Beam deflection
El hazel Aymen
Lead Mechanical Engineer | API 510 Inspector&API 570 Inspector | Oil and Gas |Asset integrity specialist| welding inspection |QA/QC engineer
1.Scope
This?work?presents?how?mechanical?engineer?classifies?beams?and?justify?their?use.?
Beam?deflection?is?the?work?subject?of?this?paper
2.Introduction
A beam is a member subjected to loads the type of the principal load is what define each beams, in mechanical engineering:
In this paper I’m going to present the engineering features that define the high performance of I beam in deflection resistance. In the first section we write the deflection and bending moment (figure 1) on the basis of elastic theory. Secondly we present how does beam section geometry improve deflection resistance then how to achieve the same goal through material intrinsic proprieties.
3. Beam deflection equations
Figure 1:
3.1General equations
Based on the elastic beam theory, a beam deflection δ or bending angle θ:
3.2 Failure equations
The critical force ( moment) to failure is given by:
4. Beam section geometry (shape)
4.1" I" beam section
Here we want to emphasis analytically how does beam section geometry affect it’s deflection (bending) resistance and performance. we made the below hypothesis:
-Elastic limit σy considered constant
-t?thickness is constan
-ym distance from section central axis
Increasing the critical force to failure equivalent to
maximize the function g.
Increasing the critical force F(or critical moment M) is equivalent to maximize the function g(b,h). Without diving in mathematical analysis , g is polynomial monotonic function of h and b so increasing any of them should increase the critical load to failure, however since h is squared ,and for practical engineering purpose, ?increasing h is the optimum choice for improving the beam deflection resistance. It's seems obvious that to improve bending resistance increase the web dimension. nevertheless practical engineering aspect limit the h value.
Although there's a mathematical justification for increasing h and similar for the flange width b, the last one determined by other factors like:
-Beam with small?(b<<<) had no engineering uses
-The value of b determine bending resistance &buckling on the X axis?
"I" beam deflection resistance is proportional to the beam web dimension h squared
4.2Beam with different sections: comparison
In table 3 we writes?the specification of different beam section:?
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Consider the function g(h) that maximize the critical force to failure:
Nevertheless the limited use of those sections (apart from the "I" beam) as deflection resistant beam attributed to other factors :
- Weight for solid (plain) section the overweight undesirable
-Manufacturing cost: hollow section has higher manufacturing cost
Two more parameter help us to evaluate the performance different beam section loaded in deflection (see table 4)
- α characterize the section (relative to simple rectangular section) and therefore the beam weight
-β characterize the modulus of Flexural rigidity (EI) [E Young’s modulus is constant]
Evaluating the performances of different beam section (for deflection and bending): reveal why the "I" beam supersede all other beams in deflection resistance.
5.Material proprieties to improve deflection resistance
Other options to?improve?deflection resistance of beam is?by selecting another material with different proprieties (Young's modulus, elastic limit) :
Increasing the critical force or minimize deflection δ? , for invariable geometry section , equivalent to increase the Young's modulus E and\or the elastic limit.
-For the elastic limit there many engineering ways to increase it
-E the Young's modulus is intrinsic material proprieties, changing the value of E is equivalent to choose another material
5.1 Improving the elastic limit
Improving the mechanical proprieties?of alloy is done using
- Heat treatment (quenching tempering, annealing)
- Thermo-chemical treatment (alloying...)
or a?combination of the above treatment can rise the elastic limit up to up 100 ksi (700 Mpa) and the tensile strength up to 120 ksi (830 Mpa).The cost of those treatment should be justifiable by the desired performance. Here some examples.
The cost of the above treatment are high and justifiable form engineering application like: HY - 130 for nuclear submarine.
5.2 Higher Young's Modulus material
Table 6 present material with different Young's modulus and other specifications that determine their performance for mechanical engineering design?
I intentionally added the row of relative cost, which was by far the number one factor in selecting material for any applications.The most used metallic engineering material is steel (carbon steel) due to relatively lower cost and mass production the higher cost justifiable only for some application :
- Aircraft design manufacturing: Aluminum alloy \& titanium alloy
-Extreme corrosion : Platinum, iridium
It's clear that the margin of improving by choosing another material with higher Young's Modulus is limited.