BE Civil Engineering (IOE, TU) Design of Steel and Timber Structures (IOE, CE 655) Question Paper 2078 Nepal

Given: $T_u=180\ \text{kN}$, plate $12\ \text{mm}\times120\ \text{mm}$, M20 (4.6), hole $d_0=22\ \text{mm}$, $f_{ub}=400\ \text{MPa}$, $f_u=410\ \text{MPa}$, $f_y=250\ \text{MPa}$, $\gamma_{mb}=1.25$, $\gamma_{m0}=1.10$, $\gamma_{m1}=1.25$.

(a) Bolt shear strength (single shear, $n_n=1$, $n_s=0$):

$$A_{nb}=0.78\cdot\frac{\pi}{4}\,d^2=0.78\times\frac{\pi}{4}\times20^2=245.0\ \text{mm}^2$$ $$V_{dsb}=\frac{f_{ub}}{\sqrt3\,\gamma_{mb}}\,A_{nb}=\frac{400}{\sqrt3\times1.25}\times245.0=\frac{400\times245.0}{2.1651}=45.26\times10^3\ \text{N}=45.26\ \text{kN}$$

Bearing strength: edge $e=33$, pitch $p=50$, $d_0=22$.

$$k_b=\min\left(\frac{e}{3d_0},\ \frac{p}{3d_0}-0.25,\ \frac{f_{ub}}{f_u},\ 1.0\right)$$ $$\frac{e}{3d_0}=\frac{33}{66}=0.500,\quad \frac{p}{3d_0}-0.25=\frac{50}{66}-0.25=0.758-0.25=0.508,$$ $$\frac{f_{ub}}{f_u}=\frac{400}{410}=0.976,\quad \Rightarrow k_b=0.500$$ $$V_{dpb}=\frac{2.5\,k_b\,d\,t\,f_u}{\gamma_{mb}}=\frac{2.5\times0.500\times20\times12\times410}{1.25}=\frac{123000}{1.25}=98.40\times10^3\ \text{N}=98.40\ \text{kN}$$

Governing bolt value $=\min(45.26,\ 98.40)=\mathbf{45.26\ kN}$ (shear governs).

Number of bolts: $n=\dfrac{T_u}{V_{db}}=\dfrac{180}{45.26}=3.98\Rightarrow \mathbf{4\ bolts}$.

(b) Plate tension strength:

Gross-section yielding:

$$T_{dg}=\frac{A_g\,f_y}{\gamma_{m0}}=\frac{(120\times12)\times250}{1.10}=\frac{1440\times250}{1.10}=327.3\times10^3\ \text{N}=327.3\ \text{kN}$$

Net-section rupture (one hole in critical section, single line):

$$A_n=(b-d_0)\,t=(120-22)\times12=98\times12=1176\ \text{mm}^2$$ $$T_{dn}=\frac{0.9\,A_n\,f_u}{\gamma_{m1}}=\frac{0.9\times1176\times410}{1.25}=\frac{433944}{1.25}=347.2\times10^3\ \text{N}=347.2\ \text{kN}$$

Design tensile strength $=\min(327.3,\ 347.2)=\mathbf{327.3\ kN}$.

Since $327.3\ \text{kN}>180\ \text{kN}$, the plate is adequate and gross-section yielding governs. Use 4 M20 bolts.

Given: $L_e=4000\ \text{mm}$, $A=6971\ \text{mm}^2$, $r_{min}=r_{yy}=53.4\ \text{mm}$, $f_y=250\ \text{MPa}$, $\alpha=0.49$, $\gamma_{m0}=1.10$, $E=2\times10^5\ \text{MPa}$.

Step 1 — Slenderness ratio:

$$\lambda_{slen}=\frac{L_e}{r_{min}}=\frac{4000}{53.4}=74.9$$

Step 2 — Euler (elastic critical) stress:

$$f_{cc}=\frac{\pi^2 E}{\lambda_{slen}^2}=\frac{\pi^2\times2\times10^5}{74.9^2}=\frac{1.9739\times10^6}{5610.0}=351.9\ \text{MPa}$$

Step 3 — Non-dimensional slenderness:

$$\bar\lambda=\sqrt{\frac{f_y}{f_{cc}}}=\sqrt{\frac{250}{351.9}}=\sqrt{0.7104}=0.843$$

Step 4 — Factor $\phi$:

$$\phi=0.5\left[1+\alpha(\bar\lambda-0.2)+\bar\lambda^2\right]=0.5\left[1+0.49(0.843-0.2)+0.843^2\right]$$ $$=0.5\left[1+0.49(0.643)+0.7106\right]=0.5[1+0.3151+0.7106]=0.5\times2.0257=1.0129$$

Step 5 — Stress reduction factor $\chi$:

$$\chi=\frac{1}{\phi+\sqrt{\phi^2-\bar\lambda^2}}=\frac{1}{1.0129+\sqrt{1.0129^2-0.7106}}=\frac{1}{1.0129+\sqrt{1.0260-0.7106}}$$ $$=\frac{1}{1.0129+\sqrt{0.3154}}=\frac{1}{1.0129+0.5616}=\frac{1}{1.5745}=0.6351$$

Step 6 — Design compressive stress:

$$f_{cd}=\frac{\chi\,f_y}{\gamma_{m0}}=\frac{0.6351\times250}{1.10}=\frac{158.78}{1.10}=144.3\ \text{MPa}$$

Step 7 — Design compressive strength:

$$P_d=A\,f_{cd}=6971\times144.3=1.006\times10^6\ \text{N}=\mathbf{1006\ kN}$$

Check: $P_d=1006\ \text{kN} < 1100\ \text{kN}$ applied.

The section is NOT adequate (capacity is about 9% short). A heavier section (e.g. ISHB 300) should be adopted. This illustrates the design-check verdict: when $P_d < P_u$, the trial section must be revised.

Given: $L=6\ \text{m}$, $w_u=40\ \text{kN/m}$, laterally restrained, $f_y=250\ \text{MPa}$, $\gamma_{m0}=1.10$.

(a) Design forces:

$$M_u=\frac{w_u L^2}{8}=\frac{40\times6^2}{8}=\frac{40\times36}{8}=180\ \text{kN·m}$$ $$V_u=\frac{w_u L}{2}=\frac{40\times6}{2}=120\ \text{kN}$$

(b) Bending strength of ISMB 400 (plastic section, laterally restrained $\Rightarrow$ no LTB):

$$M_d=\frac{\beta_b\,Z_p\,f_y}{\gamma_{m0}}=\frac{1.0\times1176.18\times10^3\times250}{1.10}=\frac{294.045\times10^6}{1.10}=267.3\times10^6\ \text{N·mm}=267.3\ \text{kN·m}$$

Limit to avoid excessive deformation (simply supported): $1.2\,Z_e f_y/\gamma_{m0}$:

$$1.2\times\frac{1020.0\times10^3\times250}{1.10}=1.2\times231.8=278.2\ \text{kN·m}>267.3\ \text{kN·m}\ \text{(OK, governing }M_d=267.3\text{)}$$

Check: $M_d=267.3\ \text{kN·m}>M_u=180\ \text{kN·m}$ $\Rightarrow$ bending is adequate.

(c) Shear strength:

$$A_v=D\,t_w=400\times8.9=3560\ \text{mm}^2$$ $$V_d=\frac{f_y\,A_v}{\sqrt3\,\gamma_{m0}}=\frac{250\times3560}{\sqrt3\times1.10}=\frac{890000}{1.9053}=467.2\times10^3\ \text{N}=467.2\ \text{kN}$$

Check: $V_d=467.2\ \text{kN}>V_u=120\ \text{kN}$ $\Rightarrow$ shear is adequate.

Also $V_u=120<0.6\,V_d=280.3\ \text{kN}$, so it is a low-shear case and no reduction in $M_d$ for shear is required.

Conclusion: ISMB 400 is safe in both flexure and shear.

Given: $M_u=2800\ \text{kN·m}$, $V_u=700\ \text{kN}$, $f_y=250\ \text{MPa}$.

(a) Economical depth of web. The economical depth by the optimum-depth formula:

$$d=\left(\frac{M_u}{f_b\,t_w}\right)^{1/2}\ \text{(approx.)},\quad\text{but commonly } d\approx\left(\frac{M}{\sigma_{bc}}\cdot k\right).$$

A standard practical estimate: $d=(3\ \text{to}\ 5)\times\sqrt[3]{\dfrac{M_u}{f_y}}\times10$ or simply choose $d\approx \dfrac{L}{10}$. Taking a practical economical depth using $d=\sqrt[3]{\dfrac{M_u\cdot k}{f_y}}$, we adopt a trial web depth $d = 1200\ \text{mm}$.

Web thickness from shear (using allowable $\tau=100\ \text{MPa}$):

$$t_w=\frac{V_u}{d\,\tau}=\frac{700\times10^3}{1200\times100}=\frac{700000}{120000}=5.83\ \text{mm}\Rightarrow \text{adopt } t_w=8\ \text{mm}.$$

Serviceability/minimum thickness without stiffeners requires $d/t_w$ limited; with $d/t_w=1200/8=150$, intermediate stiffeners will be needed.

Adopt web plate $1200\times8\ \text{mm}$.

(b) Flange area (flange-area method): lever arm $\approx d=1200\ \text{mm}$. Design bending stress $f_{bd}=f_y/\gamma_{m0}=250/1.10=227.3\ \text{MPa}$.

$$A_f=\frac{M_u}{f_{bd}\cdot d}=\frac{2800\times10^6}{227.3\times1200}=\frac{2800\times10^6}{272760}=10\,266\ \text{mm}^2$$

Provide a flange plate, say $400\times26\ \text{mm}$ giving $A_f=10\,400\ \text{mm}^2>10\,266\ \text{mm}^2$. OK.

(Check flange outstand $b/2t_f=200/26=7.7<8.4\varepsilon$, plastic — satisfactory.)

(c) Stiffeners. Intermediate transverse stiffeners are required when: (i) the web slenderness $d/t_w$ is large (here $150 > 67\varepsilon$) so the web would buckle in shear before reaching yield; and (ii) the applied shear exceeds the shear-buckling capacity of the unstiffened web, i.e. tension-field/post-critical action must be mobilised.

Bearing (load-carrying) stiffeners are provided at supports and under concentrated point loads to transmit the reaction/load into the web, preventing web crippling and vertical buckling of the web beneath the load.

Given: two vertical welds, each length $l=300\ \text{mm}$; horizontal spacing $=150\ \text{mm}$; $P=120\ \text{kN}$; $e=200\ \text{mm}$ (eccentricity from group centroid); in-plane (torsional) loading; $f_u=410\ \text{MPa}$, $\gamma_{mw}=1.25$.

Work per unit throat (treat weld as line, throat $t=1$), then size from strength.

Geometry of weld group (two vertical lines): Each line length $300\ \text{mm}$; total weld length $L_w=2\times300=600\ \text{mm}$. Centroid is at the geometric centre. The two lines are at $\pm75\ \text{mm}$ horizontally.

Polar moment of weld group (per unit throat):

$$I_x=2\times\frac{l^3}{12}=2\times\frac{300^3}{12}=2\times\frac{27\times10^6}{12}=2\times2.25\times10^6=4.5\times10^6\ \text{mm}^3$$ $$I_y=2\times l\times(75)^2=2\times300\times5625=3.375\times10^6\ \text{mm}^3$$ $$J=I_x+I_y=4.5\times10^6+3.375\times10^6=7.875\times10^6\ \text{mm}^3$$

Direct (vertical) shear per unit length:

$$q_d=\frac{P}{L_w}=\frac{120\times10^3}{600}=200\ \text{N/mm}\quad(\text{vertical, downward})$$

Twisting moment: $T=P\,e=120\times10^3\times200=24\times10^6\ \text{N·mm}$.

The most stressed point is the corner farthest from centroid: $r$ with components $x=75\ \text{mm}$, $y=150\ \text{mm}$.

Torsional shear components per unit length:

$$q_{t,x}=\frac{T\,y}{J}=\frac{24\times10^6\times150}{7.875\times10^6}=\frac{3.6\times10^9}{7.875\times10^6}=457.1\ \text{N/mm (horizontal)}$$ $$q_{t,y}=\frac{T\,x}{J}=\frac{24\times10^6\times75}{7.875\times10^6}=\frac{1.8\times10^9}{7.875\times10^6}=228.6\ \text{N/mm (vertical)}$$

Resultant per unit length: combine vertical ($q_d+q_{t,y}$) and horizontal ($q_{t,x}$):

$$q_{res}=\sqrt{(q_d+q_{t,y})^2+q_{t,x}^2}=\sqrt{(200+228.6)^2+457.1^2}$$ $$=\sqrt{428.6^2+457.1^2}=\sqrt{183698+208940}=\sqrt{392638}=626.6\ \text{N/mm}$$

Weld strength per unit length for throat $t_t=0.707\,s$ (size $s$):

$$f_{wd}=\frac{f_u}{\sqrt3\,\gamma_{mw}}=\frac{410}{\sqrt3\times1.25}=\frac{410}{2.1651}=189.4\ \text{MPa}$$

Strength per mm length $=0.707\,s\times189.4=133.9\,s\ \text{N/mm}$.

Required size:

$$133.9\,s\ge 626.6\Rightarrow s\ge\frac{626.6}{133.9}=4.68\ \text{mm}$$

Provide a 6 mm fillet weld (next standard size $\ge 4.68$ mm and satisfying minimum size for thick flange).

Answer: fillet weld size $s = 6\ \text{mm}$.

Given: $P_u=900\ \text{kN}=900\times10^3\ \text{N}$; permissible bearing pressure on concrete $w=0.45 f_{ck}=9\ \text{MPa (N/mm}^2)$.

Required bearing area:

$$A_{req}=\frac{P_u}{w}=\frac{900\times10^3}{9}=1.0\times10^5\ \text{mm}^2=100000\ \text{mm}^2$$

For a square base of side $a$:

$$a=\sqrt{A_{req}}=\sqrt{100000}=316.2\ \text{mm}$$

Provide a square slab base of $350\ \text{mm}\times350\ \text{mm}$ (rounded up to a practical size).

Check actual pressure: $w_{act}=\dfrac{900\times10^3}{350\times350}=\dfrac{900000}{122500}=7.35\ \text{MPa}<9\ \text{MPa}$. Safe.

Answer: minimum side $\approx 317\ \text{mm}$; adopt $350\times350\ \text{mm}$ slab base.

Given: $L=3\ \text{m}$, $w=4\ \text{kN/m}$, $\sigma_b=10\ \text{MPa}$, $d=2b$.

Maximum bending moment:

$$M=\frac{wL^2}{8}=\frac{4\times3^2}{8}=\frac{4\times9}{8}=4.5\ \text{kN·m}=4.5\times10^6\ \text{N·mm}$$

Required section modulus:

$$Z_{req}=\frac{M}{\sigma_b}=\frac{4.5\times10^6}{10}=4.5\times10^5\ \text{mm}^3=450000\ \text{mm}^3$$

Rectangular section ($d=2b$): $Z=\dfrac{b\,d^2}{6}=\dfrac{b(2b)^2}{6}=\dfrac{4b^3}{6}=\dfrac{2b^3}{3}$.

$$\frac{2b^3}{3}=450000\Rightarrow b^3=\frac{3\times450000}{2}=675000\Rightarrow b=\sqrt[3]{675000}=87.7\ \text{mm}$$

Then $d=2b=175.4\ \text{mm}$.

Provide a section of $b=100\ \text{mm}$, $d=200\ \text{mm}$ (practical, $d=2b$).

Check: $Z=\dfrac{100\times200^2}{6}=\dfrac{4\times10^6}{6}=666667\ \text{mm}^3>450000\ \text{mm}^3$. Safe.

Answer: $Z_{req}=4.5\times10^5\ \text{mm}^3$; adopt $100\times200\ \text{mm}$ timber beam.

Given: span $=12\ \text{m}$, truss spacing $s=4\ \text{m}$, intensity $q=0.9\ \text{kN/m}^2$ (on plan), $6$ panels, panel length $a=2\ \text{m}$.

Load per metre length of truss (per unit span):

$$w=q\times s=0.9\times4=3.6\ \text{kN/m of span}$$

Intermediate (interior) panel point — tributary length $=$ one full panel ($a=2\ \text{m}$, half panel on each side):

$$W_{int}=w\times a=3.6\times2=7.2\ \text{kN}$$

End panel point — tributary length $=$ half a panel ($a/2=1\ \text{m}$):

$$W_{end}=w\times\frac{a}{2}=3.6\times1=3.6\ \text{kN}$$

Verification (total): $5$ interior points $+$ $2$ end points:

$$5\times7.2+2\times3.6=36+7.2=43.2\ \text{kN}$$

Total roof load $=q\times \text{span}\times s=0.9\times12\times4=43.2\ \text{kN}$. Checks out.

Answer: interior panel-point load $=\mathbf{7.2\ kN}$; end panel-point load $=\mathbf{3.6\ kN}$.

Block shear failure: In a bolted tension connection, a block of material at the end of the member can tear out as a single unit. Failure occurs along a combined path — shear yielding/rupture along the bolt line(s) parallel to the load together with tension yielding/rupture across a plane perpendicular to the load connecting the bolt holes. It is a tearing-out of the bolted block rather than a clean net-section failure.

IS 800:2007 block shear strength — the design strength $T_{db}$ is the smaller of the two modes:

Mode 1 — shear yield + tension rupture:

$$T_{db1}=\frac{A_{vg}\,f_y}{\sqrt3\,\gamma_{m0}}+\frac{0.9\,A_{tn}\,f_u}{\gamma_{m1}}$$

Mode 2 — shear rupture + tension yield:

$$T_{db2}=\frac{0.9\,A_{vn}\,f_u}{\sqrt3\,\gamma_{m1}}+\frac{A_{tg}\,f_y}{\gamma_{m0}}$$

where $A_{vg},A_{vn}$ = gross/net area in shear (along bolt line) and $A_{tg},A_{tn}$ = gross/net area in tension (perpendicular to load).

Why it governs for single-leg-connected angles: When an angle is connected through only one leg, the load is transmitted eccentrically and concentrated near the connected leg. Combined with the short end distance and limited bolt arrangement, the small block of material around the bolts is prone to tearing out. The reduced effective net area and stress concentration make block shear frequently the critical (lowest) limit state, often governing over gross-section yielding and net-section rupture.

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

Three key points of difference:

1. WSM uses a single factor of safety on material strength, whereas LSM uses separate partial safety factors for loads and materials.
2. WSM is based purely on elastic stress limitation; LSM checks multiple limit states (strength + serviceability) and exploits plastic reserve.
3. WSM uses service loads; LSM uses factored loads, giving more rational and generally more economical designs.

IS 800:2007 adopts LSM as the primary method (WSM is retained as an alternative).

Lateral-torsional buckling (LTB): When a slender beam is loaded about its major (strong) axis, the compression flange behaves like a column and tends to buckle sideways. Because the tension flange resists this movement, the cross-section simultaneously deflects laterally and twists. This combined out-of-plane lateral deflection and rotation, occurring before the in-plane plastic moment is reached, is lateral-torsional buckling. It reduces the moment capacity below the full plastic moment $M_p$, so a bending reduction factor $\chi_{LT}$ (or $\beta_b<1$) must be applied for laterally unrestrained beams.

Factors influencing LTB strength:

1. Unsupported (effective) length of the compression flange — larger $L_{LT}$ lowers strength.
2. Cross-sectional shape and properties — torsional constant $I_t$, warping constant $I_w$, and minor-axis moment of inertia $I_{yy}$.
3. Type and position of loading — point vs UDL, load applied at top flange (destabilising) vs at shear centre.
4. End/support conditions and restraints (lateral and torsional restraint at supports).
5. Moment gradient along the span (uniform moment is most critical).
6. Material yield strength and residual stresses / imperfections.

Two practical measures to prevent LTB:

1. Provide adequate lateral restraint to the compression flange — e.g. cross beams, bracing, or a connected concrete/steel deck that holds the flange in position (reduces effective length).
2. Use sections with high torsional/lateral stiffness (e.g. wider flanges, or encasing/box sections), or reduce the unsupported length by introducing intermediate supports.