How to Calculate Maximum Load for a Cantilevered Beam?

[gk1989]

I have a college HNC engineering question which I am having problems with, I am OK with calculating the I values, and deflection (if needed). But am wondering on what formulae are required and how to use them, to calculate the maximum loading mass in Kg for a cantilevered beam.

The question is as follows;
A steel cantilever beam extends 2m from the side of a building. The beam has a rectangular cross section of height 100mm and width of 60mm (outside dimensions) and a wall thickness of 10mm.

Calculate the second moment of area, I, for the beam bending in a vertical plane.
I = 1/12bh³ = (1/12x60x100³) - (1/12x40x80³)
I = 3.293x10⁶ M⁴

What is the maximum load which can be lifted by an electric hoist attached to the outer end of the steel beam, if the tensile stress in the beam due to bending is not to exceed 60MN/m². The mass of the hoist itself is 50kg.

modulus of elasticity of steel, E, = 210GN/m²
modulus of rigidity for steel, G, = 80GN/m²


Thanks in advance for any help given.

Greg

 

[minger]

Firstly: units, units, units! You need to convert the dimensions of the beam into meters before you calculate.

As for the stress calculation, the easiest way is just to assume all the stresses are bending. From there, the stress value is:
$$\sigma_{max} = \frac{M_{max}y}{I}$$
Where $$M_{max}$$ is the maximum moment, y is the distance from the centroid to the outer edge, and I is the moment of inertia.

The max moment is easy, it should be located at the "wall". The value should be easy enough for you to calculate. Basically, you can just solve for the max moment based on geometry and material properties, then back out the force applied at the end.

 

[gk1989]

I have an answer, but have not included any detail of E modulus of elasticity for steel, where is this used in the equation? You haven't shown E in the previous posts formulae.

 

[nvn]

gk1989: E is used to compute deflection but not stress on this beam. GN/m^2 is called GPa. And there should always be a space between the numeric value and its following unit symbol. E.g., 210 GPa, not 210GPa. See international standard for writing units, ISO 31-0. The unit symbol for metre or meter is spelled m, not M. And the unit symbol for kilogram is spelled kg, not Kg. See NIST for the correct spelling of any unit symbol.

 

[mathmate]

For a cantilever of length L, the moment at the support of a cantilever beam due to a uniform load w (kg/m)is:
M1=wL^2/2
The moment at the support due to a point load P (m) at the end is
M2=PL
add up the two moments and use the formula supplied by Minger
$$\sigma_{max} = \frac{M_{max}y}{I}$$
to get maximum stress, where
M=M1+M2,
I=second moment of area of the cross section
$$\sigma$$ = stress
From the maximum allowable stress, you will be able to solve for P.

 

[gk1989]

In response to minger where do i add the details of the material properties, (such as rigidity, & strength) into the equation? At the moment i have an answer of a max load of ~200kg - 50kg for hoist = ~150kg. But this does not take into consideration what the beam is made out of.

M = Sigma x I / y
I have calculated the moment using the max stress (60 MN), I and y (distance from neutral axis)
then divided M by beam length (2m) to get the force at end point in Newtons. Then converted the force in Newtons to kg, by using 1kg = 9.80665 N

 

[minger]

It does not matter what the material is made out of. nvm was trying to explain this. In the most simple case, normal stress can be defined as:
$$\sigma = \frac{F}{A}$$
There is no mention of material here.

If I apply 10 lbf of force on a 1 sq. in. area, then I am putting 10 psi of stress into that part. It doesn't matter if its steel, aluminum or marshmallow. Stress is stress. Now, if you want to find the normal strain, then you need the Young Modulus, and
$$\epsilon = \frac{\sigma}{E}$$
If you want to know if the part will break, then you need the material's strength. However, for simple stress calculations, the stress is material independent.

 

[gk1989]

I want to know what the maximum end point load is in kg, which the cantilever beam can withstand without exceeding a stress of 60MN/m2 due to bending, within the beam. Do i need to include any detail of the material (steel) in the equation?

 

[minger]

I want to know what the maximum end point load is in kg, which the cantilever beam can withstand without exceeding a stress of 60MN/m2 due to bending, within the beam. Do i need to include any detail of the material (steel) in the equation?

No.

What does 60MN/m² mean? Its just a number isn't it? You only need the material if you want to compare this stress value to see if the part will fail.

 

[mathmate]

qk1989, the formulae you need should be found in post #5.
All you need to do is to substitute the numerical values, but you have to calculate (which I believe you did) the valiues of I and y (second moment of area and distance from neutral axis).

I paraphrase from post #5:

For a cantilever of length L, the moment at the support of a cantilever beam due to a uniform load w (kg/m)is:
M1=wL^2/2
The moment at the support due to a point load P (m) at the end is
M2=PL
add up the two moments and use the formula supplied by Minger

$$\sigma_{max} = \frac{M_{max}y}{I}$$

The loads are related by M1 and M2, and M in the last equation is the sum of the two, namely
M=M1+M2

Be very careful with units and g when you substitute numerical values.

 

[gk1989]

Many thanks for all your contributions, the answer of 151kg max load using the formula posted by minger allowed me to calculate the correct answer.