Give the approximate temperature at which creep deformation becomes an important consideration for each of the following metals:
1 answer:
Answer:
The creep temperature for different materials is as follows.
A. <u>Nickel :- </u> = 690.4 K
B. <u>Copper :- </u> = 542.8 K
C.<u> Iron :-</u> = 723.6 K
D.<u> Tungsten :-</u> = 1469.2 K
E. <u>Lead :- </u> = 240.2 K
F. <u>Aluminium :- </u> = 373.2 K
Explanation:
Creep is the slow plastic deformation occur at high elevated temperature.
The approximate temperature for different metals is as follows at which creep becomes considerable.
A. <u>Nickel :-</u> The relation between melting temperature & creep temperature is given by the formula.
The melting temperature of nickel = 1726 K
Creep temperature = 0.4 × 1726
= 690.4 K
B. <u>Copper :- </u>The relation between melting temperature & creep temperature for copper is given by the formula.
The melting temperature of copper = 1357 K
Creep temperature = 0.4 × 1357
= 542.8 K
C.<u> Iron :- </u>The relation between melting temperature & creep temperature for iron is given by the formula
The melting temperature of iron = 1809 K
Creep temperature = 0.4 × 1809
= 723.6 K
D.<u> Tungsten :- </u>The relation between melting temperature & creep temperature for tungsten is given by the formula
The melting temperature of iron = 3673 K
Creep temperature = 0.4 × 3673
= 1469.2 K
E. <u>Lead :-</u>The relation between melting temperature & creep temperature for lead is given by the formula
The melting temperature of lead = 600.5 K
Creep temperature = 0.4 × 600.5
= 240.2 K
F. <u>Aluminium :- </u>The relation between melting temperature & creep temperature for aluminium is given by the formula
The melting temperature of lead = 933 K
Creep temperature = 0.4 × 933
= 373.2 K
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Answer:
F = 9.11 x 10³ N = 9.11 KN
Explanation:
The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:
ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper
ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)
where,
ΔL = Equivalent Change in Length = ?
ΔT = Change in Temperature = 25°C - 12°C = 13°C
Ls = Length of Steel Segment = 300 mm = 0.3 m
Lb = Length of Brass Segment = 200 mm = 0.2 m
Lc = Length of Copper Segment = 100 mm = 0.1 m
Therefore,
ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)
ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m
ΔL = 123.5 x 10⁻⁶ m ----------------------- equation (1)
Now, we calculate this deflection in terms of an applied force (F):
ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)
ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)
ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)
ΔL = F(13.55 x 10⁻⁹ m/N) --------------------- equation (1)
Comparing equation (1) and equation (2):
123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)
F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)
<u>F = 9.11 x 10³ N = 9.11 KN</u>
