--- title: "Bar fixed at both ends: load shares, displacements" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:circular-bar-fixed-both-ends-50kn" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/circular-bar-fixed-both-ends-50kn site: BCT Engineering --- # Bar fixed at both ends: load shares, displacements Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2076 Chaitra, 10 marks - Video: [Problem 4 bar subjected to axial forces with rigid fixed support at ends, simple stresses & strain](https://www.youtube.com/watch?v=6W0-Dr_bOuk) — Edutech Guru Engineering Learning-By Priyanka ma'am - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/circular-bar-fixed-both-ends-50kn - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question A Circular bar ABCD, rigidly fixed at A and D is subjected to axial loads of 50 KN and 100 KN at B and C as shown in the figure. Find the loads shared by each part of the bar and displacements of the points B and C. Take E for the steel as 200GPa. [Figure: bar fixed at A and D; segment AB diameter 25 mm length 300 mm; segment BC diameter 50 mm length 400 mm; segment CD diameter 75 mm length 500 mm; 50 kN load at B and 100 kN load at C in opposite directions.] ## Short answer By compatibility and equilibrium, axial forces are N_AB = 4.88 kN, N_BC = -45.12 kN, N_CD = 54.88 kN. Displacement at B is 0.0149 mm, at C is -0.0311 mm (positive towards D). ## Answer Assumed loading: the 50 kN load at B acts towards D (→) and the 100 kN load at C acts towards A (←). Let the axial force in AB be $x$ (tension +ve). #### Equilibrium of joints $$ N_{AB} = x,\qquad N_{BC} = x - 50,\qquad N_{CD} = N_{BC} + 100 = x + 50\ \ (\text{kN}) $$ #### Compatibility A and D are fixed, so the total change in length is zero: $$ \sum \frac{N_iL_i}{A_iE} = 0 $$ | Part | d (mm) | A (mm²) | L (mm) | $L/A$ (mm⁻¹) | |---|---|---|---|---| | AB | 25 | 490.87 | 300 | 0.6112 | | BC | 50 | 1963.50 | 400 | 0.2037 | | CD | 75 | 4417.86 | 500 | 0.1132 | $$ x\,(0.6112+0.2037+0.1132) - 50(0.2037) + 50(0.1132) = 0 \;\Rightarrow\; 0.9281\,x = 4.525 $$ so $x = N_{AB} = 4.878$ kN (the common factor $1/E$ cancels). #### Loads shared by each part $$ N_{AB} = 4.88\ \text{kN},\quad N_{BC} = -45.12\ \text{kN},\quad N_{CD} = 54.88\ \text{kN} $$ Positive is tension, negative is compression. #### Displacements ($E = 200\times10^3$ N/mm²) $$ \begin{aligned} \delta_B &= \frac{N_{AB}L_{AB}}{A_{AB}E} = 0.0149\ \text{mm}\ (\text{towards D if +ve})\\ \delta_C &= \delta_B + \frac{N_{BC}L_{BC}}{A_{BC}E} = -0.0311\ \text{mm} \end{aligned} $$ Check: $\delta_D = \delta_C + N_{CD}L_{CD}/(A_{CD}E) = 0.000000\approx0$. **Answer: $N_{AB} = 4.88$ kN, $N_{BC} = -45.12$ kN, $N_{CD} = 54.88$ kN; $\delta_B = 0.0149$ mm, $\delta_C = -0.0311$ mm (positive towards D).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is an extra compatibility equation needed? A. Both ends are fixed, so one unknown reaction cannot be found by statics alone B. The bar has three diameters C. The loads are unknown D. Gravity acts on it Answer: A. Statics gives one equation for two reactions; the total length change being zero supplies the other. (Key point: Indeterminacy) 2. What is the compatibility condition? A. N_AB = N_BC = N_CD B. Sum of N L/(A E) over the three parts = 0 C. Deflection at B equals that at C D. Sum of forces is zero in each part Answer: B. A and D do not move, so total elongation is zero. (Key point: Compatibility) 3. What forces result? A. AB 50 kN, BC 0, CD 100 kN B. AB 25, BC 25, CD 25 kN C. AB 4.88 kN, BC -45.12 kN, CD 54.88 kN D. AB -4.88, BC 45.12, CD -54.88 kN Answer: C. x = 4.878 kN from compatibility, then N_BC = x - 50, N_CD = x + 50. (Key point: Forces) 4. How is the displacement of B found? A. delta_B = N_AB L_AB /(A_AB E) from fixed end A B. delta_B from D with N_BC C. delta_B = 50 kN L / A E D. delta_B = zero Answer: A. B moves by the elongation of AB alone, 0.0149 mm. 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