## [help]bending of beam(urgent)

can anyone pls tell me how do i explain that why I beam is used for civil engineering?
--> how do i explain that a beam with rectangular cross section is easier to bend when weight is added on it when it is placed horizontally(bigger area facing up) than when it is placed vertically.

thanks
 Recognitions: Gold Member I'm no expert on this, so don't take this as your final answer. Basically, it comes down to the resistance of the sides to both compressive and stretching forces. For a beam to bend, the 'outer' section has to stretch to create the new radius, while the 'inner' one gets squished. An I-beam is ideally suited for that purpose.
 Recognitions: Gold Member Science Advisor This is homework, so it's in the wrong place. Firstly, why do you think an I beam is the shape it is?

## [help]bending of beam(urgent)

erm...
i thk i has the shape like that because the middle part (vertical beam between I) prevent the top and the bottom part from bending in order to strengthen the beam. is that correct?
but i need a more formal explanation which i cant figure out.
i am new to this forum, sorry that having this post on wrong board.
(this is the title of my project)

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 Quote by jiayen87 erm... i thk i has the shape like that because the middle part (vertical beam between I) prevent the top and the bottom part from bending in order to strengthen the beam. is that correct? but i need a more formal explanation which i cant figure out. i am new to this forum, sorry that having this post on wrong board. (this is the title of my project)
Generally, it's because they possess a greater axial moment of inertia, which is an important term in the stress expression. For example, if you look at the cross section of an I-beam, and place the coordinate axis y and z so that the origin is at the centroid, z is 'vertical', y 'horizontal', and x is pointing 'towards us', then the expression for stress along the x-direction is $$\sigma_{x}=\frac{N}{A} + \frac{M_{y}}{I_{y}}\cdot z$$. (*) So, it is obvious that for a bigger axial moment of inertia (which is a geometrical property of the cross section) you get a smaller value of stress, which is one of the important goals in civil engineering while dimensioning elements and cross sections.

(*) P.S. We assumed that there is a normal force N and a bending moment M acting on the beam, so that's what the stress expression looks like in that case.
 i cant figure out the directions. the cross section is a I shape right?y the stress will be pointing toward us? >.< hmm...sory i a bit slow learner. because i juz enroll into university and din learn these in my secondary sch. another thing, may i know wat is a bending moment?

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 Quote by jiayen87 i cant figure out the directions. the cross section is a I shape right?y the stress will be pointing toward us? >.< hmm...sory i a bit slow learner. because i juz enroll into university and din learn these in my secondary sch. another thing, may i know wat is a bending moment?
Yes, the cross section is an I shape, and the stress is pointing towards us (i.e. in the x-direction). A bending moment is just a simple moment, but I wanted to point out that it causes bending with respect to the y-axis. I have a lot of these diagrams in my books, but I don't have the time to draw a sketch, unfortunately. But I'm sure you can find something on the net. Anyway, the answer to your question is that the geometrical configuration of I-shaped cross sections is such that it provides great moments of inertia of the area (http://www.efunda.com/math/areas/MomentOfInertia.cfm), which is in direct connection with the magnitude of stress in the cross section, i.e. in the whole beam. That's why I-shaped beams are a good choice when the loads cause moments.
 for example from the efunda after getting Ix and Iy , the total moment of inertia is the sumation of the two I?if weight is added on top of the beam in this example, what will be the direction for the 'stress'? with given mass, if i want to maximize the moment of inertia mass of an I beam, is it true that the distance between 2 flanges( top and bottom part of I beam) have to be as far as possible? thanks...

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 for example from the efunda http://www.efunda.com/math/areas/ima...rtiaExpOne.gif after getting Ix and Iy , the total moment of inertia is the sumation of the two I?
No, that'll be the polar moment of inertia. You have to understand moment of inertia is about an axis.

 if weight is added on top of the beam in this example, what will be the direction for the 'stress'?
You mean the bending stress? Draw a moment diagram for this. Along the axis of the beam, there can be both negative moment and positive, which will determine either a clockwise or counterclockwise "direction" for the bending stress along the beam axis. Beams are designed with Max Moments (negatives and positives are taken in consideration).

 with given mass, if i want to maximize the moment of inertia mass of an I beam, is it true that the distance between 2 flanges( top and bottom part of I beam) have to be as far as possible? thanks...
Mass? If you want to reduce the bending stress, you'll need to increase the area moment of inertia of cross section of the beam with respect to the axis the moment is acting. For an I beam this could be achieved by increasing said distance, but you have to watch out for buckling of the web or crushing due to shear stress.