BE Civil Engineering (IOE, TU) Design of Steel and Timber Structures (IOE, CE 655) Question Paper 2076 Nepal

Given: ISA $90\times90\times8$, $A_g = 1379\ \text{mm}^2$, leg thickness $t = 8$ mm, bolts $d=20$ mm, hole $d_h = 22$ mm, $n = 4$ bolts, pitch $p = 50$ mm, $f_y = 250$, $f_u = 410\ \text{N/mm}^2$.

Partial safety factors: $\gamma_{m0} = 1.10$, $\gamma_{m1} = 1.25$.

(a) Yielding of gross section

$$T_{dg} = \frac{A_g f_y}{\gamma_{m0}} = \frac{1379 \times 250}{1.10} = 313{,}409\ \text{N} = \mathbf{313.4\ kN}$$

(b) Rupture of net section (angle connected by one leg) Net area in connected leg: the leg with the bolt line. Connected leg width portion area $A_{nc} = (\text{leg width} - d_h - \tfrac{t}{2})\,t$. For an equal angle, connected-leg area $= (90 - \tfrac{8}{2})t = 86\times8 = 688\ \text{mm}^2$ gross; deduct one hole: $A_{nc} = (86 - 22)\times 8 = 64\times8 = 512\ \text{mm}^2$. Outstanding leg gross area $A_{go} = (90 - \tfrac{8}{2})\times 8 = 86\times 8 = 688\ \text{mm}^2$.

Using IS 800 cl. 6.3.3:

$$T_{dn} = \frac{0.9\, A_{nc} f_u}{\gamma_{m1}} + \frac{\beta A_{go} f_y}{\gamma_{m0}}$$

where $\beta = 1.4 - 0.076\left(\frac{w}{t}\right)\left(\frac{f_y}{f_u}\right)\left(\frac{b_s}{L_c}\right)$.

Here $w = 90$ mm (leg width), $t = 8$, $b_s = w + w_1 - t$ with edge distance $w_1$. Take edge distance $\approx 35$ mm so $b_s = 90 + 35 - 8 = 117$ mm. Connection length $L_c = (n-1)p = 3\times 50 = 150$ mm.

$$\beta = 1.4 - 0.076\times\frac{90}{8}\times\frac{250}{410}\times\frac{117}{150} = 1.4 - 0.076\times11.25\times0.6098\times0.78 = 1.4 - 0.407 = 0.993$$

Limits: $0.7 \le \beta \le 1.0\cdot\frac{f_u\gamma_{m0}}{f_y\gamma_{m1}} = 1.44$, and $\beta \ge 0.7$. So $\beta = 0.993$.

$$T_{dn} = \frac{0.9\times512\times410}{1.25} + \frac{0.993\times688\times250}{1.10}$$ $$= \frac{188{,}928}{1.25}\times1 + \frac{170{,}796}{1.10} = 151{,}142 + 155{,}269 = 306{,}411\ \text{N} = \mathbf{306.4\ kN}$$

(c) Block shear (IS 800 cl. 6.4.1) Gross shear area along bolt line $A_{vg} = L_v \cdot t$, with $L_v =$ end distance $+ (n-1)p = 35 + 150 = 185$ mm $\Rightarrow A_{vg} = 185\times8 = 1480\ \text{mm}^2$. Net shear area $A_{vn} = (185 - 3.5\,d_h)\,t = (185 - 3.5\times22)\times8 = (185-77)\times8 = 108\times8 = 864\ \text{mm}^2$. Tension area perpendicular: edge distance $e_2 = 35$ mm. $A_{tg} = 35\times8 = 280\ \text{mm}^2$; $A_{tn} = (35 - 0.5\times22)\times8 = (35-11)\times8 = 192\ \text{mm}^2$.

$$T_{db1} = \frac{A_{vg} f_y}{\sqrt3\,\gamma_{m0}} + \frac{0.9 A_{tn} f_u}{\gamma_{m1}} = \frac{1480\times250}{\sqrt3\times1.10} + \frac{0.9\times192\times410}{1.25}$$ $$= \frac{370{,}000}{1.905} + \frac{70{,}848}{1.25} = 194{,}220 + 56{,}678 = 250{,}898\ \text{N} = 250.9\ \text{kN}$$ $$T_{db2} = \frac{0.9 A_{vn} f_u}{\sqrt3\,\gamma_{m1}} + \frac{A_{tg} f_y}{\gamma_{m0}} = \frac{0.9\times864\times410}{\sqrt3\times1.25} + \frac{280\times250}{1.10}$$ $$= \frac{318{,}816}{2.165} + \frac{70{,}000}{1.10} = 147{,}259 + 63{,}636 = 210{,}895\ \text{N} = 210.9\ \text{kN}$$

Block shear $T_{db} = \min(250.9, 210.9) = \mathbf{210.9\ kN}$.

Governing design strength

$$T_d = \min(313.4,\ 306.4,\ 210.9) = \boxed{\mathbf{210.9\ kN}\ (\text{block shear governs})}$$

Given: $A = 5626\ \text{mm}^2$, $r_{min} = r_{yy} = 28.4$ mm, $L = 4000$ mm, $K=1.0$, $f_y=250$, $\gamma_{m0}=1.10$, $\alpha = 0.49$ (curve c), $E = 2\times10^5\ \text{N/mm}^2$.

Step 1 — Effective slenderness (about weaker yy axis):

$$\lambda = \frac{KL}{r_{min}} = \frac{1.0\times4000}{28.4} = 140.8$$

Step 2 — Euler buckling stress:

$$f_{cc} = \frac{\pi^2 E}{\lambda^2} = \frac{\pi^2 \times 2\times10^5}{140.8^2} = \frac{1{,}973{,}921}{19{,}825} = 99.57\ \text{N/mm}^2$$

Step 3 — Non-dimensional slenderness:

$$\bar\lambda = \sqrt{\frac{f_y}{f_{cc}}} = \sqrt{\frac{250}{99.57}} = \sqrt{2.511} = 1.585$$

Step 4 — Factor $\phi$:

$$\phi = 0.5\left[1 + \alpha(\bar\lambda - 0.2) + \bar\lambda^2\right] = 0.5\left[1 + 0.49(1.585-0.2) + 1.585^2\right]$$ $$= 0.5[1 + 0.49\times1.385 + 2.512] = 0.5[1 + 0.6787 + 2.512] = 0.5\times4.191 = 2.095$$

Step 5 — Stress reduction factor $\chi$:

$$\chi = \frac{1}{\phi + \sqrt{\phi^2 - \bar\lambda^2}} = \frac{1}{2.095 + \sqrt{2.095^2 - 2.512}} = \frac{1}{2.095 + \sqrt{4.389 - 2.512}}$$ $$= \frac{1}{2.095 + \sqrt{1.877}} = \frac{1}{2.095 + 1.370} = \frac{1}{3.465} = 0.2886$$

Step 6 — Design compressive stress:

$$f_{cd} = \frac{\chi f_y}{\gamma_{m0}} = \frac{0.2886\times250}{1.10} = \frac{72.15}{1.10} = 65.6\ \text{N/mm}^2$$

Check $f_{cd} \le f_y/\gamma_{m0} = 227.3\ \text{N/mm}^2$ — OK.

Step 7 — Design compressive strength:

$$P_d = A f_{cd} = 5626 \times 65.6 = 369{,}066\ \text{N} = \boxed{\mathbf{369.1\ kN}}$$

The column carries about $369\ \text{kN}$; the high slenderness ($\lambda \approx 141$) about the minor axis drastically reduces the capacity from the squash load $A f_y/\gamma_{m0} = 1278\ \text{kN}$.

Given: $L=6$ m, $w=45\ \text{kN/m}$, $Z_{pz}=1533.36\times10^3\ \text{mm}^3$, $Z_{ez}=1350.7\times10^3\ \text{mm}^3$, $h=450$, $t_w=9.4$, $f_y=250$, $\gamma_{m0}=1.10$.

Step 1 — Design actions:

$$M_u = \frac{wL^2}{8} = \frac{45\times6^2}{8} = \frac{1620}{8} = 202.5\ \text{kN·m}$$ $$V_u = \frac{wL}{2} = \frac{45\times6}{2} = 135\ \text{kN}$$

Step 2 — Bending (plastic, laterally supported), IS 800 cl. 8.2.1.2:

$$M_d = \frac{\beta_b Z_{pz} f_y}{\gamma_{m0}},\quad \beta_b = 1.0\ (\text{plastic section})$$ $$M_d = \frac{1.0\times1533.36\times10^3\times250}{1.10} = \frac{383.34\times10^6}{1.10} = 348.49\times10^6\ \text{N·mm} = 348.5\ \text{kN·m}$$

Check upper limit to avoid plastic deformation under service loads:

$$1.2\,Z_{ez}\,f_y/\gamma_{m0} = \frac{1.2\times1350.7\times10^3\times250}{1.10} = \frac{405.21\times10^6}{1.10} = 368.4\times10^6\ \text{N·mm} = 368.4\ \text{kN·m}$$

Since $348.5 < 368.4$, governing $M_d = 348.5\ \text{kN·m}$. $M_d = 348.5\ \text{kN·m} > M_u = 202.5\ \text{kN·m}$ ✓ Safe in bending.

Step 3 — Shear capacity, IS 800 cl. 8.4: Shear area $A_v = h\,t_w = 450\times9.4 = 4230\ \text{mm}^2$.

$$V_d = \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{4230\times250}{\sqrt3\times1.10} = \frac{1{,}057{,}500}{1.905} = 555{,}118\ \text{N} = 555.1\ \text{kN}$$

$V_d = 555.1\ \text{kN} > V_u = 135\ \text{kN}$ ✓ Safe in shear.

Step 4 — Low-shear / high-shear check: $0.6\,V_d = 0.6\times555.1 = 333.1\ \text{kN}$. Since $V_u = 135 < 333.1$, it is a low shear case; no moment reduction needed.

Conclusion: With $M_d = 348.5\ \text{kN·m} > 202.5$ and $V_d = 555.1\ \text{kN} > 135$, the ISMB 450 is adequate (safe in both bending and shear).

Given: $M_u = 4500\ \text{kN·m}$, $V_u = 900$ kN, $d = 1400$ mm, $f_y = 250$, $\gamma_{m0}=1.10$.

(a) Web thickness Serviceability/minimum unstiffened limit $d/t_w \le 200$:

$$t_{w,min} = \frac{d}{200} = \frac{1400}{200} = 7\ \text{mm}$$

Check against shear (simple post-elastic, $V_d \ge V_u$):

$$V_d = \frac{d\,t_w\,f_y}{\sqrt3\,\gamma_{m0}} \ge V_u \Rightarrow t_w \ge \frac{\sqrt3\,\gamma_{m0}\,V_u}{d\,f_y} = \frac{1.732\times1.10\times900\times10^3}{1400\times250}$$ $$= \frac{1{,}714{,}680}{350{,}000} = 4.9\ \text{mm}$$

Governing $t_w = 8$ mm (adopt, $> 7$ and $>4.9$). Check $d/t_w = 1400/8 = 175 \le 200$ ✓. Adopt web $1400 \times 8$ mm.

(b) Flange area Lever arm $\approx d = 1400$ mm. Required flange force:

$$F_f = \frac{M_u}{d} = \frac{4500\times10^6}{1400} = 3.214\times10^6\ \text{N}$$

Flange area (allowing flange stress $f_y/\gamma_{m0}$):

$$A_f = \frac{F_f\,\gamma_{m0}}{f_y} = \frac{3.214\times10^6\times1.10}{250} = \frac{3.536\times10^6}{250} = 14{,}143\ \text{mm}^2$$

Try flange $400$ mm wide: $t_f = 14143/400 = 35.4$ mm → adopt flange $400 \times 40$ mm ($A_f = 16{,}000\ \text{mm}^2$). Outstand check: $b_{outstand} = (400-8)/2 = 196$ mm; $b/t_f = 196/40 = 4.9 < 8.4\varepsilon\,(=8.4)$ → plastic flange, OK.

(c) Moment capacity (elastic) Overall depth $D = d + 2t_f = 1400 + 80 = 1480$ mm. Moment of inertia (web + flanges):

• Web: $I_{web} = \dfrac{8\times1400^3}{12} = \dfrac{8\times2.744\times10^9}{12} = 1.829\times10^9\ \text{mm}^4$
• Flanges: $I_f = 2\left[\dfrac{400\times40^3}{12} + 16000\times720^2\right]$ where centroid distance $= (1400+40)/2 = 720$ mm. $= 2[2.133\times10^6 + 16000\times518{,}400] = 2[2.133\times10^6 + 8.294\times10^9] = 2\times8.296\times10^9 = 1.659\times10^{10}\ \text{mm}^4$
• $I_{total} = 1.829\times10^9 + 1.659\times10^{10} = 1.842\times10^{10}\ \text{mm}^4$ Elastic modulus:

$$Z_e = \frac{I}{D/2} = \frac{1.842\times10^{10}}{740} = 2.489\times10^7\ \text{mm}^3$$

Moment capacity:

$$M_d = \frac{Z_e f_y}{\gamma_{m0}} = \frac{2.489\times10^7\times250}{1.10} = \frac{6.223\times10^9}{1.10} = 5.66\times10^9\ \text{N·mm} = \mathbf{5657\ kN·m}$$

$M_d = 5657\ \text{kN·m} > M_u = 4500\ \text{kN·m}$ ✓ Section adequate.

Given: $P_u = 1600$ kN, bearing strength of concrete $\sigma_{br} = 9\ \text{N/mm}^2$, plate $f_y = 250$, $\gamma_{m0}=1.10$, column $\approx 300\times300$ mm.

(a) Required area of base plate

$$A_{req} = \frac{P_u}{\sigma_{br}} = \frac{1600\times10^3}{9} = 177{,}778\ \text{mm}^2$$

Side of square plate $L = \sqrt{A_{req}} = \sqrt{177778} = 421.6$ mm → adopt $L = 450$ mm (square $450\times450$). Provided area $A_p = 450\times450 = 202{,}500\ \text{mm}^2$.

(b) Actual bearing pressure & projection

$$w = \frac{P_u}{A_p} = \frac{1600\times10^3}{202500} = 7.90\ \text{N/mm}^2 \ (< 9\ \text{OK})$$

Projection of plate beyond column on each side:

$$a = b = \frac{L - 300}{2} = \frac{450 - 300}{2} = 75\ \text{mm}$$

(c) Thickness of base plate (IS 800 cl. 7.4.3.1)

$$t_s = \sqrt{\frac{2.5\,w\,(a^2 - 0.3\,b^2)\,\gamma_{m0}}{f_y}}$$

With $a = b = 75$ mm (greater and smaller projections equal):

$$a^2 - 0.3b^2 = 75^2 - 0.3\times75^2 = 5625 - 1687.5 = 3937.5\ \text{mm}^2$$ $$t_s = \sqrt{\frac{2.5\times7.90\times3937.5\times1.10}{250}} = \sqrt{\frac{85{,}547}{250}} = \sqrt{342.2} = 18.5\ \text{mm}$$

Adopt base plate thickness $t_s = 20$ mm (> column flange thickness, OK).

Summary: Slab base plate $450\times450\times20$ mm on M20 concrete; bearing pressure $7.9\ \text{N/mm}^2 < 9\ \text{N/mm}^2$. Nominal anchor bolts (e.g., 4 nos. 20 mm) provided for stability since the column is axially loaded with no net uplift.

Given: weld size $s = 6$ mm, two welds each $L = 250$ mm, load $P = 300$ kN, $f_u = 410$, $\gamma_{mw} = 1.25$.

Step 1 — Throat thickness:

$$t_t = 0.7\,s = 0.7\times6 = 4.2\ \text{mm}$$

Step 2 — Effective length: Two welds: $L_{eff} = 2\times250 = 500$ mm (deducting end craters is sometimes done; here use full effective length).

Step 3 — Design strength of weld per unit length:

$$f_{wd} = \frac{f_u}{\sqrt3\,\gamma_{mw}} = \frac{410}{\sqrt3\times1.25} = \frac{410}{2.165} = 189.4\ \text{N/mm}^2$$

Step 4 — Weld capacity:

$$P_d = f_{wd}\times t_t\times L_{eff} = 189.4\times4.2\times500 = 397{,}740\ \text{N} = 397.7\ \text{kN}$$

Step 5 — Check: $P_d = 397.7\ \text{kN} > P = 300\ \text{kN}$ ✓

The $6$ mm fillet weld over $2\times250$ mm is adequate (capacity $397.7\ \text{kN} > 300\ \text{kN}$, utilization $= 300/397.7 = 75\%$).

Given: $f_{ub}=400$, $A_{nb}=245\ \text{mm}^2$, $A_{sb}=314\ \text{mm}^2$, $\gamma_{mb}=1.25$, one shear plane through threads $\Rightarrow n_n=1,\ n_s=0$.

Step 1 — Nominal shear capacity (IS 800 cl. 10.3.3):

$$V_{nsb} = \frac{f_{ub}}{\sqrt3}\left(n_n A_{nb} + n_s A_{sb}\right) = \frac{400}{\sqrt3}\left(1\times245 + 0\right) = \frac{400}{1.732}\times245 = 230.9\times245 = 56{,}580\ \text{N}$$

Step 2 — Design shear strength:

$$V_{dsb} = \frac{V_{nsb}}{\gamma_{mb}} = \frac{56{,}580}{1.25} = 45{,}264\ \text{N} = \mathbf{45.3\ kN}$$

Step 3 — Bearing strength (IS 800 cl. 10.3.4):

$$V_{dpb} = \frac{2.5\,k_b\,d\,t\,f_u}{\gamma_{mb}} = \frac{2.5\times0.49\times20\times10\times410}{1.25}$$ $$= \frac{2.5\times0.49\times20\times10\times410}{1.25} = \frac{100{,}450}{1.25} = 80{,}360\ \text{N} = \mathbf{80.4\ kN}$$

Result: Bolt value governed by shear $= \min(45.3,\ 80.4) = \mathbf{45.3\ kN}$ per bolt.

Given: span $= 12$ m, spacing $= 4$ m c/c, DL $= 0.40\ \text{kN/m}^2$, LL $= 0.75\ \text{kN/m}^2$, plan area per truss $= 12\times4 = 48\ \text{m}^2$.

Step 1 — Service loads on one truss:

$$\text{Dead load} = 0.40\times48 = 19.2\ \text{kN}$$ $$\text{Live load} = 0.75\times48 = 36.0\ \text{kN}$$

Step 2 — Factored total load (load factor $1.5$ for DL+LL):

$$W_u = 1.5\,(DL + LL) = 1.5\,(19.2 + 36.0) = 1.5\times55.2 = 82.8\ \text{kN}$$

Step 3 — Panel point (nodal) load: Distributing over $6$ equal nodal loads:

$$P = \frac{W_u}{6} = \frac{82.8}{6} = 13.8\ \text{kN}$$

(End nodes carry half panel each, but for uniform-panel modelling the interior nodal load $= \mathbf{13.8\ kN}$, end nodes $\approx 6.9\ \text{kN}$.)

Results:

• Total factored gravity load on truss $= \mathbf{82.8\ kN}$.
• Factored panel point load (interior) $= \mathbf{13.8\ kN}$.

These nodal loads are then applied at the top-chord joints to find member forces by the method of joints/sections.

Given: $b=100$ mm, $d=200$ mm, $L=3.0$ m $=3000$ mm, $f_b=10\ \text{N/mm}^2$, $\tau_{perm}=0.6\ \text{N/mm}^2$.

Step 1 — Section modulus:

$$Z = \frac{b d^2}{6} = \frac{100\times200^2}{6} = \frac{100\times40{,}000}{6} = \frac{4{,}000{,}000}{6} = 666{,}667\ \text{mm}^3$$

Step 2 — Moment of resistance:

$$M = f_b \cdot Z = 10\times666{,}667 = 6{,}666{,}670\ \text{N·mm} = 6.667\ \text{kN·m}$$

Step 3 — Safe UDL from bending ($M = wL^2/8$):

$$w = \frac{8M}{L^2} = \frac{8\times6.667\times10^6}{3000^2} = \frac{53.33\times10^6}{9{,}000{,}000} = 5.93\ \text{N/mm} = \mathbf{5.93\ kN/m}$$

Step 4 — Shear check at this load: Maximum shear $V = wL/2 = 5.93\times3000/2 = 8895\ \text{N} = 8.9\ \text{kN}$. Maximum horizontal shear stress for rectangular section:

$$\tau_{max} = \frac{3V}{2bd} = \frac{3\times8895}{2\times100\times200} = \frac{26{,}685}{40{,}000} = 0.667\ \text{N/mm}^2$$

Since $\tau_{max}=0.667 > \tau_{perm}=0.6\ \text{N/mm}^2$, shear governs.

Step 5 — Safe UDL from shear:

$$V_{safe} = \frac{2}{3}\,\tau_{perm}\,b\,d = \frac{2}{3}\times0.6\times100\times200 = 8000\ \text{N} = 8.0\ \text{kN}$$ $$w_{shear} = \frac{2V_{safe}}{L} = \frac{2\times8000}{3000} = 5.33\ \text{N/mm} = 5.33\ \text{kN/m}$$

Conclusion: Safe UDL from bending $= 5.93\ \text{kN/m}$, from shear $= 5.33\ \text{kN/m}$. Shear governs; safe UDL $= 5.33\ \text{kN/m}$.

Lateral-torsional buckling (LTB): When a beam bent about its major (strong) axis is not laterally restrained along the compression flange, at a critical load the compression flange tends to buckle sideways while the tension flange stays in place. This causes the cross-section to simultaneously deflect laterally and twist — a combined lateral deflection plus torsion known as lateral-torsional buckling. It is a stability failure that occurs before the full plastic moment is reached.

Factors affecting LTB:

1. Unsupported length $L_{LT}$ — longer unbraced length lowers $M_{cr}$ (most significant factor).
2. Cross-sectional shape — torsional rigidity $GJ$ and warping rigidity $EI_w$; deep narrow sections (large $h/b_f$) are more prone.
3. Lateral bending stiffness $EI_y$ (minor-axis stiffness).
4. End/support conditions and restraint to twist and warping.
5. Type and position of loading (load applied above shear centre is destabilizing).
6. Moment gradient along the span (uniform moment is the most severe case).
7. Material yield strength $f_y$ and residual stresses/imperfections.

IS 800:2007 treatment (cl. 8.2.2): For a laterally unsupported beam the design bending strength is

$$M_d = \beta_b\, Z_p\, f_{bd}$$

where $f_{bd} = \dfrac{\chi_{LT}\, f_y}{\gamma_{m0}}$ is the design bending compressive stress and $\chi_{LT}$ is the bending stress reduction factor:

$$\chi_{LT} = \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \bar\lambda_{LT}^2}} \le 1.0$$ $$\phi_{LT} = 0.5\left[1 + \alpha_{LT}(\bar\lambda_{LT} - 0.2) + \bar\lambda_{LT}^2\right]$$ $$\bar\lambda_{LT} = \sqrt{\frac{\beta_b Z_p f_y}{M_{cr}}}$$

Terms: $\beta_b = 1.0$ for plastic/compact, $= Z_e/Z_p$ for semi-compact; $Z_p$ = plastic modulus; $\alpha_{LT}$ = imperfection factor ($0.21$ rolled, $0.49$ welded); $\bar\lambda_{LT}$ = non-dimensional LTB slenderness; $M_{cr}$ = elastic critical moment depending on $EI_y$, $GJ$, $EI_w$ and effective length $L_{LT}$. Shorter unbraced lengths give low $\bar\lambda_{LT}$, $\chi_{LT}\to1$ (no LTB reduction); long lengths give large $\bar\lambda_{LT}$ and small $\chi_{LT}$, sharply reducing capacity.

Section classification (IS 800:2007, cl. 3.7): Sections are classified by the susceptibility of their compression elements (flange outstand, web) to local buckling before reaching a target stress/strain state. Classification is based on the width-to-thickness ($b/t$, $d/t_w$) ratio of the elements, compared against limiting values that depend on $\varepsilon = \sqrt{250/f_y}$.

Limiting $b/t$ examples (outstanding flange of rolled I-section): plastic $\le 9.4\varepsilon$, compact $\le 10.5\varepsilon$, semi-compact $\le 15.7\varepsilon$; web in bending: plastic $\le 84\varepsilon$, etc. (with $\varepsilon=1$ for Fe 410).

Importance:

1. It determines whether the full plastic moment $M_p$ or only the elastic moment $M_y$ can be mobilized — directly fixing the design bending strength $M_d$.
2. Plastic/compact sections permit plastic analysis and moment redistribution; slender sections cannot, and need reduced/effective-section design.
3. It guards against premature local buckling of thin plate elements.

Role of $b/t$: A small $b/t$ (thick, stocky elements) resists local buckling, allowing the section to yield fully and form a plastic hinge (Class 1/2). As $b/t$ increases, local buckling occurs earlier — first limiting capacity to the elastic moment (Class 3), then below yield (Class 4). Hence increasing $b/t$ progressively lowers the usable moment capacity, which is why classification precedes any member strength calculation in IS 800.