Statics engineering: Beam bending problem

[kelvin scott]

Homework Statement

a beam is 10 m long with point loads of 10kN at 3m from left and 20kN 10m from left.

R1 and R2 these supports are 2m in from either end of the beam. The beam has a UDL of 5kN/m-1 along its length.

Homework Equations

(a) determine the vertical reaction at the support(b) sketch the shear force diagram for the beam

The Attempt at a Solution

(10*1)+(10*6)+(20*10) = (R2*6)+(10*2)

im not sure at all

Homework Equations

The Attempt at a Solution

 

[berkeman]

b) sketch the shear force diagram for the beam

Please post your diagram for this. Thanks.

 

[kelvin scott]

ok

 

[Dr.D]

Start by writing a moment sum about each support. That should get you to the reactions rather quickly..

 

[kelvin scott]

ok so the moment sum should be R1
2R1 = (10*1) +(50*6)+(20*8)
2R1= 10 + 300 +160
R1 = 470/2
R1 =235

and R2

8R2 = (10*5) +(50*8) +(20*8)
8R2 =50 +400+160
R2 = 610/8
R2 = 76.25

not sure I am right on the length, and sorry for the delay

 

[Master1022]

ok so the moment sum should be R1
2R1 = (10*1) +(50*6)+(20*8)
2R1= 10 + 300 +160
R1 = 470/2
R1 =235

and R2

8R2 = (10*5) +(50*8) +(20*8)
8R2 =50 +400+160
R2 = 610/8
R2 = 76.25

not sure I am right on the length, and sorry for the delay

A quick check would be to add up R1 and R2 to see whether all the forces in the vertical direction sum to 0, which I don't think is true here...

Also, with your moment equation for R1, you seem to be defining the distances from different spots within the same equation. The general advice when doing these moment equations is to take moments about the supports. Why? If we take moments about the supports, then the reaction at that support will equal \vec 0 as \vec Moment = \vec r \times \vec F and thus we will only have 1 unknown in the equation (the other support). For example, in your equation for R1, you should take moments about support 2. The equation should read something like (defining clockwise as positive):

6R1 - (5*10) - (5 * 10 * 3) + (20 * 2) = 0 (P.s. you need to remember to turn the UDL into a force and treat it as a force acting in the centre of the beam).

Do the same for R2.

For the shear force diagram, I like to start from the left and move left to right. Conceptually, you should be asking yourself: "If I was to consider everything to the left of where I am looking at now, what force in what direction would I need to provide to keep the beam stable?" That is basically what the shear force diagram is.

If you have further questions, I am happy to respond, but I think this should be enough to make some good progress.

 

[kelvin scott]

ok so

(10*1)+(50*3)+(20*8)=R2*6
10+150+160=r2*6
320 =R2*6
320/6 =R2= 53.33

R1*6=(10*5)+((50*3) - (20*2)
R1*6 =50+150-40
R1 =160/6 = 26.67

my attempt

 

[Master1022]

ok so

(10*1)+(50*3)+(20*8)=R2*6
10+150+160=r2*6
320 =R2*6
320/6 =R2= 53.33

R1*6=(10*5)+((50*3) - (20*2)
R1*6 =50+150-40
R1 =160/6 = 26.67

my attempt

Seems right as we would expect R2 to be greater from looking at the diagram. Furthermore, the forces now add up to 80 kN.

Now, to do the shear force diagram, I would suggest you go away and try to learn about it and make an attempt. It would be quite a bit of information for me to type up here and you would be better served by a youtube video. Simply searching 'shear force diagram' into youtube should yield many good videos (note: the convention used in the videos may be the opposite to that taught in your school, this means that you would flip the graph over the x axis, f(-x) -- this was the case for me)

Once you have made an attempt, I am happy to answer further questions.

 

[kelvin scott]

ok thank you ,.