How is the brick compression test done (IS 3495)?

Bricks immersed 24 hours, tested flat between plywood with uniform load until failure

Dry bricks tested on edge at a sudden load

What is the strength of the brick that failed at 205 kN?

205 000/(110 x 75) = 24.8 N/mm^2

205 000/(230 x 75) = 11.9 N/mm^2

What class is given and why?

Class 7.5, since the average 7.81 is >= 7.5 but < 10

Class 10, since one brick exceeds 8

Fails, since one brick is under 7.5

How is water absorption computed?

(W_w - W_d)/W_d x 100 = 16.1 percent

Practice · 6 marks

Compressive strength test and classification of bricks

Short answer

Strengths: 8.10, 7.51, 8.62, 6.96, 7.87 N/mm²; average 7.81 N/mm² (s=0.62), class 7.5. Water absorption = (3.54-3.05)/3.05×100 = 16.1%.

Briefly describe the procedure for the compressive strength test of a brick (IS 3495). Five bricks of size 230 mm × 110 mm × 75 mm were tested flat with the following failure loads: 205, 190, 218, 176 and 199 kN. The bearing area is the full 230 × 110 mm bed. Find the compressive strength of each brick, the average strength, the standard deviation and the strength class (class 7.5 requires an average of at least 7.5 N/mm²; class 10 requires at least 10 N/mm²). A brick weighing 3.05 kg when dry weighed 3.54 kg after 24 hours in water: find the water absorption.

Test procedure (IS 3495, Part 1)

Select five representative bricks. Remove unevenness and fill frogs/depressions with 1:1 cement–sand mortar.
Immerse the bricks in water for 24 hours, then drain surface water.
Place the specimen flat (largest face) between plywood sheets or plates in the compression testing machine.
Apply load axially at a uniform rate of about 14 N/mm² per minute until failure.
Compressive strength $= \dfrac{\text{maximum load}}{\text{loaded area}}$ ; the average of the five bricks is reported.

Calculation

Area $A = 230 \times 110 = 25\,300$ mm².

Brick	Load (kN)	Strength = $P/A$ (N/mm²)
1	205	8.10
2	190	7.51
3	218	8.62
4	176	6.96
5	199	7.87

\begin{aligned} \bar f &= \frac{8.10 + 7.51 + 8.62 + 6.96 + 7.87}{5} = 7.81\ \text{N/mm}^2 \\ s &= \sqrt{\frac{\sum (f_i - \bar f)^2}{n-1}} = 0.62\ \text{N/mm}^2 \quad (\text{coefficient of variation} = 8\%) \end{aligned}

Classification

Average strength $7.81 \ge 7.5$ N/mm² but $< 10$ N/mm², so the bricks are class 7.5. The classification is based on the average strength of the five bricks.

Water absorption

\text{Absorption} = \frac{W_w - W_d}{W_d}\times100 = \frac{3.54 - 3.05}{3.05}\times100 = 16.1\%

This is below the 20 percent permitted for common bricks of this class, so it is acceptable.

Answer: Strengths 8.10, 7.51, 8.62, 6.96, 7.87 N/mm²; average 7.81 N/mm² (s = 0.62), class 7.5; water absorption 16.1%.

✓ Answer checked by Claude Sonnet 5.5

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All Design of Timber and Masonry Structures chapter 8 questions →

Compressive strength test and classification of bricks

Test procedure (IS 3495, Part 1)

Calculation

Classification

Water absorption

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