User:Eas4200c.f08.blue.a/Lecture 5


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 Problem 1.1 



The problem is to find the optimal $$b/a$$ in a thin walled beam to maximize the load-bearing capacity of a beam while assuming that the moment (M) is equal to the torsion (T).This is done by optimizing the cross section that carries the maximum bending moment and the maximum torsion.

Finding the Solution begins with the following assumptions: 1. The wall thickness is much less than the width and height of the beam (t<<a, t<<b)  2. The Moment on the beam is equal to the Torsion on the beam (M=T)  3. The allowable stress is equal to twice the allowable shear ($$\sigma _{allowable}= 2 \tau _{allowable}$$)  4. The beam is rectangular in shape, ie. the perimeter of the beam face (L) is equal to twice the sum of length a and b (L=2(a+b)).

Next, the shear stress ($$\tau$$) due to Torsion (T) is found. First, the total Torsion is found by separating the beam into four different elements, finding the Torsion in each, and adding them together.



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

Knowing that $$\tau =\frac{\nu }{t}$$, we are able to solve for Torsion in each element of the area of the beam.

$$T_{AB}=\frac{b}{2}\times (\nu \times a)=\frac{1}{2}\tau abt$$

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

$$T_{CD}=T_{BC}$$

$$T_{DA}=T_{CD}$$

$$T=2 \tau abt$$

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

We consider two cases for the solution.

Case1

In Case 1 we assume that the bending normal stress ($$\sigma $$) reaches the allowable stress ($$\sigma _{allowable}$$). The goal is to verify that the shear stress in the beam ($$\tau $$) is less than or equal to the allowable shear stress ($$\tau _{allowable}$$). The first step in this case is to define the bending moment as follows:

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

Where M is the bending moment, z is the ordinate of a point adn I is the second area moment of inertia defined as: $$I=\int \int_{A}^{}{z^{2}dydz}$$