User:Eas4200c.f08.spars.lee/Homework 2

=Problem 1.1=

"1.1 The beam of a rectangular thin-walled section (i.e. t is very small) is designed to carry both bending moment M and torque T. If the total wall contour length L = 2(a + b) (see Fig. 1.16) is fixed, find the optimum b/a ratio to achieve the most efficient section if M = T and $\sigma$allowable = 2$\tau$allowable. (Sun p. 17)"

Assumptions
There are four assumptions made in the problem statement.

Assumption 1:   $$\mathbf{L} = 2\mathbf{(a + b)} $$

Assumption 2:      $$\mathbf{M} = \mathbf{T} $$

Assumption 3:     $$\ \sigma_{allowable} = 2\tau_{allowable} $$

Assumption 4:     $$\ \mathbf{t<<a},\mathbf{t<<b}$$

Shear stress

 * The first step is to determine the relationship between the shear stress, $$\tau$$, and the torque, T. In actuality the shear stress takes a parabolic shape as shown in figure, but in this case since the thickness is very thin (assumption 4) the shear force can be approximated by a uniformly distributed shear flow q.


 * q = $$\tau$$t (3.47)


 * Where the shear flow is related to torque by,


 * dT = ρq ds


 * where ρ is equal to the lever arm and ds is the differential length of the very thin bar. Integrating, this equation becomes


 * T = ρqS


 * The torque in this problem can be broken up into the sum of the torques in each side, T = TAB + TBC + TCD + TDA. By symmetry TAB = TCD and TBC = TDA which simplifies the equation.


 * TAB = $$\frac{b}{2} (q\cdot a) = \frac{\tau abt}{2} $$


 * TAB = $$\frac{a}{2} (q\cdot b) = \frac{\tau abt}{2} $$


 * Therefore


 * T = $$ \mathbf{2\tau a b t} $$


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


 * This is identical to to equation provided in the problem statement with no explaination.

Solution

 * In order to solve this problem it is necessary to consider two cases.

Case One

 * Assume the bending normal stress, $$ \sigma $$, reaches $$ \sigma_{allowable} $$ before shear stress, $$ \tau $$, reaches $$ \tau_{allowable} $$.
 * Recalling that the bending normal stress is related to the bending moment by ordinate of a point on an axis perpendicular to the bending neutral axis over the second moment of inertia according to the equation


 * $$ \sigma = \frac{Mz}{I} $$


 * From which the following relationship is derived


 * $$ \sigma_{allowable} = \frac{M_{max}b}{2I} $$ → $$ M_{max} = \frac{2I\sigma_{allowable}}{b} = 2\sigma_{allowable} (\frac{I}{b})_{max} $$


 * The second moment of inertia is represented by


 * $$ I = \sum_{i=1}^{4}{[\frac{b_i(h_i)^3}{12} + A_id_i^2]} = 2[\frac{tb^3}{12} + \frac{at^3}{12} + \frac{atb^2}{4}] = 2[\frac{tb^3}{12} + \frac{at}{12}(t^2 + 3b^2)]$$


 * Assuming that t is very small compared to b this equation can be approximated by


 * $$ I = 2[\frac{tb^3}{12} + \frac{3atb^2}{12}] = \frac{tb^2}{6}[b + 3a] $$


 * Recalling assumption 1 ($$ L = 2(a + b) $$) the variable a can be stated as


 * $$ a = \frac{L}{2} - b $$


 * Which can then be used to express the second moment of inertia in terms of a single unknown variable b


 * $$ \frac{tb^2}{6}[3(\frac{L}{2} - b) + b] = \frac{tb^2}{6}[\frac{3L}{2} - 2b] = \frac{tb^2(3L - 4b)}{12} = \frac{1}{12}[-4tb^3 + 3Ltb^2]$$


 * Returning to the equation for Mmax it is evident that $$ 2\sigma_{allowable}$$ is a constant and therefore Mmax is maximized when $$ (\frac{I}{b}) $$ is maximized.


 * $$ f(b) := \frac{I}{b} = \frac{1}{12}[-4tb^2 + 3Ltb]$$
 * $$ f'(b) = -\frac{8tb}{12} + \frac{3Lt}{12} $$
 * $$ f''(b) = -\frac{8t}{12} $$


 * The negative second derivative indicates the function $$ f(b) $$ represents a concave parabola which reaches a maximum when it's slope, or first derivative, is equal to zero.


 * $$ f'(b^{(1)}) = 0 = -\frac{8tb^{(1)}}{12} + \frac{3Lt}{12}$$ → $$\frac{8tb^{(1)}}{12} = \frac{3Lt}{12} $$ → 8b(1) = 3L  → $$ b^{(1)} = \frac{3L}{8} $$


 * $$ b^{(1)} = \frac{3Lt}{12}$$ → $$ a^{(1)} = \frac{L}{2} - b^{(1)} = \frac{L}{2} - {3L}{8} = {L}{8} $$


 * $$ f(b^{(1)}) = f(\frac{3L}{8}) = \frac{1}{12}[-4t(\frac{3L}{8})^2 + 3Lt(\frac{3L}{8})] = \frac{1}{12} [\frac{9tL^2}{8} - \frac{36tL^2}{64}] = \frac{1}{12}[\frac{36tL^2}{64}] = \frac{3tL^2}{64}$$


 * Therefore


 * $$ M_{max} = 2\sigma_{allowable} \frac{3tL^2}{64}$$


 * Recalling assumption 2 ($$ M = T $$) and the equation relating the shear force to the torque ($$ \tau = \frac{T}{2abt} $$)


 * $$ \tau^{(1)}_{max} = \frac{T^{(1)}_{max}}{2a^{(1)}b^{(1)}t} = \frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t} = \frac{2\sigma_{allowable}{\frac{3tL^2}{64}}} {2\frac{L}{8}\frac{3L}{8}t} = \sigma_{allowable}$$


 * Recall from assumption 3 ($$ \sigma_{allowable} = 2\tau_{allowable}) $$


 * $$ \tau^{(1)}_{max} = \sigma_{allowable} = 2\tau_{allowable} $$


 * $$ \tau^{(1)}_{max} > \tau_{allowable} $$


 * The maximum shear stress exceeds the allowable shear stress and therefore this solution is unacceptable.

Case Two

 * Assume the shear stress, $$ \tau $$, reaches $$ \tau_{allowable} $$ before the bending normal stress, $$ \sigma $$, reaches $$ \sigma_{allowable} $$.


 * Recall the equation determined earlier relating shear stress to torque ($$ \tau = \frac{T}{2abt} $$)


 * $$ \tau^{(2)}_{max} = \frac{T}{2abt} $$ → T(2)max = $$ 2\tau_{allowable} a^{(2)} b^{(2)} t $$


 * It's clear that maximizing the torque is equivalent to maximizing the product a(2) b(2) which recalling assumption 1 ($$ L = 2(a + b) $$) that product becomes


 * $$ \frac{L}{2}b^{(2)} + (b^{(2)})^2 = maximum$$


 * Taking the derivative


 * $$ \frac{L}{2} - 2b^{(2)} = 0 $$ → $$ 2b^{(2)} = \frac{L}{2} $$ → $$ b^{(2)} = \frac{L}{4} $$


 * $$ a^{(2)} = \frac{L}{2} - b^{(2)} = \frac{L}{2} - \frac{L}{4} = \frac{L}{4} $$


 * Therefore the torque is maximized when a(2) = b(2) = $$ \frac{L}{4}$$, therefore


 * T(2)max = $$ 2\tau_{allowable} (\frac{L}{4})^2 t = \frac{1}{8}tL^2\tau_{allowable}$$


 * Recalling assumption 2 ($$\mathbf{M} = \mathbf{T} $$)


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


 * Recalling assumption 3 ($$\ \sigma_{allowable} = 2\tau_{allowable} $$)


 * $$M^{(2)}_{max} = \frac{1}{16}tL^2\sigma_{allowable}$$ → $$\sigma_{allowable} = \frac{16M^{(2)}_{max}}{tL^2}$$


 * Recall from case one


 * $$ f(b) := \frac{I}{b} = \frac{1}{12}[-4tb^2 + 3Ltb]$$ → $$ f(b^{(2)}) = \frac{1}{12}[\frac{tL}{4}[3L - L] = \frac{tL^2}{24} $$


 * Recall also from case one


 * $$ \sigma_{max} = \frac{M_{max}b}{2I} $$ → $$ \sigma^{(2)}_{max} = \frac{1}{16}tL^2\sigma_{allowable}(\frac{12}{tL^2}) = \frac{12}{16}\sigma_{allowable} $$


 * $$\sigma_{max} < \sigma_{allowable}$$


 * The maximum normal stress does not exceed the allowable normal stress and therefore this solution is acceptable. Furthermore, it proves the optimal cross section is a square such that


 * $$ \frac{b}{a} = 1 $$