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Diagonal shear and push shear tests

Short answer

The diagonal shear test applies diagonal compression to a 1.2 m square panel to failure along the loaded diagonal; τ = 0.707 P/An, ft = 0.5 P/An (An = (w+h)t/2). The push shear test pushes a brick horizontally from its bed joint to sliding, giving shear strength τ = F/(2Ab) and cohesion c = τ - μσv with μ ≈ 0.4-0.65. Diagonal test is for panel shear, push test for joint shear.

Explain in detail the diagonal shear test and push shear test of masonry wall.

Diagonal shear test (of a masonry panel)

Aim: shear (diagonal tensile) strength and shear modulus of masonry.

  • A square wall panel (about 1.2 m ×\times 1.2 m) is rotated so a diagonal is vertical; steel shoes at the two opposite corners.
  • A jack applies compression along the diagonal at a steady rate to failure; the diagonal crack and the load PP are noted. LVDTs on the diagonals measure the deformation.
τ=0.707PAn,ft=0.5PAn,An=w+h2 t\tau = \frac{0.707P}{A_n},\qquad f_t=\frac{0.5P}{A_n},\qquad A_n=\frac{w+h}{2}\,t
       P
       |
    +--v----+
    |\      |
    |  \    |  crack along the
    |    \  |  loaded diagonal
    +-----^-+

Push shear test (in situ)

Aim: shear strength of bed joints of existing wall.

  • A brick is isolated by removing a head joint at one end and the brick at the other end is removed to place a small jack and load cell.
  • The jack pushes the brick horizontally along the bed joints; the load FF at first slip is recorded.
τ=F2Ab,c=τ−μ σv\tau = \frac{F}{2A_b},\qquad c=\tau-\mu\,\sigma_v

where AbA_b is the bed area of the brick (two joints), σv\sigma_v the vertical stress at the joint and μ\mu the friction coefficient (about 0.4 to 0.65).

 +----+----+----+
 |    |[jack]->[brick]   F pushes brick along bed joint
 +----+----+----+

The diagonal shear test is a laboratory test on a panel (destructive); the push-shear test is done on the actual wall with little damage.

✓ Answer checked by Claude Opus 5.5 (high effort)

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