User:Eas4200c.f08.carbon.orear/HW1.1

HW PROBLEM 1.1
BELOW IS THE SOLUTION FROM CLASS.

Consider a square tube (seen to the right)




 * wall thickness << height (a)
 * wall thichness << width (b)

ASSUMPTION 1: $$\displaystyle L = 2 * (a + b)$$

Q: Find optimal b/a ratio to maximize load bearing capability of beam assuming:

ASSUMPTION 2: $$\displaystyle M = T$$ ASSUMPTION 3: $$\displaystyle \sigma$$ allowable $$\displaystyle = 2*\tau$$ allowable

In other words, find optimal cross section carrying max bending moment and max torque:

$$\displaystyle \tau = T /(2 * a * b * t)$$

Since cross section walls are very thin, we can assume shear stress distribution to be uniform along the wall. See the wikipedia article on shear flow to learn more. Shear Flow



Proof:

$$\displaystyle T = T$$ab $$\displaystyle + T$$bc $$\displaystyle + T$$cd $$\displaystyle + T$$da $$\displaystyle$$

$$\displaystyle T$$ab $$\displaystyle= b/2 * (\tau t * a ) = \tau *a*b*t/2$$

$$\displaystyle T$$bc $$\displaystyle= a/2 * (\tau * t * b) = \tau *a*b*t/2$$

$$\displaystyle T$$cd $$\displaystyle= b/2 * (\tau * t * a) = \tau *a*b*t/2$$

$$\displaystyle T$$da $$\displaystyle= a/2 * (\tau * t * b) = \tau *a*b*t/2$$

Recalling that $$\displaystyle T = T$$ab $$\displaystyle + T$$bc $$\displaystyle + T$$cd $$\displaystyle + T$$da $$\displaystyle$$, this proves that $$\displaystyle T = 2 \tau a b t$$ and  $$\displaystyle  \tau = T/(2 a b t)$$.

$$\displaystyle $$

Case:1
Assume stress (bending normal stress) reaches stress allowable, first. What to verify that $$\displaystyle \tau << \tau$$allowable

Recall: $$\displaystyle Stress = \sigma = (M *z)/I$$

M = bending moment, Z = ordinate of a point on axis perpendicular to neutral bending axis

I = 2nd area moment of inertia

Assume $$\sigma_{max} = \sigma_{allowable}$$, that the maximum bending normal stress increases to the allowable stress first. The goal is to find the maximum allowable bending moment which corresponds to the max bending stress at z=b/2.

Recall: $$\sigma = \frac {Mz} I, ~ Z = \frac b 2$$

=> $$M = \frac {2 I \sigma_{max}} b \frac {2 I \sigma_{allowable}} b$$ because of the case 1 assumption

Put constants together and make it a function of (a,b)

To maximize the bending moment, simply maximize the ratio I/b. First, find the moment of inertia with respect to 'b',

Recall from assumption 1:

=> $$a = \frac L 2 - b$$

=> (I/b) fuction of b

Maximize M : Max = max, from assumption 2 (T=M)

According to assumption 3, the allowable shear stress is equal to half of the allowable normal stress. Therefore, if the max shear stress is less than the allowable shear stress, then the optimal ratio is acceptable. If not, then it is not acceptable.

Assume that the maximum bending normal stress reaches the allowable stress first. Then the moment of inertia of the cross section is the sum of the moments of inertia of the cross section is the sum of the moments of inertia of the four segments of the cross section. If you use the parallel axis theorem, it is as follows:





α$$=(a/12)[t^3+t(12/4)b^2] = (at/12)[t^2+3b^2]=(3ab^2t)/12$$

if t^3 << t, hence the approximation

$$ = \frac {tb(3L - 4b)} {12} = \beta_0 + \beta_1 b + \beta_2 b^2 $$

$$ \beta_0 = 0, \quad \beta_1 = \frac {3Lt} {12}, \quad \beta_2 = -\frac 4 {12} t = -\frac t 3 $$

$$ \frac {d^2 f} {db^2} (b) = 2\beta_2 = -\frac {2t} 3 < 0 $$

$$ f(b) = 0 => b ~ \begin{cases} 0 \\ \underline{3L} \\ 4 \end{cases} $$

Plot f(b) we get inverted parabola, 0 at 0 and 3L/4. max at 3L/8









We can see that this case can NOT exist! We have to violate a known equation to use this cross-section!

Below is a Mohr's Circle representation of Case 1:



Case 2:
Assume torsional shear stress reaches the allowable shear stress first. Torsional shear stress is constant along the thin wall because the cross section is negligibly thin.

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

$$\tau_{max} = \tau_{allowable} = \tau = \frac T {2~a~b~t}$$

$$T=(2t$$τallowable)$$(ab)$$ → $$T_{max}=(2t$$τallowable$$)(ab_{max})$$

Maximize T by maximizing the product of a and b. For a constant perimeter L, the cross section of the box beam is square.

The product of a and b is the area of the cross section and a=b, therefore L/4 = a = b because of four sides of the frame.

We need to check to make sure that the maximum bending stress for the cross section is below the allowable bending stress. If this is true, then the optimal ratio b/a = 1 is acceptable, otherwise it is not acceptable.

$$\frac {a^{(2)}} {b^{(2)}} = 1$$

$$T^{(2)}_{max} = (2 + \tau_{all}) * (\frac L 4) = \frac 1 8 + L_{tall} = M^{(2)}_{max}$$



$$M_{max} = 2 \sigma_{all} (\frac I b) = \frac {3 + L^2} {32} \sigma_{all}$$

$$\tau_{max} = \frac {M_{max}} {2abt} = \frac {M_{max}} {2 (\frac L 8) (3 \frac L 8) t} = \frac {32M_{max}} {3tL^2} = \sigma_{all}$$

σmax(2) = Mmax(2)b(2)/[2I(2)] = Mmax(2)12/[tL2], where Mmax(2)=tL2/16

→σmax(2) = (12/16)σallow < σallow → ACCEPTABLE