--- title: "Collapse load of propped cantilever, 6 m span" subject: "Theory of Structures II" chapter: 5 key: "semester_5/theory_of_structures_ii:collapse-load-propped-cantilever-w-6m" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-propped-cantilever-w-6m site: BCT Engineering --- # Collapse load of propped cantilever, 6 m span Theory of Structures II · Chapter 5: Introduction to plastic analysis - Asked in: 2070 Chaitra (old course), 4 marks - Video: [Collapse Load in Propped Cantilever Beam with Eccentric Point Load ( static & Kinematic Method)](https://www.youtube.com/watch?v=3nmVTE5HW0E) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-propped-cantilever-w-6m - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5 ## Question Determine the collapse load for a propped cantilever beam shown below. Plastic moment capacity is $M_p$. [Figure: beam, left end fixed, right end roller; W at 6 m from the fixed end, 4 m to the roller.] ## Short answer For the propped cantilever with load at 6m from fixed end, hinges form at fixed support and under the load. Virtual work gives W_c = 3.5M_p/6 = 0.583M_p (M_p in kN·m). ## Answer #### Set-up Propped cantilever of span $L = 10$ m (A fixed, B roller) with a point load $W$ at C, 6 m from the fixed end ($a = 6$ m, $b = 4$ m). The hinges form at A and under the load at C (the two sections of largest moment). #### Mechanism and virtual work Part AC rotates $\theta$ about A, so C deflects $\delta = a\theta = 6\theta$. Part CB rotates $\phi$ about B with $b\phi = \delta$, so $\phi = \dfrac{6\theta}{4} = 1.5\theta$. - Rotation at A $= \theta$; rotation at C $= \theta + \phi = 2.5\theta$. - Internal work $= M_p\theta + M_p(2.5\theta) = 3.5M_p\theta$. - External work $= W\delta = 6W\theta$. $$ 6W_c\theta = 3.5M_p\theta \;\Rightarrow\; W_c = \frac{3.5}{6}M_p = 0.5833\,M_p $$ (In general $W_c = \dfrac{M_p(2b + a)}{ab}$.) #### Check by equilibrium With $M_A = -M_p$ and $M_B = 0$: $M_C = \dfrac{Wab}{L} - M_p\dfrac{b}{L} = \dfrac{24W}{10} - 0.4M_p$. Setting $M_C = M_p$ gives $W = \dfrac{1.4M_p}{2.4} = 0.5833M_p$, and the moment nowhere exceeds $M_p$. **Answer: $W_c = 0.583\,M_p$ (kN, with $M_p$ in kN·m).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Where do the plastic hinges form? A. At A and at the roller B B. At A and under load W at C C. At C only D. At B and C Answer: B. These sections carry the largest moments. (Key point: Hinge locations) 2. AC (6 m) rotates theta, so C deflects 6 theta. Part CB (4 m) rotates: A. theta B. 4 theta C. 1.5 theta D. 2.5 theta Answer: C. 4 phi = 6 theta gives phi = 1.5 theta. (Key point: Rotation compatibility) 3. What is the rotation of the hinge at C and the total internal work? A. 1.5 theta; 2.5 Mp theta B. 2.5 theta; 3.5 Mp theta C. 2.5 theta; 2.5 Mp theta D. theta; 2 Mp theta Answer: B. Rotation at C = theta + 1.5 theta = 2.5 theta; work = Mp theta + 2.5 Mp theta = 3.5 Mp theta. (Key point: Internal work) 4. From 6 W theta = 3.5 Mp theta, the collapse load is: A. 1.71 Mp B. 0.583 Mp C. 0.35 Mp D. 2.1 Mp Answer: B. Wc = 3.5/6 Mp = 0.583 Mp. 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