--- title: "Strut loads in braced cut" subject: "Foundation Engineering" chapter: 4 key: "semester_5/foundation_engineering:strut-loads-bracing-system" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/strut-loads-bracing-system site: BCT Engineering --- # Strut loads in braced cut Foundation Engineering ยท Chapter 4: Arching in Soils and Braced Cuts - Asked in: 2080 Bhadra, 4 marks - Video: [Numerical On Strut ๐Ÿ”ฅ| Braced Cuts|Retaining Structure|Anchored Sheet Pile|Foundation](https://www.youtube.com/watch?v=mxEUMg8ja_0) โ€” DIPAK MANDAL - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/strut-loads-bracing-system - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-4 ## Question Determine the forces in the struts for the bracing system shown in figure below. Assume hinges at the levels B and C. Take the spacing of the struts in each strut level as 2.0 m. [Figure: braced cut 10.0 m deep with four strut levels A, B, C, D numbered (1) to (4) at depths 1.5 m, 4.0 m, 6.5 m and 9.0 m (spacings 1.5 m, 2.5 m, 2.5 m, 2.5 m, 1.0 m); soil $\gamma_t = 18\ \text{kN/m}^3$, $c = 30\ \text{kN/m}^2$.] ## Short answer Using Peck's soft-to-medium clay diagram, apparent pressure pa = 60 kN/mยฒ. Hinge method gives reactions per metre: A=97, B=143, C=138, D=147 kN/m. With 2 m spacing, strut loads are A=194 kN, B=286 kN, C=276 kN, D=294 kN. ## Answer **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). ```text 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. | Level | Reaction (kN/m) | Strut load $= R\times2.0$ (kN) | |---|---|---| | A | 97.0 | 194 | | B | 68.0 + 75.0 = 143.0 | 286 | | C | 75.0 + 63.0 = 138.0 | 276 | | D | 147.0 | 294 | 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the soft-to-medium clay envelope used here? A. Because gamma*H/c = 6, which is greater than 4 B. Because the cut is deeper than 8 m C. Because struts are at four levels D. Because c is less than 30 kN/m^2 Answer: A. Peck's soft to medium clay envelope applies when gamma*H/c exceeds 4; here 18x10/30 = 6. (Key point: Choosing Peck's clay envelope) 2. What is the maximum apparent pressure p_a used? A. 180 kN/m^2 B. 54 kN/m^2 C. 120 kN/m^2 D. 60 kN/m^2 Answer: D. p_a = gamma*H(1 - 4c/(gamma*H)) = 180(1 - 120/180) = 60 kN/m^2, which exceeds 0.3*gamma*H = 54. (Key point: Apparent pressure value) 3. What does the hinge assumption at B and C do to the analysis? A. It makes the wall a continuous beam over all struts B. It removes the need for any struts C. It splits the wall into separate statically determinate segments D. It doubles the pressure at B and C Answer: C. Hinges split the wall into top, middle and bottom parts, each solved by statics; hinge reactions are then added at B and C. (Key point: Hinge method) 4. How is the load in a strut obtained from the reaction per metre run, and what is strut A's load? A. R x 2.0 spacing; A = 194 kN B. R / 2.0; A = 48.5 kN C. R x 2.0; A = 286 kN D. R x 1.5; A = 145.5 kN Answer: A. Strut load = reaction per metre x horizontal spacing (2.0 m); R_A = 97 kN/m gives 194 kN. 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