Working Stress Method of Design

Given: $b=350$ mm, $d=650$ mm, 5-25 mm dia, $A_{st}=5\times\frac{\pi}{4}(25)^2 = 2454.4$ mm$^2$, $M_{service}=150$ kN·m. M25: $\sigma_{cbc}=8.5$ N/mm$^2$; Fe415: $\sigma_{st}=230$ N/mm$^2$; $m=10.98$.

Step 1: Constants (M25: $\sigma_{cbc}=8.5$, Fe415: $\sigma_{st}=230$ N/mm$^2$, $m=280/(3\times8.5)=10.98$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{10.98\times 8.5}{10.98\times 8.5 + 230} = 0.289\\ x_c &= k d = 0.289\times 650 = 187.62\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{350 x^2}{2} = 10.98\times 2454.4\,(650-x)$$ $$x = 248.62\ \text{mm}$$

Since $x = 248.62 > x_c = 187.62$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=8.5$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 8.5\times 350\times 248.62\times\left(650-\frac{248.62}{3}\right)\\ &= 209.73\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 150.68$ N/mm$^2$ $<$ 230 N/mm$^2$.

Safety check: $M_r = 209.73$ kN·m $>$ 150 kN·m. The beam is safe.

Stresses at the service moment of 150 kN·m (neutral axis $x=248.62$ mm):

$$\sigma_c=\frac{M}{\frac12 b x(d-x/3)} = 6.08\ \text{N/mm}^2,\qquad \sigma_{s}=\frac{M}{A_{st}(d-x/3)}=107.76\ \text{N/mm}^2$$

Stresses at the limiting (failure) state as per WSM: the section is over-reinforced, so concrete reaches its permissible stress first. At this state concrete stress $= 8.5$ N/mm$^2$ and steel stress $= 150.68$ N/mm$^2$ (less than 230 N/mm$^2$).

Answer: Beam is safe ($M_r=209.73$ kN·m $>150$ kN·m). At failure state: $\sigma_c = 8.5$ N/mm$^2$, $\sigma_s = 150.68$ N/mm$^2$. (At 150 kN·m: $\sigma_c=6.08$, $\sigma_s=107.76$ N/mm$^2$.)

Given: $b=300$ mm, $d=550$ mm, 4-12 mm dia, $A_{st}=4\times\frac{\pi}{4}(12)^2=452.4$ mm$^2$. M20: $\sigma_{cbc}=7$ N/mm$^2$, $m=13.33$; Fe415: $\sigma_{st}=230$ N/mm$^2$.

Step 1: Constants (M20: $\sigma_{cbc}=7$, Fe415: $\sigma_{st}=230$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 550 = 158.73\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{300 x^2}{2} = 13.33\times 452.4\,(550-x)$$ $$x = 129.95\ \text{mm}$$

Since $x = 129.95 < x_c = 158.73$ mm, the section is under-reinforced: steel reaches $\sigma_{st}=230$ N/mm$^2$ first, so steel governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \sigma_{st}A_{st}\left(d-\frac{x}{3}\right) = 230\times 452.4\times\left(550-\frac{129.95}{3}\right)\\ &= 52.72\ \text{kN·m} \end{aligned}$$

Stress in concrete at this moment: $\sigma_{c} = \dfrac{\sigma_{st}\,x}{m(d-x)} = 5.34$ N/mm$^2$ $<$ 7 N/mm$^2$ (safe).

Answer: $M_r = 52.72$ kN·m (under-reinforced section).

Given: $b=300$ mm, $d=550$ mm, 4-16 mm dia, $A_{st}=4\times\frac{\pi}{4}(16)^2=804.2$ mm$^2$, $M=50$ kN·m. Permissible: $\sigma_{cbc}=5$, $\sigma_{st}=140$ N/mm$^2$. For M20, modular ratio $m=13.33$ (IS 456, Annex B-1.3).

Step 1: Neutral axis (elastic cracked section)

$$\frac{b x^2}{2}=m A_{st}(d-x)\ \Rightarrow\ \frac{300x^2}{2}=13.33\times 804.2(550-x)$$ $$x=165.72\ \text{mm}$$

Step 2: Lever arm

$$z=d-\frac{x}{3}=550-\frac{165.72}{3}=494.76\ \text{mm}$$

Step 3: Stresses

$$\begin{aligned} \sigma_{c}&=\frac{2M}{b\,x\,z}=\frac{2\times 50\times10^6}{300\times 165.72\times 494.76}=4.07\ \text{N/mm}^2\\ \sigma_{s}&=\frac{M}{A_{st}\,z}=\frac{50\times10^6}{804.2\times 494.76}=125.66\ \text{N/mm}^2 \end{aligned}$$

Both are less than the permissible values (5 and 140 N/mm$^2$), so the section is safe.

Answer: stress in extreme compression concrete $=4.07$ N/mm$^2$; stress in tension steel $=125.66$ N/mm$^2$.

Assumption: the section is given as $b\times d$, so $b=250$ mm and effective depth $d=450$ mm. For M15 concrete $\sigma_{cbc}=5$ N/mm$^2$, $m=\dfrac{280}{3\times5}=18.67$; for mild steel $\sigma_{st}=140$ N/mm$^2$.

Step 1: Balanced section constants

$$\begin{aligned} k&=\frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}}=\frac{18.67\times5}{18.67\times5+140}=0.400\\ j&=1-\frac{k}{3}=0.867\\ Q&=\frac12\sigma_{cbc}\,k\,j=\frac12\times5\times0.400\times0.867=0.867\ \text{N/mm}^2 \end{aligned}$$

Step 2: Moment of resistance (balanced section, maximum for singly reinforced)

$$M_r=Q\,b\,d^2=0.867\times250\times450^2=43.88\ \text{kN·m}$$

Step 3: Area of steel

$$A_{st}=\frac{M_r}{\sigma_{st}\,j\,d}=\frac{43.88\times10^6}{140\times0.867\times450}=803.7\ \text{mm}^2$$

Minimum steel check (IS 456, cl. 26.5.1.1): $0.85bd/f_y=0.85\times250\times450/250=382$ mm$^2$ $<A_{st}$, OK.

Provide 4-16 mm dia bars ($804$ mm$^2$).

Answer: $A_{st}=803.7$ mm$^2$ (say 4-16 mm bars) and $M_r=43.88$ kN·m.

Given: $b=230$ mm, $d=380$ mm, $A_{st}=580$ mm$^2$, $\sigma_{cbc}=5$, $\sigma_{st}=140$ N/mm$^2$ (M15 concrete, mild steel), $m=18.67$.

Step 1: Constants ($\sigma_{cbc}=5$, $\sigma_{st}=140$ N/mm$^2$, $m=\dfrac{280}{3\times5}=18.67$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{18.67\times 5}{18.67\times 5 + 140} = 0.400\\ x_c &= k d = 0.400\times 380 = 152.02\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{230 x^2}{2} = 18.67\times 580.0\,(380-x)$$ $$x = 147.85\ \text{mm}$$

Since $x = 147.85 < x_c = 152.02$ mm, the section is under-reinforced: steel reaches $\sigma_{st}=140$ N/mm$^2$ first, so steel governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \sigma_{st}A_{st}\left(d-\frac{x}{3}\right) = 140\times 580.0\times\left(380-\frac{147.85}{3}\right)\\ &= 26.85\ \text{kN·m} \end{aligned}$$

Stress in concrete at this moment: $\sigma_{c} = \dfrac{\sigma_{st}\,x}{m(d-x)} = 4.78$ N/mm$^2$ $<$ 5 N/mm$^2$ (safe).

Actual stresses at $M_r$: concrete $\sigma_c=4.78$ N/mm$^2$, steel $\sigma_s=140.00$ N/mm$^2$.

Answer: $M_r=26.85$ kN·m; actual stresses are $\sigma_c=4.78$ N/mm$^2$ and $\sigma_s=140.00$ N/mm$^2$.

Assumptions (IS 456:2000, Annex B-1.2)

• Plane sections remain plane after bending.
• Concrete is elastic in compression and carries no tension.
• Perfect bond between steel and concrete; stress in steel $=$ modular ratio $\times$ stress in surrounding concrete.
• For compression steel, effective modular ratio is $1.5m$ (to allow for creep).

Notation: $b$ = width, $d$ = effective depth, $d'$ = effective cover to compression steel, $x$ = depth of neutral axis, $A_{st}$ = tension steel, $A_{sc}$ = compression steel, $m$ = modular ratio.

   Section    Strain      Stresses/forces
   +------+   |\         Cc  (concrete)
 d'| Asc  |   | \ x      Cs  (comp. steel)
   |      |  -+--- NA
   |      |   | /
   | Ast  |   |/         T   (tension steel)
   +------+

Stress in concrete at the level of compression steel

By similar triangles of the strain diagram:

$$\sigma_c' = \sigma_{cbc}\,\frac{x-d'}{x}$$

Stress in compression steel $\sigma_{sc} = 1.5\,m\,\sigma_c'$. The concrete displaced by the steel is allowed for by using $(1.5m-1)A_{sc}$ as the equivalent concrete area.

Neutral axis (first moment of the equivalent area about the neutral axis is zero)

$$\frac{b x^2}{2} + (1.5m-1)A_{sc}(x-d') = m A_{st}(d-x)$$

Moment of resistance

Taking moments about the tension steel, with concrete compression $C_c = \tfrac12\sigma_{cbc}\,b\,x$ acting at $x/3$ from the top:

$$M_r = \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) + (1.5m-1)A_{sc}\,\sigma_{c}'\,(d-d')$$

Substituting $\sigma_c'$:

$$M_r = \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) + (1.5m-1)A_{sc}\,\sigma_{cbc}\,\frac{x-d'}{x}\,(d-d')$$

The first term is the moment of the singly reinforced (balanced) part $M_1 = Q\,b\,d^2$ when $x = x_c$, and the second is the additional moment $M_2$ from the compression steel. The tension steel then follows from $A_{st}=\dfrac{M_1}{\sigma_{st}jd}+\dfrac{M_2}{\sigma_{st}(d-d')}$.

Checks: $\sigma_{c}\le\sigma_{cbc}$, $\sigma_{st}\le\sigma_{st,perm}$ and $\sigma_{sc}=1.5m\sigma_c'\le\sigma_{sc,perm}$ (190 N/mm$^2$ for Fe415).

Assumption: clear cover 25 mm, so $d=475-25-\dfrac{16}{2}=442$ mm. $b=250$ mm, $A_{st}=3\times\frac{\pi}{4}(16)^2=603.2$ mm$^2$.

Step 1: Constants (M20: $\sigma_{cbc}=7$, Fe415: $\sigma_{st}=230$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 442 = 127.56\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{250 x^2}{2} = 13.33\times 603.2\,(442-x)$$ $$x = 139.49\ \text{mm}$$

Since $x = 139.49 > x_c = 127.56$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 250\times 139.49\times\left(442-\frac{139.49}{3}\right)\\ &= 48.27\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 202.35$ N/mm$^2$ $<$ 230 N/mm$^2$.

Answer: $M_r=48.27$ kN·m (over-reinforced, concrete governs).

Given: $b=250$ mm, $D=600$ mm, $d=600-50=550$ mm, 4-20 mm dia, $A_{st}=4\times\frac{\pi}{4}(20)^2=1256.6$ mm$^2$, span $L=6.10$ m, $\sigma_{cbc}=7$, $\sigma_{st}=230$ N/mm$^2$, $m=13.33$ (modular ratio for M20).

Step 1: Constants ($\sigma_{cbc}=7$, $\sigma_{st}=230$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 550 = 158.73\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{250 x^2}{2} = 13.33\times 1256.6\,(550-x)$$ $$x = 212.63\ \text{mm}$$

Since $x = 212.63 > x_c = 158.73$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 250\times 212.63\times\left(550-\frac{212.63}{3}\right)\\ &= 89.14\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 148.05$ N/mm$^2$ $<$ 230 N/mm$^2$.

Step 4: Safe UDL on the simply supported span

$$M_r=\frac{wL^2}{8}\ \Rightarrow\ w=\frac{8M_r}{L^2}=\frac{8\times89.14}{6.10^2}=19.16\ \text{kN/m}$$

Self weight $=0.25\times0.60\times25=3.75$ kN/m, so the superimposed load is $19.16-3.75=15.41$ kN/m.

Answer: total safe UDL $=19.16$ kN/m (including self weight), i.e. superimposed load $=15.41$ kN/m.

Requirement of steel as reinforcement

Concrete is strong in compression but weak in tension (tensile strength is about 10% of compressive strength), so it cracks at low loads. Steel is therefore provided as reinforcement for the following reasons.

• Tension: steel carries the tensile force in beams, slabs, footings and ties, with concrete protecting it from corrosion and fire.
• Shear and torsion: stirrups and bent-up bars resist diagonal tension.
• Compression: longitudinal bars share the load in columns and raise capacity of beams (doubly reinforced section).
• Control of cracks and shrinkage: distribution and temperature steel limit crack width.
• Bond: steel has a similar coefficient of thermal expansion to concrete ($\approx 1.2\times10^{-5}$ per $^\circ$C), and ribs/deformations give good bond.
• Ductility: steel gives warning before failure.

Moment of resistance of doubly reinforced section

A beam is doubly reinforced when the applied moment exceeds the balanced moment $Q\,b\,d^2$ of the singly reinforced section, or when depth is restricted, or for reversal of moment. Compression steel $A_{sc}$ is added at effective cover $d'$ from the compression face. The moment of resistance is

$$M_r = M_1 + M_2$$

where $M_1 = Q\,b\,d^2$ is the balanced-section moment and $M_2$ the additional moment taken by the compression steel and the additional tension steel.

Derivation

Assumptions (IS 456:2000, Annex B-1.2)

• Plane sections remain plane after bending.
• Concrete is elastic in compression and carries no tension.
• Perfect bond between steel and concrete; stress in steel $=$ modular ratio $\times$ stress in surrounding concrete.
• For compression steel, effective modular ratio is $1.5m$ (to allow for creep).

Notation: $b$ = width, $d$ = effective depth, $d'$ = effective cover to compression steel, $x$ = depth of neutral axis, $A_{st}$ = tension steel, $A_{sc}$ = compression steel, $m$ = modular ratio.

   Section    Strain      Stresses/forces
   +------+   |\         Cc  (concrete)
 d'| Asc  |   | \ x      Cs  (comp. steel)
   |      |  -+--- NA
   |      |   | /
   | Ast  |   |/         T   (tension steel)
   +------+

Stress in concrete at the level of compression steel

By similar triangles of the strain diagram:

$$\sigma_c' = \sigma_{cbc}\,\frac{x-d'}{x}$$

Stress in compression steel $\sigma_{sc} = 1.5\,m\,\sigma_c'$. The concrete displaced by the steel is allowed for by using $(1.5m-1)A_{sc}$ as the equivalent concrete area.

Neutral axis (first moment of the equivalent area about the neutral axis is zero)

$$\frac{b x^2}{2} + (1.5m-1)A_{sc}(x-d') = m A_{st}(d-x)$$

Moment of resistance

Taking moments about the tension steel, with concrete compression $C_c = \tfrac12\sigma_{cbc}\,b\,x$ acting at $x/3$ from the top:

$$M_r = \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) + (1.5m-1)A_{sc}\,\sigma_{c}'\,(d-d')$$

Substituting $\sigma_c'$:

$$M_r = \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) + (1.5m-1)A_{sc}\,\sigma_{cbc}\,\frac{x-d'}{x}\,(d-d')$$

The first term is the moment of the singly reinforced (balanced) part $M_1 = Q\,b\,d^2$ when $x = x_c$, and the second is the additional moment $M_2$ from the compression steel. The tension steel then follows from $A_{st}=\dfrac{M_1}{\sigma_{st}jd}+\dfrac{M_2}{\sigma_{st}(d-d')}$.

Checks: $\sigma_{c}\le\sigma_{cbc}$, $\sigma_{st}\le\sigma_{st,perm}$ and $\sigma_{sc}=1.5m\sigma_c'\le\sigma_{sc,perm}$ (190 N/mm$^2$ for Fe415).

Assumptions: effective cover $d'=40$ mm (top and bottom), so $d=425-40=385$ mm. M20: $\sigma_{cbc}=7$ N/mm$^2$, $m=13.33$. Fe500: $\sigma_{st}=275$ N/mm$^2$ (IS 456 Table 22 / Annex B). $M=64$ kN·m.

Step 1: Balanced moment of the singly reinforced section

$$\begin{aligned} k&=\frac{13.33\times7}{13.33\times7+275}=0.2533,\quad j=1-\frac{k}{3}=0.9156\\ Q&=\tfrac12\sigma_{cbc}kj=0.8118\ \text{N/mm}^2\\ M_{bal}&=Qbd^2=0.8118\times230\times385^2=27.68\ \text{kN·m} \end{aligned}$$

Since $M=64>M_{bal}=27.68$ kN·m, a singly reinforced section is not enough. The beam must be doubly reinforced.

Step 2: Moment shared by compression steel

$$M_2=M-M_{bal}=64-27.68=36.32\ \text{kN·m}$$

Step 3: Compression steel

Depth of neutral axis (balanced) $x_c=kd=97.54$ mm. Stress in concrete at the compression steel level:

$$\sigma_c'=\sigma_{cbc}\frac{x_c-d'}{x_c}=7\times\frac{97.54-40}{97.54}=4.13\ \text{N/mm}^2$$ $$A_{sc}=\frac{M_2}{(1.5m-1)\,\sigma_c'\,(d-d')}=\frac{36.32\times10^6}{(1.5\times13.33-1)\times4.13\times(385-40)}=1342\ \text{mm}^2$$

Stress in compression steel $\sigma_{sc}=1.5m\sigma_c'=82.57$ N/mm$^2$, below the permissible value, OK.

Step 4: Tension steel

$$A_{st}=\frac{M_{bal}}{\sigma_{st}jd}+\frac{M_2}{\sigma_{st}(d-d')}=\frac{27.68\times10^6}{275\times0.9156\times385}+\frac{36.32\times10^6}{275\times(385-40)}$$ $$A_{st}=286+383=668\ \text{mm}^2$$

Answer: tensile reinforcement $A_{st}=668$ mm$^2$ (e.g. 3-20 mm bars); compression reinforcement $A_{sc}=1342$ mm$^2$ (e.g. 7-16 mm bars).

Given: $b=300$ mm, $D=500$ mm, effective cover 50 mm so $d=450$ mm, 4-25 mm dia, $A_{st}=4\times\frac{\pi}{4}(25)^2=1963.5$ mm$^2$, $L=5$ m. M20: $\sigma_{cbc}=7$; Fe415: $\sigma_{st}=230$ N/mm$^2$; $m=13.33$.

Step 1: Constants (M20: $\sigma_{cbc}=7$, Fe415: $\sigma_{st}=230$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 450 = 129.87\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{300 x^2}{2} = 13.33\times 1963.5\,(450-x)$$ $$x = 206.24\ \text{mm}$$

Since $x = 206.24 > x_c = 129.87$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 300\times 206.24\times\left(450-\frac{206.24}{3}\right)\\ &= 82.56\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 110.29$ N/mm$^2$ $<$ 230 N/mm$^2$.

Step 4: Total UDL

$$w=\frac{8M_r}{L^2}=\frac{8\times82.56}{5^2}=26.42\ \text{kN/m}$$

Self weight $=0.30\times0.50\times25=3.75$ kN/m, so the external (superimposed) load is $22.67$ kN/m.

Answer: total UDL including self weight $=26.42$ kN/m.

Given: $b=250$ mm, $D=450$ mm, effective cover 40 mm so $d=410$ mm, 4-16 mm dia, $A_{st}=4\times\frac{\pi}{4}(16)^2=804.2$ mm$^2$. M20: $\sigma_{cbc}=7$, $m=13.33$; Fe415: $\sigma_{st}=230$ N/mm$^2$.

Step 1: Constants (M20: $\sigma_{cbc}=7$, Fe415: $\sigma_{st}=230$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 410 = 118.33\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{250 x^2}{2} = 13.33\times 804.2\,(410-x)$$ $$x = 149.48\ \text{mm}$$

Since $x = 149.48 > x_c = 118.33$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 250\times 149.48\times\left(410-\frac{149.48}{3}\right)\\ &= 47.11\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 162.63$ N/mm$^2$ $<$ 230 N/mm$^2$.

Answer: $M_r=47.11$ kN·m.

Given: $b=230$ mm, $D=350$ mm, nominal cover 25 mm, so $d=350-25-\dfrac{16}{2}=317$ mm. $A_{st}=4\times\frac{\pi}{4}(16)^2=804.2$ mm$^2$. $\sigma_{cbc}=7$, $\sigma_{st}=140$ N/mm$^2$, $m=13.33$.

Step 1: Constants ($\sigma_{cbc}=7$, $\sigma_{st}=140$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 140} = 0.400\\ x_c &= k d = 0.400\times 317 = 126.78\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{230 x^2}{2} = 13.33\times 804.2\,(317-x)$$ $$x = 131.50\ \text{mm}$$

Since $x = 131.50 > x_c = 126.78$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 230\times 131.50\times\left(317-\frac{131.50}{3}\right)\\ &= 28.92\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 131.62$ N/mm$^2$ $<$ 140 N/mm$^2$.

Answer: $M_r=28.92$ kN·m.

Assumption: clear cover 25 mm, so $d=450-25-\dfrac{20}{2}=415$ mm. $b=230$ mm, $A_{st}=4\times\frac{\pi}{4}(20)^2=1256.6$ mm$^2$. Permissible stresses: M20 $\sigma_{cbc}=7$ N/mm$^2$, Fe415 $\sigma_{st}=230$ N/mm$^2$, $m=13.33$. (The span 4.2 m is not needed for $M_r$.)

Step 1: Constants (M20: $\sigma_{cbc}=7$, Fe415: $\sigma_{st}=230$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 415 = 119.77\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{230 x^2}{2} = 13.33\times 1256.6\,(415-x)$$ $$x = 183.59\ \text{mm}$$

Since $x = 183.59 > x_c = 119.77$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 230\times 183.59\times\left(415-\frac{183.59}{3}\right)\\ &= 52.29\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 117.61$ N/mm$^2$ $<$ 230 N/mm$^2$.

Answer: $M_r=52.29$ kN·m.

Given: $b=250$ mm, $d=500$ mm, 3-16 mm dia, $A_{st}=3\times\frac{\pi}{4}(16)^2=603.2$ mm$^2$, $\sigma_{cbc}=7$, $\sigma_{st}=230$ N/mm$^2$, $m=13.33$.

Step 1: Constants ($\sigma_{cbc}=7$, $\sigma_{st}=230$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 500 = 144.30\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{250 x^2}{2} = 13.33\times 603.2\,(500-x)$$ $$x = 150.04\ \text{mm}$$

Since $x = 150.04 > x_c = 144.30$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 250\times 150.04\times\left(500-\frac{150.04}{3}\right)\\ &= 59.08\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 217.65$ N/mm$^2$ $<$ 230 N/mm$^2$.

Answer: $M_r=59.08$ kN·m.