--- title: "Castigliano: moment at C of propped cantilever" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:castigliano-moment-c-propped-30knm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-moment-c-propped-30knm site: BCT Engineering --- # Castigliano: moment at C of propped cantilever Theory of Structures II · Chapter 2: Force method - Asked in: 2069 Chaitra, 10 marks - Video: [Strain Energy Method - Analysis of Propped Cantilever Beam ( Point Load in the Center )](https://www.youtube.com/watch?v=lLjjJT2QS6c) — 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/castigliano-moment-c-propped-30knm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Use Castigliano's theorem to find the moment at point C of the propped cantilever beam loaded as shown in the figure below. Take EI to be constant. [Figure: beam A-B-D-C, A hinged, B support at 2 m, D at 4 m with 30 kNm moment, C fixed end at 6 m (2 m + 2 m + 2 m).] ## Short answer Apply Castigliano's theorem with the vertical reactions at A and B as redundants. The fixed-end moment at C is 6.00 kNm hogging (tension on top). ## Answer **Assumptions:** A is a hinge at $x=0$, B a roller at $x=2$ m, the 30 kNm couple acts clockwise at D ($x=4$ m) and C is the fixed end at $x=6$ m. EI is constant. The beam has four vertical/rotational reactions ($R_A,R_B,R_C,M_C$) and two equilibrium equations, so it is indeterminate to the 2nd degree. Take $R_A$ and $R_B$ (upward) as redundants. Castigliano's theorem for redundants that do not move: $\partial U/\partial R_A=0$ and $\partial U/\partial R_B=0$. #### Bending moments (x from A, sagging +) | Portion | M | |---|---| | A to B (0 to 2) | $R_Ax$ | | B to D (2 to 4) | $R_Ax+R_B(x-2)$ | | D to C (4 to 6) | $R_Ax+R_B(x-2)+30$ | #### Compatibility equations $$ \frac{\partial U}{\partial R_A}=\frac{1}{EI}\int M\frac{\partial M}{\partial R_A}dx=0:\quad \int_0^2R_Ax\cdot x\,dx+\int_2^4[R_Ax+R_B(x-2)]\,x\,dx+\int_4^6[R_Ax+R_B(x-2)+30]\,x\,dx=0 $$ which simplifies to $$ 72R_A+37.333R_B+300=0 $$ Similarly $$ \frac{\partial U}{\partial R_B}=0:\quad \int_2^4[R_Ax+R_B(x-2)](x-2)dx+\int_4^6[R_Ax+R_B(x-2)+30](x-2)dx=0 $$ $$ 37.333R_A+21.333R_B+180=0 $$ #### Solution $$ R_A=2.250\ \text{kN}\uparrow,\qquad R_B=-12.375\ \text{kN}\ (\text{negative: acts downward}) $$ $$ R_C=-(R_A+R_B)=10.125\ \text{kN}\uparrow $$ #### Moment at the fixed end C $$ M_C=R_A(6)+R_B(4)+30=13.500+(-49.500)+30=-6.00\ \text{kNm} $$ The negative sign shows hogging (tension at the top) at C. Check: $\sum V = R_A+R_B+R_C = 2.250-12.375+10.125 = 0$ ✓. The same $M_C$ is obtained by an independent stiffness analysis. **Answer: $|M_C|=6.00$ kNm, hogging (for a clockwise 30 kNm couple); $R_A=2.25$ kN, $R_B=12.375$ kN (downward), $R_C=10.125$ kN.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the beam indeterminate to the 2nd degree? A. It has 4 unknown reactions (R_A, R_B, R_C, M_C) and only 2 equilibrium equations B. It has 3 unknown reactions and 3 equations C. It has 5 unknown reactions and 2 equations D. It has 2 unknown reactions and 2 equations Answer: A. Four unknowns minus two vertical-plane equilibrium equations leaves two redundants, taken as R_A and R_B. (Key point: Degree of indeterminacy) 2. What condition does Castigliano's theorem give for the redundants R_A and R_B? A. dU/dR = 30 kNm B. dU/dR_A = 0 and dU/dR_B = 0, because those supports do not move C. dU/dR = the reaction itself D. Total strain energy U = 0 Answer: B. The partial derivative of strain energy with respect to a force equals the displacement at that point, which is zero at a rigid support. (Key point: Castigliano condition for redundants) 3. What is the moment in the portion D to C (4 to 6 m)? A. R_A x + R_B (x - 2) + 30 B. R_A x - 30 C. R_A x + R_B x D. R_B (x - 2) + 30 Answer: A. Moments of R_A and R_B to the left are added and the couple of 30 kNm enters after D. (Key point: Bending moment expression) 4. What is the moment at the fixed end C? A. 6.00 kNm sagging B. 13.5 kNm hogging C. 6.00 kNm hogging D. 30 kNm hogging Answer: C. M_C = R_A(6) + R_B(4) + 30 = 13.5 - 49.5 + 30 = -6.00 kNm, i.e. hogging. (Key point: Result for M_C) ## Close variants set in other exams - [Castigliano: fixed-end moment, 15 kN load](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-fixed-moment-15kn-2m-3m) — Theory of Structures II, chapter 2, asked 2× ## Related in Theory of Structures II - [Castigliano: fixed moment, propped cantilever UDL](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-fixed-moment-udl-5knm-6m) — Theory of Structures II, chapter 2, asked 2× - [Castigliano: fixed-end moment with 50 kNm at overhang](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-fixed-moment-50knm-overhang) — Theory of Structures II, chapter 2 - [ILD: moment at support B, fixed-propped beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-b-propped-2l-l) — Theory of Structures II, chapter 4 - [Castigliano: fixed moment, propped cantilever](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-fixed-moment-propped-100kn-50kn) — Theory of Structures II, chapter 2 - [Castigliano: propped cantilever, moment and rotation](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-propped-30knm-50kn-moment-rotation) — Theory of Structures II, chapter 2 - [ILD: fixed-end moment of propped cantilever](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-reaction-moment-fixed-propped-10m) — Theory of Structures II, chapter 4, asked 2× ## Related in other subjects - [Propped cantilever with UDL, two elements](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Apropped-cantilever-two-elements) — Computational Techniques in Civil Engineering, chapter 4 - [Propped cantilever by strain energy method](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Apropped-cantilever-redundant-reaction) — Mechanics of Solids, chapter 5 - [Propped cantilever with UDL by integration](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Apropped-cantilever-udl-prop-reaction) — Strength of Materials, chapter 7 - [Euler's formula: one end fixed, other hinged](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-one-fixed-one-hinged) — Strength of Materials, chapter 8, asked 2×