# Material Selection

1. Mar 30, 2008

### danago

A simply supported footbridge is to span a total of 7 meters. The load on the beam is a distributed force of 5kN/m plus its weight. Taking into account the beams weight, what dimensions should be selected for the beam?

Ive been given a table of beam dimensions along with their values for elastic modulus, second moment, measurements of mass per length etc.

I started by coming up with a the bending moment as a function of x, the distance from the end of the beam:

$$M(x) = \left\{ {\begin{array}{*{20}c} {(17.5 + 0.5W)x - 2.5x^2 } & {x \in [0,3.5)} \\ {3.5W + (17.5 - 0.5W)x - 2.5x^2 } & {x \in [3.5,7]} \\ \end{array}} \right.$$

Where W is the wight of the beam.

I then found that the maximum bending moment occurs at x=3.5, so:

$$M_{\max } = 30.625 + 1.75W \] Given this, i want to choose a beam with dimensions such that the maximum stress in the beam does not exceed the yield strength, where the maximum stress is given by: [tex] \sigma _{\max } = \frac{{M_{\max } y_{\max } }}{I} = \frac{{M_{\max } }}{z}$$

Where z is the elastic section modulus. In this case, i will assume the maximum yield stress to be 60% of the actual yield stress, as in my mechanics class we are generally told to use a safety factor of 0.6.

$$\sigma _{\max } = 150MPa = 150 \times 10^6 Pa$$

With these restrictions in place, i should find an elastic section modulus such that:

$$z \ge \frac{{M_{\max } }}{{\sigma _{\max } }} = \frac{{30625 + 1750W}}{{150 \times 10^6 }}$$

Now my table gives values for z in the units mm^3, so i converted it to mm^3 by multiplying by 1000^3.

$$z \ge \frac{{612500 + 35000W}}{3}$$

Now the table gives different values for the beams mass per unit length (kg/m), so i can calculate the weight of different beams and then put it into the equation above, but everytime i do so, i get answers that seem way too high.

Does anyone have any input? Im not even sure if ive approached the question in the correct way. I think whats causing me problems is the fact that the weight isnt neglected like most other problems i need to solve.

Last edited: Mar 30, 2008
2. Mar 30, 2008

### josefs

I haven'y checked your numbers, but your calc for the moment caused by the weight of the beam is wrong. The max moment caused by the beams weight is not WL/4 as you have calculated, but rather, WL/8 (the beam's weight may be considered concentrated at mid point when determining end reactions, but when considering internal stresses and moments, you must distribute it over the length of the beam. ). Otherwise, your approach looks OK, check your numbers, the SI metric system in engineering leaves me cold , since I use only standard USA units.

3. Mar 30, 2008

### danago

So when working with the weight, should i treat the bar as being uniform, so that the weight per unit length is given by W/7 N/m ?

4. Mar 30, 2008

### josefs

Yes, where W is the total weight of the beam, in Newtons (If w is the weight per unit length of the beam, W=wL, and M_max, from the beam weight alone, is WL /8 or wL^2/8).

5. Mar 30, 2008

### danago

Ahh ok that helps :) Thanks very much.