Estimating cup diameter and earing

A steel sheet has $R$ values of 0.9, 1.3, and 1.9 for the $0°, 45°,$ and $90°$ directions to rolling, respectively. For a round blank $100$ mm in diameter, estimate the smallest cup diameter to which it can be drawn. Will ears form during this operation?

Question

Estimating cup diameter and earing

A steel sheet has [latex] R [/latex] values of 0.9, 1.3, and 1.9 for the [latex] 0°, 45°,[/latex] and [latex]90° [/latex] directions to rolling, respectively. For a round blank [latex] 100 [/latex] mm in diameter, estimate the smallest cup diameter to which it can be drawn. Will ears form during this operation?

Accepted Answer

Substituting the given values into Eq. (7.20),

[latex]\overline{R}=\frac{R_{0}+2R_{45}+R_{90}}{4},[/latex] (7.20)

[latex]\overline{R}=\frac{0.9+(2)(1.3)+1.9}{4}=1.35. [/latex]

The limiting drawing ratio (LDR) is defined as the maximum ratio of the diameter of the blank to the diameter of the punch that can be drawn without failure, that is, [latex]D_{o}/D_{p}.[/latex] From Fig. 7.56, the LDR for this steel can be estimated to be approximately 2.5.

To determine whether or not earing will occur in this operation, we substitute the [latex]R [/latex] values into Eq. (7.21).

[latex]\Delta R=\frac{R_{0}-2R_{45}+R_{90}}{4}[/latex] (7.21)

Thus,

[latex]\Delta R =\frac{0.9-(2)(1.3)+1.9}{2}=0.1.[/latex]

Ears will not form if [latex]\Delta R= 0 [/latex]. Since this is not the case here, ears will form in deep drawing this material.

Question 7.9: Estimating cup diameter and earing A steel sheet has R value...

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