Max88
Mechanical
- Jan 26, 2018
- 1
Dear All,
I have SA 333 Gr.6 2" x 5.54mm PQR coupon supposed to be impact tested at -50Deg.C asper piping service class.
The specimen size is 10mm X 3.33 mm
So,
As per B31.3 Table 323.3.4 and Clause 323.3.4(b) , i calculated the impact test temperature as -60.3 Deg.C .Please correct me if i am wrong as my mother toungue is not english, i am having a hard time to interpret the clause
My calculations as below
Actual wall thickness = 5.54 mm
Impact Specimen Width Along Notch = 3.33
So the ratio is (3.33/5.54 ) X 100 = 60.1% which is less than 80%
Now refer Table table 323.3.4 to find the temperature reduction for 5.54mm
1. Temperature reduction for 6mm = 8.3 Deg.C
2. Temperature Reduction for 5mm = 11.1 Deg.C
3. Temperature reduction for 5.54mm = (8.3*2)/3 + (11.1*1)/3 = 9.33 Deg.C (Refer Note of table 323.3.4 Straight line interpolation for intermediate values is permitted)
Temperature reduction for 3.33mm (asper PQR Impact test specimen ) from table = 19.4 Deg.C
We have to take the difference
Ie, 19.4 - 9.33 = 10.07 Deg.C
We have to minimum this value from the actual requirement which is -50Deg.C
50+10.07 = -60.07 Deg.C
I have SA 333 Gr.6 2" x 5.54mm PQR coupon supposed to be impact tested at -50Deg.C asper piping service class.
The specimen size is 10mm X 3.33 mm
So,
As per B31.3 Table 323.3.4 and Clause 323.3.4(b) , i calculated the impact test temperature as -60.3 Deg.C .Please correct me if i am wrong as my mother toungue is not english, i am having a hard time to interpret the clause
My calculations as below
Actual wall thickness = 5.54 mm
Impact Specimen Width Along Notch = 3.33
So the ratio is (3.33/5.54 ) X 100 = 60.1% which is less than 80%
Now refer Table table 323.3.4 to find the temperature reduction for 5.54mm
1. Temperature reduction for 6mm = 8.3 Deg.C
2. Temperature Reduction for 5mm = 11.1 Deg.C
3. Temperature reduction for 5.54mm = (8.3*2)/3 + (11.1*1)/3 = 9.33 Deg.C (Refer Note of table 323.3.4 Straight line interpolation for intermediate values is permitted)
Temperature reduction for 3.33mm (asper PQR Impact test specimen ) from table = 19.4 Deg.C
We have to take the difference
Ie, 19.4 - 9.33 = 10.07 Deg.C
We have to minimum this value from the actual requirement which is -50Deg.C
50+10.07 = -60.07 Deg.C