Why depth of beam is more than its width?

  • 1 Answer(s)

    Take a simple ruler. Try to bend it about the axis where the depth is smaller compared to the width like in the image below.

    You’ll notice that it’s very easily bendable. Now, continue bending it by applying more load until it breaks.

    *Snap* Wow! You are very powerful. Everybody, we have a Thor here!

    Ok, now let’s try and bend it about the axis where the depth is greater. Make sure you bend the ruler only about this axis, lest you introduce bi-axial moments which we do not want right now. Also, be careful not to twist the ruler which induces torsion, we don’t want that too. We want to keep things simple here.

    Ok, let the bending begin…

    Image credit: www.open.edu/

    What’s that? You can’t? What happened?… Come on! I thought you were Thor.

    Fine. Let’s see what happened. There are two parameters which we are concerned here- Deflection and Moment Resisting Capacity. Let’s draw the two sections in question. I’ll assume the material to be made of mild steel for the sake of simplicity. Let the dimension of the ruler be 300mmx30mmx3mm.

    DEFLECTION

    The following chart gives the value of deflection of beams for various types and loading conditions.

    Image credit: engmechanics.blogspot.in/

    I know the chart seems a bit intimidating for a beginner. Not to worry, we are not going into the detail of how the formulas are arrived at.  Just have a look at the formulas. As you can see, all the formulas for deflection has EI in its denominator, meaning, the deflection is inversely proportional to the product of Young’s Modulus (E) and Moment of Inertia (I). So, as the EI doubles, the deflection becomes half. If EI gets halved, the deflection doubles. (The product EI is known as flexural rigidity.)

    A quick note on E and I. The Young’s modulus E is a material property. It is nothing but the stress required to be applied to a member to produce a unit strain in it. Higher the value of E, lesser the deformation/deflection. E.g. The value of E for steel is 210,000 MPa. So, theoretically it means that if you take a steel rod of 1m length and pull it with a load that induces a stress of 210,000 MPa then the rod would deform to a length of 2m. i.e. length of rod gets doubled. (Of course, practically speaking the rod would fail at around 250 MPa (yield strength of steel), well before you can achieve such a deformation.) Contrast this with the value of E for rubber which is around 50 MPa, 4000 times lesser than steel. Aluminium has an E value of 70,000 MPa i.e., 1/3rd that of steel. Now, if we take a steel beam and an aluminium beam of same size, load it with same magnitude of loads, if the steel beam has a deflection of ‘x’ mm, then aluminium beam would have ‘3x’ mm deflection. Essentially, E gives us an idea of how strongly a material resists deformation/deflection.

    Coming to I. Moment of inertia ( also called second moment of area) is purely a geometric property. It has got nothing to do with the type of material. It denotes how far the area is distributed from the axis under consideration. When the word moment of inertia is mentioned, it is always accompanied by an axis. When you see a value of I, you must always know to what axis it is concerned. Steel, aluminium, copper, wood, fiber glass, etc, all can have the same value of I if the sections are similar. For the same material, value of I changes as the shape and axis are changed. I basically tells us how much resistance the section offer against bending. Higher the I (more further the material are spread from the axis), higher the resistance against bending. Now combine this with E, you’ll see the significance of EI.

    Now, coming back to our case of ruler. Why was the ruler easily bendable in the first case but not in the second case? Let’s look at the deflection formula for this loading condition.

    Image credit: Textbook of Mechanics of Materials by James M. Gere.

    The loading condition for our experiment can be depicted as a simply supported beam with moments applied at both ends. The maximum deflection for such an arrangement will happen at the center of the ruler. The deflection can be calculated from the formula: ML^2/(8EI), where M is the moment applied, L is the length of ruler (=300mm). Let’s apply same load for both the cases and see what is the deflection. Now, coming to E in the equation, steel being isotropic, it has the same material properties in all the directions. So, E is constant for both the cases. Now we are left with I. Let’s calculate I for both the cases. For a rectangular section, I is given by bd³/12.

    Case 1:

    I = (30)*(3)^3/12 = 30*27/12 = 67.5 mm^4

    Case 2:

    I = (3)*(30)^3/12 = 3*27000/12 = 6750 mm^4

    Moment of inertia in the case 2 is 100 times greater compared to case 1.

    Therefore, as all the other variables in the deflection equation are equal, the deflection of the ruler in case 1 is 100 times greater than the deflection of ruler in case 2.

    MOMENT RESISTING CAPACITY

    *Sigh* This answer is getting longer than I expected. All because of the fake Thor here. Had he been able to break the freakin’ ruler, I wouldn’t be going through all this trouble. Anyway, read along.

    Now, let’s calculate the maximum moment the ruler can carry in both the cases. It can be done using flexure formula:

    Image credit: Textbook of Mechanics of Materials by James M. Gere.

    M = f*I/y

    where, M is the moment, y is the distance of plane under consideration from the neutral axis, f is the stress developed along this plane.

    The maximum bending moment induced in a beam before failure happens when the stress ‘f’ in the extreme end exceeds the strength of the material. For a mild steel, the strength is 250 MPa. The maximum stress is achieved in the plane farthest from the neutral axis, so, ‘y’ is equal to (depth/2). We already know ‘I’ from the previous calculations. Substituting all the these in the above equation we can calculate the maximum moment the ruler can carry in each case,

    Case 1:

    M = 250*67.5/1.5 = 11250 Nm or 11.25 kNm.

    Case 2:

    M = 250*6750/15 = 112500 Nm or 112.5 kNm.

    The ruler is able to carry 10 times more moments in case 2 than in case 1.

    Now coming back to your question. What was that again? Ah! Why is the depth of beam greater than its width?

    The thing is, beams are mainly subject to bending moments about the lateral axis. We want beams to have less deflections and carry more loads. That’s the whole idea, right? You don’t want beams in your house to sag down and creep you out. Also, it would be nice to have beams not collapse on you while you are asleep.

    As demonstrated above, beams with higher moment of inertia will have lesser deflection and carry more loads. Furthermore, we know that the value of I for a rectangular section is given by bd³/12. It is evident that I increases linearly with width but increases by three orders of magnitude with depth. Hence, obviously, it is more beneficial to have more depth than width.

    But, why don’t we make the widths even slimmer? That’s because there may be moments about the perpendicular axis too which can’t be ignored. A thin beam would fail due to buckling. Also, in an RCC beam, you need to be able to place the steel rebars as far away from the neutral axis as possible to have high resisting moments. Here if you don’t have a wide enough beam, there won’t be enough space available to place all the required rebars effectively.

    Image credit: constructionduniya.blogspot.in/

    I hope the explanation was clear.

    Now, how is our Thor doing after failing miserably at a simple task?

    Hey, don’t be mad at me. I am not responsible for your failure, I is.

    Answered by Prathyaksh Shetty

    Novice Answered on January 7, 2017.
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