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/sci/ - Science & Math

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File: 65 KB, 886x609, area under shear diagram.jpg [View same] [iqdb] [saucenao] [google]
2751792 No.2751792 [Reply] [Original]

how do I find the area under the shear diagram? Should I be trying to integrate something?

>> No.2751835

bump

>> No.2751858
File: 205 KB, 550x400, 128728194734.gif [View same] [iqdb] [saucenao] [google]
2751858

Mathematics

>> No.2751867

Count the pixels.

But seriously, looks like integration.

>> No.2751893

Yes. Didn't your professor cover this in class?
http://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_equation

>> No.2751895

>>2751867
integrating what though?

>> No.2751897

Nah you don't need integration for this.

6 * 12 = load

>> No.2751901

>>2751897
where does the 12 come from?

>> No.2751945 [DELETED] 

M = xw^2/2

>> No.2751952

Look at the load diagram. Then look at the shear diagram. See anything interesting? Then compare the shear with moment.

>> No.2751958

>>2751792
bitches don't know about my simply supported beam with distributed triangular load

>> No.2751966

>>2751952

It exponentially increases?

>> No.2752008

>>2751966
The point where the load peaks is where the slope of the shear is greatest. Where the shear is 0, the moment reaches a maxima. What do these imply?

>> No.2753030
File: 26 KB, 400x447, 1298117688515.jpg [View same] [iqdb] [saucenao] [google]
2753030

>>2752008

The maximum load is exactly in the middle.

You will see maximum deflection at that point.

I know with a constant UDL the slope will be a nice straight line.

In this case I can see the load goes from 0 - maximum over 6 meters, with each step having an exponential increase but how would I mathematically say that?

>> No.2753884

>>2753030
Firstly, the load graphic is wrong. The slope of the distributed load can't possibly be 6 kN/m, or else your system wouldn't be at equilibrium. What they meant is that the load PEAKS at 6 kN, so it increases at a rate of 1 kN/m. I was beating myself up for not being able to do such a basic problem until I realized that.

For the first 6 meters, the shear at any given point is:
18 - (x^2)/2 kN. Second term is area under the distributed load, which is a triangle.
Integrate for the moment.