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Strut loads, 8 m cut in sand

Short answer

Peck's sand diagram gives uniform pressure p = 0.65 Ka γ H = 27.73 kN/m². Tributary area method yields reactions: A=83.2, B=55.5, C=83.2 kN/m. With 4 m spacing, design strut loads are A=333 kN, B=222 kN, C=333 kN.

A 8 m deep cut in sand with a cut width of 5 m is braced at equal distance of 2 m from the surface at three locations. In the plan the struts are placed at a spacing of 4 m center to center. Using the empirical pressure diagram, calculate the design strut loads if the properties of sand is, angle of shearing resistance of 30∘30^\circ and unit weight of 16 kN/m316\ \text{kN/m}^3. (Also explain Terzaghi's trap door experiment with neat sketch.)

Strut loads

Data: H=8H = 8 m, ϕ=30∘\phi = 30^\circ, γ=16\gamma = 16 kN/m3^3, three struts at 2, 4 and 6 m below the ground, horizontal spacing s=4s = 4 m. The width of the cut (5 m) does not affect the Peck envelope.

Apparent pressure (Peck, sand):

Ka=tan⁡2(45∘−15∘)=0.3333,p=0.65KaγH=0.65×0.3333×16×8=27.73 kN/m2K_a = \tan^2(45^\circ-15^\circ) = 0.3333,\qquad p = 0.65K_a\gamma H = 0.65\times0.3333\times16\times8 = 27.73\ \text{kN/m}^2

The pressure is uniform over the full depth of 8 m (total 221.9 kN/m).

   0  |          |
   2m |--A-------|  p = 27.73
   4m |--B-------|
   6m |--C-------|
   8m |__________|

Reactions (tributary-area method, each strut takes the load halfway to the next support; the bottom strut takes the load to the base):

StrutTributary heightReaction R=p×hR = p\times h (kN/m)Strut load P=R×4P = R\times4 (kN)
A (2 m)0 to 3 m = 3 m83.2332.8
B (4 m)3 to 5 m = 2 m55.5221.9
C (6 m)5 to 8 m = 3 m83.2332.8

Check: ∑R=221.9\sum R = 221.9 kN/m.

Answer: design strut loads A = 333 kN, B = 222 kN, C = 333 kN (the same maximum, 333 kN, can be used for the top and bottom struts).

Terzaghi's trap door experiment

A box of dry sand has a base strip (trap door) that can be lowered.

        sand surface
   ~~~~~~~~~~~~~~~~~~~~~~
   |  \              /  |
   |   \  yielding  /   |
   |____\__ wedge __/____|
   fixed |  trap  | fixed
         | door  |
           v moves down

When the door moves down slightly the pressure on it falls sharply to about half of the overburden pressure, and the pressure on the neighbouring fixed base rises by the same total amount. Shear stresses along the nearly vertical planes carry part of the weight: this is arching. It explains why the earth pressure on braced cuts is not triangular but nearly uniform (Peck's envelope).

✓ Answer checked by Claude Sonnet 5.5

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