A 100-m wire is strung horizontally between 2 poles?

Question

The middle of the wire sags 0.25m at temperature 0C and 0.5m at 35C.
Linear density of the wire is 2kg/m.
Elastic koeeficient of the wire is k=600,000 N/m x m = 6e5N

What is coefficient of linear thermal expansion of the wire?

I mixed up the numbers. Should be:

Elastic koeeficient of the wire is k=6,000,000,000 N/m x m = 6e9N

Answer 1

Let x is the horizontal coordinate, z is the vertical coordinate, and the coordinate origin is placed at the lowest (central) point of the wire. Due to symmetry, it is sufficient to consider only the right half of the wire. The horizontal component of the wire tension T_x is constant,

Eq.(1): T_x = const,

since the horizontal external forces are absent. The total tension T is directed along the wire. This gives the condition on the vertical component T_z:

Eq.(2): T_z = z' T_x, (z'=dz/dx).

Consider the equilibrium of the wire element of length dL=√(1+z'^2) dx in the vertical direction. The gravitational force ρ g dL is balanced by the difference of T_z at the element boundaries dT_z = T_x dz':

Eq.(3): ρ g dL = T_x dz'.

Equation (3) is integrated to give the catenary solution. Applying this solution to the point x=S/2, z=d, we have

Eq.(4): d / L = (T_x / W) { cosh [ S W / (2 L T_x ) ] - 1},
Eq.(5): sinh [ S W / (2 L T_x ) ] = W / (2 T_x).

Here, S is the distance between poles, d is the wire sag, L is the wire length, and W= ρ g L is the wire weight. Substituting Eq. (4) into Eq. (5) one finds the horizontal wire tension

Eq.(6): T_x / W = L / (8 d) (1 - 4 d^2 2 / L^2).

Equation (6) shows that W / T_x is about d / L, which is about 10^(-2). For this reason one can expand cosh and sinh in Eqs. (4), (5) in Taylor series. Keeping two terms of expansion cosh(a) = 1+a^2/2 and sinh(a) = a + a^3 / 6, we reduce Eqs. (4), (5) to

Eq. (7): T_x = S^2 W / (8 d L),
Eq. (8): (L - S) / S = (8 / 3) (d/ S)^2.

Let L_0 is the unstreched wire length at given temperature. To evaluate the wire stretching,
one can replace T by T_x, because T_z/T_x is of order of d/S. Then, T_x = k Δ L/L_0. Since L and L_0 are approximately equal to S, from Eq.(7) it follows

Eq.(9): Δ L/S = (S / 8d) (W / k) .

Substituting Eq.(9) into Eq.(8) we have

Eq.(10): (L_0 - S)/S = (8 / 3) (d/ S)^2 - ( S / 8d) (W / k).

Apply Eq.(10) to the first case, d=1/4 m and temperature T=0C (S = 100m, W =2E3 N, k = 6E9 N)

(L_0 - S) / S = (1/6)*10^(-4) - (1/2)10^2 *(1/3) 10^(-6) = 0.

It means that L_0 =S at T=0C, and all the difference between L_0 and S at T=35C is due to thermal expansion of the wire. Substitute d=1/2 m in Eq. (10):

(L_0 - S) / S = (2/3)*10^(-4) - (1/4)10^2 *(1/3) 10^(-6) = (7/12)10^(-4).

Divide this on 35 to get

α = (1/6)*10^(-5) =1.67*10^(-6) /C.

Looks quite low, if I did not make a mistake somewhere. What is the wire made of?