Answer: Symbol rate, Fs = 0.846
Explanation:
The attachment below shows the calculations
Asolid rectangular rodhas a length of 90mm, made of steel material (E =207,000 MPa, Syield= 300 MPa), the cross section of the r
od is shownbelow.If the boundary conditions are fixed-freeand a safety factor of 2 on loadis required, (1)Find themaximumallowable load (Pallowable)on the rod. [9points]{hint: Pallowable= Pcr/required safety factor}(2)If the same critical load (Pcr) calculated from (1)above(Pcrisalready taken into accounta safety factor of 2)is applied to a solid round rodunderthe same boundary conditions, what is the minimum required diameter of the rod against buckling? [9points]
1 answer:
Answer:
13.6mm
Explanation:
We consider diameter to be a chord that runs through the center point of the circle. It is considered as the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. That is, it divides it into two equal parts, each of which is a radius of the circle. The radius is half the diameter.
See attachment for the step by step solution of the problem
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Answer:
86 mm
Explanation:
From the attached thermal circuit diagram, equation for i-nodes will be
Equation 1
Similarly, the equation for outer node “o” will be
Equation 2
The conventive thermal resistance in i-node will be
Equation 3
The conventive hermal resistance per unit area is
Equation 4
The conductive thermal resistance per unit area is
Equation 5
Since is given as 100,
is 40
is 300
is 25
Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain
Equation 6
Equation 7
From equation 6 we can substitute wherever there’s with 3000L+40 as seen in equation 7 hence we obtain
The above can be simplified to be
-3000L=1.665-260
Therefore, insulation thickness is 86mm
