--- title: "Rigid bar AB: vertical displacement of F" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:rigid-bar-ab-steel-aluminium-rods" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/rigid-bar-ab-steel-aluminium-rods site: BCT Engineering --- # Rigid bar AB: vertical displacement of F Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2069 Asar, 8 marks - Video: [4-16 Determine the vertical displacement of F | Mechanics of materials rc hibbeler](https://www.youtube.com/watch?v=4h8c8RHjJPs) — Engr. Adnan Rasheed Mechanical - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/rigid-bar-ab-steel-aluminium-rods - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question Determine vertical displacement of point F, if AB is a rigid bar and remains horizontal, AC and BD are rods made of steel and aluminium having diameter 20mm and 40mm respectively. E for steel and aluminium are 200GPa and 70GPa respectively. [Figure: rigid horizontal bar AB with a 90 kN load applied at F, 200 m from A (as marked), rods AC and BD vertical, 300 m high, fixed at C and D at the base; AB spans 400 m.] ## Short answer For AB to stay horizontal, both rods shorten equally. Forces: 37.5 kN in steel, 52.5 kN in aluminium. Shortening δ = PL/(AE) gives 0.179 mm. Vertical displacement of F is 0.179 mm downward (load must act 233.3 mm from A). ## Answer **Assumptions:** the dimensions marked "m" in the figure are read as millimetres (rod length 300 mm, AB = 400 mm, AF = 200 mm). The rods AC (steel, 20 mm dia.) and BD (aluminium, 40 mm dia.) are vertical posts under the rigid bar, so both are in compression. #### Rod properties $$ A_{st} = \frac{\pi}{4}(20)^2 = 314.16\ \text{mm}^2, \qquad A_{al} = \frac{\pi}{4}(40)^2 = 1256.64\ \text{mm}^2 $$ $$ A_{st}E_{st} = 314.16\times200\,000 = 6.283\times10^{7}\ \text{N}, \qquad A_{al}E_{al} = 1256.64\times70\,000 = 8.796\times10^{7}\ \text{N} $$ #### Bar remains horizontal (as stated in the question) If AB stays horizontal, both rods shorten by the same amount $\delta$, and this is also the vertical displacement of F. Each rod force is $P = AE\delta/L$, and together they carry the 90 kN load: $$ \frac{\delta}{L}\left(A_{st}E_{st} + A_{al}E_{al}\right) = 90\,000 \Rightarrow \delta = \frac{90\,000\times300}{6.283\times10^{7}+8.796\times10^{7}} = 0.179\ \text{mm} $$ Loads in the rods: $P_{st} = \dfrac{6.283\times10^{7}\times0.179}{300} = 37.5$ kN and $P_{al} = 90 - 37.5 = 52.5$ kN. For the bar to stay horizontal, the load must act where these rod forces balance it. Taking moments about A: $90\,x = 52.5\times400$, so $x = 233.3$ mm from A. #### If the load is exactly at the mid-point (AF = 200 mm) Then $P_A = P_B = 45$ kN, and $$ \delta_A = \frac{45\,000\times 300}{314.16\times 200\,000} = 0.2149\ \text{mm}, \qquad \delta_B = \frac{45\,000\times 300}{1256.64\times 70\,000} = 0.1535\ \text{mm} $$ $$ \delta_F = \frac{\delta_A + \delta_B}{2} = \frac{0.2149+0.1535}{2} = 0.1842\ \text{mm} $$ In this case the bar tilts slightly (by $0.0614/400$ rad). **Answer:** with AB remaining horizontal, the vertical displacement of F is $\delta_F = 0.179\ \text{mm}$ downward (steel 37.5 kN, aluminium 52.5 kN, load 233.3 mm from A). If the load is taken exactly at the mid-point, $\delta_F \approx 0.184$ mm and the bar tilts slightly. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why do both rods carry 45 kN? A. F is at the mid-point of AB, so the load divides equally B. Rods have equal stiffness C. The bar is hinged D. Steel is stronger Answer: A. Moments about B give P_A x 400 = 90 x 200, so P_A = 45 kN. (Key point: Rod forces) 2. Which formula gives each rod's shortening? A. delta = P A / (L E) B. delta = P L / (A E) C. delta = P L E / A D. delta = P / A Answer: B. Each rod is an axially loaded member with its own A and E. (Key point: Shortening) 3. How is delta_F found? A. Average of delta_A and delta_B since F is the mid-point B. delta_A only C. Sum of delta_A and delta_B D. delta_B only Answer: A. A rigid bar's movement varies linearly, so at the mid-point it is the mean. (Key point: Displacement of F) 4. What is delta_F? A. 0.215 mm B. 0.369 mm C. 0.184 mm D. 0.154 mm Answer: C. (0.2149 + 0.1535)/2 = 0.1842 mm. 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