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Single-storey external wall: stress, brick, mortar

Short answer

Maximum stress 0.66 N/mm². Required basic compressive stress 0.57 N/mm². Use bricks of crushing strength not less than 7.5 N/mm² with M2 mortar (cement:sand 1:6), giving permissible stress 0.59 N/mm².

An external wall of a single storeyed building is 23 cm thick and carries 100 kN/m load at the top of the wall with an eccentricity of 12 mm. The plinth level is 1.5 m above the top of foundation footing and floor to ceiling height is 3.0 m. The RCC slab bears on the wall and is 12 cm thick. Determine the maximum stress in the wall and calculate the strength of brick and grade of mortar required for the wall. Assume necessary data if any required.

Data and assumptions

  • Wall 230 mm thick, modular bricks (kp=1k_p=1); load at top 100 kN/m with eccentricity 12 mm, so e/t=12/230=0.0522e/t=12/230=0.0522 (>1/24>1/24).
  • Plinth 1.5 m above top of footing, ceiling height 3.0 m, 12 cm RCC slab bearing on the wall. Walls longer than hefh_{ef} (no cross walls given), so height governs.
  • Brick masonry 19 kN/m3^3. Design by IS 1905:1987; NBC 109 follows the same method. Critical section: base of wall at the top of the footing.

Step 1: Effective height and slenderness ratio

Height from top of footing to the centre of slab H=1.5+3.0+0.12/2=4.56H = 1.5+3.0+0.12/2 = 4.56 m (IS 1905 Table 4 Note 1). The footing and the RCC slab give full restraint, so hef=0.75Hh_{ef}=0.75H:

hef=0.75×4.56=3.420 mSR=3.4200.23=14.87 (<27)\begin{aligned} h_{ef} &= 0.75\times4.56 = 3.420\ \text{m}\\ SR &= \frac{3.420}{0.23} = 14.87\ (<27) \end{aligned}

Step 2: Load and maximum stress at the base

Self weight=0.23×4.5×19=19.67 kN/mP=100+19.67=119.67 kN/mM=100×0.012=1.20 kNm/mσmax=PA+6Mt2=119.67×103230×103+6×1.20×1061000×2302=0.520+0.136=0.656 N/mm2\begin{aligned} \text{Self weight} &= 0.23\times4.5\times19 = 19.67\ \text{kN/m}\\ P &= 100+19.67 = 119.67\ \text{kN/m}\\ M &= 100\times0.012 = 1.20\ \text{kNm/m}\\ \sigma_{max} &= \frac{P}{A}+\frac{6M}{t^2} = \frac{119.67\times10^3}{230\times10^3}+\frac{6\times1.20\times10^6}{1000\times230^2} = 0.520+0.136 = 0.656\ \text{N/mm}^2 \end{aligned}

(At the top of the wall, under the load alone, σmax=100×103230×103(1+6×12230)=0.571\sigma_{max}=\frac{100\times10^3}{230\times10^3}(1+\frac{6\times12}{230}) = 0.571 N/mm2^2.) The maximum stress in the wall is 0.656 N/mm2^2 at the base (compression; no tension since e<t/6e<t/6).

Step 3: Factors and required basic stress

  • ksk_s (Table 9, SR=14.87SR=14.87, e/t=0.052e/t=0.052, interpolated) =0.732= 0.732
  • ka=1.0k_a=1.0 (area 0.23 m2^2/m), kp=1.0k_p=1.0
  • e/t>1/24e/t>1/24: 25 percent increase allowed (cl. 5.4.1.4 a)
fb≥σkskakp×1.25=0.5200.732×1.25=0.569 N/mm2f_b \ge \frac{\sigma}{k_sk_ak_p\times1.25} = \frac{0.520}{0.732\times1.25} = 0.569\ \text{N/mm}^2

Step 4: Brick and mortar (IS 1905 Table 8)

Mortar (IS 1905 Table 1)Min. unit strength (N/mm²)Table 8 fbf_b (N/mm²)
M2 (cement:sand 1:6)7.50.59
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: maximum stress = 0.656 N/mm2^2. Required fb=0.569f_b = 0.569 N/mm2^2, so use bricks of crushing strength not less than 7.5 N/mm2^2 with M2 mortar (1:6, fb=0.59f_b=0.59 N/mm2^2). A richer mortar (M1, H2, H1) also needs 7.5 N/mm2^2 units, so there is no gain from a stronger mortar here.

✓ Answer checked by Claude Opus 5.5 (high effort)

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