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!
@sashank143114 жыл бұрын
Watch this video for 1st time and instantly became a fan of this channel. Great content and description. Thank you
@DrStructure4 жыл бұрын
Thanks and welcome
@joelestoesta5768 жыл бұрын
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!!!
@khinchiaaliya25254 жыл бұрын
same here
@venkatesh22852 жыл бұрын
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.
@Abdisabir8 жыл бұрын
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!
@marwatalal14508 жыл бұрын
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 ..
@ahamadromen6 жыл бұрын
I wish all instructors are such great in explanation and help.
@rubaalsaeed57905 жыл бұрын
kzbin.info/www/bejne/Y2OXXmdpmbOSaMU could you support me
@douglaso93979 жыл бұрын
Short. Fast. Effective. Thank you!
@wassunoor86788 жыл бұрын
finally I got good marks in structural analysis.. awesome teaching... plz make videos which are specifically for civil engineering...
@swapnil987810 жыл бұрын
great information about trusses so far....excellent..thanks from india
@DrStructure9 жыл бұрын
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.
@farahahmed82258 жыл бұрын
thnk U sir may U explain us RCC concrete design
@zinetkaplan7995 жыл бұрын
please get in Turkish subtitles
@vincenguyen61373 жыл бұрын
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.
@jabertagi38573 жыл бұрын
I thought so, she got 5 and 6 sins and cosins mixed up, but doesn't affect answer
@DrStructure3 жыл бұрын
Correct. Please see the updated version of the lecture in the course referenced in the video description field.
@andjelaandjela5838 жыл бұрын
Great explanations and lectures!
@YASHWANTKUMAR-nf6gn5 жыл бұрын
Amazing....
@bunhengpreap56456 жыл бұрын
Thanks for your lesson...👍👍👍
@jacobedwards97068 жыл бұрын
@12:37, i think you have your sin and cos mixed... not that it makes a difference with 45 degrees though.
@jacobedwards97068 жыл бұрын
unless your angle is between the members instead of what your diagram is showing
@abdisalammohamedelmi8547 жыл бұрын
jacob edwards Yeah sure agree with ya.
@ricardofreshley31455 жыл бұрын
I was over here bugging out over this lol
@samh44664 жыл бұрын
Yeah i have the same issue too
@rustystove84104 жыл бұрын
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!
@supertv27839 жыл бұрын
Thank you Dr. Structure it is so great videos...
@Katie-uh8vg6 жыл бұрын
Why do you have forces at the supports in the y-direction when you don't have any applied forces in the y-direction?
@DrStructure6 жыл бұрын
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.
@TheDeftonesmusic9 жыл бұрын
extremely ultra helpful.thanks.i appreiciate your work
@sandeepdeepu61886 жыл бұрын
how did you label tension and compression forces ??
@DrStructure6 жыл бұрын
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).
@sandeepdeepu61886 жыл бұрын
@@DrStructure Thanks a lot
@MissPiggyM976 Жыл бұрын
Thank you very much, it's so useful for me!
@mmdandf9 жыл бұрын
Thanks for your work, it really makes things easier.
@quemaspana10 жыл бұрын
I appreciate the film. Very nicely done. It taught me well.
@fathereshark84212 жыл бұрын
How do you know if the Fx/Fy force is cos or sin?
@DrStructure2 жыл бұрын
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.
@moh.faizaljalani40744 жыл бұрын
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 :(
@DrStructure4 жыл бұрын
Yes, you are correct.
@binhtrinh99387 жыл бұрын
Have you video about analyst internal forces of members by Bow's notation methods?
@DrStructure7 жыл бұрын
No, we don't, not at the present time.
@umerjavaid62617 жыл бұрын
thank you for saving my life doc
@royarmstrong6469 жыл бұрын
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.
@DrStructure9 жыл бұрын
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.
@NeonPinkClouds8 жыл бұрын
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:
@DrStructure8 жыл бұрын
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.
@NeonPinkClouds7 жыл бұрын
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
@DrStructure7 жыл бұрын
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.
@NeonPinkClouds7 жыл бұрын
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
@DrStructure7 жыл бұрын
Yes, you got it!
@travisearnshaw82874 жыл бұрын
How do you get 3.55 from equations 5 and 6?
@DrStructure4 жыл бұрын
It should have been written as 3.54, not 3.55. 5/(2*0.7071) = 3.5355 ~ 3.54
@travisearnshaw82874 жыл бұрын
@@DrStructure thank you alot, that makes sense now
@twentyonetwos39679 жыл бұрын
Very informative, but I struggle to see how you arrive at the 3.55NMs in equation 5 and 6.
@DrStructure9 жыл бұрын
+Twentyone twos Yes, the value is a bit off. It should be 3.54 ( 5/(2*0.707)).
@eng7635 жыл бұрын
Thinks my dear
@amitbhatte56218 жыл бұрын
nice explanation, excellent way you teach. pl. continue with diff analytical. do you have any analytical on finding natural frequencies of structures. pl. guide.
@DrStructure8 жыл бұрын
Not at the present time.
@luisalves196910 жыл бұрын
which sofwtare name used?
@DrStructure10 жыл бұрын
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.
@zaidradwan40005 жыл бұрын
Thanx
@ChuTheHuan10 жыл бұрын
great instruction, thank you very much !
@MrKINGKIDDO10 жыл бұрын
Great video thank you for this
@Komain7210 жыл бұрын
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! :)
@bahaahc6189 жыл бұрын
Tnx alottt u r my statics hero
@lordbatuhansyn8 жыл бұрын
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
@karenmatimura48215 жыл бұрын
Ohhhh I love you 😩😩😩
@kunal_chand8 жыл бұрын
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.
@kunal_chand8 жыл бұрын
sorry about the spelling mistake.
@DrStructure8 жыл бұрын
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.
@simrandhonchic61388 жыл бұрын
It helped me a lot. Thank you very much :)
@explorecivilengineering98357 жыл бұрын
Dear sir how can we know that which member is in tension and which member is in compression?
@DrStructure7 жыл бұрын
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.
@rajakana320210 жыл бұрын
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 ..
@DrStructure10 жыл бұрын
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.
@husnainhyder67139 жыл бұрын
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
@DrStructure9 жыл бұрын
+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).
@akuznet37988 жыл бұрын
I am also confused by this. What would Fbc be if the angle shown was not 45?
@DrStructure8 жыл бұрын
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.
@tonyt508 жыл бұрын
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
@DrStructure8 жыл бұрын
Since we have an isosceles right triangle here, sine and cosine can be used interchangeably.
@mmdandf9 жыл бұрын
I do not get one thing? Why do you say it automatically satisfies the moment equilibrium when you have a vertical reaction there?
@DrStructure9 жыл бұрын
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.
@donsamson40377 жыл бұрын
In the last example, why AC in tension while BC in compression?
@DrStructure7 жыл бұрын
Intuitively, because the applied horizontal force tends to push on BC and pull on AC.
@Hobbit1836 жыл бұрын
That was super neat, thank you 🙃
@accessuploads78346 жыл бұрын
thanks a lot
@abdullahalsalboukh74389 жыл бұрын
You are the best. Thank you a lot :)
@hiporikonachi8 жыл бұрын
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?
@DrStructure8 жыл бұрын
VideoScribe for text animation CrazyTalk for voice/face sync
@ameyakamat49859 жыл бұрын
Hello Dr., I can always work out the reactions before hand and then start working out from the corner right?
@DrStructure9 жыл бұрын
Ameya Kamat Yes, you can certainly approach the problem that way. It is generally a faster technique, if you know what you are doing.
@olajideolagunju90389 жыл бұрын
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
@tinashematimba68239 жыл бұрын
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
@DrStructure9 жыл бұрын
+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).
@tinashematimba68239 жыл бұрын
thank you so much Dr Structure.looking forward to your posts
@nnixxpalomares47959 жыл бұрын
can i use this method if the bottom member is not in 180 deg ?
@nnixxpalomares47959 жыл бұрын
+Nnixx Palomares there are 2 members at the bottom but they are not in 180 deg.
@DrStructure9 жыл бұрын
+Nnixx Palomares Yes, regardless of the geometry of the truss, the analysis technique works as long as the structure is statically determinate.
@kimerriobaugh779 жыл бұрын
what if u do not have right angle triangles can the same approach taken?
@DrStructure9 жыл бұрын
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.
@kishore10a8 жыл бұрын
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?
@DrStructure8 жыл бұрын
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.
@michaelpolowyszak1228 жыл бұрын
If a joint has more than 2 unknowns it cant really be solved.
@DrStructure8 жыл бұрын
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.
@nnixxpalomares47959 жыл бұрын
Dr. Structure how i know if i should use sin of cos in Fx and Fy?
@DrStructure9 жыл бұрын
+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).
@nnixxpalomares47959 жыл бұрын
+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.
@nnixxpalomares47959 жыл бұрын
+Dr. Structure it confuses me because in this example problem you have 45deg angle but now i am enlightened haha =))
@accessuploads78346 жыл бұрын
It is clarified
@PBSumabat1710 жыл бұрын
may i ask, what if the distance of x's and y's are given, instead of using cos and sin, ?
@DrStructure10 жыл бұрын
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
@PBSumabat1710 жыл бұрын
Thanks :)
@mohamedalwrfi93058 жыл бұрын
شكرا جزيلا
@christialjewelvergara91468 жыл бұрын
How do you get Fac=3.55N and Fbc=3.55N?
@DrStructure8 жыл бұрын
5 / (2 cos(45)) = 3.54 There discrepancy is due to round off error.
@wklzai9 жыл бұрын
can we calculate the support reaction forces first before we get into each of the joints?
@DrStructure9 жыл бұрын
+Frez Wong Yes.
@yukinatakashi617210 жыл бұрын
Thank you very much
@yenumovie30959 жыл бұрын
how do we know the langthe of the two sides are equal .w/h is "l"?
@DrStructure9 жыл бұрын
+yenus aminu That information is given to us.
@slimmy6668 жыл бұрын
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
@eng7635 жыл бұрын
Good afternoon my dear can make KZbin taking about finite element method
@gurwindersingh33749 жыл бұрын
how do we know the member is in tension or compression?
@DrStructure9 жыл бұрын
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.
@gurwindersingh33749 жыл бұрын
Thank you dr structure . .this helped me a lot..great explanation..
@awstout110 жыл бұрын
@ 16.22, how do you figure .707Fbc-.707Fac + 5 = 0 becomes 3.55 N for both FAC & FBC how?
@DrStructure10 жыл бұрын
@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.
@BigOlRub10 жыл бұрын
Dr. Structure Aren't they supposed to be 3.53606... the same with Fbd, it should be 2.83...
@DrStructure10 жыл бұрын
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
@maxameddahir47298 жыл бұрын
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
@DrStructure8 жыл бұрын
Russell Hibbeler's books on statics and structural analysis are quite good.
@aroyb-TECH8 жыл бұрын
This is Brilliant work. Do you take donations for support?
@DrStructure8 жыл бұрын
Thank you for your post. Currently, there is no mechanism in place for accepting donations in support of the work.
@eduforalledu46517 жыл бұрын
I need to know the name of this program please
@DrStructure7 жыл бұрын
I am not sure what you mean by the program. Please elaborate.
@carlson25459 жыл бұрын
shouldnt there be a zero force member at D
@DrStructure9 жыл бұрын
+Carlson Ngolah Yes, CD is a zero-force member.
@uayemyintbago44968 жыл бұрын
Grate Thanks
@nequefruto77139 жыл бұрын
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.
@DrStructure9 жыл бұрын
+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.
@Alalpog9 жыл бұрын
thank you madam.
@astafzciba8 жыл бұрын
the the cartoon character looks so high, haha thank you so much for the explanation
@duckyprime8338 жыл бұрын
how to get ac=3.55n and bc too
@DrStructure8 жыл бұрын
Please elaborate. Not sure what you are asking.
@y_p77 жыл бұрын
Thank you so much! !! You're a lifesaver
@eng.elhosary879011 жыл бұрын
Thanks
@masoodzafar97318 жыл бұрын
thanks sir
@engrxiddig3 жыл бұрын
B+r=2j 3:00
@chruntola31814 жыл бұрын
Hello Teacher please drop file pdf all lesson .Thank you very much Teacher.🙏❤