Design of Roof Trusses

Assumptions (data not given in the question). $V_b$ is not stated; take 47 m/s (Kathmandu region, IS 875 Part 3 Fig. 1). The building is in a core city, so terrain Category 4 (large city centre with obstructions). Height 36 m, so the greatest dimension is 36 m and the structure is Class B (20 m to 50 m). Design life 50 years gives $k_1 = 1.0$. The roof truss is taken on a plan width $w = 30$ m with a low roof pitch of $\alpha = 10^\circ$, truss spacing 6 m, normal permeability. $k_3 = 1$ (flat site).

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 47$ m/s.
• Risk coefficient $k_1 = 1.00$ (design life 50 years, Table 1; Table 1).
• Terrain, height and size factor $k_2$: Category 4, Class B, at height 36 m: interpolating between 30 m (0.93) and 50 m (1.05): $k_2 = 0.966$ (Table 2).
• Topography factor $k_3 = 1.000$ (flat site, slope less than 3°).

$$\begin{aligned} V_z &= 47\times1.00\times0.966\times1.000 = 45.40\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times45.40^2 = 1237\ \text{N/m}^2 = 1.237\ \text{kN/m}^2 \end{aligned}$$

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 1.20$, roof angle $\alpha = 10.0^\circ$. Values are interpolated between $\alpha = 5^\circ$ and $10^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3.

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Windward slope EF (wind 0°), with $C_{pe} - C_{pi} = -1.30$, so $p_d = -1608$ N/m² $= -1.608$ kN/m².

Step 3: Wind force on the roof / truss

$F = (C_{pe} - C_{pi})\,A\,p_z$ acting normal to the roof surface. The most severe net coefficient $-1.30$ is used for the maximum force.

• One slope of a truss (span 30 m, slope length 15.23 m, spacing 6 m): $A = 91.4$ m², $F = 1.30\times91.4\times1.237 = 147.0$ kN (suction, i.e. acting away from the roof).

Answer

• Design wind speed $V_z = 45.4$ m/s; design wind pressure $p_z = 1237$ N/m² (1.24 kN/m²).
• Net design pressure on the roof: up to $1608$ N/m² (1.61 kN/m²), suction.
• One slope of a truss (span 30 m, slope length 15.23 m, spacing 6 m): 147.0 kN.

Data. Span 10 m, pitch $1/4$ (rise/span), so rise $= 2.5$ m and $\tan\alpha = 2.5/5 = 0.5$, $\alpha = 26.57^\circ$. Eaves height $h = 5$ m; take the lesser plan dimension $w = 10$ m, so $h/w = 0.5$. Delhi: $V_b = 47$ m/s (IS 875 Part 3, Appendix A). Given: $k_1 = 1$, $k_2 = 0.8$, $k_3 = 1$. Permeability normal.

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 47$ m/s.
• Risk coefficient $k_1 = 1.00$ (given; Table 1).
• Terrain, height and size factor $k_2$: at height eaves: given, $k_2 = 0.8$ (Table 2).
• Topography factor $k_3 = 1.000$ (given).

$$\begin{aligned} V_z &= 47\times1.00\times0.800\times1.000 = 37.60\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times37.60^2 = 848\ \text{N/m}^2 = 0.848\ \text{kN/m}^2 \end{aligned}$$

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 0.50$, roof angle $\alpha = 26.6^\circ$. Values are interpolated between $\alpha = 20^\circ$ and $30^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3.

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Slope EG (wind 90°), with $C_{pe} - C_{pi} = -0.90$, so $p_d = -763$ N/m² $= -0.763$ kN/m².

Answer (design wind pressure)

• Design wind speed $V_z = 37.6$ m/s; design wind pressure $p_z = 848$ N/m² (0.85 kN/m²).
• Net design pressure on the roof: up to $763$ N/m² (0.76 kN/m²), suction.

Data and assumptions. Span 10 m, pitch $1/4$: rise 2.5 m, $\alpha = 26.57^\circ$. Eaves height $h = 6$ m, $w = 10$ m, $h/w = 0.6$. Madras: $V_b = 50$ m/s (IS 875 Part 3, Appendix A). Terrain Category 2, Class A (building below 20 m); mean roof height $= 7.25$ m, below 10 m. $k_1 = 1$, $k_3 = 1$ (not given). Normal permeability: $C_{pi} = \pm0.2$.

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 50$ m/s.
• Risk coefficient $k_1 = 1.00$ (50-year life; Table 1).
• Terrain, height and size factor $k_2$: Category 2, Class A, at height 7.25 m: $k_2 = 1.00$ (constant up to 10 m) (Table 2).
• Topography factor $k_3 = 1.000$ (flat site).

$$\begin{aligned} V_z &= 50\times1.00\times1.000\times1.000 = 50.00\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times50.00^2 = 1500\ \text{N/m}^2 = 1.500\ \text{kN/m}^2 \end{aligned}$$

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 0.60$, roof angle $\alpha = 26.6^\circ$. Values are interpolated between $\alpha = 20^\circ$ and $30^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3.

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Slope EG (wind 90°), with $C_{pe} - C_{pi} = -1.00$, so $p_d = -1500$ N/m² $= -1.500$ kN/m².

Answer

• Design wind speed $V_z = 50.0$ m/s; design wind pressure $p_z = 1500$ N/m² (1.50 kN/m²).
• Net design pressure on the roof: up to $1500$ N/m² (1.50 kN/m²), suction.

Wind load on a sloping roof truss is calculated by IS 875 (Part 3):1987 as follows.

Steps

1. Basic wind speed $V_b$ for the site from the wind map (Fig. 1) or Appendix A (3-second gust at 10 m height, Category 2, 50-year return period).
2. Design wind speed at the roof height: $V_z = V_b\,k_1k_2k_3$.
   • $k_1$ risk factor from the design life (Table 1; 1.0 for 50 years).
   • $k_2$ terrain, height and structure-size factor (Table 2): choose terrain category 1 to 4 and building class A (below 20 m), B (20 to 50 m) or C (above 50 m) for the height of the roof.
   • $k_3$ topography factor (1.0 for flat sites; up to 1.36 for hills and ridges, Appendix C).
3. Design wind pressure: $p_z = 0.6V_z^2$ (N/m², $V_z$ in m/s).
4. Pressure coefficients for the roof from Table 5 (pitched roofs of rectangular clad buildings): external coefficients $C_{pe}$ for wind angles 0° and 90°, depending on the roof angle $\alpha$ and height ratio $h/w$ (interpolate between roof angles). Internal coefficient $C_{pi} = \pm0.2$ for normal permeability (openings up to 5%), $\pm0.5$ for medium openings (5 to 20%), $\pm0.7$ for large openings.
5. Net pressure on the roof surface: $p_d = (C_{pe} - C_{pi})\,p_z$. Take the combination ($C_{pi}$ positive or negative) giving the more severe effect.
6. Wind force on a roof surface (or truss): $F = (C_{pe} - C_{pi})\,A\,p_z$, where $A$ is the surface area (slope length × truss spacing for one truss). The force acts normal to the roof surface.
7. Joint loads: divide the total normal force on a slope among the purlin positions (panel points of the rafter), the end joints getting half.
8. Combine with dead and live loads using the limit state load combinations (IS 800 Table 4) and take the worst case, including uplift (suction) where the net wind load is upward.

   wind ->     /\   windward slope: Cpe (EF)
              /  \  leeward slope:  Cpe (GH)
      pressure/    \suction  (net p = (Cpe - Cpi) pz)

Assumptions. $V_b = 47$ m/s (given). Kathmandu high-rise: terrain Category 4 (city centre). Height 50 m, so the greatest dimension is 50 m, which is not greater than 50 m: Class B. Life 50 years: $k_1 = 1.0$; flat site: $k_3 = 1$. Plan 40 m × 30 m, so $w = 30$ m and $h/w = 50/30 = 1.67$. The roof truss spans the 30 m width with a low slope $\alpha = 10^\circ$, truss spacing 5 m, normal permeability.

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 47$ m/s.
• Risk coefficient $k_1 = 1.00$ (design life 50 years, Table 1; Table 1).
• Terrain, height and size factor $k_2$: Category 4, Class B, at height 50 m: interpolating between 30 m (0.93) and 50 m (1.05): $k_2 = 1.050$ (Table 2).
• Topography factor $k_3 = 1.000$ (flat site).

$$\begin{aligned} V_z &= 47\times1.00\times1.050\times1.000 = 49.35\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times49.35^2 = 1461\ \text{N/m}^2 = 1.461\ \text{kN/m}^2 \end{aligned}$$

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 1.67$, roof angle $\alpha = 10.0^\circ$. Values are interpolated between $\alpha = 5^\circ$ and $10^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3.

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Slope EG (wind 90°), with $C_{pe} - C_{pi} = -1.00$, so $p_d = -1461$ N/m² $= -1.461$ kN/m².

Step 3: Wind force on the roof / truss

$F = (C_{pe} - C_{pi})\,A\,p_z$ acting normal to the roof surface. The most severe net coefficient $-1.00$ is used for the maximum force.

• One slope of one truss (slope length 15.23 m, spacing 5 m): $A = 76.2$ m², $F = 1.00\times76.2\times1.461 = 111.3$ kN (suction, i.e. acting away from the roof).

• Whole truss (both slopes): $A = 152.3$ m², $F = 1.00\times152.3\times1.461 = 222.5$ kN (suction, i.e. acting away from the roof).

Answer

• Design wind speed $V_z = 49.4$ m/s; design wind pressure $p_z = 1461$ N/m² (1.46 kN/m²).
• Net design pressure on the roof: up to $1461$ N/m² (1.46 kN/m²), suction.
• One slope of one truss (slope length 15.23 m, spacing 5 m): 111.3 kN.
• Whole truss (both slopes): 222.5 kN.

A purlin is a beam spanning between roof trusses that carries the roof sheeting and transfers the loads to the truss panel points. Its design is done as a beam under bi-axial bending (IS 800:2007).

Procedure

1. Spacing and span. Span $L$ = truss spacing (usually 4 to 6 m). Purlin spacing $s$ along the rafter is fixed by the roofing sheet (e.g. 1.2 to 1.8 m for GI/AC sheets) so that purlins sit at the truss panel points.
2. Loads per metre of purlin.
   • Dead load: weight of sheeting $\times s$ + self-weight of the purlin (assumed, then revised).
   • Live load (IS 875 Part 2): $0.75 - 0.02(\alpha - 10)$ kN/m² on plan for roof slopes above 10° (minimum 0.4 kN/m²), applied on the plan projection $s\cos\alpha$.
   • Wind load (IS 875 Part 3): $p_d = (C_{pe} - C_{pi})p_z$ acting normal to the roof, $\times s$.
3. Resolve vertical loads into components normal to the roof ($w\cos\alpha$) and parallel to the roof ($w\sin\alpha$). Wind acts normal to the roof.
4. Load combinations (IS 800 Table 4): $1.5(DL + LL)$, $1.2(DL + LL + WL)$, $1.5(DL + WL)$, and uplift $0.9DL + 1.5WL$ (check the reversal of stress).
5. Bending moments. $M_z = w_nL^2/8$ (simply supported; $wL^2/10$ if continuous). $M_y = w_t\,(L')^2/8$ about the minor axis, where $L'$ is the span between sag rods (a sag rod at mid-span gives $L' = L/2$).
6. Select a trial section (ISMB, ISLB, channel or angle) from $Z_{pz,req} = M_z\gamma_{m0}/f_y$ (add about 20 to 30% for the minor-axis moment).
7. Classification and capacity. $M_{dz} = Z_{pz}f_y/\gamma_{m0}$ and $M_{dy} = Z_{py}f_y/\gamma_{m0}$ (each not more than $1.2Z_ef_y/\gamma_{m0}$); check lateral-torsional buckling for the uplift case when the bottom flange is in compression.
8. Interaction (biaxial bending). $\dfrac{M_z}{M_{dz}} + \dfrac{M_y}{M_{dy}} \le 1$ (cl. 9.3.1, simplified).
9. Shear $V \le V_d = A_vf_y/(\sqrt3\gamma_{m0})$.
10. Deflection under unfactored loads, limit span/150 for elastic (GI) sheeting and span/180 for brittle (AC) sheeting (Table 6).
11. Sag rods and connections. Provide sag rods (12 to 16 mm) at mid-span or third points to carry the tangential component, and cleat angles (with two bolts) to connect the purlin to the rafter; check the bolts for shear.

   truss           truss
    |<----- L ----->|
    |==== purlin ====|   sag rod at mid-span

Assumptions (not all data are given). $V_b$ is not stated; 47 m/s is assumed. Dense forest with closely spaced trees is terrain Category 3. Greatest dimension $= 30$ m, so Class B. Design life taken as 50 years ($k_1 = 1.0$). Roof: width $w = 10$ m, rise 3.5 m so $\tan\alpha = 3.5/5 = 0.7$, $\alpha = 35.0^\circ$; $h/w = 10/10 = 1.0$. Height for $k_2$ = mean roof height $= 10 + 3.5/2 = 11.75$ m. Normal openings: $C_{pi} = \pm0.2$.

Topography factor $k_3$ (Appendix C). Hill slope 1:4, so $\theta = \tan^{-1}(1/4) = 14.0^\circ$, which is between 3° and 17°. Height of hill $Z = 450$ m, so the upwind slope length $L = Z/\tan\theta = 1800$ m and $L_e = L = 1800$ m. $C = 1.2Z/L = 1.2\times450/1800 = 0.30$. The building is at mid-height, i.e. $X = -L/2 = -900$ m (upwind of the crest), $X/L_e = -0.5$, and $H/L_e \approx 12/1800 \approx 0.007$. From Fig. 15 (hill and ridge) $s \approx 0.15$ (read from the chart, approximate). So $k_3 = 1 + Cs = 1 + 0.30\times0.15 = 1.045$.

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 47$ m/s.
• Risk coefficient $k_1 = 1.00$ (50-year life; Table 1).
• Terrain, height and size factor $k_2$: Category 3, Class B, at height 11.75 m: interpolating between 10 m (0.88) and 15 m (0.94): $k_2 = 0.901$ (Table 2).
• Topography factor $k_3 = 1.045$ (hill, Appendix C).

$$\begin{aligned} V_z &= 47\times1.00\times0.901\times1.045 = 44.25\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times44.25^2 = 1175\ \text{N/m}^2 = 1.175\ \text{kN/m}^2 \end{aligned}$$

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 1.00$, roof angle $\alpha = 35.0^\circ$. Values are interpolated between $\alpha = 30^\circ$ and $45^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal opening), cl. 6.2.3.

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Slope EG (wind 90°), with $C_{pe} - C_{pi} = -1.00$, so $p_d = -1175$ N/m² $= -1.175$ kN/m².

Step 3: Wind force on the roof / truss

$F = (C_{pe} - C_{pi})\,A\,p_z$ acting normal to the roof surface. The most severe net coefficient $-1.00$ is used for the maximum force.

• Windward slope (slope length 6.10 m × 30 m): $A = 183.1$ m², $F = 1.00\times183.1\times1.175 = 215.1$ kN (suction, i.e. acting away from the roof).

Answer

• Design wind speed $V_z = 44.3$ m/s; design wind pressure $p_z = 1175$ N/m² (1.17 kN/m²).
• Net design pressure on the roof: up to $1175$ N/m² (1.17 kN/m²), suction.
• Windward slope (slope length 6.10 m × 30 m): 215.1 kN.

Assumptions (not all data are given). Basic wind speed is not stated, so $V_b = 47$ m/s is assumed. Building: dimension over 50 m, so Class C; height 15 m. Terrain: open with scattered obstructions 1.5 to 10 m high, so Category 2. Design life 100 years. Large openings (more than 20% of wall area).

Step 1: Design wind speed

• $k_1$: for a life of 100 years and $V_b = 47$ m/s, Table 1 gives $k_1 = 1.07$.
• $k_2$: Category 2, Class C, at 15 m: interpolating between 10 m (0.93) and 15 m (0.97): $k_2 = 0.970$, i.e. $0.97$ (Table 2).
• $k_3$ (Appendix C): hill height $Z = 160$ m, slope 1 vertical to 3 horizontal, $\theta = \tan^{-1}(1/3) = 18.4^\circ > 17^\circ$, so $C = 0.36$ and $L_e = Z/0.3 = 533$ m. The building is on the downward (leeward) slope at 100 m above the base, i.e. 60 m below the crest, so the horizontal distance from the crest is $X = 60\times3 = 180$ m downwind: $X/L_e = 0.337$; $H/L_e = 15/533 = 0.028$. From Fig. 15, $s \approx 0.65$ (read from the chart; approximate). $k_3 = 1 + Cs = 1 + 0.36\times0.65 = 1.234$ (within the limit 1.36).

$$\begin{aligned} V_z &= V_bk_1k_2k_3 = 47\times1.07\times0.97\times1.234 = 60.2\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times60.2^2 = 2174\ \text{N/m}^2 = 2.17\ \text{kN/m}^2 \end{aligned}$$

Step 2: Net pressure with large openings

For openings above 20% of the wall area, $C_{pi} = \pm0.7$ (cl. 6.2.3.2). Taking the external coefficients of a rectangular clad building (Table 4, $h/w < 1/2$, wind 0°): windward wall $C_{pe} = +0.7$, leeward wall $-0.5$.

Answer

Design wind speed $V_z = 60.2$ m/s; design wind pressure $p_z \approx 2174$ N/m² ($\approx 2.17$ kN/m²). The largest net pressure on the walls is $1.4p_z \approx 3.04$ kN/m² (inward on the windward wall with the internal suction case) and the largest net suction is $1.2p_z \approx 2.61$ kN/m².

Approach. The purlin is a simply supported beam subjected to bending about both axes because the loads are inclined to the principal axes of the purlin. Design to IS 800:2007 (cl. 8.2 and 9.3), loads from IS 875.

Data

• Steel Fe 410 (E250): $f_y = 250$ N/mm², $\gamma_{m0} = 1.10$. Purlin span $L = 5.0$ m (truss spacing); inclination of rafter $\alpha = 30.00^\circ$ ($\cos\alpha = 0.866$, $\sin\alpha = 0.500$).
• Sag rods are provided at mid-span to restrain the purlin in the plane of the roof. The 1 kN/m² wind load is taken as the net design pressure normal to the roof. Spacing of purlins along the slope = 1.5 m, trusses at 5 m.

Loads per metre run of purlin

• Sheeting: $0.130\times1.5 = 0.195$ kN/m (dead, vertical), plus the self-weight of the purlin (trial section below).
• Live load (IS 875 Part 2, roof slope above 10°): $0.75 - 0.02(\alpha - 10) = 0.35$ kN/m², minimum 0.40 kN/m², on plan: use $0.40$ kN/m². Per metre of purlin: $0.40\times1.5\cos\alpha = 0.520$ kN/m (vertical).
• Wind: given wind pressure $1.0$ kN/m² $\times 1.5$ m $= 1.500$ kN/m, acting normal to the roof.

Trial section and capacities

Try ISMB 125: $D = 125$, $b_f = 70$, $t_w = 5.0$, $t_f = 8.0$ mm, $I_{zz} = 445$ cm⁴, $I_{yy} = 38.5$ cm⁴, $Z_{ez} = 71.2$ cm³, mass 13.3 kg/m (self-weight $0.130$ kN/m).

$$\begin{aligned} Z_{pz} &= b_ft_f(D - t_f) + t_w(D - 2t_f)^2/4 = 80.4\ \text{cm}^3,\quad M_{dz} = \frac{Z_{pz}f_y}{\gamma_{m0}} = 18.27\ \text{kN·m} \\ Z_{py} &= 2\cdot\frac{t_fb_f^2}{4} + \frac{(D - 2t_f)t_w^2}{4} = 20.3\ \text{cm}^3,\quad M_{dy} = \min(Z_{py}, 1.2Z_{ey})\frac{f_y}{\gamma_{m0}} = 3.00\ \text{kN·m} \end{aligned}$$

Section class: $b/t_f$ and $d/t_w$ are within the plastic limits ($9.4$ and $84$).

Load combinations and moments (IS 800 Table 4)

Normal component $w_n$ and tangential component $w_t$ (kN/m), $M_z = w_nL^2/8$ and $M_y = w_t(L')^2/8$ with $L' = L/2 (one sag rod at mid-span)$:

For the uplift case the bottom flange is in compression and unrestrained between sag rods ($L_{LT} = L/2$), so the lateral-torsional buckling capacity is used for $M_{dz}$. The greatest ratio is 0.59 $\le 1.0$. Safe in bi-axial bending.

Shear: $V_{max} = 6.70$ kN $\ll V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 82.0$ kN. OK.

Deflection (unfactored, limit span/150 for elastic sheeting, Table 6): $\delta = 5wL^4/(384EI) = 13.72$ mm $< L/150 = 33.3$ mm. OK.

Result

Provide ISMB 125 purlin, simply supported on the rafters with a sag rod (12 mm) at mid-span and cleat connections with two M12 bolts at each end.

Assumptions (data not fully given). Kathmandu: $V_b = 47$ m/s. "Highly dense city" is terrain Category 4. Plan 20 m × 30 m, so the largest dimension is 30 m: Class B. Life 50 years: $k_1 = 1.0$. Roof: span 20 m, rise 2 m, so $\tan\alpha = 2/10$, $\alpha = 11.3^\circ$ and $h/w = 5/20 = 0.25$. Normal permeability ($C_{pi} = \pm0.2$). Height for $k_2$: eaves 5 m plus half the rise = 6 m, which is below 10 m, so $k_2$ is the value at 10 m. The site is 15 m above the mean ground level and 100 m from the crest of a ridge 200 m high. The slope of the ridge is not given and is assumed steeper than 17°, so (Appendix C) $C = 0.36$ and $L_e = Z/0.3 = 667$ m. The site is upwind of the crest: $X/L_e = -100/667 = -0.15$; $H/L_e = 15/667 = 0.022$; from Fig. 15, $s \approx 0.7$ (read from the chart, approximate). Truss spacing is not given; 5 m is assumed (6 bays along 30 m).

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 47$ m/s.
• Risk coefficient $k_1 = 1.00$ (50-year life; Table 1).
• Terrain, height and size factor $k_2$: Category 4, Class B, at height 6 m (use 10 m): $k_2 = 0.76$ (constant up to 10 m) (Table 2).
• Topography factor $k_3 = 1.252$ (ridge, $C = 0.36$, $s = 0.7$).

$$\begin{aligned} V_z &= 47\times1.00\times0.760\times1.252 = 44.72\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times44.72^2 = 1200\ \text{N/m}^2 = 1.200\ \text{kN/m}^2 \end{aligned}$$

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 0.25$, roof angle $\alpha = 11.3^\circ$. Values are interpolated between $\alpha = 10^\circ$ and $20^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3.

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Windward slope EF (wind 0°), with $C_{pe} - C_{pi} = -1.30$, so $p_d = -1554$ N/m² $= -1.554$ kN/m².

Step 3: Wind force on the roof / truss

$F = (C_{pe} - C_{pi})\,A\,p_z$ acting normal to the roof surface. The most severe net coefficient $-1.30$ is used for the maximum force.

• Whole roof, 20 m × 30 m plan (slope length 10.20 m, both slopes): $A = 611.9$ m², $F = 1.30\times611.9\times1.200 = 951.0$ kN (suction, i.e. acting away from the roof).

• One truss at 5 m spacing (both slopes): $A = 102$ m², $F = 1.30\times102\times1.200 = 158.5$ kN (suction, i.e. acting away from the roof).

Answer

• Design wind speed $V_z = 44.7$ m/s; design wind pressure $p_z = 1200$ N/m² (1.20 kN/m²).
• Net design pressure on the roof: up to $1554$ N/m² (1.55 kN/m²), suction.
• Whole roof, 20 m × 30 m plan (slope length 10.20 m, both slopes): 951.0 kN.
• One truss at 5 m spacing (both slopes): 158.5 kN.

Assumptions (data not fully given). $V_b = 47$ m/s (given), terrain open industrial area: Category 2 ("open terrain with scattered obstructions"); Class A (as stated). Life 50 years: $k_1 = 1.0$; flat site: $k_3 = 1$. Roof slope is not given, so $\alpha = 20^\circ$ is assumed (rise $= 20\tan20^\circ = 7.3$ m). Eaves height 10 m, so $h/w = 10/40 = 0.25$; mean roof height $= 10 + 7.3/2 = 13.6$ m. Normal permeability ($C_{pi} = \pm0.2$); GI sheeting. Truss spacing assumed 5 m.

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 47$ m/s.
• Risk coefficient $k_1 = 1.00$ (50-year life; Table 1).
• Terrain, height and size factor $k_2$: Category 2, Class A, at height 13.6 m: interpolating between 10 m (1.00) and 15 m (1.05): $k_2 = 1.036$ (Table 2).
• Topography factor $k_3 = 1.000$ (flat site).

$$\begin{aligned} V_z &= 47\times1.00\times1.036\times1.000 = 48.71\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times48.71^2 = 1424\ \text{N/m}^2 = 1.424\ \text{kN/m}^2 \end{aligned}$$

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 0.25$, roof angle $\alpha = 20.0^\circ$. Values are interpolated between $\alpha = 10^\circ$ and $20^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3.

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Slope EG (wind 90°), with $C_{pe} - C_{pi} = -0.90$, so $p_d = -1281$ N/m² $= -1.281$ kN/m².

Step 3: Wind force on the roof / truss

$F = (C_{pe} - C_{pi})\,A\,p_z$ acting normal to the roof surface. The most severe net coefficient $-0.90$ is used for the maximum force.

• One truss (both slopes, slope length 21.28 m each, spacing 5 m): $A = 212.8$ m², $F = 0.90\times212.8\times1.424 = 272.7$ kN (suction, i.e. acting away from the roof).

• Whole roof, 40 m × 100 m plan: $A = 4256.7$ m², $F = 0.90\times4256.7\times1.424 = 5454.0$ kN (suction, i.e. acting away from the roof).

Answer

• Design wind speed $V_z = 48.7$ m/s; design wind pressure $p_z = 1424$ N/m² (1.42 kN/m²).
• Net design pressure on the roof: up to $1281$ N/m² (1.28 kN/m²), suction.
• One truss (both slopes, slope length 21.28 m each, spacing 5 m): 272.7 kN.
• Whole roof, 40 m × 100 m plan: 5454.0 kN.

Assumptions. Span 10 m, pitch $1/4$, so rise 2.5 m, $\tan\alpha = 0.5$, $\alpha = 26.57^\circ$. Eaves at 9 m: $h/w = 9/10 = 0.9$; mean roof height $9 + 1.25 = 10.25$ m. Kathmandu: $V_b = 47$ m/s (given). Terrain is not stated: Category 2 (open terrain), Class A (span below 20 m); $k_1 = 1$, $k_3 = 1$. Permeability normal: $C_{pi} = \pm0.2$.

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 47$ m/s.
• Risk coefficient $k_1 = 1.00$ (50-year life; Table 1).
• Terrain, height and size factor $k_2$: Category 2, Class A, at height 10.25 m: interpolating between 10 m (1.00) and 15 m (1.05): $k_2 = 1.002$ (Table 2).
• Topography factor $k_3 = 1.000$ (flat site).

$$\begin{aligned} V_z &= 47\times1.00\times1.002\times1.000 = 47.12\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times47.12^2 = 1332\ \text{N/m}^2 = 1.332\ \text{kN/m}^2 \end{aligned}$$

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 0.90$, roof angle $\alpha = 26.6^\circ$. Values are interpolated between $\alpha = 20^\circ$ and $30^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3.

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Slope EG (wind 90°), with $C_{pe} - C_{pi} = -1.00$, so $p_d = -1332$ N/m² $= -1.332$ kN/m².

Answer

• Design wind speed $V_z = 47.1$ m/s; design wind pressure $p_z = 1332$ N/m² (1.33 kN/m²).
• Net design pressure on the roof: up to $1332$ N/m² (1.33 kN/m²), suction.

Assumptions. Span 20 m, roof slope $30^\circ$ (rise $= 10\tan30^\circ = 5.77$ m). Height "of the truss roof above the support" is taken as the eaves height $h = 10$ m, so $h/w = 10/20 = 0.5$ and mean roof height $= 10 + 5.77/2 = 12.9$ m. Kathmandu: $V_b = 47$ m/s; town terrain with closely spaced obstructions: Category 3; greatest dimension 20 m, so Class B (20 m to 50 m) is taken. Life 50 years ($k_1 = 1$), flat site ($k_3 = 1$), normal permeability.

Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$V_z = V_b\,k_1k_2k_3$$

• Basic wind speed $V_b = 47$ m/s.
• Risk coefficient $k_1 = 1.00$ (50-year life; Table 1).
• Terrain, height and size factor $k_2$: Category 3, Class B, at height 12.9 m: interpolating between 10 m (0.88) and 15 m (0.94): $k_2 = 0.915$ (Table 2).
• Topography factor $k_3 = 1.000$ (flat site).

$$\begin{aligned} V_z &= 47\times1.00\times0.915\times1.000 = 42.99\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times42.99^2 = 1109\ \text{N/m}^2 = 1.109\ \text{kN/m}^2 \end{aligned}$$

Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 0.50$, roof angle $\alpha = 30.0^\circ$. Values are interpolated between $\alpha = 20^\circ$ and $30^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3.

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Slope EG (wind 90°), with $C_{pe} - C_{pi} = -0.90$, so $p_d = -998$ N/m² $= -0.998$ kN/m².

Answer

• Design wind speed $V_z = 43.0$ m/s; design wind pressure $p_z = 1109$ N/m² (1.11 kN/m²).
• Net design pressure on the roof: up to $998$ N/m² (1.00 kN/m²), suction.