L4 Static Equilibrium

本站添加: 2026-05-30

[00:00:01]okay Lesson Four breaking into structural analysis I'm going to try to go a little easy on you guys with this one because I know that the content for this week was a little heavy with those lateral loads so let's jump into it this lesson and the next two weeks leading up to our midterm exam we'll cover a few basic principles of structural analysis today we will focus on static equilibrium and axial forces we will briefly discuss material properties in response to these forces to prepare us for a final section of the course which will cover the four main structural building materials as we've mentioned before structures undergo Dynamic motion when subjected to Dynamic loads so structures are not completely static all the time however for most buildings we can simplify our analysis to static procedures static means that the structure does not move in other words the structure is in equilibrium when a structure is in equilibrium that means that all forces in all directions are equal the loads that we apply must then have equal and opposite reactions for the building to be in static equilibrium hopefully this sounds familiar from structures one so for a two-dimensional body there are three ways that this body can move it can move up and down left to right and it can twist we can also say that a two-dimensional body has three degrees of freedom in order for this body to be in equilibrium the summation of all forces applied to it in these three directions must be equal to zero a three-dimensional body has six degrees of freedom and for this body to remain in static equilibrium again all of these equations must equal zero foreign now let's look at an example of a wall that has two forces applied to it and three reaction forces two vertical and one horizontal what would these reaction forces need to be in order for this wall to be in static equilibrium first let's start by defining our coordinate system for this we'll follow the typical convention with up being the positive y direction to the right being the positive X Direction and counterclockwise being the positive direction about the z-axis next let's write out our equations we have sum of forces in the x equals zero sum of forces in the y equals zero and sum of moments about the z-axis equals zero let's start with the easy one we have only one force in the X Direction and One reaction in the X Direction so sum of forces in the x is equal to 10 caps plus f x and that is equal to zero now with some simple algebra we have that f x is equal to negative 10 kips and this negative just means that the direction that we drew this Arrow pointed to the right is actually incorrect and this force will actually be in that direction the negative X Direction so this FX is negative 10 kips next let's look at our sum of forces in the y direction in this direction we have 20 Kips down so negative 20 kips Plus fy1 and fy2 and that is equal to zero again we apply some simple algebra FY 1 plus FY two is equal to 20 kips now we can't solve this equation because we have two unknowns so we'll put this to the side and move on to Our Moment equation and I'm going to try to shrink all this actually so I can fit it all on one page recall that moment is basically just a fancy word for twisting and we Define it in terms of a force times a distance is equal to a moment and note that with this equation the sum of moments in the Z direction we can apply this equation at any single point and where we choose to use this equation is going to be very important so for example if we were to apply this equation at the center of our wall here this 20 Kip force would be applied directly in line with that Central Point and so if we were to take this 20 Kips times its perpendicular distance to this central point we would get 20 Kips times zero equals zero so this is going to be crucial for how we solve our problem recall that our previous equation had two unknowns in it if we can pick a point that removes one of these two unknowns from our new equation it will be easy to solve for the other unknown and then come back to this equation to find our final unknown so let's pick this point here to sum our moments and so we get sum of moments in the Z Direction is going to equal let's start with this 10 kips this 10 Kips is going to cause a rotation in the clockwise Direction and so it will be negative 10 kips times its distance three feet next we have this 20 Kip Force again it will cause a rotation about this point in the clockwise Direction and so it will also be negative 20 Kips times its perpendicular distance to that point four feet next we have our fy2 Force and it will cause a rotation in the positive direction and so we have plus fy2 times its perpendicular distance eight feet now these two remaining forces this FX and fy1 are acting through the point that we've Chosen and so the moment that they create at this point will be zero and so that's all of the forces that we need for this equation and this will also be equal to zero and so again applying some simple algebra we end up with fy2 equals 110 foot kips divided by eight feet and that gives us 13.75 kips now we can take this value and plug it back into this formula to get our final fy1 value so fy1 plus 13.75 kips equals 20 kips which gives us fy1 equals 6.25 kips so to summarize our work we have f x equal to negative 10 kips fy1 equal to 6.25 Kips and fy2 equal to 13.75 kips note that if we added an additional reaction say here or anywhere else another unknown we would no longer be able to solve for our reactions with these given equations the system would become statically indeterminate and we would deal with this condition by handing it off to a structural engineer