The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the longest in the world. Imagine that a steel wire with this length and a cross-sectional area of 4.00 x 10-6 m^2 is laid on the bridge deck with its ends attached to the towers of the bridge, on a summer day when the temperature of the wire is 35.0°C.
(a) When winter arrives, the towers stay the same distance apart and the bridge deck keeps the same shape as its expansion joints open. When the temperature drops to −10.0°C, what is the tension in the wire? Take Young's modulus for steel to be 2.00 10^11 N/m2. (The average linear expansion coefficient for steel is 1.1 10-5°C−1.)
(b) Permanent deformation occurs if the stress in the steel exceeds its elastic limit of 3.00 x 10^8 N/m2. At what temperature would the wire reach its elastic limit?
To solve this problem, we need to use the concept of thermal expansion and stress-strain relationship for the steel wire.
(a) First, let's calculate the change in temperature. The initial temperature is 35.0°C, and the final temperature is -10.0°C. The change in temperature is given by:
ΔT = T_final - T_initial
ΔT = -10.0°C - 35.0°C
ΔT = -45.0°C
Next, we need to calculate the change in length of the wire due to thermal expansion. The average linear expansion coefficient for steel is given as α = 1.1 × 10^-5°C^-1.
ΔL = αLΔT
ΔL = (1.1 × 10^-5°C^-1)(1.28 km)(-45.0°C)
ΔL = -0.0616 km = -61.6 m
Since the change in length is negative, it means that the wire contracts.
Now, we can calculate the tension in the wire using Hooke's Law, which states that the stress (σ) is directly proportional to the strain (ε). The stress can be calculated as follows:
σ = Y * ε
Where:
σ is the stress
Y is Young's modulus
ε is the strain
To find the strain, we can use the formula:
ε = ΔL / L
Where:
ε is the strain
ΔL is the change in length
L is the initial length
Let's substitute the values into the equations:
ε = (-61.6 m) / (1.28 km)
ε = -4.81 × 10^-2
Now we can find the tension (F) using Hooke's Law:
σ = Y * ε
F / A = Y * ε
F = A * Y * ε
Where:
F is the tension in the wire
A is the cross-sectional area of the wire (4.00 × 10^-6 m^2)
Substituting the values:
F = (4.00 × 10^-6 m^2) * (2.00 × 10^11 N/m^2) * (-4.81 × 10^-2)
F ≈ -3.87 × 10^7 N
Therefore, the tension in the wire when the temperature drops to -10.0°C is approximately -3.87 × 10^7 N.
(b) To determine the temperature at which the wire reaches its elastic limit, we need to find the strain at the elastic limit using Hooke's Law:
σ = Y * ε
Rearranging the equation:
ε = σ / Y
Now substitute the values:
ε = (3.00 × 10^8 N/m^2) / (2.00 × 10^11 N/m^2)
ε = 1.5 × 10^-3
Next, we find the change in temperature (ΔT):
ΔL = ε * L
ΔL = (1.5 × 10^-3)(1.28 km)
ΔL = 1.92 m
Now, we can calculate the change in temperature (ΔT) using the formula:
ΔL = αLΔT
Substituting the values:
1.92 m = (1.1 × 10^-5°C^-1)(1.28 km)(ΔT)
ΔT = 1.92 m / (1.1 × 10^-5°C^-1)(1.28 km)
ΔT ≈ 131°C
Therefore, the wire would reach its elastic limit at a temperature approximately 131°C.
To solve this problem, we can use the formula for thermal stress:
ΔL = αLΔT
where:
ΔL is the change in length,
α is the linear expansion coefficient,
L is the original length, and
ΔT is the change in temperature.
(a) We can start by calculating the change in length of the wire.
ΔL = αLΔT
ΔL = (1.1x10^-5 °C^-1)(1.28 km)(-45.0 °C - 35.0 °C)
ΔL = (1.1x10^-5 °C^-1)(1.28 km)(-80.0 °C)
ΔL = -1.4048 m
The negative sign indicates that the wire contracts when the temperature decreases by 80.0 °C.
Next, we can calculate the tension in the wire.
Tension = Young's modulus x (ΔL / L) x (cross-sectional area)
Tension = (2.00x10^11 N/m^2) x (-1.4048 m / 1.28 km) x (4.00x10^-6 m^2)
Tension ≈ -924,375 N
Therefore, the tension in the wire when the temperature drops to -10.0 °C is approximately -924,375 N.
(b) To determine the temperature at which the wire reaches its elastic limit, we can rearrange the equation for tension:
Tension = Young's modulus x (ΔL / L) x (cross-sectional area)
Using the maximum allowable stress of 3.00x10^8 N/m^2, the equation becomes:
3.00x10^8 N/m^2 = (2.00x10^11 N/m^2) x (ΔL / L) x (4.00x10^-6 m^2)
We need to solve for ΔT:
ΔT = (3.00x10^8 N/m^2) / [(2.00x10^11 N/m^2) x (ΔL / L) x (4.00x10^-6 m^2)]
Substituting the values:
ΔT = (3.00x10^8 N/m^2) / [(2.00x10^11 N/m^2) x (-2ΔL / 1.28 km) x (4.00x10^-6 m^2)]
Note: We multiply by -2 in the denominator because the wire would be under tension and elongate twice as much as in part (a).
Simplifying:
ΔT ≈ -0.586 °C
Therefore, the wire would reach its elastic limit at a temperature of approximately -0.586 °C.
delta L/L = (1.1*10^-5)(45)
= 49.5 * 10^-5 meters/meter of hypothetical stretch
that delta L/L is strain
delta L/L = sigma/E which is stress/young's
stress = 4.95*10^-4 * 2*10^11
= 9.9 *10^7 N/m^2
tension = stress * area
=9.9*10^7 * 4*10^-6 = 396 N
b)
well we got 9.9*10^7 N/m^2 for a 45 degree drop
so to get 3 * 10^8 N/m^2
we need 45*30/9.9 = 136 degree drop
that is 35 - 136 = -101 deg C
(unlikely in San Francisco)