Write a loop that prints the first 128 ASCII values followed by the corresponding characters (see the section on characters in C
hapter 2). Be aware that most of the ASCII values in the range "0..31" belong to special control characters with no standard print representation, so you might see strange symbols in the output for these values.
1 answer:
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Answer:
A) Ductility = 11% EL
B) Radius after deformation = 4.27 mm
Explanation:
A) From equations in steel test,
Tensile Strength (Ts) = 3.45 x HB
Where HB is brinell hardness;
Thus, Ts = 3.45 x 250 = 862MPa
From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.
Also, from image 2,at CW of 27%,
Ductility is approximately, 11% EL
B) Now we know that formula for %CW is;
%CW = (Ao - Ad)/(Ao)
Where Ao is area with initial radius and Ad is area deformation.
Thus;
%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100
%CW = [1 - (rd)²/(ro)²]
1 - (%CW/100) = (rd)²/(ro)²
So;
(rd)²[1 - (%CW/100)] = (ro)²
So putting the values as gotten initially ;
(ro)² = 5²([1 - (27/100)]
(ro)² = 25 - 6.75
(ro) ² = 18.25
ro = √18.25
So ro = 4.27 mm
Answer:
The minimum required thickness for a steel pressure vessel (t) is 2.275 mm
Explanation:
Minimum required thickness is the thickness of a material without corrosion allowance for each component based on the appropriate design that consider pressure, mechanical and structural loading.
Given that:
radius (r) = 65 mm = 65 × 10⁻³ m
Factor of safety (N) = 3.5
Inside presssure () = 11 MPa
Outside pressure () = 1 MPa
Yield strength () = 1000 MPa
Therefore:
, Substituting values,
The minimum required thickness for a steel pressure vessel (t) is given by the equation:
. Substituting values
The minimum required thickness for a steel pressure vessel (t) is 2.275 mm