--- title: "Mortar for 1.5 m³ brickwork numerical" subject: "Building Technology" chapter: 3 key: "semester_6/building_technology:mortar-1p5m3-1to4" source: https://www.bctengineer.com/q/semester_6/building_technology/mortar-1p5m3-1to4 site: BCT Engineering --- # Mortar for 1.5 m³ brickwork numerical Building Technology · Chapter 3: Mortars & Masonry works - Asked in: 2080 Chaitra, 3+5 marks - Video: [how to calculate mortar in brick masonry | quantity of mortar in brickwork | mortar in brickwork](https://www.youtube.com/watch?v=SDSjG62i5oc) — Civil Engineering ZoNe - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/building_technology/mortar-1p5m3-1to4 - Whole chapter: https://www.bctengineer.com/md/semester_6/building_technology/chapter-3 ## Question Define with sketch and labelling the various faces of a brick. Calculate the quantity of mortar required for the construction of 1.5 m³ first class brick work using 1:4 cement mortar. ## Short answer Brick faces: stretcher (240×57 mm), header (115×57 mm), bed (240×115 mm), frog, arris. For 1.5 m³ brickwork with 1:4 mortar: wet mortar 0.372 m³, dry 0.472 m³; cement 0.094 m³ (2.7 bags) and sand 0.378 m³; bricks 717 nos. ## Answer #### Faces of a brick A brick (standard 240 x 115 x 57 mm) has the following faces and parts: - **Stretcher (side) face:** the longest vertical face, 240 x 57 mm; it shows in a stretcher course. - **Header (end) face:** the small end face, 115 x 57 mm; it shows in a header course. - **Bed (top / bottom) face:** the horizontal face, 240 x 115 mm, on which mortar is spread. - **Frog:** a depression (about 10 mm deep) in the top bed face which holds mortar. - **Arris:** the edge where two faces meet. ```text 240 (length) +----------------+ /| /| / | top (bed) / | 57 +----------------+ | (height) | | header face | | | +-------------|--+ | / stretcher | / 115 (width) +----------------+ ``` #### Mortar for 1.5 m³ brickwork in 1:4 **Given:** brickwork volume $= 1.5\ \text{m}^3$; brick size 57 mm x 115 mm x 240 mm (actual size, without mortar); mortar joint $= 10$ mm; mortar 1:4 (cement : sand). **Step 1: Size of brick with mortar** $$ (0.25 \times 0.125 \times 0.067) = 0.002094\ \text{m}^3 $$ **Step 2: Number of bricks** $$ N = \frac{1.5}{0.002094} = 716.4 \approx 717\ \text{bricks} $$ **Step 3: Volume of bricks without mortar** $$ 717 \times 0.24 \times 0.115 \times 0.057 = 1.128\ \text{m}^3 $$ **Step 4: Wet volume of mortar** $$ 1.5 - 1.128 = 0.372\ \text{m}^3 $$ **Step 5: Dry volume of mortar** (add 27% for dry bulk, voids in sand, wastage and shrinkage) $$ 0.372 \times 1.27 = 0.472\ \text{m}^3 $$ **Step 6: Cement and sand** for 1:4 (sum of ratio $= 5$) $$ \text{Cement} = \frac{1}{5} \times 0.472 = 0.094\ \text{m}^3 = 136\ \text{kg} \approx 2.7\ \text{bags (50 kg, 0.0347 m}^3\text{ each)} $$ $$ \text{Sand} = \frac{4}{5} \times 0.472 = 0.378\ \text{m}^3 $$ **Answer: Mortar = 0.372 m³ (wet), 0.472 m³ (dry); cement 0.094 m³ and sand 0.378 m³; bricks = 717 nos.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the mortar calculation, what brick size is used to find the number of bricks in 1.5 m^3 of brickwork? A. 0.240 x 0.115 x 0.057 m B. 0.25 x 0.125 x 0.067 m C. 0.25 x 0.115 x 0.057 m D. 0.24 x 0.125 x 0.067 m Answer: B. Each dimension gets a 10 mm mortar joint added: 250 x 125 x 67 mm, giving 0.002094 m^3. (Key point: Brick size with mortar) 2. How is the wet volume of mortar obtained? A. Volume of bricks multiplied by 1.27 B. Volume of brickwork minus volume of bricks without mortar C. Volume of brickwork divided by 5 D. Number of bricks times 0.002094 Answer: B. Wet mortar = 1.5 - 1.128 = 0.372 m^3, i.e. total masonry minus the solid bricks. (Key point: Wet volume of mortar) 3. Why is the wet mortar volume multiplied by 1.27? A. To include the volume of the frog B. To add the volume of the bricks C. To allow for dry bulk, voids in sand, wastage and shrinkage D. To convert the 1:4 ratio into cement bags Answer: C. Dry materials occupy more volume than wet mortar, so 27% is added: 0.372 x 1.27 = 0.472 m^3. (Key point: Dry volume factor) 4. For 1:4 mortar with dry volume 0.472 m^3, what are the cement and sand volumes? A. 0.118 m^3 and 0.354 m^3 B. 0.094 m^3 and 0.378 m^3 C. 0.236 m^3 and 0.236 m^3 D. 0.094 m^3 and 0.472 m^3 Answer: B. Sum of ratio is 5, so cement = 0.472/5 = 0.094 m^3 and sand = 4/5 x 0.472 = 0.378 m^3. 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