Find the lenghth of ( final) the one metor iron rod due to expansion temperature from 20 c to 120 c ( of rod -6* 10 / cl)
Answers
Answer:
Free expansion of the rod =αLΔθ
=15×10
−60
C×2m×(50−20)
0
C
=9×10
−
4m=0.9mm
If the expansion is fully prevented, then
Strain=
2
9×10
−4
=4.5×10
−4
Temperature Stress = Strain ×Y
=4.5×10
−4
×2×10
11
=9×10
7
N/m
2
If 0.4 mm expansion is allowed, then length restricted to expand
=0.9−0.4=0.5mm
Strain=
2
5×10
−4
=2.5×10
−4
Temperature stress = Strain ×Y=2.5×10
−4
×2×10
11
=5×10
7
N/m
2Free expansion of the rod =αLΔθ
=15×10
−60
C×2m×(50−20)
0
C
=9×10
−
4m=0.9mm
If the expansion is fully prevented, then
Strain=
2
9×10
−4
=4.5×10
−4
Temperature Stress = Strain ×Y
=4.5×10
−4
×2×10
11
=9×10
7
N/m
2
If 0.4 mm expansion is allowed, then length restricted to expand
=0.9−0.4=0.5mm
Strain=
2
5×10
−4
=2.5×10
−4
Temperature stress = Strain ×Y=2.5×10
−4
×2×10
11
=5×10
7
N/m
2
Explanation: