Use the results of Prob. 5–82 for plane strain near the tip with u 5 0 and n 5 13. If the yieldstrength of the plate is Sy, what
is s1 when yield occurs?(a) Use the distortion-energy theory.(b) Use the maximum-shear-stress theory. Using Mohr’s circles, explain your answer
1 answer:
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Answer:
R min = 28.173 ohm
R max = 1.55 × ohm
Explanation:
given data
capacitor = 0.227 μF
charged to 5.03 V
potential difference across the plates = 0.833 V
handled effectively = 11.5 μs to 6.33 ms
solution
we know that resistance range of the resistor is express as
V(t) = ...........1
so R will be
R = ....................2
put here value
so for t min 11.5 μs
R =
R min = 28.173 ohm
and
for t max 6.33 ms
R max =
R max = 1.55 × ohm
Answer:
critical stress required for the propagation is 27.396615 × N/m²
Explanation:
given data
specific surface energy = 0.90 J/m²
modulus of elasticity E = 393 GPa = 393 × N/m²
internal crack length = 0.6 mm
to find out
critical stress required for the propagation
solution
we will apply here critical stress formula for propagation of internal crack
( σc ) = .....................1
here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 × m
so now put value in equation 1 we get
( σc ) =
( σc ) =
( σc ) = 27.396615 × N/m²
so critical stress required for the propagation is 27.396615 × N/m²