An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tens
ile force of 1 kN is applied axially to this heated rod, what is the expected mechanical response
2 answers:
Answer: the metal will experience a strain of approximately 2.037Mpa
This strain is lesser than if the force was applied at room temperature.
This will reduce internal stress and increase some mechanical properties of the aluminum such as mechanical hardening.
Explanation:
Detailed explanation and calculation and comparison with equivalent tensile stretching at ordinary room temperature is compared.
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Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae ------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
Answer:
The value of Modulus of elasticity E = 85.33 ×
Beam deflection is = 0.15 in
Explanation:
Given data
width = 5 in
Length = 60 in
Mass of the person = 125 lb
Load = 125 × 32 = 4000
We know that moment of inertia is given as
I = 1.40625
Deflection = 0.15 in
We know that deflection of the beam in this case is given as
Δ =
E = 85.33 ×
This is the value of Modulus of elasticity.
Beam deflection is = 0.15 in