Lesson 6 Shear and Bending Capacity

Ajouté : 2026-05-30

[00:00:01]now that we have defined our demand loads and the subsequent stresses in Our member we can start to talk about how we resist these sheer and bending moment forces in other words we can find the capacity that we need to design Our member for so let's start with shear and recall that we had a formula for the average shear stress where we took the maximum Shear force and divided it by the area of the cross section and that needs to be less than or equal to our design yield stress multiplied by a factor of safety and we can rearrange this formula to get V times FY times area must be greater than or equal to our demand load so this is our demand load this is our capacity so let's look at this example where we have a maximum shear stress in the beam of 60 kips and just by looking at this Shear diagram we can also infer that this is a simply supported Beam with a uniformly distributed load so Shear capacity is going to be equal to our fee Factor which for shear is equal to 0.75 times our design yield stress FY which is given as 50 Kips per square inch multiplied by our cross-sectional area of 12 inches squared and so this gives us 450 Kips of capacity which is greater than our demand load of 60 Kips and therefore our design is okay now with bending again we had a formula for our maximum bending stress which was equal to our maximum bending moment divided by our plastic section modulus and that also needs to be less than or equal to our design yield stress multiplied by a factor of safety and once again we can rearrange this formula to give us our demand and our capacity note here that with bending we are using a fee factor a factor of safety of 0.9 and compare that to our factor of safety for Shear which was 0.75 so for Shear we are reducing our capacity by 25 for bending we are reducing only by 10 percent and that's because Shear is a much more sudden and catastrophic failure than bending so we need to apply a factor of safety for Shear that will further reduce our capacity whereas with moment a reduction of ten percent is permissible so let's calculate the moment capacity for this beam here we'll have our fee factor of 0.9 times our design yield stress 50 Kips per inch multiplied by our plastic section modulus 15 inches cubed and so we get a value of 675.

[00:03:45]now can we compare this value 675 to 150 and say that our design is okay well not quite because if we look at our units here we'll see that we've ended up with kips times inches and we were given foot kips so we need to convert to foot clips by multiplying by one foot over 12 inches and so we get 56.25 foot caps Which is less than 150 footcaps and so this design would be no good would not be acceptable and so we would need to choose a different beam section with a higher plastic section modulus and so that's how we calculate shear and moment capacities for beams but it's important to remember that yielding is not the only mode of failure when a member is subjected to bending moments and we learned that in our foam demonstration recall that when we bent the member too much it started to pop out of plane and we called that lateral torsional buckling so if a beam has a significantly long unbraced leg then it may Buckle out of plane before it actually starts to yield for this course we will not be checking this mode of failure but it's important to note that it can occur especially when you have a relatively large unbraced leg so if we have a long Beam with nothing framing into it on either side the structural engineer may consider cross bracing to brace that member against lateral torsional buckling and you can see this done very often in wood framed houses for example my own apartment where we have these X pieces and these cross braces are here to prevent the beam from failing due to lateral torsional buckling now finally when we talk about bending we need to also talk about deflection and stiffness stiffness is of course the member's ability to resist deflection and the member's stiffness is based on two primary characteristics first it's geometry or it's moment of inertia next we must also consider the material properties of the member when defining stiffness as we saw with our stress strain curve last week we defined a variable called the modulus of elasticity and this modulus of elasticity is equal to the stress over the strain of the member in other words it provides us with a means to relate the deformation of a member to the load applied and so these are the two variables that Define stiffness mathematically e and I so whenever you see these two variables together in an equation you should think stiffness and if you see these two variables along with some type of Applied loading and a length you should think of deflection and we saw these two variables previously in our buckling formula and buckling of course is a failure mode for compression where the member bends out of plane and so stiffness plays a role in calculating the critical buckling load so now let's look at some equations for calculating deflection and for this course we are not going to derive these formulas we're simply going to list them out and note that we have e times I in all of these equations so let's try writing out an example using this condition here and say we have a uniformly distributed load of seven Kips per foot length of 22 feet an e of 29 000 Kips per square inch which is a typical modulus of elasticity for steel and an i of 212 inches to the fourth so our maximum deflection which occurs at mid-span will be equal to 5 times 7 Kips per foot times 22 feet to the fourth power over 384 times 29 000 Kips per square inch times 212 inches to the fourth now let's take a closer look at units here you'll see that both in the numerator and the denominator we have Kips and so those will cancel out just fine but you can see here that in the numerator we have feet and in the denominator we have inches so we're going to end up with units of feet cubed over inches squared but we want to end up with a value that is expressed in inches so we need to do something about this numerator here to do that we need a conversion factor and so we would need to multiply by 12 inches the quantity cubed over one foot that quantity cubed and so that will be our conversion factor that we apply so now plugging all of that into a calculator we get about six inches of deflection and if this load here this seven Kips per foot if that were our unfactored or service level live load our deflection limit would be L over 360.

[00:10:49]and so we would have a deflection limit of 0.73 inches so our maximum deflection here exceeds our limit and this design would be no good we would need a deeper Beam with a larger moment of inertia here in order to control this deflection now finally let's talk about how we can improve our efficiency of our structural system Structural Engineers are always looking for ways to be more efficient and by efficiency we mean using less material using less material means less cost to the owner directly as well as indirectly through labor costs and transportation of the material less material also generally means more room for the architect and finally less material means less cost to the environment so let's think about how we can be more efficient with our geometric shapes of our beams when resisting bending forces and recall that with bending we are primarily resisting these stresses at the outermost fibers of the member and so if we can group more area at those locations we can likely improve efficiency and so let's take a look at that mathematically let's first consider a simple rectangular shape its area will be equal to the width times the height or 24 inches squared and its moment of inertia which we now know we use to calculate deflection is equal to the width times the height cubed over 12.

[00:12:50]and for this rectangular section we get 288 inches to the fourth now let's take a look at an eye shaped section we calculate its area here by breaking it down into three separate rectangular components and that adds up to an area of 16.25 inches squared now let's try to calculate the moment of inertia for this section and to do so we will need the parallel axis theorem as we did with calculating area we will break down our eye shape into three separate rectangular portions and each of these portions will have its own moment of inertia calculated by BH cubed over 12.

[00:13:45]but we want to consider the moment of inertia of the total section and parallel axis theorem is what allows us to do so by taking the area of each of these components and relating them to the total section by the distance from their individual centroids to the total sections centroid and so our total moment of inertia will be equal to the sum of the BH cubed over 12 of each of these components plus the sum of their areas times the distance from their centroid to the centroid of the total section squared so let's try to calculate that with this example here we will start with our first rectangular section which has a width of six inches times a height of one inch cubed over 12.

[00:14:51]its area is six inches times one inch and its distance from its centroid to the centroid of the total section is going to be 4.75 inches squared Plus let's look at our second rectangular section here the web of the member it has a width of half an inch times a height of eight and a half inches cubed over 12.

[00:15:30]it has an area of 0.5 inches times 8.5 inches and its distance from its centroid to the centroid of the total section is zero and so this term becomes zero Plus now let's look at the third rectangular section which is actually going to be the same as the first because this is a symmetric shape so six inches times one inch cubed over 12. plus six inches times one inch times 4.75 inches squared and be sure to be very careful about which terms you are cubing and which terms you are squaring and so here we get 297 inches to the fourth now if you don't like this parallel axis theorem equation and this 80 squared term there's also another way that we can calculate this moment of inertia for the total section and that is by considering the total area bounded by the extreme dimensions of the object and then subtracting out the negative spaces and so the total moment of inertia calculated this way will be six inches times 10 and a half inches cubed over 12.

[00:17:13]minus these two rectangles on either side are going to be the same so minus 2 times the width of one of these rectangles which will be 2.75 inches times the height of one of those rectangles 8.5 inches cubed over 12.

[00:17:41]and now we get out our calculator and the nice thing about doing it this way is that both terms are cubed so we don't have to worry about mixing up cubes and squares and we see here once again we get 297 inches to the fourth note that for our eye shape we ended up with a slightly larger moment of inertia but we used less area so the eye shape is more efficient for resisting bending forces than the solid rectangular shape and that's the last video for this week I know there was quite a lot of math this week so I expect a lot of questions this week and next week are probably two of the most math intensive weeks of the course but once we get past this structural analysis portion of the course which is mainly directed toward helping you out with the are we can get back to discussing broader concepts of structural engineering and structural systems so with that good morning good afternoon and good night and I'll see you next time