--- title: "Timber-steel plate composite member: permissible load" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:timber-steel-composite-permissible-load" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/timber-steel-composite-permissible-load site: BCT Engineering --- # Timber-steel plate composite member: permissible load Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2078 Kartik, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/timber-steel-composite-permissible-load - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question Two $150\ \text{mm}\times75\ \text{mm}\times4\ \text{m}$ long timber members are reinforced with a steel plate $150\ \text{mm}\times6\ \text{mm}\times4\ \text{m}$ long as shown in figure. The three members are adequately bolted together. The permissible stresses for the timber and the steel members are $6\ \text{N/mm}^2$ and $130\ \text{N/mm}^2$ respectively. E for timber is $8.4\ \text{GN/m}^2$ and for steel is $210\ \text{GN/m}^2$. Calculate the permissible tensile load for the composite member and the amount of elongation due to this load. [Figure: cross-section 150 mm deep: timber 75 mm, steel plate 6 mm, timber 75 mm side by side.] ## Short answer The steel plate governs (130 N/mm²) with timber at5.2 N/mm². Permissible load =234.0 kN; elongation =2.476 mm. ## Answer The members are bolted together, so they extend equally: $\varepsilon_t = \varepsilon_s$ and $\sigma_s/E_s = \sigma_t/E_t$. $$ m = \frac{E_s}{E_t} = \frac{210}{8.4} = 25.0,\qquad \sigma_s = m\,\sigma_t $$ #### Areas $$ A_t = 2\times(150\times75) = 22500\ \text{mm}^2,\qquad A_s = 150\times6 = 900\ \text{mm}^2 $$ #### Which material governs? - If timber reaches 6 N/mm², steel would be at $25\times6 = 150$ N/mm² $> 130$ N/mm² (not permitted). - If steel reaches 130 N/mm², timber is at $130/25 = 5.2$ N/mm² $< 6$ N/mm² (safe). Hence the steel plate governs: $\sigma_s = 130$ N/mm² and $\sigma_t = 5.2$ N/mm². #### Permissible load $$ P = \sigma_s A_s + \sigma_t A_t = 130\times900 + 5.2\times22500 = 117000 + 117000 = 234000\ \text{N} = 234.0\ \text{kN} $$ #### Elongation (L = 4 m = 4000 mm) $$ \delta = \frac{\sigma_s}{E_s}L = \frac{130}{210\times10^3}\times4000 = 2.476\ \text{mm} $$ **Answer: permissible tensile load = 234.0 kN; elongation = 2.476 mm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What relation holds between stresses in steel and timber? A. sigma_s = m sigma_t with m = E_s/E_t = 25 B. sigma_s = sigma_t C. sigma_t = 25 sigma_s D. sigma_s = sigma_t / 25 Answer: A. Equal strain gives sigma_s/E_s = sigma_t/E_t. (Key point: Compatibility) 2. Why does the steel plate govern? A. Steel has the larger area B. Timber at 6 N/mm^2 would put steel at 150 > 130 N/mm^2, exceeding its limit C. Steel has lower E D. Timber has higher permissible stress Answer: B. When steel hits 130, timber is only 5.2 < 6 (safe), so steel limits the load. (Key point: Governing material) 3. How is the permissible load computed? A. 130 x A_s + 6 x A_t B. 130 x (A_s + A_t) C. sigma_s A_s + sigma_t A_t using 130 and 5.2 D. 6 x (A_s + A_t) Answer: C. The stresses must be compatible values: 130 and 5.2 N/mm^2 giving 117 + 117 = 234 kN. (Key point: Load) 4. How is the elongation found? A. delta = (sigma_s/E_s) L = 2.476 mm B. delta = sigma_t L / E_s C. delta = P L / A_t E_s D. delta = 6 x 4000 / 8400 mm using the full 6 Answer: A. Both materials extend equally: 130/210000 x 4000 = 2.476 mm. 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