SA04: Truss Analysis: Method of Joints
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- čas přidán 6. 08. 2024
- This lecture is a part of our online course on introductory structural analysis. Sign up using the following URL: courses.structure.education/
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Thank you so much, it amazes me i can get these sort of resources online for free! If it wasn't for people like you i would have never got into university, thanks again!
This is the synthetic meaning of Engineering! I do Thank you- Whoever prepared and sacrificed his time for this valuable asset. We do solve the problems not for ourselves only, but for the Mankind too. You've reminded me of the Engineering aesthetics. Keep on keeping on!
Short. Fast. Effective. Thank you!
Thank you so much madam , it's been a year since i started studying engineering mechanics and i swear i haven't understood a word , but now i got it.. thanks alot ..
Watch this video for 1st time and instantly became a fan of this channel.
Great content and description.
Thank you
Thanks and welcome
I was expecting a hard time reviewing method of joints, but when I saw this vid, I am grateful, thanks a lot Dr. structure. This 20 min tutorial is far more helpful than my instructor's one and a half lecture with sarcasm!!!
same here
Don't judge a book by its cover. Dr.structure channel is excellent one no doubt at all. At the same time you can't except the same quality of lecture delivery in classroom as it depends on lot of factors like time constraints, non availability of resources like smartboard etc., in board presentation it is very difficult to make understand the concepts through visualisations it is all because of lack of resources in education institutions.
Thanks for your work, it really makes things easier.
Great explanations and lectures!
I wish all instructors are such great in explanation and help.
czcams.com/video/-24-43dN_2c/video.html
could you support me
extremely ultra helpful.thanks.i appreiciate your work
great instruction, thank you very much !
great information about trusses so far....excellent..thanks from india
Thank you Dr. Structure it is so great videos...
I appreciate the film. Very nicely done. It taught me well.
thank you for saving my life doc
There are lots of pin jointed strictures out there. The most famous is there new World Trace Center building in the US. It has steel and concrete pin joints! The CN Tower in Torontoe CANDA has something lie 300 concrtee pin jpoints and theyt solev it use matrices for the deflectrions to prevent crackign in tension!
It helped me a lot.
Thank you very much :)
finally I got good marks in structural analysis.. awesome teaching... plz make videos which are specifically for civil engineering...
Thanks for your lesson...👍👍👍
Jide Olagunju You probably have made a sign error in your solution. BC is a compressive member and AC and AB are tensile members. The sign does matter as the design of a compression member is different than that of a tension member.
thnk U sir may U explain us RCC concrete design
please get in Turkish subtitles
Great video thank you for this
Thanx
You are the best. Thank you a lot :)
Thank you very much, it's so useful for me!
Amazing....
Thank you very much
I knew in this case cos 45 = sin 45. However, to be consistent in derivering a equation, equation (5) maybe wriiten as Sum Fx=Fbc cos45 - Fac cos45 + 5 = 0.
I thought so, she got 5 and 6 sins and cosins mixed up, but doesn't affect answer
Correct.
Please see the updated version of the lecture in the course referenced in the video description field.
This was extremely helpful. I have a final on Saturday and although this is more work than I was taught in lectures, I feel as though this way of doing it makes way more sense than what I was taught and is fool-proof (if you can do basic equation solving :P) Thanks you very much! :)
thanks a lot
Tnx alottt u r my statics hero
Hi Dr. Structure, I tried solving the problem assuming tension for AC, AB and BC. after I finished solving, I got tension for AC, and BC but compression for AB. I got all the same values you got but that was the only difference. Is my answer wrong for this? Or does it matter? Thanks for your reply in advance
Why do you have forces at the supports in the y-direction when you don't have any applied forces in the y-direction?
Generally speaking, the absence of an applied force in a direction (say, the y-direction) does not imply reaction forces in that direction don't develop. In fact, in most cases couple of vertical reaction forces need to develop in order to ensure that the sum of the moments adds up to zero.
thank you madam.
nice explanation, excellent way you teach. pl. continue with diff analytical. do you have any analytical on finding natural frequencies of structures. pl. guide.
Not at the present time.
Hello Dr., I can always work out the reactions before hand and then start working out from the corner right?
Ameya Kamat Yes, you can certainly approach the problem that way. It is generally a faster technique, if you know what you are doing.
Thank you so much! !! You're a lifesaver
شكرا جزيلا
That was super neat, thank you 🙃
@12:37, i think you have your sin and cos mixed... not that it makes a difference with 45 degrees though.
unless your angle is between the members instead of what your diagram is showing
jacob edwards Yeah sure agree with ya.
I was over here bugging out over this lol
Yeah i have the same issue too
which sofwtare name used?
Several software systems are being used.The main ones are:
Camtesia Studio for compiling and creating final videos, VideoScribe for generating text (writing) animations and CrazyTalk for character animation and lip-synching.
Thanks
It is clarified
Have you video about analyst internal forces of members by Bow's notation methods?
No, we don't, not at the present time.
Very informative, but I struggle to see how you arrive at the 3.55NMs in equation 5 and 6.
+Twentyone twos Yes, the value is a bit off. It should be 3.54 ( 5/(2*0.707)).
There was a mistake
When you find Equilibrium of x you wrote the equation =sin ( 45 ) it must be cos (45) .
For this question sin45 = cos45 these are same and it change nothing but when people try to understand they will confuse
can we calculate the support reaction forces first before we get into each of the joints?
+Frez Wong Yes.
at the time 12.38 if u c u mention Fbc is sin but it should be cosine for x direction and same goes to Fac ..
Here, the angle and its complementary angle are both 45 degrees. Depending on which angle you use, you get either sine or cosine of 45 degrees. Both formulations are correct.
Thinks my dear
Im confused at joint c (at 12:37) where angle 45 above line Fac, so the angle of Fac is also 45 since they are perpendicular 45+45=90. is this correct? I need to know if my assumption is right. please response :(
Yes, you are correct.
Ohhhh I love you 😩😩😩
hi Dr. structure I would like to say thank u your tremendous effort showing as can you give me some name of books about mechinecs and structure that helping me to be expert structure
Russell Hibbeler's books on statics and structural analysis are quite good.
Grate Thanks
i have enjoyed alot and learnt alot
but please clearfy Fbc part is it in x direction forces are those with sin(45) or Cos(45)
Thankyou
+Husnain Hyder If we are using the angle that is facing the x-component of the force, then it is Sin(angle), otherwise it is Cos(angle).
I am also confused by this. What would Fbc be if the angle shown was not 45?
Starting @12:34 The equation summing the forces in X direction is given as:
Fbc sin(45) - Fac sin(45) + 5 = 0
And the equation for Y direction is given as:
-Fac cos(45) - Fbc cos(45) - Fcd = 0
An there is a 45-degree angle labeled on the diagram.
The angles used in the above two equations is not the angle labeled on the diagram. It is its complementary angle, which is also 45 degrees.
If that complementary angle is anything but 45, say it is 30 degrees, then we need to used 30 degrees in the above equations.
Why are you using SIN for the forces in the X direction here, when in the example a few seconds before you were using COS in the X direction? Thank you
Since we have an isosceles right triangle here, sine and cosine can be used interchangeably.
what if u do not have right angle triangles can the same approach taken?
There is always a right triangle.
If we have an inclined truss member with know end positions, we can easily construct the right triangle needed for calculating the sine and cosine.
If we know the end positions of the member, then we know the member's length. Make this length the hypotenuse of a right triangle. Since we know the end coordinates of the hypotenuse, we very easily determine the height and the base of the right triangle.
Say a truss member has end coordinates (5,4) and (8,0). Then, the hypotenuse of the right triangle has a length of 5. The height of the triangle is 4 and the base of it is 3.
thanks sir
How do you know if the Fx/Fy force is cos or sin?
It depends on the angle we are using to determine Fx and Fy.
Let’s refer to the side of the right triangle along the x-axis as X and call the side along the y-axis Y.
If we are using the angle facing X to do the calculations, we need to use the sine of the angle to determine Fx and cosine of the angle to determine Fy.
If we are using the angle facing Y to do the calculations, we need to use cosine of the angle to determine Fx and sine of the angle to determine Fy.
At the time of 12:50 of the video, why do you use sin with the sum of all forces in the x-direction and cos with the sum of all forces in the y-direction? I thought it was the other way around.
It depends on the angle being used for writing the equilibrium equations. We either use the angle between the member and the x-axis or the angle between the member and the y-axis. Here, the angle between the force and the y-axis was used.
how did you label tension and compression forces ??
When drawing a truss member in tension, we always show the internal (tension) force acting along the center-line of the member, pointing away from the member (pulling it). Since the member has two ends, we show the (pulling) force acting at both ends.
The same (tension) force, when drawn at the joint, also needs to be shown pointing away from the joint.
The sum of the tension force shown at the end of the member and the same force shown acting at the joint should be zero. Meaning, the two forces need to point toward each other (in the case of tension) or away from each other (in the case of compression).
@@DrStructure Thanks a lot
Hi, what software are you using to make the video? that is the pen cartoon as you write and the cartoon face with the voice?
VideoScribe for text animation
CrazyTalk for voice/face sync
Oh I got confused !! in the beginning you made joint A Hinged and Joint B Pinned, then you switched it
Thanks for this demonstration anyway :D
This is Brilliant work. Do you take donations for support?
Thank you for your post. Currently, there is no mechanism in place for accepting donations in support of the work.
Good afternoon my dear can make CZcams taking about finite element method
How do you get 3.55 from equations 5 and 6?
It should have been written as 3.54, not 3.55.
5/(2*0.7071) = 3.5355 ~ 3.54
@@DrStructure thank you alot, that makes sense now
In the last example, why AC in tension while BC in compression?
Intuitively, because the applied horizontal force tends to push on BC and pull on AC.
the F(bc) tho in x direction should be cos(45). but the answer is correct since cos and sin are equal in 45deg angle.
+ml12 There are two acute angles in a right triangle. We can use either angle to determine the x and y components of a force. If we use the angle between the hypotenuse and the base (along x axis), cosine is used for Fx and sine is used for Fy. However, if we use the acute angle between the hypotenuse and the vertical side of the triangle, then cosine has to be used for Fy and sine has to be used for Fx.
the the cartoon character looks so high, haha thank you so much for the explanation
how do we know the langthe of the two sides are equal .w/h is "l"?
+yenus aminu That information is given to us.
I do not get one thing? Why do you say it automatically satisfies the moment equilibrium when you have a vertical reaction there?
mmdandf Yes, an (isolated) truss joint could be subjected to multiple forces (vertical, horizontal or inclined), but since they all pass through the joint no bending moment develops about the joint. Therefore, the moment equation for a typical truss joint becomes: 0 = 0. The same is true for an (isolated) truss member.
Obviously if we did not isolate the truss joints, if we consider the truss as a whole, then the moment equilibrium equation could have non-zero terms.
thank u very much...
Why should not we apply force to the truss member itself . I think that should make us do the analysis of each whole member ,that is we will then solve the problem using Method Member . Please make video of Method of Members as well. Nice explanation given in thee rest of the videos.
sorry about the spelling mistake.
By definition, truss members carry axial loads only. This means loads are applied only at the joints. If a member is directly subjected to transversal loads where a bending moment develops, it needs to be considered and analyzed as a beam.
Dear sir how can we know that which member is in tension and which member is in compression?
Before the analysis is done, we don't know which ones are in tension and which are in compression. By convention, we assume all the members to be in tension. That is why we show the force vectors pointing away from the members and the joints.
If a computed member force turns out to be negative, we know that member is in compression. If the computed value is positive, the member indeed is in tension.
dr Structure is it possible that we directed a force to the x axis and use sin.l only know that sin is for the y axis ,cos for the x axis
+Tinashe Matimba It depends on the angle you are using for calculating the x and y component of the force.
Given a force vector, you can always construct a right triangle that has the force as its hypotenuse. The right triangle has three interior angles: a 90-degree angle and two smaller angles.
Let's refer to the interior angle between the hypotenuse and the vertical side of the triangle as A and the interior angle between the hypotenuse and the horizontal side of the triangle as B.
You can use either A or B to find the x and y components of the force (F).
Using angle A:
Fx = F sin(A)
Fy = F cos(A).
Using angle B:
Fx = F cos(B)
Fy = F sin(B).
thank you so much Dr Structure.looking forward to your posts
may i ask, what if the distance of x's and y's are given, instead of using cos and sin, ?
Sine and cosine of an angle are related to the sides of the right triangle that defines the angle. If x is the base of a right triangle and y is the height of it. Then the hypotenuse (R) is the square root of the sum of the squares of the side. Or,
R*R = x*x + y*y
If you know x and y, you can calculate R.
If the angle of interest is between the base and the hypotenuse, then
sine of the angle = y/R
cosine of the angle = x/R
If the angle is between the height and the hypotenuse, then
sine of the angle = x/R
cosine of the angle = y/R
Thanks :)
You have explained for a joint with only 2 unknowns .What will happen if a joint has 3 or more unknowns?Can we also solve the problem using the moment?
If the truss is statically determinate, the system of equilibrium equations, when solved simultaneously, gives the values for all the member forces regardless of how many members are connected to a joint.
You can always use the moment equilibrium equation to solve for one or more unknowns, but that is not called the method of joints. We can view it as a hybrid approach which, by the way, may prove to be faster for hand calculations in some cases.
If a joint has more than 2 unknowns it cant really be solved.
Not true. If a joint has more than two unknowns, but the truss is statically determinate, we can calculate all the member forces by solving the system of (joint) equilibrium equations.
@ 16.22, how do you figure .707Fbc-.707Fac + 5 = 0
becomes 3.55 N for both FAC & FBC how?
@16.22
There are two equilibrium equations for joint C of the truss: Equation 5 (sum of the forces in the x direction = 0) and Equation 6 (sum of the forces in the y
direction = 0). Here are the two equations:
Equation 5: 0.707 Fbc - 0.707 Fac + 5 = 0
Equation 6: -0.707 Fbc - 0.707 Fac = 0
From Equation 6 we get: Fbc = -Fac.
Substituting -Fac for Fbc in Equation 5, we get:
-0.707 Fac - 0.707 Fac + 5 = 0.
Solving this equation for Fac, we get (approximately) 3.55 N for Fac.
Dr. Structure
Aren't they supposed to be 3.53606...
the same with Fbd, it should be 2.83...
Joshua de Jesús Yes, more accurately Fbc and Fac each should have a magnitude of 5/2 cos 45 = 3.54.
And, Fbd = 3.54 cos 45 = 2.5
Dr. Structure how i know if i should use sin of cos in Fx and Fy?
+Nnixx Palomares It depends on the angle you are using for calculating the x and y component of the force.
Given a force vector, you can always construct a right triangle that has
the force as its hypotenuse. The right triangle has three interior
angles: a 90-degree angle and two smaller angles.
Let's refer to the interior angle between the hypotenuse and the
vertical side of the triangle as A and the interior angle between the
hypotenuse and the horizontal side of the triangle as B.
You can use either A or B to find the x and y components of the force
(F).
Using angle A:
Fx = F sin(A)
Fy = F cos(A).
Using angle B:
Fx = F cos(B)
Fy = F sin(B).
+Dr. Structure ahhhhhh... okay =D that was very helpful =D
because i have a problem here where i am solving for an scalene kind of triangle. BTW thanks a lot.
+Dr. Structure it confuses me because in this example problem you have 45deg angle but now i am enlightened haha =))
How do you get Fac=3.55N and Fbc=3.55N?
5 / (2 cos(45)) = 3.54
There discrepancy is due to round off error.
I need to know the name of this program please
I am not sure what you mean by the program. Please elaborate.
how to get ac=3.55n and bc too
Please elaborate. Not sure what you are asking.
can i use this method if the bottom member is not in 180 deg ?
+Nnixx Palomares there are 2 members at the bottom but they are not in 180 deg.
+Nnixx Palomares Yes, regardless of the geometry of the truss, the analysis technique works as long as the structure is statically determinate.
shouldnt there be a zero force member at D
+Carlson Ngolah Yes, CD is a zero-force member.
how do we know the member is in tension or compression?
Gurwinder singh At the beginning we assume all the members to be in tension, the internal force pointing away from the member's ends. If a calculated member force is negative, then in actuality the member is in compression. Otherwise, our initial assumption was correct, the member is indeed in tension.
Thank you dr structure . .this helped me a lot..great explanation..
I need portable method of truss
Hello Teacher please drop file pdf all lesson .Thank you very much Teacher.🙏❤
at 12:25 why did you write a positive "FbcSin45", arent you assuming that member is in compression so it should be "-FbcSin45", I've always had MAJOR difficulties with the sign of members, please help! D:
Two comments here:
1. All the members are assumed to be in tension. That is why the forces are pointing away from the joint. If the force was assumed to be compressive, it would had pointed the other way, toward joint C.
2. When we are writing the equilibrium equations, we don't pay attention to tension or compression issue. We pay attention only to the direction of the force based on the established coordinate system. Here, we assumed x is positive to the right. This is not to be confused with a tension force. Here, we are just establishing the positive direction of our coordinate system. We are basically saying, "for the purpose of writing the equilibrium equation, any force that points to the right is positive, and any force pointing to the left is negative. This way, we know what to add and what to subtract when we are finding the algebraic sum of the forces at the joint.
Don't you mean all members pointing away from the point are in compression and all members pointing towards the point are in tension? i'm so confused, sorry
No.
If a member (say, AB) is in tension, then the axial force shown at ends of the member are pulling away, pointing away from the ends. That is how we show a tensile force in a member.
But, we need to draw a different free-body diagram for the end joints. So, the member is on a different free-body diagram than its end joints.
Looking at the free-body diagram of joint A:
The axial force in AB acting at the joint must have a direction opposite to the force drawn on the member itself. So, if the force on the member is pulling away (if it is a tension force), then the same force acting at joint A must also pull away from the joint.
Tension forces are always pull away from the ends of the member, and the joints. Compression forces are opposite in direction to tension forces, at the ends of the members and at the joints.
Dr. Structure oh so in simple words, you were drawing the connection between the joints and the members. and if a is in tension, the direction of the force is going towards the joint but at the joint the direction of the force is pulling away, therefore connecting with the member so it balances out? ive never seen this method before, sorry
Yes, you got it!
B+r=2j 3:00
I'm confused, where did 0.707 come from?
+cumpleted mczella cos(45) = sin(45) = 0.707
Thanx