Question

The center rod CD of the assembly shown in the figure is heated from T₁ = 30^oC to T₂ = 180^oC using electrical resistance heating. The two end rods AB and EF are heated from T₁ = 30^oC to  T₂ = 50^oC. At the lower temperature T₁, the gap between C and the rigid bar is 0.7 mm. Rods AB and EF are made of steel and each have a cross-sectional area of 125 mm². CD is made of aluminum and has a cross-sectional area of 375 mm². E_Steel = 200 GPa, E_Aluminum = 70 GPa,    α_steel = 12x10^-6/^oC, α_Aluminum = 23 x10^-6 /^oC.

Calculate the force and the stress in rods AB, EF and CD caused by the increase in temperature.  

Answer 1

FBD of rigid bar is as follows

$∆T_{AB}=T_2-T_1$

=50°C-30°C=20°C

$∆T_{CD}=T_4-T_3$

=180°C-30°C=150°C

$A_{AB}=125*10^{-6}m^{2}$ , L_AB=0.3m

$A_{EF}=125*10^{-6}m^{2}$ , L_EF=0.3m

$A_{CD}=375*10^{-6}m^{2}$ , L_CD=0.24m

$\alpha_{AB}=12*10^{-6}/°C, E_{AB}=200GPa$

$\alpha_{EF}=12*10^{-6}/°C, E_{EF}=200GPa$

$\alpha_{CD}=23*10^{-6}/°C, E_{CD}=70GPa$

$\sum M_C=0$

F_ABy-F_EFy=0

F_AB=F_EF

$\sum F_y=0$

$F_{AB}+F_{EF}-F_{CD}=0$

$F_{AB}+F_{AB}-F_{CD}=0$

$2F_{AB}=F_{CD}=2F_{EF}$

$\delta_{AB}=(\delta_{AB})_T+(\delta_{AB})_F$

$=\alpha_{AB}∆T_{AB}L_{AB}+\frac{F_{AB}L_{AB}}{A_{AB}E_{AB}}$

$=(12*10^{-6})(20)(0.3)+\frac{F_{AB}*0.3}{(125*10^{-6})(200*10^{9}}$

$=72*10^{-6}+(0.012*10^{-6})F_{AB}$

$\delta_{CD}=(\delta_{CD})_T-(\delta_{CD})_F$

$=\alpha_{CD}∆T_{CD}L_{CD}-\frac{F_{CD}L_{CD}}{A_{CD}E_{CD}}$

$=(23*10^{-6})(150)(0.25)+\frac{F_{CD}*0.24}{(375*10^{-6})(70*10^{9}}$

$=828*10^{-6}-(9.143*10^{-9})F_{CD}$

$\delta_{AB}=\delta_{CD}-∆gap$

$72*10^{-6}+(0.012*10^{-6})F_{AB}=$

$828*10^{-6}-(9.143*10^{-9})F_{CD}-0.7*10^{-3}$

$(0.012*10^{-6})F_{AB}+(0.01828*10^{-6}F_{AB}=128*10^{-6}-72*10^{-6}$

$F_{AB}=1849.41N$

=1.849kN

F_EF=F_AB=1.849kN

F_CD=2F_AB=2*1.849kN

=3.698kN

$\sigma_{AB}=\sigma_{EF}=\frac{F_{AB}}{A_{AB}}$

$=\frac{1.849*10^{3}}{125*10^{-6}}$

=14.79MPa

$\sigma_{CD}=\frac{F_{CD}}{A_{CD}}$

$=\frac{3.698*10^{3}}{375*10^{-6}}$

=9.86MPa