Question

During open die forging process using two flat and parallel dies, a solid disc of initial radius $\left(R_{In}\right)$ 200mm and initial height $\left(H_{In}\right)$ 50 mm attains a height ($H_{FN}$) of 30 mm and radius of $R_{FN}$. Along the die-disc interfaces.

(i) The coefficient of friction $\left(\mu \right)$ is :

$\mu =0.35[1+e^{-R_{IN/R_{FN}}}]$

(ii) In the region $R_{SS}\le r\le R_{FN}$

Sliding friction prevalis and

$P=\sqrt{3}K.e^{2\mu \left(R_{F{N}^{-r}}\right)/H_{FN}}$ and $\tau =\mu p$, sliding friction prevalis and

Where P and $\tau$ are the normal and the shear stresses respectively; K is the shear yield srength of steel and r is the radial distance of any point.

(iii) In the region $0\le r\le R_{SS}$ sticking condition prevalis.

The value of $R_{SS}\left(inmm\right)$, where sticking condition changes to sliding friction is

• A. 241.76
• B. 254.55
• C. 265.45
• D. 278.20

Answer 1

The correct option is B 254.55

At $R_{S{S}^{\prime }}$

$\frac{\pi }{4}{d}_0^2\times h_0=\pi R_{FN}^2\times h_1$

$R_{FN}=258m$

Shear stress sticking (k)

= Shear stress in sliding $\left(\mu P_{ss}\right)$

$\therefore RSS=R_{FN}-\frac{h}{2\mu }In\left(\frac{1}{\sqrt{3}\mu }\right)$

$=258.2-\frac{30}{2\times 0.51}In\left(\frac{1}{\sqrt{3}\times 0.51}\right)$

= 254.55 mm