BE Civil Engineering (IOE, TU) Design of Steel and Timber Structures (IOE, CE 655) Question Paper 2080 Nepal

Step 1 — Bolt shear capacity (single shear, threads in shear plane)

Net tensile stress area of M20 bolt: $A_{nb} = 245\,\text{mm}^2$. For class 4.6, $f_{ub} = 400\,\text{MPa}$.

$$V_{dsb} = \frac{f_{ub}}{\sqrt{3}\,\gamma_{mb}}\,(n_n A_{nb}) = \frac{400}{\sqrt{3}\times1.25}\times(1\times245) = \frac{400}{2.165}\times245\,\text{N}$$ $$V_{dsb} = 184.75 \times 245 = 45{,}264\,\text{N} = 45.26\,\text{kN}$$

Step 2 — Bolt bearing capacity (on $12\,\text{mm}$ plate, $f_u = 410\,\text{MPa}$)

$$k_b = \min\left(\frac{e}{3d_0},\;\frac{p}{3d_0}-0.25,\;\frac{f_{ub}}{f_u},\;1.0\right)$$ $$\frac{e}{3d_0}=\frac{33}{3\times22}=0.500,\quad \frac{p}{3d_0}-0.25=\frac{50}{66}-0.25=0.508,\quad \frac{f_{ub}}{f_u}=\frac{400}{410}=0.976$$

So $k_b = 0.500$.

$$V_{dpb}=\frac{2.5\,k_b\,d\,t\,f_u}{\gamma_{mb}}=\frac{2.5\times0.5\times20\times12\times410}{1.25}\,\text{N}=\frac{1{,}230{,}000}{1.25}=98{,}400\,\text{N}=98.4\,\text{kN}$$

Step 3 — Governing bolt value = $\min(45.26,\,98.4)=45.26\,\text{kN}$ (shear governs).

Step 4 — Number of bolts

$$n=\frac{P_u}{V_{db}}=\frac{220}{45.26}=4.86 \Rightarrow \textbf{5 bolts}$$

Step 5 — Check net section tension capacity of plate (single line, 5 bolts; net width = gross − one hole)

$$A_n=(b-d_0)\,t=(160-22)\times12=1656\,\text{mm}^2$$ $$T_{dn}=\frac{0.9\,A_n\,f_u}{\gamma_{m1}}=\frac{0.9\times1656\times410}{1.25}\,\text{N}=\frac{611{,}064}{1.25}=488{,}851\,\text{N}=488.9\,\text{kN}$$

Gross yielding: $T_{dg}=\dfrac{A_g f_y}{\gamma_{m0}}=\dfrac{(160\times12)\times250}{1.1}=436{,}364\,\text{N}=436.4\,\text{kN}$.

Plate design strength $=\min(488.9,\,436.4)=436.4\,\text{kN} > 220\,\text{kN}$. Plate is safe.

Conclusion: Provide 5 nos. M20 (4.6) bolts at pitch $50\,\text{mm}$, edge distance $33\,\text{mm}$ in a single line. Connection capacity $=5\times45.26=226.3\,\text{kN} > 220\,\text{kN}$. Joint is safe.

  e=33  p=50  p=50  p=50  p=50  e=33
 |---|----|----|----|----|----|
  (o)  (o)  (o)  (o)  (o)        <- 5 bolts, single line
 ==== plate 160 x 12 (Fe410) ====

(a) Gross-section yielding

$$T_{dg}=\frac{A_g f_y}{\gamma_{m0}}=\frac{1336\times250}{1.10}=303{,}636\,\text{N}=303.6\,\text{kN}$$

(b) Net-section rupture (IS 800 cl. 6.3.3 for angles)

Net area of connected leg: deduct one hole from connected leg. Connected leg area at centre line $A_{nc}=768-(18\times8)=768-144=624\,\text{mm}^2$.

Outstanding leg gross $A_{go}=568\,\text{mm}^2$.

Shear-lag factor $\beta=1.4-0.076\left(\dfrac{w}{t}\right)\left(\dfrac{f_y}{f_u}\right)\left(\dfrac{b_s}{L_c}\right)$.

Here outstanding leg width $w=75\,\text{mm}$, $t=8$, $b_s=w+w_1-t=75+ (50-? )$ — for a single line use $b_s = w + g - t$. Taking gusset edge so $b_s = 75 + 40 - 8 = 107\,\text{mm}$ is conservative; use the simplified $b_s = w = 75\,\text{mm}$ giving:

Bolt-line length $L_c=3\times40=120\,\text{mm}$ (between first and last bolt).

$$\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{75}{120}=1.4-0.076\times9.375\times0.610\times0.625$$ $$\beta=1.4-0.272=1.128$$

Bounds: $0.7 \le \beta \le \dfrac{f_u\gamma_{m0}}{f_y\gamma_{m1}}=\dfrac{410\times1.10}{250\times1.25}=1.443$. So $\beta=1.128$ is valid.

$$T_{dn}=\frac{0.9\,A_{nc}\,f_u}{\gamma_{m1}}+\frac{\beta\,A_{go}\,f_y}{\gamma_{m0}}$$ $$=\frac{0.9\times624\times410}{1.25}+\frac{1.128\times568\times250}{1.10}$$ $$=\frac{230{,}256}{1.25}+\frac{160{,}176}{1.10}=184{,}205+145{,}615=329{,}820\,\text{N}=329.8\,\text{kN}$$

(c) Block shear (single line)

Shear plane along bolt line: $L_v=e+3p=30+120=150\,\text{mm}$ (one face). $A_{vg}=150\times8=1200\,\text{mm}^2$; $A_{vn}=(150-3.5\times18)\times8=(150-63)\times8=696\,\text{mm}^2$. Tension plane (edge to bolt line): take $e_2=35\,\text{mm}$. $A_{tg}=35\times8=280\,\text{mm}^2$; $A_{tn}=(35-0.5\times18)\times8=(35-9)\times8=208\,\text{mm}^2$.

$$T_{db1}=\frac{A_{vg}f_y}{\sqrt3\,\gamma_{m0}}+\frac{0.9 A_{tn}f_u}{\gamma_{m1}}=\frac{1200\times250}{\sqrt3\times1.10}+\frac{0.9\times208\times410}{1.25}$$ $$=\frac{300{,}000}{1.905}+\frac{76{,}752}{1.25}=157{,}480+61{,}402=218{,}882\,\text{N}=218.9\,\text{kN}$$ $$T_{db2}=\frac{0.9 A_{vn}f_u}{\sqrt3\,\gamma_{m1}}+\frac{A_{tg}f_y}{\gamma_{m0}}=\frac{0.9\times696\times410}{\sqrt3\times1.25}+\frac{280\times250}{1.10}$$ $$=\frac{256{,}824}{2.165}+\frac{70{,}000}{1.10}=118{,}625+63{,}636=182{,}261\,\text{N}=182.3\,\text{kN}$$

$T_{db}=\min(218.9,\,182.3)=182.3\,\text{kN}$.

Design tensile strength $=\min(303.6,\,329.8,\,182.3)=\textbf{182.3 kN}$ (block shear governs).

Step 1 — Slenderness ratio (governed by minimum radius of gyration $r_y$)

$$\lambda=\frac{KL}{r_{min}}=\frac{3500}{53.4}=65.54$$

Step 2 — Euler / non-dimensional slenderness

$$f_{cc}=\frac{\pi^2 E}{\lambda^2}=\frac{\pi^2\times2\times10^5}{65.54^2}=\frac{1{,}973{,}921}{4295.5}=459.5\,\text{MPa}$$ $$\bar\lambda=\sqrt{\frac{f_y}{f_{cc}}}=\sqrt{\frac{250}{459.5}}=\sqrt{0.544}=0.738$$

Step 3 — Imperfection factor & design stress (buckling class b, $\alpha=0.34$)

$$\phi=0.5\,[1+\alpha(\bar\lambda-0.2)+\bar\lambda^2]=0.5\,[1+0.34(0.738-0.2)+0.738^2]$$ $$\phi=0.5\,[1+0.183+0.545]=0.5\times1.728=0.864$$ $$\chi=\frac{1}{\phi+\sqrt{\phi^2-\bar\lambda^2}}=\frac{1}{0.864+\sqrt{0.864^2-0.545}}=\frac{1}{0.864+\sqrt{0.747-0.545}}$$ $$=\frac{1}{0.864+\sqrt{0.202}}=\frac{1}{0.864+0.449}=\frac{1}{1.313}=0.762$$ $$f_{cd}=\frac{\chi f_y}{\gamma_{m0}}=\frac{0.762\times250}{1.10}=\frac{190.4}{1.10}=173.1\,\text{MPa}$$

Step 4 — Design compressive strength

$$P_d=A_e\,f_{cd}=6971\times173.1=1{,}206{,}680\,\text{N}=1206.7\,\text{kN}$$

Step 5 — Check

$$P_d=1206.7\,\text{kN} > P_u=1100\,\text{kN}\quad\checkmark$$

Slenderness $65.5 < 180$ (limit for members carrying compression).

Conclusion: ISHB 250 @ 54.7 kg/m is adequate with a utilisation of $1100/1206.7 = 0.912$ (about 91%). The section provides an economical solution; no heavier section is required. Provide ISHB 250.

Step 1 — Design bending moment and shear

$$M_u=\frac{w_u L^2}{8}=\frac{45\times6^2}{8}=\frac{45\times36}{8}=202.5\,\text{kN·m}$$ $$V_u=\frac{w_u L}{2}=\frac{45\times6}{2}=135\,\text{kN}$$

Step 2 — Design bending strength (laterally restrained, plastic section, $\beta_b=1.0$)

$$M_d=\frac{\beta_b Z_p f_y}{\gamma_{m0}}=\frac{1.0\times1533.36\times10^3\times250}{1.10}=\frac{383.34\times10^6}{1.10}=348.5\times10^6\,\text{N·mm}=348.5\,\text{kN·m}$$

Check limit to avoid excessive plasticity: $1.2\,Z_e f_y/\gamma_{m0}=\dfrac{1.2\times1350.7\times10^3\times250}{1.10}=368.4\,\text{kN·m}$. Since $348.5 < 368.4$, OK.

$$M_d=348.5\,\text{kN·m} > M_u=202.5\,\text{kN·m}\quad\checkmark$$

Step 3 — Design shear strength

$$V_d=\frac{f_{yw}\,(D\,t_w)}{\sqrt3\,\gamma_{m0}}=\frac{250\times(450\times9.4)}{\sqrt3\times1.10}=\frac{250\times4230}{1.905}=\frac{1{,}057{,}500}{1.905}=555{,}118\,\text{N}=555.1\,\text{kN}$$ $$V_d=555.1\,\text{kN} > V_u=135\,\text{kN}\quad\checkmark$$

Step 4 — High-shear / low-shear interaction

$0.6V_d=0.6\times555.1=333.1\,\text{kN} > V_u=135\,\text{kN}$, hence it is a low-shear case and no moment-shear interaction reduction is required.

Conclusion: Flexure utilisation $=202.5/348.5=0.58$; shear utilisation $=135/555.1=0.24$. ISMB 450 is adequate and safe (governed by flexure at ~58% utilisation).

Step 1 — Required area of base plate

$$A_{req}=\frac{P_u}{0.45 f_{ck}}=\frac{1600\times10^3}{9}=177{,}778\,\text{mm}^2$$

For a square plate side $L=\sqrt{177{,}778}=421.6\,\text{mm}$. Adopt $430\times430\,\text{mm}$ square base plate.

Provided area $A=430\times430=184{,}900\,\text{mm}^2$.

Step 2 — Actual bearing pressure

$$w=\frac{P_u}{A}=\frac{1600\times10^3}{184{,}900}=8.65\,\text{N/mm}^2 \;(<9\,\text{N/mm}^2)\quad\checkmark$$

Step 3 — Projections beyond column

Larger projection $a=\dfrac{L-D}{2}=\dfrac{430-300}{2}=65\,\text{mm}$ (along depth)

Smaller projection $b=\dfrac{L-B}{2}=\dfrac{430-250}{2}=90\,\text{mm}$ (along flange width)

Governing (larger) projection $=90\,\text{mm}$.

Step 4 — Required thickness (IS 800 cl. 7.4.3.1 for slab base)

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

Using $a=90$ (larger projection), $b=65$ (smaller projection):

$$a^2-0.3b^2=90^2-0.3\times65^2=8100-0.3\times4225=8100-1267.5=6832.5$$ $$t_s=\sqrt{\frac{2.5\times8.65\times6832.5\times1.10}{250}}=\sqrt{\frac{162{,}536}{250}}=\sqrt{650.1}=25.5\,\text{mm}$$

Adopt base-plate thickness $= 26\,\text{mm}$ (must also be $\ge$ column flange thickness).

Summary

Provide weld/cleat connection of column to base plate and four anchor bolts (nominal, e.g. 4 nos. M20) for stability. Design complete.

Step 1 — Effective throat thickness

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

Step 2 — Design strength per unit length

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

Strength per mm length $=f_{wd}\times t_t=189.4\times4.2=795.5\,\text{N/mm}=0.795\,\text{kN/mm}$.

Step 3 — Required effective length

$$L_{eff}=\frac{P_u}{0.795}=\frac{90}{0.795}=113.2\,\text{mm}$$

Add $2s=12\,\text{mm}$ for end returns/start-stop: total length $\approx 113.2+12=125\,\text{mm}$.

Final: design strength $=0.795\,\text{kN/mm}$; provide weld effective length $\approx \textbf{114 mm}$ (say 125 mm including end returns).

(a) Intermediate transverse stiffeners

• Increase the shear buckling resistance of the web by reducing the effective panel aspect ratio, allowing tension-field action to develop.
• Subdivide the web into panels so each panel can carry shear beyond the elastic critical buckling load.
• Prevent/limit out-of-plane buckling of the slender web under shear.

(b) Bearing (load-carrying) stiffeners

• Transfer concentrated loads or reactions (at supports and under point loads) from the flange into the web without local crushing or web crippling.
• Prevent local buckling of the web directly beneath concentrated loads.
• Provide a stiff load path and act with a length of web as an effective column section.

Web slenderness limit (IS 800:2007, cl. 8.6.1):

• If $\dfrac{d}{t_w} \le 67\varepsilon$ (where $\varepsilon=\sqrt{250/f_y}$), the web is thick and need not be checked for shear buckling — no intermediate stiffeners required on that account.
• If $\dfrac{d}{t_w} > 67\varepsilon$, the web is thin and must be designed for shear buckling (simple post-critical or tension-field method), typically requiring intermediate stiffeners.

For Fe 410 ($f_y=250$, $\varepsilon=1$) the limit is $d/t_w = 67$.

Step 1 — Tributary plan area per truss

Spacing $=4\,\text{m}$, span $=12\,\text{m}$. Total plan area per truss $=12\times4=48\,\text{m}^2$.

Step 2 — Service loads

Dead load $=0.20\times48=9.6\,\text{kN}$

Live load $=0.75\times48=36.0\,\text{kN}$

Step 3 — Factored total load on truss

$$W_u=1.5(DL+LL)=1.5(9.6+36.0)=1.5\times45.6=68.4\,\text{kN}$$

Step 4 — Panel point load

With $6$ equal panels there are $5$ intermediate top-chord joints and $2$ end joints (each carrying half a panel). Number of equivalent full panels $=6$.

Load per full panel $=\dfrac{W_u}{6}=\dfrac{68.4}{6}=11.4\,\text{kN}$.

An intermediate panel point carries one full panel load:

$$\boxed{P_{intermediate}=11.4\,\text{kN}}$$

(End/support joints carry half: $5.7\,\text{kN}$ each, plus any wall/cladding load.)

Section properties ($b=100$, $d=200\,\text{mm}$)

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

Cross-section area $A=100\times200=20{,}000\,\text{mm}^2$.

(a) Bending governs load

Moment capacity $M=\sigma_b Z=11\times666{,}667=7.333\times10^6\,\text{N·mm}=7.333\,\text{kN·m}$.

For UDL: $M=\dfrac{wL^2}{8}\Rightarrow w=\dfrac{8M}{L^2}=\dfrac{8\times7.333}{3^2}=\dfrac{58.67}{9}=6.52\,\text{kN/m}$.

(b) Shear governs load

Max horizontal shear stress for rectangle $\tau_{max}=\dfrac{1.5 V}{A}$.

Allowable shear force $V=\dfrac{\tau A}{1.5}=\dfrac{0.9\times20{,}000}{1.5}=12{,}000\,\text{N}=12\,\text{kN}$.

For UDL: $V=\dfrac{wL}{2}\Rightarrow w=\dfrac{2V}{L}=\dfrac{2\times12}{3}=8.0\,\text{kN/m}$.

Governing safe load

$$w_{safe}=\min(6.52,\,8.0)=\textbf{6.52 kN/m}\;(\text{bending governs}).$$

Section classification (IS 800:2007, cl. 3.7) classifies cross-sections according to the susceptibility of their plate elements (flange/web) to local buckling, measured by the width-to-thickness ratio $b/t$ compared against limiting values (in terms of $\varepsilon=\sqrt{250/f_y}$).

The four classes:

1. Class 1 — Plastic: Can form a plastic hinge and undergo large rotation (rotation capacity required for plastic analysis). Lowest $b/t$. Reaches plastic moment $M_p$.

2. Class 2 — Compact: Can develop plastic moment $M_p$ but has limited rotation capacity (insufficient for plastic redistribution). Slightly higher $b/t$.

3. Class 3 — Semi-compact: Extreme fibre can reach yield stress $f_y$, but local buckling prevents development of the full plastic moment. Reaches only yield (elastic) moment $M_y$.

4. Class 4 — Slender: Local buckling occurs before yield is reached at extreme fibre; an effective (reduced) cross-section must be used.

Influence on bending strength:

$$M_d=\frac{\beta_b Z f_y}{\gamma_{m0}}$$

• Class 1 & 2 (plastic/compact): $\beta_b=1.0$, use plastic modulus $Z_p$ → $M_d=Z_p f_y/\gamma_{m0}$.
• Class 3 (semi-compact): $\beta_b=Z_e/Z_p$, effectively use elastic modulus $Z_e$ → $M_d=Z_e f_y/\gamma_{m0}$.
• Class 4 (slender): use effective elastic modulus $Z_{eff}$ of the reduced section.

Thus classification directly selects whether $Z_p$, $Z_e$ or $Z_{eff}$ is used in the moment-capacity equation.

Comparison of WSM and LSM

Additional points

• WSM checks one condition (stress); LSM checks two categories — ultimate limit state (strength, stability) and serviceability limit state (deflection, vibration).
• LSM gives a more uniform margin of safety across different load combinations.

Recommended method: IS 800:2007 primarily recommends the Limit State Method (LSM); the Working Stress Method is retained only as an alternative where LSM cannot be conveniently applied.