
#37
Mar2110, 09:16 PM

P: 87

yes i understand it.




#38
Mar2110, 09:24 PM

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You can't pass the module with a 1 hour lecture. Unless you studied it on your own for many many hours after that. I don't know what you were expecting.




#39
Mar2110, 09:27 PM

P: 87

it is not just bending on beams. There are many others such as second moment of area, kinetics, kinematics...




#40
Mar2110, 09:33 PM

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#41
Mar2110, 09:35 PM

P: 87

Hello. I did some work and i need some tips for the next stage:




#42
Mar2210, 12:41 PM

P: 87

Hello. I did some work and i need some tips for the next stage:
I found the P1, P2, Mx, Qx P1=20*2.15=43 N and it acts (2.15/2)+0.1=1.175 from the right end P2=0.5*2.15*20=21.5 N and it acts (2.15/3)+0.1=0.816 from the right end therefore P1+P2=Qx therefore Qx=64.5 N so Mx=(43*1.175)+(21.5*0.816)= 68 Nm therefore R=Qx=64.5 so Rcos15=62.3 and Rsin15=16.69 is that right? Cheers 



#43
Mar2210, 09:28 PM

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I'm glad you did not give up. Those values are correct, but i'm not sure why you call it Qx and Mx; the value you calculated for Qx is the vertical upward internal reaction at the right end of the horizontal beam (call it Q), and the value for Mx (call it M_right) is the internal clockwise moment at the right end of the horozontal beam. You still need to calculate the moment at the support. Now you need to draw the shear and moment doagrams as requested iin the problem..




#44
Mar2210, 10:04 PM

P: 87

yes... here is the trouble. how can i found the moment on the fixed support(i think that is the same as the upward but in different direction)?




#45
Mar2210, 10:44 PM

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l already gave you the equation for the moment at the fixed base support, in a sketch many posts ago. That is one way of doing it. But you need to calculate the values of x1 and x2, using basic geometry and trig.




#46
Mar2310, 12:33 AM

P: 87

I did it. I found 64.5. I have to use the Rcos15 to find the x1 and x2?




#47
Mar2310, 05:45 AM

P: 87

I use trigonometry and i found the x1 and x2 and i found the Mx at the fixed end of the framework which is : 134.805. ??




#48
Mar2310, 05:49 AM

P: 87

What is the total plan and the next stages to solve this question?




#49
Mar2310, 06:17 AM

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#50
Mar2310, 06:22 AM

P: 87

i know that we have to cut the beam in x values to find in each point of x the value of moment and force. ?
for AB the Mb and Qb will zero then if i start from the left of the beam then for BC part i will cut the just after the first force which is the force of triangular the problem i found there is that is that i cant find many unknows x to have quadtrative equation because always the unknows x is just one like 8x3 and this is a straight linear equation. ?? 



#51
Mar2310, 06:38 AM

P: 1

We've got a similar problem(the exact same) to this one for our assignment and we were wundering if the strength of the tubeing by workin out the external area then taking the internal area away from it to get the area of the piping and then divide the values of the forces we got from the wieght of the billboard. we did this but the values we got were extremly small compared to the yeild stress of the low carbon steel. the value we got was 0.2MNm^2 but the yeild stress is 300MNm^2 thanks for any help you can give




#52
Mar2310, 10:24 AM

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#53
Mar2310, 10:27 AM

P: 87

This is an assignment of a UK university. As you can there are many person with difficulties on this assignment as we teached only 2 hours on this module for the beams..




#54
Mar2310, 10:51 AM

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