Deflection of beam. Efect of % errors in span and load data.

OK so I'm practicsing some of the questions out of my text book for when I start my course in 2 weeks and finding this one pretty difficult, but here it goes.

Bestend Properties is refurbishing an old hospital that was built in the 1950's. The building's main structure is a steel frame. The company's structural engineers are trying to determine the strength of the existing floor beams by loading them up and measuring the deflection.

The Formula that relates to deflection to the strength and loadings on the beam is:

E=5wL^4/384ly

Where:

E - Is Young's Modulus (strength of material)

w - Is the measured uniformly distributed load on the beam

l - Is the moment of inertia of the beam (a constant due to the cross sectional shape)

L - Is the measured span of the beam.

y - Is the deflection of the loaded beam

After the on-site testing was done, it was discovered that some of the measuring equipment was wrongly calibrated. The errors that were discovered were as follows:

The span L was 5% too long.

The load w was 3% too low.

Using you knowledge of binomial theory, produce a formula to show the net percentage effect of these calibration errors on the value of E (Young's Modulus).

I'm not even sure where to start, if anyone could help me I would really appreciate it.

Thanks

Re: Can someone help me with this, I have no clue?

Hi Chip6891,

I think you would do this beam to beam and calculate wrong values then right values. Find % difference. You need size and weight of beam, span , uniform load in kips and midpoint deflection in inches.Also the moment of inertia of cross section

Re: Can someone help me with this, I have no clue?

There are several methods for dealing with this sort of problem, but since the binomial theorem is mentioned I suppose that it has to be this one ...

Replace w with w + $\displaystyle \delta $w and L with L + $\displaystyle \delta$L. Those are the 'original' values together with there respective (actual) errors.

As a result of these changes, E will change to E + $\displaystyle \delta$E.

Now multiply out the RHS (using the binomial theorem for the L to the power four term), retaining the term for E and first order terms only. (i.e ignore any terms involving the products of two or more $\displaystyle \delta$'s.)

The E's on both sides will cancel.

Now divide both sides of the resulting equation by 100E (replacing E with its original expression on the RHS) and you should arrive at a result that says that the percentage error in E is approximately equal to the sum of the percentage error in w plus 4 times the percentage error in L.

Re: Deflection of beam. Efect of % errors in span and load data.

Hi, thanks for getting back.

So when you say "Replace w with w + http://www.mathhelpforum.com/math-he...7d1eba1e18.pngw and L with L + http://www.mathhelpforum.com/math-he...e15357ad29.pngL."

Should step one in solving the equation now look like this?

E=5(w + http://www.mathhelpforum.com/math-he...7d1eba1e18.pngw)(L + http://www.mathhelpforum.com/math-he...e15357ad29.pngL^4)/384ly

or is this wrong? It's ok maths isn't my strongest point :(

Re: Deflection of beam. Efect of % errors in span and load data.

Hi Chips891,

I thought about my first answer and I see a simplification.

E= 5 wl* l^3/ 384 I y where wl is uniform load in kips ( lb per inch) y is deflection in inches

E1/E2 = wl1 (l1)^3/wl2 (l2)^3

wl1/wl2 =1.05 (l1)^3 /(l2)^3= (.95)^3

E2/E1= 1.05* (.95)^3 = .88 so the first E would be lowered by 12%

bjh

Re: Deflection of beam. Efect of % errors in span and load data.

The first line should look like this ...

E + $\displaystyle \delta$E = 5(w + $\displaystyle \delta$w)(L + $\displaystyle \delta$L$\displaystyle )^4$/384ly

The changes in w and L bring about a change $\displaystyle \delta$E in E.

Re: Deflection of beam. Efect of % errors in span and load data.

Hi BobP,

The formula for the deflection of a beam is as given by theOP and is used to calculate this amount using 29* 10^6 psi as E. It can be used to calculate E given a deflection by inserting the other values.In addition to the simplification I show I applied this to a real beam with load and span.If you have an interest I would post it

bjh

Re: Deflection of beam. Efect of % errors in span and load data.

OK, so I have done that.

So for the next step do I have to use Binomial Theorem and expand (L+&L)^4 ?

Re: Deflection of beam. Efect of % errors in span and load data.

Hi Chip6891,

What does your book ask you to do? Are you asked for a statistical analysis after testing all beams in the structure? What course are you going to take?

bjh

Re: Deflection of beam. Efect of % errors in span and load data.

Hi bjh

I'm starting the BTEC National in Construction in about 2 weeks time. I'm doing a practice question out of the BTEC National in Construction and Civil Engineering text book, its a distinction question.

The question is asking me to produce a formula using Binomial Theorem to show the net percentage of the errors to the origional question asked above. I've never had to learn maths on this scale before, but I've got to learn it for this course if I want to get the distinction questions.

Re: Deflection of beam. Efect of % errors in span and load data.

Hello again Chip6891,

I read what I could find on line for BTECand see that it is a2year course in Construction and Civil Engineering.I don't understand what the Binomial Theorem would have to do with this.If it said use basic algebra I have done that.I did make a mistake in post 5.E1 in this question has a higher modulus than E2(post revised)where E2 is the one with corrections.

It is very important in similar problems that all units are consistant.

bjh

Re: Deflection of beam. Efect of % errors in span and load data.

Hi bjh

Yea to be honest I'm a little stumped to how to apply binomial theorem to this equation.

I'm only doing one year of the BTEC Natational as a bridge to HNC in construction. This is because I am already a qualified tradesmen (Carpenter) to level 3. Also because of this I don't think I have to do the maths unit, but I thought it would help as this leads to more professional positions in construction.

It does show me a worked example in the textbook which I do not fully understand.

Find the approximate percentage error in the calculated volume of a right circular cone if the radius is taken as 2% too small and the hight as 3% too large.

Volume (V) of a right cone = 1/3(Pir^2h)

The error in height (http://latex.codecogs.com/png.latex?%5Cdeltah) = h3/100

The error in radius (http://latex.codecogs.com/png.latex?%5Cdeltar) = r2/100

V+http://latex.codecogs.com/png.latex?%5CdeltaV = 1/3(Pi(r-r*2/100)^2(h+h*3/100)

V+http://latex.codecogs.com/png.latex?%5CdeltaV = 1/3(Pir^2h(1-2/100)(1-3/100))

Since the errors in height and its radius are small when compared to the origional lengths, we can approximate using binomial theory that:

V+http://latex.codecogs.com/png.latex?%5CdeltaV = 1/3Pir^2h(1-2*2/100)(1+3/100)

Multiplying out the brackets:

V+http://latex.codecogs.com/png.latex?%5CdeltaV = 1/3Pir^2h(1-1/100) approximate, ignoring the last small term.

V+http://latex.codecogs.com/png.latex?%5CdeltaV = V(1-1/100)

http://latex.codecogs.com/png.latex?%5CdeltaV = -V(1/100)

Therefore. for a right circular cone with a radius taken as 2% too small and a height taken as 3% too large, the calculated volume will be 1% too small.

Re: Deflection of beam. Efect of % errors in span and load data.

We seem to have two threads running wrt this problem, I confess to knowing very little about the 'practical' side of the problem, I'm simply looking at it from a maths point of view.... given percentage changes in the values of w and L, derive a formula for the resulting percentage change in E.

The binomial expansion isn't difficult when you see the pattern, look at the expansions for 2, 3 and 4.

$\displaystyle (a+b)^2 = a^2 + 2ab + b^2,$

$\displaystyle (a+b)^3 = a^3 + 3a^2b+3ab^2+b^3,$

$\displaystyle (a+b)^4 = a^4 + 4a^3b+6a^2b^2+4ab^3+b^4.$

You get the coefficients from Pascal's triangle.

If you see the pattern and you have Pascal's triangle it's easy to write out the expansions for 5, 6, ...

Only the first two terms in the expansion of (L + $\displaystyle \delta$L$\displaystyle )^4$ are required, because we are going to ignore terms involving the products of two $\displaystyle \delta$'s. (We assume that one $\displaystyle \delta$ is small so that the product of any two of them will be small enough to be negligible.)

The next step is to multiply out the brackets on the RHS.

Re: Deflection of beam. Efect of % errors in span and load data.

Hi Chip,

I agree with your answer which I calculated using basic algebra.Good luck with your course and future in construction

Re: Deflection of beam. Efect of % errors in span and load data.

Hi guys

Well anyways I had a little time the weekend to play around with what you have told me, and this is what I have got so far to the original question, I'm just wondering if it is right?

E+&E = (w+&w)(L+&L)^4/384ly

E+&E = (w+&w)(L^4+4L^3&L)/384ly

E+&E = (w+&w*3/100)(L^4+4L^3*5/100)

E+&E = [5wL^4(&w*3/100)(4L^3*5/100)]

E+&E = [5wL^4(1*3/100)(12*5/100)]

I was just wondering if this was right so far, and if so what would I have to do next? I'm guessing I have to multiply out the brackets, but would I have to ignore any terms?