Answer: 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.
