(a) Sabbir usually (sit)______ in the front bench.
1 answer:
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Answer:
critical stress = 595 MPa
Explanation:
given data
fracture toughness = 74.6 MPa-
crack length = 10 mm
f = 1
solution
we know crack length = 10 mm
and crack length = 2a as given in figure attach
so 2a = 10
a = 5 mm
and now we get here with the help of plane strain condition , critical stress is express as
critical stress = ......................1
put here value and we get
critical stress =
critical stress = 595 MPa
so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.
plain stress condition occur in thin body where stress through thickness not vary by the thinner section.
Explanation:
Consider a fluid of density, ρ moving with a velocity, U over a flat plate of length, L.
Let the Kinematic viscosity of the fluid be ν.
Let the flow over the fluid be laminar for a distance x from the leading edge.
Now this distance is called the critical distance.
Therefore, for a laminar flow, the critical distance can be defined as the distance from the leading edge of the plate where the Reynolds number is equal to 5 x
And Reynolds number is a dimensionless number which determines whether a flow is laminar or turbulent.
Mathematically, we can write,
Re =
or 5 x =
( for a laminar flow )
Therefore, critical distance
So x is defined as the critical distance upto which the flow is laminar.