This is Antarctica, nearly 6 million square miles of ice-covered wasteland, larger than the United States and Europe combined. Less than one-third of this great continent has been explored. Previous expeditions have discovered indications of coal, iron, and other minerals in this last frontier. Its lean elevation is 6,000 feet, but some of its mountain peaks rise above 15,000 feet. It is the highest and coldest continent in the world. In Antarctica, scientists hope to find the answers to many questions, which will unlock secrets of the universe. The International Geophysical Year is a combined effort of scientists of more than 60 nations to gain knowledge of the Earth and related phenomena by worldwide simultaneous observations. One of the most extensive of all international geophysical year investigations is taking place in the Antarctic regions, with 11 nations participating. In an effort to help find answers to these questions, the United States Antarctic programs was conceived, and by direction of the President of the United States, Rear Admiral Richard E. Byrd, Veteran Polar Explorer, was named Officer in Charge. The Navy was assigned the task of logistic support for the program. Task Force 43 was organized under Commander in Chief Atlantic Fleet to implement the operation, which was assigned the codename Operation Deep Freeze. Rear Admiral George Dufek was appointed Commander Task Force 43, with the responsibility to construct bases, operate them, supply and resupply them, and transport scientists to and from Antarctica. The major objectives of Operation Deep Freeze I were to establish a main IGY base near the site of the former Little Americas and an air facility at McMurdo Sound, investigate Marie Birdland and the South Pole for suitable scientific observation posts, to support a planned U.S. scientific program in the Antarctic, and provide logistic support for the United States participation in IGY. The USS Arneb, a.k.a. 56 Admiral Dufek's flagship, was one of nine ships constituting Task Force 43's ship group. The others were USS Wyandotte, a.k.a. 92, USS Greenville Victory, USS Nespolin, A.O.G. 55, and the Icebreakers, USS Edisto, AGB 2, USS Glacier, AGB 4, and the East Wind, a Coast Guard Icebreaker. And finally, two-yard oilers, the Y.O.G. 70 and 34, were each loaded with 270,000 gallons of aviation fuel to be used as Antarctic filling stations for the air unit. The Task Force Air Unit, Air Development Squadron 6, commanded by Commander Gordon Evie, consisted of two R5D Skymasters, four R4D Skytrains, two P2B Neptune's, as well as four De Havilland Otters, and six H.O.4S3 helicopters carried aboard the Task Force's ships. At the Atlantic Leeds Construction Battalion Center, Davisville, Rhode Island, a special Mobile Construction Battalion was commissioned by Admiral Dufek with Commander Herbert Whitney, commanding officer of the unit. These CDs have the responsibility of constructing all of the bases in the Antarctic. They were trained in cold weather construction, but the Antarctic, notorious for its unpredictability, will pose many problems for them. Here, Commander Whitney reads his orders, assigning him as commanding officer. Mrs. Dufek is accorded the honor of carrying old glory in the changing of the color ceremony. Julia Hawks, selected by the crew as co-sponsor, representing the women who must remain behind, carried the battalion flag. Families and friends are afforded an opportunity to see some of the unique equipment that will be carried on the expedition. It's a 10,000-mile trip down there. And in three months, this weasel will be moving across the polar plateau. Junior checks it out. While Mom inspects this tent, the roller will be used to build airstrips on snow-covered ice fields, a winterized portable building. This will be used on the trail for a mess hall and radio communication. In August 1955, the task of receiving, packaging and loading the ships of Task Force 43 began at Davisville, Rhode Island. Here were loaded an estimated 20,000 measurement tons of cargo, bulk petroleum products, and the Antarctic life preservers, sled dogs. The first departure at Davisville in early September was marked by the excitement and sadness of farewell. Many of these families will be separated for over a year, and the separation will be spanned only by intermittent mail and wireless communication. From September until mid-November, this scene was repeated, as Task Force ships departed from various East and West Coast ports. On November 14th at Norfolk, Admiral Byrd spoke at departure ceremonies for the Arneb, Admiral Dufek's flagship. On route to Fuller Waters, the icebreaker glacier passed through the Panama Canal, where she took aboard a load of bamboo. Later, at sea, the bamboo was cut into strips. It will be used as trail and supply dump markers in the deep snow. Other preparations for the expedition were carried on at sea. Here, sleds are being rigged. In equatorial waters, the routine was interrupted by the traditional ceremonies conducted by King Neptune and his court. The polywogs were put through the appropriate ritual of initiation, required of all sailors crossing the equator for the first time. Among the Operation Deep Freeze personnel were several foreign officers. Here, they inspect the Navy's newest and most powerful icebreaker. In warm tropical waters, religious services were conducted on the open deck. On one occasion, the ship hauled far out into Pacific, while crew members took a dip and acquired tans that they hoped would last out the Antarctic weather. The crew of the little Y-O-G's were deprived of many of the comforts and conveniences offered by the bigger ships. Nine are shot to the vessel, and the high line is rigged to send supplies across. Two of these vessels were towed to the Antarctic. Life aboard them was rough, wet and lonely. Thanksgiving Day aboard the ships was celebrated in the South Pacific, far from a supermarket. But the ship's mess tables were amply provided with a traditional bird and trimming. The USS Glacier arrived at Port Littleton, New Zealand, December the 7th, 1955. Here, she was joined by Captain Gerald L. Ketchum, Deputy Task Force Commander, and Admiral Bird. On December 10th, the Glacier began the final leg of its journey to Antarctica, as hundreds of New Zealanders gathered at the pier to bid the ship and its crew a bon voyage. Six days later, the Glacier received a baptism of ice near Scott Island when she entered the ice bank. December 18th, the Task Force sighted the Antarctic continent, when Mount Erebus, the only known active volcano in the Antarctic, towering 13,000 feet above McMurdo Sound, came into view. The Glacier moored to the ice, and its helicopter flew a survey party ashore to find an area on the bay ice suitable for a long, flat runway that would support the weight of the large planes of the air unit. Commander Abbey and Commander Whitney surveyed an 8,000-foot strip of ice at the south end of McMurdo Sound. They tested its thickness with a gas-powered chainsaw to determine if the ice would support heavy planes without skis. This site was found to be suitable. It was located immediately north of the 19-2 expedition camp of Royal Navy Captain Robert Scott, leader of the second party to reach the pole. Food stores left here by the Scott party over a half-century ago were found to be perfectly preserved and quite edible. The Glacier then unloaded vehicles and materials for the establishment of a temporary base camp to support seabee and air contingents until permanent buildings were completed. Some permanent inhabitants arrived to discharge their ambassadorial duties by greeting and entertaining the visiting delegation from the United States. The emperors are the best known year-round inhabitants of the Antarctic. Penguins are equipped with built-in toboggan for traveling over the snow. This Adelaide penguin ripperie was located less than one half mile from the Glacier's mooring. With the completion of the airstrip, Captain Ketchum and Party transferred to the ice to await the arrival of the Edisto. From this ship, he would direct operations in the air. The USS Edisto arrived soon after the Glacier departed McMurdo Sound to furnish logistic support for air operations and construction parties. The offloading began immediately. Supplies, equipment and materials were unloaded for transportation to the temporary camp and airstrip sites. The sled dogs were happy to be in their element and were eager to go. Surface vehicles began moving materials over the day ice to the airstrip and permanent base site. December 20, the aircraft made their fly-in, guided along the 2,200 miles by ships stationed at 300-mile intervals. Four aircraft arrived without misshapes. Headwinds forced four of the smaller aircraft to turn back to New Zealand. Aircraft had now flown nonstop from a large landmass into the Antarctic continent for the first time in history. The Glacier proceeded from its ocean station to Scott Island and on Christmas Day, rendezvoused with the other ships of the Task Force north of the ice pack. Rear Admiral Dupec greeted Admiral Byrd as he transferred from the Arneb to the Glacier to command the transit of the Task Force. The mighty Glacier led the column through the ice pack. Clearing the pack four days later, the Task Force was met by the Edisto which led the Wyandotte and Nespolin into McMurdo Sound. The Glacier escorted the Arneb and Greenville victory on to Canaan Bay, 450 miles to the east. December 28, the Wyandotte and Nespolin arrived in McMurdo Sound and sent a mooring party ashore to bury timbers called Dead Men for attaching the ship's mooring lines. After the long flight from New Zealand, an even more hazardous task was undertaken. Aircraft taxied 30 miles directly to the Tanker Nespolin to expedite flight operations and take full advantage of the little remaining good weather. Refueling at the ship, the aircraft were on ice only four feet thick, over water 1800 feet deep. A temporary camp was established and air operations commenced immediately. An R5D using wheels rather than skis makes a Genesis take-off, beginning a long-range exploratory mission. In less than one month, pilots of EX-6 mapped and photographed over a million square miles of territory, never before seen by human eyes. Later, Admiral Byrd, Dr. Paul Seipel and other Antarctic gold timers paid a visit to the site of Little America 1 and 2. They located this radio tower built during Byrd's 1929 expedition. When first erected, it stood 75 feet above the surface of the snow. On January 4th, 1956, Little America was officially commissioned by Admiral Byrd and Admiral Dufek. The glacier, after carving out a mooring in Canine Bay for the cargo ship, was ordered by Admiral Dufek to McMurdo Sound to assist Captain Ketchum. For 10 days, the Edisto and the Coast Guard east wind attempted to cut a channel through the ice of McMurdo to within a reasonable distance of the base site at that point. During this time, only 14 miles had been gained. The glacier went to work on the ice. It takes immense power and tons of weight to accomplish the feat of crushing acres of ice ranging from 10 to 12 feet thick. These scenes taken from the ice show you the power of the glacier. An icebreaker, contrary to popular idea, does not break ice by cutting through it, but is so designed that the bow will ride upon the ice. And in a porpoising action, the sheer weight of the ship crushes and cleaves through the ice. In 48 hours, she came within six miles of the base and established an offloading point and a turnaround circle. The 30-mile artery through the ice was named Glacier Channel. Since Bay Ice did not move out into the open sea as anticipated, it was too risky to bring the cargo ships into the channel. Therefore, the three icebreakers were required to shuttle supplies from the ships through the channel to the offloading point. Here at Glacier Point, the cargo is unloaded for transfer to the base site. The unloading operation involved a calculated risk. The Glacier cargo handling equipment, designed for only 12 tons, lowered a 24-ton V8 tractor. The risk paid off. Cargo moved to six miles to the camp loaded in tractor trains. This operation involved many hazards because of the treacherous Bay Ice. Finally, on January 14th, the first load of building materials and supplies arrived at the base site, and one of the major objectives of Operation Deep Freeze had been accomplished. The indispensable Glacier served not only as a cargo transport, but also as an aircraft tender. By January 18th, above freezing temperatures had caused ice runways to melt and crack. The aircraft, after receiving their final refueling and maintenance from the Glacier, took off for New Zealand. 450 miles east, near Little America, supplies were shuttled to a temporary cache on Bay Ice, approximately halfway to the base. The warm weather caused melting and weakening of the ice, until finally the temporary cache was in danger. Admiral Dufek ordered everything transported to the firmer shelf ice on the double. Every available tractor, carry lift, and helicopter was pressed into use. In 48 hours, all of the cargo had been moved to Little America 5, only a short time before the Bay Ice broke up and floated out the seat. In the construction of bases, the CDs lived up to their can-do reputation. With the building area leveled, foundations were laid for the special type buildings, which would be the living, operating, and recreational facilities for a long, long winter night. Floors were laid, and walls erected. They looked thin, but special insulation and expert construction make them snow-type. Roofs must be especially strong to support the heavy accumulation of snow throughout the three years' occupancy. Tractors cleared the area between the buildings of snow and ice so a tunnel system could be built. An elaborate tunnel system was built to connect all of the buildings and serve as storage spaces. Both here and at McMurdo Sound, 250,000-gallon fuel tanks were constructed to serve as storage for aviation fuel to be used in Operation Deep Freeze II. When completed, these tanks were filled from the USS Nespolet, board about four miles away. The problem of transporting fuel four miles was met by using several collapsible fuel hose storage tanks provided by the Marine Corps. This tank, when unrolled and filled, holds 10,000 gallons. Gasoline was pumped to these tanks, and by increasing the pressure at booster stations with auxiliary pumps, was finally brought to the huge fuel tank at Hut Point through portable pipelines. This experimental crevasse detector, designed to give a signal when a crevasse was approached, proved not completely reliable when used by trail parties. The loss of an otter-supporting aircraft and the failure of the crevasse detector delayed the Marie Birdland Trail Expedition. With winter weather setting in, the project was postponed until Deep Freeze II. To climax the activities of Operation Deep Freeze I, the glacier with Admiral Dootheck aboard made a notable cruise around the Antarctic continent. The purpose of this voyage was to locate suitable sites for more I-GY stations to be established in later operation. Survey crews returning from surveying a possible site on the Knox coast experienced the sudden fury of an Antarctic blizzard, a vanguard of the fast approaching polar weather. Their small craft fought through the ICCs to the side of the glacier and was finally hoisted safely aboard. The glacier's crews around the continent brought Deep Freeze I to a successful conclusion, and she pointed her bow north, leaving behind 73 men at the Little America base and 93 sailors and scientists at the Naval Air Facility in McMurdo Sound. With the wintering over party were 1500 tons of supplies and materials to sustain them through the coming year of isolation from the rest of the work. Some of these materials will go into the construction of bases at the South Pole and Marie Birdland before ships return to the Antarctic. At the base of Observatory Hill, men prepared for the bitter cold and the violent winds of the long winter night. And at sea, the glacier plunged northward into warmer ice-free water, her destination, the United States, where she would be fitted out for Operation Deep Freeze II. Thank you for watching.


Welcome back everyone. Today we are going to discuss the idea of impulse and how the response of a system subject to pulse forces can be approximated using the same expression which only depends on the area of the loading and not the shape of the loading for a specific value of pd by Tn ratio. So, let us see how we can get solution for the response of a single degree of freedom system subject to impulsive forces. So, today we are going to see how to obtain the response of a single degree of freedom system subject to different type of pulse excitation. Till now what we have done last class we found out what is the response subject to step force or step excitation in which a force is applied over a duration of td ok and then it is taken off and then we found out how to what is the response of a single degree of freedom system. So, basically the procedure that we describe that we have to consider the forced duration or forced phase of the response and then free vibration phase ok. So, for any pulse type motion ok we can divide our response into forced response and free vibration response ok basically to represent when the force is applied on the single degree of freedom system and when the force is removed. Now today we are going to discuss different type of pulse excitations ok examples of which could be observed in real life ok. So, one of those pulse excitation is triangular pulse ok. So, basically you have a force of amplitude p naught ok and then it decreases with time and then goes to 0 over the time duration td. So, as I said this is a force that is applied suddenly and then it goes to 0 over time duration td ok. Basically this type of like one of the examples where this type of pulse excitation could be observed or where the loading could be represented as a triangular pulse is basically blast loading ok. So, if you have a surface blast what happens the shock wave actually travels like this ok and when it hits a structure it applies a pressure loading on that structure and that pressure loading the time variation of that pressure loading can actually be approximated using a triangular pulse ok alright. So, let us see how do we find out response to triangular pulse excitation. So, we are going to divide our force pt or the excitation pt and two phases first phase is the forced phase ok which we are going to write as p naught 1 minus t by td. So, this is 1 here ok. This is for time duration smaller than td and for greater than td this is basically 0. Now, to get the response ok we are going to first find out the response during the forced vibration phase and then during the free vibration phase using the initial conditions at the end of forced vibration phase here ok. So, that would be ut and u dot of td ok. So, the procedure remains same ok you could either use to Hamilton integral to find out the response ok or you could just solve like you know utilize the classical method of solving differential equations ok. So, what I am going to do I am just going to write down the final solution ok. So, let us first consider the forced vibration case ok. So, I can write down my ut if I consider as dohamel integral I can write this down as uts p naught by m omega m 0 to t this load would be 1 minus tau by td times sin omega n t minus tau d tau remember I am not doing anything new I am just finding out what we have discussed that ut due to an arbitrary force can be found out using this expression here ok this dohamel integral here in which we write down p tau and then h of t minus tau which is the unit impulse response function and which for this case ok would become something like sin omega t minus tau ok d tau. So, this we had derived in the first lecture of this chapter. So, we are just basically utilizing that. So, once we substitute the values ok we can go ahead and find out the response as ut equal to us t naught which is basically p naught by k ok times 1 minus t by td minus cos omega n t plus sin omega n t by omega n td ok. So, this is for the forced vibration phase now what will happen at the end of the forced vibration phase it would have acquired certain displacement and velocity. So, when the load becomes 0 it still has that velocity and displacement ok. So, it is going to go into free vibration without any application of external load ok and it will keep on vibrating with that because there is no damping in the system ok. So, basically what I am saying for the free vibration phase let us let us get the response. So, for the free vibration case I can just write down my responses u of t is equal to u of td cos omega n t minus td plus u dot of td divided by omega n sin omega n t minus td. This is the free vibration response with initial conditions are now provided at t equal to td not t equal to 0. So, remember the original equation used to be u 0 cos omega n t and then u dot 0 divided by omega n sin omega n t and this was the expression for ut ok when the initial conditions were provided at t equal to 0 ok. Now, the initial conditions are at t equal to td. So, basically I am going to find out my response using this expression here ok. So, we can substitute the value of u of td and u dot of td from this expression here substituting it here and then we can simplify it and get the final response of the system which I can write it as u of t is equal to q s t naught times sin omega n t by omega n td minus sin omega n t minus td divided by omega n td and then we have a cos term here ok alright. So, now, we have obtained the response for forced vibration phase and the free vibration phase ok. So, one of the things that we need to find out is actually subject to a triangular pulse let us say p naught n td ok. What is the peak response of the single degree of freedom system ok. So, this is an important parameter that we need to find out and to find out first we first need to consider whether the peak response occurs during the forced vibration phase or whether it occurs during the free vibration phase because depending upon that I will have to find out the maximum value of u during the forced vibration phase or maximum value of u during the free vibration phase ok. So, that we need to do ok. So, let us see how do we do that ok. So, for the free for the forced vibration phase basically what we need to do we have we can define our r t s ok let us call it r 1 ok we will call this r 1 r 1 of t is basically u of t where r 1 is basically the response modification factor u of t divided by u s t naught ok. So, we will get this expression here which I have here. So, I am just going to write it again ok alright. So, this expression I get and to find out at what time ok this is maximum basically I need to differentiate it with respect to time and find out and basically equate it to 0 and then find out t equal to t max ok from that expression when I substitute it equate it to 0 ok. And then I am going to substitute this time back to this expression here to find out the r 1 max ok and basically what do I get r 1 max as in this situation ok 2 minus t m by t d where t m is the value of the t that I get when I substitute d r by dt equal to 0 and the value of t m that I actually get I am going to write that as well ok. So, t m basically I get it as t n by pi and then tan inverse 2 pi t d by t n. So, this expression here becomes 2 minus t n by t d ok and then tan inverse 2 pi t d by t n. Now, as you can see from this expression here my r 1 is only a function of t d by t n which is the same conclusion that we had obtained for other type of pulse motion which was the step force. So, even for the sine force again t d by t n is the important parameter based on which my peak response depends ok. So, this is for the case when I am assuming that my maximum response occur due to during the forced vibration phase ok. Let us say that is not the case ok. So, what I am saying ok what I am saying here basically let me say this is I am representing the displacement on y axis and time on x axis ok and I am also trying the static displacement curve which is basically starts from p naught by k the peak value. So, this is basically u s t naught equal to p t divided by k. So, this is the time variation ok. So, it might happen that the maximum response can occur during the forced vibration phase or it can occur during the free vibration phase ok. So, in that case let us say it occurs during the free vibration phase the response would still be increasing something like this ok and somewhere during the free vibration phase I get the maximum response. So, at at the end of the forced response whatever response that I will get could not be the maximum value ok that would be the value ok which would further be increasing due to the during the free vibration phase ok. So, in this case basically T d is smaller than the T m and T m is basically obtained during the free vibration phase ok. This is opposed to the case when my maximum response occur during the forced vibration phase in this basically what happens it goes something like this ok. So, during the forced vibration itself the maximum response occurs ok and it keep on going like this. So, again this is my u s t of t equal to p t by k this is p naught by k and this is the u t curve ok and this is my T m and this is T d ok. So, in this case which is the case that we have just discussed the maximum response occurs during the forced vibration phase if that is not the case then let us see what happens if the maximum response occurs during the free vibration phase ok. So, to find that out what we need to do ok for the free vibration r a t I need to write is as u t divided by u s t naught ok this is for the free vibration phase ok. So, let me write it here and I can substitute the expression here ok to find out the expression which I get it as nothing but omega T d and this expression here ok. So, let me just write down the expression for your reference alright. So, this is the expression that you get. Now, again even for this we are going to follow the same procedure ok d r by d t for the maximum response during the free vibration ok I am going to equate it to 0 then find out the value of t equal to t m at which the maximum response occurs and then substitute it back to this expression here to find out the r max ok. So, the t m I will just give you the answer the t m that we get here basically t n by 2 pi and then tan inverse 1 minus cos omega n t d ok times sin omega n t d minus omega n t d ok. So, if I substitute this t equal to t m back in this expression r t ok I would get the sorry this should be here r 2 max I would get the r 2 max here ok. So, and you can go ahead and then again plot basically for each phase ok. So, for forced vibration phase and then the free vibration phase ok both condition you can find out what is the r 1 max and r 2 max and then to find out the overall response you can take the maximum both value. So, let me first draw it the r value ok for both of these expressions you know you can do it numerically or you can do it analytically ok. So, this I am going to write down the horizontal axis as let me draw here this is basically t d by t n because that is the parameter on which my r max depends on ok. So, what happens in this case ok for the first case I am first going to plot the forced vibration response if the my maximum occurs during the forced vibration response I will see that the curve looks like something like this ok and for the free vibration response ok if I try to find out it would look like something like this ok something like this. So, this is basically r 1 max and r 2 max and this point at which this crossover basically is you can find that as the crossover point as t d by t n equal to 0.371. So, when t d by t n is less than 0.371 the maximum response is governed by r 2 max ok when it is less than 0.371 when it is greater than 0.371 it is governed by the r 1 max the forced vibration response ok and then you can find out the overall maximum response of the system. Now, note that this is different from the case for the step excitation for which the t d by t n if it was smaller than 0.5 then my free vibration response governed ok. So, in that case my r 2 governed and when it was greater than then the forced vibration governed ok. So, this value is actually different ok. So, that you need to keep in mind ok. So, once you have the r value basically given the plot of r as a function of t d by t n ok and r is nothing, but what r is basically your peak dynamic displacement divided by the peak static displacement which is p naught by k and if we have this r curve for any pulse pulse type excitation ok given the property t d and the property of the system t n t n we can find out what is the r value and then we can find out the what is the maximum dynamic displacement ok. So, there are several curves like this ok. So, some of those curves like you know that you can encounter in real life are let us say sine pulse excitation ok you could also have double triangular excitation something like this or you could have like you know different type of pulse excitation ok. And like it is not possible to do all find out the response to all of those excitation, but it is worth noting that how to develop the solution. So, you understand the procedure. So, we have discussed step force excitation and we have discussed triangular ok. So, we are not going to discuss these, but you can refer to any standard textbook to find out the basically r for each ok. So, response modification factor is a function of t d ok. So, this is p here, this is p naught, this is p naught, this is p t here ok. Again this is well I should not say like this, let us say it is like this ok. So, it is p t and this is t d here ok. So, for all of these cases you can find out what is the value of r versus t d by t n ok. And it could be of any general shape and that is what is important to understand and also how the response would differ for different different type of pulse excitation ok. Response that is known then the procedure is similar to find out the maximum dynamic response in the system alright. Now, we are going to switch over to different a special situation of these pulse excitations ok and then see how does that actually affect the response. Let us we saw that for all these pulse excitation when t d by t n ok was greater than let us say some value ok, let us say it was greater than 0.5 ok. So, in those cases we saw that the maximum or the peak response ok peak response occurs during the forced vibration phase during the forced vibration phase ok. And if it occurs during the forced vibration phase ok. So, if it occurs during this phase here or this phase here or this phase here then it what is the shape of the pulse it matters a lot it plays a big role in finding out the peak dynamic displacement ok. And that we can demonstrate by comparing the curves for ok. So, let us say I have three type of excitation ok. One type of excitation is the step excitation alright in which the forces p naught and duration is t d. Second type of excitation is let us say sine excitation ok in which again the peak is p naught and this is t d and the third type of excitation ok is actually triangular excitation which is t d and peak p naught ok. Now, before getting into the mathematics of it can you imagine if you have been given these three type of pulse excitation just by looking at the nature of the curve which is going to provide you the maximum response ok. Think about it for a second and then see how to find out well if my t d is greater than 0.5 or let us say the peak response occurs during the forced vibration phase. Then the peak response depends on the shape of the curve ok and which would be higher among three depends on the how fast the load is applied for each of these cases. So, if you look at it for the step type excitation the load is applied here suddenly compared to the sine half sine pulse excitation in which it is little bit more gradual compared to this one. And then if you consider this triangular pulse then the peak excitation or the forces actually the rate of increase of force is again smaller than these two. So, if the peak is say then analytically I can say that my step excitation the step force would provide the maximum response in the system ok. And you can go ahead and like you know compare the actually the curves of r versus t d by t n and that would be pretty much evident. So, if you try to plot this let us say this is the r value and this is t d by t n. So, we know that for a step excitation the r or the maximum response can reach up to 2 or the dynamic displacement could be up to 2 times the static displacement ok. So, I am going to plot it here ok and this looks like something like this alright. And this value here is actually 0.5 ok. Let us say this is 1, this is 2, this is 3 like that ok. And this is here 2 and let me show this one here is 1 ok. So, this curve we had already derived and you can go ahead and see that if that is the case or not ok. And if you consider the sine pulse excitation or triangular excitation again it would increase like this with smaller initial slope ok. And it would give you a dynamic response which is higher than static something like this. And for triangular it will start again like this with again little bit smaller slope and then it would give like something like this ok. So, both of them are smaller ok. And this I am just like you know drawing it from the expressions that you would get and you can refer that to any textbook ok. But this is typically how it looks like ok. So, you can see that depending upon the shape of the curve it matters a lot ok. If the forced vibration response occurs during the forced vibration phase or for case where T d is greater than 0.5. Now, let us consider a case in which T d by T n is actually very very small ok. In that case basically what I am saying that either it is so small that the. So, in that case when it is very small what will happen? My peak response would occur during the free vibration phase. So, the peak response would occur during free vibration phase ok alright. And if the duration T d is very small compared to T n then what we can write the total response u t as whatever the response due to unit pulse times the whatever impulse due to the load or the pulse or the force that you are applying the pulse force you are applying. So, let us say I have some random distribution ok. So, let me say something like this it does not look very good let us say something like this ok of T d here and peak is P naught this is sub general shape ok. If T d is very less what will happen? The response would actually occur much after the time T d during the free vibration phase ok. So, in that case let us first calculate what is the area of this P naught versus T d curve ok. So, the and that we define as an impulse. So, I can say this would be P d P t from 0 to T d that would be the impulse ok. Now, if the duration is very small we know that it can be treated as an impulse ok. And the response can be obtained as I times the response due to unit impulse function which is nothing but m omega n times sin omega n T or you can use the like you know same expression if you assume that due to impulse you get initial velocity ok. So, but no initial displacement remember impulse only provides initial velocity. So, this gives you u dot 0 is equal to i by m and if you consider the free vibration response with 0 initial displacement it would again give you the same expression ok. So, my ut is nothing but i divided by m omega n sin omega n T ok alright. So, for cases where the pulse duration is very small compared to the time period of the system the response can be represented as this ok. And in this case if you look at this does it matter whether it is a let us say a sign pulse or a triangular pulse or a stiff force as long as my i is same for all 3 cases or the area under the force time curve is same or the impulse total impulse is same for all 3 cases. It does not matter whether it increasing at a slower rate or whether it is increasing at faster rate or basically what is the variation of the or the shape of the pulse ok. So, that is a very important conclusion ok. So, you can go ahead and you can find out the response using this and this is basically the upper bound of the response considering the assumption that all the force is actually concentrated at T equal to 0 which is ok if T d is very small ok. Then we do not have to consider time variation of this one I can just consider this whole impulse to be situated at T equal to 0 and this is the response that we get. So, again we can go ahead and we can compare the responses ok to different type of pulse. Now, let us again consider 3 type of pulses which is basically let us say this is a rectangular pulse P naught by 2 and this is T d and this one is a sign pulse which is here is basically pi P naught pi 4 and this is here is T d ok and then I have a triangular pulse which is P naught T d. Now, if you look at carefully all three curves the total area or the impulse for all of these are same as half P naught times T d ok. This is same for all three. So, the impulse is same for all three ok and if you try to plot let us say U naught divided by P naught by k ok. Let us see what do you get ok. So, if I try to plot this as a function of T by T d what you will see ok the curve actually looks like something like this here ok this is 2 here. So, the initial line is actually same for all of them ok this curve is actually also touches this one. So, it goes something like this here ok this might not be like an exact representation, but this is let us say for comparison purposes. So, what do we see when T d by T n is up to like it is a very small this is actually 0.5 ok when it is half of this let us say 0.25 ok in that situation the U naught or P naught by k actually it is same for all the pulse type motion. So, when T d by T n is smaller than let us say 0.25 ok then the shape of the curve does not matter ok as long as the area under the force variation is same for all 3 pulses ok and the total response only depends on the total impulse ok. So, as long as you can calculate impulse as a time integral of the force ok and if this situation is satisfied here then your response only depends on the impulse ok the area and not the shape of the pulse ok. So, remember that alright. So, this is an important conclusion. So, and this is like an extended. So, I need every time you are given a problem in which you have been asked to find out what is the response to a any arbitrary impulse ok let us say this is here T d you first find out what is T d by T n ok if it is a smaller than let us say 0.5 then you can just go ahead and use this expression you do not have to utilize the expression for R d from like you know different curve ok to find out what is the response this would be a reasonable approximation ok sorry this curve here for this condition alright ok. So, these basically we have discussed response to single degree of freedom system to different type of pulses and the methodology to get the response for force vibration and free vibration and then the peak response alright. Now, let us consider a case which is all too common in reality ok cases in which you have a ground excitation ok. So, when you have ground excitation alright being applied as a pulse force. So, let us see what happens in those cases and one of the very common example is actually when you have a sign type curve here ok. So, and this could represent like you know simple bumper on a road or it could represent like you know any type of non-continuous or like you know non-periodic curvature in the road ok. So, let us say this is basically represented as u g t. So, the acceleration due to ground excitation if it is represented as u g of t equal to peak ground acceleration times sin ok. We are going to represent this pi t d by t m because this is how we represent a half sin pulse excitation. This is for t smaller than t d and this is equal to 0 for t greater than t d where t d is basically time taken ok to cross this ground curvature. So, let us say if I have a vehicle which is moving over with velocity v ok and this whole length is given to me as l then t d simply becomes l by v ok. The length of the this curvature here divided by the velocity of the vehicle v alright and I can represent in terms of excitation something like this oh pardon me this is wrong actually this should be pi t by t d ok not t d by t n is the way ok. So, this is how we represent the this curve here ok alright. So, if the ground excitation is given and remember that here this is in terms of displacement ok this is in terms of displacement and we can differentiate and find out what is the ground excitation ok in terms of time, but let us say this for this case this is how it is ok and we could have this available the ground excitation ok and then let us say this is some function of ok u g naught times sin of some function we do not care ok. Now, we know that for this type of excitation alright what is the peak response ok. So, u naught which is the peak dynamic displacement times the peak static displacement we know that for pulse type excitation something like this is represented as r a d ok. Now, u s t naught is nothing, but peak force due to ground excitation or ground excitation divided by k. Now, we have derived if there is a ground excitation then how can we write down the effective force as minus m u g naught this is the peak value of the ground excitation ok divided by k and this can be further written as if I drop the well let us just keep it like this and denominator I would have k by m which can be further write down written down as omega n square ok and we can drop the negative term it is inconsequential here.  So, if r d is given for any type of pulse excitation that is being applied through the ground ok with peak ground acceleration of u g naught double dot ok I can go ahead and find out what is the peak displacement in the any vehicle that is going over that ground excitation or like in any system to which that ground excitation is being applied ok utilizing the r d for that particular pulse type ground excitation ok. So, it could be like you know something like this or it could be also like this we do not know it yet ok, but this expression is for all the general any type of ground excitation which is which can be approximated as a pulse type excitation ok. One more thing we can see here if we have an undamped system ok for an undamped system we know that ok I can write down this expression here equal to 0 ok remember this is for undamped system for the ground excitation ok and this is the same expression that we utilize when we write this as u t as relative acceleration times the ground excitation and which comes on the right hand side to give the minus m times u g double dot ok, but I can write this for any type of single degree of freedom system subject to ground excitation and if you utilize this I can say that minus my total acceleration as minus k by m times u of t ok which is nothing but minus omega n square u of t. So, the peak value would also be related of both these quantities the total acceleration and the relative displacement ok. So, I can further write this as peak value of the total acceleration equal to omega n square times u 0 and I have dropped basically the negative term here ok. So, this is the peak dynamic displacement this is the peak total acceleration ok and I can go ahead and substitute this expression here.  So, if the ground excitation is given in terms of any of the pulses ok this like you know the sign pulse or the triangular pulse and if we know the r for any of those ok and it could be f n a general shape I do not have to go ahead and solve the differential equation I can just utilize the r value to get the these quantity peak dynamic displacement ok and then peak value of total acceleration in the system alright and this is something similar to what we did for earthquake excitation only thing in this one is that my r d has changed now to this type ok. One more thing to remember you might get confused that we used to write down transmissibility as u t not ok sorry this should be here u t not divided by u g not ok how come this is now r d well remember when zeta is 0. So, we have talked about undamped system my t r is basically equal to r d ok and that you can get from the expression remember t r is nothing, but 1 plus 2 psi omega by remember what was the expression for t r in the numerator I had this term ok I had that term here and then divided was r d basically if damping is 0 then this term is 0 and then basically this becomes equal to sorry r d is not in the numerator this is r d here this becomes r d here ok. So, that is why we get the same expression ok. So, with this the theoretical discussion on the non-periodic excitations are now concluded what we are going to do now we are going to discuss two small problems ok and then we are going to discuss the solution of those problems ok. So, in the first case ok so let us say this is a example one here what do I have basically I have a building alright which is a one story building ok and it has been idealized as a 4 meter high frame ok. So, this is a building here which I am idealizing as a single story frame this is total as 4 meter height is 4 meter ok and it is given that the beam is almost rigid and these columns are actually pinned at the base ok. So, this is like a fixed connection and pin connection at the bottom ok and the properties of the columns are provided ok. So, the properties are columns ok are given as this is as 2772 ok the section modulus which is basically the moment of inertia divided by the distance of neutral axis ok is given as 252 centimeter cube the elastic modulus of steel material for these columns which are these columns are made up of is 200,000 mpa ok and the system basically has natural time period of 0.5 second that is also given to you. So, what has been asked that a pulse type of square pulse load of 20 peak is let us say this is PT here 20 kilo Newton is applied ok the time duration of this is 0.2 second and what do you need to find out the response quantities which are basically the peak displacement of this due to this loading ok and also the peak stress in one of these columns ok. So, the peak stress in those columns alright. So, these are all the data that have been given to you and you need to find out the response of the single story frame subject to this square type pulse motion rectangular pulse motion ok. So, pause here for a second ok and then try to solve this problem alright alright. Let us now discuss the solution for this problem remember as we discussed whenever a pulse type excitation is given ok and the property of the structure is given the first step is to find out what is the value of T d by T n because based on T d by T n we might decide not to utilize the R d at all ok because if T d by T n is very very small let us say smaller than 0.25 then we can just assume it to be a an impulse ok and then calculate the area and find out the response using the expression I by m omega n times sin omega n t ok and if it is greater than that then we will have to resort to using the R d versus the response the response spectra ok the R d versus T d by T n curve. So, T d by T n here is 0.2 times 0.5 which is 0.4 and this is greater than 0.25 second sorry 0.25 not second this is the ratio ok. So, 0.25. So, we cannot assume that the applied forces as a behaves like an impulse and we need to find out what is the R d value. Now, you can go to the chart or the response spectra that we have or for rectangular pulse we know that when T d by T n is smaller than half then the R d is basically given by this expression ok R d is given by this expression 2 sin pi T d by T n and if you substitute all values you will get this one as 1.902 ok. Now, to find out the value of static displacement so that we can find out the dynamic displacement first I need to find out the equivalent lateral stiffness of the system ok lateral stiffness of the system and that I can find out as k equal to remember I have two columns and the columns have boundary condition as fixed here and pinned here ok. So, for fixed pin condition we know that it is 3 EI by L cube and you can substitute all values of EI and L cube and you can find out the this one this comes out to be around 260 kilo Newton per meter ok and we have to well this would be 2 multiplied with this and if you multiply this this comes out to be 520 kilo Newton per meter ok. So, you can now go ahead and find out the peak static displacement which is P naught pi k I know that 20 kilo Newton load was applied. So, this divided by 520 times 10 to the power 3 which basically gives me a value of 3.85 centimeter ok. Now, we know that if I have a single degree of freedom system represented through this frame and if I have a displacement let us say something like this, then the force equivalent to static force is nothing, but the lateral stiffness times the. So, the peak value of the lateral forces lateral stiffness times the peak dynamic displacement. So, the dynamic displacement here is R d times UST naught which is 1.902 times 3.85 and this is equal to 7.32 centimeter ok and we can use that to find out what is the maximum ok what is the maximum lateral force in the system ok that is not difficult to do. Let us see how do we do that ok. So, either we can do this or what we can also do remember if we have something like situation something like this where this one is getting displaced by U naught. So, basically I am saying this is getting displaced by U naught ok. The moment at this point can be written as whatever the lateral stiffness times basically U naught ok. So, this is the sorry this is the rotational stiffness that I need to write down as U naught. So, I can go ahead and substitute the value and then I can find out what is the maximum response of the system. Now, this theta k theta is nothing, but the lateral stiffness times L ok. So, you can go either of these approaches you can also you can either go ahead and find out what is the lateral stiffness times k U naught or you can based on moment you can also find out remember that you will get exactly the same value because k theta ok it is nothing, but k times L ok alright. So, when you do that ok when you do that the value that you would get let me write it down here m you will get as 76.1 kilo Newton per meter and the force F s naught you will get as basically whatever the P naught times R d ok which is 20 times 1.902 ok and this comes out to be approximately as 38.0 kilo Newton ok. Now, remember that this is the force in the total. So, this is the store or this is the force in both columns. So, if you want to consider force in one column you will have to divide it by 2 ok. So, this you need to divide it by 2 ok to find out basically whatever the force you get and that would be 19 kilo Newton ok and same for the moment, moment you can also get as 19. So, the moment now at the top of one of the column ok would be 19 times whatever the length of that column is which is basically again 76.1 kilo Newton per meter which will basically same as this one ok alright. So, this is 76 here ok. So, once you know the moment in any section ok how do you find out the stress ok the stress is nothing, but moment times c divided by i where c is the distance from the neutral axis and if you want to find out the maximum basically stress then this c becomes the half of the section depth ok. So, in this case that would be the section modulus. So, I can directly write this as this value and this gives me a stress of 309.301.9 mega Pascal ok and this is the procedure that we basically utilize. So, this is a very simplistic version of what is actually done during the analysis and design practice, but remember if you have given something like this and you have been told that the yield strength of the steel is 450 MPa just for example ok and these steel columns this steel building is actually subjected to that pulse load. Basically you are going to follow this procedure you are going to find out what is the maximum stress due to the applied load and then you are going to compare with respect to yield stress and then you are going to make the conclusion whether the structure is safe or not whether it is going to yield or not ok. So, this is the procedure that we follow alright ok. Now, let us consider a second example ok. The second example that we have is basically let me just write down here ok. We have an overhead tank alright. So, this is an overhead tank on which a force is being applied at the top and the properties are given this is 20 meters ok and the mass is also being given as ok. So, the mass is given as 50132 kg lateral stiffness is given as 1600 kilo Newton per meter ok and you can calculate the time period as 2 pi m by k equal to 1.11 second ok and the damping is also given 1.24 percent ok. Now, the Pt is basically an arbitrary force for which the variation is given to me. So, it is something like this ok a linear variation up to a peak value of 200. So, this is Pt and the units are in kilo Newton ok. So, it goes to 200 over a time duration of 0.02 second and then it is given like this at 0.05 this is 80 and then 0.06 this is 20 and this Td here at which it becomes 0 is actually 0.08 second ok. So, basically what we need to find out is the maximum force in this alright in the system and the basically maximum base here and the maximum moment for this overhead tank alright. Now, for this type of arbitrary force ok as we discussed the first step is basically find out what is your Td by Tn ok. So, Td is 0.08 Tn is 1.1. So, definitely this is much smaller than 0.25 ok. So, this can be treated as an impulse load ok and we can find out the area under this curve ok and there are multiple ways to do that you can consider to be made up of this triangle and then some of the area like this or you can like you know consider to be like you know made up of multiple using trapezoidal loop rule you can find out the total area. So, I am going to write it as ok remember this is nothing, but 0s to 8 second 0.8 second Pt dt ok and this you can write down first ok all these separations are equal 0.02. So, I can take that out and then I can write it as 0 plus 2 times 20 plus 2 times 80 ok and then 2 times remember this is 220 plus 0 and this gives me as 6 kilo Newton second all right. So, now remember what is the what was the expression if it is a pure impulse ut is equal to i by m omega n times sin omega n t. So, the peak dynamic response is nothing, but i by m omega n which I can write it as i by k and omega n I can write it as 2 2 pi by t n here all right. So, we are going to substitute the values ok and then see what do we get ok. So, 6 into 10 to the power 3 times 2 pi times 1600 times 10 to the power 3 into 1.11 and this gives me a value of 2.12 centimeter. So, once we have the dynamic displacement remember this is the water attack let us say it is deforming like this by u naught. So, the total force at the base would be the lateral stiffness times the dynamic displacement ok and we can substitute those values here ok times 0.2 1 2 ok and this we will get as 33.9 kilo Newton. And once we have the basically the base shear ok we can also find out our moment as ok moment can could be easily found out remember how would it look like if you have this force being applied here your shear force is actually varies like this all right ok. And the moment would actually vary start from 0 and with a constant slope due to constant value of shear force increases up to value m here which is nothing, but whatever the shear force is times the height of this ok which is 20. So, moment is 33.9 times 20 which gives me a value of 678 kilo Newton per meter ok. So, for this arbitrary excitation using the principle of dynamics we have found out what is the total base shear and total base moment ok and then we can go ahead and like you know design the system subject to these forces and moments ok all right. So, with these two examples we are going to conclude this chapter ok all right. Thank you very much.


So the role of the 21st TSC is to set conditions for the success of this exercise. Currently we have soldiers, teams deployed from Estonia all the way down to Greece. A central point of our operation is here at Duris in Albania. So the 104th, like all of our NAID allies, is one of our critical capabilities and teaming with them to take the skill sets they have where they're strong in, couple them with our skill sets to make operations happen such as here on the beach and then particularly down in Alexandria where we have a task force running the port operation in the LSA from the 104 UK embedded into our task force for the 21st TSC. All this gives us the ability to demonstrate that we can project forces, that we work together, we have a strong alliance and strong bonds so that we can demonstrate defeat aggression or deter aggression and demonstrate our capabilities to project combat power anywhere within the European continent. For us, JLOTS is about being able to go into multiple different types of ports. The harbor here is not deep enough for us to bring one of our larger ships in like we would transport an armored brigade combat team. So what do we do? We moor that ship offshore. We set up a mobile dock if you will and do transloads to smaller army watercraft and then bring it the last tactical mile to the actual port. The ability to project fuel for heavy armor formations and to provide and sustain those heavy armor formations is absolutely critical. This is an exceptional capability that shows that we can take a fuel barge, have a floating line and then we can push that fuel into a beach head and then transport it from here inland on army fuel tankers. Now the other piece is that it demonstrates the great partnership that we have between the United States Navy and the United States Army because without the Navy system which basically transfers with the floating hose if you will all the way across the beach we would not be able to make this happen. Everywhere we turn we are demonstrating our ability to project force, deter aggression and then train with our NATO partners. We are in Estonia, Poland, Germany, Albania, Croatia, Greece, North Macedonia and we're not only working with partners like the 104 UK but we're working with those host nation forces as well. So this demonstrates the power and the strength of our alliance, the fact that we're living together, training together and making operations happen together.


Are there any specific tools or software available for publishers to catch ethical concerns or any kind of scientific misconduct in a paper? So there's no specific tools for misconduct first day but we do have tools to use with Adobe Photoshop that allow us to get a more forensics look at photographs and the tools are freely available by the Office of Research Integrity by the NIH on their website and they're just kind of macros that go into Photoshop that help us look more closely at photographs. We also use Identicate for plagiarism but those really are just the two main tools that we use in my office for addressing, looking to see whether a concern is real or not.


Hey there friends. Thanks for checking in once again by popular demand We have Vincent Jackson joining the channel, and we're going to talk about the greatest Mistake that concealed carriers make Vincent. Thanks for joining once again. Thanks for having back band So we want to talk about the concealed carrier and then the most overlooked Mistake or habit or skill that they make what is it the most overlooked skill? It's not going back for additional training Additional training after They get their permit is so important and that's actually how we met is because Vincent's a police officer. He also does fire arm training courses We met because I intended one of his classes and it really was outstanding We're just talking off-camera how we need to do that again But the average concealed carry person which you train quite a few Don't come back for any additional training. Is that true? That's true They don't know do they not come back to me not like I don't say that you have to come back to me And I told them that in that class you don't have to come back to me But go somewhere and get more training from somebody right more than just standing at a range like we are right now Still shooting at a target something that involves more self-defense Situations, yes professional instructions like to call it. Yeah professional instruction now When when you see these people you encourage them to come back How many would you say out of ten actually go acquire additional training one one out of ten? So ten percent of people what go back for this one training whether I see them or I talked to when they say they Went somewhere else one so so this is by far the most overlooked Concealed carry habit. There is is not getting that additional training one of the things I found is that People are they get their permit they are so consumed with the gun How they carry it the holster Spare mag and that's all good and that's all important But that that's not even nearly important as additional training after the permit I saw a post online actually once and it says most people don't need a $2,000 gun They need a $300 gun and $1,700 with the training Then and that's a good point now One of the benefits that you don't hear anybody talk about is that if you have to use that gun and self-defense You can expect to get arrested you can expect to spend the night in jail you can expect to be in front of a jury and a judge and The greatest amount of evidence that you knew what you were doing our certificates that that Show the additional training that you had after your CCW would you agree with that 100% Dan See the thing about it is people get so caught up if you said earlier on the gun It's not the gun even if you take a class and it's a seminar class or lecture class on the law I don't know how you how you're supposed to conduct yourself and things you're supposed to do all that goes Into an account when your attorney is presenting that to a prosecutor or in front of a jury that looks better for you It looks way better because now it's it says if you are The expert you have the training you made an educated choice Based on your training and the one thing that we all fall back on is Training is our training and that's the one thing that very few people want to do one out of ten is like like you said now The main thing that I see you can tell me if I'm wrong is that people don't want to put forth the additional money That's true. Okay, and we all have bills to pay and that's totally understandable You know that we want to put our money in where we need to but if you're going to take the class and you're going to get The permit and you're going to carry a gun which is an incredible responsibility Definitely, I can't think of a better investment than into yourself So that you have the skills necessary in case you need to defend yourself or a loved one The thing about the money factor is people look at it at oh it costs 200 300 400 500 dollars to take a class But we're not saying to take a class every month if you say one structured class once a year It's a big difference and think of all those McDonald's Dollar menu items you buy over the course of 365 days that you can put that money to the side and take that course once a year The people that didn't want to put the money if they found themselves in a self-defense situation They would pay an incredible amount of money to have the skills necessary to be successful. Definitely So any last words for the people out there who maybe have their permit and thinking yeah I heard I should do this, but I haven't yet Wait, what would you tell them if you're waiting to get your training? Don't wait you longer summer times here Go out get the training I agree 100% it's overlooked and that needs to change. Hey, thanks for joining once again Always a pleasure Like I like making videos with this guy and sharing the good information that you always have appreciate it I appreciate you watching and you guys be safe


Hey guys, Justin here from Razorback Offroad. Today we're going to be going over the install of our cargo rack specifically designed for this new model XT. So stick with me and we'll do a quick overview of all the parts and hardware and tools necessary to get this build underway. Next, we'll go over step-by-step process of how to get it on your machine as quickly and as easily as possible. Let's get started. So to start we've got our tray assembly. It comes with three pieces, a left and right side and a center support bracket. Next, we've got the left and right sides of the rack itself with an outer cover plate and an inner mounting bracket.  Okay, let's get started with the driver side. First thing I'm going to do is going to grab one of these inner brackets and four of my carriage bolts with the nyloc nuts and go put them where I can reach them once I get the side on. Next, we'll grab the side panel and get it set in place. So before you go to set your side panel, you need to make sure that this dump support bracket is out of the way. So move that up and you'll see there's a hole in the bottom of the channel designed to fit right over that pinion support. So what you're going to want to do is hold this on kind of a 45-degree angle and watch as you fit it right over that pinion bracket and lift it up into place as it sits right over the top of that bottom bar. Now you'll want to reach in and grab your inner bracket. As we go to install the center bracket, you'll notice that you've got one leg that's longer than the other. It's important that you get the longer leg positioned vertically in the channel. So we're going to set that in place, line up the bolt holes and we're just going to set the bolts in place and get the nuts started just by hand before we start tightening down the inside bolts first. It's important when you go to tighten these bolts that you start with the inner brackets. We'll tighten those down and what that does is helps us make sure that our outer cover plate goes on a lot more smoothly. Now that we've got the four bolts and nuts loosely in place, we're going to go ahead and tighten the inner bolts first. It's important that you do that so that it helps us when we go to install the cover plate on the outside. But before we start tightening, you need to make sure that the side panel is slid all the way forward and as you tighten you want to be lifting up tight on the inner bracket itself. It would be easiest if you hold the bolt in place with the carriage bolt, get that started to where it's starting to snug up and then you can reach in here and hold up tight on that bracket and tighten it the rest of the way down. Okay, now that we've got these inner ones tight, we're going to leave these upper bolts loose and go to the cover plate bracket and once we have the cover plate on, we'll come back and tighten those later. Next, we'll move on to the outer cover plate. If you look closely, you'll notice it has a little tail on the end. That goes towards the rear side of the machine. So we set this in place, slide it all the way to the back to where the plates touch and that will make your holes line up perfectly. Now that we've got our cover plate in place, we're going to put the bolts in.  Again, we're going to want to put these in place and then just get them started and we'll tighten them up in a minute. A good way to tell which bolts you need to be using is to look at the hole that you're going into. You'll notice that these particular holes are round, so they'll take the hex head flange bolts, not the carriage bolts. Now that we've got our cover plate bolts from the outside started, we want to make sure we come in from the top and get our carriage bolts started from the top. So when we go to tighten these down, you're going to need the 7-16 box-end wrench to go on the inside of the channel and the 3-8 wrench to tighten on the outside. It can seem a little tricky, but if you just reach the wrench in this channel from the left-hand side and kind of snake it over to the right bolt, it actually fits in there pretty easily. For the other bolt, same thing, we're just going to reach through this right-hand channel hole, reach over to the other side and get that one next. This front bolt is the easiest to get to because you can reach into it right from the top. This last bolt is going to be the hardest one to get to, but again, if you kind of snake your wrench in there and reach in there and spin this bolt until that latches up, it's not too bad. Now that we've got our four lower bolts tightened up, we'll get to these other bolts and finish those out. Just don't forget to go back and tighten up the first two bolts from your inner bracket. Once you've got those four bolts tightened up, don't forget to go back and get these upper bolts with your 7-16 wrench. Okay, now that we've got all of our bolts tightened up, we'll move on to the other side of the machine and do the other side following the exact same process. Now I'm going to send you over to Andrew. He's going to help you put the tray together. He's already gone through this before and done a really good job of showing how to make sure we get this tray all tightened down together in order and then we'll get it put on the machine. Once you have your sides installed on your machine, the next thing you're going to assemble is the tray. So what I like to start with, I like to start with one side of the tray. And what I'll do is I'll hang one end of the tray off the table a little bit. That way you have plenty of access to install your hardware. You're going to grab your center bracket. So this center bracket is going to be installed right on this side. And there's kind of a bent part in the tray here. And what you're going to do is you're going to slide it in there first. And then you're going to slide it down in there. See, if you put this in in first, it won't fit in there. So you've got to slide it in just like this. And you're going to grab your carriage bolts. And it's going to be 10 carriage bolts and you're going to grab your nuts here. So I'll usually like to get one started, so I start in the middle. Okay, you're going to grab your nut. Okay, and you're going to start there. And I usually do the bottoms first and then I do the sides, just the bottom sides here last. So once you have these all installed, you're going to put it in through the side. And you'll notice that on the carriage you'll have this square portion of it. And we've actually cut this into the bracket and also on the tray. So you're just going to match this to where you'll see the carriage bolt sit flushly against the metal here. Because you can install it backwards, but it won't sit right. Okay, so once you have all the bolts installed on the first half of the tray, what you're going to want to do is you're going to want to orient the trays. And I like to hang, again, one part of it kind of off, like so. And what you're going to want to do is you're going to want to install it. Like I said, with the bent piece in, you're going to want to put this end in first. And then this should slide right down into this tray. And hanging it off a little bit, you can get at least the first one here started. Put this next one down. You may need to kind of prop it up a little bit so you can hold it up like so. And now do the middle bolt, middle bolt here. So once you have the bottom ones in, you've got to put the last ones in on the ends. Okay, and it might take some shimming in here, but it will go in on this side. So that should be all for the carriage bolts. So now you're going to want to grab the flange bolts and you're going to need three of them. There's one bolt that has to go on the front of the tray, and then there's going to be two that are going to need to be installed on the back. So we start with the front one. The front one you may need to just kind of tilt it up to kind of get clear access to it. Okay, and install the two in the back. You may need to tilt it up. It just makes it easier for access. And if you notice, and I am leaving all the bolts loose, it just makes it go together a lot better. Keep the tray together, or make it easier to install the tray. So now once you kind of have all of your hardware in place, now you're going to take your 7-16ths socket and your ratchet, and you're just going to go through all of them and just make sure that they're tight now, now that you kind of have them all in there. So we're going to go ahead and just tighten everything. Okay, and I'll kind of flip it over just to show you. So since it's just go through and make sure all of them are tight, you're going to also need a 3-8 socket and a 7-16ths box wrench. You're going to tighten all the bolts in the back and do the one in the front. Okay, and don't forget the ones in the side, feet this side. Now that you have all the hardware installed on your tray and tightened down, we're now going to take the tray and install it on the side. Okay, so I'm going to set the tray in place with the angled portion facing towards me. Go ahead and set it on the top brackets and slide it all the way to the back until it stops. From here, we'll just go set all of our bolts in place. Again, not tightening down, just get them all started and then we'll go through and tighten them all down. All right, now that we've got all of our bolts in place, I'm going to start tightening them down. I'm going to start with these two bolts here on each side so that it will pull the tray tight all the way to these brackets. As you tighten those down, it's important that you want to leave a little bit of a gap in between the tray and the side itself so that you don't get those rubbing together as you ride. Again, so as you're tightening these down, make sure that you're getting a little bit of a gap and even gap along the side of the tray and the panel of the rack. Now we'll go ahead and finish up the same thing on the other side. So there you have it. The rack is on and ready to go. Keep in mind that this specific model is designed for the Kawasaki Mule Model XT. If you're unclear about whether or not that fits your machine, go to razorbackoffroad.com. There you'll find a detailed fitment list. While you're there, don't forget to check out some of the different accessories that we have that fit all of our racks as well as don't forget to like, subscribe, or comment on this video. And again, hope to see you out on the trail.