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

Nemo1
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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 $$Maximum Deflection = \frac{wl^{3}}{48EI}$$
$W$ $=$ $weight$ $in$ $Kg$ $times$ $9.81mm/s^2$, $l$ $=$ $length$, $E$ $=$ $Flexure$ $Modulus$ & $I$ $=$ $Moment$ $of$ $inertia$

This bit I get:)

The next question has me stumped:confused:; 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.
 
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Nemo said:
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 $$Maximum Deflection = \frac{wl^{3}}{48EI}$$
$W$ $=$ $weight$ $in$ $Kg$ $times$ $9.81mm/s^2$, $l$ $=$ $length$, $E$ $=$ $Flexure$ $Modulus$ & $I$ $=$ $Moment$ $of$ $inertia$

This bit I get:)

The next question has me stumped:confused:; 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)
 
Yes, sorry, it was ten Kilograms as the load.

Cheers Nemo
 
Nemo said:
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)
 
I like Serena said:
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:
$$\sigma_{w}=\frac{\sigma_{y}}{n}=\frac{40MPa}{2}=20MPa$$

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 $$Maximum$ $Deflection=\frac{Wl^{3}}{48EI}$$ 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
 

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

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 $$Maximum$ $Deflection=\frac{Wl^{3}}{48EI}$$ 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.
 
So I think I have solved this:

Using the flexural Strength Formula:

$$\sigma_{fs}=\frac{3F_{f}L}{2bd^{2}}$$

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

$$20=\frac{3 \cdot 10 \cdot 9.81 \cdot L}{2 \cdot 12 \cdot 12^{2}}$$

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

Thanks again for your help Serena.

Cheers Nemo
 
Nemo said:
So I think I have solved this:

Using the flexural Strength Formula:

$$\sigma_{fs}=\frac{3F_{f}L}{2bd^{2}}$$

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

$$20=\frac{3 \cdot 10 \cdot 9.81 \cdot L}{2 \cdot 12 \cdot 12^{2}}$$

$$L=\frac{25600}{109}=234.8623853mm$$ 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)
 
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