# Shear force diagrams simple stuff hey?

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

Related Engineering and Comp Sci Homework Help News on Phys.org
radou
Homework Helper
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.

heres the pic of everything

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

#### Attachments

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

Graphing the piecewise function and/or the area method?

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 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 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

Last edited by a moderator:
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
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

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??

radou
Homework Helper
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. )

Last edited by a moderator:
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

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:

radou
Homework Helper
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.

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
Homework Helper
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.

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

:p

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