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Most of the real life structures are statically indeterminate.

What is the benefit of designing a statically indeterminate structure instead of a statically determinate one?

I don't mean only beams and trusses.Per example, the shaft of a merchant vessel is considered as a typical statically indeterminate structure.

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  • $\begingroup$ Possible duplicate of Why the design of continuous beam is always more economical than beam with supports? $\endgroup$ – Wasabi Nov 26 '17 at 19:11
  • $\begingroup$ @Wasabi Sorry I don't know much about statics, I can't understand why this question is a duplicate of the nice link you give $\endgroup$ – veronika Nov 26 '17 at 20:33
  • $\begingroup$ You'll see the question states that continuous beams are always more economical than isostatic (statically determinate) ones, and the answers mostly agree with that statement and explain why that happens. That's the advantage of statically indeterminate structures: they're cheaper (or, put another way, stronger for the "same price"). $\endgroup$ – Wasabi Nov 27 '17 at 1:34
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    $\begingroup$ If stress is the limit you go with statically determinant, and if deflection/stiffness is the limit then you go with indeterminant. $\endgroup$ – ja72 Nov 27 '17 at 3:07
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    $\begingroup$ @ja72 You could write a nice answer $\endgroup$ – veronika Nov 27 '17 at 5:43
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For example, take a look at the following static systems. Assume they have the same length and the same (constant) cross-section. Thus an equal allowed bending moment $M_u$.

The first system is statically determinate, as it is supported only by simple supports. The maximum moment developing within the beam is $M=\frac{QL}{4}$, thus the load under which the beam fails is $$Q_u=4\frac{M_u}{L}$$ This coincides with the formation of a plastic hinge at point B, which leaves the systems statically under-determinate, a mechanism.

In the second system, the beam is supported by two clamped supports, which both reduce the maximum moment at the point where the load is applied. If you determine the static forces within the beam you will find that, neglecting residual stresses, the maximum moment $M=\frac{QL}{8}$. Thus, $$ Q_u=8\frac{M_u}{L} $$ To turn into a mechanism, three plastic hinges have to be formed, which requires more work.

Hence the second system theoretically can bear twice the load of the first system.

Statically indeterminate systems in general are more stable and more rigid, but harder to calculate. Another important factor are the above-mentioned residual stresses, resulting amongst others from thermal expansion or the manufacturing process (e.g. inhomogeneous cooling of steel).

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