Cantilevered Beam Question: HELP!!

[Monarch]

Hello!

I am having some trouble with an integral calculus question, can anyone help with some simplification or explanations?

The question goes as follows:

"The vertical deflection y mm of a cantilevered beam is given by:

y= [integral] (F/EI)(Lx-1/2x^2)dx
(sorry I can't copy in the integral sign or the x squared (x^2) symbol)

Where F=load at the free end (N)
E= Young's Modulus of Elasticity (MPa)
I = second moment of area of the beam (mm^4 - again, that's ^4 to the power of 4)
L = length of the beam (mm)
x = distance from the fixed end (mm)

Given that deflection at the fixed end is zero:

(i) Find the deflection of the beam if F=4N, E=2x10^5 MPa, I=1x10^4 mm^4, L=1500mm and x=300mm.

(ii) Write a formula (without integrals) for the maximum deflection of the beam. Hint: A cantilevered beam is like a diving board, fixed at one end. Where on the beam is the maximum deflection?

For the part (i) I simply plugged in the information to the formula given and I get the integral of a number dx. However, the integral of a number is that number times x. But I have no idea what to make of that.

And the second part I have figured that you would replace L with x as the distance from the fixed end is the same as the length of the beam.

Am I on the right track? I would really appreciate some third party insight!

- Monarch.

 

[Deleted member 4993]

Hello!

I am having some trouble with an integral calculus question, can anyone help with some simplification or explanations?

The question goes as follows:

"The vertical deflection y mm of a cantilevered beam is given by:

y= [integral] (F/EI)(Lx-1/2x^2)dx
(sorry I can't copy in the integral sign or the x squared (x^2) symbol)

The way you wrote it - it translates to

\(\displaystyle y \, = \, \int(L\cdot x - \frac{1}{2}\cdot x^2) dx\)

Is that the correct equation.

If not, please write the correct equation with proper grouping symbols (parenthesis - keep PEMDAS in mind)

Where F=load at the free end (N)
E= Young's Modulus of Elasticity (MPa)
I = second moment of area of the beam (mm^4 - again, that's ^4 to the power of 4)
L = length of the beam (mm)
x = distance from the fixed end (mm)

Given that deflection at the fixed end is zero:

(i) Find the deflection of the beam if F=4N, E=2x10^5 MPa, I=1x10^4 mm^4, L=1500mm and x=300mm.

(ii) Write a formula (without integrals) for the maximum deflection of the beam. Hint: A cantilevered beam is like a diving board, fixed at one end. Where on the beam is the maximum deflection?

For the part (i) I simply plugged in the information to the formula given and I get the integral of a number dx. However, the integral of a number is that number times x. But I have no idea what to make of that.

And the second part I have figured that you would replace L with x as the distance from the fixed end is the same as the length of the beam.

Am I on the right track? I would really appreciate some third party insight!

- Monarch.

 

[Monarch]

Yes that is exactly correct except after the integral sign and before Lx-1/2x^2 there is a F/EI (ratio) mulitplied.

 

[BigGlenntheHeavy]

\(\displaystyle y \ = \ f(x) \ = \ \frac{F}{EI}\int (Lx-\frac{x^{2}}{2})dx \ = \ \frac{F}{EI}(\frac{Lx^{2}}{2}-\frac{x^{3}}{6}) \ + \ C.\)

\(\displaystyle f(0) \ = \ 0 \ = \ 0+C \ = \ 0.\)

\(\displaystyle Hence, \ f(x) \ = \ y \ = \ \frac{F}{EI}(\frac{Lx^{2}}{2}-\frac{x^{3}}{6}).\)

\(\displaystyle For \ (i) \ plug \ in \ the \ values.\)

\(\displaystyle For \ (ii), \ f \ ' \ (x) \ = \(\frac{F}{EI}( Lx-\frac{x^{2}}{2}) \ = \ 0, \ x \ = \ 0 \ or \ x \ = \ 2L\)

\(\displaystyle x \ = \ 0 \ is \ a \ rel \ min \ and \ x \ = \ 2L \ is \ a \ rel \ max \ (checked \ graph).\)

\(\displaystyle Ergo, \ f(2L) \ = \ \frac{2FL^{3}}{3EI}.\)

[attachment=0:k1158qaf]xyz.jpg[/attachment:k1158qaf]

 

[wjm11]

(i) Find the deflection of the beam if F=4N, E=2x10^5 MPa, I=1x10^4 mm^4, L=1500mm and x=300mm.

(ii) Write a formula (without integrals) for the maximum deflection of the beam. Hint: A cantilevered beam is like a diving board, fixed at one end. Where on the beam is the maximum deflection?

For part one, be sure to keep track of your units.

For part two, consider that 0 < x < L. Maximum deflection of any cantilever with a single point load is always at the free end of the cantilever. Deflection is in the direction of the applied load.

Further, a cantilever with a point load on it will have a “curved” shape between the fixed end and the applied load. Beyond the load point, the beam will be straight, but rotated through some angle. The angle of the beam at the point of the applied load can be calculated. Using this angle and the length of the free end of the beam, an additional deflection can be calculated. Add this deflection to the deflection calculated at the load point to obtain total deflection of the end of the beam.