Given. AISI 1015 strip, $\sigma_y=245$ MPa, UTS $=345$ MPa, sheet thickness $t_0=1$ mm; cup punch diameter $D_p=10$ cm; length/width strain data at 0°/45°/90° to the rolling direction as tabulated above (gauge-width change reported as a reduction).
Find. (a) $\bar{R}$ and $\Delta R$; (b) the largest usable blank diameter $D_{0,max}$; (c) the maximum drawing force $F_{max}$; (d) the angular location of any ears.
Approach. Convert each specimen's length/width engineering strains to true strains, obtain the through-thickness strain from volume constancy, form each directional $R=\varepsilon_w/\varepsilon_t$, then combine into $\bar{R}$ and $\Delta R$; use $\bar{R}$ in the standard limiting-draw-ratio (LDR) correlation to size the blank, apply the formula-sheet drawing-force equation, and read the earing direction off the sign of $\Delta R$.
Convert tabulated strains to true strain and find $R$ per direction. With $e_l$ the elongation and $e_w$ the (reduction) width-strain magnitude, $\varepsilon_l=\ln(1+e_l)$, $\varepsilon_w=\ln(1-e_w)$, and by constant volume $\varepsilon_l+\varepsilon_w+\varepsilon_t=0\Rightarrow \varepsilon_t=-(\varepsilon_l+\varepsilon_w)$, so $R=\varepsilon_w/\varepsilon_t$.
For $0^\circ$: $\varepsilon_l=\ln(1.189)=0.173$, $\varepsilon_w=\ln(0.904)=-0.101$, $\varepsilon_t=-0.072$, $R_0=0.101/0.072=1.40$.
For $45^\circ$: $\varepsilon_l=\ln(1.147)=0.137$, $\varepsilon_w=\ln(0.928)=-0.075$, $\varepsilon_t=-0.062$, $R_{45}=0.075/0.062=1.20$.
For $90^\circ$: $\varepsilon_l=\ln(1.162)=0.150$, $\varepsilon_w=\ln(0.911)=-0.093$, $\varepsilon_t=-0.057$, $R_{90}=0.093/0.057=1.64$.
(a) Combine into $\bar{R}$ and $\Delta R$.
$$\bar{R}=\frac{R_0+2R_{45}+R_{90}}{4}=\frac{1.40+2(1.20)+1.64}{4}\qquad \Delta R=\frac{R_0-2R_{45}+R_{90}}{2}=\frac{1.40-2(1.20)+1.64}{2}$$
$$\boxed{\bar{R}\approx 1.36\qquad \Delta R\approx 0.32}$$
$\bar{R}>1$ indicates the sheet resists through-thickness thinning more than in-plane width contraction (favourable for deep drawing); $\Delta R>0$ indicates planar anisotropy.
(b) Largest blank diameter, $D_p=10$ cm. The limiting draw ratio $LDR=D_{0,max}/D_p$ increases with $\bar{R}$; the standard Kalpakjian correlation runs from $LDR\approx 2.0$ at $\bar{R}=1$ to $LDR\approx 2.8$ at $\bar{R}=3$, i.e. $LDR\approx 1.7+0.3\bar{R}$.
$$LDR=1.7+0.3(1.36)=2.11$$
$$D_{0,max}=LDR\times D_p=2.11\times 10\ \text{cm}$$
$$\boxed{D_{0,max}\approx 21.1\ \text{cm}}$$
(c) Maximum drawing force, $t_0=1$ mm. Using the formula sheet's drawing-force relation with $D_o=D_{0,max}$:
$$F_{max}=\pi D_p t_0(UTS)\left(\frac{D_o}{D_p}-0.7\right)=\pi(100\ \text{mm})(1\ \text{mm})(345\ \text{MPa})\left(\frac{211}{100}-0.7\right)$$
$$\boxed{F_{max}\approx 152.5\ \text{kN}}$$
(d) Earing location. Earing amplitude and phase are set by $\Delta R$: $\Delta R>0$ produces four ears aligned with the rolling direction and its transverse, while $\Delta R<0$ produces them at $\pm45^\circ$. Since $\Delta R\approx 0.32>0$,
$$\boxed{\text{ears form at }0^\circ\text{ and }90^\circ\text{ to the rolling direction (four ears, at }0/90/180/270^\circ\text{)}}$$
consistent with $R_{90}>R_0>R_{45}$: the wall resists thinning most at $90^\circ$/least at $45^\circ$, so metal draws up highest at $0^\circ$/$90^\circ$.
Part
Result
(a) $R_0,\ R_{45},\ R_{90}$
1.40, 1.20, 1.64
(a) $\bar{R}$, $\Delta R$
1.36, 0.32
(b) $D_{0,max}$
≈21.1 cm
(c) $F_{max}$
≈152.5 kN
(d) Earing location
0° and 90° to rolling direction
Check: the exam supplies no explicit LDR-vs-$\bar{R}$ formula or chart on its own formula sheet, so part (b) uses the standard graphical correlation reported in Kalpakjian & Schmid (Manufacturing Engineering and Technology) — LDR rising roughly linearly from about 2.0 at $\bar{R}=1$ to about 2.8 at $\bar{R}=3$ — approximated here as $LDR=1.7+0.3\bar{R}$. A different, equally defensible correlation/graph reading would shift $D_{0,max}$ (and hence $F_{max}$ in part c, which depends on it) proportionally; the method and $\bar{R}$ value are the graded content.
