Determine load carried by each cylinder:
A hollow steel cylinder of cross-sectional area 2000 mm^{2} concentrically surrounds a solid aluminium cylinder of cross-sectional area 6000 mm^{2}. Both cylinders have the same length of 500 mm before a rigid block weighing 200 kN is applied at 20^{o}C as shown in Figure. Determine
(a) The load carried by each cylinder at 60^{o}C.
(b) The temperature rise required for the entire load to be carried by the aluminium cylinder alone.
Figure
For computation purposes, take following values :
E_{steel} = 210 GN/m^{2} and E_{aluminium} = 70 GN/m^{2}
σ_{steel }= 12 × 10^{- 6} K^{-1} and α_{aluminium} = 23 × 10^{- 6} K^{-1}
Figure shows the free thermal expansions Δ_{a} and Δ_{s }together with the common expansion Δ under the load of 200 kN (the subscripts a and s standing for aluminium and steel respectively).
For a temperature rise of ΔT K,
We have,
Δ_{a} = 500 × 23 × 10^{-6} × ΔT = 11.5 × 10^{-3} ΔT mm
Δ_{s} = 500 × 12 × 10^{-6} × ΔT = 6 × 10^{-3} ΔT mm
Under load, the strains are
ε_{α} = Δ_{a} -Δ /500
and ε_{σ} = Δ_{s} -Δ /500
and the corresponding stresses are as follows :
σ_{α} =70 × 10^{3}/500 (Δ_{α} - Δ) = 140 (Δ_{α} - Δ) N mm^{-2}
σ_{s} = 210 × 10^{3}/500 (Δ_{s} - Δ) = 420 (Δ_{s} - Δ) N mm^{-2}
For equilibrium of vertical forces,
σ_{a} × 6000 + σ_{s} × 2000 = 200 × 10^{3} N
Substituting for σ_{a}, σ_{s}, Δ_{a} and Δ_{s}, we get
(11.5 × 10^{- 3} × ΔT - Δ) + (6 × 10^{-3} ΔT - Δ) = 5/21
Hence,
Δ= 8.75 × 10^{-3} ΔT - 5/42
The loads taken by the aluminium and the steel are therefore,
P_{a} = σ_{a} × 6000 N
= 840 ( 2.75 × 10^{-3} ΔT +5/42) kN
P_{s} = σ_{s} × 2000 N
= 840 ((5/42) - 2.75 × 10^{- 3} ΔT) kN
These equations will be valid as long as Δ is less than Δ_{s}. the load will be completely carried by aluminium when Δ_{s} becomes equal to Δ.
(a) at 60^{o}C,
ΔT = 60 - 20 = 40 K
P_{α} = 840 ( 2.75 × 40 × 10^{-3} +5/42)
= 192.4 kN
P_{s }= 200 - 192.4 = 7.6 kN
(b) The load will be carried completely by aluminium when
6 × 10^{-3} × ΔT = 8.75 × 10^{- 3} × ΔT - 5/42
or ΔT = 5 × 10^{3}/2.75 × 42 = 43.3^{o} C
i.e. at a temperature of (20 + 43.3) = 63.3^{o}C.