Question

A 20-gage (0.091186 cm) steel sheet is bent to a radius of 1.27 cm. Yield stress (Y) is 275.79 MPa. Calculate (i) the radius of the part after it is bent and (ii) the required bend angle to achieve a 1.57 rad bend after spring back has occurred .You may use E = 199,947.96 MPa.

b. Explain why and how various factors influence springback in bending of sheet metals

Answer 1

Sol: giren t = 0.091186 cm R; = d.27 cm Y = 275.79 MPa E= 199 947.96 MPa . Factor Rir. Et d. 27 X 275.79 199947.96X.091186 = 0.01921043 Et 1.27 -46.01921043)2- 30.019210437+1 -0.942397 Rf 1. 34 76 cm de di = ( 2 Rf1t) + 1 (2 Rilt)t + Qi: 1.57 (2X1.34767.01921043) + 1 ( 2x127/.01921043) +) ♡ Qi: 1.6652 radian

& Spring Back is Recovery in Bending. & when Bend is completed spring back. angle makes larger after the bend , load removed is called Negative factors 1. Over Bending 2. Coining 3. carried out at elevated 4. Stretch Bending. temprature
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