A large part in a turbine-generator unit operates near room temperature and is made of ASTM A470-8 steel ( ). A surface crack has been found that is roughly a semi-ellipse, with surface length 2c= 50 mm and a depth a= 15 mm. the stress normal to the plane of the crack is 250 MPa, and the member width and thickness are large compared to the crack size. What is the safety factor against brittle fracture? Should the power plant

safety factor= 1.26

safety factor does not exceed 2. Therefore, it is not safe to operate.

Explanation:

Express the stress infinity factor (k) that relate to the crack length (Q) applied stress for the centre crack of the plate.

K= Fsq √ πa/Q

Here,

the dimension less function quality is F

Crack length is a

Q= 1 + 1.48(a/c)^1.65

The surface length is c

c= 25mm

a= 15mm

Substitute c= 25mm and a=15mm into the equation

Q= 1 + 1.48(15mm/25mm)^1.65

Q= 1 + 1.48(0.6)^1.65

Q= 1 + 0.630

Q= 1.630

Find the stress intensity factor of the corresponding ratio of 0.4 and 0.12

Where

F= 1.12

S= 250mpa

Q= 1.63

a= 15mm

Substitute into the qequation

K= (1.12) (250mpa) √π(15mm× (1/1000mm))/1.63

K= (1.12) (250mpa)√47.1mm× (1/1000mm))/1.63

K= 2.80mpa √0.0471× (1m)/1.63

K= 280mpa × 0.170√m

K= 476mpa√m

Calculate the safety

Xk= Kk/K

Where fraction toughness is Kk

From the table of fraction toughness, corresponding tensile property for metal at room temperature, select this fraction toughness for ASTM A470-8 steel.

Substitute 47.6mpa√m for k

60mpa√m for Kk

Xk= 60mpa√m/47.6mpa√m

Xk= 1.26

In conclusion, from the above result, safety factor does not exceed 2. Therefore, it is not safe to operate.

answer:

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explanation:

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