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Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for while a is taken as 0.003m and Y is already known
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material
Answer:
a)Δs = 834 mm
b)V=1122 mm/s
Explanation:
Given that
a)
When t= 2 s
s= 114 mm
At t= 4 s
s= 948 mm
So the displacement between 2 s to 4 s
Δs = 948 - 114 mm
Δs = 834 mm
b)
We know that velocity V
At t= 5 s
V=1122 mm/s
We know that acceleration a
a= 90 t
a = 90 x 5
Answer: Current = 2A
Explanation:
Inductor L = 10mH
E.m.f = 1 V
Time t = 20 ms
Using
V = Ldi/dt
di/dt = V/ L
di/dt = 1/10×10^-3
di/dt = 100
di = 100dt
Integrate both sides
I = 100(t1 - to)
Assume that initially there is no current through the inductor.
to = 0
I = 100 × t
I = 100 × 20×10^-3
I = 2A