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Cross wall design, 120 mm slab, 3 m wide

Short answer

230 mm thick wall; ground storey base stress 0.290 N/mm², required basic compressive stress 0.328 N/mm², slenderness ratio 10.17 (ks=0.886). Use bricks of at least 5 N/mm² with M2 (1:6) cement mortar.

Design an interior cross wall of a two storeyed building to carry 120 mm thick RCC slab with 3.0 m ceiling height. The wall is unstiffened and supports a 3.0 m wide slab on both sides. Assume necessary data relevant to Nepal. Live load on roof = 2 kN/m2^2; live load on floor = 2.5 kN/m2^2; floor finishing = 1.5 kN/m2^2. [Figure: slab, floor and brick wall in cross section.]

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 120 mm, ceiling 3.0 m, tributary width 3.0 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 + finish25×0.120+1.5=4.50025\times0.120 + 1.5 = 4.500 kN/m2^2-
Roof slab reaction(4.500+2.0)×3.0(4.500 + 2.0)\times3.019.50
Floor slab reaction(4.500+2.5)×3.0(4.500 + 2.5)\times3.021.00
Wall per storey0.23×3.00×190.23\times3.00\times1913.11

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=3.00+0.120=3.120H=3.00+0.120=3.120 m.

hef=0.75×3.120=2.340 mSR=2.3400.23=10.17 (<27)\begin{aligned} h_{ef} &= 0.75\times3.120 = 2.340\ \text{m}\\ SR &= \frac{2.340}{0.23} = 10.17\ (<27) \end{aligned}

ks=0.886k_s = 0.886 (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 base32.610.1420.160
Ground storey base66.720.2900.328

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 base0.1603.53.53.5
Ground storey base0.3283.53.53.5

Answer: 230 mm thick wall; the ground storey governs, requiring 3.5 N/mm2^2 brick with M2 (1:6) mortar (required fb=0.328f_b=0.328 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 = 10.17 < 27, OK).

✓ Answer checked by Claude Opus 5.5 (high effort)

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