Question on strength of materials asked at a phd interview

[chiraganand]

Consider a cantilever beam of rectangular cross section made up of two materials steel and aluminium. an axial force is applied at one end. the question asked was which material should make up the upper part and which should make up the lower part of the beam?

Could someone please help me in this?

 

[SteamKing]

Is the beam horizontal or vertical?

 

[chiraganand]

The beam is horizontal

 

[pukb]

for a horizontal beam loaded axially, it is either in pure tension or pure compression until the interface is perfect. so it really doesnot matter which material is put above or below. the displacement is going to be same for both materials but stresses developed will be different based on ratio of youngs modulus of materials.

 

[AlephZero]

for a horizontal beam loaded axially, it is either in pure tension or pure compression

That is wrong, unless you have discovered a way to make aluminum and steel with the same modulus of elasticity.

The OP's question seems to have some information missing, and doesn't make much sense as it stands IMO - apart from the conclusion that the OP has never come across a bimetallic strip before.

 

[Studiot]

This is clearly a question about prestressing, although there is not enough information supplied to provide a definitive answer since the axial force direction is not specified.

I shall assume that the nonsensical statement "an axial force is applied at one end" really means that an axial force is applied somehow to both ends, eg as a reaction at one end.

The axial prestressing increases either the horizontal tension or compression according to its direction.

Since steel is stronger than aluminium in both tension and compression it should be placed so as to resist the enhanced force.

That is

in the tension zone (bottom) for an axial tensile force

in the compression zone (top) for an axial compressive force

This further assumes true composite action and no mention has been made of the necessary shear connections to achieve this.

 

[nvn]

chiraganand: (1) Are you sure the applied force is axial, and not transverse? In other words, is the applied force horizontal (axial), or vertical (transverse)?

(2) If the applied force is axial, is it a tensile force, or a compressive force?

(3) If the applied force is transverse, is it pointing upward, or downward?

 

[chiraganand]

Guys this was all the info i had been supplied with when the interview had taken place. The force is axial (horizontal) and is applied at the free end of the beam. the only other things provided were the length of the beam and the value of the axial force. i thought if we consider the whole beam as steel or aluminium then the cross section of the beam would change accordingly right cause of the ratio of E1/E2?

 

[pukb]

That is wrong, unless you have discovered a way to make aluminum and steel with the same modulus of elasticity.

The OP's question seems to have some information missing, and doesn't make much sense as it stands IMO - apart from the conclusion that the OP has never come across a bimetallic strip before.

I don't think there would be any other kind of reaction developed other than tension or compression when a beam is loaded axially. Now, if you see as I mentioned earlier, IF INTERFACE IS ASSUMED TO BE PERFECT, there would be equal elongation of both materials leading to different stresses in the ratio of youngs modullus of the material. so it really does not matter which material is placed where.
Please correct me if i am wrong.
In case of a bimetallic strip, coeff of thermal exp of two mat is different. hence strains are different but in this case, strains are going to be equal.

 

[Chestermiller]

It is hard to imagine how it could possibly matter which material is on the top or bottom, unless the weight of the beam were important, and you wanted to minimize the amount of bending. Then, if the free end of the beam were initially tipping downward under its own weight (prior to application of the tension), then putting the steel on the top would tend to create a moment when the tension is applied that would tip the end up in the opposite direction.

 

[nvn]

chiraganand: Notice question 2 in post 7: (2) Is the applied axial force on your cantilever tension, or compression?

 

[chiraganand]

The force applied is tensile force

 

[nvn]

chiraganand: You know the steel will be subjected to higher stress than the aluminum, regardless of whether the steel is placed on the top or bottom. Therefore, to minimize the stress on the steel (and thus maximize the beam axial tensile load capability), place the steel on the bottom. If you place the steel on the bottom, then the compressive stress in the steel due to gravity counteracts the steel applied axial tensile stress, thereby reducing the steel stress, which increases the beam axial tensile load capacity.

 

[chiraganand]

nvn: how can we tell that the steel would be exposed to higher stress as the cross section of both the sections is same.

 

[nvn]

chiraganand: In your beam, strain, eps, is (almost) uniform across the entire cross section. Therefore, sigma1 = E1*eps is much greater than sigma2 = E2*eps, where eps = epsilon = strain, sigma1 = steel stress, sigma2 = aluminum stress, E1 = steel tensile modulus of elasticity, and E2 = aluminum tensile modulus of elasticity.

 

[pongo38]

Can I suggest that if you apply a tensile force to the geometric centre of each end, then, because the materials are different, there will be a secondary effect of bending in addition to the predominantly axial effect? Just as a symmetrical object subject to an off-centre load will bend, so also an unsymmetrical object (as this is) subject to a symmetrical load, will bend.

 

[Chestermiller]

Can I suggest that if you apply a tensile force to the geometric centre of each end, then, because the materials are different, there will be a secondary effect of bending in addition to the predominantly axial effect? Just as a symmetrical object subject to an off-centre load will bend, so also an unsymmetrical object (as this is) subject to a symmetrical load, will bend.

This is essentially what I was getting at in my (admittedly poorly worded) post #10, which went largely disregarded. The unequal tensile stresses will create a bending moment.

Chet

 

[chiraganand]

Chet: as you said because its unsymmetric as two materials are used i get it that it will bend but will the positioning of the different metals make a difference to the bending?

 

[Chestermiller]

Sure it will make a difference. Just calculate the bending moment to see which way it will bend. If you switch the two materials (assuming equal thicknesses) the bending moment will switch signs.