Design of Masonry Walls for Gravity Loads

IS 1905 (Code of practice for structural use of unreinforced masonry) uses the working stress method. The permissible stress is the basic compressive stress reduced by factors for slenderness, area and shape:

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

Basic compressive stress $f_b$

It is the compressive stress that a short, concentrically loaded masonry element can carry safely. It is given in a table against the compressive strength of the masonry unit and the grade of mortar (cement–sand, cement–lime–sand or lime mortars). Higher unit strength and stronger mortar give higher $f_b$. For ordinary brick masonry it is of the order of 0.5 to 2 N/mm².

Modification factors

1. Stress reduction factor $k_s$: accounts for slenderness ($SR$) and eccentricity of load. It is 1.0 for low slenderness (about 6 or less) and reduces as $SR$ rises. Typical values for axial load:

2. Area factor $k_a$: small sections have a higher chance of defects. For loaded area $A < 0.2$ m²,

$$k_a = 0.7 + 1.5A$$

($A$ in m²); $k_a = 1.0$ for $A \ge 0.2$ m². 3. Shape modification factor $k_p$: applies to units whose height-to-width ratio differs from that of standard brick; for bricks $k_p = 1.0$. Squat blocks give higher values and are taken from tables.

Increase for wind and earthquake

For combined gravity and wind or earthquake loads, the permissible stresses may be increased by one-third (33 percent), provided the section is also safe for the gravity load alone.

Other points

• The permissible stress is checked against the actual stress $P/A$ (plus bending effects if eccentric).
• The slenderness ratio of a load-bearing wall should not exceed 27.
• Permissible tensile and shear stresses are given separately and are small; masonry is not designed to take tension.

Definitions

• Effective height ($H_{eff}$): the height of the wall that is considered to buckle, obtained by multiplying the actual (unsupported) height $H$ by a factor depending on end restraint.
• Effective length ($L_{eff}$): the horizontal length considered to buckle between cross walls or other vertical supports, modified for the edge restraint.
• Effective thickness ($t_{eff}$): the thickness used in slenderness calculation. For a solid single leaf wall it is the actual thickness $t$; for a wall with piers or cross walls a stiffness factor increases it (1.0 to 2.0); for a cavity wall it is taken as two-thirds of the sum of the leaf thicknesses.
• Slenderness ratio ($SR$):

$$SR = \frac{H_{eff}}{t_{eff}} \quad \text{or} \quad \frac{L_{eff}}{t_{eff}}, \text{ whichever is smaller}$$

For a column, $SR$ is the effective height over the least lateral dimension.

Effective height (IS 1905)

 slab  ======  top fixed       free top -> 1.5H / 2.0H
        ||                         |
        ||  H                      |  H
        ||                         |
 ground ======  base fixed      #########
 0.75 H                          cantilever

Importance of slenderness

• A slender wall fails by buckling at a stress much lower than the crushing strength. The permissible stress is therefore reduced by $k_s$, which decreases as $SR$ rises.
• The code limits $SR$ (27 for load-bearing walls), so walls cannot be made taller or thinner than a certain limit.
• Slenderness can be reduced by increasing the thickness, adding piers or cross walls, or providing restraint at the top.

Step 1: Slenderness

Lateral and rotational restraint at both ends gives $H_{eff} = 0.75H$:

$$\begin{aligned} H_{eff} &= 0.75 \times 3200 = 2400\ \text{mm} \\ SR &= \frac{2400}{230} = 10.43 \end{aligned}$$

$SR < 27$, so OK.

Step 2: Stress reduction factor

Interpolate between $SR = 10$ ($k_s = 0.89$) and $SR = 12$ ($k_s = 0.84$):

$$k_s = 0.89 - \frac{0.43}{2}\times 0.05 = 0.879$$

Step 3: Permissible stress

Per metre length, area $A = 0.23 \times 1.0 = 0.23$ m² $> 0.2$ m², so $k_a = 1.0$; for bricks $k_p = 1.0$.

$$f_c = 1.0 \times 0.879 \times 1.0 \times 1.0 = 0.879\ \text{N/mm}^2$$

Step 4: Actual stress at base

$$\begin{aligned} W_{wall} &= 19 \times 0.23 \times 3.2 = 13.98\ \text{kN/m} \\ P &= 120 + 13.98 = 133.98\ \text{kN/m} \\ f &= \frac{133\,980}{230 \times 1000} = 0.583\ \text{N/mm}^2 \end{aligned}$$

$0.583 < 0.879$, so the wall is safe (utilisation 66 percent).

Step 5: Maximum load

$$P_{max} = f_c\,A = 0.879 \times 230 \times 1000 = 202\,170\ \text{N/m} \approx 202\ \text{kN/m}$$

Allowable additional slab load $= 202.2 - 14.0 = 188\ \text{kN/m}$.

Answer: Actual stress 0.58 N/mm² < 0.879 N/mm², safe. Maximum total load = 202 kN/m (slab load about 188 kN/m).

Principle

The load over a window opening is carried by the lintel/band and shared by the piers on either side, so the piers carry their own length of wall load plus half of each adjacent opening. Area factor $k_a$ is applied to the smaller piers.

Step 1: Slenderness and $k_s$

$H_{eff} = 0.75 \times 3000 = 2250$ mm, $SR = 2250/230 = 9.78 < 27$.

$$k_s = 0.89 + \frac{10 - 9.78}{2}\times 0.06 = 0.897$$

Step 2: Load on piers

Check: $2 \times 127.5 + 255.0 = 510 = 6.0 \times 85$ kN. OK.

Step 3: Self-weight (to base)

• End pier: $19 \times 0.23 \times 0.75 \times 3.0 = 9.83$ kN, total $P = 137.3$ kN.
• Middle pier: $19 \times 0.23 \times 1.5 \times 3.0 = 19.67$ kN, total $P = 274.7$ kN.

Step 4: Area factor and permissible stress

• End pier: $A = 0.23 \times 0.75 = 0.1725$ m² $< 0.2$, so $k_a = 0.7 + 1.5 \times 0.1725 = 0.959$.
• Middle pier: $A = 0.345$ m² $> 0.2$, $k_a = 1.0$.

$$f_{c,end} = 1.0 \times 0.897 \times 0.959 = 0.860\ \text{N/mm}^2, \qquad f_{c,mid} = 1.0 \times 0.897 = 0.897\ \text{N/mm}^2$$

Step 5: Actual stress

$$\begin{aligned} f_{end} &= \frac{137\,330}{0.1725\times10^{6}} = 0.796\ \text{N/mm}^2 < 0.860 \\ f_{mid} &= \frac{274\,670}{0.345\times10^{6}} = 0.796\ \text{N/mm}^2 < 0.897 \end{aligned}$$

The end pier has the smallest margin (93 percent utilisation); it should not be reduced in width further. Pier width must also satisfy seismic opening rules.

Answer: Both piers are safe: stress 0.80 N/mm² against permissible 0.86 N/mm² (end pier) and 0.90 N/mm² (middle pier).

Middle-third rule

If the resultant load acts within the middle third of the thickness ($e \le t/6$), the whole section is in compression. If $e > t/6$, tension appears on the far face, and masonry (weak in tension) may crack.

Data

$P = 60$ kN/m, $t = 230$ mm, $e = 45$ mm, $b = 1000$ mm (per metre).

$$\frac{t}{6} = \frac{230}{6} = 38.3\ \text{mm} < e = 45\ \text{mm}$$

The load is outside the middle third, so some tension is expected.

Stresses

$$\begin{aligned} A &= 230 \times 1000 = 2.3\times10^5\ \text{mm}^2 \\ Z &= \frac{1000 \times 230^2}{6} = 8.817\times10^{6}\ \text{mm}^3 \\ M &= Pe = 60\,000 \times 45 = 2.7\times10^{6}\ \text{N·mm} \\ \frac{P}{A} &= \frac{60\,000}{2.3\times10^5} = 0.261\ \text{N/mm}^2 \\ \frac{M}{Z} &= \frac{2.7\times10^6}{8.817\times10^6} = 0.306\ \text{N/mm}^2 \end{aligned}$$

Or, using the formula $\sigma = \dfrac{P}{A}\left(1 \pm \dfrac{6e}{t}\right)$ with $6e/t = 1.174$:

$$\begin{aligned} \sigma_{max} &= 0.261\,(1 + 1.174) = 0.567\ \text{N/mm}^2 \ (\text{compression}) \\ \sigma_{min} &= 0.261\,(1 - 1.174) = -0.045\ \text{N/mm}^2 \ (\text{tension}) \end{aligned}$$

Checks

• Compression: $0.567 < 0.90$ N/mm². Safe.
• Tension: $0.045 < 0.07$ N/mm². Within the permissible flexural tension.

Both limits are satisfied. If tension were more than allowed, the eccentricity must be reduced (for example by making the beam bearing central, using a bearing plate) or the wall thickness increased or reinforced.

Answer: $\sigma_{max} = 0.567$ N/mm² (compression) and $\sigma_{min} = -0.045$ N/mm² (tension); both within permissible values, so the wall is safe.

Slenderness (column)

Slenderness is based on the least lateral dimension:

$$H_{eff} = 3500\ \text{mm}, \qquad SR = \frac{3500}{340} = 10.29$$ $$k_s = 0.89 - \frac{0.29}{2}\times 0.05 = 0.883$$

Area factor

$A = 0.34 \times 0.46 = 0.1564$ m² $< 0.2$ m²

$$k_a = 0.7 + 1.5\times0.1564 = 0.935$$

$k_p = 1.0$ for brick.

Permissible stress and safe load

$$\begin{aligned} f_c &= f_b\,k_s\,k_a\,k_p = 1.1 \times 0.883 \times 0.935 \times 1.0 = 0.907\ \text{N/mm}^2 \\ P &= f_c A = 0.907 \times 340 \times 460 = 141\,900\ \text{N} \approx 142\ \text{kN} \end{aligned}$$

Comparison with 460 × 460 mm column

$SR = 3500/460 = 7.61$, $k_s = 0.95 + (8 - 7.61)/2 \times 0.05 = 0.96$; $A = 0.2116$ m² $> 0.2$, so $k_a = 1.0$.

$$f_c = 1.1 \times 0.96 \times 1.0 = 1.056\ \text{N/mm}^2, \qquad P = 1.056 \times 460 \times 460 = 223\,400\ \text{N} = 223\ \text{kN}$$

The larger column carries 57 percent more load because of its larger area and lower slenderness. Self-weight of the column ($19 \times 0.1564 \times 3.5 = 10.4$ kN) should be deducted from the safe load to get the superimposed load, about 131 kN for the 340 × 460 column.

Answer: Safe axial load ≈ 142 kN for 340 × 460 mm (131 kN superimposed after self-weight) and ≈ 223 kN for 460 × 460 mm.

When an opening is formed in a wall, the load of the masonry above it is not carried fully by the lintel. The masonry above acts as an arch (arching action) when it is strong enough, and the load tends to flow around the opening to the adjacent piers.

Arching action

Masonry over an opening forms a natural arch if there is sufficient height of masonry and well-bonded abutments (piers) at both sides, so that horizontal thrust can be resisted. A small triangular zone above the opening loads the lintel; the rest of the load arches to the sides.

        load spreads at 45 degrees to piers
          \          |           /
           \         |          /
   pier     \    masonry above /    pier
  ########   \_______|_______/    ########
  ########   |_____lintel_____|   ########
  ########   ||    opening   ||   ########

Load on the lintel (common rules)

1. Masonry load: only the weight of masonry within an equilateral triangle on the opening span $L$ (height $= 0.866L$) is taken on the lintel. Masonry above this triangle is ignored.
2. Floor / roof loads: if a floor slab or beam lies within the triangle (within about $L$ height above the opening), its load is added; if the floor lies above this level, no floor load is taken, because the arching action carries it.
3. Concentrated loads: loads within the triangle, and loads above it that fall within the span, are included; their dispersal is taken at 45 degrees.
4. The bearing of the lintel on masonry should be at least 100 mm or $L/12$.

Effect on piers

The load of the masonry above the opening is transferred to the piers on both sides. Each pier therefore carries its own wall load, the half-span loads from the adjacent openings, and any concentrated loads, and is checked as a short column for the area factor. Narrow piers between openings are critical, so the code limits their minimum width.

Step 1: Permissible stress

Each storey is fixed at both ends: $H_{eff} = 0.75 \times 3000 = 2250$ mm.

$$SR = \frac{2250}{230} = 9.78, \quad k_s = 0.89 + \frac{0.22}{2}\times 0.06 = 0.897$$

Area per metre $= 0.23$ m² $> 0.2$, so $k_a = 1.0$, $k_p = 1.0$:

$$f_c = 0.5 \times 0.897 = 0.448\ \text{N/mm}^2$$

Step 2: Self-weight per storey

$$W_{s} = 19 \times 0.23 \times 3.0 = 13.11\ \text{kN/m}$$

Step 3: Loads and stresses at each base level

Stress $= \text{load}/(230 \times 1000)$.

The ground storey stress exceeds the permissible value by about 1 percent, so it is not acceptable.

Step 4: Redesign of the ground storey with 345 mm wall

Self-weight of 345 mm ground wall $= 19 \times 0.345 \times 3.0 = 19.67$ kN/m. Load at base:

$$P = 15 + 50 + 2 \times 13.11 + 19.67 = 110.89\ \text{kN/m}$$ $$\begin{aligned} SR &= \frac{2250}{345} = 6.52, \quad k_s \approx 0.99 \\ f_c &= 0.5 \times 0.99 = 0.49\ \text{N/mm}^2 \\ f &= \frac{110\,890}{345 \times 1000} = 0.321\ \text{N/mm}^2 < 0.49 \end{aligned}$$

A 345 mm wall at ground storey is safe (stress ratio 0.65). The 230 mm upper walls rest on the 345 mm wall; the load from above passes through a 115 mm offset which is kept to the inner face (centrally aligned if possible) so that the eccentricity is small.

Answer: Stresses at base: top 0.122, middle 0.288, ground 0.454 N/mm² (permissible 0.448). The ground storey needs a 345 mm wall (stress 0.321 N/mm² < 0.49).