--- title: "Cylindrical shell with joint efficiencies: change in dimensions" subject: "Strength of Materials" chapter: 5 key: "semester_3/strength_of_materials:cyl-shell-joint-efficiency" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/cyl-shell-joint-efficiency site: BCT Engineering --- # Cylindrical shell with joint efficiencies: change in dimensions Strength of Materials · Chapter 5: Thin Walled Vessels - Asked in: 2075 Chaitra, 6 marks - Video: [Efficiency of joints and change the dimensions of the shell](https://www.youtube.com/watch?v=tKzRVHykM_s) — CRR MECHANICAL DIPLOMA - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/cyl-shell-joint-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5 ## Question A cylindrical shell of 260mm external diameter 2.5 m length and 5mm wall thickness is subjected to internal pressure of 1.60 MPa. Calculate the change in diameter, length and volume of the cylinder if the cylinder has a longitudinal joint (85% efficiency) and circumferential joint (65% efficiency). Take Young's modulus = 200GPa and Poisson's ratio = 0.3 ## Short answer Deformation uses solid plate stresses: σ_c=40 MPa, σ_l=20 MPa. Joint efficiencies increase local stresses but not strain. δd=0.0425 mm, δL=0.100 mm, δV=46.63 cm³. Joint stresses 47.06 and 30.77 MPa. ## Answer **Data:** external diameter 260 mm, $t = 5$ mm, so internal diameter $d = 260 - 2(5) = 250$ mm; $L = 2500$ mm, $p = 1.6\ \text{MPa}$, $E = 2\times10^5\ \text{MPa}$, $\nu = 0.3$, longitudinal joint efficiency $\eta_l = 0.85$, circumferential joint efficiency $\eta_c = 0.65$. #### Stresses In the solid plate (away from the joints): $$ \sigma_c = \frac{pd}{2t} = \frac{1.6(250)}{2(5)} = 40.0\ \text{MPa}, \qquad \sigma_l = \frac{pd}{4t} = 20.0\ \text{MPa} $$ At the joints (for strength, the efficiency reduces the effective resisting area): $$ \sigma_{c,joint} = \frac{pd}{2t\eta_l} = \frac{40.0}{0.85} = 47.06\ \text{MPa}, \qquad \sigma_{l,joint} = \frac{pd}{4t\eta_c} = \frac{20.0}{0.65} = 30.77\ \text{MPa} $$ #### Changes in dimensions The joint efficiencies affect only the local stress at the riveted or welded seams. The overall deformation of the shell is governed by the stresses in the solid plate, so the efficiencies are not used in the strain calculation. $$ \varepsilon_c = \frac{40.0-0.3(20.0)}{2\times10^5} = 1.700\times10^{-4}, \qquad \varepsilon_l = \frac{20.0-0.3(40.0)}{2\times10^5} = 4.000\times10^{-5} $$ $$ \delta d = \varepsilon_c d = 0.04250\ \text{mm}, \qquad \delta L = \varepsilon_l L = 0.10000\ \text{mm} $$ $$ \varepsilon_v = 2\varepsilon_c + \varepsilon_l = 3.800\times10^{-4}, \qquad V = \frac{\pi}{4}(250)^2(2500) = 1.2272\times10^{8}\ \text{mm}^3 $$ $$ \delta V = \varepsilon_v V = 46633.0\ \text{mm}^3 = 46.63\ \text{cm}^3 $$ **Answer:** $\delta d = 0.0425$ mm, $\delta L = 0.1000$ mm, $\delta V = 46.63$ cm³ (all increases). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What diameter is used for d in the stress formulas? A. Internal diameter, 250 mm (260 - 2 x 5) B. External diameter, 260 mm C. Mean diameter, 255 mm D. Internal radius, 125 mm Answer: A. Pressure acts on the inside, so d = external diameter minus twice the thickness = 250 mm. (Key point: Internal diameter) 2. Which stresses are used to find the changes in diameter and length? A. Joint stresses (47.06 and 30.77 MPa) B. Stresses in the solid plate (40 and 20 MPa); efficiencies are not used C. Hoop stress divided by 0.85 only D. Average of plate and joint stresses Answer: B. Joint efficiency only raises the local stress at seams; overall deformation is governed by the solid plate stresses. (Key point: Role of joint efficiency) 3. Which efficiency is applied to the hoop stress at the joint? A. Circumferential joint efficiency 0.65, giving 61.5 MPa B. Both multiplied, 0.85 x 0.65 C. Longitudinal joint efficiency 0.85, giving 47.06 MPa D. None; hoop stress is unchanged at joints Answer: C. Hoop stress tends to split the shell along a longitudinal seam, so the longitudinal joint efficiency reduces its resisting area. (Key point: Joint stress) 4. Which expression gives the circumferential strain? A. sigma_c/E = 2e-4 B. (sigma_c + 0.3 sigma_l)/E C. (sigma_l - 0.3 sigma_c)/E = 4e-5 D. (sigma_c - 0.3 sigma_l)/E = 1.7e-4 Answer: D. Each strain includes the lateral Poisson contraction from the other principal stress: eps_c = (40 - 6)/2e5 = 1.7e-4. 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