Find Optimal Beam Dimensions for Supported Footbridge | Material Selection

AI Thread Summary
The discussion focuses on determining optimal beam dimensions for a simply supported footbridge spanning 7 meters, subjected to a distributed load of 5 kN/m plus the beam's weight. The maximum bending moment is calculated as a function of distance along the beam, with the maximum occurring at the midpoint. To ensure safety, the maximum stress must not exceed 150 MPa, factoring in a safety margin of 60% of the yield strength. Participants highlight the importance of correctly calculating the moment caused by the beam's weight, clarifying that it should be treated as uniformly distributed for accurate stress analysis. Overall, the approach is validated, but attention to detail in calculations is emphasized to avoid errors.
danago
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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:

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

Where W is the wight of the beam.

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

<br /> M_{\max } = 30.625 + 1.75W<br /> \]<br /> <br /> 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:<br /> <br /> &lt;br /&gt; \sigma _{\max } = \frac{{M_{\max } y_{\max } }}{I} = \frac{{M_{\max } }}{z}&lt;br /&gt;<br /> <br /> 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. <br /> <br /> &lt;br /&gt; \sigma _{\max } = 150MPa = 150 \times 10^6 Pa&lt;br /&gt;<br /> <br /> With these restrictions in place, i should find an elastic section modulus such that:<br /> <br /> &lt;br /&gt; z \ge \frac{{M_{\max } }}{{\sigma _{\max } }} = \frac{{30625 + 1750W}}{{150 \times 10^6 }}&lt;br /&gt;<br /> <br /> Now my table gives values for z in the units mm^3, so i converted it to mm^3 by multiplying by 1000^3.<br /> <br /> &lt;br /&gt; z \ge \frac{{612500 + 35000W}}{3}&lt;br /&gt;<br /> <br /> 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. <br /> <br /> Does anyone have any input? I am not even sure if I&#039;ve approached the question in the correct way. I think what&#039;s causing me problems is the fact that the weight isn&#039;t neglected like most other problems i need to solve.
 
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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.
 
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 ?
 
danago said:
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 ?
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).
 
Ahh ok that helps :) Thanks very much.
 
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