Lesson 1 Math Review

Added: 2026-05-30

[00:00:00]okay so let's get into this math review and I apologize if this stuff seems basic and boring I promise we will get into more interesting and challenging topics soon also if you haven't already I would highly recommend that you open up homework number one and browse a few of those questions before you jump into this video just look over each question think about how you would solve that question and what information if any you are lacking to provide an answer then as you listen to this video you'll be able to fill in any blanks you may have as you follow along and you won't have to come back to try to find those answers later so please pause this video if you haven't already look over the homework for about five to ten minutes and then come back so let's start with unit conversions in Structural Engineering defining units and using those units consistently in calculations is crucial mixing up units in calculations can lead to drastically over or under designing a structural component of course over designing your component is okay so long as the owner has a flexible budget which is never so it's actually not really okay but underestimating is a lot less okay as it can lead to structural failures including collapse so units are very very important in simple terms the process of converting units consists of taking a given value defined by a starting unit and multiplying that value by a ratio of a desired new unit to the starting unit to produce the new value defined by the new unit so that might sound really complicated so it's actually probably easier just to look at some examples with numbers so for example if you are given a starting value of two feet and you want to express that value in inches you take the two feet and multiply it by the ratio of inches to feet which is 12.

[00:02:12]to produce 24 inches simple but now let's look at a little more challenging example say we are given a value of 4 000 pounds per square inch so when you see this unit of measure squared we are no longer talking about a distance but we're talking about an area so this would be inches this area would be inches squared so we're talking about a force pounds applied over an area of inches squared and say we want to Define this value of 4 000 pounds per square inch in units of Kips per square foot so we are starting with four thousand pounds per square inch converting to Kips is simple we know that the ratio of pounds per Kip is a thousand and so we have one thousand pounds for every one tip and so our pounds will count cancel out and we'll be left with four tips per square inch now looking at the denominator we know that the ratio of inches to feet is 12.

[00:04:01]but we are starting with inches squared and we need to end with feet squared therefore we need to square both the numerator and denominator of our ratio so we have our ratio of 12 inches for every one foot but we are dealing with area here and our starting unit is squared and so our ratio needs to be squared and so we will end up with four kips per square inch multiplied by 144 inches squared per foot squared and again these inches squared will cancel out and we will get 576 tips per square foot for our final answer so that covers unit conversions now our second fundamental mathematical principle that we will review is force vectors a force Vector as you may recall is a force with both magnitude and direction so often in Structural Engineering we deal with non-orthogonal forces and we need to break down the components of these Force vectors into their orthogonal components there are two primary ways of doing this first through the use of angles and trig functions and second is through similar triangle relationships so let's look at examples of both say we have a force vector defined by a magnitude of 10 Kips and an angle of 53 degrees if we want to find the X component of this Force vector then we need to use appropriate trig functions to find the x or horizontal component so what exactly are trig functions and how do we select the appropriate one to use here well trig functions are basically just simple functions that we use to relate angles and lengths of sides of triangles so you can see here on the right I've drawn a triangle with an angle defined Theta and the sides of the triangle are listed in relation to this given angle we have an opposite side an adjacent side and lastly we have the hypotenuse of the triangle so trig functions provide relationships between these sides and are given angle Theta the sine function will relate the opposite side to the hypotenuse the cosine function relates the adjacent side to the hypotenuse and lastly the tan function or tangent relates the opposite side to the adjacent side you may recall the acronym Soka TOA so how does this apply to force vectors well if we scroll down here can see that if we draw an imaginary line from the end of this Force vector we can see that we've now drawn a triangle with that angle enclosed and so we have a hypotenuse we have an opposite side and we have our adjacent side and it's this adjacent side that lies on our x-axis and so the equation that we need here will relate this angle that's given to the hypotenuse and our adjacent side and of course the equation that does that is this cosine equation so cosine 53 degrees be equal to the adjacent side which is what we want the X component over the hypotenuse which is the magnitude of that Force Vector 10 kips and so we get that the X component we'll call it f x is equal to 10 kips times cosine of 53 degrees which comes out to about six kips now note that this does not mean that all X components will be solved with the cosine function that is not true and that is often where a lot of people get mixed up say for example we were not given this 53 degrees but instead we were given this angle on this side and of course that angle would be 37 degrees now if we enclose our triangle note that we have this side as the opposite and our vertical side as the adjacent and so if we were given this 37 degree angle here to find the X component we would need an equation that relates this angle and our given hypotenuse to the opposite side of the triangle and that equation would be the sine function so sine of 37 would be equal to the X component over 10 kips and we would get that f x is equal to 10 Kips times sine 37.

[00:10:56]should also equal six kips so definitely pay attention to which angle you're given complete the triangle and visualize whether you're looking at the opposite side of the angle or the adjacent side so that you can select the appropriate trig function now if you don't like trig functions there's actually another way to break down Force components with a force vector and that is through similar triangle relationships and often it is this method that is more useful to Structural Engineers especially with truss analysis and that's because the dimensions of the truss are usually given and the lengths of each of the members are given and so rather than looking at the angles of the members and using trig functions we can actually just use the dimensions of these members directly so let's say we have a diagonal structural member and we know that its slope is defined by four over three a rise of 4 over a run of three so let's say that this 10 Kip force is an axial Force that is acting along the length of the member or the hypotenuse of our triangle because everything is based on given ratios the process of solving Force components with similar triangles is actually similar to unit conversions we know that the ratio of the vertical to the horizontal component is four over three and we currently have the magnitude along the hypotenuse therefore we need to start by solving for the hypotenuse of our similar triangle in order to establish a relationship between the given vector and the horizontal and vertical components so to do this we start with Pythagorean theorem so we have 3 squared plus 4 squared equals 5.

[00:13:21]and so the length of the hypotenuse of our similar triangle is 5.

[00:13:27]now that we have all three sides of our similar triangle defined we can break apart our magnitude value into its components so let's say we want the horizontal component first we're starting with our given value of 10 kips and we want the horizontal component so we need to set up that ratio so the force Vector along the hypotenuse is related to the horizontal component with a ratio of 3 over 5.

[00:14:08]so it's exactly like converting units where in a sense taking away the hypotenuse in the denominator and replacing it with the horizontal component in the numerator and so we get six kips for our horizontal component note that we could also solve very easily for the vertical component we took 10 kips by the ratio of the hypotenuse to the vertical which would be four and we would get eight kips or you could even take the horizontal component sex caps and multiply by a ratio of the horizontal component to the vertical component and again you get eight kips so if you see a problem set up like this where you have a diagonal member with some kind of similar triangle relationship set up and you need to solve for the components what you don't want to do is try to solve for an angle and then use trig functions you will just get yourself very confused what I would personally recommend that you do is that you solve for the similar triangle relationship and use this kind of mathematical procedure but of course it's up to you whatever you are most comfortable with whether it's trig functions or similar triangles ultimately it's up to you whichever you prefer