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Load bearing wall: crushing strength and mortar

Short answer

Required basic compressive stress 0.366 N/mm²; bricks of crushing strength not less than 5 N/mm² with cement mortar M2 (1:6) are adequate.

A load bearing brick masonry wall of a building is 250 cm [sic] thick, is laterally supported by RCC slabs at top and bottom, which are 13 cm thick each, and the clear height between slabs is 3.5 m. If the wall has an axial load of 79.5 kN/m at the base, inclusive of self weight, what should be the crushing strength of bricks and grade of mortar for the wall? The wall is 5 m long between cross walls and bricks used are of modular size. Assume suitable data if any required.

Data and assumptions

  • "250 cm" is read as 250 mm (25 cm) thick, since a 2.5 m thick load-bearing wall is not meaningful. Modular bricks, kp=1k_p=1.
  • Axial load 79.5 kN/m at the base inclusive of self weight (e=0e=0). Wall 5 m long between cross walls; clear height 3.5 m between 13 cm RCC slabs, so H=3.5+0.13=3.63H=3.5+0.13=3.63 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=3.63H = 3.63 m. Effective height hef=0.75H=0.75×3.63=2.723h_{ef} = 0.75 H = 0.75\times3.63 = 2.723 m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) Lef=1.0L=5.00L_{ef} = 1.0L = 5.00 m. The lesser value governs:

SR=heft=2.7230.250=10.89 (<27, Table 7, cement mortar)SR = \frac{h_{ef}}{t} = \frac{2.723}{0.250} = 10.89\ (<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=79.50P = 79.50 kN/m.

Stress on the wall:

σ=PA=79.50×103250×103 mm2/m=0.318 N/mm2\sigma = \frac{P}{A} = \frac{79.50\times10^3}{250\times10^3\ \text{mm}^2/\text{m}} = 0.318\ \text{N/mm}^2

Step 3: Reduction factors

  • Stress reduction factor ksk_s (Table 9, SR=10.89SR=10.89, e/t=0.000e/t=0.000, by linear interpolation) =0.868= 0.868
  • Area reduction factor kak_a (cl. 5.4.1.2): area per metre 0.25 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)
  • e/t≤1/24e/t \le 1/24: no increase

Step 4: Required basic compressive stress

σ≤fb ks ka kp×1.00⇒fb≥0.3180.868×1.0×1.0×1.00=0.366 N/mm2\sigma \le f_b\,k_s\,k_a\,k_p\times1.00 \Rightarrow f_b \ge \frac{0.318}{0.868\times1.0\times1.0\times1.00} = 0.366\ \text{N/mm}^2

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

Smallest standard unit strength giving fb≥0.366f_b \ge 0.366 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 0.366). 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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