3 moment equation example #1 (part 1/3)

[@triplep4567]

From a fluids guy, thank you so much! Very clear

[@Engineer4Free]

You’re welcome!! 🙂

[@marcapaulajos]

Is there a way to use this method if support "B" was a spring support rather than a rigid support?

[@Engineer4Free]

Yeah it's possible to introduce springs to complicate things. I don't have any statically indeterminate videos involving springs, but if you just google "3 moment equation spring" some examples come up!

[@marcapaulajos]

I see that you add 6EI * (h1/L1 + h2/L2). I'm not finding any examples though.

[@marcapaulajos]

I see that to the RHS of the equation you add 6EI*(h1/L1 + h2/L2) to account for deflection. I'm not seeing any examples though.

[@marcapaulajos]

Does it not just double the amount of unknowns using this method and become unsolvable? I've found some examples but if they are statically indeterminate with deflection there's always something they conveniently know before starting to solve the problem.

[@manishkumar-xq6xn]

you are getting negative value of vb1, does it not mean that we have to reverse the direction we have assumed?

[@r.a2229]

Please can you make a video that quickly (shortcut) solves the centroid of different shapes. Thank you

[@maliknirvana1890]

www.mathalino.com/reviewer/strength-materials/three-moment-equation here you will find different shapes with quick expressions

[@moocow7191]

can you please make a video solving for reactions and drawring shear and moment diagram?

[@Engineer4Free]

Yes see engineer4free.com/statics and check the chapters on 2D static equilibrium and also the chapter on SFD and BMD 👍

[@deepkallakuri7569]

Can i take Ma clockwise at 3:58?

[@Engineer4Free]

You could, but you would need to change its sign to negative. You'll notice that I drew it counterclockwise which opposite the positive sign convention, but wrote the magnitude as positive. Thats the same as drawing it to match the positive sign convention but indicating that the magnitude is negative.

[@creativemonk6721]

thank you bro. excellent vid

[@Engineer4Free]

You're welcome, thanks for watching! 😁

[@dhoult]

You should have wL^2/8 for the uniform load, not wL/8. That would make the units for "A2", kN-m^2, not kN-m.

[@Engineer4Free]

Hey Daniel, thanks for noticing and mentioning the typo, I am aware of this one, and have acknowledged it in the video description.

[@dhoult]

@@Engineer4Free Leave it to me to comment before reading.

[@nellvincervantes3223]

Hi. Why we do not have an area for overhang span? I know it will not be included in the formula but it is really dont have an area?

[@Engineer4Free]

In the 3 moment method, we draw those areas for spans as if they were simply supported. Basically, for an area to be modelled as simply supported, it needs a connection on each end. So the overhang is outside of the realm of the 3 moment equation. No big deal though, because we can very easily determine the internal shear and bending moment at all parts of the overhang using basic SFD/BMD statics.

[@nellvincervantes3223]

@@Engineer4Free thank you sir

[@maliknirvana1890]

Let me get this straight : the Ma and Mb are negative , but you took their values while summing moments according to the conventionnal drawing right ?!

[@Engineer4Free]

The internal bending moment at a is negative. The internal bending moment at b is also negative. Both are in accordance with the positive sign convention for beam bending. Ma was found using a virtual cut on a free body diagram of the overhanging part, which gives us enough information in this case to solve it. Mb requires the use of the three moment equation in this case.

[@shilobajohnpishikeni8424]

hello, please why is Ma not zero since it is a pin/roller support?

[@Engineer4Free]

The roller just means that the support reaction is not capable of applying a moment to the beam. It doesn't mean that the member won't have any internal bending moment at that location due to other things acting on the beam.

[@r.a2229]

Overhang will produce a moment at A

[@shilobajohnpishikeni8424]

@@r.a2229 yeah. Realized that after typing the question. Thanks dear. I never solved a problem with overhang that's why.

[@Engineer4Free]

Thanks for helping out Rita! And yeah Shiloba, check out videos 7 and 8 here: engineer4free.com/structural-analysis for some simple examples with overhang. It will help to get some practice!

[@choysakanto6792]

If a free end downwards force occurs on the extreme right, would the substitute moment be positive as well just as the extreme left free end downwards force had a negative moment?

[@Engineer4Free]

In a case like that, your draw a FBD with a virtual cut to the left of the free (right) end. The shear on this FBD would be pointing up on the left side where the cut is. The force couple that you get would induce a clockwise rotation, so the moment at the cut (left side) needs to fight that, so it is counterclockwise. A counter clockwise moment on the left of a cut is a negative moment according to the sign convention that we are using. In both cases, (overhanging left or overhanging right) the point force alone would cause a downward deflection at the free end that results in a concave down curvature. Concave down curvature is indicative of a negative moment in that region. That's a nice quick check to do too make sure you are using the right signs, but be careful because it's only considering the curvature as if that's the only load affecting that region, which is often not true and likely wont reflect the actual curvature of the fully loaded structure.

[@mohamedthasneem7327]

thank you very much

[@Engineer4Free]

you're welcome Mohamed :)

[@azizuke]

how did you calculat the area of the deformation plzz !!

[@Engineer4Free]

Hey I'm not sure what you mean by area of deformation. Can you be more specific or point to the time in the video that you're referring to?

[@azizuke]

Oh thanks for reponding bro, well her an example of a unifome distributed load, and then you calculated A2 = 2/3 bh, lets say instad of that we had a triangular load; i apreaciat your help bro

[@Engineer4Free]

Oh I see. A UDL that covers a whole span creates a parabolic BMD, and the area of a parabola is always (2/3)bh, a nice easy thing to remember for basic problems that show up often, and it's useful to use as an example because the calculation of area and centroid for a problem like this doesn't take so long that it detracts from the actual method that is trying to be conveyed. I don't think I have any videos on triangular distributed loads, so I can't remember off the top of my head if there is a quick formula to memorize in the case of a triangular load that starts with a magnitude of zero and increases constantly across a whole span. Triangular distributed loads are inherently more complicated, because it might not start at zero, and/or if it doesn't span the whole span, it's going to be a lot more complicated. If someone else reading this knows the short answer then please share!

[@安然-x9t]

how is Vc equal to 41.5625??

[@Engineer4Free]

The image on the right is the free body diagram of span BC if were virtually sectioned it infinitesimally close to the ends of the span. We take the sum of moments about some point (I pick point B) and set it to zero because the structure is in static equilibrium. The sum of moments about B is the sum of moments about B so we need to write an expression with all of the moments about B that is equal to zero. We pick a sense to be the positive one and sum them all to zero, or put all of the CCW on one side of the equation and all of the CW ones on the other side (its just a rearrangement of putting them all on one side with appropriate signs and setting to zero, which is how I organized the equation in this video). The only CCW moment about B on span BC is the bending moment Mb = 84.375 kNm. The other moments on the span are all CW about point B. One is caused by the distributed load, and to find it we take the resultant force (10kN/m)*(10m)=100kN times the perpendicular distance from the line of action of the force to the point we are taking moment about (5m - it's in the middle of the uniform distributed load) and that's where the (10)(10)(5) term comes from on the right hand side. The next moment is caused by Vc which is 10m away from B acting at 90 degrees, so the moment it causes about B is the magnitude (Vc) times the distance (10m). That's where the Vc(10) term comes from in the right hand side. Mc, the bending moment at point C is also labelled on the diagram,, so I include it in the expression, but we have already determined it to be 0 kNm, so that's where the 0 comes from in the right hand side. 84.375kNm = (10kN/m)(10m)*(5m) + Vc(10m) + 0 is one equation with one unknown. We just rearrange to find Vc. It's worth mentioning that the units check out and Vc will indeed be in kN. Hope that clears it up for you!

[@generalezaknenou]

ありがとうございます!

[@SoundlyBytes]

good but your words aint so clear when explaining the quick calculations

[@Engineer4Free]

Check out this video: www.engineer4free.com/4/3-moment-equation-explained where I introduce and discuss the concept, rather than work through it

[@jimmybob767]

Hello, I am trying to solve a question were there is a UDL across the whole beam and a point load acting directly above the middle support. What would the bending moment diagram look like for this and what equations would i use to get force and area of each of the shapes. any help is greatly appreciated thank you.

[@maliknirvana1890]

www.mathalino.com/reviewer/strength-materials/three-moment-equation here you will find different shapes with quick expressions