--- title: "Brick and mortar for wall with timber flooring" subject: "Concrete Technology and Masonry Structure" chapter: 8 key: "semester_5/concrete_technology_and_masonry_structure:wall-eccentric-230mm-135kn-timber" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wall-eccentric-230mm-135kn-timber site: BCT Engineering --- # Brick and mortar for wall with timber flooring Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads - Asked in: 2076 Chaitra, 12 marks - 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/wall-eccentric-230mm-135kn-timber - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8 ## Question A wall 230 mm thick, using local bricks, carries 135 kN/m load at top of wall having eccentricity ratio of 1/12. The wall is 4 m long between cross walls and is 3 m clear height between RCC slab at bottom and timber flooring top. What shall be the strength of brick and grade of mortar? Assume necessary data if any required. ## Short answer Required basic compressive stress 0.62 N/mm². Use bricks of crushing strength not less than 10 N/mm² with M2 mortar (cement:sand 1:6); permissible stress 0.81 N/mm². ## Answer #### Data and assumptions - Wall 230 mm thick, local bricks (assumed $230\times110\times55$ mm, $k_p=1$), $e/t=1/12$, load 135 kN/m at the top. Wall 4 m long, clear height 3 m. - The RCC slab at the bottom gives lateral and rotational restraint; the timber floor at the top is assumed to be adequately anchored but not spanning onto the wall (lateral restraint only). Hence Table 4 Sl. 2 applies: $h_{ef}=0.85H$. - Design code: IS 1905:1987; NBC 109 follows the same method. - Brick masonry 19 kN/m$^3$ (IS 875 Part 1). #### Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6) Height of the wall $H = 3.00$ m (clear height, as no slab thickness is given). Effective height $h_{ef} = 0.85 H = 0.85\times3.00 = 2.550$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 4.00$ m. The lesser value governs: $$ SR = \frac{h_{ef}}{t} = \frac{2.550}{0.230} = 11.09\ (<27,\ \text{Table 7, cement mortar}) $$ #### Step 2: Design load per metre run at the base $$ \begin{aligned} \text{Self weight} &= t\,H_c\,\gamma = 0.230\times3.00\times19 = 13.11\ \text{kN/m}\\ P &= 135.00 + 13.11 = 148.11\ \text{kN/m} \end{aligned} $$ Stress on the wall: $$ \sigma = \frac{P}{A} = \frac{148.11\times10^3}{230\times10^3\ \text{mm}^2/\text{m}} = 0.644\ \text{N/mm}^2 $$ #### Step 3: Reduction factors - Stress reduction factor $k_s$ (Table 9, $SR=11.09$, $e/t=0.083$, by linear interpolation) $= 0.837$ - Area reduction factor $k_a$ (cl. 5.4.1.2): the area per metre, 0.23 m$^2$, is not less than 0.2 m$^2$, so $k_a = 1.0$ - Shape factor $k_p = 1.0$ (brick height/width $\le 0.75$, Table 10) - Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25 #### Step 4: Required basic compressive stress $$ \sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.644}{0.837\times1.0\times1.0\times1.25} = 0.615\ \text{N/mm}^2 $$ (Without the 25 percent increase the requirement would be $f_b \ge 0.769$ N/mm$^2$.) #### Step 5: Strength of brick and grade of mortar (IS 1905 Table 8) Smallest standard unit strength giving $f_b \ge 0.615$ N/mm$^2$: | Mortar (IS 1905 Table 1) | Min. unit strength (N/mm²) | Table 8 $f_b$ (N/mm²) | |---|---|---| | M2 (cement:sand 1:6) | 10 | 0.81 | | M1 (cement:sand 1:5) | 7.5 | 0.74 | | H2 (cement:sand 1:4) | 7.5 | 0.74 | | H1 (cement:sand 1:3) | 7.5 | 0.75 | **Answer: use bricks of crushing strength not less than 10 N/mm$^2$ with mortar M2 (cement:sand 1:6)** ($f_b = 0.81$ N/mm$^2$ $\ge$ 0.615). The other combinations in the table are also acceptable; a stronger mortar allows a weaker brick (e.g. 7.5 N/mm$^2$ bricks with M1 mortar). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With RCC slab at the bottom and timber floor at the top (lateral restraint only), h_ef is: A. 0.75H B. 0.85H C. 1.0H D. 2.0H Answer: B. Table 4 Sl. 2 (lateral but no rotational restraint at one end) gives h_ef = 0.85H = 2.55 m. (Key point: Effective height factor) 2. SR with h_ef = 2.55 m and t = 0.23 m is: A. 11.09 B. 13.04 C. 8.3 D. 22.2 Answer: A. SR = 2.55/0.23 = 11.09, less than 27. (Key point: Slenderness) 3. Design load at the base is: A. 135 kN/m B. 135 + 13.11 = 148.11 kN/m C. 135 - 13.11 kN/m D. 13.11 kN/m Answer: B. Load acts at top, so self weight 0.23 x 3 x 19 = 13.11 kN/m is added for the base check. (Key point: Design load) 4. With k_s = 0.837 and the 25 percent increase, f_b is 0.615. The choice with M2 is: A. 7.5 N/mm^2 units B. 10 N/mm^2 units (f_b 0.81) C. 5 N/mm^2 units D. 3.5 N/mm^2 units Answer: B. 7.5 N/mm^2 with M2 gives 0.59 < 0.615, so 10 N/mm^2 is required. 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