Thanks and welcome

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.

kzfaq.info/get/bejne/Y5hkYJdkybTDYpc.html could you support me

thnk U sir may U explain us RCC concrete design

please get in Turkish subtitles

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.

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.

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

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.

+Twentyone twos Yes, the value is a bit off. It should be 3.54 ( 5/(2*0.707)).

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.

No, we don't, not at the present time.

Not at the present time.

VideoScribe for text animation CrazyTalk for voice/face sync

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

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.

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

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.

+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.

Ameya Kamat Yes, you can certainly approach the problem that way. It is generally a faster technique, if you know what you are doing.

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!

Yes, you are correct.

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.

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.

Intuitively, because the applied horizontal force tends to push on BC and pull on AC.

+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.

5 / (2 cos(45)) = 3.54 There discrepancy is due to round off error.

Russell Hibbeler's books on statics and structural analysis are quite good.

Thank you for your post. Currently, there is no mechanism in place for accepting donations in support of the work.

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.

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.

+yenus aminu That information is given to us.

Please elaborate. Not sure what you are asking.

+Frez Wong Yes.

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.

+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

@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

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 :)

I am not sure what you mean by the program. Please elaborate.

+Carlson Ngolah Yes, CD is a zero-force member.

+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.

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..

+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 =))

+cumpleted mczella cos(45) = sin(45) = 0.707