---
title: "Interior cross wall design: stiffened, 130 mm slab"
subject: "Concrete Technology and Masonry Structure"
chapter: 8
key: "semester_5/concrete_technology_and_masonry_structure:interior-cross-wall-stiffened-130mm"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/interior-cross-wall-stiffened-130mm
site: BCT Engineering
---

# Interior cross wall design: stiffened, 130 mm slab

Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads

- Asked in: 2074 Chaitra, 9 marks
- Video: [Masonry Wall Design - Part 2 || Design of Interior Wall with Pier || IS 1905:1987 || Nepali Language](https://www.youtube.com/watch?v=AhzBHwTw6O8) — Sumit Paudel
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/interior-cross-wall-stiffened-130mm
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8

## Question

Design an interior cross-wall of a two-storeyed building to carry 130 mm thick RCC slab with ceiling height of 2.8 m, and the wall is 3.2 m long which is stiffened and supports a 2 m slab on both sides as shown in the figure. Assume all necessary data relevant to Nepal code. [Figure: cross section of wall - 2000 mm slab span on each side, 130 mm thick RCC slabs, 2.8 m storey height for two storeys, on foundation.]

## Short answer

230 mm thick wall; slenderness ratio 9.55 (ks=0.903), ground storey stress 0.211 N/mm², required basic compressive stress 0.233 N/mm². Use bricks of at least 5 N/mm² with M2 (1:6) mortar.

## Answer

#### 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, $k_p=1$. Brick masonry 19 kN/m$^3$, RCC 25 kN/m$^3$.
- 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 3.2 m long and stiffened by cross walls at its ends (effective length $1.0L=3.2$ m, Table 5), which is greater than $h_{ef}$, so height governs the slenderness. Slab 130 mm, ceiling 2.8 m, slab span 2.0 m each side (tributary 2.0 m).
- Finish 1.0 kN/m$^2$, live load 1.5 kN/m$^2$ (roof) and 2.0 kN/m$^2$ (floor) assumed as per IS 875 Part 2 for residential use.

#### Step 1: Loads per metre run of wall

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

| Item | Working | kN/m |
|---|---|---|
| Slab dead + finish | $25\times0.130 + 1.0 = 4.250$ kN/m$^2$ | - |
| Roof slab reaction | $(4.250 + 1.5)\times2.0$ | 11.50 |
| Floor slab reaction | $(4.250 + 2.0)\times2.0$ | 12.50 |
| Wall per storey | $0.23\times2.80\times19$ | 12.24 |

#### Step 2: Slenderness ratio and factors

Interior wall with slabs on both sides: load is axial ($e=0$, IS 1905 App. A-4). RCC slab bearing on the wall gives full restraint (Table 4 Sl. 1): $h_{ef}=0.75H$, with $H=2.80+0.130=2.930$ m.

$$
\begin{aligned} h_{ef} &= 0.75\times2.930 = 2.197\ \text{m}\\ SR &= \frac{2.197}{0.23} = 9.55\ (<27) \end{aligned}
$$

$k_s = 0.903$ (Table 9, $e=0$, interpolated), $k_a = 1.0$ (area 0.23 m$^2$/m $\ge 0.2$), $k_p=1.0$.

#### Step 3: Stress and required $f_b$ at the base of each storey

| Section | Load $N$ (kN/m) | $\sigma=N/(t\times1000)$ (N/mm²) | Required $f_b=\sigma/(k_sk_ak_p)$ (N/mm²) |
|---|---|---|---|
| Top (2nd) storey base | 23.74 | 0.103 | 0.114 |
| Ground storey base | 48.47 | 0.211 | 0.233 |

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

Smallest standard unit strength for each section:

| Section | Required $f_b$ | M2 (1:6) | M1 (1:5) | H2 (1:4) |
|---|---|---|---|---|
| Top (2nd) storey base | 0.114 | 3.5 | 3.5 | 3.5 |
| Ground storey base | 0.233 | 3.5 | 3.5 | 3.5 |

**Answer: 230 mm thick wall; the ground storey governs, requiring 3.5 N/mm$^2$ brick with M2 (1:6) mortar** (required $f_b=0.233$ N/mm$^2$; $f_b$ provided 0.35). The same brick and mortar can be used throughout. As a practical minimum, adopt bricks of not less than 5 N/mm$^2$ with 1:6 cement mortar (SR = 9.55 < 27, OK).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Roof slab reaction per metre run of wall is found by:
   A. (slab dead + finish + live) x tributary width
   B. (slab dead + live) x ceiling height
   C. Slab dead load only x tributary width
   D. Wall self weight x tributary width
   Answer: A. Reaction = (dead + finish + live, kN/m^2) x tributary width, giving 11.50 kN/m for the roof. (Key point: Slab load per metre)

2. For an RCC slab bearing on the wall (full restraint), the effective height is:
   A. h_ef = 0.75H with H = clear height + slab thickness
   B. h_ef = 2.0H
   C. h_ef = 1.0H using clear height only
   D. h_ef = 0.5H
   Answer: A. Full restraint gives h_ef = 0.75H, so H = 2.93 m, h_ef = 2.197 m and SR = 9.55 (< 27). (Key point: Effective height)

3. Which section governs the brick strength, and why?
   A. Top storey base, since it is thinner
   B. Ground storey base, since it carries both slabs and both wall weights (48.47 kN/m)
   C. Both give identical loads
   D. Roof level, since live load is least there
   Answer: B. Load accumulates downward, so the ground storey base has the larger stress; required f_b = 0.233 N/mm^2. (Key point: Governing section)

4. How is the required f_b found and what is the result?
   A. f_b = sigma x k_s x k_a x k_p, giving 10 N/mm^2 bricks
   B. f_b = sigma/(k_s k_a k_p) with k_s = 0.903; Table 8 gives 3.5 N/mm^2 brick with M2 (1:6) mortar
   C. f_b = N x t; H1 mortar is needed
   D. f_b = sigma/1.25; no brick strength is needed
   Answer: B. f_b = sigma/(k_s k_a k_p) = 0.233 N/mm^2, which Table 8 satisfies with 3.5 N/mm^2 brick and M2 mortar (5 N/mm^2 adopted in practice). (Key point: Required f_b and result)

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