User:EAS4200c.f08.blue.c

User page for Tom Sexton, member of Team Blue.

Because the shear stress is greatest at the top and bottom of the beam section (where z = b/2), the equation of shear stress can be rewritten as follows: $$\sigma = \frac{Mz}{I}  \Rightarrow   M=\frac{2I\sigma\ _{allowable}}{b} $$

Thus, it is necessary to find the value of I in terms of b (recalling that a can be written as a function of b: a=L/2-b). From the problem parameters we know that for this instance, Mmax = Tmax. Because Mmax has been determined, it must only be put into the relation:

$$\tau _{max}=\frac{T^{(1)}_{max}}{2a^{(1)}b^{(1)}t}=\frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t}$$

Now check the values of $$\tau _{max}$$ with the value of $$\tau _{allowed}$$.If $$\tau _{max}$$ is larger than $$\tau _{allowable}$$ the optimal ratio of $$\frac{b^{(1)}}{a^{(1)}}$$would be acceptable. But $$\tau _{max}$$ is smaller than $$\tau _{allowable}$$, then the optimal ratio of $$\frac{b^{(1)}}{a^{(1)}}$$ would be unacceptable

Case 2
For this case we assume that the shear stress is eqaul to the maximum allowable shear stress in the section. Then we will test to see if the normal stress falls within acceptable tolerances (notably, less than $$\sigma _{max}$$)

By the equation given earlier in the problem the value of T becomes:

$$T=(2t\tau _{allowable})(ab)$$

Where the dimensions of the rectangle ab are equal to one another, hence a square. Therefore if the total perimeter of the square is L and all sides are equal, both a and b are equal to L/4. Plugging this quantity into the previous equation and using the given relation that M=T, we have:

$$M^{(2)}_{max}=\frac{1}{8}tL^{2}\tau _{allowable}$$

Using this relation and the previously defined equation for normal stress:

$$\sigma ^{(2)}_{max}=\frac{M^{(2)}_{max}b^{(2)}}{2I^{(2)}}$$

We can find $$\sigma ^{(2)}_{max}$$ and compare it with the allowable normal stress. If the maximum normal stress is less than the allowable shear stress, then the ratio of b to a for case 2 is acceptable. But, if it is greater, then that ratio is not correct.

Solving Case 1
First the moment of inertia, I, must be computed. This is done by summing the moments of inertia of each of the four parts of the section. In order to do this, the Parallel axis theorem must be applied. Solving for the horizontal axis yields:

$$I=\frac{tb^{2}}{6}(3a+b)$$

And with the relation a=L/2-b, the plot of f(b)=I/b is shown to the right.

Taking the derivative of the function f(b) yields the maximum value of the curve to be b(1)=3L/8. Thus the ratio of $$\frac{b^{(1)}}{a^{(1)}}$$ is equal to 3 and the value of a(1) is found to be L/8. Solving for $$\tau _{max}$$ with the values of a(1) and b(1) yield:

$$\tau _{max}=\frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t}=\frac{32M^{(1)}_{max}}{3tL^{2}}$$

but because this larger than half of $$\tau _{allowable}$$ this case is not possible

HW2 Contribution


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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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