BE Civil Engineering (IOE, TU) Design of Steel and Timber Structures (IOE, CE 655) Question Paper 2079 Nepal

(a) Design strength of one bolt

Bolt geometry: $d = 20\ \text{mm}$, $d_0 = 22\ \text{mm}$, net (stress) area $A_{nb} = 0.78 \times \frac{\pi}{4}d^2 = 0.78 \times \frac{\pi}{4}(20)^2 = 245\ \text{mm}^2$.

Grade 4.6 $\Rightarrow f_{ub} = 400\ \text{MPa}$.

Shear strength (single shear, threads in plane):

$$V_{dsb} = \frac{f_{ub}}{\sqrt{3}\,\gamma_{mb}}\,(n_n A_{nb}) = \frac{400}{\sqrt3 \times 1.25}\times (1 \times 245)$$ $$= \frac{400}{2.1651}\times 245 = 184.75 \times 245 = 45{,}264\ \text{N} = 45.26\ \text{kN}$$

Bearing strength: governing thickness $t = 12\ \text{mm}$ (single plate in single shear).

$$k_b = \min\left(\frac{e}{3d_0},\ \frac{p}{3d_0}-0.25,\ \frac{f_{ub}}{f_u},\ 1.0\right)$$ $$\frac{e}{3d_0} = \frac{33}{66} = 0.50,\quad \frac{p}{3d_0}-0.25 = \frac{55}{66}-0.25 = 0.833-0.25 = 0.583$$ $$\frac{f_{ub}}{f_u} = \frac{400}{410} = 0.976,\quad \Rightarrow k_b = 0.50$$ $$V_{dpb} = \frac{2.5\,k_b\,d\,t\,f_u}{\gamma_{mb}} = \frac{2.5 \times 0.50 \times 20 \times 12 \times 410}{1.25}$$ $$= \frac{1{,}230{,}000}{1.25} = 98{,}400\ \text{N} = 98.4\ \text{kN}$$

Bolt value $V_{db} = \min(45.26,\ 98.4) = \mathbf{45.26\ kN}$ (shear governs).

(b) Number of bolts

$$n = \frac{180}{45.26} = 3.98 \Rightarrow \mathbf{4\ bolts}$$

Provide 4 bolts in a single line at pitch $55\ \text{mm}$, edge distance $33\ \text{mm}$. Capacity $= 4 \times 45.26 = 181.0\ \text{kN} > 180\ \text{kN}$. OK.

(c) Plate tension check

Width $b = 150\ \text{mm}$, $t = 12\ \text{mm}$.

Gross-section yielding:

$$A_g = 150 \times 12 = 1800\ \text{mm}^2$$ $$T_{dg} = \frac{A_g f_y}{\gamma_{m0}} = \frac{1800 \times 250}{1.10} = 409{,}091\ \text{N} = 409.1\ \text{kN}$$

Net-section rupture (one bolt hole in the critical line):

$$A_n = (b - d_0)\,t = (150 - 22)\times 12 = 1536\ \text{mm}^2$$ $$T_{dn} = \frac{0.9\,A_n f_u}{\gamma_{m1}} = \frac{0.9 \times 1536 \times 410}{1.25} = \frac{566{,}784}{1.25} = 453{,}427\ \text{N} = 453.4\ \text{kN}$$

Design tensile strength $= \min(409.1,\ 453.4) = \mathbf{409.1\ kN} > 180\ \text{kN}$. The plate is safe.

Step 1 — Governing slenderness

Weaker axis governs: $r_{min} = r_{yy} = 53.4\ \text{mm}$, $KL = 4000\ \text{mm}$.

$$\lambda = \frac{KL}{r_{min}} = \frac{4000}{53.4} = 74.9$$

Step 2 — Non-dimensional slenderness

$$f_{cc} = \frac{\pi^2 E}{\lambda^2} = \frac{\pi^2 \times 2\times10^5}{(74.9)^2} = \frac{1.9739\times10^6}{5610} = 351.9\ \text{MPa}$$ $$\bar\lambda = \sqrt{\frac{f_y}{f_{cc}}} = \sqrt{\frac{250}{351.9}} = \sqrt{0.7104} = 0.843$$

Step 3 — Imperfection factor & $\phi$

$$\phi = 0.5\left[1 + \alpha(\bar\lambda - 0.2) + \bar\lambda^2\right] = 0.5\left[1 + 0.49(0.843-0.2) + 0.843^2\right]$$ $$= 0.5\left[1 + 0.49(0.643) + 0.711\right] = 0.5\left[1 + 0.3151 + 0.711\right] = 0.5(2.0261) = 1.0130$$

Step 4 — Stress reduction factor $\chi$

$$\chi = \frac{1}{\phi + \sqrt{\phi^2 - \bar\lambda^2}} = \frac{1}{1.0130 + \sqrt{1.0130^2 - 0.711}}$$ $$= \frac{1}{1.0130 + \sqrt{1.0262 - 0.711}} = \frac{1}{1.0130 + \sqrt{0.3152}} = \frac{1}{1.0130 + 0.5614} = \frac{1}{1.5744} = 0.6352$$

Step 5 — Design compressive stress

$$f_{cd} = \frac{\chi f_y}{\gamma_{m0}} = \frac{0.6352 \times 250}{1.10} = \frac{158.8}{1.10} = 144.4\ \text{MPa}$$

Step 6 — Design capacity

$$P_d = A_e\, f_{cd} = 6971 \times 144.4 = 1{,}006{,}612\ \text{N} \approx \mathbf{1006.6\ kN}$$

The design axial compressive capacity of the column $\approx 1006\ \text{kN}$.

(a) Design actions

$$M_u = \frac{wL^2}{8} = \frac{40 \times 6^2}{8} = \frac{40 \times 36}{8} = 180\ \text{kN·m}$$ $$V_u = \frac{wL}{2} = \frac{40 \times 6}{2} = 120\ \text{kN}$$

(b) Moment capacity (plastic section, laterally restrained)

For a plastic section the design bending strength:

$$M_d = \frac{\beta_b\,Z_p\,f_y}{\gamma_{m0}},\quad \beta_b = 1.0$$ $$M_d = \frac{1.0 \times 1533.36\times10^3 \times 250}{1.10} = \frac{383.34\times10^6}{1.10} = 348.5\times10^6\ \text{N·mm} = 348.5\ \text{kN·m}$$

Limit check to avoid excessive plasticity at SLS:

$$1.2\,\frac{Z_e f_y}{\gamma_{m0}} = \frac{1.2 \times 1350.7\times10^3 \times 250}{1.10} = \frac{405.21\times10^6}{1.10} = 368.4\ \text{kN·m}$$

Since $348.5 < 368.4$, governing $M_d = 348.5\ \text{kN·m}$.

$M_d = 348.5\ \text{kN·m} > M_u = 180\ \text{kN·m}$ → safe in flexure.

(c) Shear capacity

Shear area (rolled I-section) $A_v = h\,t_w = 450 \times 9.4 = 4230\ \text{mm}^2$.

$$V_d = \frac{f_y\,A_v}{\sqrt3\,\gamma_{m0}} = \frac{250 \times 4230}{\sqrt3 \times 1.10} = \frac{1{,}057{,}500}{1.9053} = 555{,}030\ \text{N} = 555.0\ \text{kN}$$

$V_d = 555.0\ \text{kN} > V_u = 120\ \text{kN}$ → safe in shear.

Low-shear check: $V_u = 120 < 0.6\,V_d = 333\ \text{kN}$, so no moment–shear interaction reduction is needed.

The ISMB 450 section is adequate for both flexure and shear.

Step 1 — Weld group geometry (treat throat = unit, line method)

Two vertical lines, each length $l = 250\ \text{mm}$, horizontal spacing $b = 200\ \text{mm}$.

Total weld length $L_w = 2 \times 250 = 500\ \text{mm}$.

Centroid lies midway between the two lines. Each line is at $\pm 100\ \text{mm}$ horizontally.

Polar moment of weld group per unit throat (treating welds as lines):

$$I_x = 2 \times \frac{l^3}{12} = 2 \times \frac{250^3}{12} = 2 \times 1.3021\times10^6 = 2.6042\times10^6\ \text{mm}^3$$ $$I_y = 2 \times l \times \left(\frac{b}{2}\right)^2 = 2 \times 250 \times 100^2 = 5.0\times10^6\ \text{mm}^3$$ $$I_p = I_x + I_y = 2.6042\times10^6 + 5.0\times10^6 = 7.6042\times10^6\ \text{mm}^3$$

Step 2 — Critical point

Farthest weld point: $r_x = 100\ \text{mm}$ (horizontal), $r_y = 125\ \text{mm}$ (top of weld). Applied torsion $T = P\,e = 120\times10^3 \times 150 = 18\times10^6\ \text{N·mm}$.

Step 3 — Stresses per unit throat

Direct (vertical) shear spread over total length:

$$q_1 = \frac{P}{L_w} = \frac{120\times10^3}{500} = 240\ \text{N/mm}$$

Torsional shear (per unit throat), components:

$$q_{2x} = \frac{T\,r_y}{I_p} = \frac{18\times10^6 \times 125}{7.6042\times10^6} = 295.9\ \text{N/mm}\ \text{(horizontal)}$$ $$q_{2y} = \frac{T\,r_x}{I_p} = \frac{18\times10^6 \times 100}{7.6042\times10^6} = 236.7\ \text{N/mm}\ \text{(vertical)}$$

Step 4 — Resultant force per unit throat

Vertical total $= q_1 + q_{2y} = 240 + 236.7 = 476.7\ \text{N/mm}$. Horizontal $= q_{2x} = 295.9\ \text{N/mm}$.

$$q_R = \sqrt{476.7^2 + 295.9^2} = \sqrt{227{,}243 + 87{,}557} = \sqrt{314{,}800} = 561.1\ \text{N/mm}$$

Step 5 — Required throat & weld size

Design strength of weld per unit throat:

$$f_{wd} = \frac{f_u}{\sqrt3\,\gamma_{mw}} = \frac{410}{\sqrt3 \times 1.25} = \frac{410}{2.1651} = 189.4\ \text{N/mm}^2$$

Required throat thickness $t_t$:

$$t_t = \frac{q_R}{f_{wd}} = \frac{561.1}{189.4} = 2.96\ \text{mm}$$

Weld size $s = \dfrac{t_t}{0.7} = \dfrac{2.96}{0.7} = 4.23\ \text{mm}$.

Provide a $6\ \text{mm}$ fillet weld (next practical size above $4.23\ \text{mm}$, also satisfying the minimum size for the connected thickness).

(a) Plan area of base plate

Required bearing area:

$$A_{req} = \frac{P}{\sigma_{br}} = \frac{1600\times10^3}{9.0} = 177{,}778\ \text{mm}^2$$

For a square base, side $= \sqrt{177{,}778} = 421.6\ \text{mm}$.

Provide a $450\ \text{mm} \times 450\ \text{mm}$ base plate. Provided area $= 450 \times 450 = 202{,}500\ \text{mm}^2$.

Actual bearing pressure:

$$w = \frac{P}{A_{prov}} = \frac{1600\times10^3}{202{,}500} = 7.90\ \text{N/mm}^2 \ (< 9.0,\ \text{OK})$$

(b) Cantilever projections

Projection beyond column footprint:

$$a = \frac{450 - 300}{2} = 75\ \text{mm (larger projection, along depth)}$$ $$b = \frac{450 - 250}{2} = 100\ \text{mm (larger projection, along width)}$$

Governing (larger) projection: $b = 100\ \text{mm}$ ; smaller $a = 75\ \text{mm}$.

(c) Thickness from plate bending (IS 800 slab-base formula)

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

Take the larger projection as $a = 100\ \text{mm}$ and smaller as $b = 75\ \text{mm}$:

$$a^2 - 0.3\,b^2 = 100^2 - 0.3(75)^2 = 10{,}000 - 0.3(5625) = 10{,}000 - 1687.5 = 8312.5\ \text{mm}^2$$ $$t_s = \sqrt{\frac{2.5 \times 7.90 \times 8312.5 \times 1.10}{250}} = \sqrt{\frac{180{,}589}{250}} = \sqrt{722.4} = 26.9\ \text{mm}$$

Provide a base plate thickness of $28\ \text{mm}$ (next standard plate).

Summary: Base plate $450 \times 450 \times 28\ \text{mm}$ on M20 concrete; bearing pressure $7.9\ \text{N/mm}^2 < 9.0\ \text{N/mm}^2$.

Limit State Method (LSM)

LSM is a semi-probabilistic design philosophy in which a structure is designed so that it does not reach any of the specified limit states — conditions beyond which it ceases to satisfy the requirements for which it was built. The design ensures an acceptably low probability of failure by applying partial safety factors to both loads (increasing them) and material strengths (reducing them).

The basic design inequality:

$$\text{Design action } (\sum \gamma_f\,Q_k) \ \le\ \text{Design resistance } \left(\frac{R_k}{\gamma_m}\right)$$

Limit State of Strength (Ultimate)

Concerns the safety of the structure under maximum (factored) loads. Includes:

• Yielding / rupture, buckling (member or local), fracture due to fatigue.
• Loss of stability (overturning, sway), brittle fracture, plastic collapse.

For this limit state, loads are factored UP (e.g. $1.5\,DL + 1.5\,LL$) and material strength divided by $\gamma_m$ ($\gamma_{m0}=1.10$ for yielding, $\gamma_{m1}=1.25$ for ultimate/rupture, $\gamma_{mb}=1.25$ bolts, $\gamma_{mw}=1.25$ welds).

Limit State of Serviceability

Concerns the performance/comfort under working (unfactored) loads. Includes:

• Deflection limits (e.g. span/300 for live load on beams).
• Vibration, durability/corrosion, and excessive local damage.

Here load factors are usually $1.0$ and the structure is checked for usability, not collapse.

Role of partial safety factors

• Partial load factor $\gamma_f$ accounts for variability/uncertainty in loads and load combinations (overestimates actions).
• Partial material factor $\gamma_m$ accounts for variability in material strength, fabrication tolerances and modelling uncertainty (underestimates resistance).

Together they provide a reliable margin of safety while keeping the design economical.

Block shear geometry (along connected leg)

Three bolts in a line: gross shear length $= 2 \times 50 + 35 = 135\ \text{mm}$.

• Gross shear area $A_{vg} = 135 \times t = 135 \times 8 = 1080\ \text{mm}^2$.
• Net shear area $A_{vn} = (135 - 2.5\,d_0)\,t = (135 - 2.5\times18)\times8 = (135-45)\times8 = 720\ \text{mm}^2$.

Tension path (edge distance of connected leg, take $e_2 = 35\ \text{mm}$ to bolt line):

• Gross tension area $A_{tg} = 35 \times 8 = 280\ \text{mm}^2$.
• Net tension area $A_{tn} = (35 - 0.5\,d_0)\times8 = (35 - 9)\times8 = 26\times8 = 208\ \text{mm}^2$.

Block shear strength — two modes

Mode 1 (gross shear yield + net tension rupture):

$$T_{db1} = \frac{A_{vg} f_y}{\sqrt3\,\gamma_{m0}} + \frac{0.9\,A_{tn} f_u}{\gamma_{m1}}$$ $$= \frac{1080\times250}{\sqrt3\times1.10} + \frac{0.9\times208\times410}{1.25}$$ $$= \frac{270{,}000}{1.9053} + \frac{76{,}752}{1.25} = 141{,}710 + 61{,}402 = 203{,}112\ \text{N} = 203.1\ \text{kN}$$

Mode 2 (net shear rupture + gross tension yield):

$$T_{db2} = \frac{0.9\,A_{vn} f_u}{\sqrt3\,\gamma_{m1}} + \frac{A_{tg} f_y}{\gamma_{m0}}$$ $$= \frac{0.9\times720\times410}{\sqrt3\times1.25} + \frac{280\times250}{1.10}$$ $$= \frac{265{,}680}{2.1651} + \frac{70{,}000}{1.10} = 122{,}710 + 63{,}636 = 186{,}346\ \text{N} = 186.3\ \text{kN}$$

Result

$$T_{db} = \min(203.1,\ 186.3) = \mathbf{186.3\ kN}$$

The block shear strength of the angle is 186.3 kN (Mode 2 governs).

Components of a welded plate girder

A plate girder is a built-up flexural member fabricated from steel plates welded together to form a deep I-section.

        flange plate (top, compression)
   ===================================
   |   ||        ||         ||      |   <- bearing stiffener (at supports)
   |   ||  web    ||         ||      |   <- intermediate stiffeners
   |   ||  plate  ||         ||      |
   ===================================
        flange plate (bottom, tension)

Main parts: top & bottom flange plates, a deep thin web plate, transverse (intermediate) stiffeners, bearing/load-bearing stiffeners over supports and under point loads, and (occasionally) longitudinal stiffeners and flange splice/weld connections.

(a) Intermediate transverse stiffeners

• Divide the web into panels and increase the shear buckling resistance of the thin web by reducing the panel aspect ratio $c/d$.
• Permit tension-field action, allowing a thinner, more economical web to carry higher shear.
• They are not designed for direct load but stabilize the web against diagonal buckling.

(b) Bearing stiffeners

• Placed at supports and under concentrated loads to transfer vertical reactions/loads from flange into the web without web crippling or buckling.
• Designed as a short column (stiffener + effective web length) to carry the bearing load.

(c) Role of the web in shear

• The web carries almost the entire shear force ($V_d = f_{yw}A_v/(\sqrt3\,\gamma_{m0})$), while flanges resist most of the bending moment.
• A slender web may buckle in shear before yielding; stiffeners and tension-field action raise this capacity.

Why plate girders are preferred for large spans/loads

• Section depth and flange/web proportions can be tailored to the moment & shear demand (rolled sections come in fixed sizes).
• They achieve greater depth and section modulus than the largest rolled beams, giving high flexural efficiency and stiffness for long spans (bridges, crane gantries) and heavy loads with optimal material use.

(a) Panel point (nodal) load — dead + live

Each intermediate node carries the load from a tributary plan area:

$$A_{trib} = \text{panel length} \times \text{spacing} = 2\ \text{m} \times 4\ \text{m} = 8\ \text{m}^2$$

Dead load (covering + purlins):

$$DL = 0.18 \times 8 = 1.44\ \text{kN}$$

Add an allowance for self-weight of truss (≈ taken as part of DL); using only given covering value here.

Live load:

$$LL = 0.75 \times 8 = 6.00\ \text{kN}$$

Factored (DL + LL) nodal load (LSM, $1.5(DL+LL)$):

$$P = 1.5\,(1.44 + 6.00) = 1.5 \times 7.44 = 11.16\ \text{kN}$$

Working (unfactored) DL + LL nodal load $= 7.44\ \text{kN}$; factored $= 11.16\ \text{kN}$ per intermediate node.

(End nodes carry half this tributary area, i.e. about $3.72\ \text{kN}$ working.)

(b) Critical load combination for the bottom tie

The bottom chord (tie) is normally in tension under gravity (DL + LL). The critical combinations are:

1. $1.5\,(DL + LL)$ → maximum gravity, gives the maximum tension in the bottom tie. This usually governs the tie design.
2. $1.2\,DL + 1.2\,LL + 1.2\,WL$ or $1.5\,(DL + WL)$ with wind suction → wind uplift can reverse the force in the bottom tie, putting it into compression (a length normally designed for tension), so the stress reversal combination must also be checked for buckling/slenderness.

Governing for strength: $1.5(DL+LL)$ for maximum tension; check the $DL+WL$ (uplift) case for possible force reversal.

Section properties

Breadth $b = 100\ \text{mm}$, depth $d = 200\ \text{mm}$.

$$Z = \frac{b\,d^2}{6} = \frac{100 \times 200^2}{6} = \frac{4{,}000{,}000}{6} = 666{,}667\ \text{mm}^3$$ $$A = b\,d = 100 \times 200 = 20{,}000\ \text{mm}^2$$

(1) From bending

Moment capacity:

$$M = \sigma_{b,perm}\,Z = 10 \times 666{,}667 = 6{,}666{,}670\ \text{N·mm} = 6.667\ \text{kN·m}$$

For a simply supported UDL: $M = \dfrac{wL^2}{8}$

$$w = \frac{8M}{L^2} = \frac{8 \times 6.667}{3.0^2} = \frac{53.34}{9} = 5.93\ \text{kN/m}$$

(2) From shear

Max horizontal shear stress in rectangular section: $\tau_{max} = \dfrac{1.5\,V}{A}$

$$V_{allow} = \frac{\tau_{perm}\,A}{1.5} = \frac{0.85 \times 20{,}000}{1.5} = \frac{17{,}000}{1.5} = 11{,}333\ \text{N} = 11.33\ \text{kN}$$

For UDL: $V = \dfrac{wL}{2}$

$$w = \frac{2V}{L} = \frac{2 \times 11.33}{3.0} = \frac{22.66}{3} = 7.55\ \text{kN/m}$$

Governing safe load

$$w_{safe} = \min(5.93,\ 7.55) = \mathbf{5.93\ kN/m}\quad(\text{bending governs})$$

The maximum safe uniformly distributed load is $\approx 5.93\ \text{kN/m}$, limited by bending.

Section classification (IS 800:2007)

When a steel section is loaded in compression/bending, its thin plate elements (flange outstands, web) may buckle locally before the whole section yields. To account for this, IS 800 classifies cross-sections into four classes based on the width-to-thickness ratio ($b/t$, $d/t$) of their elements compared with limiting values (functions of $\varepsilon = \sqrt{250/f_y}$). Classification controls how much of the moment capacity can be used and whether plastic analysis is allowed.

The four classes

Design implication

• Plastic & compact sections → design on plastic modulus $Z_p$; plastic sections additionally allow plastic methods of analysis.
• Semi-compact → design on elastic modulus $Z_e$.
• Slender → use effective (reduced) properties to allow for local buckling.

Thus, more compact (stocky) plate elements allow a greater fraction of the section's strength to be mobilized in bending.