User:Eas4200c.f08.blue.a/Lecture 13


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! Homework Solution: Triangle Area Proof

Problem: prove that $$A_{BEC}=\frac{1}{2}(ED)(BC)$$

Solution: The area of the triangle BED is calculated from the formula for the area of a right triangle: $$A=\frac{1}{2}bh$$ b= lenght of horizontal base h= length of vertical height Thus, the area of triangle BED is

$$A_{BED}=\frac{1}{2}(ED)(BD)$$

and the area of the interior right triangle CED is

$$A_{CED}=\frac{1}{2}(ED)(CD)$$

Therefore, the area of triangle BEC can be found by subtracting these two areas:

$$A_{BEC}=\frac{1}{2}(ED)(BD)-frac{1}{2}(ED)(CD)=\frac{1}{2}(ED)(BC)$$
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For the open thin walled cross section:



This means that $$T=2q\bar{A}$$ and that the reationJ=\frac{1}{2}\pi a^{4}

Uniform Bar with Circular Cross Section
Nonwarping case: Cross Section behaves as rigid disks



For this case, Torque is calculated thus:

$$T=\int_{A}^{}{\int_{}^{}{}r\tau dA}$$  with   $$\tau =G\gamma $$

and define $$\gamma =\frac{rd\alpha }{dx}$$

Therefore, $$T=\int_{A}^{}{rG(r\theta )dA}$$ where G and $$\theta$$ are independent of variables y and z

This means that the above formula can be written as: $$T=G\theta (\int_{A}^{}{r^{2}dA})$$. The integral enclosed in brackets is known as the polar moment of inertia, J. if the radius of a circular cross section is defined as a, then J can written as:

$$J=\frac{1}{2}\pi a^{4}$$

Now consider the case of a hollow circular cross section with thin walls (t<<a)

Let: $$r_{inner}=a$$ and $$r_{outer}=b$$

This yields a polar moment of inertia of:

$$J=\frac{1}{2}\pi (b^{4}-a^{4})$$

and with some manipulation:

$$J=\frac{1}{2}\pi(b-a)(b+a) (b^{2}-a^{2})$$

Where (b-a) can be approximated as t and (b+a) is approximately $$2r\bar{}$$.

in addition we assume:

$$b^{2}\cong r\bar{}^{2}$$ $$a^{2}\cong r\bar{}^{2}$$

Substituting yields the eqution:

$$J=2\pi tr\bar{}^{3}$$

rearranging gives the equation:

$$J=2\pi ^{-\frac{1}{2}}t(\pi r\bar{}^{2})^{\frac{3}{2}}$$

where $$\pi r\bar{}^{2}=A\bar{}$$. Therefore, J is proportional to $$A\bar{}^{\frac{3}{2}}$$ with $$(2\pi ^{-\frac{1}{2}}t)$$ as the proportionality factor.