Skip to main content

Brick and mortar for 23 cm wall (165 kN/m)

Short answer

Required basic compressive stress 0.67 N/mm². Use bricks of crushing strength not less than 10 N/mm² with M2 mortar (cement:sand 1:6); permissible stress 0.81 N/mm².

A brick wall 23 cm thick using modular brick carries an eccentric load of 165 kN/m at base (eccentricity ratio at 1/12). The wall is 4.5 m long between cross walls. The clear height of the wall is 3.1 m between RCC slabs of 10 cm thick at top and bottom. What should be the strength of brick and grade of mortar? Assume that joints are not raked.

Data and assumptions

  • Wall 23 cm thick, modular bricks, joints not raked. Load 165 kN/m at base inclusive of self weight, e/t=1/12e/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=3.2H=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/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=3.20H = 3.20 m. Effective height hef=0.75H=0.75×3.20=2.400h_{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) Lef=1.0L=4.50L_{ef} = 1.0L = 4.50 m. The lesser value governs:

SR=heft=2.4000.230=10.43 (<27, Table 7, cement mortar)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=165.00P = 165.00 kN/m.

Stress on the wall:

σ=PA=165.00×103230×103 mm2/m=0.717 N/mm2\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 ksk_s (Table 9, SR=10.43SR=10.43, e/t=0.083e/t=0.083, by linear interpolation) =0.857= 0.857
  • Area reduction factor kak_a (cl. 5.4.1.2): area per metre 0.23 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≥0.7170.857×1.0×1.0×1.25=0.670 N/mm2\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 fb≥0.837f_b \ge 0.837 N/mm2^2.)

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

Smallest standard unit strength giving fb≥0.670f_b \ge 0.670 N/mm2^2:

Mortar (IS 1905 Table 1)Min. unit strength (N/mm²)Table 8 fbf_b (N/mm²)
M2 (cement:sand 1:6)100.81
M1 (cement:sand 1:5)7.50.74
H2 (cement:sand 1:4)7.50.74
H1 (cement:sand 1:3)7.50.75

Answer: use bricks of crushing strength not less than 10 N/mm2^2 with mortar M2 (cement:sand 1:6) (fb=0.81f_b = 0.81 N/mm2^2 ≥\ge 0.670). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Concrete Technology and Masonry Structure chapter 8 questions →