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Single-room wall: brick strength and mortar

Short answer

Required basic compressive stress 0.385 N/mm²; use bricks of crushing strength not less than 5 N/mm² with cement mortar M2 (1:6).

A brick masonry wall of a single room building is 20 cm thick and is supported by 10 cm thick RCC slab at its top and bottom. The wall carries a vertical load (inclusive of its own weight) of 8000 kg/m at the base at an eccentricity ratio of 0.1. The length of the wall is 3 m between cross walls. The clear height of storey is 3 m. Determine the required crushing strength of bricks and the type of mortar to be used. Use modular bricks.

Data and assumptions

  • Wall 20 cm thick, modular bricks, 10 cm RCC slabs at top and bottom. Load inclusive of self weight =8000 kg/m×9.81=78.48=8000\ \text{kg/m}\times9.81 = 78.48 kN/m, e/t=0.1e/t=0.1.
  • Wall 3 m long between cross walls, clear height 3 m, so H=3+0.10=3.1H=3+0.10=3.1 m.
  • Design code: IS 1905:1987 (Tables 4 to 10); NBC 109 follows the same permissible-stress method.
  • Brick masonry unit weight 19 kN/m3^3 (IS 875 Part 1); modular brick, height/width ≤0.75\le 0.75, so kp=1k_p=1.

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

Height between slab centres H=3.10H = 3.10 m. Effective height hef=0.75H=0.75×3.10=2.325h_{ef} = 0.75 H = 0.75\times3.10 = 2.325 m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) Lef=1.0L=3.00L_{ef} = 1.0L = 3.00 m. The lesser value governs:

SR=heft=2.3250.200=11.62 (<27, Table 7, cement mortar)SR = \frac{h_{ef}}{t} = \frac{2.325}{0.200} = 11.62\ (<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=78.48P = 78.48 kN/m.

Stress on the wall:

σ=PA=78.48×103200×103 mm2/m=0.392 N/mm2\sigma = \frac{P}{A} = \frac{78.48\times10^3}{200\times10^3\ \text{mm}^2/\text{m}} = 0.392\ \text{N/mm}^2

Step 3: Reduction factors

  • Stress reduction factor ksk_s (Table 9, SR=11.62SR=11.62, e/t=0.100e/t=0.100, by linear interpolation) =0.816= 0.816
  • Area reduction factor kak_a (cl. 5.4.1.2): area per metre 0.20 m2^2 is not less than 0.2 m2^2, so ka=1.0k_a = 1.0
  • Shape factor kp=1.0k_p = 1.0 (modular bricks, height/width ≤0.75\le 0.75, Table 10)
  • Since e/t>1/24e/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

σ≤fb ks ka kp×1.25⇒fb≥0.3920.816×1.0×1.0×1.25=0.385 N/mm2\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.392}{0.816\times1.0\times1.0\times1.25} = 0.385\ \text{N/mm}^2

(Without the 25 percent increase the requirement would be fb≥0.481f_b \ge 0.481 N/mm2^2.)

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

Smallest standard unit strength giving fb≥0.385f_b \ge 0.385 N/mm2^2:

Mortar (IS 1905 Table 1)Min. unit strength (N/mm²)Table 8 fbf_b (N/mm²)
M2 (cement:sand 1:6)50.44
M1 (cement:sand 1:5)50.50
H2 (cement:sand 1:4)50.50
H1 (cement:sand 1:3)50.50

Answer: use bricks of crushing strength not less than 5 N/mm2^2 with mortar M2 (cement:sand 1:6) (fb=0.44f_b = 0.44 N/mm2^2 ≥\ge 0.385). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.

✓ Answer checked by Claude Opus 5.5 (high effort)

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