Design of masonry walls for gravity loads

Distances are measured from the faces. The 160 mm distance is across the 400 mm side (from the 800 mm face), and the 350 mm distance is along the 800 mm side (from the 400 mm face). Hence the eccentricities from the centroid are $e_1 = 200-160 = 40$ mm and $e_2 = 400-350 = 50$ mm.

Data

Section 400 mm $\times$ 800 mm, $P=250$ kN. The load acts 160 mm from the 800 mm face and 350 mm from the 400 mm face.

Eccentricities

$$\begin{aligned} e_{1} &= 40\ \text{mm (across the 400 mm side)}\\ e_{2} &= 50\ \text{mm (along the 800 mm side)} \end{aligned}$$

Section properties

$$\begin{aligned} A &= 400\times800 = 320000\ \text{mm}^2\\ Z_1 &= \frac{800\times400^2}{6} = 21333333\ \text{mm}^3\\ Z_2 &= \frac{400\times800^2}{6} = 42666667\ \text{mm}^3 \end{aligned}$$

Stresses

$$\begin{aligned} \sigma_0 &= \frac{P}{A} = \frac{250\times10^3}{320000} = 0.7812\ \text{N/mm}^2\\ \sigma_{b1} &= \frac{Pe_1}{Z_1} = \frac{250\times10^3\times40}{21333333} = 0.4688\ \text{N/mm}^2\\ \sigma_{b2} &= \frac{Pe_2}{Z_2} = \frac{250\times10^3\times50}{42666667} = 0.2930\ \text{N/mm}^2 \end{aligned}$$

Corner stresses $\sigma=\sigma_0\pm\sigma_{b1}\pm\sigma_{b2}$ (positive = compression, N/mm²):

Check: $\sigma_0$ = average of the four corners = 0.7812. All four corners are in compression, because the load lies inside the kernel ($6e_1/400+6e_2/800 = 0.975 < 1$).

Answer: A = 1.543, B = 0.957, C = 0.605, D = 0.020 N/mm² (compression positive); maximum 1.543 N/mm² at A.

Data and assumptions

• Wall 230 mm thick, modular bricks, joints not raked, so the full thickness 230 mm is effective (IS 1905 cl. 5.5.1.1).
• Load 100 kN/m at top, resultant $e/t = 1/12$. Wall 5 m long between cross walls, clear height 3.5 m between RCC slabs (slab thickness not given, so $H = 3.5$ m taken).
• RCC slabs give lateral and rotational restraint at top and bottom.
• Design code: IS 1905:1987 (Tables 4 to 10); NBC 109 follows the same permissible-stress method.
• Brick masonry unit weight 19 kN/m$^3$ (IS 875 Part 1); modular brick, height/width $\le 0.75$, so $k_p=1$.

Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6)

Height between slab centres $H = 3.50$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.50 = 2.625$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 5.00$ m. The lesser value governs:

$$SR = \frac{h_{ef}}{t} = \frac{2.625}{0.230} = 11.41\ (<27,\ \text{Table 7, cement mortar})$$

Step 2: Design load per metre run at the base

$$\begin{aligned} \text{Self weight} &= t\,H_c\,\gamma = 0.230\times3.50\times19 = 15.30\ \text{kN/m}\\ P &= 100.00 + 15.30 = 115.30\ \text{kN/m} \end{aligned}$$

Stress on the wall:

$$\sigma = \frac{P}{A} = \frac{115.30\times10^3}{230\times10^3\ \text{mm}^2/\text{m}} = 0.501\ \text{N/mm}^2$$

Step 3: Reduction factors

• Stress reduction factor $k_s$ (Table 9, $SR=11.41$, $e/t=0.083$, by linear interpolation) $= 0.828$
• Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.23 m$^2$ is not less than 0.2 m$^2$, so $k_a = 1.0$
• Shape factor $k_p = 1.0$ (modular bricks, height/width $\le 0.75$, Table 10)
• Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25

Step 4: Required basic compressive stress

$$\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.501}{0.828\times1.0\times1.0\times1.25} = 0.485\ \text{N/mm}^2$$

(Without the 25 percent increase the requirement would be $f_b \ge 0.606$ N/mm$^2$.)

Step 5: Strength of brick and grade of mortar (IS 1905 Table 8)

Smallest standard unit strength giving $f_b \ge 0.485$ N/mm$^2$:

Answer: use bricks of crushing strength not less than 7.5 N/mm$^2$ with mortar M2 (cement:sand 1:6) ($f_b = 0.59$ N/mm$^2$ $\ge$ 0.485). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.

Data and assumptions

• Wall 20 cm thick, modular bricks, joints not raked. Load 100 kN/m at top, $e/t=1/12$.
• Wall 5 m long between cross walls, clear height 3.4 m between RCC slabs (slab thickness not given, $H=3.4$ m).
• RCC slabs give lateral and rotational restraint (Table 4 Sl. 1).
• Design code: IS 1905:1987 (Tables 4 to 10); NBC 109 follows the same permissible-stress method.
• Brick masonry unit weight 19 kN/m$^3$ (IS 875 Part 1); modular brick, height/width $\le 0.75$, so $k_p=1$.

Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6)

Height between slab centres $H = 3.40$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.40 = 2.550$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 5.00$ m. The lesser value governs:

$$SR = \frac{h_{ef}}{t} = \frac{2.550}{0.200} = 12.75\ (<27,\ \text{Table 7, cement mortar})$$

Step 2: Design load per metre run at the base

$$\begin{aligned} \text{Self weight} &= t\,H_c\,\gamma = 0.200\times3.40\times19 = 12.92\ \text{kN/m}\\ P &= 100.00 + 12.92 = 112.92\ \text{kN/m} \end{aligned}$$

Stress on the wall:

$$\sigma = \frac{P}{A} = \frac{112.92\times10^3}{200\times10^3\ \text{mm}^2/\text{m}} = 0.565\ \text{N/mm}^2$$

Step 3: Reduction factors

• Stress reduction factor $k_s$ (Table 9, $SR=12.75$, $e/t=0.083$, by linear interpolation) $= 0.784$
• Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.20 m$^2$ is not less than 0.2 m$^2$, so $k_a = 1.0$
• Shape factor $k_p = 1.0$ (modular bricks, height/width $\le 0.75$, Table 10)
• Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25

Step 4: Required basic compressive stress

$$\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.565}{0.784\times1.0\times1.0\times1.25} = 0.576\ \text{N/mm}^2$$

(Without the 25 percent increase the requirement would be $f_b \ge 0.720$ N/mm$^2$.)

Step 5: Strength of brick and grade of mortar (IS 1905 Table 8)

Smallest standard unit strength giving $f_b \ge 0.576$ N/mm$^2$:

Answer: use bricks of crushing strength not less than 7.5 N/mm$^2$ with mortar M2 (cement:sand 1:6) ($f_b = 0.59$ N/mm$^2$ $\ge$ 0.576). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.

Data and assumptions

• Wall 23 cm thick, modular bricks, joints not raked. Load 165 kN/m at base inclusive of self weight, $e/t=1/12$.
• Wall 4.5 m long between cross walls, clear height 3.1 m between 10 cm RCC slabs, so $H=3.1+0.10=3.2$ m (centre to centre).
• Design code: IS 1905:1987 (Tables 4 to 10); NBC 109 follows the same permissible-stress method.
• Brick masonry unit weight 19 kN/m$^3$ (IS 875 Part 1); modular brick, height/width $\le 0.75$, so $k_p=1$.

Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6)

Height between slab centres $H = 3.20$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.20 = 2.400$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 4.50$ m. The lesser value governs:

$$SR = \frac{h_{ef}}{t} = \frac{2.400}{0.230} = 10.43\ (<27,\ \text{Table 7, cement mortar})$$

Step 2: Design load per metre run at the base

Load at the base (self weight already included) $P = 165.00$ kN/m.

Stress on the wall:

$$\sigma = \frac{P}{A} = \frac{165.00\times10^3}{230\times10^3\ \text{mm}^2/\text{m}} = 0.717\ \text{N/mm}^2$$

Step 3: Reduction factors

• Stress reduction factor $k_s$ (Table 9, $SR=10.43$, $e/t=0.083$, by linear interpolation) $= 0.857$
• Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.23 m$^2$ is not less than 0.2 m$^2$, so $k_a = 1.0$
• Shape factor $k_p = 1.0$ (modular bricks, height/width $\le 0.75$, Table 10)
• Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25

Step 4: Required basic compressive stress

$$\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.717}{0.857\times1.0\times1.0\times1.25} = 0.670\ \text{N/mm}^2$$

(Without the 25 percent increase the requirement would be $f_b \ge 0.837$ N/mm$^2$.)

Step 5: Strength of brick and grade of mortar (IS 1905 Table 8)

Smallest standard unit strength giving $f_b \ge 0.670$ N/mm$^2$:

Answer: use bricks of crushing strength not less than 10 N/mm$^2$ with mortar M2 (cement:sand 1:6) ($f_b = 0.81$ N/mm$^2$ $\ge$ 0.670). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.

Data and assumptions

• Wall 250 mm thick, modular bricks, joints not raked. Load 350 kN/m at top, $e/t=1/12$.
• Wall 5 m long between cross walls, clear height 3.5 m between RCC slabs (slab thickness not given, $H=3.5$ m).
• Note: a 350 kN/m load is high for brick masonry, so strong units are needed.
• Design code: IS 1905:1987 (Tables 4 to 10); NBC 109 follows the same permissible-stress method.
• Brick masonry unit weight 19 kN/m$^3$ (IS 875 Part 1); modular brick, height/width $\le 0.75$, so $k_p=1$.

Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6)

Height between slab centres $H = 3.50$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.50 = 2.625$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 5.00$ m. The lesser value governs:

$$SR = \frac{h_{ef}}{t} = \frac{2.625}{0.250} = 10.50\ (<27,\ \text{Table 7, cement mortar})$$

Step 2: Design load per metre run at the base

$$\begin{aligned} \text{Self weight} &= t\,H_c\,\gamma = 0.250\times3.50\times19 = 16.62\ \text{kN/m}\\ P &= 350.00 + 16.62 = 366.62\ \text{kN/m} \end{aligned}$$

Stress on the wall:

$$\sigma = \frac{P}{A} = \frac{366.62\times10^3}{250\times10^3\ \text{mm}^2/\text{m}} = 1.466\ \text{N/mm}^2$$

Step 3: Reduction factors

• Stress reduction factor $k_s$ (Table 9, $SR=10.50$, $e/t=0.083$, by linear interpolation) $= 0.855$
• Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.25 m$^2$ is not less than 0.2 m$^2$, so $k_a = 1.0$
• Shape factor $k_p = 1.0$ (modular bricks, height/width $\le 0.75$, Table 10)
• Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25

Step 4: Required basic compressive stress

$$\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{1.466}{0.855\times1.0\times1.0\times1.25} = 1.372\ \text{N/mm}^2$$

(Without the 25 percent increase the requirement would be $f_b \ge 1.715$ N/mm$^2$.)

Step 5: Strength of brick and grade of mortar (IS 1905 Table 8)

Smallest standard unit strength giving $f_b \ge 1.372$ N/mm$^2$:

Answer: use bricks of crushing strength not less than 17.5 N/mm$^2$ with mortar H1 (cement:sand 1:3) ($f_b = 1.45$ N/mm$^2$ $\ge$ 1.372). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.

Distances are measured from the faces. The 160 mm distance is across the 400 mm side (from the 800 mm face) and the 350 mm distance is along the 800 mm side (from the 400 mm face). Hence $e_1 = 200-160 = 40$ mm and $e_2 = 400-350 = 50$ mm.

Data

Section 400 mm $\times$ 800 mm, $P=200$ kN. The load acts 160 mm from the 800 mm face and 350 mm from the 400 mm face.

Eccentricities

$$\begin{aligned} e_{1} &= 40\ \text{mm (across the 400 mm side)}\\ e_{2} &= 50\ \text{mm (along the 800 mm side)} \end{aligned}$$

Section properties

$$\begin{aligned} A &= 400\times800 = 320000\ \text{mm}^2\\ Z_1 &= \frac{800\times400^2}{6} = 21333333\ \text{mm}^3\\ Z_2 &= \frac{400\times800^2}{6} = 42666667\ \text{mm}^3 \end{aligned}$$

Stresses

$$\begin{aligned} \sigma_0 &= \frac{P}{A} = \frac{200\times10^3}{320000} = 0.6250\ \text{N/mm}^2\\ \sigma_{b1} &= \frac{Pe_1}{Z_1} = \frac{200\times10^3\times40}{21333333} = 0.3750\ \text{N/mm}^2\\ \sigma_{b2} &= \frac{Pe_2}{Z_2} = \frac{200\times10^3\times50}{42666667} = 0.2344\ \text{N/mm}^2 \end{aligned}$$

Corner stresses $\sigma=\sigma_0\pm\sigma_{b1}\pm\sigma_{b2}$ (positive = compression, N/mm²):

Check: $\sigma_0$ = average of the four corners = 0.6250. All four corners are in compression, because the load lies inside the kernel ($6e_1/400+6e_2/800 = 0.975 < 1$).

Answer: A = 1.234, B = 0.766, C = 0.484, D = 0.016 N/mm² (compression positive); maximum 1.234 N/mm² at A.

Distances are measured from the faces. The 170 mm distance is across the 500 mm side (from the 900 mm face) and the 360 mm distance is along the 900 mm side (from the 500 mm face). Hence $e_1 = 250-170 = 80$ mm and $e_2 = 450-360 = 90$ mm.

Data

Section 500 mm $\times$ 900 mm, $P=300$ kN. The load acts 170 mm from the 900 mm face and 360 mm from the 500 mm face.

Eccentricities

$$\begin{aligned} e_{1} &= 80\ \text{mm (across the 500 mm side)}\\ e_{2} &= 90\ \text{mm (along the 900 mm side)} \end{aligned}$$

Section properties

$$\begin{aligned} A &= 500\times900 = 450000\ \text{mm}^2\\ Z_1 &= \frac{900\times500^2}{6} = 37500000\ \text{mm}^3\\ Z_2 &= \frac{500\times900^2}{6} = 67500000\ \text{mm}^3 \end{aligned}$$

Stresses

$$\begin{aligned} \sigma_0 &= \frac{P}{A} = \frac{300\times10^3}{450000} = 0.6667\ \text{N/mm}^2\\ \sigma_{b1} &= \frac{Pe_1}{Z_1} = \frac{300\times10^3\times80}{37500000} = 0.6400\ \text{N/mm}^2\\ \sigma_{b2} &= \frac{Pe_2}{Z_2} = \frac{300\times10^3\times90}{67500000} = 0.4000\ \text{N/mm}^2 \end{aligned}$$

Corner stresses $\sigma=\sigma_0\pm\sigma_{b1}\pm\sigma_{b2}$ (positive = compression, N/mm²):

Check: $\sigma_0$ = average of the four corners = 0.6667. Corner D is in tension (-0.3733 N/mm²) because $6e_1/500+6e_2/900 = 1.560 > 1$, i.e. the load lies outside the kernel (middle-third) of the section.

Answer: A = 1.707, B = 0.907, C = 0.427, D = -0.373 N/mm² (compression positive); maximum 1.707 N/mm² at A.

Data and assumptions

• Design code: IS 1905:1987; NBC 109 follows the same permissible-stress method.
• Interior cross wall 230 mm thick (one brick, Nepal practice), modular bricks, $k_p=1$. Brick masonry 19 kN/m$^3$, RCC 25 kN/m$^3$.
• Stated slab width is taken as the tributary width of slab per metre run of wall (each side contributing half of its span). Finish applies to roof and floors; live load not reduced.
• Sections are checked at the base of each storey (wall height for weight = clear ceiling height).
• Wall unstiffened, so slenderness is governed by height. Slab 120 mm, ceiling 3.0 m, tributary width 3.0 m.

Step 1: Loads per metre run of wall

Unit weights (IS 875 Part 1): RCC 25 kN/m$^3$, brick masonry 19 kN/m$^3$.

Step 2: Slenderness ratio and factors

Interior wall with slabs on both sides: load is axial ($e=0$, IS 1905 App. A-4). RCC slab bearing on the wall gives full restraint (Table 4 Sl. 1): $h_{ef}=0.75H$, with $H=3.00+0.120=3.120$ m.

$$\begin{aligned} h_{ef} &= 0.75\times3.120 = 2.340\ \text{m}\\ SR &= \frac{2.340}{0.23} = 10.17\ (<27) \end{aligned}$$

$k_s = 0.886$ (Table 9, $e=0$, interpolated), $k_a = 1.0$ (area 0.23 m$^2$/m $\ge 0.2$), $k_p=1.0$.

Step 3: Stress and required $f_b$ at the base of each storey

Step 4: Brick strength and mortar (IS 1905 Table 8)

Smallest standard unit strength for each section:

Answer: 230 mm thick wall; the ground storey governs, requiring 3.5 N/mm$^2$ brick with M2 (1:6) mortar (required $f_b=0.328$ N/mm$^2$; $f_b$ provided 0.35). The same brick and mortar can be used throughout. As a practical minimum, adopt bricks of not less than 5 N/mm$^2$ with 1:6 cement mortar (SR = 10.17 < 27, OK).

Data and assumptions

• Design code: IS 1905:1987; NBC 109 follows the same permissible-stress method.
• Interior cross wall 230 mm thick (one brick, Nepal practice), modular bricks, $k_p=1$. Brick masonry 19 kN/m$^3$, RCC 25 kN/m$^3$.
• Stated slab width is taken as the tributary width of slab per metre run of wall (each side contributing half of its span). Finish applies to roof and floors; live load not reduced.
• Sections are checked at the base of each storey (wall height for weight = clear ceiling height).
• Wall unstiffened, so slenderness is governed by height. Slab 125 mm, ceiling height 3.2 m, tributary width 2.5 m.

Step 1: Loads per metre run of wall

Unit weights (IS 875 Part 1): RCC 25 kN/m$^3$, brick masonry 19 kN/m$^3$.

Step 2: Slenderness ratio and factors

Interior wall with slabs on both sides: load is axial ($e=0$, IS 1905 App. A-4). RCC slab bearing on the wall gives full restraint (Table 4 Sl. 1): $h_{ef}=0.75H$, with $H=3.20+0.125=3.325$ m.

$$\begin{aligned} h_{ef} &= 0.75\times3.325 = 2.494\ \text{m}\\ SR &= \frac{2.494}{0.23} = 10.84\ (<27) \end{aligned}$$

$k_s = 0.869$ (Table 9, $e=0$, interpolated), $k_a = 1.0$ (area 0.23 m$^2$/m $\ge 0.2$), $k_p=1.0$.

Step 3: Stress and required $f_b$ at the base of each storey

Step 4: Brick strength and mortar (IS 1905 Table 8)

Smallest standard unit strength for each section:

Answer: 230 mm thick wall; the ground storey governs, requiring 3.5 N/mm$^2$ brick with M2 (1:6) mortar (required $f_b=0.292$ N/mm$^2$; $f_b$ provided 0.35). The same brick and mortar can be used throughout. As a practical minimum, adopt bricks of not less than 5 N/mm$^2$ with 1:6 cement mortar (SR = 10.84 < 27, OK).

The permissible compressive stress in masonry cannot be taken equal to the strength of unit or prism, because the real wall differs from the tested prism. IS 1905 therefore starts from the basic compressive stress $f_b$ (Table 8; for a short, axially loaded wall of standard unit shape) and multiplies it by reduction factors:

$$f_c = f_b \times k_s \times k_a \times k_p$$

Role of each factor (IS 1905 cl. 5.4.1)

• Stress reduction factor $k_s$ (Table 9): accounts for slenderness ratio and eccentricity of load. A tall slender wall buckles at a lower stress, and an eccentric load causes bending; $k_s$ falls from 1.0 (SR $\le 6$, $e=0$) to about 0.2 for high SR and large $e/t$.
• Area reduction factor $k_a$: for small sections (area $A < 0.2$ m$^2$, such as columns and piers) the effect of local weakness and workmanship is larger; $k_a = 0.7+1.5A$ ($A$ in m$^2$).
• Shape modification factor $k_p$ (Table 10): units with larger height-to-width ratio give a higher strength in test than flat units; for units with $h/w \le 0.75$, $k_p = 1$; taller units get a factor above 1 (up to 1.8, only for units up to 15 N/mm$^2$).

Together they provide the factor of safety (about 3 on the average prism strength) and make the stress allowed depend on how the wall is shaped, how slender it is, and how it is loaded. In design the factors are also used with the 25 percent increase when $e/t > 1/24$ (cl. 5.4.1.4).

Data and assumptions

• Design code: IS 1905:1987; NBC 109 follows the same permissible-stress method.
• Interior cross wall 230 mm thick (one brick, Nepal practice), modular bricks, $k_p=1$. Brick masonry 19 kN/m$^3$, RCC 25 kN/m$^3$.
• Stated slab width is taken as the tributary width of slab per metre run of wall (each side contributing half of its span). Finish applies to roof and floors; live load not reduced.
• Sections are checked at the base of each storey (wall height for weight = clear ceiling height).
• Mortar 1:4 cement:sand = grade H2 (IS 1905 Table 1) in all storeys.

Step 1: Loads per metre run of wall

Unit weights (IS 875 Part 1): RCC 25 kN/m$^3$, brick masonry 19 kN/m$^3$.

Step 2: Slenderness ratio and factors

Interior wall with slabs on both sides: load is axial ($e=0$, IS 1905 App. A-4). RCC slab bearing on the wall gives full restraint (Table 4 Sl. 1): $h_{ef}=0.75H$, with $H=3.50+0.150=3.650$ m.

$$\begin{aligned} h_{ef} &= 0.75\times3.650 = 2.737\ \text{m}\\ SR &= \frac{2.737}{0.23} = 11.90\ (<27) \end{aligned}$$

$k_s = 0.842$ (Table 9, $e=0$, interpolated), $k_a = 1.0$ (area 0.23 m$^2$/m $\ge 0.2$), $k_p=1.0$.

Step 3: Stress and required $f_b$ at the base of each storey

Step 4: Brick strength and mortar (IS 1905 Table 8)

Mortar fixed as H2 (cement:sand 1:4):

Answer: 230 mm wall, mortar H2 (1:4). Minimum brick strength: top storey 3.5 N/mm$^2$, second storey 5 N/mm$^2$, ground storey 7.5 N/mm$^2$ (slenderness ratio 11.90 < 27 in all storeys). For simplicity the whole wall may be built with 7.5 N/mm$^2$ bricks (or 7.5 N/mm$^2$ for 10 per cent extra safety).

Data and assumptions

• Design code: IS 1905:1987; NBC 109 follows the same permissible-stress method.
• Interior cross wall 230 mm thick (one brick, Nepal practice), modular bricks, $k_p=1$. Brick masonry 19 kN/m$^3$, RCC 25 kN/m$^3$.
• Stated slab width is taken as the tributary width of slab per metre run of wall (each side contributing half of its span). Finish applies to roof and floors; live load not reduced.
• Sections are checked at the base of each storey (wall height for weight = clear ceiling height).
• Wall unstiffened, so slenderness is governed by height. Slab 120 mm (60 mm screed with finish = 1.2 kN/m$^2$), ceiling 3.0 m, tributary width 2.5 m.

Step 1: Loads per metre run of wall

Unit weights (IS 875 Part 1): RCC 25 kN/m$^3$, brick masonry 19 kN/m$^3$.

Step 2: Slenderness ratio and factors

Interior wall with slabs on both sides: load is axial ($e=0$, IS 1905 App. A-4). RCC slab bearing on the wall gives full restraint (Table 4 Sl. 1): $h_{ef}=0.75H$, with $H=3.00+0.120=3.120$ m.

$$\begin{aligned} h_{ef} &= 0.75\times3.120 = 2.340\ \text{m}\\ SR &= \frac{2.340}{0.23} = 10.17\ (<27) \end{aligned}$$

$k_s = 0.886$ (Table 9, $e=0$, interpolated), $k_a = 1.0$ (area 0.23 m$^2$/m $\ge 0.2$), $k_p=1.0$.

Step 3: Stress and required $f_b$ at the base of each storey

Step 4: Brick strength and mortar (IS 1905 Table 8)

Smallest standard unit strength for each section:

Answer: 230 mm thick wall; the ground storey governs, requiring 3.5 N/mm$^2$ brick with M2 (1:6) mortar (required $f_b=0.275$ N/mm$^2$; $f_b$ provided 0.35). The same brick and mortar can be used throughout. As a practical minimum, adopt bricks of not less than 5 N/mm$^2$ with 1:6 cement mortar (SR = 10.17 < 27, OK).

Design of masonry (IS 1905) is based on working stresses, so the following factors are to be considered:

1. Strength of masonry unit and mortar grade: the basic compressive stress $f_b$ depends on both (Table 8); stronger mortar and unit give higher $f_b$.
2. Slenderness ratio: effective height or length divided by effective thickness, whichever is less; limited to 27 (cement mortar) for walls and 12 for columns (Table 7).
3. Effective height and length: depend on end restraint, such as RCC slab (0.75 H), timber floor (0.85 H or 1.0 H), free top (1.5 H) (Tables 4 and 5).
4. Eccentricity of loading: eccentricity ratio $e/t$ reduces the permissible stress (Table 9); App. A gives guidance.
5. Area of section: for sections smaller than 0.2 m$^2$ the area reduction factor $k_a$ applies.
6. Shape of masonry unit: height-to-width ratio of units gives shape factor $k_p$ (Table 10).
7. Type of loads: vertical load, wind, earthquake; no tension in masonry normally; limited flexural tension (0.07 N/mm$^2$ for M1 mortar) for lateral loads.
8. Openings and stiffening: openings reduce the effective area; stiffening by cross walls, piers or buttresses increases effective thickness (Table 6).
9. Workmanship and quality control: raked joints reduce thickness; curing, bond and supervision.
10. Seismic zone: buildings in zones III to V need special strengthening (IS 4326 / NBC 105, 109).

The compressive capacity of masonry (wall, pier or column) is a function of the strength of the unit and the grade of the mortar, but the unit has the greater influence.

• Masonry fails when the units split in tension: the mortar is softer and tends to expand laterally under load more than the brick, which induces lateral tension in the unit. So prism strength rises with unit strength.
• A stronger mortar gives smaller lateral expansion and a higher masonry strength. A very weak mortar (e.g. lime) gives low strength. But strength increase from mortar is small compared to the increase from unit strength; e.g. doubling mortar strength raises masonry strength only about 10 to 20 percent.

IS 1905 Table 8 shows this directly. Basic compressive stress $f_b$ (N/mm$^2$):

For a given mortar, $f_b$ increases almost in proportion to unit strength up to about 10 N/mm$^2$; for a given unit, going from M2 to H1 raises $f_b$ by about 25 to 40 percent. Hence the designer usually raises unit strength first and uses a standard mortar, and does not use mortar stronger than needed. Mortar must also bond well, be workable and be properly cured.

Data and assumptions

• Design code: IS 1905:1987; NBC 109 follows the same permissible-stress method.
• Interior cross wall 230 mm thick (one brick, Nepal practice), modular bricks, $k_p=1$. Brick masonry 19 kN/m$^3$, RCC 25 kN/m$^3$.
• Stated slab width is taken as the tributary width of slab per metre run of wall (each side contributing half of its span). Finish applies to roof and floors; live load not reduced.
• Sections are checked at the base of each storey (wall height for weight = clear ceiling height).
• Wall 3.2 m long and stiffened by cross walls at its ends (effective length $1.0L=3.2$ m, Table 5), which is greater than $h_{ef}$, so height governs the slenderness. Slab 130 mm, ceiling 2.8 m, slab span 2.0 m each side (tributary 2.0 m).
• Finish 1.0 kN/m$^2$, live load 1.5 kN/m$^2$ (roof) and 2.0 kN/m$^2$ (floor) assumed as per IS 875 Part 2 for residential use.

Step 1: Loads per metre run of wall

Unit weights (IS 875 Part 1): RCC 25 kN/m$^3$, brick masonry 19 kN/m$^3$.

Step 2: Slenderness ratio and factors

Interior wall with slabs on both sides: load is axial ($e=0$, IS 1905 App. A-4). RCC slab bearing on the wall gives full restraint (Table 4 Sl. 1): $h_{ef}=0.75H$, with $H=2.80+0.130=2.930$ m.

$$\begin{aligned} h_{ef} &= 0.75\times2.930 = 2.197\ \text{m}\\ SR &= \frac{2.197}{0.23} = 9.55\ (<27) \end{aligned}$$

$k_s = 0.903$ (Table 9, $e=0$, interpolated), $k_a = 1.0$ (area 0.23 m$^2$/m $\ge 0.2$), $k_p=1.0$.

Step 3: Stress and required $f_b$ at the base of each storey

Step 4: Brick strength and mortar (IS 1905 Table 8)

Smallest standard unit strength for each section:

Answer: 230 mm thick wall; the ground storey governs, requiring 3.5 N/mm$^2$ brick with M2 (1:6) mortar (required $f_b=0.233$ N/mm$^2$; $f_b$ provided 0.35). The same brick and mortar can be used throughout. As a practical minimum, adopt bricks of not less than 5 N/mm$^2$ with 1:6 cement mortar (SR = 9.55 < 27, OK).

Data and assumptions

• Thickness not given: Nepal standard brick $230\times110\times55$ mm, so a one-brick wall of 230 mm is adopted (joints not raked). Load 175 kN/m inclusive of self weight, $e/t=1/12$.
• Wall 4.2 m long between cross walls, clear height 3.2 m between 12 cm RCC slabs, so $H=3.2+0.12=3.32$ m.
• Design code: IS 1905:1987 (Tables 4 to 10); NBC 109 follows the same permissible-stress method.
• Brick masonry unit weight 19 kN/m$^3$ (IS 875 Part 1); modular brick, height/width $\le 0.75$, so $k_p=1$.

Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6)

Height between slab centres $H = 3.32$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.32 = 2.490$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 4.20$ m. The lesser value governs:

$$SR = \frac{h_{ef}}{t} = \frac{2.490}{0.230} = 10.83\ (<27,\ \text{Table 7, cement mortar})$$

Step 2: Design load per metre run at the base

Load at the base (self weight already included) $P = 175.00$ kN/m.

Stress on the wall:

$$\sigma = \frac{P}{A} = \frac{175.00\times10^3}{230\times10^3\ \text{mm}^2/\text{m}} = 0.761\ \text{N/mm}^2$$

Step 3: Reduction factors

• Stress reduction factor $k_s$ (Table 9, $SR=10.83$, $e/t=0.083$, by linear interpolation) $= 0.845$
• Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.23 m$^2$ is not less than 0.2 m$^2$, so $k_a = 1.0$
• Shape factor $k_p = 1.0$ (modular bricks, height/width $\le 0.75$, Table 10)
• Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25

Step 4: Required basic compressive stress

$$\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.761}{0.845\times1.0\times1.0\times1.25} = 0.720\ \text{N/mm}^2$$

(Without the 25 percent increase the requirement would be $f_b \ge 0.900$ N/mm$^2$.)

Step 5: Strength of brick and grade of mortar (IS 1905 Table 8)

Smallest standard unit strength giving $f_b \ge 0.720$ N/mm$^2$:

Answer: use bricks of crushing strength not less than 10 N/mm$^2$ with mortar M2 (cement:sand 1:6) ($f_b = 0.81$ N/mm$^2$ $\ge$ 0.720). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.

Data and assumptions

• Wall 230 mm thick, local bricks (assumed $230\times110\times55$ mm, $k_p=1$), $e/t=1/12$, load 135 kN/m at top. Wall 4 m long, clear height 3 m.
• RCC slab at the bottom (lateral and rotational restraint) and timber floor at the top assumed adequately anchored but not spanning on the wall (lateral restraint only): Table 4 Sl. 2, $h_{ef}=0.85H$.
• Design code: IS 1905:1987; NBC 109 follows the same method.
• Brick masonry 19 kN/m$^3$ (IS 875 Part 1).

Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6)

Height between slab centres $H = 3.00$ m. Effective height $h_{ef} = 0.85 H = 0.85\times3.00 = 2.550$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 4.00$ m. The lesser value governs:

$$SR = \frac{h_{ef}}{t} = \frac{2.550}{0.230} = 11.09\ (<27,\ \text{Table 7, cement mortar})$$

Step 2: Design load per metre run at the base

$$\begin{aligned} \text{Self weight} &= t\,H_c\,\gamma = 0.230\times3.00\times19 = 13.11\ \text{kN/m}\\ P &= 135.00 + 13.11 = 148.11\ \text{kN/m} \end{aligned}$$

Stress on the wall:

$$\sigma = \frac{P}{A} = \frac{148.11\times10^3}{230\times10^3\ \text{mm}^2/\text{m}} = 0.644\ \text{N/mm}^2$$

Step 3: Reduction factors

• Stress reduction factor $k_s$ (Table 9, $SR=11.09$, $e/t=0.083$, by linear interpolation) $= 0.837$
• Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.23 m$^2$ is not less than 0.2 m$^2$, so $k_a = 1.0$
• Shape factor $k_p = 1.0$ (modular bricks, height/width $\le 0.75$, Table 10)
• Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25

Step 4: Required basic compressive stress

$$\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.644}{0.837\times1.0\times1.0\times1.25} = 0.615\ \text{N/mm}^2$$

(Without the 25 percent increase the requirement would be $f_b \ge 0.769$ N/mm$^2$.)

Step 5: Strength of brick and grade of mortar (IS 1905 Table 8)

Smallest standard unit strength giving $f_b \ge 0.615$ N/mm$^2$:

Answer: use bricks of crushing strength not less than 10 N/mm$^2$ with mortar M2 (cement:sand 1:6) ($f_b = 0.81$ N/mm$^2$ $\ge$ 0.615). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.

Data and assumptions

• Wall 230 mm thick, modular bricks ($k_p=1$); load at top 100 kN/m with eccentricity 12 mm, so $e/t=12/230=0.0522$ ($>1/24$).
• Plinth 1.5 m above top of footing, ceiling height 3.0 m, 12 cm RCC slab bearing on the wall. Walls longer than $h_{ef}$ (no cross walls given), so height governs.
• Brick masonry 19 kN/m$^3$. Design by IS 1905:1987; NBC 109 follows the same method. Critical section: base of wall at the top of the footing.

Step 1: Effective height and slenderness ratio

Height from top of footing to the centre of slab $H = 1.5+3.0+0.12/2 = 4.56$ m (IS 1905 Table 4 Note 1). The footing and the RCC slab give full restraint, so $h_{ef}=0.75H$:

$$\begin{aligned} h_{ef} &= 0.75\times4.56 = 3.420\ \text{m}\\ SR &= \frac{3.420}{0.23} = 14.87\ (<27) \end{aligned}$$

Step 2: Load and maximum stress at the base

$$\begin{aligned} \text{Self weight} &= 0.23\times4.5\times19 = 19.67\ \text{kN/m}\\ P &= 100+19.67 = 119.67\ \text{kN/m}\\ M &= 100\times0.012 = 1.20\ \text{kNm/m}\\ \sigma_{max} &= \frac{P}{A}+\frac{6M}{t^2} = \frac{119.67\times10^3}{230\times10^3}+\frac{6\times1.20\times10^6}{1000\times230^2} = 0.520+0.136 = 0.656\ \text{N/mm}^2 \end{aligned}$$

(At the top of the wall, under the load alone, $\sigma_{max}=\frac{100\times10^3}{230\times10^3}(1+\frac{6\times12}{230}) = 0.571$ N/mm$^2$.) The maximum stress in the wall is 0.656 N/mm$^2$ at the base (compression; no tension since $e<t/6$).

Step 3: Factors and required basic stress

• $k_s$ (Table 9, $SR=14.87$, $e/t=0.052$, interpolated) $= 0.732$
• $k_a=1.0$ (area 0.23 m$^2$/m), $k_p=1.0$
• $e/t>1/24$: 25 percent increase allowed (cl. 5.4.1.4 a)

$$f_b \ge \frac{\sigma}{k_sk_ak_p\times1.25} = \frac{0.520}{0.732\times1.25} = 0.569\ \text{N/mm}^2$$

Step 4: Brick and mortar (IS 1905 Table 8)

Answer: maximum stress = 0.656 N/mm$^2$. Required $f_b = 0.569$ N/mm$^2$, so use bricks of crushing strength not less than 7.5 N/mm$^2$ with M2 mortar (1:6, $f_b=0.59$ N/mm$^2$). A richer mortar (M1, H2, H1) also needs 7.5 N/mm$^2$ units, so there is no gain from a stronger mortar here.