Answer:
t= 4.5 mm
Explanation:
Given that
P = 520 KPa ( gauge)
Maximum allowable normal stress ,σ= 150
d= 2.6 m
Wall thickness = t
The normal stress for pressure vessel given as
( hoop stress)
We always take maximum stress for safe design.
Now by putting the values
t= 4.5 mm
So the minimum thickness, t, of the wall is 4.5 mm
Answer:
Explanation:
The given equation is :
Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:
ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂
θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂
=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain
=3.335 *10^-4
ε(avg) =150 *10^-6
orientation of γmax
θ = 31.71 or -58.29
To determine the direction of γmax
= 1.67 *10^-4
Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.
<h3>Wat is the max stress theory?</h3>The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”
Stress states at various critical locations are f= 2.662.
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