1. May 19, 2007

### Double A

1. The problem statement, all variables and given/known data

See attachement titled p8.jpg

2. Relevant equations
The following equations apply to a cantilevered beam held fixed on the left end:

$$\delta_{max} = -\frac{PL^3}{3EI}$$

$$\theta = -\frac{PL^2}{2EI}$$

$$\delta = \frac{P}{6EI}(x^3-3Lx^2)$$

The following equations apply to a simply supported beam with a constant distrubuted load:

$$\delta_{max} = -\frac{5\omega L^4}{384EI}$$

$$\theta = -\frac{\omega L^3}{24EI}$$

$$\delta = -\frac{\omega}{24EI}(x^4-2Lx^3+L^3x)$$

3. The attempt at a solution

I started my superposition solution as shown in attachment s8.jpg. I wanted find out if the deflection caused by the distributed load between points A and B is equivilent to the amount of deflection that would occur at point C in the first case. For the second case is the deflection caused by the load occur between points B and C or accross the entire beam? I'm having a difficult time trying to find the best equation to discribe each case.

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• ###### s8.JPG
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Last edited: May 19, 2007
2. May 21, 2007

### PhanthomJay

You can't apply the fixed cantilever deflection for the point load case, because there is no fixity at point B (the beam can rotate at B). You apparently (?) have access to tables for determining deflections and slopes for various loading and support conditions. For the uniform loading between A and B, you can calculate the deflection at C geometrically by using the slope of the deflection curve at B as the slope of the straight line deflection betwen B and C. Then, separately, find the deflection at C , from tables, for the point load case of a beam overhanging 2 supports with a load P at the free end, and set the deflections equal to solve for P. If you don't have tables, I guess you'll have to grind it out using calculus or shear-moment-slope-deflection diagrams. Did I understand your question correctly?

3. May 22, 2007

### Double A

Your response gets me going in the right direction. I'll try what you mentioned and see what I turn up with. If I run into difficulties I'll look for further consultation. Thanks for leading me in the right direction. I was not exactly sure how to find the deflection occuring at C based upon the loadings.

4. May 26, 2007

### haynewp

Use virtual work. Set the deflections equal for each case and solve for P. I would guess this is how your instructor wants it to be solved and not just by manipulating existing tables.

5. Jan 27, 2009

### deviferr

I agree with haynewp

6. Jan 28, 2009

### PhanthomJay

I disagree with haynewp. The problem statement lists table deflections and slopes for certain cases. I would guess this is how the instructor wants it to be solved, not by using tedious calculations. Only the instructor and Double A knows for sure, and he or she has probably graduated by now.

7. Jan 28, 2009

Just my feelings, but in an exam you'll not have tables and other aids to help, so your better of going through the numerical calculation method with calculus.

You can solve this using maculay method, well as far as i can see.

I have not really looked at it, so maybe no help. But seems to me you need to calculate the deflection at c when the load is applied between a and b.

I would then calculate how much load i would need to apply to c in order to deflect c downwards by the value of the deflection upwards, thus deflecting c back to zero.

Done something similar to this, if i find the notes i will post them.

8. Dec 24, 2009

### bcoub

hey guys~
i think i have found something,
1- deflection caused by the uniform load = qL^3/24EI
where L is 10, as i remember..
2- deflection caused by p =2pL^3/3EI

^_^