# Forge-flattening

Is there an easy way to estimate how much energy is needed to forge a cube of iron into a thin square sheet with a hammer? I suppose i should integrate the force $$F = \text{yield_strength} \times \text{area(h)}$$ through the height, $h_0 = \text{height of the initial cube}$, $h_1$ -- resulting sheet's thickness. Am I right? Also, where to look for compressive yield strength of iron at high temperatures? I've read its better to forge it at $1000-1100 °\text{C}$

• I'm sure the vast majority of the energy is dissipated as vibration. Jul 20, 2018 at 20:47
• I doubt the energy is ever calculated in the real world. The shop determines from experience how much force is needed for the material, area , temperature, the extent of upset and the desired number of blows. Jul 21, 2018 at 16:27
• blacksmith37 from your experience, how much time do you think would it take to flattern a 2 kg billet into a 1.5 mm sheet of low carbon steel? is there maybe a book or article on productivity of hand forging? Jul 23, 2018 at 8:40

Your suggestion is reasonable as an estimate for pressing the metal; hammering would require significantly more energy (see below). The volume $V=Ah$ of metal is constant so as its thickness $h$ decreases its area $A$ increases. A graph of the temperature dependence of Yield Strength for several metals is given in Engineering ToolBox.

At room temperature the Yield Strength is about $p=50MPa$. The work required to compress a cube of side $h_0$ to a plate of thickness $h_1$ is $ph_0^3 \ln{\frac{h_0}{h_1}}$. For $h_0=100mm$ and $h_1=1mm$ this work is approx. $230kJ$.

An upper bound is the energy required to heat the metal to its melting point then change it from solid to liquid. Gravity would then re-shape the metal without further input of energy. For $h_1^3=0.001m^3$ of iron ($\approx 0.8kg$) this is about $375kJ$ when starting from $1100^{\circ}C$ and $760kJ$ when starting from room temperature $25^{\circ}C$. This is the same order of magnitude as the Yield Strength calculation.

For a lower bound estimate you could calculate the energy of the bonds which need to be broken in order to increase the surface area of the metal between its initial and final shapes. Bond energy for iron is about $120 kJ/mol$.

Hammering also dissipates energy through heating with a much smaller amount lost as sound. Whether this non-useful energy is significant depends on whether the hammer blows exceed the yield strength of $50MPa$ and by how much. The proportion of energy dissipated as sound increases as the surface area of the metal plate increases but remains small. Vibration energy is eventually lost mostly through heating of dense metal rather than radiation of sound in air.

For an interesting discussion of hammering see Can hammer blows increase workpiece temperature? which states that $90-95 \text{%}$ of the impact energy goes into raising temperature. It concludes that manual hammering does not cause any significant rise in temperature because the metal anvil quickly conducts away the heat generated. But power hammering can cause a significant increase in appropriate circumstances.

• 230 kJ is quite small, about 100 strikes of a sledgehammer or 2 minutes of work. Is it really that easy? Or maybe the most of the applied energy is lost? (goes to the temperature, 90-95%?) Jul 23, 2018 at 9:43
• 230kJ assumes pressing rather than hammering. As the article explains, with a metal anvil most of the heat generated by hammering would be conducted away, and as the piece gets thinner and the contact area increases the rate of heat flow increases. If the workpiece were insulated about 375/0.9=420kJ would be required. I think your figure of 100 strikes assumes a hammer of about 25kg which is a power hammer - wikipedia article quotes 9kg as the practical limit for manual work, and even at that weight I doubt any blacksmith would sustain a rate of about 1 strike/second for 100 seconds. Jul 23, 2018 at 10:40
• i have corrected for high temperature forging (divided by 10 = 23 kJ), and afaik the striker does not only rise a hammer above his head to create potential energy but also pulls it down when it falls, adding to its speed. a man can provide about 400 watts of work for a short time (about a hour), e.g. every second rising a 10 kg hammer to the height of 2 meters (200J) and pulling it down with additional force of 100N, for the total of ~200N -> 400J blow.. Jul 23, 2018 at 11:25
• OK. At 900C the yield strength is approx 10% that at room temp. so only 23kJ required. If only 10% of the impact energy is doing useful work when hammering (90% goes into heating, which is conducted away) then 230kJ input is still required, which is 575 blows at 400J per blow. Jul 23, 2018 at 12:03
• Kevin Peffers has written other articles, eg Blacksmithing : It's all about energy and power and Heat Energy : on the move and fast. Jul 23, 2018 at 12:12