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

# • Notched bar in plane bending

#### Lower-Bound:

The area immediately under the notch, above the neutral axis is in tension σ = 2k. The area below the neutral axis is in compression σ = 2k. where:

h = thickness of slab beneath the notch.

$$2k.$$ $$\frac{h}{2}$$$$.b$$ = magnitude of forces in tensile and compressive regions.
$$\frac{h}{2}$$ = distance between the two.

Equating the couples, $$M =$$ $$\left( {2k.\frac{h}{2}.b} \right)$$ $$\frac{h}{2}$$ = $$0.5k{h^2}b$$

#### Upper-Bound:

Assume failure occurs by sliding around a ‘plastic hinge’ along a circular arc of length l and radius r. If the rotation is δθ, the internal work done $$= k.lb.r\delta \theta$$ along one arc.

External work = Mδθ by one moment.

$\Rightarrow M = klbr$

where no assumptions have been made regarding l and r.

The upper bound theorem states that whatever values are taken for l and r will lead to an upper bound. Clearly we wish to find the lowest possible value. From the above geometry,

$$l = r\alpha$$    and     $$r =$$ $$\frac{h}{{2\sin {\raise0.7ex\hbox{\alpha } \!\mathord{\left/ {\vphantom {\alpha 2}}\right.\kern 2pt} \!\lower0.7ex\hbox{2}}}}$$

$\Rightarrow M = \frac{{k{h^2}b}}{4}\frac{\alpha }{{{{\sin }^2}{\raise0.7ex\hbox{\alpha } \!\mathord{\left/ {\vphantom {\alpha 2}}\right.\kern 1pt} \!\lower0.7ex\hbox{2}}}}$

and so to find the lowest possible value of M, we minimise the function $$\frac{\alpha }{{{{\sin }^2}{\raise0.7ex\hbox{\alpha } \!\mathord{\left/ {\vphantom {\alpha 2}}\right.\kern 1pt} \!\lower0.7ex\hbox{2}}}}$$

Let  $$Y =$$ $$\frac{\alpha }{{{{\sin }^2}{\raise0.7ex\hbox{\alpha } \!\mathord{\left/ {\vphantom {\alpha 2}}\right.\kern 1pt} \!\lower0.7ex\hbox{2}}}}$$

$$\frac{{{\rm{d}}Y}}{{{\rm{d}}\alpha }} = \frac{1}{{{{\sin }^4}{\alpha \mathord{\left/ {\vphantom {\alpha 2}} \right. \kern 1pt} 2}}}\left\{ {{{\sin }^2}\frac{\alpha }{2} - 2\frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \frac{\alpha }{2}} \right\}$$

= 0      when    $$\sin \frac{\alpha }{2} = \alpha \cos \frac{\alpha }{2}$$

$\Rightarrow \tan \frac{\alpha }{2} = \alpha$

$\Rightarrow M = \frac{{k{h^2}b}}{4}.\frac{1}{{\sin {\raise0.5ex\hbox{\alpha } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}\cos {\raise0.5ex\hbox{\alpha } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}}} = {\raise0.7ex\hbox{1} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern 1pt} \!\lower0.7ex\hbox{2}}\frac{{k{h^2}b}}{{\sin \alpha }} \cong 0.69k{h^2}b$

Taking the lower bound and the upper bound as limits, we therefore find

$\Rightarrow 0.5 \le \frac{M}{{k{h^2}b}} \le 0.69$

This forms a good example of constraining the value of the external force between lower bound and upper bound. It is also a good example of how to produce a lower limit on an upper bound calculation.