valuation) methods discovers a damage: a half-circular surface crack of size a = 5 mm . ngineer Ben wants to replace e damaged part with a new one, ut a replacement part is not nmediately available. Hence, ne cracked part is still being sed in the machine to support e tensile load P. Engineer Ben K = FS/na For tensile load P For bending load M P 3M S = S, = 2bt S = Sp : bt² %3D F = F, = 0.728 F = F, = 0.728 rants to make sure that the Stress intensity factor for half-circular surface crack (textbook, Fig racked part is safe, so he adds a
valuation) methods discovers a damage: a half-circular surface crack of size a = 5 mm . ngineer Ben wants to replace e damaged part with a new one, ut a replacement part is not nmediately available. Hence, ne cracked part is still being sed in the machine to support e tensile load P. Engineer Ben K = FS/na For tensile load P For bending load M P 3M S = S, = 2bt S = Sp : bt² %3D F = F, = 0.728 F = F, = 0.728 rants to make sure that the Stress intensity factor for half-circular surface crack (textbook, Fig racked part is safe, so he adds a
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Fracture Mechanics
![A machine part (see Figure) with outer dimension of tx 2b (t = 50 mm, b = 25 mm) is designed to
support a tensile force P = 70 kN in the machine. In a safety inspection, NDE (Nondestructive
Evaluation) methods discovers a damage: a half-circular surface crack of size a = 5 mm .
Engineer Ben wants to replace
the damaged part with a new one,
but a replacement part is not
immediately available. Hence,
the cracked part is still being
used in the machine to support
the tensile load P. Engineer Ben
K = FSTA
For tensile load P
For bending load M
3M
S = S, =
2bt
S = S, =
bt2
F = F; = 0.728
F = F, = 0.728
wants to make sure that the
Stress intensity factor for half-circular surface crack (textbook, Fig 8.17)
cracked part is safe, so he adds a
bending moment M of negative
magnitude, which bends the part in the direction to press the two crack surfaces to close (as against to
open the crack to cause fracture).
Determine the magnitude of M that will give a safety factor against fracture XK = 2.3 when the
cracked part is subjected to both the tensile force P = 70 kN and the bending moment M.
The material is ASTM A470-8 steel which has fracture toughness Kic
= 60 MPaym.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0be26343-f96f-4a20-afce-426c5bd9a1f8%2F98bb2a04-fe6b-4569-bc3f-7f3cb5ae2e0a%2F5rq4b_processed.png&w=3840&q=75)
Transcribed Image Text:A machine part (see Figure) with outer dimension of tx 2b (t = 50 mm, b = 25 mm) is designed to
support a tensile force P = 70 kN in the machine. In a safety inspection, NDE (Nondestructive
Evaluation) methods discovers a damage: a half-circular surface crack of size a = 5 mm .
Engineer Ben wants to replace
the damaged part with a new one,
but a replacement part is not
immediately available. Hence,
the cracked part is still being
used in the machine to support
the tensile load P. Engineer Ben
K = FSTA
For tensile load P
For bending load M
3M
S = S, =
2bt
S = S, =
bt2
F = F; = 0.728
F = F, = 0.728
wants to make sure that the
Stress intensity factor for half-circular surface crack (textbook, Fig 8.17)
cracked part is safe, so he adds a
bending moment M of negative
magnitude, which bends the part in the direction to press the two crack surfaces to close (as against to
open the crack to cause fracture).
Determine the magnitude of M that will give a safety factor against fracture XK = 2.3 when the
cracked part is subjected to both the tensile force P = 70 kN and the bending moment M.
The material is ASTM A470-8 steel which has fracture toughness Kic
= 60 MPaym.
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