--- title: "Diagonal shear test and flat-jack test calculations" subject: "Design of Timber and Masonry Structures" chapter: 8 key: "semester_5/design_of_timber_and_masonry_structures:diagonal-shear-flat-jack-numerical" source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/diagonal-shear-flat-jack-numerical site: BCT Engineering --- # Diagonal shear test and flat-jack test calculations Design of Timber and Masonry Structures · Chapter 8: Testing of Masonry Elements - Asked in: year not given, 6 marks - Video: [Flat Jack Test | How to perform Flat Jack Test on Stone and Concrete Masonry | Jack Flat Test |](https://www.youtube.com/watch?v=RLiDBKoyZFM) — Construction Factory - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/diagonal-shear-flat-jack-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-8 ## Question (a) A 1.2 m × 1.2 m square brick masonry panel, 230 mm thick, is tested in diagonal compression and fails at a load of 160 kN along the diagonal. Calculate the shear strength of the masonry. (b) In a single flat-jack test, the pressure needed to restore the original distance between the gauge points was 0.62 N/mm². The jack constants are $K_m = 0.74$ and $K_a = 0.85$. Find the in-situ stress in the masonry. (c) In a push shear test, a brick is pushed horizontally and fails at 14 kN; two bed joints, each 230 mm × 110 mm, carry the shear. Find the shear stress. ## Short answer Diagonal: f_v = 0.707×160000/(1200×230)=0.41 N/mm²; flat-jack: σ_m=0.74×0.85×0.62=0.39 N/mm²; push: τ=14000/(2×230×110)=0.28 N/mm². ## Answer #### (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 fraction of solid area (1 for a 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 an overburden stress $\sigma_v$, the initial shear strength is corrected as $f_{v0} = \tau - \mu\,\sigma_v$ (with friction coefficient $\mu \approx 0.4$–$0.6$). | Test | Result | |---|---| | Diagonal shear | $f_v = 0.41$ N/mm² | | Flat jack | $\sigma_m = 0.39$ N/mm² | | Push shear | $\tau = 0.28$ N/mm² | **Answer: (a) 0.41 N/mm²; (b) 0.39 N/mm²; (c) 0.28 N/mm².** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is shear strength found from the diagonal compression test? A. f_v = P / (w x t) B. f_v = 0.707 P / A_n, with A_n = (w + h)/2 x t x n, giving 0.41 N/mm^2 C. f_v = 2P / A D. f_v = P x 0.707 x A_n Answer: B. The diagonal load gives pure shear on the other diagonal; A_n = 276 000 mm^2. (Key point: Diagonal compression test) 2. How does the single flat-jack test find in-situ stress? A. Measure crushing load of the jack B. A slot relieves stress and closes; the jack pressure is raised until the original gauge distance is restored C. Measure the pressure when the wall cracks D. Jack pushes a brick sideways Answer: B. The pressure restoring the original distance, corrected by constants, equals the existing stress. (Key point: Flat jack principle) 3. What is the in-situ stress formula? A. sigma = p / (K_m K_a) B. sigma = p + K_m + K_a C. sigma = K_m K_a p = 0.74 x 0.85 x 0.62 = 0.39 N/mm^2 D. sigma = K_m p / K_a Answer: C. The pressure is multiplied by the jack stiffness constant and the area ratio. (Key point: Flat jack formula) 4. What area is used in the push shear test with two bed joints? A. One joint 230 x 110 B. Brick face 230 x 75 C. Head joint area D. Two bed joints, 2 x 230 x 110 = 50 600 mm^2, giving 0.277 N/mm^2 Answer: D. The brick shears along two bed joints, so tau = 14 000/50 600. (Key point: Push shear test) ## Related in Design of Timber and Masonry Structures - [Compressive strength test and classification of bricks](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Abrick-compressive-strength-test-classification) — Design of Timber and Masonry Structures, chapter 8 - [Safe load on a brick masonry column](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Abrick-masonry-column-safe-load) — Design of Timber and Masonry Structures, chapter 5 - [Check of a solid brick wall under gravity load](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Asolid-masonry-wall-gravity-check) — Design of Timber and Masonry Structures, chapter 5 - [Masonry pier under axial load and in-plane moment](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Amasonry-pier-axial-moment-check) — Design of Timber and Masonry Structures, chapter 7 - [Distribution of lateral force among masonry walls](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Alateral-load-sharing-among-piers) — Design of Timber and Masonry Structures, chapter 6 - [Out-of-plane bending of a masonry wall](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Aout-of-plane-wall-flexure-check) — Design of Timber and Masonry Structures, chapter 6 ## Related in other subjects - [Flat-jack test for brick masonry wall](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Aflat-jack-test) — Concrete Technology and Masonry Structure, chapter 10, asked 2× - [Brick and mortar for 230 mm wall (100 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-230mm-100kn-5m) — Concrete Technology and Masonry Structure, chapter 8, asked 7× - [Diagonal shear test of masonry wall](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Adiagonal-shear-test) — Concrete Technology and Masonry Structure, chapter 10, asked 6× - [Flat-jack and push-shear tests](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Aflat-jack-and-push-shear) — Concrete Technology and Masonry Structure, chapter 10, asked 2×