Statics- beams internal effects i dont get it

[Fat Ryan]

Homework Statement

this is a picture of an example from the book. I am going off of it to solve a problem that will be on my test tomorrow. i can figure out how to calculate the shear force and moment of each section (though i don't fully understand it), but I am just not getting how they decide on these sections. sorry for the crappy photo, all i have it my camera phone.

the sections are labeled like 0<x<4, 4<x<8 etc. but when they draw the 4<x<8 section in the example they include the 0<x<4 section...? why? and how come x changes with every section. that's confusing the hell outta me. oh and on the first section diagram to the left, how did they come up with (x/4)*100? and 12.5x^2?

thanx

http://img150.imageshack.us/img150/2323/staticsko8.th.jpg

 

[PhanthomJay]

Between 0 and 4, the loading is linearly varying; betweem 4 and 8, it is uniform; no distributed load the rest of the way. The sections are chosen to reflect these different loading variations. Between 0 and 4, the load varies fron 0 to 100lb/ft over the 4 foot length; thus, by taking a free body diagram of this section that includes the left support and cuts through the beam at a distance x from the left, the total loading is the area of the triangular distribution which has a length x, and a height of 100x/4 (from the basic geometry of slopes), thus the area is 1/2 the base times height = 12.5x^2, that's the total of the distributed load at that point. Now when you look at the section between 4 and 8, your free body diagram includes the known reaction at the left support, and you determine shears and moments in accordance with the translational (sum of fy =0) and rotational (sum of moments = 0) equilibrium equations.

 

[Fat Ryan]

thanx, i understand that part, its just the whole x thing that confuses me. i know I am not explaining myself very well, sorry. my dad just explained to me where 100x/4 and 12.5x^2 come from so i got that part now. you know what, screw it. i think i know it enough to be able to solve the problem. i think i just need help with the final part of calculating the max moment and its location x from the left.

my dad pointed out that if you take the derivative of the moment from the 4<x<8 section and set it equal to 0, you get the answer for the distance x from the left...that being 4.47ft. and if you take 4.47ft and plug it back into the original moment equation, thatll give you the maximum moment. what i don't get it how are you supposed to know to use the second section's moment? intuitively it makes sense, but what about on a more difficult problem where its not quite so obvious? how do you choose which moment to use? or is there some way to use them all to get the answer?

 

[PhanthomJay]

thanx, i understand that part, its just the whole x thing that confuses me. i know I am not explaining myself very well, sorry. my dad just explained to me where 100x/4 and 12.5x^2 come from so i got that part now. you know what, screw it. i think i know it enough to be able to solve the problem. i think i just need help with the final part of calculating the max moment and its location x from the left.

my dad pointed out that if you take the derivative of the moment from the 4<x<8 section and set it equal to 0, you get the answer for the distance x from the left...that being 4.47ft. and if you take 4.47ft and plug it back into the original moment equation, thatll give you the maximum moment. what i don't get it how are you supposed to know to use the second section's moment? intuitively it makes sense, but what about on a more difficult problem where its not quite so obvious? how do you choose which moment to use? or is there some way to use them all to get the answer?

Points of maximum moments occur at points where there is no shear (i.e., where dM/dx = V = 0, as your dad pointed out). I general, there may be more than one location where the moments are at a 'maximum' in a particular section, so it is wise to check all sections where the shear in that section is zero (draw a good shear diagram, and note where the shear crosses the x-axis (V=0), then draw a moment diagram). A 'maximum' moment in one section may be greater than a 'maximum' moment in another.

 

[Fat Ryan]

ok thanx. well i def failed the test today. i don't think i did too bad on the beam problem, but i did HORRIBLE on the centroid problem :( i think my prof picked the hardest possible centroid problem and then made it even harder.

 

[PhanthomJay]

ok thanx. well i def failed the test today. i don't think i did too bad on the beam problem, but i did HORRIBLE on the centroid problem :( i think my prof picked the hardest possible centroid problem and then made it even harder.

Yes, i'll have to agree that your prof is not giving you the easy ones... If you can handle the shear and moment diagram for the example you presented, you can handle 'em all...

 

[Fat Ryan]

the one on the test was a little easier, but i know i screwed up at the points where the sections end. i am still confused about that and i think i ended up leaving an entire force out, lol. i was also rushed because i wasted too much time on the centroid problem. but i will see tomorrow what i got. just got to worry about the final now :(

 

[Fat Ryan]

btw, thank you for the help :)