 In this problem we have an high steel beam which needs to be reinforced by bonding to plates of the same material to the top and bottom flanges as you can see here, so this one and this one are the reinforcements and we need to determine what is the maximum bending moment and shear force that can be carried by both the unreinforced and the reinforced beam we have here that the maximum normal stress allowed is equal to 160 MPa the maximum shear stress is equal to 75 MPa and in the case of the adhesive that we are going to use here in order to join these parts the maximum shear stress is equal to 5 MPa so we can start calculating what are the shear force and the bending moment the maximum ones that we have for the reinforced and reinforced case we know that the bending stress is given by m times y divided by the moment of inertia so from this formula we have that the maximum bending moment is equal to the maximum stress times the moment of inertia divided by y so from this equation we have that note that we already know what is this moment of inertia of the section it's already given and this distance is the maximum distance then from here to the top part because of course the maximum stress occurs at this and this surface now we can calculate what is the maximum shear force we know that the shear stress is given by v times q divided by the moment of inertia times t then from this formula we have that v is equal to we also know that for a nice section like this one the shear stress profile is something like this then tau is maximum at the neutral axis q is also maximum at the neutral axis then we have to calculate q at the neutral axis so I will draw here the cross section again we know that q is equal to y bar prime times a prime and in this case what I'm going to do is I will divide this area here into two different parts this region here and this region here a1 and a2 then q is equal to a1 prime plus y1 prime bar plus a2 prime plus times y2 prime bar then we have that a1 prime is equal to 20 times 160 times the distance from the centroid of this section to the neutral axis so this is equal to 200 plus this is small distance 10 millimeters and a2 is equal to 15 millimeters here times 200 and y2 prime bar is the distance from the centroid of a2 to the neutral axis so this is equal to 100 and the final result is that q max is equal to then as I said before v max is equal to tau max moment of inertia thickness divided by q max we already know what is the moment of inertia the thickness at the neutral axis is equal to 15 millimeters then this is equal to then this is the maximum shear force for the unreinforced beam and this is the maximum bending moment for the unreinforced beam now we can solve the reinforced case and first we need to determine what is the new moment of inertia so we know that the new moment of inertia is equal to the original moment of inertia plus the moment of inertia of this part plus the moment of inertia of this part so they are symmetric they are the same so 2 times the moment of inertia of the rectangular part and we can use Steiner's theorem or Parallel Axis theorem to translate this moment of inertia to the neutral axis plus a v squared so this is the area of these 2 reinforcements and this is the distance from this axis of symmetry to the neutral axis so if we apply this formula we have that the new moment of inertia is equal to and this is equal to note that to calculate the moment of inertia of the rectangle we have used the formula and this distance as I said is equal to the distance from here to the top part so 220 millimeters plus this small distance which is this side of the rectangle 15 millimeters divided by 2 then now we can calculate what is the maximum bending moment as before we know that the stress is equal to the moment times y divided by the moment of inertia so the maximum bending moment is equal to sigma max times the moment of inertia divided by y max and in this case we have that y is the maximum distance from the neutral axis to the outermost point of the structure so in this case y max is equal to 220 plus 15 millimeters then the maximum bending moment for this reinforced case is equal to 405 kNm now we need to determine what is the maximum shear force we know that the shear force the maximum one is equal to tau times moment of inertia times t divided by q maximum q here and maximum shear force as well but in this case we need to consider both we have two critical cases because we don't know if the whole structure is failing or if the adhesive is failing so we need to calculate what is the maximum shear force for the adhesive and we need to know what is the maximum shear force for the failure of the whole section and we need to determine which of them is the most critical one so we can start for example with the adhesive so this is the cross section and I have here the reinforcement then the adhesive is supplied here so we have that the maximum shear stress for this glue is equal to 5 MPa the thickness of the beam at this point at this plane is equal to 150 mm and the moment of inertia is equal to the new moment of inertia that we calculated before then in order to apply this formula the maximum shear force for the adhesive this is equal to tau max for the adhesive times the moment of inertia times the thickness divided by q of course for the reinforcement part so what is q here in order to calculate q we know that q is equal to y' bar times a' so in this case it is the area of this reinforcement multiplied times the distance from the centroid of this section to the neutral axis of the structure this distance here then this is equal to and now we can calculate what is this maximum shear force so if we substitute these results here we have that Vmax is equal to then this is the maximum shear force 872 kN that this adhesive part can resist now we need to consider the failure of the beam material so we have here beam material as we did before we need to calculate first q and once we know q we know that the maximum shear force is equal to the maximum shear stress for the material in this case times the moment of inertia times the thickness divided by q max in this case we have that the maximum shear stress of the material is equal to 75 MPa the thickness of the material at the point where q is maximum is I'm calling this d is equal to 15 mm and yeah of course the moment of inertia is the moment of inertia of the reinforced beam then let's calculate what is q max for the reinforced case of course we have that q max is equal to the q max of the unreinforced plus the q max of the reinforcement of the reinforcement part so basically q is equal to the q of this part plus q of this part and we already know them we had that q of the unreinforced part the unreinforced section is equal to 9.72 times 10 to the power of minus 4 and we have that q for the reinforcement it was equal to 5.19 to the power of minus 4 so if we sum them we have that is equal to 1.484 times 10 to the minus 3 then we have that the maximum shear force for the reinforcement is equal to then this is the shear force required for the beam to fail and if we compare this result here with this one we can see that this case is the most critical so we take this one and for us this is the maximum shear force for the reinforced beam then as a brief summary of our results we have that for the unreinforced case maximum moment was equal to 264 kNm and the maximum shear force was equal to 420 kN and for the reinforced case we have that the maximum moment it is equal to 405 kNm and the maximum shear force is equal to 451 kN and in the case that the adhesive fails so it fails at 872 kN