Question
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?
(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?
Answers
Damon
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)
= 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)