man keep up what you do you have almost all the statics stuff i need on youtube thanks.......

Your videos make me like (and understand) statics. Thank you! :)

God bless you.... I'm glad I found your channel, this is the best and keep it up!

Thank you so much for your consistent nice work!!!

Thank you tonnes for existing !

hey do you have a video of a cart on a incline . that you find the forces of the wheels and the reaction forces supporting the beam that is on an incline. thank you

you are wonderful! Thank you so much!!

Hi, I have a software program which I use for this work. The simple beam end reactions are equal for any incline, but when a gantry is considered, the lower support takes much more load than the top. Even to the point at which the top support, when inclided at 36 deg. the reaction becomes negative. I find this very confusing. Can you throw light on this please?

I have since my last post done another test where the top support is only constrined vertically and laterally. The horizontal support is considerd sliding. This sorts out the problem. I can also try a support which has some friction there too as for real life.

Which software/tablet did you use to write this example? The result is very good.

How would you approach this one if it were pinned on each side, i.e. stable and statically indeterminate?

Hello, I didn't fully understand this problem. Is it possible is you can do another one that is similar. I like your style of teaching.

More statics videos are on my channel!

Sir one thing which i didn't understand is that when u were to take moment about point B u multiplied the horizontal distance that is Ay , u did multply by the cosine of the distance . My understanding is that when we are taking a moment we should see the direct distance between the support . I didn't. get the multiplication of the distance of 10 feet by cosine of 30

How to calculate if are both pin support

structurefree How come you used the horizontal distance between A and B in writing the moment equation for the reaction at A and you didn't use the horizontal distance for the other two forces (you used 20/3' and 5')?

@structurefree

10 жыл бұрын

since the arm is the perpendicular distance with respect to the line of action of the force....or shortest distance to the point where you are taking moments about.

@jacinyan3893

8 жыл бұрын

It is exactly where I am confused about at first

@troyajohnson26

7 жыл бұрын

Its the perpendicular distance. The two applied forces are perpendicular to the beam angle, so just take the distance along the beam, 20/3' and 5' from point B.

From where i can get that book u solved question

I’m not geeting the sign convention, I mean how do u put (+) and (-) signs ???

can you upload arches & cables plz?

Here's an example problem on calculating reactions for an angled beam.

@ajmarks100

10 жыл бұрын

hello sir.. well i have ran into another problem again. anyway, here it is.. "A reinforced concrete beam of rectangular section is 25 cm wide and 50 cm deep, steel reinforcement of A=11 cm^2 is palced at 5 cm above the tension face, the maximum compressive stress in concrete is 4.2 N/mm^2, modular ratio m=15, calculate the moment of resistance M and the stress in the steel.[[ANS. 38kNm,89N/mm^2]]" according to strength of materials by GH Ryder. NB: i did the problem and got 45 and 95 respectively. i would be greatful if u could do the problem and see if u get 38 and 89 like the book says and mayb left a brief explanation. BTW i have to submit the solution in 5 days time. (march 25)

Can you please make a Shear/Moment diagram video session for this very problem? That would be awesome!! StructureFree *chika-chika*

I was introduced to the channel recently through my search for refresher materials on structural analysis and am in love with the videos. I have 2 issues and require your assistance for this video, 1. The example are not in SI. As i and many more have studied in the UK, Europe or India, It would be great if the video you have posted in non-SI units to be provided in SI units. 2. The sin and cos conventions for resolving the angled forces are always tricky and confusing for me. Can you help me understand it better? in both 2D and 3D Your assistance will be much appreciated.

@ourhub9388

5 жыл бұрын

To better anderstand the sin and cos for angled forces you just have to draw the Vertical and Horizontal Axis at 4:32 and divide the force into its own Vertical and Horizontal components. Or you just have to look for Vectors and trigonometry. I hope it was helpfull, good luck

How do you draw their shear force and bending moment diagrams? Would it be the same as if it was just a regular beam?

@structurefree

Жыл бұрын

Yes

thanku so much

How do you know when to take cos and sin. I thought cos was always the angle from the horizontal to the x-axis.

@muhammedkoya258

7 жыл бұрын

cos is related to the side touching the angle, sin is for the side away (opposite) the angle

how did you come up with the by= 6.73.. i keep on calculating but i didnt get that ans. thanks

@kevinmasicampo1366

10 жыл бұрын

pls reply asap..

Here,we are calculating support reactions which are in y direction,so we should consider y component of load acting on beam.But why u considered total load on beam?reply structurefree

sir plz explain how you resolved the forces

@structurefree

8 жыл бұрын

+akshay pande For the distributed loading, I just broke it up into the area of a triangle and rectangle.

Hi there, do you mind to do an example for vertical loading on the beam with two pin supports? Is there a way to prove that the horizontal reactions are zero?

@structurefree

10 жыл бұрын

For this statically determinate, angled beam, if the loading were entirely vertical you could apply equilibrium in the horizontal = 0 and it would tell you that the horizontal reaction at B is zero.

@marcapaulajos

4 жыл бұрын

I think she means if the two supports are pinned, therefore there would be 2 horizontal reactions. Therefore you'd have 4 unknowns with vertical loading.

i love you bro

Please use SI units ... cheers :)

Sir, can I please get clarity on why when we calculate moment at B, we do not use the vertical component of 10k and 20k. What I mean is, why don't we say 10k sin 60 degrees time 20'/3 and the same thing with 20k sin 60 degrees multiply by the distance of 5

Please pray for me.

Al right :D

@structurefree

3 жыл бұрын

alright, alright, alright

Big mistake: You assumed the load to the left of midpoint of the beam would be handled by the left support & load to the right by right support, which is ONLY true when the beam is horizontal. Otherwise, more of the load is ALWAYS born by the lower support, depending on the angle, increasing with the angle of the beam to horizontal.

@ourhub9388

5 жыл бұрын

That's only valid when you think of it as in real life because the load will point downwards. But when the force is also applied with the same angle as the member it might not be the case.

what you are doing is really interesting;but not visible.make sure next time u do visible actions.

Hey you messed up, when you're findingmoments at B, It's 10feet/3 for the triangle, not 20feet/3.

@structurefree

10 жыл бұрын

the distance from the resultant to point B is 20ft/3. the 10ft/3 is the distance to point A.

@afaq494

10 жыл бұрын

structurefree im gng to have to watch the video again lol

@offroger76

8 жыл бұрын

+structurefree Right, it's 6.66 feet.

I do not believe these answers. I haven't done them on paper, but you need to do the 20'/3/(SIN30 or COS30) as it is closer to the vertical moment of Ay. Unless I haven't done these in a while, I am mistaken, but please give me the answer as it would be vertical beam to prove the numbers are correct. Thanks, and sorry for questioning without doing the work, but I'm a little busy. Thanks in advance, and I hope I'm not starting trouble.

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@structurefree

3 жыл бұрын

I hope this means my videos are legit!