Solving Max Distance Between Supports for 2m Beam w/ Yield & Tensile Strengths

[Nemo1]

Hi Community,

I have this question which I have been able to partially answer.
Part 1: From samples, we measured the Flexure Modulus of a material and was asked to solve for the deflection of the material at a $1m$ span and $1kg$ which I used \(\displaystyle Maximum Deflection = \frac{wl^{3}}{48EI}\)
$W$ $=$ $weight$ $in$ $Kg$ $times$ $9.81mm/s^2$, $l$ $=$ $length$, $E$ $=$ $Flexure$ $Modulus$ & $I$ $=$ $Moment$ $of$ $inertia$

This bit I get:)

The next question has me stumped; Given a beam $2m$ long with a cross section of 12x12mm with a yield stress of $40MPa$ and a tensile Strength of $70MPa$ with a safety factor of two what is the maximum distance between the supports?

I am after a pointer in the right direction, can I somehow derive the formula above to solve?

Cheer Nemo.

 

[I like Serena]

Hi Community,

I have this question which I have been able to partially answer.
Part 1: From samples, we measured the Flexure Modulus of a material and was asked to solve for the deflection of the material at a $1m$ span and $1kg$ which I used \(\displaystyle Maximum Deflection = \frac{wl^{3}}{48EI}\)
$W$ $=$ $weight$ $in$ $Kg$ $times$ $9.81mm/s^2$, $l$ $=$ $length$, $E$ $=$ $Flexure$ $Modulus$ & $I$ $=$ $Moment$ $of$ $inertia$

This bit I get:)

The next question has me stumped; Given a beam $2m$ long with a cross section of 12x12mm with a yield stress of $40MPa$ and a tensile Strength of $70MPa$ with a safety factor of two what is the maximum distance between the supports?

I am after a pointer in the right direction, can I somehow derive the formula above to solve?

Cheer Nemo.

Hi Nemo! ;)

It seems some information is missing.
To figure out any boundaries, we need to know the expected load (or point force) on the beam.
Is that given? (Wondering)

 

[Nemo1]

Yes, sorry, it was ten Kilograms as the load.

Cheers Nemo

 

[I like Serena]

Yes, sorry, it was ten Kilograms as the load.

Cheers Nemo

With a yield stress of $40 \text{ MPa}$, it takes a force of $40\text{ MPa} \times (12\text{ mm})^2 = 5760\text{ N}$ before permanent deformation occurs.
That corresponds to $576\text{ kg}$.

A point load of $10 \text{ kg}$ is nowhere near.
Something is missing here. (Worried)

 

[Nemo1]

With a yield stress of $40 \text{ MPa}$, it takes a force of $40\text{ MPa} \times (12\text{ mm})^2 = 5760\text{ N}$ before permanent deformation occurs.
That corresponds to $576\text{ kg}$.

A point load of $10 \text{ kg}$ is nowhere near.
Something is missing here. (Worried)

This is the question:
View attachment 5589

I have been able to find the safety factor by:
\(\displaystyle \sigma_{w}=\frac{\sigma_{y}}{n}=\frac{40MPa}{2}=20MPa\)

I looked at trying to find the Young's Modulus from the tensile and yield stress given via the secant modulus, but I don't think that is the right approach.

If I had the Young's Modulus then I would use \(\displaystyle Maximum$ $Deflection=\frac{Wl^{3}}{48EI}\) Formula and set the deflection to $0$ and solve for the length $l$.

Here is where I am confused, there must be a method to solve using only the information shown.

P.s. Many thanks for your help so far Serena.:D

Cheers Nemo

 

[I like Serena]

This is the question:I have been able to find the safety factor by:
\(\displaystyle \sigma_{w}=\frac{\sigma_{y}}{n}=\frac{40MPa}{2}=20MPa\)

I looked at trying to find the Young's Modulus from the tensile and yield stress given via the secant modulus, but I don't think that is the right approach.

If I had the Young's Modulus then I would use \(\displaystyle Maximum$ $Deflection=\frac{Wl^{3}}{48EI}\) Formula and set the deflection to $0$ and solve for the length $l$.

Here is where I am confused, there must be a method to solve using only the information shown.

P.s. Many thanks for your help so far Serena.:D

Cheers Nemo

You should have a formula for the bending stress $\sigma_{bend}$ in a beam.
Can you find one?

It should be something like:
$$\sigma_{bend} = \frac{M}{S}$$
where $M$ is the moment in the beam, and $S$ depends on the profile of the beam.

The maximum moment in a beam under a point load $W$ in the center is:
$$M_{max} = \frac{WL}{4}$$
where $L$ is the length of the beam.

For a rectangular beam, as we have here, we have:
$$S = \frac{bh^2}{6}$$
where $b$ is the width of the beam and $h$ is the height of the beam.

 

[Nemo1]

So I think I have solved this:

Using the flexural Strength Formula:

\(\displaystyle \sigma_{fs}=\frac{3F_{f}L}{2bd^{2}}\)

I was able to then plugin my numbers and solve for $L$

\(\displaystyle 20=\frac{3 \cdot 10 \cdot 9.81 \cdot L}{2 \cdot 12 \cdot 12^{2}}\)

\(\displaystyle L=\frac{25600}{109}=234.8623853mm\) which is the distance they have to be apart.

Thanks again for your help Serena.

Cheers Nemo

 

[I like Serena]

So I think I have solved this:

Using the flexural Strength Formula:

\(\displaystyle \sigma_{fs}=\frac{3F_{f}L}{2bd^{2}}\)

I was able to then plugin my numbers and solve for $L$

\(\displaystyle 20=\frac{3 \cdot 10 \cdot 9.81 \cdot L}{2 \cdot 12 \cdot 12^{2}}\)

\(\displaystyle L=\frac{25600}{109}=234.8623853mm\) which is the distance they have to be apart.

Thanks again for your help Serena.

Cheers Nemo

Yep. That's the correct formula and it matches what I wrote. (Nod)