SA10: Truss Analysis: Method of Sections

[@DrStructure]

+Jascha Annogui wrote: Ho do we find the force in member BN in Example 3? Answer: This can be done in a few steps: 1. Calculate Fgh. This is done using the fbd @10:38. Taking moments about Point A, we get: 2(2) - 4 Fgh = 0. Or, Fgh = 1 kN. 2. Calculate Fmh. This is done using the fbd @9:28. Taking moments about Point B, we get: 2(4) - Fgh(4) - Fmh sin(45)(4) = 0. Since Fgh = 1, we get: Fmh = 1.414 kN. 3. Now, summing the forces in the y direction in the above fbd, we get: Fbn + Fmh cos(45) = 2. Since Fmh = 1.414, we get: Fbn = 1 kN.

[@DrStructure]

+Mayed Amwahi: asked: @5:00 why multiply 0.25 by 5? Answer: Since we are taking moments about B, we need to multiply the 0.25 kN force by its moment arm (the shortest distance from the line of action of the force to Point B). This moment arm is the horizontal distance between A and C which is 5m.

[@cglick3811]

Thank you for a clear, concise, and consistent approach to the method of sections.

[@kimerriobaugh77]

best structure video on youtube keeo it up

[@umindujayabandu8954]

I feel like Alfred is teaching me Trusses in Batman School

[@DewiMadden]

Structures have never seemed so easy! Thanks!

[@Komain72]

I wish you guys did a bunch of different subjects related to physics because it would help me immensely. I've watched two videos about this and understand it more than I ever have from my lectures. Thank you!!

[@sushmitadebnathswarna4515]

I wish my teacher would explain the topic like this... Really helpful for my course... :)

[@DrStructure]

+Jim Law asked: How can I solve for support reactions in Example 2? To do so, consider the entire structure. It is subjected to three known forces and has three unknown support reactions. To find the reactions, write, and simultaneously solve, three equilibrium equations: Sum of the moments about Point J must be zero. Sum of the forces in the x direction must be zero. Sum of the forces in the y direction must be zero. Or, 5(10) + 3(8) + 2(2) - Ky (2) = 0 5 + 3 + Jx = 0 Jy + Ky + 2 = 0 where Ky is the vertical reaction at K and Jx and Jy are the horizontal and vertical reactions at J. This gives: Jx = - 8 kN, Ky = 39 kN, Jy = -41 kN.

[@argeliasimon9047]

You made everything so easy to understand! Thanks a lot!

[@nimdanilox7783]

Great Work. Lectures Make It sound like rocket science but you make it sound like ABC.

[@SeekTheTruthAndTruth]

You really are doctor of structure .

[@dz9001]

Nicely presented, very understandable.

[@Mohammed-fn9og]

Thank you Dr. Structure this video was very helpful to me as it illustrated the method of sections very simply.

[@jakeisrael7899]

you did a hell of a good job on this video. Thanks a lot!

[@marcobicodo]

Very well presented in a way that even a toddler would understand lmao. Thank you for this! *subscribed*

[@aaronleong2425]

thanks you , Ill have my final exam tomorrow, and this video saved my life LOL

[@neerajpavaskar8501]

the explanation was awesome 👍

[@nandha6197]

DR Structure thanks for crystal clear explanation.

[@amarildokuhn]

Hi Dr. Structure,@7:11 you wrote the equation of moment about point D. According to you, the forces that make moment at point D are: 5kN at point A, 3kN at point B, 2kN at point C and the force Fce in the line of action CE. My question though is: why the force Fcd does not make moment at point D too? Because the components of that force would be Fcd (x) and Fcd (y), in a distance of 2 and 2 meters, respectively. Then our equation of moment would be a little bit different, adding those two forces.

[@DrStructure]

Amarildo Kuhn Okay, let's write the moment of Fcd about point D. Fcd is an inclined force, it has an x component and a y component. Since the angle of inclination is 45 degrees, the component forces are: x-component of Fcd = 0.707 Fcd y-component of Fcd = 0.707 Fcd Now let's take the moment of Fcd about point D. Assume clockwise moment to be positive. The moment is: 0.707Fcd (2) - 0.707Fcd (2) = 0. That is, the x and y components have the same moment value about D, but in opposite directions, so they cancel each other out. Alternatively, if you examine the line of action of Fcd, you can see that it passes through point D. Therefore, its moment arm about D is zero. This means the moment of the force about D is zero.

[@amarildokuhn]

Dr. Structure Thanks for your answer. I was really confused about this, but now it's clear. They both cancel each other. However, let's suppose that the length of square was 3 in the x-axis and 2 in the y-axis. We would ended up finding an angle of 33.7 degrees. In this case, we would have: x-component of Fcd = 0.83 Fcd y-component of Fcd = 0.55 Fcd In this assumption, even though the line of action of Fcd passes through point D, the x and y components would have different moment values about D, is that correct?

[@DrStructure]

Amarildo Kuhn That is right, the components would be different if the angle is different than 45 degrees, but so would the two moment arms. For your example we get: (0.83 Fcd)(2) - (0.55 Fcd)(3) = 0.

[@vilhoo3053]

Where do you get the (2)(2) from at 10:40?

[@DrStructure]

That is the moment of the 2 kN force about point A. The moment arm for the force (the shortest distance from the point of application of the force to point A is 2 m).

[@JesusBeatlejuice]

Great video, very concise! Doesn't waste time solving equilibrium equations we already know how to do and provides a broad range of examples.

[@wms3974]

Could please explain how did you get the 3 equations especially the third one 2:30 ?

[@DrStructure]

Summing the moments about point H, clockwise assumed positive: Three forces have a non-zero moment about H: The forces and their moment arms are: Force: Ay Moment arm: 20 m Force: 3 kN (horizontal force at B) Moment arm: 5 m (the shortest distance from the line of action of the force to H. This is the height of the truss). Force: 4 kN (the vertical force applied at F) Moment arm: 5 m (this is the distance between G and H). Equation: (20)(Ay) + (5)(3) - (5)(4) = 0 .

[@DrStructure]

The moment arm for the 3kN force is the shortest (perpendicular) distance between the force and point H. In order to determine the distance, extend the line of action of the force indefinitely in both directions. Here the line is going to be horizontal, since the force is horizontal, going from negative infinity to positive infinity in the x direction. The shortest (perpendicular) distance between the force and H is the same as the perpendicular distance between H and this infinitely long line. All we need to do is to draw a line from H that makes a 90-degree angle with the line of action of the force. Basically, that would be the dimension line, labeled 5 m, that appear at the right side of the truss drawing @3:04

[@yashtibrewal4259]

Awesome.... This video will surely make me pass today!

[@glendylmontales4394]

Your videos are awesome Dr. Structure! Keep up the good work. Thank you so much. :)

[@VIVEKSINGH-bp6fu]

+Dr. Structure In exercise problem (A) I am getting Fce as 0 kN but Fbd as 11.25 kN (T) . please help........ Thanks.

[@DrStructure]

+VIVEK SINGH Yes, your answers are correct. There should be a youtube annotation showing the correct value for Fbd.

[@khalidosama8624]

Thank you so much. A lifesaver.

[@esvw1986]

All your videos are so awesome.

[@Josephandrade1145]

why do we ignore the force acting on BC @4:05?

[@DrStructure]

When writing the equilibrium equations we only consider the forces that appear (are drawn) on the free-body diagram including any support reactions, applied loads and the internal force in each cut truss members. If a member is not cut, its internal force is not going to show up on the free-body diagram. Here, AB, AC and BC are not cut, so they are not considered when writing the equilibrium equations.

[@aveira.a4916]

@@DrStructure @13:06 5(8) +3(6) +2(2) +fgi(2), where did you get 2(2) value? And fgi(2) is the force multiplied by the distance 2m right?

[@DrStructure]

@@aveira.a4916 There is a vertical force of 2 kN at joint C. The line of action of that force is an infinitely long vertical line that passes through joint G. The shortest distance (the moment arm) for that force, therefore is the horizontal distance between joint H and that vertical line. That distance is 2 meters. Hence, the 2 kN force creates a clockwise moment of (2 kN)(2 m) about H. And, yes, that moment arm for Fgi is 2 m, the same distance mentioned above. It is the same distance since both 2 kN force at C and Fgi act along the same vertical line.

[@andreasandreou4863]

How do you know when to use cos or sin @4:50

[@DrStructure]

Generally, you can use either sine or cosine. It is more a matter of preference or convenience.

[@zoelimj]

Hello Dr Structure, thank you very much for the great video! I have a question, though. Why is F(ac) not included in the equation of the sum of moment about D at 7:00?

[@DrStructure]

Zoe L. Because the internal force in member AC does not appear in the free-body diagram (FBD). Since the force is not shown on the FBD, it does not play a role in the equilibrium equations. The xxternal loads applied to a structure are always shown on the free-body diagram. The force internal to a member, however, appear in the diagram only if the member is "cut" exposing the (internal) force. In the example, we cut through three truss members, but AC is not one of them. Therefore, the force in member AC has no impact on the equilibrium equation.

[@dibyendusarkar3622]

There was a mistake at 2:39!! It should be Hx+3=0.

[@usmanjaved6116]

Great demonstration !

[@alex-tt4lw]

So helpful. Thank you!

[@shaunlooweiting6404]

best explanation ever !!! u r awesome....

[@supertv2783]

Thank you Dr. Structure it is so great videos...

[@SAbassi1992]

Thanx soo much from SAUDI ARABIA

[@DrStructure]

Sultan Abbasi You are welcome!

[@Titanic12344]

at 4:58 when you work out the sum of moments, why would you consider Fce when there is no force and even if you did, why not Fbd? Shouldn't be 0.25*5-Fbd*5-Fce*5

[@DrStructure]

+Susku Hud The moment equation given in the solution is for point B, we are taking sum of the moments about B. The equation consists of one unknown force only (Fce), so we can solve for it directly. Fce = 0.25 kN. Alternatively, we can use another point for writing the moment equation. Say, we decide to use point C. Now we get: 0.25 (5) + Fbd (5) + Fbe (5)cos(45) + 3(5) = 0. But this one contains two unknowns (Fbd and Fbe), so we cannot use it to solve for any of the unknowns directly. We have to use it in conjunction with the other equilibrium equations to find the unknowns. Generally, you want to use a moment equation that embodies the least number of unknowns, preferably only one unknown that can be found directly from the equation.

[@Titanic12344]

+Dr. Structure Thank you but i have another question. at 6:56 you take the moment about D? what you didn't include Fcd? Why did you ignore it?

[@DrStructure]

+Susku Hud Since we are taking sum of the moments about D, and Fcd passes through D, the moment arm for the force is zero. That is, the contribution of Fcd to the total moment about D is (Fcd)(0) which is zero. That is why it was not explicitly written in the moment equation.

[@TheTwigz19]

@5:52 im trying to figure out why we did the moments minus Fce multiplied by 5. Isnt Fce in the x direction? isnt moments focused on forces in the y direction. So why was negative Fce multiplied by 5 used? And should that be positive Fce multiplied by 5 since the force is to the right? Alsl I learned that clockwise is negative so i am so confused.

[@DrStructure]

+Peta-anne D Here, we are writing the equilibrium equations for the free-body diagram consisting of joints A, B and C. I am assuming your question(s) pertain to the third equilibrium equation:(0.25)(5) - Fce (5) = 0. This is a moment equation, we are taking sum of the moments about joint B. You asked: Isn't Fce in the x direction? Isn't moment focused on forces in the y direction? The answer to the first question is YES, Fce is in the x direction. The answer to the second question is NO. Regardless of its direction, any force that does not pass through joint B has a moment about B. Here, five forces are shown on the free-body diagram (Fbd, Fbe, Fce, the 3kN force and the 0.25kN force). Three of these forces (Fbd, Fbe and the 3kN force) pass through joint B, therefore their moment about B is zero. The remaining two forces (Fce and the 0.25kN force) however have non-zero moments about B. The moment of the 0.25kN force about B equals to the magnitude of the force times its moment arm (the shortest distance from the line of action of the force to B). This distance is 5 m (the horizontal distance between A and C). Similarly, the moment of Fce about B is Fce times the shortest distance from the line of action of the force to B (the moment arm). This distance is 5m (the distance between C and B). As for the sign convention for moment, you can assume either clockwise or counterclockwise to be positive, it does not make any difference which direction is considered positive as long as it is done consistently. If we assume counterclockwise is positive, we get: (5)(Fce) - (0.25)(5) = 0. If we assume clockwise is positive, we get: (0.25)(5) - Fce (5) = 0. Either way, we get Fce = 0.25 kN.

[@TheTwigz19]

Ohhhhhhh wow!!! Great explanation thank u so much!!!

[@taanyasachdeva1880]

+Dr. Structure can I learn forming the moment equation from any of your other videos? I Could vaguely understand the (Fce)(5) part.

[@DrStructure]

+Taanya Sachdeva We are summing the moments about B. The 0.25kN reaction force creates a clockwise moment (we assumed clockwise to be positive). But Fce creates a counterclockwise moment, hence the negative sign. The moment arm for the 0.25kN is 5m. This is the horizontal distance between A and B. The moment arm for Fce is also 5m. This is the vertical distance between C and B. So, we get: +0.25(5) - Fce(5) = 0.

[@taanyasachdeva1880]

Dr. Structure thank you so much :)

[@gurwindersingh3374]

It is awesome but got a question- at duration 13.55 of this video, in second scenario why did you not include 2KN force acting at point C and why ( Fdf *2) is negative whilst in first scenario all forces are positive..

[@DrStructure]

Gurwinder singh In the second scenario we are taking sum of the moments about point G. Since the 2 kN force passes through G, it has a zero moment arm, meaning its moment about the point is zero. The equation assumes the moments that have a clockwise rotation about G to be positive. Here, Fdf causes a counterclockwise moment about G, hence the negative sign.

[@oa793]

best trus video on youtube

[@user-ny7pk2fv6o]

Thanks a lot men u are perfect. You really helped me thanks

[@Anonymous-vh9tc]

It helped me a lot! Appreciate it

[@vernardoortega6822]

Awesome video, very clear explanation. Thanks.

[@NaturalDesignChick]

@2:38, the second equation is written as HY+3=0. Should this not have been HX instead? Thanks for the awesome videos!

[@DrStructure]

+Julie C Yes, the correction was made using KZbin annotation.

[@NaturalDesignChick]

+Dr. Structure Thank you for the response. I have annotations turned on, but don't see any corrections. I wanted to be certain I understood the process properly.

[@DrStructure]

+Julie C Thanks for letting me know, I'll look into it.

[@ahmadsalmankhan3200]

Great Job.. ! Awesome .... :)

[@kevinmasicampo1366]

i hope there is a video of solution of the excercise problems

[@neo-koder]

extremely helpful!

[@PashtunUniverse]

I wish u taught in our university

[@kevinding9328]

where can i find the solution for the exercise.

[@DrStructure]

The answers are given at the end of the lecture. Review comments for additional information about the individual problems.

[@debasishmajhi687]

ooo it's how we can easily find the force in arms of given structure...... thanks 4 the vdo sir.and also thanks for the practice questions //

[@jcannogui2118]

Hi .. how do i find the force BN in example 3 ?

[@Jarlieee]

Thank you so much, Doc. Great vid! :)

[@miguelsady5135]

Do you have any video on Trusses - Cremona Method?

[@DrStructure]

Unfortunately, no.

[@romeotorongo735]

ths is jus awesome thank u

[@dace8060]

our instructor taught us that in method of sections we can only bisect 3 members but on the book i read it says the same but some of the examples bisect 4 members//confused

[@DrStructure]

Thanks for the question. There are generally no rules that prevent us from cutting through more than, or less than, three members. In this method we use the conditions of static equilibrium (the equilibrium equations) to find unknown member forces. Since there are three equilibrium equations (assuming a planar truss) then we cannot solve for more than three unknown forces at a time. Under the assumption that you wish to determine the axial force in ALL the cut members, then we should not cut through more than three members. If we cut through say 4 members, then we end up having 4 unknowns but only 3 equations. But what if we don't necessarily want to calculate all the unknown forces that result from cutting the truss? What if we want to calculate the force in a single member only? Do we still need to cut through three members? The answer is no. The number of members that we cut through is rather irrelevant here. What is relevant? That the cut enables us to find the unknown force using one of the equilibrium equation (generally the moment equation). More specifically, say we cut through N members exposing N unknown forces one of which being the force we wish to determine. If all the forces except our target force pass through a single point, then we can solve for the target force by writing the moment equilibrium equation about that point. That is since most of the forces pass through the point, their moment about the point would be zero. The only non-zero term in the moment equation would be due the target force. We can then use the equation (which contains only one variable) to solve for the target member force. This is how a K-truss generally analyzed using the Method the Sections.

[@dace8060]

Dr. Structure thanks a lot.

[@derikdecastro]

Life Saver! Just one request, maybe if you could possibly create a tutorial in analysis of structure using method of members (pin connected structures like frames). Thank you and more power :)

[@jimlaw1023]

how can i solve the example 2 support reactions?

[@DrStructure]

+Vishean Veloo Question: Hi, may I know why do you get '15' in the equation of Moment in 2.43? I dont understand that part. Answer: The moment equation is obtained by taking sum of the moments about point H. Three forces produce non-zero moment about H: the left reaction (Ay), the applied 3kN horizontal force and the applied 4kN vertical force. The moment arm for Ay is 20m, the moment arm for the 3kN force is 5m (the height of the truss), and the moment arm for the 4kN force is 5m. So, we get: 20(Ay) + 5(3kN) - 5(4kN) = 0. Hence, 20 Ay + 15 - 20 = 0.

[@kerryaziz7380]

+Dr. Structure Hi I had this problem too, however it still doesn't explain how the value is 15. Is it not supposed to be 15 (distance from H) x 3 (force) which would be 60? thanks

[@DrStructure]

+Kerry Aziz The moment arm for the horizontal 3kN force is not 15, it is 5. The moment arm of a force about a point is the shortest distance from the line of action of the force to that point. This would be the length of a perpendicular line from that point (here, Point H) to the line of action of the force. What you are doing is taking the horizontal distance from Point B to Point H. That distance, 15 m, is not the moment arm for the horizontal force at B. Since it is not perpendicular to the line of action of the force. To determine the moment arm for the 3 kN force, draw a line over the force that is an infinitely long in both directions. This is the line of action of the force. The shortest distance from point H to this line is the moment arm for the force about H. This distance is 5m, the height of the truss.

[@kerryaziz7380]

+Dr. Structure ooooh I understand now, thanks a lot!! :)

[@vishean9743]

Hi, may I know why do you get '15' in the equation of Moment in 2.43? I dont understand that part.

[@peterscott5814]

are you taking moments of counter clockwise as positive?

[@peterscott5814]

forget it ive answered my question when you said assume clockwise is positive. this makes up for the reaction forces being different

[@rinilbabu3911]

hi Dr.structre i am unable to get the answer for exercise problem 3..can u help me with giving some hints..

[@DrStructure]

Rinil Babu Cut through members CE, CD and BD. Draw the free-body diagram for the left substructure. Three unknown member forces are present in the diagram. Two of the unknown forces pass through a point, you want to write the moment equation about that point. The equation would contain only one unknown, Fbd.....

[@rinilbabu3911]

Dr. Structure thanku i got the correct answer..its always been a tough topic for me sometime..now u made it very easy for us..a lot of thanks

[@SuperTaiko95]

thank you for the video. I just want to ask, fir the questions given at the end of the video, why is the value for F(ce)=0 for question A). Your reply is greatly appreciated

[@DrStructure]

+SuperTaiko95 Cut the truss vertically through members BD, BE and CE. Now, consider the left segment of the structure. Its free-body diagram consists of four forces: the vertical reaction at the roller support, Fce, Fbe and Fbd. Notice that three of these forces (the support reaction, Fbd and Fbe) pass through point B. So, if we take sum of the moments about B, we get: 4 Fce = 0 This means for the equilibrium equation to be satisfied, Fce has to be zero.

[@SuperTaiko95]

Ok. That makes sense. But what if I was to consider the right hand-side of the structure? Do I have to take into account the reaction forces at point E as well as the other forces?

[@DrStructure]

+SuperTaiko95 Yes, we can take the right substructure. But then its free-body diagram would consist of Fbd, Fbe, Fce and the horizontal and vertical reactions at E as well as the applied loads at G and H. Now we have five unknown forces meaning we need to do a bit of calculations before we can show that Fce is zero.

[@Raianrod]

+Dr. Structure My answer is Fce=0 and Fbd=-11.25. But the correct answer is Fbd=1.25 Kn. Why? please healp me.

[@DrStructure]

+kazuto kirigaya Yes, the correct answer is 11.25 kN, the member is in tension. There is a (youtube annotation) note placed on the video clip @15:10 that shows the correct value.

[@warrensze321]

thank you so much!

[@johnfox137]

Awesome refresher

[@JOYANTASAHA2001]

SEARCHING EXACTLY THE TYPE OF PROBLEM TO SOLVE🧡🧡🧡

[@TejSwaroop]

I have a question about 2nd equilibrium equation @2:38 ... it's Summation F(X) and hence equation i was expecting was -H(X)+3 = 0 , but the equation says H(Y) ! Can someone explain. Thanks

[@DrStructure]

Yes, the correct equation is Hx + 3 = 0, NOT Hy + 3. The correction was made using KZbin annotation which you should be able to see on a desktop. Unfortunately, KZbin annotations don't show up on mobile devices.

[@waseemahmad6895]

Where frm can i get answers to the end problems to cross check plz do reply

[@DrStructure]

We will provide the solution for each problem in pdf format within 24 hours. You will be able to find links in the video description field then.

[@DrStructure]

The exercise problem solutions are now available. See links in the video description field.

[@humblegod13]

Can you please explain the problem no. 2 in exercise part?

[@DrStructure]

Do you mean Problem (B)? Cut member AB, then examine the free-body diagram of the right segment of the structure which would be the entire structure minus the support at A. The applied load (2 kN) at D creates a bending moment about C. The moment arm is: 5 + 5 sin(45). The axial force in AB (the cut member) also creates a moment about C. This moment arm is: 5 m (since the angle between AB and BC is 90 degrees). Since sum of the moments about C must be zero, we can write: 5 Fab = 2 (5 + 5 sin(45)) Now solve for Fab.

[@humblegod13]

I have a question.. how can I know the values of DE and EC?

[@humblegod13]

oh sorry got it.. thank you for being this supportive

[@wilmdrdo1228]

In example 3. what if we want to find the force MB? Should we still use the method of sections? Thanks!

[@DrStructure]

Yes, I would use the Method of Sections to find the force in member MB. But this involves two steps: First, determine the force in member AB. Then, cut the truss through members AB, MB, MH and GH. Now set sum of the moments about point H for the right substructure to zero. The moment equation contains only one unknown, the force in member MB. Solve the equation for the unknown force.

[@mayedanwahi4112]

I would like to thank you for this great explanation, but i'm not quite sure why did you multiply 0.25 by 5 (@5.00)

[@wrenchesinthegears]

is there a reason this video represents clockwise moments as positive instead of negative?

[@DrStructure]

When summing moments about a point, the assumed sign convention is rather arbitrary. We can use either clockwise or counterclockwise as the positive direction.

[@wrenchesinthegears]

Yeah, I guess what im asking is, why don't you follow the right hand rule?

[@cedricmwendanga5803]

I need also help with the truss D

[@DrStructure]

Cut through AC and AB, take the left segment of the truss, and sum moments about B.

[@shahoamedi2700]

hello dr. what if we want 4 forces or more ?

[@DrStructure]

Assuming the structure is statically determinate, you maybe able to use the method of sections repeatedly by finding 1, 2 or 3 forces at a time until all the unknowns are determined. If that is not an option, use method of joints.

[@shahoamedi2700]

Can u solve am example that we should use the method of section twice

[@DrStructure]

Shaho Amedi Consider Example 3 (starting @7:51). Suppose we are asked to determine member forces BC, KC, KI and HI. If we cut the truss vertically through all 4 members, we end up with four unknown forces which cannot be determined using the three equilibrium equations. Instead, we can cut the k-truss in the manner shown in the example to find the force in members BC and HI. Then, cut the truss vertically through all four members to find the remaining two unknown forces.

[@lakegraham9007]

When you took your equilibrium equations for the first truss example, you get H_x to be -3kN, however it looks like you assumed to the right as positive, and if you got a negative value, why not draw the arrow pointing to the left?

[@DrStructure]

+Lake Graham Sure. You can draw the force in the other direction and show its magnitude to be positive. It is correct either way.

[@teamnasirsarah7809]

When taking moment about point B in example 1, why isn't " +Fbesin45" considered? Would that distance no also be considered perpendicular to B?

[@DrStructure]

Since Fbe passes through B, it creates no moment about B. Simply put, if you draw the line of action of the force and it passes through a point, the moment of the force about the point is zero.

[@ajj7794]

finding out Fgi where your working shows 5x8+3x6+2x2+Fgi x2 = 0 where do you get the 2X2 when the distance to H is 6 metres from C. also how do you know to ignore all the joints except A, B, C and Fgi is it because you do not know the forces at the other joints. thanks.

[@DrStructure]

+xganh zu There is an external 2kN force applied at C, the shortest distance from C to H is 2m, not 6m. So, the moment that the force creates about H is 2x2. The reason we ignore all the intermediate joints in writing the equilibrium equation is that there are no external forces applied at those joints. The applied loads are only at A, B and C.

[@ajj7794]

ohh ok thx

[@zahidahmadzai8023]

at 5.08, why did u take fce as negative, shouldn't it be positive since its going to the right??

[@DrStructure]

When writing the moment equilibrium equation, we need to assume either clockwise or counterclockwise as the positive direction. In this case, we assumed clockwise moment to be positive. We are writing the sum of the moments about point B. The support reaction (0.25 kN) created a clockwise moment of (0.25 kN)(5 m) about point B. And Fce creates a counterclockwise moment of (5 m) (Fce). So, the moment equilibrium equation needs to be written as: +(0.25)(5) - (Fce)(5) = 0

[@salmanjalali]

Could you explain what the equations are in exercise A? ive got Fce as 0 but Fbd as 11.25.... Thanks

[@jeffriefendi5381]

I got the same answer i think they mistyped it

[@nanakyei5209]

Dr. Structure, thanks alot...I have shorter process now(only moment equilibrium eqn can be used to find all member forces through method of sections

[@johnmichaelsimon1300]

Hey Dr. Structure. Thank you for the great video it was really helpful. I have a question regarding example 3. How do you compute the reactions for BN and NH? Or is it since the trusses are symmetrical then the reactions would be zero? Thank you in advance for your reply.

[@DrStructure]

The support reactions for the truss are computed before we start cutting the truss. And, they remain the same regardless of how we cut the truss. In this case, the left and right support reactions are 2 kN each, as long as we don’t change the applied loads. That is, as long as the truss is subjected to a downward 4 kN force at joint I, the reactions at joints F and L remain the same.

[@darkozy298]

Hi, I'm not very good in this mateer, and I would like to know if it was possible to have a solution for the problem "A" please! Thanks

[@DrStructure]

+Darkozy dofus I think the solution that you are looking for was discussed in this comment section, take a look and see if you can find the answer to your question.

[@MsFairysmile]

Thank you this really helping us. how could you draw the professor btw?

[@DrStructure]

MsFairysmile VideoScribe software.

[@rajsah8487]

Dr.Str.,In the exercise problem 1,2 & 3 I have got the answers in -ve,i.e. all compression,is it correct????also I didn't get Fce(problem 1) = 0.

[@DrStructure]

You are probably making a sign mistake in your calculations. When drawing a truss free body diagram, we always show the force in a cut member, at the joint, as an arrow pointing away from the joint. This means we assume the member to be in tension. If the (member) force arrow is drawn pointing toward the joint, then that signifies a compressive force in the member. In Problem A, if we cut the truss vertically through BD, BE and CE, three member forces are exposed, Fbd, Fed and Fce. Take the right substructure, and take the sum of the moments about E. The only unknown force in that moment equation is Fbd, since the other two unknown forces pass through E. The equation can be written this way: 5(6) + 5(3) - Fbd (4) = 0. This gives Fbd = 11.25 kN. The positive sign here indicates BD indeed is in tension, as was initially assumed. Now take the left substructure, the one that includes joints A, B, C, the roller at C, and the three cut members. Take the sum of the moments about B, all the forces (including the support reaction at C which is unknown to us at this point) except for Fce pass through B, so their moments about the point is zero. The equation then becomes: Fce (4) = 0. Or, Fce = 0.

[@rajsah8487]

@@DrStructure Thanks Dr.Your dedication to the work is awesome.

[@stanzinnorboo1336]

In the video it shows 1.25 for the problem A and I got 11.25 , I got clear in this reply

[@mustang1479]

Ur a legend 👏🏼

[@subhajitfitness3201]

for prob no. 2: Is member BD a zero force member..??

[@DrStructure]

No! BD is not a zero-force member in exercise problem 2.

[@noxmylove8747]

how to solve exercise b? i tried cutting it in ab but needed value for ay.

[@DrStructure]

You don't need Ay. Cut AB, take the right substructure (that would be everything except A and the lower part of AB), then sum the moments about C. Two forces cause moment about C in that substructure, the applied load and the force in AB. That means the moment equation has only one unknown in it, the force in AB. Solve the equation for the unknown force.

[@itstwinklerkate]

how come you didnt consider the zero force members?

[@DrStructure]

Not sure what you mean by that, please elaborate.

[@itstwinklerkate]

In 5:37, would I get a different answer if I were to remove zero- force members HI , FG and DE and then section the truss afterwards?

[@itstwinklerkate]

Dr. Structure Thank you for replying by the way!!

[@DrStructure]

No, the analysis results would be the same. In the free-body diagram that we used to determine Fce, none of the zero-force members are present. If they were, we could have removed them in order to simplify the analysis.

[@itstwinklerkate]

Dr. Structure Oh I get it now. Thank you so much!!!!

[@syedaghashahali5317]

Thank you for this wonderful tutorial but i encounetred a problem in exercise C i cant manage to figure it out how is it solved

[@DrStructure]

Cut through CE, CD and BD. Take the left substructure, and take sum of the moments about C. Only the 5 kN force and the force in BD appear in the moment equation. You need to use basic geometry to figure out the two moment arms.

[@syedaghashahali5317]

thank you so much

[@DrStructure]

You're welcome.

[@Ashishkumar-id1nn]

Answer of exercise problem A should be Fbd=11.25 kN.

[@jasonfrigillana9218]

2:43 hx looks like hy, that should be hx

[@azadahoramazda3775]

in problem A the support reaction C is negative. it means that the support should have a downward reaction. but it's a roller, how can a roller have a downward reaction? i dont think the truss is statically determinant in this way. please answer

[@DrStructure]

Roller supports are often found in heavy structures like bridges where dead load is significantly larger than any load that could result in an uplift force at the support. In structural analysis we often calculate reaction and member forces for individual loads separately, then add them up according to design equations to determine design loads. Although the roller could be subjected to an uplift force under, say, the load due to a moving vehicle, but the overall force on the support is going to be downward when all the other loads are added to the mix. In other types of (lighter) structural system where an uplift force could develop at a roller support, various mechanisms (e.g., sliding bearings) can be used to ensure that the support remain attached to its base.

[@azadahoramazda3775]

wow, that was a good explanation. i didnt think of that.

[@janebo-o2804]

Where did the "20Ay +15 -20=0" came from?☺ (In the beginning of the example)

[@DrStructure]

The equation represents the sum of the moments about H. There are three forces that create a moment about H: The vertical reaction at A. Its moment about H is: 20 Ay (clockwise) The horizontal (3 kN) force at B. Its moment about H is: (3)(5) = 15 (clockwise) The vertical (4 kN) force at F. Its moment about H is (4)(5) = 20 (counterclockwise). Adding these three terms two of which is clockwise and one counterclockwise gives us: 20Ay + 15 - 20 = 0

[@nadeemkuskiwala6220]

in example-2 can we take moment about B to find out force in member Fce???? similar to example 1. i took this but did not get wright answer.

[@DrStructure]

+Nadeem Kuskiwala Not directly. The moment equation about B includes both Fce and Fcd. Unless we know Fcd already, the equation has too many unknown for determining Fce.

[@joshuariddellski]

awesome video however the audio is extremely quiet

[@humblegod13]

Can you please do a video explaining the problems please??? I really need it

[@ghibligirl]

How would we solve if we know the lengths but only have a center force?

[@DrStructure]

What is a center force? Please elaborate.

[@a.t.a.9044]

Hello, thanks very much for these great lectures.. Please, can I get the solution of B & C problems ?

[@DrStructure]

PROBLEM B: Cut through member AB exposing the internal force in the member. Let's label the force Fb. Fb has two components: an x component (Fb cos45) and a y component (Fb sin45). Since member AB is cut, we can isolate substructure BDEC (the main structure minus the pin at A and the lower part of member AB). Take sum of the moment about point C in this substructure. Three forces cause moment about C: the applied force and the x and y components of Fb. The moment arm for the applied force is (5 + 5 sin45). This is the vertical distance from Point D to the bottom of the structure. The moment arm for the x component of Fb is (5 sin45). The moment arm for the y component of Fb is also (5 sin45). So the entire moment equation (about C) can be written as: 2(5 + 5 sin45) - (Fb cos45)(5 sin45) - (Fb sin45)(5 sin45) = 0. Solving the equation for Fb, we get: 3.41 kN PROBLEM C: Cut through members CE, CD and BD exposing three internal forces: Fce, Fcd and Fbd. Now consider the left substructure, the one consisting of nodes A,C, and B only (and the applied load as well as the three exposed internal forces). Sum of the moments about C must be zero. Note that two of the unknown forces (Fce and Fcd) pass through C therefore they don't contribute to the total moment at C. Only two forces cause bending moment about C: the applied force and Fbd. Therefore the moment equation has only two terms: the moment of the 5 kN force and the moment of Fbd about C. The interior angle at A is 36.86 degrees. This is calculated using inverse tangent of 18/24. This makes the vertical distance from C to the bottom of the truss 10 sin36.86 = 6 m. This is the moment arm for Fbd. The moment arm for the applied load (the horizontal distance between A and C) can be calculated using : (10)^2 = (6)^2 + (arm)^2. Or, arm = 8 m. Now, let's write the moment equation: 5(8) + Fbd (6) = 0. Or, Fbd = -6.67 kN

[@a.t.a.9044]

Dr. Structure Thank you very much ..

[@janayool1001]

+Dr. Structure sir in problem(B) you missed up the (Ay)force from the hinge support and also you should say Fb(5sin(45)*2) because Fbx does not make moment at C . correct me if i am wrong please !

[@DrStructure]

+jana yool When the truss is cut through member AB, two substructures result: the one to the left consisting of joint A and part of member AB, and the part to the right. If we consider the right substructure for further analysis, then Ay does not appear on the free-body diagram since point A is not a part of the right substructure. Further, neither the x component nor the y component of the force in member AB passes through point C. Taking sum of the moments about C, the moment arm for the x component of the force is 5 sin(45), and the moment arm for the y component of the force is 5 cos(45).

[@janayool1001]

You are right thank you sir

[@plutocash1150]

how do you know when to neglect inner members when cacuating moments

[@DrStructure]

+Brian Benz If a member is not cut, if its internal force does no appear on the free-body diagram, it does no contribute to the equilibrium equations.

[@plutocash1150]

+Dr. Structure thanks a lot,very helpful. Maybe I will stop crying now after I manage to solve (hopefully) even one question succesfully

[@accessuploads7834]

many thanks