Recent content by gunta

but if they said that there is a space of 0.7mm wouldn't that mean that that is the expansion when the temp rise before it goes under stress at the bar ?

i think i figured it out... i used the 30 deg temp for the equation to find the temp change using the elongation... that is 0.7 = 23x10^-6 x 70 x (T - 30) and then solved for T and then from that temp i used the long equation from above and solved F ... you think that is correct?

i tried using the equation where (alpha of aluminium - alpha of steel) x (change in temp) = F [(1/(EaxAa) + (1/(EsxAs)] and i did that for both temp changes and then used the equation stress = F/A to find both stresses please tell me if this is correct please... i cannot take a pic of my work...

1.The center rod CD of the assembly shown in the figure is heated from T1 = 30oC to T2 = 180oC using electrical resistance heating. The two end rods AB and EF are heated from T1 = 30oC to T2 = 50oC. At the lower temperature T1, the gap between C and the rigid bar is 0.7 mm. Rods AB and EF are...