Strut loads in braced cut

Data: cut depth $H = 10$ m, $\gamma = 18$ kN/m$^3$, $c = 30$ kN/m$^2$ (clay), struts at A, B, C and D at 1.5, 4.0, 6.5 and 9.0 m, horizontal spacing 2.0 m, hinges at B and C.

1. Apparent pressure (Peck, clay)

$$\frac{\gamma H}{c} = \frac{18\times10}{30} = 6 > 4\ \Rightarrow\ \text{soft to medium clay}$$ $$p_a = \gamma H\left(1 - \frac{4c}{\gamma H}\right) = 180\left(1 - \frac{120}{180}\right) = 60.0\ \text{kN/m}^2\ (> 0.3\gamma H = 54)$$

The diagram rises from 0 at the top to 60 kN/m$^2$ at $0.25H = 2.5$ m, then stays at 60 kN/m$^2$ to the base (total load $= 525$ kN/m).

 0   |
     |\
 2.5 | 60
 A 1.5
 B 4.0  ---- 60 constant
 C 6.5
 D 9.0
10   |_____ 60

2. Reactions per metre run (hinge method)

• Top part (0 to 4.0 m): load $= 165.0$ kN/m. Moments about A give the hinge reaction at B: $R_{B1} = 68.0$ kN/m, so $R_A = 165.0 - 68.0 = 97.0$ kN/m.
• Middle part (B to C, 2.5 m), simply supported with uniform pressure 60: $R_{B2} = R_{C1} = 60\times2.5/2 = 75.0$ kN/m.
• Bottom part (6.5 to 10 m): load $= 210$ kN/m. Moments about D give $R_{C2} = 63.0$ kN/m, so $R_D = 210 - 63.0 = 147.0$ kN/m.

Check: $\sum R = 97 + 143 + 138 + 147 = 525$ kN/m, which equals the total load of 525 kN/m.

Answer: strut loads A = 194 kN, B = 286 kN, C = 276 kN, D = 294 kN.