A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa√m. If it is known that the largest surface crack is 0.
1 answer:
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To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.
By definition the entropy change would be defined as
Using the Boyle equation we have
Where,
= Specific heat at constant pressure
= Initial temperature of gas
= Final temprature of gas
R = Universal gas constant
= Initial specific Volume of gas
= Final specific volume of gas
According to the statement, it is an isothermal process and the tank is therefore rigid
The equation would turn out as
<em>Therefore the entropy change of the ideal gas is 0</em>
Into the surroundings we have that
Where,
Q = Heat Exchange
T = Temperature in the surrounding
Replacing with our values we have that
<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>
Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
= = 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2