Arod is 2m long at temperature of 10oc. find the expansion of the rod, when the temperature is raised to 80oc. if this expansion is prevented, find the stress induced in the material of the rod. take e = 1.0 x 105 mpa and α = 0.12 per degree centigrade.

The expansion of the rod is 0.00168 meters and the stress induced is 84MPa

Linear thermal expansion, ΔL = α(ΔT)L

Length of rod, L = 2m

Change in temperature, ΔT = (T2 - T1) = (80°C - 10°C) = 70°C

Coefficient of expansion, α = 0.000012

Plugging values into the formula :

ΔL = 0.000012 × 70 × 2 = 0.00168 m

The stress induced, σ; if expansion is prevented is defined as :  α(ΔT)E ; where E = modulus of elasticity :

σ = 0.000012 × 70 × 1.0 x 10^5

σ = 0.000012 × 70 × 100000

σ = 84 MPa

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ΔL = 1.68 mm

σ = 84 MPa

Explanation:

Thermal expansion is:

ΔL = α ΔT L

Thermal stress is:

σ = α ΔT E

Given:

α = 1.2×10⁻⁵ /°C

E = 1.0×10⁵ MPa

ΔT = 80°C − 10°C = 70°C

L = 2 m

ΔL = (1.2×10⁻⁵ /°C) (70°C) (2 m)

ΔL = 0.00168 m

ΔL = 1.68 mm

σ = (1.2×10⁻⁵ /°C) (70°C) (1.0×10⁵ MPa)

σ = 84 MPa

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )

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