BE Civil Engineering (IOE, TU) Design of RCC Structures (IOE, CE 702) Question Paper 2076 Nepal

Given data

• Clear span $L_c = 5.6\ \text{m}$, $b = 250\ \text{mm}$, $D = 500\ \text{mm}$, cover $= 40\ \text{mm}$
• $f_{ck} = 20\ \text{N/mm}^2$, $f_y = 415\ \text{N/mm}^2$
• Effective depth $d = D - \text{cover} = 500 - 40 = 460\ \text{mm}$

Effective span (IS 456 cl. 22.2) — least of (clear span + d) and (centre to centre of supports). Take $L_{eff} = L_c + d = 5.6 + 0.46 = 6.06\ \text{m}$.

(a) Factored design moment

Total characteristic load $w = 12 + 9 = 21\ \text{kN/m}$.

Factored load $w_u = 1.5 \times 21 = 31.5\ \text{kN/m}$.

$$M_u = \frac{w_u L_{eff}^2}{8} = \frac{31.5 \times 6.06^2}{8} = \frac{31.5 \times 36.7236}{8} = \frac{1156.79}{8} = 144.6\ \text{kN·m}$$

Factored design moment $M_u = 144.6\ \text{kN·m}$.

(b) Limiting moment of resistance (Fe415, $x_{u,max}/d = 0.48$)

$$M_{u,lim} = 0.36 f_{ck} b\, x_{u,max}(d - 0.42 x_{u,max}) = 0.138 f_{ck} b d^2$$ $$M_{u,lim} = 0.138 \times 20 \times 250 \times 460^2 = 0.138 \times 20 \times 250 \times 211600$$ $$= 0.138 \times 1.058 \times 10^9 = 146.0 \times 10^6\ \text{N·mm} = 146.0\ \text{kN·m}$$

Since $M_u = 144.6\ \text{kN·m} < M_{u,lim} = 146.0\ \text{kN·m}$, the section can be designed as singly reinforced and will be under-reinforced.

Tension steel — solve $M_u = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right)$.

Using the standard quadratic form:

$$A_{st} = \frac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \frac{4.6\,M_u}{f_{ck} b d^2}}\right] b d$$

Compute the term: $\dfrac{4.6 \times 144.6\times10^6}{20 \times 250 \times 460^2} = \dfrac{665.16\times10^6}{1.058\times10^9} = 0.6288$

$$A_{st} = \frac{0.5 \times 20}{415}\left[1 - \sqrt{1 - 0.6288}\right]\times 250 \times 460$$ $$= 0.024096\left[1 - \sqrt{0.3712}\right]\times 115000 = 0.024096(1 - 0.6093)\times115000$$ $$= 0.024096 \times 0.3907 \times 115000 = 1082.6\ \text{mm}^2$$

Required $A_{st} = 1083\ \text{mm}^2$.

Provide 4–20 mm bars $= 4 \times 314.16 = 1256.6\ \text{mm}^2 > 1083\ \text{mm}^2$. OK.

Minimum steel check: $A_{st,min} = \dfrac{0.85 b d}{f_y} = \dfrac{0.85 \times 250 \times 460}{415} = 235.5\ \text{mm}^2 < 1083$. OK.

Maximum steel check: $A_{st,max} = 0.04 b D = 0.04\times250\times500 = 5000\ \text{mm}^2 > 1256.6$. OK.

(c) Cross-section sketch

   250 mm
  +---------+
  | o   o   |   <- 2-12 mm hanger bars (top)
  |         |
  |         |  500 mm (D)
  |         |
  | o o o o |   <- 4-20 mm main bars (bottom), d = 460 mm
  +---------+
  8 mm stirrups @ nominal spacing

Provide 2–12 mm anchor bars at top and 8 mm two-legged stirrups.

Step 1 — Trial depth

Short span $l_x = 4.0\ \text{m}$. For two-way simply supported slab, span/depth $\approx 28$ (with modification).

$d \approx \dfrac{4000}{28} = 143\ \text{mm}$. Adopt $d = 130\ \text{mm}$, with 20 mm cover and 10 mm bars → $D = 130 + 20 + 5 = 155$, round to $D = 160\ \text{mm}$, $d_x = 160 - 20 - 5 = 135\ \text{mm}$, $d_y = 135 - 10 = 125\ \text{mm}$.

Step 2 — Effective span

$l_x = 4.0 + 0.135 = 4.135\ \text{m}$ (clear span + d), $l_y = 5.0 + 0.125 = 5.125\ \text{m}$.

$l_y/l_x = 5.125/4.135 = 1.24 \approx 1.25$. OK to use given coefficients.

Step 3 — Loads

• Self weight $= 0.16 \times 25 = 4.0\ \text{kN/m}^2$
• Floor finish $= 1.0\ \text{kN/m}^2$
• Live load $= 3.0\ \text{kN/m}^2$
• Total $w = 8.0\ \text{kN/m}^2$; Factored $w_u = 1.5 \times 8.0 = 12.0\ \text{kN/m}^2$

Step 4 — Design moments (per metre width)

$$M_x = \alpha_x w_u l_x^2 = 0.099 \times 12.0 \times 4.135^2 = 0.099 \times 12.0 \times 17.098 = 20.31\ \text{kN·m}$$ $$M_y = \alpha_y w_u l_x^2 = 0.051 \times 12.0 \times 17.098 = 10.46\ \text{kN·m}$$

Step 5 — Check depth

$d_{req} = \sqrt{\dfrac{M_x}{0.138 f_{ck} b}} = \sqrt{\dfrac{20.31\times10^6}{0.138\times20\times1000}} = \sqrt{7359} = 85.8\ \text{mm} < 135\ \text{mm}$. OK.

Step 6 — Reinforcement (short span, $d_x = 135$)

$$A_{st,x} = \frac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \frac{4.6 M_x}{f_{ck} b d_x^2}}\right] b d_x$$

Term $= \dfrac{4.6 \times 20.31\times10^6}{20\times1000\times135^2} = \dfrac{93.43\times10^6}{364.5\times10^6} = 0.2563$

$$A_{st,x} = 0.024096(1-\sqrt{0.7437})\times1000\times135 = 0.024096(1-0.8624)\times135000$$ $$= 0.024096 \times 0.1376 \times 135000 = 447.6\ \text{mm}^2$$

Provide 10 mm @ 175 mm c/c $\left(\dfrac{1000\times78.54}{175}=448.8\ \text{mm}^2\right)$. OK.

Long span, $d_y = 125$

Term $= \dfrac{4.6 \times 10.46\times10^6}{20\times1000\times125^2} = \dfrac{48.12\times10^6}{312.5\times10^6} = 0.1540$

$$A_{st,y} = 0.024096(1-\sqrt{0.846})\times1000\times125 = 0.024096(1-0.9198)\times125000$$ $$= 0.024096\times0.0802\times125000 = 241.6\ \text{mm}^2$$

$A_{st,min} = 0.0012\times1000\times160 = 192\ \text{mm}^2 < 241.6$. Provide 8 mm @ 200 mm c/c $\left(\dfrac{1000\times50.27}{200}=251.3\ \text{mm}^2\right)$. OK.

Summary

Provide torsion reinforcement at corners (corners not held down → not mandatory, but provide nominal corner mesh). Overall slab depth D = 160 mm.

Given: $P_u = 1800\ \text{kN}$, $l = 3.0\ \text{m}$, $f_{ck}=25$, $f_y=500$.

(a) Slenderness check

End condition: held in position at both ends, not restrained against rotation (pinned-pinned) → effective length factor $\approx 1.0$, so $l_{eff} = 1.0 \times 3.0 = 3.0\ \text{m}$.

Assume section $b = D = 350\ \text{mm}$. Slenderness ratio $= \dfrac{l_{eff}}{D} = \dfrac{3000}{350} = 8.57 < 12$.

Column is short. (Design by cl. 39.3 with minimum eccentricity check.)

(b) Section and longitudinal steel

For a short axially loaded column (cl. 39.3):

$$P_u = 0.4 f_{ck} A_c + 0.67 f_y A_{sc}$$

Let $A_g =$ gross area, $A_{sc} = p\% A_g$, $A_c = A_g - A_{sc}$.

Try $A_g = 350 \times 350 = 122500\ \text{mm}^2$.

Solve for $A_{sc}$:

$$1800\times10^3 = 0.4\times25\times(122500 - A_{sc}) + 0.67\times500\times A_{sc}$$ $$1800000 = 10(122500 - A_{sc}) + 335 A_{sc}$$ $$1800000 = 1225000 - 10 A_{sc} + 335 A_{sc}$$ $$1800000 - 1225000 = 325 A_{sc}$$ $$A_{sc} = \frac{575000}{325} = 1769\ \text{mm}^2$$

Steel percentage $= \dfrac{1769}{122500}\times100 = 1.44\%$ (between 0.8% min and 6% max). OK.

Provide 4–25 mm bars $= 4\times490.87 = 1963.5\ \text{mm}^2 > 1769$. OK.

Minimum eccentricity check (cl. 25.4):

$$e_{min} = \frac{l}{500} + \frac{D}{30} = \frac{3000}{500} + \frac{350}{30} = 6 + 11.67 = 17.67\ \text{mm}$$

Limit $0.05 D = 0.05\times350 = 17.5\ \text{mm}$. Since $e_{min} (17.67) > 17.5$, slightly exceeds; the formula $P_u = 0.4f_{ck}A_c+0.67f_yA_{sc}$ is valid only when $e_{min} \le 0.05D$. Increase $D$ to 375 mm: $e_{min} = 6 + 12.5 = 18.5$, $0.05\times375 = 18.75 > 18.5$. OK.

Recompute with $A_g = 375\times375 = 140625\ \text{mm}^2$:

$$1800000 = 10(140625 - A_{sc}) + 335 A_{sc} = 1406250 + 325 A_{sc}$$ $$A_{sc} = \frac{393750}{325} = 1212\ \text{mm}^2$$

$p = 1212/140625\times100 = 0.86\% > 0.8\%$ min. Provide 4–20 mm $= 1256.6\ \text{mm}^2 > 1212$. OK.

Adopt 375 × 375 mm section with 4–20 mm bars.

(c) Lateral ties (cl. 26.5.3.2)

Tie diameter $\ge$ greater of $\dfrac{1}{4}\times$(largest bar) $= \dfrac{20}{4}=5\ \text{mm}$, and $6\ \text{mm}$ → use 8 mm ties.

Spacing $\le$ least of:

• least lateral dimension $= 375\ \text{mm}$
• $16\times$ smallest longitudinal bar $= 16\times20 = 320\ \text{mm}$
• $300\ \text{mm}$

Provide 8 mm ties @ 300 mm c/c.

Final design: 375 × 375 mm column, 4–20 mm longitudinal bars, 8 mm ties @ 300 mm c/c.

Given: Column $400\times400$ mm, service load $P = 1000\ \text{kN}$, SBC $q_s = 200\ \text{kN/m}^2$, $f_{ck}=20$, $f_y=415$.

(a) Plan size

Add 10% for self weight of footing: total service load $= 1.1\times1000 = 1100\ \text{kN}$.

Required area $= \dfrac{1100}{200} = 5.5\ \text{m}^2$. Side $= \sqrt{5.5} = 2.345\ \text{m}$. Provide square footing $2.4\ \text{m}\times2.4\ \text{m}$ (area $= 5.76\ \text{m}^2$).

(b) Factored net upward soil pressure (use column load only, soil pressure from factored load)

$$q_u = \frac{1.5\times1000}{2.4\times2.4} = \frac{1500}{5.76} = 260.4\ \text{kN/m}^2$$

(c) Thickness from shear

Assume effective depth $d$. Try $d = 450\ \text{mm}$.

One-way (flexural) shear — critical section at distance $d$ from column face.

Cantilever projection $= \dfrac{2400 - 400}{2} = 1000\ \text{mm}$.

Distance of critical section from edge $= 1000 - 450 = 550\ \text{mm} = 0.55\ \text{m}$.

Shear force per metre width $V_u = q_u \times 0.55 \times 1.0 = 260.4\times0.55 = 143.2\ \text{kN}$.

Nominal shear stress $\tau_v = \dfrac{V_u}{b d} = \dfrac{143.2\times10^3}{1000\times450} = 0.318\ \text{N/mm}^2$.

Assume $p_t \approx 0.25\%$ → permissible $\tau_c \approx 0.36\ \text{N/mm}^2$ (M20, Table 19). Since $\tau_v = 0.318 < 0.36$. One-way shear OK.

Two-way (punching) shear — critical section at $d/2 = 225\ \text{mm}$ from column face. Perimeter side $= 400 + d = 400 + 450 = 850\ \text{mm}$.

Punching perimeter $b_0 = 4\times850 = 3400\ \text{mm}$.

Punching shear force $V_p = q_u(A_{footing} - A_{crit}) = 260.4\times(2.4^2 - 0.85^2) = 260.4\times(5.76 - 0.7225) = 260.4\times5.0375 = 1311.8\ \text{kN}$.

$\tau_v = \dfrac{V_p}{b_0 d} = \dfrac{1311.8\times10^3}{3400\times450} = 0.857\ \text{N/mm}^2$.

Permissible punching stress $\tau_{c,p} = k_s\times0.25\sqrt{f_{ck}}$, with $k_s = 1$ for square column, $= 0.25\sqrt{20} = 0.25\times4.472 = 1.118\ \text{N/mm}^2$.

Since $\tau_v = 0.857 < 1.118$. Punching shear OK.

Adopt $d = 450\ \text{mm}$, overall depth $D = 450 + 50 + 8 = 508 \approx$ 510 mm.

(d) Flexural reinforcement

Bending moment at column face (cantilever):

$$M_u = q_u \times \frac{(\text{projection})^2}{2}\times B = 260.4\times\frac{1.0^2}{2}\times2.4 = 312.5\ \text{kN·m (total)}$$

Per metre width: $M_u' = 260.4\times\dfrac{1.0^2}{2} = 130.2\ \text{kN·m/m}$.

$$A_{st} = \frac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \frac{4.6 M_u'}{f_{ck} b d^2}}\right]b d$$

Term $= \dfrac{4.6\times130.2\times10^6}{20\times1000\times450^2} = \dfrac{598.9\times10^6}{4.05\times10^9} = 0.1479$

$$A_{st} = 0.024096(1-\sqrt{0.8521})\times1000\times450 = 0.024096(1-0.9231)\times450000$$ $$= 0.024096\times0.0769\times450000 = 833.9\ \text{mm}^2/\text{m}$$

$A_{st,min} = 0.0012\times1000\times510 = 612\ \text{mm}^2 < 833.9$. OK.

Provide 16 mm @ 225 mm c/c $\left(\dfrac{1000\times201.06}{225}=893.6\ \text{mm}^2\right)$ both ways. OK.

Summary: 2.4 m × 2.4 m × 0.51 m footing, 16 mm @ 225 mm c/c each way.

Given: $b_f = 1000\ \text{mm}$, $D_f = 120\ \text{mm}$, $b_w = 250\ \text{mm}$, $d = 520\ \text{mm}$.

Tension steel $A_{st} = 6\times\dfrac{\pi}{4}\times20^2 = 6\times314.16 = 1884.96\ \text{mm}^2$.

Step 1 — Assume NA within flange. Treat as rectangular beam of width $b_f$.

Depth of NA from $C = T$:

$$0.36 f_{ck} b_f x_u = 0.87 f_y A_{st}$$ $$0.36\times20\times1000\times x_u = 0.87\times415\times1884.96$$ $$7200\, x_u = 680545$$ $$x_u = \frac{680545}{7200} = 94.5\ \text{mm}$$

Since $x_u = 94.5\ \text{mm} < D_f = 120\ \text{mm}$, the neutral axis lies within the flange. The section behaves as a rectangular beam of width $b_f = 1000$ mm.

Step 2 — Check $x_{u,max}$ (Fe415): $x_{u,max} = 0.48 d = 0.48\times520 = 249.6\ \text{mm}$.

$x_u = 94.5 < 249.6$ → section is under-reinforced, steel yields. OK.

Step 3 — Moment of resistance

$$M_u = 0.87 f_y A_{st}(d - 0.42 x_u)$$ $$= 0.87\times415\times1884.96\times(520 - 0.42\times94.5)$$ $$= 680545\times(520 - 39.69) = 680545\times480.31$$ $$= 326.87\times10^6\ \text{N·mm}$$

Ultimate moment of resistance $M_u = 326.9\ \text{kN·m}$.

Verification with concrete couple:

$$M_u = 0.36 f_{ck} b_f x_u(d - 0.42 x_u) = 0.36\times20\times1000\times94.5\times480.31$$ $$= 680400\times480.31 = 326.8\times10^6\ \text{N·mm} = 326.8\ \text{kN·m}.$$

Both approaches agree (small rounding). $M_u \approx 327\ \text{kN·m}$.

Step 1 — Nominal shear stress

$$\tau_v = \frac{V_u}{b d} = \frac{250\times10^3}{300\times550} = 1.515\ \text{N/mm}^2$$

Step 2 — Check max shear stress. For M20, $\tau_{c,max} = 2.8\ \text{N/mm}^2$. Since $\tau_v = 1.515 < 2.8$, section size is adequate.

Step 3 — Compare with $\tau_c$. $\tau_c = 0.56\ \text{N/mm}^2 < \tau_v$, so shear reinforcement is required.

Step 4 — Shear to be carried by stirrups

$$V_{us} = (\tau_v - \tau_c)b d = (1.515 - 0.56)\times300\times550 = 0.955\times165000 = 157575\ \text{N} = 157.6\ \text{kN}$$

Step 5 — Stirrup spacing. Use 8 mm two-legged stirrups, $A_{sv} = 2\times50.27 = 100.53\ \text{mm}^2$.

$$s_v = \frac{0.87 f_y A_{sv} d}{V_{us}} = \frac{0.87\times415\times100.53\times550}{157575\times1}$$ $$= \frac{0.87\times415\times100.53\times550}{157575} = \frac{19{,}962{,}000}{157575} = 126.7\ \text{mm}$$

Step 6 — Maximum spacing limits. Least of $0.75 d = 412.5\ \text{mm}$ and $300\ \text{mm}$. Both exceed 126.7 mm.

Provide 8 mm two-legged stirrups @ 125 mm c/c near the support.

Development length formula (IS 456 cl. 26.2.1):

$$L_d = \frac{\phi\, \sigma_s}{4\, \tau_{bd}}$$

where $\sigma_s = 0.87 f_y$ (stress at yield in design).

Step 1 — Design bond stress for deformed bars.

$$\tau_{bd} = 1.2\times1.6 = 1.92\ \text{N/mm}^2$$

(60% increase for deformed bars).

Step 2 — Steel stress.

$$\sigma_s = 0.87\times415 = 361.05\ \text{N/mm}^2$$

Step 3 — Development length (tension).

$$L_d = \frac{20\times361.05}{4\times1.92} = \frac{7221}{7.68} = 940.2\ \text{mm}$$

$L_d \approx 940\ \text{mm} = 47\phi$ (tension).

Step 4 — Compression. For bars in compression, $\tau_{bd}$ is increased by 25%:

$$\tau_{bd,comp} = 1.92\times1.25 = 2.40\ \text{N/mm}^2$$ $$L_d = \frac{20\times361.05}{4\times2.40} = \frac{7221}{9.6} = 752.2\ \text{mm} \approx 752\ \text{mm} = 37.6\phi$$

Compression development length $\approx 752\ \text{mm}$, i.e. about 80% of the tension value.

Why doubly reinforced beams?

When the factored moment exceeds the limiting moment of resistance $M_{u,lim}$ of a singly reinforced section (and the section dimensions cannot be increased due to architectural/headroom constraints), compression steel is added in addition to extra tension steel. The compression steel resists the excess moment $\Delta M = M_u - M_{u,lim}$. Doubly reinforced sections also improve ductility, reduce long-term deflection and creep, and help anchor stirrups.

Strain:        Stress block:
  0.0035          0.446 fck
   |--->          |==== Cc (concrete)
   |  d'  --> Asc  --> Csc (comp. steel)
  -+-- NA
   |
   |  --> Ast --> T = 0.87 fy Ast (tension)

Additional tension steel for $\Delta M = 40\ \text{kN·m}$

The excess moment is resisted by the couple of compression steel and the corresponding additional tension steel acting at lever arm $(d - d')$.

$$\Delta M = 0.87 f_y A_{st2}(d - d')$$ $$40\times10^6 = 0.87\times415\times A_{st2}\times(450 - 50)$$ $$40\times10^6 = 0.87\times415\times400\times A_{st2} = 144420\, A_{st2}$$ $$A_{st2} = \frac{40\times10^6}{144420} = 276.9\ \text{mm}^2$$

Additional tension steel $A_{st2} = 277\ \text{mm}^2$.

This is added to $A_{st,lim}$ (the steel for $M_{u,lim}$) to get the total tension steel. The compression steel $A_{sc}$ is found from $0.87 f_y A_{st2} = (f_{sc} - 0.446 f_{ck})A_{sc}$, where $f_{sc}$ depends on the strain at the compression-steel level.

Step 1 — Effective depth from deflection (span/depth) criterion.

$$\left(\frac{l}{d}\right)_{allowed} = 20\times1.4 = 28$$ $$d_{req} = \frac{3200}{28} = 114.3\ \text{mm}$$

Adopt $d = 120\ \text{mm}$, with 20 mm cover + 5 mm (half bar) → $D = 145\ \text{mm}$.

Step 2 — Design moment per metre width.

$$M_u = \frac{w_u l^2}{8} = \frac{10\times3.2^2}{8} = \frac{10\times10.24}{8} = 12.8\ \text{kN·m/m}$$

Step 3 — Check depth for moment.

$$d_{req} = \sqrt{\frac{M_u}{0.138 f_{ck} b}} = \sqrt{\frac{12.8\times10^6}{0.138\times20\times1000}} = \sqrt{4638} = 68.1\ \text{mm} < 120\ \text{mm}$$

Deflection governs. OK.

Step 4 — Main reinforcement ($d = 120$ mm). Term $= \dfrac{4.6\times12.8\times10^6}{20\times1000\times120^2} = \dfrac{58.88\times10^6}{288\times10^6} = 0.2044$

$$A_{st} = 0.024096(1-\sqrt{0.7956})\times1000\times120 = 0.024096(1-0.8920)\times120000$$ $$= 0.024096\times0.1080\times120000 = 312.3\ \text{mm}^2/\text{m}$$

$A_{st,min} = 0.0012\times1000\times145 = 174\ \text{mm}^2 < 312.3$. OK.

Provide 10 mm @ 250 mm c/c $\left(\dfrac{1000\times78.54}{250}=314.2\ \text{mm}^2\right)$.

Distribution steel $= 0.0012\times1000\times145 = 174\ \text{mm}^2$ → provide 8 mm @ 280 mm c/c $\left(\dfrac{1000\times50.27}{280}=179.5\ \text{mm}^2\right)$.

Summary: $D = 145\ \text{mm}$, main 10 mm @ 250 c/c, distribution 8 mm @ 280 c/c.

Step 1 — Load on going (per m² of plan/horizontal area).

Waist slab is inclined. Slope length per tread $= \sqrt{R^2 + T^2} = \sqrt{150^2 + 250^2} = \sqrt{22500 + 62500} = \sqrt{85000} = 291.5\ \text{mm}$.

Waist slab self weight (on plan) $= \gamma\times t\times\dfrac{\sqrt{R^2+T^2}}{T} = 25\times0.18\times\dfrac{291.5}{250} = 25\times0.18\times1.166 = 5.247\ \text{kN/m}^2$.

Step (triangular) self weight $= \gamma\times\dfrac{R}{2} = 25\times\dfrac{0.15}{2} = 1.875\ \text{kN/m}^2$.

Finish $= 1.0\ \text{kN/m}^2$; Live load $= 3.0\ \text{kN/m}^2$.

Total characteristic load $= 5.247 + 1.875 + 1.0 + 3.0 = 11.12\ \text{kN/m}^2$.

Factored load $w_u = 1.5\times11.12 = 16.68\ \text{kN/m}^2$.

Per metre width of stair, $w_u = 16.68\ \text{kN/m}$.

Step 2 — Maximum design bending moment.

Treating the flight as a simply supported slab of effective span $3.0\ \text{m}$:

$$M_u = \frac{w_u l^2}{8} = \frac{16.68\times3.0^2}{8} = \frac{16.68\times9}{8} = \frac{150.12}{8} = 18.77\ \text{kN·m/m}$$

Total factored load $= 16.68\ \text{kN/m}^2$; maximum design moment $M_u = 18.8\ \text{kN·m}$ per metre width.

(a) Working Stress Method (WSM) vs Limit State Method (LSM)

(b) Characteristic strength and partial safety factors

• Characteristic strength: the value of material strength below which not more than 5% of test results are expected to fall (mean − 1.65 × standard deviation).
• Partial safety factor for materials (cl. 36.4.2): concrete $\gamma_m = 1.5$; steel $\gamma_m = 1.15$.
• Partial safety factors for loads (Table 18) at ultimate limit state:
  • DL + LL: $1.5\,(DL) + 1.5\,(LL)$
  • DL + WL: $1.5(DL + WL)$ or $0.9\,DL + 1.5\,WL$
  • DL + LL + WL: $1.2(DL + LL + WL)$
• For serviceability limit state, load factors are generally $1.0$.

(c) Four detailing requirements for an RC beam

1. Minimum tension steel $A_{st,min} = \dfrac{0.85 b d}{f_y}$ and maximum steel $\le 0.04 bD$.
2. Minimum clear cover (nominal) of 20–25 mm for beams for durability and fire resistance.
3. Development length / anchorage: bars must extend $L_d$ beyond the point of maximum stress; at simple supports $\dfrac{M_1}{V} + L_0 \ge L_d$.
4. Shear stirrups: maximum spacing $\le 0.75 d$ (or 300 mm); minimum shear reinforcement $\dfrac{A_{sv}}{b s_v} \ge \dfrac{0.4}{0.87 f_y}$. Also maintain minimum side-face reinforcement for deep beams ($D > 750$ mm).