So, today's session, as I already told you, the first session will be by Dr. Alok Jadusar, who is the Program Director for Pediatric Neuro Radiology Fellowships and he is a radiologist at Annan Robert, Durich Children's Hospital, Chicago. He is Assistant Professor at Northern Western University, Freeberg School of Medicine. So, let us start with Sir's session for today. Hello everyone. It's an absolute pleasure to be speaking at this MRI conference today and I would like to thank Dr. Bhatkar and the organizing team for the invitation. Today I'm going to talk about Pediatric Posteria Fossa Tumors. It's something which we encounter fairly often in our clinical practice, however, it's also a topic that's been evolving very rapidly over the last few years. So, for this talk, I'm going to talk about the classic imaging features of the Posteria Fossa Tumors. I'll provide an overview of the new molecular classification and I'll also discuss how these molecular subgrouping impacts the clinical management as well as the imaging protocols and then I'll briefly touch upon some future directions. To help you have a system tumor for the needing cause of cancer related mortality in children, these are the most common toilet tumors in children, second only to leukemia and lymphoma and they are found for 20 to 25% of all pediatric cancer. Of the CNS tumors, the Posteria Fossa Tumors is a particularly important subgroup which accounts for about 50% of all intracranial pediatric tumors. So, the four big Posteria Fossa Tumors, which we always talk about are measures of last two miles, very better pyrocytic astrocytome miles, Brinsen Pio miles and Epidemo miles and these would be the focus of my talk today. There are certain less common Posteria Fossa Tumors such as ATRT and Choroid Plexus Tumors which should be considered in the differential diagnosis in the appropriate scenarios and I will discuss them briefly. Then there are certain other tumors which are seen only in the setting of genetic tumor. Let's start with Metroloblastoma. Metroloblastoma is the most common malignant brain tumor of childhood. It's also the most common Posteria Fossa Tumor accounting for about 40% of Posteria Fossa Tumors. It's a WHO grade 4 tumor and it belongs to the category of embryology. The age for Metroloblastoma is 4 to 10 years, although it can be seen from instances to a hotel's court. About two-thirds of these tumors are located in the midline involving the court ventricle, although they are centered more towards the group near the word mess. Although one-third of these patients, one-third of these tumors can also be eccentric under the CT-angle system or along the cerebellar peduncle and we'll talk about them in a bit. The emerging appearance of Posteria Fossa Metroloblastoma on the non-contrast head CT we see a hyperdense mat centered in the court ventricle which on MRI demonstrate diffusion restriction, ISO to hyper intensity signal and mild but diffuse post-contrast enhancement. The most important emerging features that help you distinguish Metroloblastoma from other Posteria Fossa Tumor is the higher hyperdensity on CT and the diffusion restriction on MRI. Both of these are features of high singularity and high nuclear sacrifice ratio which is characteristic of these high-grade tumors.  This was an intermediate CE MOSFET cross-copy. However, at a low TEMRS, presence of Torine P has been described as a more specific finding of Metroloblastomas. On the protrusion imaging we see diffusion increase protrusion throughout the tumor. So this Metroloblastoma was a classic subtype based on histopathology. The traditional Metroloblastomas have been classified on the basis of ecology with the classic subtype being most common accounting for about 70% of cases. The second most common is this more plastic or modular which accounts for about 15% of cases although it's the most common subtype in general. And the last subtype, anaplastic or large-cell Metroloblastomas accounts for about 10 to 15% of cases but it has the most common. More recently based on advances in methylation techniques and genetic profiling of these tumors four very distinct subgroups of Metroloblastomas have been described. These are wingless or the rinse which is the least common accounting for about 10% of cases. It's seen in slightly older children with a pK of 10 to 12 years are typically corresponds to the classic histopathological subtype and it has the best problem of the four Metroloblastomas subgroup. Next is the sonic hedgehog which accounts for about 30% of these tumors. This subgroup has a bi-modulate distribution and it's seen in infants and then another pK is seen in adolescents and adults. Histopathological rate corresponds to the anaplastic variant of the Metroloblastoma and it carries an intermediate problem of this. Further the problem of the sonic hedgehog group is dependent on the status of T53 mutation with the T53 gene mutation corresponding to a worst problem of this compared to the wild type. Then we have the group 3 Metroloblastoma which accounts for about 25% of cases typically seen in infants and young children and it has the worst problem of all the four subgroups and finally group 4 which is the most common subgroup accounting for 35% of cases. It's seen in slightly older children with a pK of 10 years and it carries an intermediate. The importance of these molecular subgroups has significantly increased in the recent years and now these are being used for risk stratification and treatment planning. This is also reflected in the most recent WHO classification from 2016 which now divides Metroloblastoma along according to genically defined subgroups as well as histologically defined subgroups. For an individual case the owners are on the pathologist to take into account all the histological immunohistochemical and genetic features and come up with the unifying diagnosis as part of the process. Imaging implications of these subgroups these subgroups tend to have slightly different imaging appearance and also clinical behavior in terms of metastatic spread and recurrence. So the wingless subtype like I said has the best diagnosis and these new opiumes are typically located eccentrically either in the CP angle or in the cerebellar potential region. These are typically diffusely enhancing lesions and metastatic spread is there in these tumors. Tonic hedgehogs like I said corresponds to that there's no plastic variant and accordingly it is seen more peripherally along the cerebellar hemisphere may emerge the duro along the centurium. These are typically enhancing tumors and metastasis is relatively less common in these tumors and recurrence is also uncommon for these tumors but when it occurs it's more likely to be in the local surgical bed. For group 3 and group 4 tumors both of these tumors occur in the midline the big difference is that the group 3 tumors are typically enhancing but the group 4 tumors are non-enhancing. Metastatic disease is very common with both of these subgroups at presentation and when they're recurrent the recurrence is almost always at a distance side rather than in the local surgical bed. Two different patients with different subgroups of mesozoblastoma the patient on the left side has a heterogeneously hyperdense mass centered in the right cerebellar peduncle with factory diffusion restriction and post contrast enhancement also on the T2 weighted images. So based on the location this is most likely a wing subtype which it turned out to be. The tumor on the right side is a lobular hyperdense mass which is along the lateral aspect of the cerebellar hemisphere and it's everything particular. The mass amounts to diffusion restriction, low signal on T2 weighted images and post contrast enhancement. Morphologically this is a plastic subtype and it belongs to the molecular group of chronic hedgehog T53 by subtype. These are two midline of the tumor. So the tumor on the left is a diffusion restricting T2 hyperdense mass involving the post ventricles. The mass also denotes diffuse post contrast enhancement. Also there is some modular enhancement along the cerebellar folia as we can see on this T1 post contrast as well as the post contrast T2 flare imaging. So this is a group 3 mesoblastoma with associated leptomanageal spread at the time of diagnosis. This tumor on the right of the screen is again a diffusion restricting T2 hyperdense mass centered in the post ventricles. But this tumor shows minimal tumor enhancement and this is the corresponding group 4. Certain advanced imaging techniques have been used to try and predict these molecules subgroup. This study was able to distinguish between chronic hedgehog and group 3 flash group 4 tumors using low TE and not spectroscopy. What they found was that chronic hedgehog typically have a prominent hole in peak and a prominent lipid lactate peak but lack of a torian peak while group 3 and group 4 tumors readily demonstrate presence of a torian peak with relatively small lipid lactate peak. Also there are multiple attempts being made to use machine learning and artificial intelligence techniques to distinguish between different molecules of groups of mesoblastomas as well as to predict or respond to treatment. One such study which used radiomic feature extraction technique was able to distinguish between chronic hedgehog and group 4 mesoblastoma. Cs of dissemination like I've said is fairly common in mesoblastoma is seen in about 11 to 43% of the patients. It can be intrapanial and or intrapanial. This is most common in group 3 and group 4 and red in the other two molecules as well. The matter is that metastasis to bone and liver has been described but is a disseminated metastatic disease. The patient on the left side has a small diffusion restricting and enhancing force ventricular mass but extensive leptomanin gel metastatic disease along the cerebellum as well as throughout the spinal cord including the fecal factor may have. Again this is one of the common imaging appearance of group 3 mesoblastoma that the primary tumor can be relatively small even in the presence of extensive metastatic disease. The patient on the right side has a size ventricular tumor which only denotes mild and patchy enhancement but there is extensive metastatic disease towards the cerebellum folia and also surface of the spinal cord. In addition there is this enhancing E2 hyperintense intramedular lesion between the lower spinal cord response to a drop metastasis within the central canal of this spinal cord. As we said group 4 tumors can often be non-enhancing or minimally enhancing and that is also reflected in the metastatic disease. Oftentimes the mess can be non-enhancing for example in this patient it's very difficult to identify the metastatic disease on this post-confrost even weighted imaging. However we can already see this modular mess along the cerebellum folia on the decretion weighted and the post-confrost weighted images. It's very important to obtain the clear imaging post-confrost for these tumor patients and it's also important to look at the diffusion weighted and the post-confrost images very carefully not just in the resection bed but elsewhere to look for the combination. Another technique that can be helpful in problem solving could be diffusion weighted imaging in the spine. For example in this patient there is this modular metastatic focus along the aspect of the thoracic cord which is seen more complicatedly on the diffusion weighted images. Similarly there are two additional modules along the cord ishiana which has been slightly better. Another technique which we have started using routinely in all of the drop metastasis MRF protocol is a 3D high-resolution T2 sequence like a KISS-TSR space. For example in this case we see these two modular foci along the ventral and the topical metastatic which we can also see on these coronal reconstructed images and I think one of the advantages of this technique is that it is 3D and you can get nice axial and coronal reconstructions and personally I have found coronal reconstructions to be very helpful in many of these cases. So these two modules are almost impossible to identify upon the post-confrost T1 weighted images and we can painfully see them on the diffusion weighted images as well. Moving on to next post-confrost tumor which is a cerebellar phylocitic astrocytoma this is the second most common post-confrost tumor accounting for 30-35% of cases. These are benign astrocytic tumors that go at your grade 1 and they typically have an excellent prognosis with resection only without the need for chemotherapy or radiation. The PAH is 5-13 years equally common in male and female 50% of these occur in midline location and 50% are hemispheric. The typical classic imaging appearance of a JPA which is assisted in accounting with a module is seen in about 50% of cases. In about 40-45% cases there is a larger solid component with a central necrosis or cystic area and less than 10% of these tumors can be completely solid and non-necrophic. In sample of a typical JPA we see a mixed solid and cystic mass in the right cerebellar hemisphere with associated base with active hydrocephalus. The fluid component does not suppress completely on flares such as the proteinaceous composition and the solid component is hyper intense on T2 and lack diffusion restriction which are both useful feature in distinguishing it from higher grade tumor. The solid component also demonstrate decreased post-confrost enhancement. This is the second morphological type of JPA where we have a predominantly solid mass with a central necrotic area against the solid masses. Hyper intense on T2 does not demonstrate diffusion restriction as there is diffuse and having post-confrost enhancement. This is a midline posture for the mass which is mixed solid and cystic. The solid portion is hyper intense on T2 does not demonstrate diffusion restriction. I put them on CT and has diffuse post-confrost enhancement. If you just look at T2 and the post-confrost imaging this can be very difficult to distinguish from a major blast trauma however the diffusion rate is and the non-confrost CT images can be very helpful in making that distinction. Brief word about the molecular alterations that can be seen in JPAs. This is the micro-conactivated protein kinase or the MAPK pathway which is probably one of the most important on-prosinic pathway in human cancer. It is implicated in pediatric low grade tumors some of the adult tumors as well as in many cancer predispositions like NF1, tuberculosis, and hematoma syndrome. As far as JPAs are concerned the most important genetic alteration is seen in DPF molecules. So there are two types of B-Ref mutations identified. One is the B-Ref fusion which is seen in about 70 to 80 percent of the co-flip of the JPAs and some authors now equate the presence of B-Ref mutation to the diagnosis of JPA and this is typically associated with better prognosis. Another mutation is the B-Ref M618 point mutation which is seen in only a small percent of the co-flip of the JPA, maybe a slightly higher percent of the super-tentorial JPA like those involving the optic pathway and the hypothalamus. And also this mutation is seen in various other glial and clio-neuromus tumors. This is associated with work prognosis. The important of B-Ref testing is that there are a lot of molecules along the pathway which are targets for chemo-therapeutic agents. For a cerebral JPA, it is sometimes less important because the tumor is restricted completely and may not need chemotherapy but for no great tumors in non-receptible locations like brain stem or discoteltalysis, the importance of B-Ref testing cannot be emphasized enough. Moving on to the next foot-tumor type which is the brain stem clio-neuromus. This is the third most common tumor, accounting for 50 to 30 percent of cases. This is not one tumor entity but it's a heterogeneous group of atrocytic tumors of course, histological subtype and grade is predicted by their location within the brain. So briefly let's talk about the various anatomical locations in the brain stem where the tumor can occur. So the brain stem is divided into mid-brain, pons and medula with the pons being separated from the mid-brain by the pontomism to palyx alkyl and from the medula by the pontum, medula is alkyl. Each of these regions is divided into a ventral or a basal portion and a dorsal portion called tegment and the tegmentum forms the floor of the cerebral aqueduct and the ventricle while the root is formed by the tecton and the cerebral aqueduct. So the most important brain stem tumor is the one that arrives in the basal or the ventral aspect of the palm and is typically called PGR diffused intrinsic point in glioma. These are high-grade tumors who are very aggressive and have an extremely poor prognosis. The second tumor type in the brain stem could involve the mid-brain either in the tecton which are typically grade one tumors, they do not require any treatment other than CSF conversion or they can arise ventrally in the mid-brain when they can also extend into thalamide. These are also typically grade one but can be more infiltrative grade two to three tumors. When these tumors occur in the brain stem, they are more common along the dorsal aspect and they can have an exothermic component extending into the overlying extractal space and can involve the cervical plantar part. And finally the least common group of these dorsal point in tumors, these are lower grade and the main importance is in distinguishing these tumors from the more aggressive ventral counterpart. So the DIPG has been renamed in the most recent WHO classification as the diffuse glioma, which has the signature genetic alteration of K-27 mutation within the HPG source. DIPGs account for 60 to 75 percent of all brain stem tumors, they are typically at-strategic tumors. On histopath they are grade two to four however for practical purposes they are all treated as grade four tumors because of the uniformly poor prognosis. The PK is three to ten years and they are equally common in male and female. The typical clinical presentation of these brain stem gliomas is an axiom cranial novelties like in the seven-year-old child and this is also the typical clinical imaging appearance on MRI. So we have this large D2 hyperintensive infiltrating mass involving the pump, the mass extends specifically into the pre-founding system, increasing the basilar artery. On the sagittal images we can see that the mass is centered more ventrally within the pump with mass effect on the floor of the post ventricle. There is no appropriated diffusion restriction. There is no post contrast enhancement in the mass and the mass also extends into the right cerebellum peduncle and can sometimes also extend into the cerebellum. A lot of imaging features have been studied as predictors of prognosis within DIPG. The two imaging features which have been consistently associated with poorer prognosis are presence of necrosis and ring enhancement and diagnosis and extra-contin extension into the cerebellum peduncle. However, vertical extension into mid-plane on medulla is typically not associated with worse prognosis. Then there are certain advanced imaging parameters which have been associated with worse prognosis which includes high relative CBV and K-tran on initial fusion imaging and lower diffusion value based on ADC histogram. Another thing to remember about these tumors is that the imaging appearance can change rapidly after radiation treatment. For example, these tumors can show increasing areas of enhancement and necrosis after radiation and that should not be confused with progressive tumor. The ADC value is typically decreased after radiation and which may be attributed to increasing rates from the radiation. However, on MR perfusion the CBV values can transiently increase after radiation again. It's something useful to know about as it should not be confused with tumor progression. One of the most useful predictor of tumor versus T2 progression is the tumor volume on the T2 weighted images and also you can use MR spectroscopy as a problem solving technique. I briefly talked about genetic mutations in TIPG so K27M is the characteristic mutation which can involve the 3.1 or 3.3 and there are slight clinical differences between these two subsides. The 3.1 mutation is seen in a slightly younger age group for the 6 years and has a longer survival while the 3.3 mutation is seen in slightly older age group and is associated with a problem. Some tumors can involve the medulla or the mid-brain as you already discussed. These are two different patients. The patient on the left is an 11 year old child who presented with a large expansile T2 apparent as a mass involving the medulla with exoperative total extension as well as extension into this cervical spinal cord. There is no diffusion restriction between the tumor and the follic portion the most heterogeneous but avid post-frontal enhancement. So this is a grade 1 JPA however it is in a non-refectible location. Similarly we have another tumor which is centered in the mid-brain involving both the tectum and the tegmentum as hyper-intensive 22 weighted images no diffusion restriction amongst this patchy post-frontal enhancement against this for the lower-grade tumor. Moving on to the last major post-refective sub-type which is the ependymoma. Ependymomas are the least common accounting for 10 to 24 percent of the tumor. These are clear tumors with ependymal differentiation and they are typically WHO grade 2 but they can be grade 3. The peak age is slightly younger than megaloblastomas and JPA seen in 1 to 5 year age group. Most of them are sporadic although they have been associated with NF2 in addition to schwannomal and meningeomal. Based on the morphology the tumors can be divided into mid-floor type, lateral type and roof type. The mid-floor type is the most common which involves the post-ventricle and it is centered in close relation to the dorsal aspect of the brain stem. Lateral sub-type is slightly less common as it is centered in the CP angle system and the roof type is the least common of the three. On imaging these tumors are typically more heterogeneous than megaloblastomas and JPAs. Calcifications are very common in about half the cases and you may also see hemorrhage and cystic change. Another important imaging feature is this plastic appearance of the tumor where it insinuates to the post-ventricular outflow foramina. On the diffusion weighted imaging these tumors have intermediate diffusion between megaloblastomas and JPAs. So this is a 4 year old child with the typical mid-floor type of ependymal mass. On CT we have a hypodense mass which is gathered from great calcifications and obstetrics as a specialist. On MR we see a T2 hyper intense mass which extends throughout to bilateral foramina of Plushka, extends dorsally to the foramina of Majanti and extends inferiorly into the typical spinal canal. The mass does not demonstrate any diffusion restriction and there is a mild patchy post-ventral. This is an example of the lateral type of ependymal mass which is centered in these about these angles. Again it is a T2 hyper intense mass without any diffusion restriction and then also diffuse both contrast and have some. The lateral type of megaloblastomas typically have a worse problems compared to a mid-floor type partly because of difficulty in obtaining a complete surgical infection given the presence of important neurovascular structures in that location but also partly because of the more aggressive genetic subtype associated with these tumors. Cs of dissemination can occur with ependymoma but it is much less common compared to ependymoma and if you see Cs spread at the time of diagnosis then you should consider the anaplastic sub. The genetic subgroups of ependymomas are not as well defined as megaloblastomas and JPAs however most recently the intratentorial ependymomas have been divided into two major groups the procephalotide A which is also the pediatric type and more commonly seen in mid-lines and the procephalotide B which is called the adult type with more commonly seen in the T3 angle system and it has a worse problem compared to type A. Also based on genetic studies it has been recognized that the intratentorial, procephalotorial as the final ependymoma are distinct disease entities and they have different genetic mutations and these subtypes of ependymoma are most probably going to be included in the next W.H.O. classification which comes out in 2020. A brief word about some of the less common procephalotumors including ATRT and Coralphexas tumors ATRTs are rare tumors seen in children less than two years of age. These are W.H.O. grade 4 tumors characterized by the presence of ratoid cells and the signature genetic mutation is the loss of I and I1 or the DRG1 gene. These can occur both to crime and intratentorially however half of them are located in the intratentorial location. These are on the differential diagnosis called methyloblastomol because of their hyperdensity or PT and presence of deficient restriction on MRI. However these are typically seen in much younger age youth as we can see this patient is probably three months old. So the young age in association with relatively heterogeneous appearance of the tumor can't help you but just the diagnosis of ATRT although you may not be able to completely distinguish it from methyloblastoma. Coralphexas papilloma these can occur in pediatric as well as the age group. Pediatric tumors are more commonly located in the natural ventricle when the adult tumors tend to be located in the forced ventricle. The key imaging features of Coralphexas papilloma is the morphology of the tumor. They have this lobular from like appearance to the margin. These are typically hyper intent demonstrate post-frontal enhancement and speckle calcification is very common in about 25 years. This is an 18 year old with Coralphexas papilloma. We have a hyper dense mass and it is a forced ventricle with multiple parts. The mass is hyper intense on T2 weighted images. There is not enough restriction. So like I said this is lobular and from like morphology of the tumor should be your clue to distinguish like JPA and ependymal mass in this case. When somebody I talked about the four major for safe of the tumors I talked about how location morphology and decrease restriction can be extremely helpful in distinguishing between these peptides. I talked about the evolving molecular classification and how it applies to imaging and finally I spoke about some diagnostic indicators based on imaging. For future even the rapid advances in the molecular subtyping and the treatment protocols of these tumors, our conventional imaging may not provide adequate information to guide the precision medicine and we need to really rethink the RADFAT approach we have for our image interpretation. It probably needs to be more a RAD molecular approach going forward. The goals of imaging in future should be to predict the molecular subtypes and the important being when the patient goes for surgery the only information surgeon has is imaging and oftentimes surgeon is faced with a dilemma between achieving a complete restriction versus risking neurological deficits especially for tumors such as modular blast tumors which can be closely related to brain steps. Having some idea about the molecular subtypes can help them decide how aggressive they need to be in trying to remove that last bit of tumor. Also molecular subtyping can take a few weeks to few months depending on your access and resources and imaging based predictions can then also be very helpful for treatment planning in terms of radiation another goal of imaging should be to identify quantitative imaging biomarkers which can help predict response to these novel therapies to this end multiple advanced imaging techniques including diffusion or fusion and MRF techniques are being looked at. However the most potential is probably with AI and machine learning techniques that you are using a radiometric extraction or a CNN based deep learning technique and it should be interesting how they plan out in the future. With that I would like to thank you for your attention if you have any questions or if you like a copy of this slide be free to email me. Thank you so much.


Hi everyone and welcome to Barns Takeout, your daily serving of art. My name is Martha Lucy and I work at the Barns. I oversee our education department and collections research and like most of you, I am at home right now. I am missing the world, I'm missing my friends and my co-workers and my extended family and I'm also missing the collection at the Barns. And I imagine that some of you are as well, so we have decided to do these daily talks. My co-workers and I, colleagues who are scholars and curators and conservators and educators, we're going to be presenting a short talk every day about some of our favorite works in the collection. So there will be a new one every day, just go to YouTube and check to see what's there. So the one that I've chosen for today is called The Luncheon and it is a painting by Pierre Auguste Renoir, the impressionist painter and it was painted in 1875. The reason that I chose this one, there are a couple reasons. One is it's just so gorgeously painted and I'm going to show you some close-ups in a minute to sort of show you what I'm talking about. Second, I like the food theme, it's kind of related to our takeout theme here. But mostly I chose this because it's got such a great human quality to it. It's really just a warm painting that's about human interaction and human interaction in a public space. So here you're looking at a photograph of the barn's galleries. This is room 13 where the painting hangs and you can see it right here hanging next to right in between other Renoirs. In fact, everything on this wall, all of the paintings that is, are by Pierre Auguste Renoir. It's the only wall in the collection that's devoted to an entire artist and that is because Albert Barnes loved Renoir. I think that's probably something of an understatement. He collected 181 works by this artist. And mostly he loved the works from later in Renoir's career, so from 1890-ish until his death in 1919. But the luncheon is a sort of rarity in the Barnes collection because it's an earlier work. It was done during the 1870s. So during kind of at the height of Renoir's impressionist period, it's a simple subject. It's two people having lunch. You can see soup terrine over here. They're having some wine. Here's the wine bottle. There's a baguette at the table on the table. And the scene takes place in a restaurant. It seems like there's probably a window that they're sitting next to because there seems to be some natural light that's falling on the table and that's especially falling on the woman. But how do we know that this is a restaurant? Maybe this is, why would we assume that? Why wouldn't we think that maybe this is people just in their home sitting down for lunch? Well the clue to that really lies in this hat, this boaters hat that's sitting in the foreground hanging on the edge of a chair. The boaters hat tells you that this is a couple that is probably on a canoe outing. They are taking a break to have lunch. They are spending a day on the river. This is probably meant to be taking place in the suburbs of Paris. We don't know where for sure but it was a really popular activity at the time for Parisians to go on these sort of boat outings and to stop for lunch. And because of the popularity of this new form of leisure, all these little restaurants kind of cropped up along the banks of the Seine in places like Chateau. And one of Renoir's most famous paintings, the Luncheon of the Boating Party, which is at the Phillips Collection, is of a similar subject. It shows a big group having, you know, breaking for lunch. And so that one, that one we know takes place in Chateau. But this one doesn't show a big group. It doesn't even show you the rest of the restaurant. It's really zoomed in on just this one couple having a private conversation. And it's a very quiet painting. And I think that Renoir really meant for it to be a quiet painting, an intimate painting. You're brought close up to them. You feel like you could kind of even maybe be at the table with them. One of the things that really draws me to this picture and that makes me appreciate it so much is the way that they're interacting. Just the little details that Renoir includes that are so just well observed from the way that people really behave. And I'm talking about, look at the man's hand and the way that he's got his fingers just kind of resting on the stem of the glass. And the way that he's gazing at her and the way that she's leaning forward, doing something with her knife and kind of leaning into the conversation, either talking and trying to make a point or listening. It's hard to tell who's talking at this moment. But Renoir, the point is that there are just these subtle little details and these little moments of body language that make the painting so relatable. Renoir was just a master of capturing all the kind of subtleties of human gesture. When this painting was first exhibited in Paris, the French critics wrote about it. And I think because it was such a painting that you can kind of enter into in a way, they love to kind of imagine what was going on here.  This is some sort of romantic trist. But just to look for a minute at like more closely at the way that it's painted, especially the still life on the table, it's pretty spectacular.  Look at that. Look at the soup terry in here when you when you get close to it. From a distance, it kind of looks like you can you sort of know, you know that it's it's silver and that the white is meant to kind of make it look shiny. But look how it's composed of so many different colors. He is just boldly kind of slapping paint on there and kind of brushing together reds and yellows. And it's so dynamic, this part and the bread, the baguette, which he which he brushes through with these red strokes that kind of bleed into the the terrain so that you're getting that so that the edges of these objects are becoming kind of they're no longer distinct. They're the color is kind of connecting them together. And that's that was a very radical thing to do at the time to not have individual objects in your painting, like bounded, you know, securely bounded look also just at these little things that he does.  I think that without that, I don't know if you would realize that she was holding a knife in her hand. Maybe you would because of the way that her that her her fingers are clenched. But it's just that little thing that that that kind of triggers that in your mind and gives you that information. What really kind of bulls me over here, though, is the wine glasses because look how he's rendering these objects that are solid objects. They are physical solid objects. But he look at this one especially, there's no outline to them. He he manages to give you the feeling and the look of a wine glass without any edges to it. The wine glass is communicated is conveyed solely through color, just dashes of paint. And I love these sort of little licks of red in the bottom of the glasses that let you know that there are just a couple of sips of wine left. Also over here, this this stroke here, just this lovely, I think it's the only stroke of green of that color green and the whole painting. And it's meant to kind of show the top of the bottle. But just look how how lively that brush stroke is. And then I want to back to that that idea. His ability to to define objects, to define dimensions and volume, only using color, is one of the things that makes Rymor so masterful. And I was just I was looking at this painting this morning, and I was I just kept staring at the take off. And at the edge of the table here and looking at how he is able to to tell you that this is a sharp corner here. Again, it's not he's not using line. He's just using color. And but then if you look up here onto the wall, it's like this amazing contrast. And I don't know if it's deliberate, but you know, the decoration, on the wallpaper or or this paneling, whatever it is, is a traditional kind of linear articulation of a corner. And then it just stands an amazing contrast to this, the articulation of this corner down here. That's done in a completely different way. Renoir was also a just a master at composition. I mean, he could use color. Really like, like nobody. But also the little tricks that he does to to make his paintings hold together. You know, the way that all the objects on the table are arranged to kind of create this X. That just gives that it makes the painting hold together. Things like that. See the the bread, this knife here, the glasses that make this diagonal, even this knife here. So all of these things are are why this is one of my favorite paintings at the barns and I can't wait to go see it again in person. I think that, you know, if this painting was was it was painted almost 150 years ago, but it's still so relatable. I think because of that. I mean, everybody can recognize the kind of conversation that these people are having and can relate to it. And Albert Barnes actually said that this was one of the things that drew him to work with. And Albert Barnes actually said that this was one of the things that drew him to Renoir was Renoir's ability to communicate kind of human emotion and human sensation. So I hope that you enjoyed this. Please stay safe and join us again tomorrow for another Barnes take out. Bye.


The patch notes. I'm so excited and worried No, well not worried, but just it's like it's that feeling where I'm too excited. I need to know something dude All right, here comes the battle system changes. This is universal for everything aerial guard cancel I want to see how this looks. I'm excited. Let's go. I hate blocking blocking in a fighting game My god That is pretty cursed anti-air attacks love that love that There is a new guard cancel in this game Can you show us the new guy? I heard rumors about this one to chat, you know, you try that for example You're gonna get punished, right? So there's a new guard cancel in town. There's a new guard cancel in town I heard rumors about this watch. Yeah, so there's a guard cancel vanish now guys.  So frame one full-screen key blasts I'm going to be with harder to you.  Oh, yeah Rejump not even but just oh, yeah rejump easier. Let's go We better players boy. This is dude. This is why I don't learn This is why I don't learn. I wait for the game to come to me baby. Let's go another system change Oh rumors about this one too. Here we go chat. This is the crazy one Here we go chat that was the one I heard The dragon ball fighters has forever changed Easy spear bombs there's so much new to learn dude. It's a new game Oh After what we got shown there's not a lot I believe you have a character to show us right now Vegeta blue. Oh, they're gonna show his leg They're gonna show his leg here. He comes. This is this is the guy you want to see two years later. Watch chat He's back It also moves forward like go to tabloos to M. It is all It's better they go now compared to before Oh my God, he's a you know much longer one now.  Why are they giving this to piccolo? Yeah, you know wouldn't be a DVMC patch without a piccolo buff. So, you know how you love to see it Hey, bro, what is this? They won't stop until piccolo is the number one character in every fighting game Walk it now.  Why they make piccolo so good. You can snap. Oh my God That's not fair There's no need There's no need There's no need for this My god Oh It crosses up, okay, that's it. That's a it's low you can also jump back what we got for trunks New super burning attack follow-up spicy. I just want to see the level to see like slice through You're not gonna show it You can't mention a new level to it Don't go black next you actually think is it time is it time?  Crouching unique attack added a new crouching unique attack.  So you Superdash brought that I guess that makes sense because it takes you into the ground.  What about few zamasu? Because we got him right here. Oh wait am I reading that right powdered shell now invincible diseases? I am a Dragon Ball fan, but am I reading that right? Broly veins win again, dude Everybody's getting buff that bet that better be true. You're better not lose. Try something. It's you know level five I don't come in now a follow-up attack can now be performed on hit by holding the button and spending one bar key He's got a level four Vigino has a level four. Please show the level four So it's literally his level three you hold the button, and then he'll do his level one the Punch super afterwards.  He has a dash when he whoa. Whoa. What he got life. What's literally anything? This is the something new I was talking about wow, that is cheap What huge winner huge winner.  My brain can't wrap my mind around what I just saw It's not a bug it's a feature He's jeering all move all grabbing moves No, really really you can still call a sister in it though. That is No, dude because yes, the change is No, dude, nobody deserves this Quite interesting one the crotch. Oh my god. All right. I'm gonna shake off the jeering nurse He just overall looks crazy. I like that. He's a immediately invincible during dimension a whole double-teleport is very cool You're in oh Jeering no, no, no No, no No, no, no, dude. No, I can't be real. No It wasn't a bug it was a feature No, no, it can't if I It can't be real if I don't click play dude. It can't be real. It's not I I can't I can't Ah It wasn't a bug. It was a feature. He's cheering. He's jeering No No It's not even jeerits video anymore.  This is Janemba now. This is Janemba video.  Oh My god, that was so fast Do it stretch it. Nah, dude. Nah, dude. Nah, what am I watching here, dude? Nah, what am I watching here, dude fake video fake video not real Not a real video. All right, here's base Goku. You got a new drop down. Oh That's the drop down.    Oh my god Why did jeering get nerve Broly, here's Broly.  Yeah, jeering was crazy. Yeah, jeering was too good Why why do I always have to lose something? All right, this is gohan. We kind of already saw but he gets He gets double dash at level three, which is crazy level four. He gets oh my cheap Also showing off reverse beat. Oh my god. Oh my god flex flex flex not by I think we saw this one He just that the side man is the real big improvement here beat fast Speed up and he's also immune to key blast end assist. So what is what is cyber been care about Broly?  Hey, how you doing, man?  There's a feature not a bug. That's the thing Feature it was a mistake. There's a feature. He's jeering nights. It's just a feature. Can we talk on topic?  This isn't correct


So welcome back, most common encountered cases of dynamic loading on a structure is the ground excitation due to an earthquake and as an instructional engineer, you will have to design structure to resist the earthquake forces generated by the ground shaking. So in today's class, we are going to learn how to set up the equation of motion of any structure subject to ground excitation and investigate further into that. Then we are going to look into a different type of loading which would demonstrate the difference between or which will demonstrate the effect of dynamic loading. So we are going to talk about how to apply a pulse load. So if a load P0 is suddenly applied, then we will see what would be the dynamic effect on the structure. And then we are going to solve some problems in which we are going to simplify our structure to a spring mass damper system and then solve for its response. So we are going to start our lecture today and in today's lecture, we are just going to be reviewing some of the concepts that we did in the last class and then we are going to be solving some problems here. So let us start with our equation of motion that we did last class. In the last class, we said that if we have let us say equation of motion as Fd over Fi plus Fd plus Fs equal to the applied force. This was our equation of motion and we considered two representation. The first representation was basically a frame representation. This representation is better suited for people with the structural engineering background and what we assume that the whole mass of the frame is concentrated at a level and then under the action of an external load Pt, it actually deforms by deformation let us say u and if it is a damped system then there was also damper connected in between. So this was the first representation and the same representation could also be understood using another representation we discussed which was basically a spring mass damper representation. The spring mass representation damper is used more frequently in mechanical vibration subject or in physics. So in this spring mass representation, again I have the same mass m here, the stiffness k damping coefficient c and under the action of this applied load here it undergoes our displacement u. So these two representation we considered and if we had let us say a linear system. So for a linear system we said that my spring force or the stiffness force would be simply k times u and for linear damper it would be simply c times velocity so that I can write my equation of motion as m u double dot plus c u dot plus k u and that should be equal to p of t and this is a second order differential equation. So this was in general when there is a force p of t that was being applied on the structure now we also saw that if we had an earthquake excitation in which there was a ground so for an earthquake excitation the same equation can be adapted to be written in a different form. So for an earthquake excitation with ground acceleration represented by u g t and displacement basically represented by u g t here and the acceleration u g double dot t. So we saw that the total displacement of the this frame that we had there would be a rigid component due to so rigid motion of the body due to ground displacement so this is u g of t and then there would be a further deformation due to the response of the structure so in this case it would be something like this because initially I had a mass here which finally shifted here. So this remember that this is u g of t here and then the displacement of the structure I have represented as representing as u of t. So the total displacement of the structure can be written as ground displacement plus the relative displacement of the structure or the single degree of freedom representation of the structure. Okay and same would be the relationship if you double differentiate it you would again get u g of double of t and then u double dot t. So this is the representation now similarly if you consider the free body diagram of this remember that there is only f i f d and f s here there is no external for as such that we had at the top of the structure p of t okay and we said that for a inertial force is always due to total acceleration however the damping and the stiffness force is due to the deformation of the structure. So the rigid motion which is basically u g of t which is the rigid offset it is not going to produce any stiffness force in the structure because it does not produce any deformation in the structure okay similarly for the damping the damping is contributed by the velocity in the structure with respect to its base and for that representation you could only have velocity with the structure is deforming and undergoing let us say a velocity u dot okay. So that is why we could write it as m of u g of t plus the relative acceleration and f d I can write it as c of u of t and then f s s a of u of t and that should be equal to 0 and I was able to write this equation as reformulated and write it as okay u t I will bring the u g double dot t on the right hand side then write it as this term here which is basically the same as the equation that we discussed earlier this equation okay or the equation here except there is a difference that now we have a p effective here which is represented by the mass of the structure times the ground acceleration and this is basically the earthquake force that acts on the structure so earthquake force on a structure is acts opposite to the direction of ground acceleration and it is mass times mass of the structure times the ground acceleration okay so just keep that in mind okay so the next step that comes that if I have a second order differential equation how do we solve it and so if I have this equation of motion here as again consider let me write that again how do I solve this so depending upon whether it is a free vibration okay free vibration so free for free vibration your p t is equal to 0 or whether it is an undamped system so for an undamped system your c is equal to 0 or a damped system okay c is not equal to 0 I could obtain different solution of the value basically the expression for u of t okay and that we are going to do in the subsequent chapter when we are going to study free vibration and the response of undamped and the damped system subject to periodic and non periodic loadings okay but what I want to show you here today is that to a simple example how a load that is applied suddenly or you can call it like you know dynamically how does it differ from the load that is applied statically and how would actually the response would differ okay so for example let me take an example here in this example I am basically applying a step force in which the load p naught is applied okay but it is applied suddenly okay so I have this time variation of the load and then I need to find out what would be the response of the system so I am considering basically an undamped system so that my c is equal to 0 just for the sake of demonstration and I want to find out what is my basically the peak displacement now if you remember from your undergraduate classes before if you were told that there is a force p naught in a spring that has a spring constant as k the deformation it was set and I am going to represent it as u static it could be simply calculated as p naught by k correct now let us see if I apply this step force suddenly and then if I have available the time variation of the PT okay then I am going to solve the equation of motion and then I am going to see whether my dynamic response is any different from this value and how much actually is that difference okay so in this case I have given the force PT which is p naught for all time greater than equal to 0 okay and my because it is an undamped system my equation of motion becomes Mu equal dot sku is equal to PT which is here is p naught so my goal is to solve this equation okay and we are going to come to like you know in subsequent chapters how do we solve this differential equation I just want to give an overview that how do we solve the second order differential equation okay so for a second order differential equation okay the total solution ut is represented by some of a particular solution and then some of complementary solution which is also called homogeneous solution okay so a particular solution is any solution that satisfies this equation okay so any solution that satisfies the equation your equation of motion can be used as a particular solution so for in this case I can directly look at that if I substitute up equal to p naught by k then the double differential of upt is actually 0 and then I get p naught equal to p naught so this could be used as a particular solution okay and then I have to find out complementary or homogeneous solution which is basically finding out the solution of this equation here k of uc of t equal to 0 so we set the right hand side equal to 0 and that is why we call it a homogeneous solution okay and then we try to find out the value of or the expression for uc of t okay so in this case for this kind of differential equation I am going to assume my uc of t as e to the power lambda t okay and when I substitute in the above equation what do I get as m e to the power or let me say m lambda square because it is a double differential e to the power lambda t okay plus k e to the power lambda t okay so it should be m lambda square plus k equal to 0 okay now I know that this term cannot be equal to 0 it does not give me any feasible solution so what I am left with is this expression here if I quit that to 0 I get lambda as plus minus under root minus k by m which can be further written as plus minus i to represent the under root minus 1 and this one I will write it as k by m again I will write as this minus i omega n omega n is basically the natural frequency of the system that you are considering we are going to come back to that in a later chapter just right now you like you know it interpret as any other constant through which I am representing this quantity under root k by m okay so I have two roots here lambda 1 which is i omega n and the lambda 2 which is minus i omega n okay so if I have two roots the solution of this complementary this equation right here okay can be written as a linear combination of both roots okay so I would write that as remember that I had assumed this has to be the solution so linear combination of the two solution a e to the power lambda 1 t and b e to the power lambda 2 t okay which would nothing but e a to the power i omega n t plus b e to the power minus i omega n t okay and I also remember from my knowledge of complex number that I can write e i to the power i x equal to cos x plus i sin x and if I replace i i x by minus x I can write this e to the power minus x s again cos x minus i sin x okay so I will do that here okay and then rearrange the terms to obtain a plus b cos omega n t plus i a minus b sin omega n t okay so this and this these two are unknown coefficients that I need to find out I am going to represent them with two unknown coefficients here and I can write it as at c cos omega n t plus d sin omega n t okay so this is the complementary or the homogeneous solution so the total solution of the differential equation I can write it as u particular plus u complementary okay and this would be p naught by k plus c cos omega n t plus d sin omega n t now you see that we consider a second order differential equation and now we have two unknown constants so we need two conditions to actually find out the specific solution to our differential equation okay and where those two conditions are going to come from well it comes from the initial conditions for a dynamic problem okay so the initial conditions in a equation of motion would be your initial velocity and initial displace sorry initial displacement and initial velocity okay and in this case because it is given that initially like you know they were at rest suddenly a force is applied so I can assume you can assume that u 0 and u dot 0 both to be equal to 0 okay for my case okay so you can substitute that in this equation here to obtain the value of the constant c and d okay and you will see that okay you will get the final solution as p naught by k 1 minus cos omega n t right so this is the solution that I have okay due to the step force that is being applied to my single degree of freedom system so let me just write it again here or redraw it again so I have p t here and then I have t here and this is p naught now remember had it been a static problem then my u static would have been just p naught by k so if this is my u of t n so this is the response this is the applied force and this is the response for a static problem I would have this displacement as p naught by k but I can see that my dynamic displacement actually is not p naught by k but it actually fluctuates due to time variation of cos of omega n t and if you try to plot that it would look like something like at t equal to 0 your displacement is 0 okay and at t equal to let us say pi what you will find your displacement is actually minus this term would be minus 1 so that we will get 2 p naught by k so if I draw this time displacement this value here is 2 p naught by k which is the maximum dynamic displacement okay which would occur when cos omega n t is minus 1 so I can write u max is equal to 2 p naught by k okay which is 2 times the u static that I had determined okay so you can see that if I had applied the load p naught statically or for a static problem I would get a displacement which is equal to p naught by k however if I apply the same load suddenly like a step force then I am getting a peak displacement which is 2 times the static displacement okay so this is the effect of the structural dynamics that is coming into the play here okay and this too is as you would study later it is called dynamic amplification of the response and there are different factors to define it okay so this example was just to demonstrate that if a load p naught is applied statically or if it is applied dynamically then your response are actually or the peak responses are actually different and for this specific case it was 2 times the static displacement okay so I hope this problem will give you a better idea of what the dynamic nature of problem could do to the response of a single degree of freedom system okay alright if that is clear to you what we are going to do now we are going to do or solve some examples related to these type of equation of motion okay so and then see how do we tackle different problems okay so in example one here what do I have here is actually a slab okay slab that is supported and four columns okay so these are the slabs that are supported on four columns okay and the dimension of these this slab is b in this direction and b along this direction okay and it is given that these four columns with respect to these axis is x and y have stiffnesses kx the lateral stiffnesses kx and the lateral stiffness ky in the two perpendicular directions so each of these columns would have these stiffnesses into direction okay and what you need to do for a small angular deformation of the slab in plane angular deformation so let us say slab is actually rotating okay like this in plane you need to set up the equation of motion for this slab here alright so I would like you to pause this and then try to solve this problem alright let us discuss the solution of this problem okay so let us consider the top view of this slab that I have here okay slab let me just draw it like this okay so this is the slab that I have and if I rotate it if it has an angular rotation let me again draw the deflected position of this slab which would be I just redraw it here okay would be something like something like this okay sorry for the bad drawing but let me I mean figuratively like it does the job okay so what is happening here it is rotating by theta here okay so that at all points it is rotating by theta with respect to its original position okay now remember as it rotates okay at any corner the horizontal displacement is the same displacement that I have here which is represented by the half of this length which is d by 2 okay times the angle so this displacement horizontal displacement is d by 2 theta similarly the vertical displacement at this point is nothing but the half of this length which is b by 2 times this angular displacement which is b by 2 times theta and same would be the displacement at all the four corners okay and like you know if you want you can do the same thing you can derive the same thing by considering the angular displacement by plotting a line and then doing like this okay this is just a simple way of considering that whatever the horizontal displacement at this point okay at this point here would be the same horizontal displacement at this point the vertical displacement of this point would be the vertical displacement of the slab at horizontal position okay now when it tries to rotate it like that what will happen yes there the column would apply a force in this direction which would be kx times the horizontal displacement which is d by 2 theta and in this direction which would be okay ky times b by 2 times theta and same would be the case at other corners as well for example here it would apply in this direction okay and in this direction okay and at this point it is going to apply in this direction and in this direction and at this point it is going to apply in this direction and in this direction and the one thing that is common that at each corner each pair of these forces are actually creating clockwise moment about this point the center of this slab. So if I want to set up the equation of motion, I need to sum up the moment due to all these forces that are being applied by the column on the slab at all of its corners. And then there is another force or the moment basically the inertial moment due to rotation of this slab which would be opposite to the direction of rotation and it would be i times theta double dot. So I can simply write it as i times theta double dot and then take the moment of these column forces that are being applied on the slab. So it would be first force would be kx times d by 2 times theta and the moment of this force about the center would be again d by 2 plus ky b by 2 times theta and the moment about center would be multiplying would be obtained by multiplying with the lever on b by 2. And this times 4 as you can see it is producing the same moment at all the four corners. So I will multiply it with 4 to get the final equation. So I can write it as final equation of motion as kx d square plus ky b square times theta equal to 0. So this is my final equation of motion for this given problem. Once that is clear let us discuss another example. Example 2 here in this example what do I have? I actually have a rigid bar here. So let me draw a rigid bar. It is connected to a pin connection, pin support at one third of its length and there are three springs each of stiffness k connected to this rigid bar at length l by 3 l by 3 and l by 3 from each other. And what I need to do again is to set up the equation of motion for an angular rotation of this bar ok. And the mass of the bar is m and as I said the total length is l ok. So take some time and solve this problem ok. Let us discuss the solution to this problem ok. First let me say let me draw the initial position and the defined position for an angular rotation. So let us say it rotates in clockwise direction by an angle theta. So let us say the rotation is in this direction by angle theta ok. Due to this rotation the springs at the leftmost end would be stretched by a deformation and I can find that out as l by 3 times theta and this second spring is going to compressed by l by 3 times theta and the third spring would be 2 l by 3 times theta ok. So in terms of forces this spring is going to apply downward force and these two springs are going to apply upward forces like this ok. And if you consider the free body diagram due to clockwise rotation there would be an inertial movement which would be applied anticlockwise ok. And that would be the I of this or the moment of inertia of this rod about the point of rotation times theta double dot ok. So if I take an moment of all these forces and equate it to 0 about the point of rotation O which is the pin support ok. Let us see what do I get? I get I theta double dot plus the moment of this force all these three forces spring forces are in the same direction the counter clockwise direction ok. So it would be k l by 3 which is the spring force times the lever arm about the point of rotation is l by 3 ok. For the second spring k l by 3 times theta times l by 3 and then the right most spring would be k 2 l by 3 times theta into 2 l by 3 and that should be equal to 0 ok. Now the I of this bar about its about the point of rotation would be I of this bar about its center of mass plus the mass of the bar times the distance of the point of rotation from the center of mass which I can calculate as this ok. And this I can write as m l square by 12 and this as m l square by 36 which eventually gives me m l square by 9 ok. So I will substitute that here m l square by 9 plus this I would get as 6 k l square by 9 times theta equal to 0 there is another theta double dot term here ok. So the final equation of motion would be m l square or I can just divide that and I can write it as theta double dot plus 6 k divided by m times theta equal to 0. So this is the final equation of motion for this system that is shown here ok alright. Let us go ahead and do example 3 ok. So in example 3 what do I have it is not a very difficult problem but I just want to demonstrate you the concept of rotation and twisting ok. So in this case what do I have I have a thin shaft here ok which is rigidly connected to a disc ok. The polar moment of inertia of this shaft ok can be calculated from its diameter d the radius of this disc is r ok the mass of this disc is m ok the shear modulus of this shaft is actually g ok let me see do you need anything else yeah you need the length of this shaft. Now can you imagine if I try to provide twist to this disc ok what is going to happen remember that disc in its own plane is very rigid compared to the twisting stiffness of the shaft. So all the twisting is going to happen in the shaft and if you remember if I have 2 spring in series let us say k1 and k2 ok where let us say k1 is less less than k2. So k equivalent is actually decided by the whatever the most flexible spring is in this case it would be approximately k1 ok and that you can find out from the expression 1 by k1 plus 1 by k2 right if k1 is very very small than k2 this quantity would be very very large to this quantity so that you can neglect it so that when you invert it k equivalent becomes equal to k1 ok. So in this case disc own stiffness torsional stiffness is much higher than the shaft stiffness ok and that is why I am just going to consider the whole resistance is coming from the shaft and if you remember from your solid mechanics class if I have a shaft of length l ok and then shear modulus g and polar moment of inertia j ok and if I apply a twisting moment on this shaft ok the relationship between this twisting moment and the twist in this shaft so if I project it here let us say this is deforming by theta this I can write it as jg by l times theta which is the basically the rotational stiffness or the twisting stiffness of this shaft right here ok. So if you consider the free body diagram of this disc and then assume that it is rotating by theta remember the same would be the rotation in the shaft ok so shaft is going to apply and equal an opposite moment which I have here which would be m of the shaft let us say it is ms ok and because I am rotating it by theta what will happen I need to again include in the free body diagram a rotational inertia ok which would be whatever the mass moment of inertia of this disc times theta double dot ok so if I take moment of all the forces about center I can write it as i theta double dot plus ms that should be equal to 0 and I know that my ms is this ok so I write it as mr square by 2 ok plus jg by l times theta that should be equal to 0 and if you remember the polar moment of inertia of a shaft is given by the formula pi d4 by 32 ok so that you can use here alright so this was our third example ok let us go through another example which is example over here ok and in this example what do I have I have two concentric discs which are rigidly connected to each other ok so I am going to draw the figure here and there is a string which is connected to this internal disc and there is a mass hanging from this disc ok and there is a spring connected to the outer disc and then the stiffness of this spring is k the internal radius is r and the external radius is 2r the mass here is m ok so you need to set up the equation of motion for this system ok so pause the video here and then take some time to attempt this problem and see what do you get as final answer ok alright let us discuss the problem ok now for this setup that I have here let us consider the motion from the initial equilibrium position remember it is not the undeformed position but the equilibrium position that I am considering alright and if you remember gravity load need not to be considered if I am considering the motion from the equilibrium position and if initially the gravity load was being balanced by a spring ok deformation in the spring which is the case here so I do not need to consider that gravity load in the equation of motion or the free body diagram ok so if I consider the free body diagram of the mass here ok alright what are the forces acting on this mass mass the mass is going down so there would be inertial force which is acting upwards which would be m u double dot and then there is a tension in this string here alright and I am saying that this is coming down by u now can I say if this string comes down by u ok the circumferential rotation of this inner disc would also be u so that the angular rotation of the inner disc ok so I am saying that this is coming here by u ok so that this theta that I get ok would be how much ok this theta would be u divided by the inner radius of the inner disc ok which is r ok and same would be the rotation of the outer disc as well however the circumferential rotation distance would be actually 2r times theta which if I substitute the value of theta it would be 2u and the same would be the extension in this spring here ok so the force in the spring would be actually k times 2u ok which is the deformation in the string and that would be the force that would be applied on this outer disc by this spring so if I consider free body diagram of the whole disc arrangement ok I would have force 2ku acting here I have force t which is acting here ok remember that this do not have any mass themselves ok so if I take the moment of all the forces about the center of the disc I get t times r is equal to 2ku times 2r so that gives me t as 4ku so now I know that what is my force in the string I can write down the equation of motion of this mass ok which would be nothing but mu double dot plus t and that should be equal to 0 which I can write it as mu double dot plus 4u and that should be equal to 0 alright ok now let us see our example 5 an example 5 I again have a spring mass system except in this case ok I do not have a block but I have a rotational disc here ok and I need to find out what is the equation of motion if there is rolling without slipping so if that is the condition and these constants are given to you ok the radius is basically r here you need to find the equation of motion of this system here alright ok so take some time and then try to attempt this problem so that we are going to come back to again this later and then solve it alright.


So I am a, so really enjoyed being part of the jury today. And I thought there were, you know, good amount of brands who have, you know, especially in the last two years and during the pandemic, et cetera. I think I did see in some of the areas some good inputs coming into it from marketing perspective, right? I also see a lot of, you know, new players coming into it, which is again interesting, right? So overall really enjoyed being there, right? And the platform over the last so many years being there has actually given ample opportunity for brands to talk about their thinking, their strategy and how they're able to, you know, really put in that the marketing entire evolution within the journey of doing some fantastic work. So yeah, so good to be associated with it. There are many, many multiple new plays, right? And we don't probably only call out a few, but I think this has been a period where a lot more thinking on the evolution within the marketing aspect, you know, has happened, right? So definitely, you know, that needs, that requires a lot more, you know, lot more such opportunities to display, reflect, demonstrate some of this great thinking and execution. So I may stand short for that. So yeah.


Metallic bonding occurs between metal atoms. The valence electrons of a metal atom have a low ionization energy and the atom itself has low electronegativity. This means that the valence electrons are not strongly tied to their parent atom and this leads us to a model of metallic bonding called the electron C model. When metal atoms are close to one another they no longer behave like neutral atoms. Instead the valence electrons can leave their parent atom and move around forming a C of delocalized electrons between the atoms. Because each atom has lost valence electrons it's now a positive ion. As a consequence there's an electrostatic attraction between the metal cations and the C of negatively charged electrons. And because the electron C is able to move freely all around the metal cations we say that this form of bonding is non-directional because the electrostatic attraction operates in all directions.