Shear force diagrams simple stuff hey?

  • #1
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
 

Answers and Replies

  • #2
radou
Homework Helper
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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.
 
  • #3
heres the pic of everything

how do i include the couples in the diagrams i havent encountered these b4
 

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  • #4
Pyrrhus
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First, what is the method you were taught to draw the diagram?

Graphing the piecewise function and/or the area method?
 
  • #5
i think ud describe it as the graphing piecewise function although i havent heard it described as tht b4 ne im from the uk so not my fault :smile:

basically this is what i use

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

is there a "better" way im sorta teachin myself this stuff really ne help would b hugely appreciated :smile: iv got another problem im stuck on but i wona give it a proper go b4 askin learn better tht way if u can solve something yourself
 
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  • #6
Gokul43201
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how do i include the couples in the diagrams i havent 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
 
  • #7
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 doesnt really touch on the diagram side of things i havent really managed to find alot 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 dont mind me asking what do all you guys do for a living??
 
  • #9
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
 
  • #10
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 doesnt 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
:grumpy:
 
  • #11
radou
Homework Helper
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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 doesnt 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
:grumpy:
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.
 
  • #12
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!
 
  • #13
radou
Homework Helper
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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.
 
  • #14
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You wouldn't happen to be doing Dr. Badi's Mechanics courswork at the University of Hertfordshire would you, studentlife?

:p
 
  • #15
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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
 

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