Shear force diagrams simple stuff hey?

[studentlife]

shear force diagrams simple stuff hey!?

hey guys this is probably very straight forward stuff for u but i have an example question on a loaded beam. there is a UDL, 2 pure couples and 2 fixed points.

now how do i do the shear force and bendin moment diagram for this

really lookin forward to your ideas thanks

 

[radou]

hey guys this is probably very straight forward stuff for u but i have an example question on a loaded beam. there is a UDL, 2 pure couples and 2 fixed points.

now how do i do the shear force and bendin moment diagram for this

really lookin forward to your ideas thanks

A sketch of the idea would do some good.

In general, the relation dM(x)/dx = T(x) can be very useful when constructing shear force diagram. (T represents the shear force, and M the bending moment.)

Set up the equations of equilibrium for the beam at some point x to get the bending moment function M(x), after calculating the reactions.

 

[studentlife]

heres the pic of everything

how do i include the couples in the diagrams i haven't encountered these b4

 

[Pyrrhus]

First, what is the method you were taught to draw the diagram?

Graphing the piecewise function and/or the area method?

 

[studentlife]

i think ud describe it as the graphing piecewise function although i haven't heard it described as tht b4 ne I am from the uk so not my fault

basically this is what i use

http://www2.umist.ac.uk/construction/intranet/teaching/ul222/exp/sfbmdex.htm"

is there a "better" way I am sort of teachin myself this stuff really ne help would b hugely appreciated iv got another problem I am stuck on but i wona give it a proper go b4 askin learn better tht way if u can solve something yourself

 

[Gokul43201]

how do i include the couples in the diagrams i haven't encountered these b4

So, if you're looking at piecewise section from the left end, you would, as usual have M(x) = sum of reaction to moments from -4kN, R(Ay) and the distributed load, for 0<=x<8, and beyond that point, you also include the reactions to the pure couples, so that M(x) = all these contributions + couple at C, for 8<=x<18, and M(x) = all of previous contributions + couple at E, for x>=18

 

[studentlife]

thanks for such quick responses!

i think i need to practice these diagrams alot! are there any good websites out there or books, i have a mechanics book but doesn't really touch on the diagram side of things i haven't really managed to find a lot on the web.

so basically i jus add the couples to the other loads at C and E, how would this look visually?

this is a pretty awesome forum i read tons of threads already learned alot!
if you don't mind me asking what do all you guys do for a living??

 

[radou]

Here are qualitative sketches of what your http://usera.imagecave.com/polkijuhzu322/mdiag.jpg" diagrams should look like. (The pictures look terrible, but I guess you'll live on it. )

 

[studentlife]

wow those 2 images are great, the shear force was exactly how i pictured it and the bending moment explained the couples

thanks for all ur help

 

[studentlife]

Moments about a Clockwise:

(10 x 5) + [(0.5 x 10) x 5] = (4x3) + (3x15) x (R/b x20)

75 = 57+(R/b x 20)

R/b= 18 / 20

R/b = -0.9

This doesn't seem right, what have I done wrong??

to find reaction R/a using all Upward=Downward

R/a + 4 + 3 = 5 + 10 + 0.9

R/a = 8.9

help

 

[radou]

Moments about a Clockwise:

(10 x 5) + [(0.5 x 10) x 5] = (4x3) + (3x15) x (R/b x20)

75 = 57+(R/b x 20)

R/b= 18 / 20

R/b = -0.9

This doesn't seem right, what have I done wrong??

to find reaction R/a using all Upward=Downward

R/a + 4 + 3 = 5 + 10 + 0.9

R/a = 8.9

help

I'm too lazy to check your calculation, but I can demonstrate how it should look like. Let's set the sum of moments around point B to equal zero, to get the reaction Ra at point A (which is assumed to point 'upwards'). Assume clockwise moments are positive. We have:

-3 + 10 - 0.5*10*15 + Ra*20 - 4*23 = 0.

 

[studentlife]

so ur asuming that Rb will equal 0? or is there another way of findin Rb

this is real interesting really making me think logically about things, i love learning this stuff!

 

[radou]

so ur asuming that Rb will equal 0? or is there another way of findin Rb

this is real interesting really making me think logically about things, i love learning this stuff!

I'm not assuming Rb will equal 0. We only set the sum of moments about the point B to equal 0, right? So, the reaction Rb does not produce a moment at this point. Thus, we have one equation with one unknown, Ra, which we can solve easily.

Further on, you can easily google-up some basic information on the construction of internal force diagrams. If there's something specific you wish to know about, I'll be glad to help.

 

[piguy]

You wouldn't happen to be doing Dr. Badi's Mechanics courswork at the University of Hertfordshire would you, studentlife?

:p

 

[snowJT]

this stuff is probably a little too late for me, but its all very useful, I'm done all those classes for now where I have to draw those damn shear force diagrams and bending moments