User:Eas4200c.f08.radsam/Example: Problem 1.1

=Statement:= A beam (Figure 1) is designed to carry both a moment $$M\!$$ and torque $$T\!$$. If the perimeter of the cross section is $$L=2(a+b)\!$$ and is held constant, find the optimal $$b/a\!$$ ratio if $$M=T\!$$ and $$\sigma_{allowable}=2\tau_{allowable}\!$$. Also note: $$\tau=\frac{T}{2abt}\!$$.

=Proof:=  Error 1: Your diagram on shear flow is incomplete. You should also redraw the figure instead of scanning the hand-drawn figure and upload the scan. Eml4500.f08 19:13, 21 September 2008 (UTC)

Error corrected. Eas4200c.f08.radsam.a. 15:59, 23 September 2008 (UTC)

The figure looks better, but still incomplete; review your classnotes and check with your teammates. Eml4500.f08 13:15, 24 September 2008 (UTC)

Corrected again: if this is wrong I'll eat my Wikiversity page. ;) Eas4200c.f08.radsam.a. 00:38, 4 October 2008 (UTC)


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!Problem Explanation and Proof of Shear Stress The closed-walled beam under scrutiny is one of negligible thickness, thus one can assume that the shear stress distribution is of a uniform nature...

Therefore adding respectively the torque on each side of the box...

$$T= T_{AB}+T_{BC}+T_{CD}+T_{DA}$$

...where torque equals the lever arm times the shear stress by the wall thickness... $$T_{AB}=\frac{b}{2}(va) = \frac{1}{2}\tau(abt)$$

$$T_{BC}=\frac{a}{2}(vb) = \frac{1}{2}\tau(abt)$$

$$T_{CD}=\frac{b}{2}(va) = \frac{1}{2}\tau(abt)$$

$$T_{DA}=\frac{a}{2}(vb) = \frac{1}{2}\tau(abt)$$

and recalling...

$$T= T_{AB}+T_{BC}+T_{CD}+T_{DA}$$

$$T=2\tau(abt)$$

or

$$\tau=\frac{T}{2abt}$$

Contribution by radsam.z
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! Allowable versus Yield Stress

As explained in Section 1.2 in "Aircraft Structures and Materials," yield stress is the stress point where a material begins plastic deformation. The stress/strain relation prior to this yield point is relatively linear for most materials; this means that the strain is proportional to the stress and that the material will go back to its original state after the applied stress is released. However, if the applied stress is greater than the yield stress, the material will deform permanently. The allowable stress refers to a point which is lower than the yielding limit of the material. This stress is calculated to be less than the yield stress for safety reasons and due to uncertainties in the expected properties of the material. When designing a structure, engineers let stresses reach the allowable limit because the structure would still withstand the applied forces, but it is crucial for stresses not to reach the yield stress since this will cause permanent deformation, and ultimately, failure of the structure.

Contribution by radsam.d
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=Solution:= Due in Homework #2