Finding the deflection using the unit load method.

1. Mar 4, 2011

willnich35

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

A uniform beam, supported by a uniform strut, is loaded as shown in Figure (Attached). Using the unit load method, determine the value of the vertical deflection at C.
BEAM: E = 205 kN/mm2, I = 90 x 106 mm4.
STRUT: E = 205 kN/mm2, A = 1500 mm2.

2. Relevant equations

The virtual work equation, which is.

1.$$\Delta$$=$$\sum$$p$$\frac{PL}{AE}$$+$$\int$$$$\frac{mM}{EI}$$dx

3. The attempt at a solution

REAL Reactions at A and B in the x and y directions:

Fx = Ax + Bx = 0
Fy = Ay + 15 + (4*2) - By = 0
Mb = (15*2) - 8 - 2Ay = 0

Ay = 11 kN
By = 34 kN
Ax = Bx = 17 * SqRt(2)

Axial Force (P) in member AB = 17 * SqRt(2)
DB = 34 * SqRt(2)

VIRTUAL Reactions

Apply a force of 1 kN downwards at C

Axial Force (p) in member AB = -2
DB = 2 * SqRt(2)

I then attempted to solve the virtual work equation, however the value of A for member AC is not given.

So i guess my questions are, have I started this correctly, and how do I solve the problem?

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2. Mar 4, 2011

pongo38

I think you must be worrying about the axial shortening of DB, whose A is not given. Even if you know it, and assuming no compression buckling, the shortening of the strut would have a relatively small contribution to make to the deflection at C. You could insert your own practical value of A to assure yourself how small it really is. One might also recognise that AB also has some axial compression, but you don't know that area either. To be pedantic, the hinge at A is off the centreline, so there actually is a small additional moment due to the eccentric axial load there, but I recommend you note this but ignore it. Incidentally I would say Ax= - 11 not + 11. I don't agree your values of Bx and Ax. Have you checked your figure for equilibrium with the reaction components you have calculated, for example by taking moments about C? Have you drawn a force diagram to scale for all the external forces on this structure? That is a good way to reveal errors. On re-reading the question, I see a value of A is given.

3. Mar 5, 2011

willnich35

Thank you very much for your quick answer. Massive help! Should be able to work it out now.

4. Mar 5, 2011

willnich35

I have an answer, which makes sense. I would greatly appreciate it if you would check it over.

Ay = -11 kN
By = 34 kN

Ax = Bx = 34 * SqRt(2)

Axial Force (P) in member DB = 34 * SqRt(3)

VIRTUAL Reactions

Axial Force (p) in member DB = 2 * SqRt(3)

$$\frac{pPL}{AE}$$ = 1.876mm (Deflection due to axial load on DB)

$$\int$$$$\frac{(-x1)(-11x1)}{EI}$$dx1 + $$\int$$$$\frac{(-x2)(-15x2)}{EI}$$dx2

Integrating between 0 and 2 for both x1 and x2

= 3.758mm (Deflection due to bending moments in beam)

TOTAL DEFLECTION = 5.634mm Downwards

5. Mar 6, 2011

pongo38

I don't agree your reactions at Ax and Bx. Have you really checked equilibrium using your values, as I suggested.
And even if I did, how do you get axial force in BD as you do? You are not showing all your working, and therefore your work is uncheckable. In your integrals, you seem to have forgotten the effect of the udl. Is the expression pPL/AE really giving you the deflection at C? I think not. So what is it saying to you?
Finally, your answer has ridiculous "accuracy". It is disproportionate to the assumptions being made. In that sense, it is wrong. Please try again. It's worth it.

6. Mar 6, 2011

willnich35

Ok, thanks again. Not my strongest subject this, il have another go.

7. Mar 7, 2011

willnich35

Ok, here's my next attempt
I've done it in word, as the workings are easier, and attached a picture.

I really hope the reactions Ax and Bx, and the force in member AD are right now, the triangles should explain how i got the numbers.

As for the virtual work equation,

dx1 works from A to B
dx2 works from C to B

M1 is the real bending moment equation of section AB
m1 is the virtual bending moment equation of section AB
M2 is the real bending moment equation of section CB
m2 is the virtual bending moment equation of section CB

M1 hopefully now incorporates the moment due to the UDL.

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8. Mar 7, 2011

willnich35

Just noticed the second line should be Fy= not Fx=

and Ax = -Bx

9. Mar 7, 2011

pongo38

I don't agree your result, doing it a different way. For example, PL/AE is the elastic shortening of DB, and I get 0.44 mm. This has a vertical component of 0.44 / sqrt 2 = 0.31 mm. So the deflection at C due to elastic shortening of DB should be 0.62 mm, not 1.25 mm (your figure). I agree your result for bending = 4.19 mm.

10. Mar 7, 2011

willnich35

Thank you, that makes sense.
However it does seem to conflict slightly with my notes, so il go have a chat with my lecturer tomorrow.
Thanks for your time and help though, huge help!