Sketching load beam forces: a point load of 500 N and a distributed load of 1000 N/m

Scott92

Scott92

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1681567715334.pngIm not sure if anyone can help me with this one,
The beam is subjected to a point load of 500 N and a distributed load of 1000 N/m, as indicated in the diagram. The weight of the beam is 2000 N and may be considered as a 2 kN point load acting downwards at the middle of the beam

To determine the equivalent force of the distributed load we simply multiply the value of the distributed load by the length of the beam over which it acts, as follows.

Fd=(1000n.m-1)(5m)=5000n or 5kn

B. Sketch the beam with all the applied forces (including reaction and equivalent forces) at the correct position on the beam. Below is what ive got but how do i work out how far across the beam and what force is being applied ? Thanks for any help
ktmmuxzh.png
 
khansaheb said:
Your diagram is incorrect.

You have NOT shown the effective distributed load (1000 N/m) and the effective WEIGHT of the in your sketch (Free Body Diagram or FBD). Showing those is an absolute necessity for solving the problem.
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Yeah sorry I should but this what I have so far I'm just unsure as where in distance the 1000n/m goes
 
Scott92 said:
Yeah sorry I should but this what I have so far I'm just unsure as where in distance the 1000n/m goes
Click to expand...
Other ones I've seen have like different measurements eg be like 4kn at 4 m 7kn at 8m etc and Im not sure how get force applied and distance so can be used on the sketch if that makes sense
 
Have you taken course in STATICS where "replacing distributed load by concentrated load" was discussed? A refresher can be had at


You replace the "uniformly" distributed load by a concentrated load of (1000 n/\(\displaystyle \cancel m\) * 5 \(\displaystyle \cancel m\) =) 5000 N. You will place this concentrated load at the center of the distribution (2.5 m away from the point B).

Similarly you'll have to replace and locate the "weight" (2000 N).
 
khansaheb said:
Have you taken course in STATICS where "replacing distributed load by concentrated load" was discussed? A refresher can be had at


You replace the "uniformly" distributed load by a concentrated load of (1000 n/\(\displaystyle \cancel m\) * 5 \(\displaystyle \cancel m\) =) 5000 N. You will place this concentrated load at the center of the distribution (2.5 m away from the point B).

Similarly you'll have to replace and locate the "weight" (2000 N).
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I haven't will have read of that thank you for your help appreciate it
 
Scott92 said:
No I haven't this part of a HNC course a science module
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If that's the case, then I very much doubt that you are expected to calculate the magnitudes of the Reaction forces, so all you need to do is complete your existing diagram by adding the equivalent point loading on the beam (due to the distributed load) at the midway position along the length of the beam over which it is acting plus the weight of the beam itself, acting downwards from point B; ie: you only need to add two further (arrowed) lines to your original sketch. ?

If you do need to determine the magnitudes of Ra & Rc, then you simply need to know that the Sum of the moments at A (& C) must equal zero (else the beam would be turning), so you just need to add up all the clockwise moments at A and the Reaction force at C (Rc) must be creating an equal & opposite moment to those.
(I trust you will already know how to calculate moments if this level of detail is required of you?)

Then do the same with the anti-clockwise moments at the other end (at C) to find Ra.

For example, if the beam had NO loadings on it, then its (2kN) weight (acting downwards at B) would produce a 10kNm clockwise moment at A, therefore Rc would need to produce a 10kNm anti-clockwise moment (at A) to balance that and so Rc would require to be 1kN acting vertically upwards (at a distance of 10m from A) to do that.

Similarly, in this (facile) example, following the same procedure at the other end of the beam (at C) would produce the same result for Ra; ie: Ra = Rc =1kN (so each support carries an equal share of the beam's 2kN weight).

In this very simple case, intuition alone would lead most people to expect that the 2kN weight of a uniform beam would be equally shared by the supports at either end (thus producing a 1kN Reaction force at each support) but when there are other forces acting at differeing lengths along the beam in addition to it's weight then, although it's no longer quite so simple (you can't now guess what Ra & Rc are) the procedure is just the same.

The Sum of the moments at each end must be zero and it is the Reaction at one end that "balances" the moments created by all the forces acting to produce moments at the other end.

Hope that helps. ?
 
1681646751599.png This is what ive got and as mentioned i do need to work out the reaction forces which i cant get to balance does that mean ive gone wrong on this diagram?
Im doing
3(500)=1500
5(2000)=10000
7.5(5000)=37500
add answers divide by 5 =9.800kn
and im confused at how to do Rc
How i did which is wrong..
2(500)=1000
5(2000)=10000
6.5(5000)=32500
add together and divide by 5 gives 8.700kn which is too high to balance so i know is wrong do i divide one by 20m instead
 
Scott92 said:
Im doing
3(500)=1500
5(2000)=10000
7.5(5000)=37500
Click to expand...
What principle of Physics are you following for calculations shown above?

If you are taking moment of a force, please specify the point/s about which you are taking these moments?
 
1681650633039.png Just trying to copy this from my workbook with different figures the points im after are Rc and Ra which is the two ends
 
Scott92 said:
add answers divide by 5 =9.800kn
Click to expand...
Why are you dividing by 5?

Read the response #8 - very carefully (with pencil and paper to work along).

Have you been taught - the principle of moment balance in static equilibrium (either in Physics or elementary Mechanics)?

If not, do a google search - tell us what you find.
 
It's all self taught
The 5 is from the equivalent force equation or should I be dividing by 20 as it's the length.
Will have a research thanks
 
So in my confused mind here bare with so the downward and upward forces must match
so downward ive got 7.5kn and if i do
3(500)=1500
5(2000)=10000
7.5(5000)=37500 added together divide by 10 =4.9kn at Ra so Rc must be 2.6kn to match the downward as the beams not moving?
 
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