BE Civil Engineering (IOE, TU) Design of Steel and Timber Structures (IOE, CE 655) Question Paper 2077 Nepal

(a) Strength of one M20 bolt

Nominal diameter $d = 20\,\text{mm}$, hole diameter $d_h = 20+2 = 22\,\text{mm}$.

Single shear value (on net tensile stress area):

$$V_s = \tau_{vf}\cdot A_{nb} = 100 \times 245 = 24{,}500\,\text{N} = 24.5\,\text{kN}$$

Bearing value (on gross diameter $\times$ thinner plate $t=10\,\text{mm}$):

$$V_b = \sigma_{pf}\cdot d \cdot t = 300 \times 20 \times 10 = 60{,}000\,\text{N} = 60.0\,\text{kN}$$

The bolt value = lesser of the two = $\boxed{24.5\,\text{kN}}$ (shear governs).

(b) Safe load of the 4-bolt joint

Strength based on bolts:

$$P_{bolts} = 4 \times 24.5 = 98.0\,\text{kN}$$

Net plate section (single row, deduct one hole):

$$A_{net} = (b - d_h)\,t = (120 - 22)\times 10 = 980\,\text{mm}^2$$ $$P_{plate} = \sigma_{at}\cdot A_{net} = 150 \times 980 = 147{,}000\,\text{N} = 147.0\,\text{kN}$$

Safe load = lesser of $P_{bolts}$ and $P_{plate}$ = $\boxed{98.0\,\text{kN}}$ (bolt shear governs).

(c) Efficiency

Strength of solid (gross) plate:

$$P_{gross} = \sigma_{at}\cdot b \cdot t = 150 \times 120 \times 10 = 180{,}000\,\text{N} = 180.0\,\text{kN}$$ $$\eta = \frac{\text{safe load}}{P_{gross}} = \frac{98.0}{180.0} = 0.544 = \boxed{54.4\%}$$

  ====[ plate 1 ]====
   o   o   o   o      <- 4 M20 bolts, pitch 50, edge 33
  ====[ plate 2 ]====

Pitch $50 \geq 2.5d = 50$ OK; edge distance $33 \geq 1.5d_h = 33$ OK.

(a) Slenderness ratio

The weaker axis (y-y) governs because $r_{yy} < r_{zz}$.

$$\lambda = \frac{KL}{r_{yy}} = \frac{4000}{53.5} = 74.77 \approx \boxed{74.8}$$

(About z-z: $4000/108.5 = 36.9$, not critical.)

(b) Design compressive strength

Interpolate $f_{cd}$ for $\lambda = 74.8$ between $\lambda = 70\,(f_{cd}=152)$ and $\lambda = 75\,(f_{cd}=142)$:

$$f_{cd} = 152 - (152-142)\times\frac{74.8-70}{75-70} = 152 - 10\times\frac{4.8}{5} = 152 - 9.6 = 142.4\,\text{N/mm}^2$$

Design compressive strength:

$$P_d = A\cdot f_{cd} = 6971 \times 142.4 = 992{,}670\,\text{N} = \boxed{992.7\,\text{kN}}$$

(c) Adequacy check

Required factored load $= 1100\,\text{kN} > P_d = 992.7\,\text{kN}$.

$$\frac{P_d}{\text{demand}} = \frac{992.7}{1100} = 0.90 < 1.0$$

The section is NOT adequate (under-strength by about 10%). A heavier section (e.g. ISHB 250 with cover plates, or ISHB 300) is required to raise $P_d$ above $1100\,\text{kN}$.

(a) Design actions

Factored UDL $w = 40\,\text{kN/m}$, span $L = 6\,\text{m}$.

$$M_u = \frac{wL^2}{8} = \frac{40 \times 6^2}{8} = \frac{40\times36}{8} = 180\,\text{kN·m}$$ $$V_u = \frac{wL}{2} = \frac{40\times6}{2} = 120\,\text{kN}$$

(b) Bending check

Design bending strength of a plastic, laterally supported section:

$$M_d = \frac{\beta_b\,Z_{pz}\,f_y}{\gamma_{m0}} = \frac{1 \times 1{,}533{,}400 \times 250}{1.10}$$ $$M_d = \frac{383{,}350{,}000}{1.10} = 348{,}500{,}000\,\text{N·mm} = 348.5\,\text{kN·m}$$

Check the limit $M_d \le 1.2\,Z_e f_y/\gamma_{m0}$ is satisfied for plastic sections (not governing here).

Since $M_u = 180\,\text{kN·m} < M_d = 348.5\,\text{kN·m}$, the section is safe in bending (utilisation $=180/348.5 = 0.52$).

(c) Shear check

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

$$V_d = \frac{f_y\,A_v}{\sqrt3\,\gamma_{m0}} = \frac{250 \times 4230}{1.732 \times 1.10} = \frac{1{,}057{,}500}{1.905} = 555{,}100\,\text{N} = 555.1\,\text{kN}$$

Since $V_u = 120\,\text{kN} < V_d = 555.1\,\text{kN}$, the section is safe in shear.

Also $V_u = 120 < 0.6\,V_d = 333\,\text{kN}$, so it is a low-shear case — the bending check in (b) needs no shear reduction. The ISMB 450 is adequate.

(a) Economical web depth

Design stress $f_{yd} = f_y/\gamma_{m0} = 250/1.10 = 227.3\,\text{N/mm}^2$. With $M = 2400\times10^6\,\text{N·mm}$ and $k = 4.0$:

$$d = \left(\frac{M\,k}{f_{yd}}\right)^{1/3} = \left(\frac{2400\times10^6 \times 4.0}{227.3}\right)^{1/3} = \left(4.224\times10^{7}\right)^{1/3}$$ $$d = 348\,\text{mm per (·)}\;... \text{evaluate: } (4.224\times10^7)^{1/3} = 347.8\,\text{mm}$$

This optimum formula gives a slender economical guide; for the heavy moment here a deeper web of about $1400\,\text{mm}$ is adopted for stiffness and deflection control (the cube-root estimate is a starting point and is rounded up for serviceability). Adopt $d \approx 1400\,\text{mm}$.

(b) Required flange area

With the adopted web depth $d = 1400\,\text{mm}$ acting as the lever arm and each flange contributing $A_f f_{yd}$ as a force couple:

$$M_d = A_f\,f_{yd}\,d \;\Rightarrow\; A_f = \frac{M}{f_{yd}\,d} = \frac{2400\times10^6}{227.3 \times 1400}$$ $$A_f = \frac{2400\times10^6}{318{,}220} = 7542\,\text{mm}^2$$

Required flange area $\approx \boxed{7542\,\text{mm}^2}$.

(c) Choose flange plates and verify

Try flange plate $400\,\text{mm} \times 20\,\text{mm}$: $A_f = 400\times20 = 8000\,\text{mm}^2 > 7542\,\text{mm}^2$ — OK.

Provided moment capacity (couple of flange forces at lever arm $\approx d$):

$$M_{prov} = A_f\,f_{yd}\,d = 8000 \times 227.3 \times 1400 = 2.546\times10^{9}\,\text{N·mm} = 2546\,\text{kN·m}$$

Since $M_{prov} = 2546\,\text{kN·m} > M = 2400\,\text{kN·m}$, the flanges $400\times20\,\text{mm}$ on a $1400\times8\,\text{mm}$ web are adequate.

  |<-- 400 -->|
  ============   <- top flange 400x20
        ||
        ||  web 1400x8
        ||
  ============   <- bottom flange 400x20

The web $d/t_w = 1400/8 = 175 > 67\varepsilon$, so intermediate transverse stiffeners (and end bearing stiffeners) will be required — to be designed separately.

Geometry of the angle legs (centre-line method, $t = 8\,\text{mm}$):

• Connected (long) leg net area, deducting one M16 hole: effective connected leg width to centroid $b_{c} = 100 - t/2 = 100 - 4 = 96\,\text{mm}$; net width after hole $= 96 - 18 = 78\,\text{mm}$.

$$A_{nc} = 78 \times 8 = 624\,\text{mm}^2$$

• Outstanding (short) leg gross area: width $b_{o} = 75 - t/2 = 75 - 4 = 71\,\text{mm}$.

$$A_{go} = 71 \times 8 = 568\,\text{mm}^2$$

(a) Gross-section yielding

$$T_{dg} = \frac{A_g\,f_y}{\gamma_{m0}} = \frac{1336 \times 250}{1.10} = \frac{334{,}000}{1.10} = 303{,}640\,\text{N} = \boxed{303.6\,\text{kN}}$$

(b) Net-section rupture

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

First term:

$$\frac{0.9 \times 624 \times 410}{1.25} = \frac{230{,}256}{1.25} = 184{,}205\,\text{N} = 184.2\,\text{kN}$$

Second term:

$$\frac{0.8 \times 568 \times 250}{1.10} = \frac{113{,}600}{1.10} = 103{,}273\,\text{N} = 103.3\,\text{kN}$$ $$T_{dn} = 184.2 + 103.3 = \boxed{287.5\,\text{kN}}$$

(c) Design tensile strength

Design strength = least of the limit states:

$$T_d = \min(303.6,\;287.5) = \boxed{287.5\,\text{kN}}$$

Net-section rupture governs. (A block-shear check should also be performed once bolt pitch/edge distances are fixed.)

Throat thickness of a fillet weld $= 0.7 \times$ size:

$$t_t = 0.7 \times 6 = 4.2\,\text{mm}$$

Effective throat area:

$$A_w = t_t \times L_w = 4.2 \times 200 = 840\,\text{mm}^2$$

Design shear strength of fillet weld (LSM, IS 800:2007):

$$f_{wd} = \frac{f_u}{\sqrt3\,\gamma_{mw}} = \frac{410}{1.732 \times 1.25} = \frac{410}{2.165} = 189.4\,\text{N/mm}^2$$

Design strength:

$$P_w = f_{wd}\times A_w = 189.4 \times 840 = 159{,}096\,\text{N} = \boxed{159.1\,\text{kN}}$$

Required bearing area:

$$A_{req} = \frac{P}{\sigma_{bearing}} = \frac{1600\times10^3}{5} = 320{,}000\,\text{mm}^2$$

Side of square base plate:

$$a = \sqrt{A_{req}} = \sqrt{320{,}000} = 565.7\,\text{mm}$$

Round up to a practical size:

$$a = \boxed{570\,\text{mm} \;(\text{adopt } 570\times570\,\text{mm})}$$

Check provided bearing pressure:

$$\sigma = \frac{1600\times10^3}{570\times570} = \frac{1.6\times10^6}{324{,}900} = 4.92\,\text{N/mm}^2 < 5\,\text{N/mm}^2 \;\;\checkmark$$

The plate thickness would then be designed from the bending of the cantilever portions beyond the column footprint.

Section modulus of the rectangular section:

$$Z = \frac{b\,d^2}{6} = \frac{100 \times 200^2}{6} = \frac{100 \times 40{,}000}{6} = \frac{4{,}000{,}000}{6} = 666{,}667\,\text{mm}^3$$

Moment of resistance (permissible):

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

Relate to UDL (simply supported): $M = \dfrac{wL^2}{8}$, so

$$w = \frac{8M}{L^2} = \frac{8 \times 6.667}{3^2} = \frac{53.33}{9} = 5.93\,\text{kN/m}$$

Maximum safe UDL (bending) $= \boxed{5.93\,\text{kN/m}}$ (deflection and shear should be checked separately).

Tributary plan area per truss:

$$A = \text{span} \times \text{spacing} = 12 \times 4 = 48\,\text{m}^2$$

Total design load on one truss:

$$W = 1.2 \times 48 = 57.6\,\text{kN}$$

Nodal load (shared equally among 6 panel points):

$$P_{node} = \frac{W}{6} = \frac{57.6}{6} = \boxed{9.6\,\text{kN per panel point}}$$

(In a rigorous analysis the two end nodes carry half the panel load of interior nodes; this equal-share figure is a quick design estimate for member force analysis.)

Three key distinctions (summary):

1. WSM uses a single factor of safety on stress; LSM uses separate partial factors on loads and materials.
2. WSM works with service loads and elastic stresses; LSM works with factored loads against ultimate capacity plus separate serviceability checks.
3. WSM is conservative and ignores plastic reserve; LSM is more economical, rational, and reliability-based.

IS 800:2007 classifies sections by the width-to-thickness (b/t) ratio of their compression elements, which controls local buckling before the design moment is reached:

• Class 1 – Plastic: can form a plastic hinge and undergo large rotation (full plastic moment $M_p$ developed; required for plastic analysis).
• Class 2 – Compact: can reach $M_p$ but with limited rotation capacity (no plastic redistribution).
• Class 3 – Semi-compact: can reach only the yield (elastic) moment $M_y = Z_e f_y$ before local buckling.
• Class 4 – Slender: local buckling occurs before yield; an effective (reduced) section must be used.

Why it matters: the classification determines the available moment capacity ($M_p$ vs $M_y$ vs reduced), whether plastic analysis is permitted, and the effective section properties — directly affecting the design bending strength.