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Brick and mortar: wall 175 kN/m, 4.2 m

Short answer

Required basic compressive stress 0.72 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 wall is 4.2 m long between cross walls and the clear height of the wall is 3.2 m between RCC slabs of 12 cm thick at top and bottom. The brick wall carries an eccentric load of 175 kN/m (eccentricity of loading at 1/12) inclusive of self weight. What should be the strength of brick and grade of mortar? Use Nepal Standard Brick and joints are not raked.

Data and assumptions

  • Thickness not given: Nepal standard brick 230×110×55230\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=1/12e/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=3.32H=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/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=3.32H = 3.32 m. Effective height hef=0.75H=0.75×3.32=2.490h_{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) Lef=1.0L=4.20L_{ef} = 1.0L = 4.20 m. The lesser value governs:

SR=heft=2.4900.230=10.83 (<27, Table 7, cement mortar)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=175.00P = 175.00 kN/m.

Stress on the wall:

σ=PA=175.00×103230×103 mm2/m=0.761 N/mm2\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 ksk_s (Table 9, SR=10.83SR=10.83, e/t=0.083e/t=0.083, by linear interpolation) =0.845= 0.845
  • 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≥0.7610.845×1.0×1.0×1.25=0.720 N/mm2\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 fb≥0.900f_b \ge 0.900 N/mm2^2.)

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

Smallest standard unit strength giving fb≥0.720f_b \ge 0.720 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 0.720). 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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