Forming Process

Original length of the billet (L_0)=100mm
Original diameter of billet (d_0)=100mm
Mean flow stress of billet material (\sigma _m)=50MPa
Die Shape factor K_S=1.05
Original Cross Sectional Area of Billet
A_0=\frac{\pi}{4}d_0^2=\frac{\pi}{4}100^2=7853.981
Cross sectional area of extruded product
A_f=(10 \times 50)+ (10 \times 50)=1000 mm^2
Extrusion ratio r=\frac{A_0}{A_f}=\frac{7853.9816}{1000}=7.854
Extrusion pressure will be maximum at the start of extrusion process where L_0=100mm
\begin{aligned} P_{max}&= K_S\sigma _m\left [ 0.8+1.5 \ln (r)+\frac{2L_0}{d_0} \right ]\\ &=1.05 \times 50 \left [ 0.8+1.5 \ln (7.854)+\frac{2 \times 100}{100} \right ]\\ &=309.305 MPa \end{aligned}
Maximum Extrusion force
\begin{aligned} F_{max}&= P_{max} \times A_0\\ &=309.305 \times \frac{\pi}{4}(100)^2\\ &=2429265.97N\\ &=2429.3kN \end{aligned}

Initial thickness h_1=4mm
Width w= 100 mm
Roll Diameter D= 200 mm
Roll Radius R= 100 mm
Coefficient of Friction \mu =0.1
Flow stress is given by \sigma =207+414\varepsilon


We have equation for maximum possible reduction is
\begin{aligned} \Delta h&=\mu ^2R\\ \Delta h&=(0.1)^2 \times 100=1mm=h_1-h_2\\ h_2&=h_1-\Delta h=4-1=3mm \end{aligned}
Now True strain \varepsilon = \ln\left ( \frac{h_1}{h_2} \right )= \ln \left ( \frac{4}{3} \right )=0.2876
Stress when no rolling \varepsilon =0, \sigma _1=207 MPa
Stress when material is rolled to maximum possible extent \varepsilon =0.2876, \sigma _2=207+414(0.2876)=326.06 MPa
Since Strain hardening characteristics is linear the average flow stress \bar{\sigma }=\frac{\sigma _1+\sigma _2}{2}=\frac{207+326.06}{2}=266.53MPa
Projected length of contact L=\sqrt{R\cdot \Delta h}=\sqrt{100 \times 1}=10mm
Rolls separation force,
\begin{aligned} F&=1.15\bar{\sigma }\left ( 1+\frac{\mu L}{2\bar{h}} \right )wL\\ &=1.15 \times 266.53\left ( 1+\frac{0.1 \times 10}{2 \times \frac{7}{2}} \right ) \times 100 \times 10\\ F&=350.3 N\\ F&=0.350kN \end{aligned}

\begin{array}{l} h_{1}=50 \mathrm{~mm} ; \quad h_{2}=0.9 h_{1} ; \\ b_{1}=250 \mathrm{~mm} ; \quad b_{2}=1.03 b_{1} ; \\ L_{1}=3600 \mathrm{~mm} ; \quad L_{2}=? \end{array}
Volume remains const. in theory of plasticity
\begin{aligned} h_{1} b_{1} L_{1} &=h_{2} b_{2} L_{2} \\ &=0.9 h_{1} \times 1.03 b_{1} \times L_{2} \\ L_{2} &=\frac{3600}{0.9 \times 1.03}=3883.5 \mathrm{~mm}=3883.5 \mathrm{~mm} \end{aligned}

Particle Size, Shape, and Distribution:
Particle size is generally controlled by screening, that is, by passing the metal powder through screens (sieves) of various mesh sizes. Several other methods also are available for particle-size analysis:
1. Sedimentation, which involves measuring the rate at which particles settle in a fluid.
2. Microscopic analysis, which may include the use of transmission and scanning- electron microscopy.
3. Light scattering from a laser that illuminates a sample, consisting of particles suspended in a liquid medium; the particles cause the light to be scattered, and a detector then digitizes the signals and computes the particle-size distribution.
4. Optical methods, such as particles blocking a beam of light, in which the particle is sensed by a photocell.
5. Suspending particles in a liquid and detecting particle size and distribution by electrical sensors.

Given: Width of plate, w = 200 mm
Thickness of plate, t_0 = 20 mm
Radius of roller, R = 300 mm
Reduced thickness of plate in one pass,
t_f = 18 mm
Rollers speed, N = 50 rpm
Strength coefficient, (K) = 300 MPa
Strain hardening exponent, n = 0.2
Coefficient of friction, \mu = 0.1
Find: Roll force (F)
\Delta h=h_0-h_f=20-18=2mm
Contact length, L=\sqrt{R\Delta h}=\sqrt{300 \times 2 }=24.49mm
True strain, \varepsilon _T=\ln \left ( \frac{t_0}{t_f} \right )=\ln\left ( \frac{20}{18} \right )=0.1054
Average Flow stress, \bar{\sigma _f}=\frac{K(\varepsilon )^n}{1+n}=\frac{300 \times 0.1054^{0.2}}{1.2}=159.4MPa
Rolling force, F=\sigma _f(Lw)=159.4 \times 24.49 \times 200=780.776kN