L5 Solving for Reactions

Hinzugefügt: 2026-05-30

[00:00:05]now that we have reviewed our boundary conditions let's try solving for some reactions before we dive in here note how we have defined our positive directions positive X is to the right positive Y is up and positive moment about the z-axis is in the counterclockwise Direction so let's start with this first example in the top left here we have a member supported by a pin on the left and a roller on the right and we have placed a point load at Mid span this member here could be a girder with a beam framed into it where we are getting our Point load from and these supports could be columns on either side so now let's solve for our reactions in terms of our Point load and we do so with our three equilibrium equations first we have sum of forces in the x equals zero and in this example and indeed as we'll see all of our examples here we only have one force in the X Direction our reaction at the PIN there is no applied force in the horizontal Direction and so as this is the only force in the horizontal Direction and the sum of all forces we know is equal to zero this Force must be equal to zero now let's sum forces in the y direction here we have our reaction in the y direction at our first support and our reaction in the y direction at our second support both in the positive y direction next we have our Point load in the negative y direction and this is equal to zero now we adjust with some slight algebra ry1 plus ry2 is equal to our Point load p and that's as far as we can get with this equation so let's jump to our third equation Our Moment equation sum of moment about the Z axis and recall that we can take this sum of moments at any location along this member so let's take it at this point here so now our sum of moments will be p times its perpendicular distance to the point L over two and this point load will cause a rotation in this direction about that point which is the clockwise Direction and so this moment will be negative next we have r y two times its perpendicular distance the total length l and ry2 is creating a moment in the counterclockwise Direction and so it will be positive is equal to zero again we apply algebra ry2 l is equal to p L over two divide by L on both sides and we get r y two is equal to P over 2.

[00:03:51]now we take this equation and plug it back into our sum of Y equation and we get r y one plus P2 is equal to p and so that means that ry1 will also be equal to P over 2.

[00:04:16]and so both of these reactions are equal to P over 2.

[00:04:22]next let's try an example where we also have a point load but this time that point load is located a distance a from the support on the left and a distance B from the support on the right and note that L is equal to a plus b similar to our last example we will have sum of forces in the Y is equal to ry1 plus ry2 minus p is equal to zero and so we will have ry1 Plus ry2 is equal to p next with our sum of moment in the Z Direction we will take the moment at the same location as our last example at our pin support on the left we will now have P times a and again that will be negative Plus ry2 times l equal to zero and here we apply simple algebra ry2 times L is equal to P times a and we get that r y two is equal to P times a over l as before we plug this equation into here and we get r y one plus P times a over l equal to p and we can arrange that so that ry1 is equal to P minus P times a over l and we can take this further with a few more algebraic steps if we say that P times 1 minus a over l so simply extracting the common factor here of P and now we can also say that this expression is equal to P times in parentheses L over l which is of course equal to one minus a over l close the parentheses and now this is equal to P times parentheses L minus a over l and now recall that L is equal to a minus B and so we can plug that back in and we get a plus B minus a over l so finally got us keep scrolling to the right here we have P times B over L is equal to our r y one and our ry2 with P times a over l here so this is p times B over L and this is p times a over l and hopefully when you first looked at this you set up the expectation that the support on the left here would be larger than the support on the right because our applied force is closer to our support on the left and so we can quickly check our equations based on that expectation we have P times the greater distance B over l for our reaction one and we have P times the shorter distance a over L for our reaction two and so yes we meet that expectation our reaction on the left will be larger than our reaction on the right now let's continue to a distributed load and the units for our distributed load will typically be in Kips per foot and the way that we will solve for our reactions in this case is we will first find an equivalent Point load at the centroid of this distributed load and so our equivalent Point load will be equal to since this is a uniform load W Times l and now you can see that this setup is exactly the same as our first diagram up here and so we can base all of our equations off of what we have found already so ry2 will be equal to P over 2 or WL over 2.

[00:09:33]and ry1 will be equal to p over 2 or again WL over 2.

[00:09:42]w l over two WL over 2.

[00:09:47]next we have a distributed load but this one is not uniform it is a triangular distributed load as in the previous problem we will find an equivalent Point load at the centroid of the shape of this distributed load and in the case of a triangle that centroid occurs at two-thirds l or on this side one-third l and the magnitude of this equivalent Point load will be equal to W Times l over 2.

[00:10:32]or the area of a triangle width times height divided by two and so now this problem looks a lot like our previous example above it and so we can use those formulas here so our Ry 1 will be equal to note that ry1 is now the further distance away from our Point load and so it will be the smaller reaction so it will be the P times a over l and we can substitute now that we know p and a in different terms for this problem p is now W Times L over two a is now L over three and the whole thing gets divided by L so now solving the algebra here one of these L's will cancel out and we will be left with W Times L over six for our Ry one now ry2 will be equal to P times the other length B the longer length over l p is of course W Times L over two B is two-thirds times l and the whole thing gets divided by L and we end up with w l over three is equal to r y two lastly let's look at these examples which have fixed conditions at either end now in addition to drawing a vertical reaction and a horizontal reaction we also draw a moment reaction and so when we sum our forces in the y direction we have ry1 minus P equals zero and we can very quickly solve for ry1 is equal to p lastly for this example let's solve for Our Moment reaction m z so we'll take our sum of moments about the z-axis at the same location as before at our support and that will be equal to our reaction moment m z which has been drawn in the counterclockwise Direction and so it will be positive and then we will have Our Moment due to our applied force which is creating a clockwise moment about our Point here so minus P times the perpendicular distance to our Point l is equal to zero and so that gives us that our reaction m z is equal to P times l I didn't really give myself much room here but MZ equals P times l now our final problem as before with our distributed load we will start by making an equivalent Point load at the centroid of the distributed load and so that equivalent Point load will be equal to the distributed load times the length now we sum forces in the Y r y one minus W Times L equals zero gives us ry1 equals W Times l for our reaction in the Y as for our moment reaction sum of moment in the Z again m z our reaction is positive here and the moment due to our applied load will be negative W Times l multiplied by its perpendicular distance to our support where we are taking our moment times L over two equals zero and that gives us m z equal to WL squared over 2.

[00:15:47]squared over 2.

[00:15:50]so now we have a few equations for reactions given various loading conditions in the next video we will use these formulas we've derived to draw shear and moment diagrams