One-dimensional, steady-state conduction with uniform internal energy generation occurs in a plane wall with a thickness of 50 m
1 answer:
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Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
=
...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
=
= 12 kpsi
and
=
...........2
put here value and we get
=
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432
Answer:
strains for the respective cases are
0.287
0.318
0.127
and for the entire process 0.733
Explanation:
The formula for the true strain is given as:
Where
True strain
l= length of the member after deformation
original length of the member
<u>Now for the first case we have</u>
l= 1.6m
thus,
<u>similarly for the second case we have</u>
l= 2.2m
(as the length is changing from 1.6m in this case)
thus,
<u>Now for the third case</u>
l= 2.5m
thus,
<u>Now the true strain for the entire process</u>
l=2.5m
thus,