 In this problem, we consider this beam AB, which is being supported at A and B, and we have also distributed load with load intensity equal to W. And we need to find what is the maximum load intensity W allowed for this beam if we know that the maximum normal stress is equal to 100 MPa. And we also know what is the shape of the cross-section, what in this case is a D-shaped cross-section. Then, what is the approach that we want to follow to solve this exercise? Well, it's quite simple. We know that the normal stress, in this case, we don't have any kind of axial load, so this is equal to the bending stress and this is equal to the moment times Y divided by the moment of inertia, right? So in this case, we want to calculate what is the maximum load intensity W. So this is given by the maximum normal stress, so the maximum normal stress is equal to M maximum, which is a function of this load intensity times Y max divided by the moment of inertia. So from this equation, we want to calculate what is this load intensity W. So basically, the maximum moment is a function of the load intensity, so this is equal to sigma max times moment of inertia divided by Y max. So we need to calculate what is this Y max. We need to calculate what is the moment of inertia of the section and we need to calculate what is this function of W. So once we know all that information, we can calculate and we can solve this problem for the load intensity. So let's first start, for instance, calculating what is this moment of inertia. Later we will calculate what is this distance Y max and finally we are going to calculate what is the function of moments for this beam AB. We're going to start calculating the moment of inertia of the section. First, we need to calculate what is the position of the center of mass, the centroid of this section. So we need to calculate what is this distance H. So we'll define here our reference. And we know that the total area times the position of the center of mass of the whole section is equal to the sum of the individual sections. Well, in order to do this, we can split this section into two different areas A1 which will be here. This is A1 and A2, which is this one, A2. Now it's more clear. So the total area is equal to A1 plus A2. So this is equal to the product A1, Y1. This is A1 is equal to 150 times 10 and the centroid of this area 1 is here. So this distance is equal to 5 millimeters. And we have here A2, Y2. This is equal to 90 times 10 times the position of the centroid here. So this is Y2. This is equal to 90 over 245 plus 10, 55 millimeters. So we have that 8 is equal to A1, Y1 plus A2, Y2 divided by the total area. So this is equal to... Now we can calculate the moment of inertia of the section. The moment of inertia is equal to the moment of inertia of the web plus the moment of inertia of the flange. So we're going to start, for example, with the web, the moment of inertia of the web. So this is the section that I have. This is the centroid of the web that we calculated before. And this is the centroid of the whole section. And we have here my dimensions. Remember this is 10 and this is 90. Then we can use the parallel axis theorem so we know that the moment of inertia of the web is equal to the moment of inertia of the web with respect to the centroid of the section. So we know that for a rectangle plus the area of the web times the distance from the centroid of the web to the centroid of the whole section, D squared. So this is equal to... B is equal to 10 times 90 squared divided by 12 plus the area of the web 90 times 10 times this distance here and this is equal to the distance this y-web. So this is equal to 55 minus h, 23.8. And now we can calculate the moment of inertia of the flange. We can use again the parallel axis theorem so the moment of inertia of the flange with respect to its centroid is equal to Be h squared over 12 plus... Now we want to express it with respect to the neutral axis plus a flange times this distance squared. So this is equal to the B squared flange is 150 times 10 squared divided by 12 plus the area of the flange 150 times 10 plus the distance from the centroid of the flange to the centroid of the section. So this is equal to 23.8 which is h minus 5. So if we combine these two results we find that now we know that in order to find this maximum intensity load w we need to know what is y-max, right? So which is the distance from the neutral axis to any of the point of the structure? So remember this is my cross-section. The center, the neutral axis is somewhere here. This is 23.8. And I have that this is 150. So the maximum distance y-max of course will be the distance from here to the bottom surface, to the bottom part. So this is y-max. Then y-max is equal to 150 minus 23.8 and this is equal to 76.2 millimeters. Okay, we already know what is y-max. The moment of inertia now we need to calculate what is the distribution of moments on the beam. So now we have to calculate and later we will calculate what is the maximum moment, okay? Remember that we need to calculate first what are the reaction forces. Then remember this is my beam I have here r a and r b and this is the distributed load and this distance is equal to 1.5 meters. Then from equilibrium I have that the sum of forces in the vertical direction is equal to 0 r a plus r b minus w times 1.5 is equal to 0 and since we have symmetry we have that r a is equal to r b and from these two equations we find that r a is equal to r b and they are equal to from this equation 1.5 divided by 2 times w so this is 0.75 times the load intensity value. Okay, we already know what are the reaction forces now we need to calculate the equilibrium at any point x. So this is my reference x I have this then what is the moment as a function of x at this point so first this reaction force is creating a moment here like this but of course the internal reaction moment it goes in the opposite direction and remember that we always define this to be positive so r a is creating a positive moment so r a times x positive then now we need to take into account the contribution of the distributed load so this is equal to the area of this load this is equal to w times x now we need to multiply times the distance in this case for rectangular distribution this is x divided by 2 x squared divided by 2 and we have here a negative sign so remember this is creating a moment here like this the internal moment distribution is like that for us this is negative this is my moment as a function of x we can develop this a bit more we know r a so this is equal to 0.75 wx minus wx squared divided by 2 now I need to find what is m max and remember that m max appears where this here force is equal to 0 so dm dx is equal to v equal to 0 so what is dm dx we need to take derivatives here in this expression we have that this is equal to w 0.75 here minus x here then from here I find that x equal to 0.75 meters so this is the mid span of the beam so exactly here at this point then of course now we know that m max is equal to the moment at 0.75 meters so if we substitute in this equation we have that then this is the maximum moment and as I said before this is a function of the load intensity w now we have all the ingredients to use the flexural formula remember from the beginning of the exercise that we said that the maximum shear stress is equal to the maximum moment which is a function of this w plus times y max divided by the moment of inertia then from here we have that m max is equal to 0.221w and this is equal to sigma max 100 mega pascals times the moment of inertia divided by y max which is equal to 76.2 meters so if we calculate w from here we have that this is equal to 9.5 kN per meter