Testing of Masonry Elements

Test procedure (IS 3495, Part 1)

1. Select five representative bricks. Remove unevenness and fill frogs/depressions with 1:1 cement–sand mortar.
2. Immerse the bricks in water for 24 hours, then drain surface water.
3. Place the specimen flat (largest face) between plywood sheets or plates in the compression testing machine.
4. Apply load axially at a uniform rate of about 14 N/mm² per minute until failure.
5. 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.

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

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%.

(a) Diagonal compression (shear) test

For a square panel (ASTM E519), the load $P$ along the diagonal produces pure shear along the other diagonal. The shear stress is

$$f_v = \frac{0.707\,P}{A_n}, \qquad A_n = \frac{w + h}{2}\,t\,n$$

where $w$ and $h$ are the width and height of the panel, $t$ the thickness, and $n$ the percentage of solid area (1 for solid wall).

$$\begin{aligned} A_n &= \frac{1200 + 1200}{2} \times 230 \times 1 = 276\,000\ \text{mm}^2 \\ f_v &= \frac{0.707 \times 160\,000}{276\,000} = 0.41\ \text{N/mm}^2 \end{aligned}$$

(b) Single flat-jack test

A slot is cut in a mortar bed joint, which relieves the vertical stress and reduces the distance between gauge points. A flat jack is inserted and pressurised until the original distance is restored. The in-situ stress is

$$\sigma_m = K_m K_a\, p$$

where $p$ is the jack pressure, $K_m$ the jack constant (stiffness of the jack) and $K_a$ the ratio of jack area to slot area.

$$\sigma_m = 0.74 \times 0.85 \times 0.62 = 0.39\ \text{N/mm}^2$$

(c) Push shear test

A brick is freed by removing the head joints, and a jack pushes it along the bed joints. The shear area is that of the two bed joints:

$$A = 2 \times 230 \times 110 = 50\,600\ \text{mm}^2, \qquad \tau = \frac{14\,000}{50\,600} = 0.277\ \text{N/mm}^2$$

If there is overburden stress $\sigma_v$, the initial shear strength is corrected as $f_{v0} = \tau - \mu\,\sigma_v$ (with friction $\mu \approx 0.4$ - 0.6).

Answer: (a) 0.41 N/mm²; (b) 0.39 N/mm²; (c) 0.28 N/mm².

Ultrasonic pulse velocity (UPV) test

A transmitter transducer generates an ultrasonic pulse (typically 50–150 kHz) at one point on the wall, and a receiver picks it up at another. The electronic unit measures the travel time $t$. The pulse velocity is

$$V = \frac{L}{t}$$

where $L$ is the path length between transducers.

Arrangement: direct (transducers on opposite faces), semi-direct (on adjacent faces), or indirect (same face). Direct is the most reliable. A couplant (grease or gel) is used for good contact.

  Transmitter ))))))))  path L  )))))))) Receiver
      [T]=========== wall ===========[R]
          \_______ time t _______/

Uses

• Compare quality: high velocity means dense, sound masonry; low velocity means voids, cracks, loose fill or damage.
• Estimate the dynamic elastic modulus: $E_d = \rho V^2\dfrac{(1+\nu)(1-2\nu)}{(1-\nu)}$.
• Locate flaws and check grout injection; estimate strength using calibrated correlations.

Limitations: results depend on moisture, mortar and unit properties, so the correlation to strength is poor; the pulse is scattered in heterogeneous masonry such as rubble stone, and the method needs access to both faces.

Elastic wave tomography

Many source and receiver points are placed around the wall section (or in a grid on both faces). The travel time of each ray is measured by hammer impact or ultrasonic transducers. A computer inversion (e.g. iterative ray tracing) divides the section into cells and finds the velocity in each cell so that the calculated travel times match the measured times. The result is a velocity map (image) showing the internal condition.

   S1  S2  S3  S4        Source points on one side
   \\\ ||| /// \\
   .-------------.       cell grid
   | 2.1  1.4 2.0|       velocity (km/s) in cells;
   | 2.2 [0.8]2.1|       low value [0.8] = void
   '-------------'
   R1  R2  R3  R4        Receivers

Uses: finding voids, cracks, delaminations, and the inner core quality of thick stone and multi-leaf walls, and checking the success of grouting.

Limitations: needs many readings and processing software; resolution is limited by wave length and number of rays; interpretation requires experience.

Both tests are non-destructive (no damage to the heritage fabric) and are used together with semi-destructive tests such as the flat-jack test for calibration.