); deflection safe.

Let $x_u$ be the depth of the neutral axis and $x_{u,max}$ its limiting value ($0.48d$ for Fe415). In the stress block, the compressive force is $C=0.36f_{ck}b\,x_u$ at lever arm $z=d-0.42x_u$.

  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 ($x_u=x_{u,max}$)

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

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

2. Under-reinforced section ($x_u<x_{u,max}$)

Steel yields first; $x_u=\dfrac{0.87f_yA_{st}}{0.36f_{ck}b}$.

$$M_u=0.87f_yA_{st}(d-0.42x_u)=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)$$

Failure is ductile.

3. Over-reinforced section ($x_u>x_{u,max}$)

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

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

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

Under-reinforced section

• Steel area is less than the balanced value, so $x_u<x_{u,max}$ and steel yields before concrete crushes.
• Failure is ductile: wide cracks and large deflection warn the occupants, and the beam can redistribute moment.
• Economical, since the high strength of steel is fully used. Moment $M_u=0.87f_yA_{st}d(1-A_{st}f_y/bdf_{ck})$.

Over-reinforced section

• Steel area is more than balanced, so $x_u>x_{u,max}$; concrete reaches 0.0035 while steel is still elastic.
• Failure is brittle and sudden, with few cracks and no warning. Steel strength is wasted.

 under-reinforced        over-reinforced
 strain: es > ey         strain: es < ey
 |\ 0.0035               |\ 0.0035
 | \ xu < xumax          | \  xu > xumax
 +--  NA                 +----  NA
 |   \ es (large)        |  \ es (small)
 Ductile                 Brittle

Significance in design

1. The code limits $x_u\le x_{u,max}$, so every designed section is under-reinforced or at most balanced.
2. If $M_u>M_{u,lim}$, increase depth/width or provide compression steel (doubly reinforced) instead of more tension steel.
3. Maximum steel is limited to 4% of the gross area; minimum steel $0.85bd/f_y$ prevents sudden failure on cracking.
4. Ductile behaviour is essential in earthquake-resistant design.

• Balanced section: a section in which the steel and concrete reach their limiting strains together, i.e. concrete reaches the ultimate strain 0.0035 just when steel reaches its yield strain. Here $x_u=x_{u,max}$ and $A_{st}=A_{st,lim}$.
• Under-reinforced section: the steel area is less than for the balanced section, so steel yields first while concrete is not yet crushed ($x_u<x_{u,max}$). Failure is ductile with warning, and this type is preferred.
• Over-reinforced section: the steel area is more than for the balanced section, so concrete crushes (strain 0.0035) before steel yields ($x_u>x_{u,max}$). Failure is sudden and brittle, and it is not permitted in design.

Limiting values of $x_{u,max}/d$ (IS 456): 0.53 for Fe250, 0.48 for Fe415, 0.46 for Fe500.

Rectangular RC beam sections are classified by the way reinforcement is provided and by the failure mode.

1. Singly reinforced sections (steel in tension zone only)

• Under-reinforced ($x_u<x_{u,max}$): steel yields first, ductile failure; used in design.
• Balanced ($x_u=x_{u,max}$): steel and concrete reach limits together; $M_{u,lim}=0.138f_{ck}bd^2$ (Fe415).
• Over-reinforced ($x_u>x_{u,max}$): concrete crushes first, brittle failure; not allowed.

2. Doubly reinforced section

Steel is provided in both the tension and compression zones. Used when $M_u>M_{u,lim}$ with a restricted depth, when moment reverses, or to reduce long-term deflection.

  Singly reinforced      Doubly reinforced
  +-----------+          +-----------+
  |           |          | o  o  Asc |  <- d'
  |           |          |           |
  |           |          |           |
  | o  o  o   |          | o  o  o   |
  +-----------+          +-----------+
     Ast                    Ast

Moment of a doubly reinforced section: $M_u=M_{u,lim}+f_{sc}'A_{sc}(d-d')$, where $f_{sc}'=f_{sc}-f_{cc}$.

Strain diagrams (all with concrete strain 0.0035 at the top):

 Under-reinforced   Balanced        Over-reinforced
 0.0035             0.0035          0.0035
 |\                 |\              |\
 | \                | \             |  \
 |  \ xu<xumax      |  \ xu=xumax   |    \ xu>xumax
 +----  NA          +----  NA      +-------  NA
 |   \              |   \           |      \
 |    \ es>ey       |    \ es=ey    |       \ es<ey

The depth of the neutral axis is found from the strain diagram: $\dfrac{x_{u,max}}{d}=\dfrac{0.0035}{0.0035+\varepsilon_y}$, which gives 0.53, 0.48 and 0.46 for Fe250, Fe415 and Fe500. Only under-reinforced (or balanced) sections are accepted in design.

A doubly reinforced beam is used when the factored moment $M_u$ exceeds the limiting moment $M_{u,lim}=0.138f_{ck}bd^2$ (Fe415) of the section whose size is restricted.

  b                strain     stress        forces
 +---------+ d'     0.0035   0.446fck
 | o  Asc  |         |\     fcc|==|     Cc = 0.36fck.b.xu
 |         |         | \xu    |  |     Cs = Asc(fsc-fcc)
 |         |        -+--NA    |  |
 |         |         |  \
 | o o  Ast|         |   \es   T = 0.87fy.Ast
 +---------+

Steps

1. Assume $b$, $D$, cover $d'$; find $d$ and factored moment $M_u$.
2. Compute $M_{u,lim}=0.36f_{ck}b\,x_{u,max}(d-0.42x_{u,max})$ using $x_{u,max}=0.53d$, $0.48d$ or $0.46d$ for Fe250, Fe415, Fe500.
3. If $M_u>M_{u,lim}$, find the extra moment $M_{u2}=M_u-M_{u,lim}$, to be carried by a steel couple.
4. Find strain in compression steel at the limiting NA: $\varepsilon_{sc}=0.0035\left(1-\dfrac{d'}{x_{u,max}}\right)$. Get the stress $f_{sc}$ from the stress-strain curve (IS 456 Fig. 23 / SP16 Table F). Concrete stress displaced by steel $f_{cc}=0.446f_{ck}$.
5. Compression steel: $$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}$$
6. Tension steel: $$A_{st}=A_{st1}+A_{st2},\quad A_{st1}=\frac{0.36f_{ck}b\,x_{u,max}}{0.87f_y},\quad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}$$
7. Check limits: $A_{st}\ge0.85bd/f_y$; $A_{st}$ and $A_{sc}$ not more than 4% each of $bD$; $d'/d\approx0.1$-$0.2$.
8. Detailing: provide stirrups spacing $\le$ the least of $16\phi$ of compression bar, 300 mm and $b$, so that compression bars do not buckle. Check shear and deflection (with compression steel factor $k_c$).

Concept

In a monolithic slab-beam system, the slab on either side of the beam web acts with the web in resisting bending, forming a T-beam (slab on both sides) or L-beam (slab on one side, e.g. edge beam). The compressive stress in the slab is not uniform across its width; it is highest at the web and falls away from it, because the shear that transfers compression from the web to the flange is carried by shear deformation of the slab (shear lag).

 compressive stress across flange      T-beam section
   ___                                 +------------------+  Df
  /   \___                             +----+      +------+
 /        \___                              |  bw  |
 |  web     slab edge                      |      |

For simple design, the actual variable stress is replaced by a uniform stress over a smaller effective flange width $b_f$ which gives the same total compression.

Effective flange width (IS 456 cl. 23.1.2)

• T-beam: $b_f=\dfrac{l_0}{6}+b_w+6D_f$
• L-beam: $b_f=\dfrac{l_0}{12}+b_w+3D_f$
• Isolated T-beam: $b_f=\dfrac{l_0}{\dfrac{l_0}{b}+4}+b_w$
• $b_f$ must not exceed the actual width of the flange, or the centre-to-centre spacing of beams.

Here $l_0$ is the distance between points of zero moment (equal to the effective span for a simply supported beam), $b_w$ the web width, $D_f$ the flange thickness and $b$ the actual flange width.

Given: $b=250$ mm, $D=300$ mm, clear span 4 m between walls 300 mm thick, DL = 10 kN/m (including self weight), LL = 8 kN/m, M20, Fe415. Assume effective cover 40 mm top and bottom, so $d=300-40=260$ mm and $d'=40$ mm.

Step 1: Effective span (IS 456 cl. 22.2) — least of centre-to-centre of supports ($4+0.3=4.3$ m) and clear span $+d$ ($4+0.26=4.26$ m). Take $L_{eff}=4.26$ m.

Step 2: Factored load and moment

$$w_u=1.5(10+8)=27.0\ \text{kN/m},\qquad M_u=\frac{w_uL^2}{8}=\frac{27.0\times4.26^2}{8}=61.25\ \text{kN·m}$$

Step 3: Limiting moment of the section

$$M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times250\times260^2=46.63\ \text{kN·m}$$

Since $M_u>M_{u,lim}$ and the depth is fixed, the beam must be doubly reinforced.

Doubly reinforced design

$$M_{u2}=M_u-M_{u,lim}=61.25-46.63=14.62\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.48d=124.8$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{40}{124.8}\right)=0.00238$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=341.6$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{14.62\times10^6}{(341.6-8.92)(260-40)}=200\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=622\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=184\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=806\ \text{mm}^2$$

Provide: tension steel 5-16 mm dia ($1005$ mm$^2$) and compression steel 2-12 mm dia ($226$ mm$^2$).

Check: $p_t=100\times1005/(250\times260)=1.55\%$ $<4\%$ of gross area OK. Stirrups of 8 mm dia (two-legged) at suitable spacing are provided for shear and to hold the compression bars (spacing $\le 16\phi$ of compression bar). Shear design to be done separately.

Answer: $M_u=61.25$ kN·m; $A_{st}=806$ mm$^2$ (5-16) and $A_{sc}=200$ mm$^2$ (2-12).

Given: $b_f=1500$ mm, $b_w=240$ mm, $D_f=150$ mm, $d=590$ mm, M20, Fe415.

Step 1: Limiting depth of neutral axis (Fe415)

$$x_{u,max}=0.48d=0.48\times590=283.2\ \text{mm}>D_f=150\ \text{mm}$$

so the neutral axis lies in the web. Also $D_f/d=150/590=0.254>0.2$, therefore the flange is not thin and IS 456 Annex G-2.2 uses the equivalent flange thickness

$$y_f=0.15x_{u,max}+0.65D_f=0.15\times283.2+0.65\times150=139.98\ \text{mm}$$

Step 2: Limiting moment of resistance

$$M_{u,lim}=0.36f_{ck}b_wx_{u,max}(d-0.416x_{u,max})+0.45f_{ck}(b_f-b_w)y_f\left(d-\frac{y_f}{2}\right)$$ $$\begin{aligned} M_{u,web}&=0.36\times20\times240\times283.2\times(590-0.416\times283.2)=231.07\ \text{kN·m}\\ M_{u,flange}&=0.45\times20\times(1500-240)\times139.98\times\left(590-\frac{139.98}{2}\right)=825.45\ \text{kN·m} \end{aligned}$$ $$M_{u,lim}=231.07+825.45=1056.52\ \text{kN·m}$$

Step 3: Limiting area of steel (from $C=T$)

$$A_{st,lim}=\frac{0.36f_{ck}b_wx_{u,max}+0.45f_{ck}(b_f-b_w)y_f}{0.87f_y}=5752\ \text{mm}^2$$

Answer: $M_{u,lim}=1056.52$ kN·m and $A_{st,lim}=5752$ mm$^2$.

Given: $b_f=1600$ mm, $D_f=110$ mm, $b_w=300$ mm, $D=460$ mm, effective cover 40 mm so $d=420$ mm; 4-20 mm bars, $A_{st}=4\times\frac{\pi}{4}(20)^2=1256.6$ mm$^2$; M20, Fe415.

Step 1: Locate neutral axis. Assume $x_u\le D_f$ (rectangular behaviour with width $b_f$). Equating compression and tension:

$$0.36f_{ck}b_fx_u=0.87f_yA_{st}\ \Rightarrow\ x_u=\frac{0.87\times415\times1256.6}{0.36\times20\times1600}=39.38\ \text{mm}$$

$x_u=39.38<D_f=110$ mm, so the neutral axis is in the flange (assumption correct).

Step 2: Check section type. $x_{u,max}=0.48d=0.48\times420=201.6$ mm $>x_u$, so the section is under-reinforced.

Step 3: Moment of resistance

$$\begin{aligned} M_u&=0.87f_yA_{st}(d-0.42x_u)\\ &=0.87\times415\times1256.6\times(420-0.42\times39.38)\\ &=183.05\ \text{kN·m} \end{aligned}$$

Answer: $M_u=183.05$ kN·m (design moment of resistance of the T-section).

Given: $b=300$ mm, $d=500$ mm, span $L=7$ m, dead load 45 kN/m (excluding self weight), M20, Fe415, effective cover to compression steel $d'=40$ mm. Limit state of collapse in flexure.

Assumption: overall depth $D=550$ mm ($d$ plus 50 mm cover); the effective span is taken as 7 m.

Step 1: Loads

$$\begin{aligned} \text{Self weight}&=0.30\times0.55\times25=4.12\ \text{kN/m}\\ w&=45+4.12=49.12\ \text{kN/m}\\ w_u&=1.5\times49.12=73.69\ \text{kN/m} \end{aligned}$$

Step 2: Design moment at mid-span

$$M_u=\frac{w_uL^2}{8}=\frac{73.69\times7^2}{8}=451.34\ \text{kN·m}$$

Step 3: Limiting moment of resistance (Fe415, $x_{u,max}=0.48d$)

$$M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times300\times500^2=206.95\ \text{kN·m}$$

Since $M_u>M_{u,lim}$ and the size is fixed, the section is doubly reinforced.

Doubly reinforced design

$$M_{u2}=M_u-M_{u,lim}=451.34-206.95=244.39\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.48d=240.0$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{40}{240.0}\right)=0.00292$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=353.2$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{244.39\times10^6}{(353.2-8.92)(500-40)}=1543\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1436\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=1471\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=2907\ \text{mm}^2$$

Provide:

• Tension steel: 6-25 mm dia ($2945$ mm$^2$), arranged in two layers.
• Compression steel: 5-20 mm dia ($1571$ mm$^2$) at $d'=40$ mm.
• Stirrups (8 mm, 2-legged) at spacing not more than $16\phi_c$, 300 mm and $b$ to hold the compression bars (shear design separately).

Checks: minimum steel $0.85bd/f_y=307$ mm$^2$ OK; $p_t=1.96\%<4\%$ and $p_c=1.05\%$ OK.

Answer: Mid-span $A_{st}=2907$ mm$^2$ (6-25 mm) and $A_{sc}=1543$ mm$^2$ (5-20 mm).

Given: $b_f=925$ mm, $d=450$ mm, $D_f=100$ mm, $b_w=250$ mm; tension steel 4-20 mm, $A_{st}=1256.6$ mm$^2$; compression steel 3-16 mm, $A_{sc}=603.2$ mm$^2$; M20, Fe415. Effective cover to compression steel is not given; assume $d'=40$ mm.

Step 1: Neutral axis. Assume $x_u\le D_f$ so the flange acts as a rectangle of width 925 mm. Force equilibrium $C=T$:

$$0.36f_{ck}b_fx_u+A_{sc}(f_{sc}-f_{cc})=0.87f_yA_{st}$$

where $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$ and $f_{sc}$ depends on the strain $\varepsilon_{sc}=0.0035\left(1-\dfrac{d'}{x_u}\right)$.

Total tension $T=0.87\times415\times1256.6=453.7$ kN.

Trial and error (checking $f_{sc}$ from the Fe415 design curve each time):

$$x_u=53.18\ \text{mm},\quad \varepsilon_{sc}=0.0035\left(1-\frac{40}{53.18}\right)=0.00087,\quad f_{sc}=173.9\ \text{N/mm}^2$$

Check: $C=0.36\times20\times925\times53.18+603.2(173.9-8.92)=453.7$ kN $=T$.

$x_u=53.18$ mm $<D_f=100$ mm (flange) and $<x_{u,max}=0.48d=216.0$ mm, so the assumption is correct and the section is under-reinforced.

Step 2: Ultimate moment (about the tension steel)

$$\begin{aligned} M_u&=0.36f_{ck}b_fx_u(d-0.42x_u)+A_{sc}(f_{sc}-f_{cc})(d-d')\\ &=0.36\times20\times925\times53.18(450-0.42\times53.18)+603.2(173.9-8.92)(450-40)\\ &=151.47+40.80=192.28\ \text{kN·m} \end{aligned}$$

Answer: $M_u=192.28$ kN·m.

Given (from figure): $b_f=2250$ mm, $D_f=150$ mm, $b_w=250$ mm, $D=450$ mm, 4-25 mm bars, $A_{st}=4\times\frac{\pi}{4}(25)^2=1963.5$ mm$^2$; M20, Fe415. Effective cover is not given; assume 50 mm, so $d=450-50=400$ mm.

Step 1: Position of neutral axis. Assume it lies in the flange ($x_u\le D_f$):

$$x_u=\frac{0.87f_yA_{st}}{0.36f_{ck}b_f}=\frac{0.87\times415\times1963.5}{0.36\times20\times2250}=43.76\ \text{mm}$$

$x_u=43.76$ mm $<D_f=150$ mm, so the assumption holds. The section behaves as a rectangular beam of width 2250 mm.

Step 2: Check for under-reinforcement

$$x_{u,max}=0.48d=0.48\times400=192.0\ \text{mm}>x_u$$

The section is under-reinforced (ductile).

Step 3: Ultimate moment capacity

$$\begin{aligned} M_u&=0.87f_yA_{st}(d-0.42x_u)\\ &=0.87\times415\times1963.5\times(400-0.42\times43.76)\\ &=270.54\ \text{kN·m} \end{aligned}$$

Answer: $M_u=270.54$ kN·m.

Given: $L=5.5$ m, $b=230$ mm, $D=550$ mm, superimposed load 50 kN/m (excluding self weight), Fe415, mild exposure, effective cover 50 mm. Grade of concrete is not stated; M20 (minimum for mild exposure, IS 456 Table 5) is adopted. $d=550-50=500$ mm.

Step 1: Loads and moment

$$\begin{aligned} \text{Self weight}&=0.23\times0.55\times25=3.16\ \text{kN/m},\quad w=53.16\ \text{kN/m}\\ w_u&=1.5w=79.74\ \text{kN/m}\\ M_u&=\frac{w_uL^2}{8}=\frac{79.74\times5.5^2}{8}=301.53\ \text{kN·m} \end{aligned}$$

Step 2: Limiting moment

$$M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times230\times500^2=158.66\ \text{kN·m}<M_u$$

The section is doubly reinforced (size is limited).

Doubly reinforced design

$$M_{u2}=M_u-M_{u,lim}=301.53-158.66=142.87\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.48d=240.0$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{50}{240.0}\right)=0.00277$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=351.9$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{142.87\times10^6}{(351.9-8.92)(500-50)}=926\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1101\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=879\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=1980\ \text{mm}^2$$

Provide: tension 5-25 mm dia ($2454$ mm$^2$), compression 5-16 mm dia ($1005$ mm$^2$), with 8 mm stirrups.

Step 3: Deflection check (IS 456 cl. 23.2)

$$p_t=\frac{100A_{st}}{bd}=2.13\%,\quad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=194.2\ \text{N/mm}^2,\quad k_t=1.55$$ $$p_c=0.87\%,\quad k_c=\frac{1.6p_c}{p_c+0.275}=1.22$$ $$\left(\frac{L}{d}\right)_{allow}=20\times k_t\times k_c=37.8,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{5500}{500}=11.00$$

The beam is safe in deflection.

Answer: $M_u=301.53$ kN·m, $A_{st}=1980$ mm$^2$ (5-25 mm), $A_{sc}=926$ mm$^2$ (5-16 mm); deflection safe.