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Cross wall design, 125 mm slab, 3.2 m

Short answer

230 mm thick interior cross wall, effective height 2.494 m, slenderness ratio 10.84. Required basic compressive stress 0.292 N/mm²; adopt bricks of crushing strength not less than 3.5 N/mm² with M2 mortar (1:6 cement:sand).

Design an interior cross wall of a two-storeyed building to carry 125 mm thick RCC slab with 3.2 m ceiling height. The wall is unstiffened and supports a 2.5 m wide slab on both sides. Assume necessary data relevant to Nepal. Live load on roof = 1.5 kN/m2^2; live load on floor = 2.0 kN/m2^2; floor finishing = 1.2 kN/m2^2. [Figure: brick wall between slabs with 3200 mm storey heights, on a floor.]

Data and assumptions

  • Design code: IS 1905:1987; NBC 109 follows the same permissible-stress method.
  • Interior cross wall 230 mm thick (one brick, Nepal practice), modular bricks, kp=1k_p=1. Brick masonry 19 kN/m3^3, RCC 25 kN/m3^3.
  • Stated slab width is taken as the tributary width of slab per metre run of wall (each side contributing half of its span). Finish applies to roof and floors; live load not reduced.
  • Sections are checked at the base of each storey (wall height for weight = clear ceiling height).
  • Wall unstiffened, so slenderness is governed by height. Slab 125 mm, ceiling height 3.2 m, tributary width 2.5 m.

Step 1: Loads per metre run of wall

Unit weights (IS 875 Part 1): RCC 25 kN/m3^3, brick masonry 19 kN/m3^3.

ItemWorkingkN/m
Slab dead + finish25×0.125+1.2=4.32525\times0.125 + 1.2 = 4.325 kN/m2^2-
Roof slab reaction(4.325+1.5)×2.5(4.325 + 1.5)\times2.514.56
Floor slab reaction(4.325+2.0)×2.5(4.325 + 2.0)\times2.515.81
Wall per storey0.23×3.20×190.23\times3.20\times1913.98

Step 2: Slenderness ratio and factors

Interior wall with slabs on both sides: load is axial (e=0e=0, IS 1905 App. A-4). RCC slab bearing on the wall gives full restraint (Table 4 Sl. 1): hef=0.75Hh_{ef}=0.75H, with H=3.20+0.125=3.325H=3.20+0.125=3.325 m.

hef=0.75×3.325=2.494 mSR=2.4940.23=10.84 (<27)\begin{aligned} h_{ef} &= 0.75\times3.325 = 2.494\ \text{m}\\ SR &= \frac{2.494}{0.23} = 10.84\ (<27) \end{aligned}

ks=0.869k_s = 0.869 (Table 9, e=0e=0, interpolated), ka=1.0k_a = 1.0 (area 0.23 m2^2/m ≥0.2\ge 0.2), kp=1.0k_p=1.0.

Step 3: Stress and required fbf_b at the base of each storey

SectionLoad NN (kN/m)σ=N/(t×1000)\sigma=N/(t\times1000) (N/mm²)Required fb=σ/(kskakp)f_b=\sigma/(k_sk_ak_p) (N/mm²)
Top (2nd) storey base28.550.1240.143
Ground storey base58.340.2540.292

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

Smallest standard unit strength for each section:

SectionRequired fbf_bM2 (1:6)M1 (1:5)H2 (1:4)
Top (2nd) storey base0.1433.53.53.5
Ground storey base0.2923.53.53.5

Answer: 230 mm thick wall; the ground storey governs, requiring 3.5 N/mm2^2 brick with M2 (1:6) mortar (required fb=0.292f_b=0.292 N/mm2^2; fbf_b provided 0.35). The same brick and mortar can be used throughout. As a practical minimum, adopt bricks of not less than 5 N/mm2^2 with 1:6 cement mortar (SR = 10.84 < 27, OK).

✓ Answer checked by Claude Opus 5.5 (high effort)

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