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TRUSSES - Method of Joints & Method of Sections in 12 Minutes! | COFYT

TRUSSES - Method of Joints & Method of Sections in 12 Minutes!

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TRUSSES - Method of Joints & Method of Sections in 12 Minutes!

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0:00 to find the stresses within a structure 0:02 or a member of a structure 0:04 we typically start by finding the forces 0:06 that they are subjected to 0:08 we find these stresses under operation 0:10 to compare them to the material 0:12 properties of the members 0:13 so that we can determine if the 0:15 structure will fail or if it will be 0:17 safe 0:17 and if the structure is in fact safe we 0:20 can even report what we call a factor of 0:22 safety 0:23 but you'll learn more about that in a 0:24 mechanics of materials course 0:26 for now that first step is being able to 0:29 figure out how external loads cause 0:31 reaction forces 0:32 and how internal forces within members 0:34 can be calculated 0:35 to do all of that we first need to 0:37 understand how force vectors are 0:39 decomposed into orthogonal components 0:41 so that we can perform basic operations 0:44 like addition subtraction 0:45 and further into the course cross 0:47 products for torques and moments 0:49 force vectors have a magnitude and a Force Vectors 0:52 direction 0:53 the notation includes a small arrow 0:55 above the load variable 0:56 like f or p and the numbers refer to 0:59 that magnitude in force units 1:01 like newtons or pounds and the direction 1:03 usually in degrees not radians 1:06 even though in math focused courses this 1:08 direction is 1:09 absolute with the positive orientation 1:11 of the horizontal line being zero 1:13 and angles going counterclockwise 1:15 increase in value 1:16 for statics the angle or direction of a 1:19 force vector 1:19 can be with respect to any linear 1:22 geometry of the setup 1:23 besides this notation vectors can be 1:26 represented with their three-dimensional 1:28 components in cartesian coordinates 1:30 and although we will cover this much 1:32 later in cylindrical or spherical 1:34 coordinates too 1:35 a force vector can therefore be 1:37 presented as its components magnitude 1:39 in the x y and z axis with the 1:42 circumflex mark above the i 1:44 j and k directional unit vectors 1:47 respectively 1:48 or separately as f sub x f sub 1:51 y and f sub z finding these components 1:55 is useful because what we will use to 1:57 find the reaction forces 1:58 and with them the internal forces is 2:01 stating that a simple sum of forces in 2:03 the x 2:04 y or z direction and later in the course 2:07 sum of moments and torques 2:08 are equal to zero since in most static 2:11 scenarios 2:12 the members of the structure are not 2:14 accelerating 2:15 when given a force vector with its 2:17 magnitude and direction 2:18 simple trig functions like sine cosine 2:21 and tangent are 2:22 used to find the vector components to 2:24 find the force vector magnitude and 2:26 direction 2:27 based on its components which is the 2:28 opposite process 2:30 we use the pythagorean theorem if you 2:32 want to review those basic concepts 2:34 you'll find some links in the 2:36 description of this video so let's start Vector Components in 2D 2:38 with the 2d force vector f 2:40 5 newtons in magnitude at 53.13 degrees 2:45 this force vector can be written as 3i 2:48 plus 4j 2:49 newtons or we can say that fx is 3 2:52 newtons 2:52 and fy is 4 newtons to find these values 2:56 here 2:56 we are using the definition of sine 2:58 which is the length of the side opposite 3:00 to the angle you know 3:02 over the hypotenuse and cosine which is 3:05 the length of the adjacent side to the 3:07 angle you know 3:08 over the hypotenuse if in this case sine 3:11 of 53.13 3:12 is fy over the magnitude of f and cosine 3:16 of 53.13 is fx over the magnitude of f 3:20 then fy is f sine of 53.13 3:24 and fx is f cosine of 53.13 3:29 now i say in this case because sine is 3:32 not necessarily the y component and 3:34 cosine the x component 3:36 if we had been given the complementary 3:38 angle 36.87 3:41 then fy would be f cosine of 36.87 3:45 and fx would be f sine of 36.87 3:50 notice that in the first case i used 3:52 sine for y 3:53 and in the second sine for x but this of 3:56 course matches our values 3:58 since the sine of theta is equal to the 4:00 cosine of the complementary angle 4:03 this means that depending on the angle 4:05 you are given or the angle you can 4:07 measure 4:07 sine and cosine can be used to find any 4:10 of the components 4:11 this is obviously especially true for 3d 4:14 vectors 4:14 as there will even be z components so of 4:17 course we cannot generalize 4:19 y and x for sine or cosine given the From Vector Components to Vector 4:22 components of a vector 4:24 we can also find the vector's magnitude 4:26 and direction 4:27 since the components are orthogonal 4:29 meaning that they form a 90 degree angle 4:32 between them 4:32 we can use pythagoras to find the 4:34 hypotenuse which would be the magnitude 4:36 of the vector 4:38 to find the direction of the vector we 4:40 would use the trig function tangent 4:42 which is equal to the length of the 4:44 opposite side of the given angle 4:45 over the length of the side adjacent to 4:48 that angle 4:49 and therefore f y over fx the angle 4:52 would therefore be 4:53 the arc tangent of f y over f x 4:56 now this angle can be tricky since if we 4:58 had for example 5:00 f x equal to minus 5 and f y equal to 5:03 minus 12 5:04 solving for the angle mathematically 5:06 would still give us a value between 5:08 0 and 90 degrees when in fact we know 5:11 that the angle from the positive 5:13 direction of the horizontal line 5:15 which i just mentioned was the absolute 5:17 reference for angles 5:18 would be between 180 and 270 degrees 5:23 you can check out a trig related video 5:25 link in the description below 5:26 if you want to brush up on this but for 5:28 static scores purposes 5:30 just remember that you need to identify 5:32 the quadrant in which the resulting 5:34 vector is pointing towards 5:36 we use these two concepts decomposing Sum of Vectors 5:38 vectors and finding the resulting vector 5:40 from its components 5:42 when adding vector forces notice that 5:44 the sine of 30 which is the angle we 5:47 were given 5:47 is opposite over hypotenuse in this case 5:50 fx over f 5:52 and cosine of 30 is adjacent over 5:54 hypotenuse in this case 5:56 f y over f so f x is using 5:59 sine instead of cosine as opposed to the 6:02 example before 6:03 doing the same for p and now with the 6:05 components of 6:06 f and the components of p we can add 6:09 forces in the x 6:10 direction and forces in the y direction 6:12 to have the components of the resulting 6:15 force 6:15 v to find the magnitude of v we use the 6:18 pythagorean theorem and with the tangent 6:20 function 6:21 we find that the resulting vector is 6:23 angled at 52 degrees 6:25 worth pointing here is that the Negative Magnitude Vectors 6:27 direction and the orientation of a 6:29 vector 6:30 are two distinct properties adding these 6:32 two vectors graphically 6:34 so that we can identify our resulting 6:36 vector v 6:37 we can see that a vector of minus 21.13 6:41 pounds 6:41 at 232 degrees is equivalent 6:45 to the force vector v we just found this 6:48 will be very important 6:50 when finding reaction and internal loads 6:52 if the direction of a vector you're 6:54 trying to find is not obvious from the 6:56 setup you're given 6:57 you can always assume a direction and if 7:00 the magnitude results in a negative 7:02 value it means that the direction was 7:04 the opposite of what was assumed 7:06 we'll talk more about which directions 7:08 to always assume to avoid extra work 7:10 or confusion in later videos these same 3D Vectors and 3D Components 7:14 trig functions and pythagorean theorem 7:16 are used when dealing with 3d vectors 7:19 the components of a given vector can be 7:21 found using sines and cosines 7:23 in this case with two angles and of 7:26 course just one magnitude value 7:28 the y component of f would be equal to 7:30 50 cosine of 30 7:32 since the y component is the adjacent 7:34 side and therefore cosine 7:36 of the 30 degree angle on the other hand 7:39 the opposite side of the 30 degrees 7:42 which is at the top 7:43 but also at the bottom that is 50 sine 7:46 of 30 7:47 would be the hypotenuse of the right 7:49 triangle that is found on the horizontal 7:51 plane 7:52 with that hypotenuse and the angle shown 7:55 we can decompose that into the x 7:57 component of f with 25 cosine of 1626 8:02 and into the z component of f with 25 8:05 sine of 1626 when finding these values 8:10 you have to remember that the component 8:12 of z is going in the negative direction 8:14 of the z 8:15 axis so if we're presenting the vector f 8:18 with the i j and k notation we would 8:21 write it as 24 8:22 i plus 25 square root of 3 j minus 8:26 seven k now if we're only writing its 8:29 components 8:30 f x f y and f z f z 8:33 can be either negative if there is no 8:35 graphical representation of where the 8:37 vectors are located 8:38 so that anyone reading the information 8:40 knows that the z 8:41 component is negative or positive if the 8:44 vector component 8:45 f z is shown like it is in this case 8:48 here 8:49 remember that a vector in one direction 8:51 with a negative sign 8:52 is the same as a vector with a positive 8:55 sign in the opposite direction Lecture Example 8:57 so now let's take a look at a brief 8:59 example where we put into use what we've 9:01 covered here today 9:02 for the force vectors f and p shown 9:05 what is the magnitude and direction of 9:08 an additional force 9:09 v so that all three forces added 9:11 together 9:12 would be equal to zero before watching 9:15 the solution to this example 9:16 pause this video and try working on it 9:19 and solving it yourself 9:21 we know that when adding vectors we add 9:23 their components 9:24 separately and if we want the resulting 9:26 vector to be 9:27 zero it means that its components are 9:30 also zero 9:31 to find the magnitude of v we will first 9:33 need to find its components 9:35 v x and v y and to do that we need to 9:38 find the components of 9:40 forces f and p the x component of 9:43 f will use the trig function sine since 9:45 that component is the side opposite to 9:48 the angle 30 9:49 and f y is equal to f cosine of 30 9:53 and since the x component of f is 9:55 pointing left 9:56 we know that f x is negative for vector 9:59 p 10:00 we will use a right triangle that has an 10:02 angle 10:03 equal to 180 minus 135 10:07 p x would be the adjacent side and p y 10:10 the side opposite to that angle and 10:12 again since the x component is pointing 10:15 in the negative x axis direction px will 10:18 be negative 10:19 substituting these values we find the 10:21 components of vector v 10:23 with pythagoras we find its magnitude 10:26 and with tangent 10:27 the angle of orientation the angle that 10:30 we define as theta 10:31 can be that between the vertical line 10:33 and the vector 10:34 or between the horizontal line and the 10:36 vector 10:37 in the first case we would have the x 10:39 component over the y component 10:41 and if we're just looking at the lengths 10:43 of a triangle 10:44 they would both be positive including v 10:47 y and in the second case 10:49 tangent of phi would be equal to v y 10:51 over v 10:52 x the overall direction meaning the 10:55 value of the angle with respect to the 10:57 horizontal line would therefore be 10:59 270 plus theta or minus 11:02 phi the process we've done here today is 11:05 very similar to what we will do to find 11:07 reaction forces 11:08 or internal forces given some external 11:11 forces 11:12 and the static or non-accelerating 11:14 structure 11:16 for more examples on force vectors and 11:18 vector components 11:19 make sure to check out the links in the 11:21 description below 11:22 the link to the next main video is also 11:25 in the description 11:28 thanks for watching

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About this video

Video Title: Force Vectors: Components & Vector Addition (Engineering Statics)
Channel: Less Boring Lectures
Speakers: [Speaker's name not available in transcript]
Duration: 00:11:28

Overview

This video explains how to find stresses within a structure by first calculating the forces acting on its members. It covers the concepts of force vectors, decomposing them into orthogonal components using trigonometry, and then recombining these components to find the resultant vector. The video demonstrates these principles with 2D and 3D examples, emphasizing their application in determining reaction and internal forces in statics.

Key takeaways

Understanding Force Vectors: Force vectors have both magnitude and direction, and can be represented numerically or with components along Cartesian axes (i, j, k).
Decomposing Vectors: Trigonometric functions (sine, cosine, tangent) are used to break down a force vector into its orthogonal components (e.g., Fx, Fy, Fz). The choice of sine or cosine depends on the angle given and which component is being calculated.
Reconstructing Vectors: The Pythagorean theorem is used to find the magnitude of a vector from its components, and the arctangent function is used to determine its direction, with careful consideration of the vector's quadrant.
Vector Addition: Vectors can be added by summing their corresponding components separately. The resulting components can then be used to find the magnitude and direction of the resultant vector.
Application in Statics: The ability to decompose, reconstruct, and add vectors is fundamental for calculating reaction forces and internal forces in static structures where members are not accelerating.

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How are force vectors represented in Cartesian coordinates?
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