the ends of a steel rod are fixed. What is the thermal stress in the rod when the tempeture decreases by 80 degrees K. Youngs modulus for steel is Y=2x10^11 N/msquared.
2006-08-10 07:48:07 · 3 answers · asked by jonessunrunner1 2 in Science & Mathematics ➔ Physics
The previous answer is not correct. The uniaxial stress in a constrained rod (i.e., the ends are fixed) undergoing thermally induced deformation is (see first source):
stress = delta-T * (coefficient of thermal expansion) * (Young's modulus)
You need to know the linear coefficient of thermal expansion (not the rod's length) to answer your question. The second source reports that for Al at 20 C, this parameter has a value of 2.3 * 10^-7/K. The stress is then given by:
(-80 K) * (2 * 10^11 N/m^2) * (2.3 * 10^-7/K) = -3.68 * 10^6 N/m^2
= -3.68 * 10^6 Pa = -36.8 bars = -533.79 psi
2006-08-10 11:45:20 · answer #1 · answered by hfshaw 7 · 0⤊ 0⤋
Young's modulus is stress/strain. It is the (length-normalized) spring constant for the steel. What is the expansion coefficient of steel? This will give you the strain because the confined rod is shorter than its unconfined length. Then use Young's modulus to get the stress.
This question is analogous to: "The spring constant of a spring is 10N/m. Find the force on the spring when it is compressed by 2m."
2006-08-10 12:31:23 · answer #2 · answered by Benjamin N 4 · 0⤊ 0⤋
you don't know how long the rod is? the formula is stress= youngs mod x change in temp x length. so you gotta know the length!!
2006-08-10 08:00:36 · answer #3 · answered by golden_everglade 2 · 0⤊ 1⤋