Where do greywater pipes generally feed into?
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.
Answer:
a) Δd(change in wood diameter) = 5%
b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter
C) new diameter (D2) = 10.5 in
Explanation:
Wood pole diameter = 10 inches
moisture content = 5%
FSP = 30%
A) The percentage change in the wood's diameter
note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter
Δd/d = 1/5(30 - 5)
Δd/d = 5%
Δd = 5%
B) would the wood swell or shrink
The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter
C) The new diameter of the wood
D2 = D + D( )
D = initial diameter= 10 in , M1 = initial moisture content = 5%
therefore D2 = 10 + 10( 5/100 )
new diameter (D2) = 10.5 in
In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.
With the given data we can proceed to calculate the compression stress:
Through Goodman's equations the combined effort by fatigue and compression is expressed as:
Where,
Fatigue limit for comined alternating and mean stress
Fatigue Limit
Mean stress (due to static load)
Ultimate tensile stress
Security Factor
We can replace the values and assume a security factor of 1, then
Re-arrenge for
We know that the stress is representing as,
Then,
Where =Max Moment
I= Intertia
The inertia for this object is
Then replacing and re-arrenge for
Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm