Internal Loadings in Structural Members | Mechanics Statics | (Solved Examples)

Learn to figure out shear forces, normal forces and bending moments with step by step examples. We go through how to solve for unknown forces, support reactions and more.
🔹 Finding moments - • Moment of a Force | Me...
🔹 2 force members - • Frames and Machines | ...
🔹 Distributed loads - • Reduction of a Simple ...
If you found these videos helpful and you really want to give something, it's very much appreciated.
Donate:
🔹 ko-fi.com/questionsolutions
🔹 www.questionsolutions.com/donate
🔹 PayPal: [e-mail]
Intro (00:00)
Determine the normal force, shear force, and moment at point C. (01:33)
Determine the normal force (03:01)
Determine the internal normal force, shear force, and moment at point D. (05:24)
Find more at www.questionsolutions.com
Book used: R. C. Hibbeler and K. B. Yap, Engineering Mechanics Statics.
Hoboken: Pearson, 2017.

[@QuestionSolutions]

Please wait, before you write a comment asking why a clockwise moment is positive, or counter-clockwise is positive, please watch this video first: kzbin.infoP029mqnp4XY thanks!

[@arentsteen5452]

There is great honor in educating. Your videos have a truly absurd quality/views ratio. I applaud your for your effort.

[@QuestionSolutions]

Thank you for your kind comment, must appreciated.

[@prabin1637]

Quality is on point for both the animations and knowledge, thank you so much

[@QuestionSolutions]

Thank you so much :)

[@551red]

thank you so much for your examples. all made sense, so clear and succinct.

[@QuestionSolutions]

I’m really glad to hear that you found the examples helpful and easy to understand! Your feedback is greatly appreciated. Happy studying!

[@mun9018]

I love how your videos are paced. You show how the solution is constructed but don't dawdle on the value.

[@QuestionSolutions]

Thank you very much! I think understanding the steps is more important than the final value. Most professors give 1/10 for the final answer, and 9/10 for the steps to get there. 😅

[@dylanr9856]

First of all, Thank you so much for this series. Because of you, I got 85% on my Mechanics midterm. I was just wondering if you could continue into chapter 8 - friction

[@QuestionSolutions]

You're very welcome! So the reason why I didn't go into friction that deep is because for first year courses, most of the time, it stops at the "coefficient of friction multiplied by the normal force." That was covered in a lot of the dynamics/statics videos already. Maybe your university goes further in for a first year course, I am not sure, but since most universities stop there for a first year course, I also stopped. I am trying to cover most of the first year courses so students get a good grasp of the fundamentals for the upper years. Currently, I am on thermodynamics, so I probably won't visit statics for some time, but when I start with more of the upper year courses, I will definitely cover it. So please accept my apologies, but I probably won't get it done in time for your final exams. On the bright side though, there are a lot of people teaching this stuff on KZbin/other resources so I think you will be good to go :) Best wishes with your exams!

[@shashikant.]

Thank you so much. Your method of explaining is amazing.

[@QuestionSolutions]

You're very welcome! I am glad these videos are helpful to you :)

[@stefanhagens671]

Amazing video, thank you so much!!!

[@QuestionSolutions]

You're very welcome!

[@manasladdha4601]

Thanks to you I got a 10 in my Mechanics course in my first semester. Back to the same channel, for my Solid Mechanics course in my fourth semester now. :D

[@QuestionSolutions]

Nice work! Keep it up and I wish you the best.

[@hemanthkumarkalidindi8580]

Explained well ,keep it up

[@QuestionSolutions]

Thank you very much!

[@ahmedelwazan9144]

Thanks a lot i appreciate your effort ☺️

[@QuestionSolutions]

Thank you very much!

[@robertosantos4730]

Your videos are GREAT. holy shit. I keep thinking you made an error but its me making stupid errors. YOu sire deserve a star.

[@QuestionSolutions]

It's better to know you made a mistake than to keep going in the same pattern. Well done on catching your own mistakes, and keep up the good work!

[@inkpaintings8316]

Thanks a lot broo

[@QuestionSolutions]

You're welcome 👍

[@user-cr5wd9rm2v]

great video

[@QuestionSolutions]

Thank you very much!

[@quailland1982]

If you consider counter-clockwise to be the positive direction for your moment instead of clockwise, would your answers be different than if you considered clockwise the positive direction for your moment calculations like you did?

[@QuestionSolutions]

No, it gives the same answer. It makes no difference.

[@atocousin3335]

Can you explain why on the second question, you wrote the two force member force pointing outwards like tension, but for the third question you had the two force member force pointing inwards?

[@QuestionSolutions]

It was actually just an assumption. You can draw it either way, if you end up with a negative answer, then it's opposite to your assumption. :)

[@almiraozek1587]

03:30 I really don't understand why the direction of Fbc is left-up arrow and what does it mean being a 2 force member. If you can explain I'll appriciate.

[@QuestionSolutions]

Please see this video: kzbin.info/www/bejne/f6nOqI2Pjap2oc0 Especially at 1:35 where I explain two force members.

[@dremr2038]

Normal force + when it causes tension. Shear force + when cause clockwise and moment is positive when moving in concave upward manner.

[@QuestionSolutions]

0:54

[@dremr2038]

@@QuestionSolutions 😁😁😁

[@zarinsabah3033]

for the second problem , when we took moment at A ,why didnt we took the reaction forces at point c?

[@QuestionSolutions]

Point C doesn't matter since we only look at member AB. We assumed a single force acting at B from the member BC.

[@neilbertdetera2126]

From 5:45, could you please tell me how come that you use 4/3 instead of 3/4? I'm usually confused about it, and I found out that it's the stuff that I must know hehe, thank you Sir! A very helpful video.

[@QuestionSolutions]

So here, we are using sine/cosine/tangent to figure out lengths. In this case, we used tan, which is opposite over adjacent. So looking from the angle, (imagine you're the angle), what is the length of the opposite side? That's 3 + 1 = 4 m. Then we look for the adjacent side. That's 3 m. So opposite over adjacent is 4/3. Does that help or maybe I didn't understand your question well. Please let me know, thanks!

[@neilbertdetera2126]

@@QuestionSolutions very helpful po, Thank you!😄

[@Jasonvoobhggg]

Thank you so much this video is very useful!! I have a question, what about fixed support only on a beam? Although it is like the pin support but I kinda have bad feeling toward it 😂🙏🏻

[@QuestionSolutions]

There isn't much of a difference, you'd have 3 reactions though. With a pin, you have x and y reactions, but in a fixed support, you'd have a moment along with x and y reactions.

[@Jasonvoobhggg]

@@QuestionSolutions When I was passing through, I found out that we took a hinge alongside a fixed support. And this needs a cut then tryna figure out each reaction in a sliced beam.. In a fixed support, there is a moment equation

[@JAMESANDREWBAUTISTA]

Hello bro, where did you get the moment arm 8 from 1200(4)-FBC sin 36.86 (8) =0. THANKS BRO

[@QuestionSolutions]

So the 8 comes from 4m + 4m = 8 m. It is the length of the beam AB. We are treating member BC as a force, so the force is applied at point B, which is 8 m away from point A.

[@uygarunlu6494]

thank you for your examples . 06:05 we cut the AE and we forget the top.Could we do it from the top?

[@QuestionSolutions]

You can, the choice is yours on which side you want to use.

[@antonymanjali3604]

5:50 how come you choose to multiply FBC cos 53.13 (4) and not (3) - only x direction forces create a reaction.

[@QuestionSolutions]

So you're right when you say only the x-component creates a moment but I'm unsure why you think 3m is the perpendicular distance. Point B is 4m in the vertical direction from point A. The x-component lies horizontal at point B. I hope that helps 👍

[@abdallahjassim672]

Minute 3:40 Why didn't you use the reactions of c? I thought you you will write all the reactions and find them. Then we can take the right or left part?

[@QuestionSolutions]

You can write them, but it doesn't matter what they are. We only care about the top beam, so we don't need them to calculate force FBC. So focus on one beam at a time, unless you consider the whole object, in which case, internal forces like FBC doesn't matter.

[@abdallahjassim672]

@@QuestionSolutions Ok, thank you.

[@peerankurdi]

Man thank you, you are amazing, may i ask where do u study or u done with uni?

[@QuestionSolutions]

You're very welcome. I don't really like sharing personal info, but I studied at University of Toronto. 👍

[@peerankurdi]

@@QuestionSolutions sorry man😅 u should definitely be a proff there 😂

[@QuestionSolutions]

@@peerankurdi 😉

[@DC-lv5hk]

at 2:59 you took all the clockwise forces as positive and all the anticlockwise as negative but isn't it the other way around

[@QuestionSolutions]

The direction you choose for moments to be positive is up to you. It's just an assumption. If you get a positive answer, your assumption was right, if it's a negative answer, it's opposite to your assumption.

[@euouae2097]

Why do i get different values of shear force in left and right segment for example no. 1

[@QuestionSolutions]

That's hard for me to answer without seeing your work 😅

[@kaeli5445]

i think there's some confusion, you moment at point a so that 10kn is going down it should be clock wise right? therfore - 10(1.5)-15(1.5)+by(6) by= 82.5

[@kaeli5445]

-10(1.5

[@QuestionSolutions]

There is no confusion. Clockwise moments were picked to be positive so the only thing negative would be B_y. You can pick counter-clockwise to be positive, you will still get the same answer. It makes no difference. You just have a lot of negatives to deal with. Do whichever method you like best.

[@yusufgungor8323]

Is it supposed to be clockwise moments are negative? How are we accept as positive?

[@QuestionSolutions]

See: kzbin.infoP029mqnp4XY?si=0byrZmV1OCKCOPgi

[@user-ob2oj3sn5n]

Why is clockwise moment positive in the first example? Isn't it meant to be negative?

[@QuestionSolutions]

Are you referring to 2:03? If yes, then you can assume clockwise or counter-clockwise to be positive. You will still get the same answer.

[@nohelyBELLY468]

At 5:21 why does the moment b equation have the 600(4) - 600 (2) - Md = 0 I thought it would be 600 (4) + 600 (2) - Md = 0. I can't see why F= 600 N isn't clockwise :/

[@QuestionSolutions]

Think about the beam rotating about point D. So it's free to move. In that case, if you apply a force from the top at the middle, how will it spin? It would spin counter-clockwise. Now imagine applying a force from the bottom, how would it spin? It would spin clockwise. To see it better, cut a small piece of paper, and hold your finger at point D, gently, so that you can spin it with your other hand. Then apply the forces one from above, and another from the bottom and see how it spins. Another way to see it is to realize that in this case, we have a force from the top and another force from the bottom, it's not possible for it to spin the same way when the forces are coming from 2 difference sides.

[@opmacace523]

In the second example you should’ve have a point between B and C that we should calculate as well

[@QuestionSolutions]

Why?

[@lumiereistaken]

hey bro, may I know how to use sin and cos in Fbc?

[@QuestionSolutions]

If you're looking for a refresh on breaking forces into components and when to use sine and cosine, see this video: kzbin.infovynnKlJD_Jo?feature=share

[@manarawad3339]

In 1:58 if we have the same question and we have a moment on B if we sperate AC why we don't take the moment with us as the moment Will influence in all frame?

[@QuestionSolutions]

I am sorry, I don't understand your question. Are you saying "why aren't we taking the moment about point B?" If so, you can, you'll get the same answers.

[@manarawad3339]

@@QuestionSolutions No No i mean if we put a moment on B in this question there isn't a moment on B

[@QuestionSolutions]

@@manarawad3339 Do you mean like an external moment applied at B?

[@manarawad3339]

@@QuestionSolutions yeah i mean that

[@QuestionSolutions]

@@manarawad3339 If you split the beam at C, then the right side will have the moment while the left side will not. So when you do the math for seperate segments, you will only care about the applied moment for the right side, not left. In essence, you think of it as 2 seperate beams

[@user-hv6ef9ie1g]

2:56 How can we decide whether we should put +Mc or -Mc? Like, at 5:20 you put Md as -Md even though the direction is the same as 2:56 (which means it should've been positive(?)) CMIIW

[@QuestionSolutions]

It's completely up to you, so at 2:56, we chose clockwise as positive. Notice that the moment arrow (the curved blue arrow is facing clockwise), so it's a positive MC. But look at 5:20, again, we chose clockwise as positive, but our moment arrow, the blue one, is counter-clockwise. So it's negative.

[@user-hv6ef9ie1g]

@@QuestionSolutions Ohh i see, thanks for the answer

[@QuestionSolutions]

@@user-hv6ef9ie1g You're very welcome!

[@khaledsh8506]

At 2:40 why didn't you take the moment at B = 0 then V works out to be 7.5. Nice video btw thanks.

[@QuestionSolutions]

You have an unknown moment at C, so you can't do that.

[@thefourthdimension5467]

?

[@kamilamustapa8074]

minute 4.08,1000 come from where? i dont understand

[@QuestionSolutions]

You write a moment equation about point A and solve for F_BC. Please look at 3:52.

[@darrylcarter3691]

Looks like something used for Civil Engineering

[@QuestionSolutions]

Most of statics is the basic foundation for civil engineering. 👍

[@hasibsami8255]

When u take moment at c 2:50.. shouldn't the Mc become 0?

[@QuestionSolutions]

I can see why this can be confusing, but this isn't the same as when you take a moment about a point and any line of action of a force that intersects it becomes zero. Instead, imagine you have an object and you have an additional moment applied to it separately. So not created by an applied force, but rather, a direct moment being applied.

[@hasibsami8255]

@@QuestionSolutions Thank u for clearing that. i was just checking section method and ur reply clear all my confusion.

[@QuestionSolutions]

@@hasibsami8255 Glad to hear :)

[@-Abdullah.]

why did you use cos here? 5:49

[@QuestionSolutions]

Cosine because we need the x-component, which is the adjacent side to the angle.

[@breezy1106]

why do we chose the side that is easier in this video but when drawing a shear diagram the right side is different from the left side

[@QuestionSolutions]

I don't understand your question. Did you look at the video covering how to draw shear and moment diagrams? kzbin.info/www/bejne/pJzcn6V6frODn8U