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Composite bar: total change in length and stresses

Short answer

Axial forces are 25, 75, and 100 kN in aluminum, copper, steel respectively (all tension). Stresses: σ_al =2.89, σ_cu =13.22, σ_st =50.93 N/mm²; total elongation =0.0709 mm.

The composite section of aluminum, copper and steel bar is rigidly held by supports on left end. Find the total change in length and stresses in each bar, if the composite bar is subjected to loads at section as shown in figure. Take Ecopper=100 kN/mm2E_{copper} = 100\ \text{kN/mm}^2, Ealuminum=70 kN/mm2E_{aluminum} = 70\ \text{kN/mm}^2 and Esteel=200 kN/mm2E_{steel} = 200\ \text{kN/mm}^2. [Figure: bar fixed at the left end; aluminium bar 105 mm ϕ105\ \text{mm}\ \phi, length 300 mm, with a 50 kN load acting to the left at its right end; copper bar 85 mm ϕ85\ \text{mm}\ \phi, length 250 mm, with a 25 kN load acting to the left at its right end; steel bar 50 mm ϕ50\ \text{mm}\ \phi, length 100 mm, with a 100 kN load acting to the right at its free end.]

Order from the fixed wall: aluminium (300 mm), copper (250 mm), steel (100 mm). Loads: 50 kN (←) at the aluminium/copper junction, 25 kN (←) at the copper/steel junction and 100 kN (→) at the free end. Check: 100 kN to the right balances 50 + 25 = 75 kN to the left plus the wall reaction of 25 kN (←), so the wall pulls on the bar with 25 kN.

Axial force in each part (tension +ve, from the free end)

|=== Al ===|=== Cu ===|== St ==>  100 kN
Wall   <-50 kN     <-25 kN
  • Steel: Ns=+100N_s = +100 kN (tension)
  • Copper: Nc=100−25=+75N_c = 100 - 25 = +75 kN (tension)
  • Aluminium: Na=100−25−50=+25N_a = 100 - 25 - 50 = +25 kN (tension)

Areas

Aa=π4(105)2=8659.0,Ac=π4(85)2=5674.5,As=π4(50)2=1963.5 mm2A_a = \frac{\pi}{4}(105)^2 = 8659.0,\quad A_c = \frac{\pi}{4}(85)^2 = 5674.5,\quad A_s = \frac{\pi}{4}(50)^2 = 1963.5\ \text{mm}^2

Stresses

σa=25×1038659.0=2.887 N/mm2σc=75×1035674.5=13.217 N/mm2σs=100×1031963.5=50.930 N/mm2\begin{aligned} \sigma_a &= \frac{25\times10^3}{8659.0} = 2.887\ \text{N/mm}^2\\ \sigma_c &= \frac{75\times10^3}{5674.5} = 13.217\ \text{N/mm}^2\\ \sigma_s &= \frac{100\times10^3}{1963.5} = 50.930\ \text{N/mm}^2 \end{aligned}

Change in length (δ=NL/AE\delta = NL/AE)

δa=25×103×3008659.0×70×103=0.0124 mmδc=75×103×2505674.5×100×103=0.0330 mmδs=100×103×1001963.5×200×103=0.0255 mmδtotal=0.0709 mm\begin{aligned} \delta_a &= \frac{25\times10^3\times300}{8659.0\times70\times10^3} = 0.0124\ \text{mm}\\ \delta_c &= \frac{75\times10^3\times250}{5674.5\times100\times10^3} = 0.0330\ \text{mm}\\ \delta_s &= \frac{100\times10^3\times100}{1963.5\times200\times10^3} = 0.0255\ \text{mm}\\ \delta_{total} &= 0.0709\ \text{mm} \end{aligned}

Answer: σAl=2.89\sigma_{Al} = 2.89, σCu=13.22\sigma_{Cu} = 13.22, σSt=50.93\sigma_{St} = 50.93 N/mm² (all tensile); total elongation = 0.0709 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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