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"yeah,all you have to do is Using equation (1), solve for YB: -63200 (k-ft3)/EI + 2666.67 (k-ft3)/EI * YB = 0 YB = 23.7 Multiply the unit load, Q, at YB by 23.7 to get the final reaction. The positive answer indicates that the reaction is in the direction of the applied unit force. simples " in metric?? lol and dont say simples, my lecturer says that all the time!!! | |||
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"yeah but to truely learn you must do it yourself so for your own good i am not going to help you " pmsl cheers for that lol remember you said that when they find me delerious and banding my head on the wall lol | |||
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"yeah,all you have to do is Using equation (1), solve for YB: -63200 (k-ft3)/EI + 2666.67 (k-ft3)/EI * YB = 0 YB = 23.7 Multiply the unit load, Q, at YB by 23.7 to get the final reaction. The positive answer indicates that the reaction is in the direction of the applied unit force. simples " All that workin out just to figure out if you have the right condom | |||
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"GOOGLE!!!!! can you tell I have no clue ( and not just about this post) xx" thats why i am stuck even GOOGLE doesnt know !!! i keep asking, i even added please google just said i was needy | |||
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"Not sure, r u talking - II theorem of Castillan - Maxwell-Mors integrals - Theorem of Vershchagin here? " can't be verschagin's because then you would have to include Logical operations and Kolmogorov complexities into the minimum length programs to translate between given strings | |||
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" Granny. Google is indeed your friend, especially if you use spell checker too. " Laine. I did not use Google .... ffs ..... I worked it all out in my head.... I think he may have been talking about Separate moment equations required for different regions of beam. e.g. After integrating equations separately to get slope and deflection equations, extra constants of integration found by equating deflections and slopes at points where 2 equations apply, e.g. at B. integrating 2 equations twice gives 4 constants of integration 2 found by using v= 0 at A and C 2 found by equating slope and deflections at B. Method becomes very complicated if more points loads - other simpler methods available This tho really ......... | |||
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"I can write ESSO OIL on my calculator... well if I turn it upseide down it reads that... does that help?" Can you do BOOBS tho ? | |||
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"Oh alright ..... I got it from the same site as Be Nice ....... " I'm still trying to get Scotty to beam up me. | |||
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"Not sure, r u talking - II theorem of Castillan - Maxwell-Mors integrals - Theorem of Vershchagin here? can't be verschagin's because then you would have to include Logical operations and Kolmogorov complexities into the minimum length programs to translate between given strings " fair comment | |||
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"I can write ESSO OIL on my calculator... well if I turn it upseide down it reads that... does that help? Can you do BOOBS tho ?" sadly...... NO | |||
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"Oh alright ..... I got it from the same site as Be Nice ....... " fair comment | |||
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" Serious questions " that rules me out..... | |||
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"Can anyone tell me the force required to hold a 10 tonne load, which is on steel wheels, on rails, on an incline of 29 degrees? Also, is elastic stretch of steel wire proportional to load applied? Serious questions " gerrof, sposed to be fun on ere! | |||
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"Can anyone tell me the force required to hold a 10 tonne load, which is on steel wheels, on rails, on an incline of 29 degrees? Also, is elastic stretch of steel wire proportional to load applied? Serious questions " Force will be load * sin theta So 10 tonne * 9.81 (Gravity) * sin29 ie. 10*10^3 * 9.81 * sin29 = 47559 kN I think Also strench is due to strain and strain is proportional to stress, stress as iduced by the load. So yes....its youngs modulus | |||
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"yeah,all you have to do is Using equation (1), solve for YB: -63200 (k-ft3)/EI + 2666.67 (k-ft3)/EI * YB = 0 YB = 23.7 Multiply the unit load, Q, at YB by 23.7 to get the final reaction. The positive answer indicates that the reaction is in the direction of the applied unit force. simples " fookin teachers pet! | |||
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