# • Notched bar in tension

The plane strain condition is satisfied when breath, b » h, the depth of the bar.

Lower-Bound:

Find a stress system, e.g. σ = 0 in the length of the bar where there is the notch, σ = 2k elsewhere.

Therefore, for a breadth b, P = 2khb = load = stress × area

Upper-Bound:

Postulate a suitable simple deformation pattern.

Assume yielding by slip on 45º shear planes with shear yield stress k. Let displacement along shear plane AB = δx.

Then internal work done = \( {k.\left| {AB} \right|b\delta x} = k\sqrt 2 bh\delta x\), where the force is \( k\left| {AB} \right|b \) acting on the shear plane AB.

Distance moved by the external load \(P = \delta x\cos {45^ \circ }\) = \(\frac{{\delta x}}{{\sqrt 2 }}\)

\( \Rightarrow P \) \(\frac{{\delta x}}{{\sqrt 2 }}\) = \(k\sqrt 2 bh\delta x\)

\( \Rightarrow P = 2kbh\)

So, here we obtain the same result for the upper bound and lower bound \( \Rightarrow P = 2kbh\) is the true failure load, the load required to cause plastic flow.