A tension test is carried out on an Al alloy specimen which has an original diameter of 0.505 in and an original gauge length of
2 in. After fracture, the diameter at the fractured neck is measured to be 0.37 in and the gauge length after fracture is 2.32 in. The maximum load during the test was 37 kN whereas at fracture the load was 34 kN.
a. What is the tensile strength?
b. What are the engineering and true stresses at fracture?
c. What are the engineering and true strain at fracture?
d. What is the ductility of this metal?
a. What is the tensile strength?
b. What are the engineering and true stresses at fracture?
c. What are the engineering and true strain at fracture?
d. What is the ductility of this metal?
1 answer:
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Answer: 383.22K
Explanation:
L = 3m, w = 1.5m
Area A = 3 x 1.5 = 4.5m2
Q' = 750W/m2 (heat from sun) ,
& = 0.87
Q = &Q' = 0. 87x750 = 652.5W/m2
E = QA = 652.5 x 4.5 = 2936.25W
T(sur) = 300K, T(panel) = ?
Using E = §€A(T^4(panel) - T^4(sur))
§ = Stefan constant = 5.7x10^-8
€ = emmisivity = 0.85
2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)
T(panel) = 383.22K
See image for further details.
Answer:
HB = 3.22
Explanation:
The formula to calculate the Brinell Hardness is given as follows:
where,
HB = Brinell Hardness = ?
P = Applied Load in kg = 500 kg
D = Diameter of Indenter in mm = 10 mm
d = Diameter of the indentation in mm = 1.55 mm
Therefore, using these values, we get:
<u>HB = 3.22 </u>