Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

# • Hodographs I

A hodograph is a diagram showing the relative velocities of the various parts of a deformation process.

To analyse a complicated deformation process with many shear planes, it is worth looking at the basic equation for the rate of energy dissipation in an upper bound situation in more detail.

ABCD is distorted into A'B'C'D' by shear along $$\overrightarrow {SS}$$ at a velocity $$\underline {{v_s}}$$ in the metal.

Suppose ABCD moves towards the shear plane SS' at a velocity $${v_1}$$ and suppose that there is a pressure P acting on the area al (where l is the dimension out of the plane of the paper) helping to cause this movement.

Rate of performance of work externally $$= pal\left| {\underline {{v_1}} } \right|$$

Rate of performance of work internally $$= k\left| {SS} \right|l\left| {\underline {{v_s}} } \right|$$

since the only internal work assumed to occur is that required to effect the shear deformation so that ABCD → A'B'C'D'.

Equating these, $$\Rightarrow pa\left| {\underline {{v_1}} } \right| = k\left| {SS} \right|\left| {\underline {{v_s}} } \right|$$

$$\Rightarrow pa = k\left| {SS} \right|\frac{{\left| {\underline {{v_s}} } \right|}}{{\left| {\underline {{v_1}} } \right|}}$$

Simple vector algebra relates v1, v2 and vs as on the diagram below:

If in a deformation process there are n such shear planes of the type SS',

then $$pa = k\sum\limits_n {\left| {SS_n^`} \right|\left| {{v_{sn}}} \right|}$$ setting $$\left| {\underline {{v_1}} } \right| = 1.0$$, i.e. unit velocity.

### Rules for constructing a hodograph

The animation below illustrates the seven rules for constructing a hodograph, for the case of a constrained punch.

An analysis of the geometry of the hodograph enables an upper bound for the applied force to be calculated.

Let Oq be a velocity vector of unit magnitude in the hodograph, i.e. νOq = 1
Due to the dead metal zone, Q and Q' move at the same velocity.
O is a stationary component of the system, anywhere in the surrounding perfectly rigid metal which has not yielded at all.
Oq and Oq' are in essence vectors defining the motion of particles in region Q'.
Or is velocity of a particle in region R.
q
'r is a vector defining the shear velocity parallel to Q'R.
Os
is velocity of a particle in region S.
rs is a vector defining the shear velocity parallel to RS.

Hence,

${v_{Or}} = \frac{1}{{\tan \theta }}{\rm{ ,\; }}{v_{q'r}} = \frac{1}{{\sin \theta }}{\rm{\; and \; }}{v_{Os}} = {v_{rs}} = \frac{1}{{2\sin \theta }}$

Using the rate of performance of work formula we have:

$p\left( {\frac{b}{2}} \right) = k\left\{ {Q'R{v_{q'r}} + OR{v_{Or}} + RS{v_{rs}} + OS{v_{Os}}} \right\}$

where Q'R is the length of the line dividing regions Q' and R, OR is the length of the line dividing regions O and R, RS is the length of the line dividing regions R and S and OS is the length of the line dividing regions O and S

$p\left( {\frac{b}{2}} \right) = kb\left\{ {\frac{1}{{2\cos \theta }}.\frac{1}{{\sin \theta }} + 1.\frac{1}{{\tan \theta }} + \frac{1}{{2\cos \theta }}.\frac{1}{{2\sin \theta }} + \frac{1}{{2\cos \theta }}.\frac{1}{{2\sin \theta }}} \right\}$

$= kb\left\{ {\frac{1}{{\sin \theta \cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right\} = kb\left\{ {\frac{{1 + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}} \right\}$

$\Rightarrow \frac{p}{{2k}} = \frac{{1 + {{\cos }^2}\theta }}{{\sin \theta \cos \theta }} = f\left( \theta \right)$

$$\frac{{df}}{{d\theta }}$$ = 0 and f is then a minimum when

$\cos 2\theta = - \frac{1}{3}{\rm{ }} \Rightarrow {\rm{ }}\theta = {54.74^ \circ }{\rm{ when }}\sin \theta = \frac{{\sqrt 2 }}{{\sqrt 3 }}{\rm{ and }}\cos \theta = \frac{1}{{\sqrt 3 }}$

minimum $$\frac{p}{{2k}}$$$$= 2\sqrt 2 = 2.83$$ from this upper bound analysis.

When indenting using a sliding (frictionless) punch, we can postulate a different deformation pattern without the dead metal zone. The system also has a plane of symmetry and a hodograph can be constructed as follows:

As before, Oq = 1.0

Material in R travels in direction shown with velocity $$Or =$$ $$\frac{1}{{\sin \theta }}$$

$$\left| {Or} \right| = \left| {rs} \right| =$$ $$\frac{1}{{\sin \theta }}$$$$= \left| {st} \right| = \left| {Ot} \right|$$

$$\left| {Os} \right| =$$ $$\frac{{2\cos \theta }}{{\sin \theta }}$$

$$\left| {qr} \right| =$$ $$\frac{1}{{\tan \theta }}$$ = $$\frac{{\cos \theta }}{{\sin \theta }}$$

Lengths in drawing of indent: $$QR = SO =$$ $$\frac{b}{2}$$

$$OR = RS = ST = OT =$$ $$\frac{b}{{4\cos \theta }}$$

We therefore have:

$$\frac{{pb}}{2}$$ = $$k\left\{ {OR{v_{Or}} + RS{v_{rs}} + OS{v_{Os}} + ST{v_{st}} + TO{v_{tO}}} \right\}$$

$$= kb$$ $$\left\{ {\frac{1}{{4\cos \theta }}.\frac{1}{{\sin \theta }} + \frac{1}{{4\cos \theta }}.\frac{1}{{\sin \theta }} + \frac{1}{2}.\frac{{2\cos \theta }}{{\sin \theta }} + \frac{1}{{4\cos \theta }}.\frac{1}{{\sin \theta }} + \frac{1}{{4\cos \theta }}.\frac{1}{{\sin \theta }}} \right\}$$

$$= kb$$ $$\left\{ {\frac{1}{{\sin \theta \cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right\}$$

$$\Rightarrow$$ $$\frac{P}{{2k}}$$ = $$\left\{ {\frac{{1 + {{\cos }^2}\theta }}{{\sin \theta \cos \theta }}} \right\}$$ as before for the case of the constrained punch.

This analysis has assumed that no friction occurs at the punch face to cause particles in R to move parallel to OR. If there is friction, we can take it to be sticking friction, so that there is a shear stress k acting and slippage velocity $$= {v_{qr}}$$

$$\Rightarrow$$ in this case $$\frac{P}{{2k}}$$ = $$\left\{ {\frac{{2 + 3{{\cos }^2}\theta }}{{2\sin \theta \cos \theta }}} \right\}$$

$$\Rightarrow$$ of the three possible upper bound solutions, the 'best' answer is $$\frac{P}{{2k}}$$ = $$2\sqrt 2 = 2.83$$

This is a 10% overestimate of the true value of $$\frac{P}{{2k}}$$ found from slip-line field theory.