--- title: "Collapse value of P: continuous beam" subject: "Theory of Structures II" chapter: 5 key: "semester_5/theory_of_structures_ii:collapse-p-continuous-beam-p-2p" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-p-continuous-beam-p-2p site: BCT Engineering --- # Collapse value of P: continuous beam Theory of Structures II · Chapter 5: Introduction to plastic analysis - Asked in: 2069 Asar, 5 marks - Video: [Plastic Analysis Problem No - 4 ( Finding Plastic Moment Capacity in a Two span continuous Beam )](https://www.youtube.com/watch?v=mqVh9Pn-GDw) — 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-p-continuous-beam-p-2p - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5 ## Question For the given continuous beam with the same plastic moment of resistance $M_P$ for all the members, calculate the value of P at collapse. [Figure: beam ABC, A fixed, B and C supports; P at 3 m from A, B at 6 m, 2P at 2 m beyond B; BC = 6 m (2 m + 4 m).] ## Short answer Continuous beam with fixed end A. Span AB mechanism gives P = 1.333M_p. Span BC mechanism (hinges at B and under 2P) gives P = 0.625M_p. Combined mechanism gives higher value. Collapse load P_c = 5/8 M_p = 0.625M_p. ## Answer #### Set-up Beam ABC: A fixed, B and C supports; AB = 6 m with $P$ at mid-span J (3 m from A); BC = 6 m with $2P$ at K, 2 m from B and 4 m from C. $M_p$ is the same throughout. Degree of indeterminacy = 2; the possible mechanisms are the failure of each span alone and their combination. #### Mechanism 1: span AB (hinges at A, J, B) Mid-span deflection $\delta = 3\theta$; hinge rotations $\theta$, $2\theta$, $\theta$. $$ P\times3\theta = M_p(\theta + 2\theta + \theta) = 4M_p\theta \;\Rightarrow\; P = \frac{4}{3}M_p = 1.333\,M_p $$ #### Mechanism 2: span BC (hinges at B and K) Part BK (2 m) rotates $\theta_1$ about B and part KC (4 m) rotates $\theta_2$ about C, with $2\theta_1 = 4\theta_2$, so $\theta_1 = 2\theta_2$. The deflection under the load is $\delta = 4\theta_2$. - Hinge at B: $\theta_1 = 2\theta_2$; hinge at K: $\theta_1 + \theta_2 = 3\theta_2$. - Internal work $= M_p(2\theta_2 + 3\theta_2) = 5M_p\theta_2$. - External work $= 2P\times4\theta_2 = 8P\theta_2$. $$ 8P\theta_2 = 5M_p\theta_2 \;\Rightarrow\; P = \frac{5}{8}M_p = 0.625\,M_p $$ #### Mechanism 3: combined (hinges at A, J and K; no hinge at B) With no hinge at B, J-B-K rotates as one rigid piece about support B, so when K moves down J moves up. Let this piece rotate $\theta$: K moves down $2\theta$ and J moves up $3\theta$. Hinge rotations: A $\theta$, J $2\theta$, K $\theta + \dfrac{2\theta}{4} = 1.5\theta$. - Internal work $= M_p(\theta + 2\theta + 1.5\theta) = 4.5M_p\theta$. - External work $= 2P\times2\theta - P\times3\theta = P\theta$. This gives $P = 4.5M_p$, much larger. (In the opposite sense the external work is negative, so no mechanism forms.) #### Collapse load The lowest value is for span BC: **Answer: $P_c = \dfrac{5}{8}M_p = 0.625\,M_p$** (hinges at B and under the $2P$ load; $2P_c = 1.25M_p$). #### Check (equilibrium) At collapse $M_B = -M_p$ and $M_K = +M_p$. For BC: $M_K = \dfrac{2P\times2\times4}{6} - \dfrac{4}{6}M_p = \dfrac{8P}{3} - \dfrac{2}{3}M_p$; setting this equal to $M_p$ gives $P = \dfrac{5}{8}M_p$. In AB, $M_A$ is not fixed by statics at collapse (span AB is still indeterminate). For example, with $M_A = -0.5M_p$ the moment under $P$ is $\dfrac{PL}{4} + \dfrac{M_A + M_B}{2} = 0.9375M_p - 0.75M_p = 0.19M_p$, well within $M_p$, so the yield condition is also satisfied. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Span AB (hinges at A, J, B) with P at the 3 m mid-span gives: A. P = 1.333 Mp B. P = 0.625 Mp C. P = 4.5 Mp D. P = 0.75 Mp Answer: A. P x 3 theta = 4 Mp theta gives P = 4/3 Mp. (Key point: Span AB mechanism) 2. In BC, BK = 2 m rotates theta1 and KC = 4 m rotates theta2. The relation is: A. theta1 = theta2 B. theta1 = 2 theta2 C. theta2 = 2 theta1 D. theta1 = 4 theta2 Answer: B. Equal deflection at K: 2 theta1 = 4 theta2, so theta1 = 2 theta2. (Key point: Span BC compatibility) 3. Internal work is 5 Mp theta2 and external work is 8P theta2. What is P? A. 1.6 Mp B. 0.625 Mp C. 1.333 Mp D. 0.8 Mp Answer: B. P = 5/8 Mp = 0.625 Mp. (Key point: Span BC result) 4. Why does span BC govern over AB? A. It has a heavier load, so it gives a higher P B. Its P (0.625 Mp) is the lowest C. It has more hinges D. AB is fixed Answer: B. Collapse occurs at the lowest load; BC gives 0.625 Mp against 1.333 Mp for AB. 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