# Calculating Axial Stress in a Concrete-Encased Steel Column

• steve321
In summary, the conversation discusses a problem involving axial stress in a concrete-encased steel column. The column is subjected to a concentric axial load of 362 kN and the goal is to find the axial stress in the steel section. The conversation includes equations for Hooke's Law and stresses, as well as discussions on the relationship between axial stress and tensile stress and the use of composite structures. Ultimately, it is determined that the axial stress is not uniform across the cross section and must be calculated separately for the concrete and steel components.
steve321

## Homework Statement

hi, I'm new to this and am trying to understand axial stress. here's the question I've been given:

"consider the 250mm x 250mm concrete-encased steel column, shown above (picture a square with an H-shaped steel beam in it). the steel section is W150 x 22, with Es = 200,000 MPa and a cross sectional area of 2850 mm2, and is centered in the column. the concrete has Ec = 25,000 MPa. The column is subjected to a concentric axial load of 362kN. What is the axial stress is the steel section?

## Homework Equations

well, from the internet so far I've found fa = Pc / A, where 'fa' is the axial stress, 'Pc' is the compressive force, and 'A' is the cross sectional area. so i guess what I'm looking for is the compressive force, since i have the cross sectional area. also it says that 'fa' is expressed in ksi, which is a unit I'm not familiar with. is that the same as psi?

'E' i think is the modulus of elasticity, something I'm reading up on right now. I've found the equation E= f/e which will help me turn the modulus of elasticity into stress (f) and strain (e), but I'm not sure if either of those help me get to compressive force.

is the compressive force 362kn? I'm guessing it'll be in Newtons or Kn. if this is the case, why are they throwing in all those numbers with the concrete? do i have to find out how much of this force the concrete is taking, and then subtract that from the overall force to find out how much the steel is being stressed?

thanks for your help. I've asked my teacher for a handful of questions that should help me get ready for the exam, and i really want to understand each one so that i can apply it to whatever's thrown my way.

Hi steve321, welcome to PF.

1. ksi is kilopounds per square inch. 1 ksi = 1000 psi.
2. E is the Young's elastic modulus, yes.
3. The overall compressive force is 362 kN, yes. This force is divided between the steel and the concrete. You can use Hooke's Law, $\sigma=E\varepsilon$, and the fact that the materials are bonded (and thus the strains are equal), to calculate the stress, and then the force, for the concrete and for the steel.

i'm afraid I'm not following

if the compressive force is 362kn, can i not just plug in the values into the original equation i googled?

fa = 362/2850

fa = .127 (i'm not sure what unit this would be in)

i looked up σ=Eε in wikipedia, and it seems as though i don't have enough known values for this. i know E is 200,000 MPa for the concrete, but do not know its strain or stress (which i thought were the σ and ε, respectively - axial stress is what I'm trying to find out)

thanks

steve321 said:
if the compressive force is 362kn, can i not just plug in the values into the original equation i googled?

No. The 362 kN load is not applied solely on the steel beam; it is applied to the entire column. One has to find the load applied on the steel, then divide it by the cross-sectional area of the steel.

steve321 said:
i looked up σ=Eε in wikipedia, and it seems as though i don't have enough known values for this.

Sure you do. The steel and the concrete are connected, so the strain in each is equal: $$\sigma_\mathrm{steel}/E_\mathrm{steel}=\sigma_\mathrm{concrete}/E_\mathrm{concrete}.$$ That's one equation. And the compressive load is divided between the steel and the concrete: $$\sigma_\mathrm{steel}A_\mathrm{steel} +\sigma_\mathrm{concrete}A_\mathrm{concrete}=362\,\mathrm{kN}.$$ That's two equations in two unknowns.

are axial stress and tensile stress the same thing?

using those two equations you suggested, i was able to plug in the following:

1. σsteel = (σconcrete/25,000)(200,000)
σsteel = σconcrete(8)

2. σconcrete(8)(2850)+σconcrete(59,650)=362kN
σconcrete = 227.76
σsteel = 1822.10

if tensile and axial stress are the same thing, then that should be my answer?

also, can you tell me how you went from σ=Eε to σ(A) = Pc [compressive force]? i can't seem to make that jump.

thank you!

steve321 said:
are axial stress and tensile stress the same thing?

Tensile stress is positive axial stress; that is, axial stress that would tend to pull the object apart rather than push it together. In contrast, compressive stress is negative axial stress.

steve321 said:
also, can you tell me how you went from σ=Eε to σ(A) = Pc [compressive force]? i can't seem to make that jump.

They're not related; the first is Hooke's Law, the second follows from the definition of axial stress (force divided by area).

hold up, can i just use f = P/A (axial stress = axial force/area)?

so f = 362/(2.5 x 2.5)
f = 57.92

and that's the axial stress for the whole column

and then

57.92/6.25 = x/2.850
x = 26.41?

still pretty confused!

Mapes said:
Tensile stress is positive axial stress; that is, axial stress that would tend to pull the object apart rather than push it together. In contrast, compressive stress is negative axial stress.

in the same way, is tensile force positive axial force, and compressive force negative axial force?

Mapes said:
No. The 362 kN load is not applied solely on the steel beam; it is applied to the entire column. One has to find the load applied on the steel, then divide it by the cross-sectional area of the steel.

wouldn't it be easier to just find the axial stress of the whole thing, and then cross multiply to find proportionate stress of just the steel? why would i need to find the load applied to the steel first? would that be disproportionate to the cross-sectional area?

steve321 said:
hold up, can i just use f = P/A (axial stress = axial force/area)

For beams of a single material. For composite structures, the best approach is to split the components up and treat them individually.

steve321 said:
in the same way, is tensile force positive axial force, and compressive force negative axial force?

Yes.

steve321 said:
would that be disproportionate to the cross-sectional area?

Right. The stress isn't uniform across the cross section. (The strain is, because the materials are attached.)

sorry about so many questions, i just picked up a book at the library that has a ton of examples using f=P/A and i don't get one thing

all the questions where P is expressed in kN and area in mm2 i get right. all the questions expressed in kips and inches i get wrong. for example:

determine average axial tensiile stress in a bar if P=50 kips and cross-sectional area is 4 in2

naturally i go 50/4 and assume my answer is 12.5ksi. look it up in the back and the correct answer is 33.9ksi. what gives?!

Mapes said:
Right. The stress isn't uniform across the cross section. (The strain is, because the materials are attached.)

thanks for this - in that case, I'm going with my first answer, although i suspect it's probably 18.22, not 1822. i haven't quite figured out where to put the decimal.

i think what threw me off is that i wasn't aware σ = +fa (hope i have this right)

steve321 said:
sorry about so many questions, i just picked up a book at the library that has a ton of examples using f=P/A and i don't get one thing

all the questions where P is expressed in kN and area in mm2 i get right. all the questions expressed in kips and inches i get wrong. for example:

determine average axial tensiile stress in a bar if P=50 kips and cross-sectional area is 4 in2

naturally i go 50/4 and assume my answer is 12.5ksi. look it up in the back and the correct answer is 33.9ksi. what gives?!

Seems like a book error.

1. the total force is 362kn, and this is spread over the steel and the concrete. so when i add up the force on each of these materials, i expect them to come back to 362kn.

2. using the two equations suggested to me:

σsteel/Esteel=σconcrete/Econcrete.

&

σsteelAsteel+σconcreteAconcrete=362kN.

i plugged in the following to get the stress on both the steel and the concrete:

a. σsteel = (σconcrete/25,000)(200,000)
σsteel = σconcrete(8)

b. σconcrete(8)(2850)+σconcrete(59,650)=362kN
σconcrete = 227.76
σsteel = 1822.10

3. now, if i apply stress = force/area to each of those answers, i get:

1822.10(2850) = 5192700
227(59,650) = 13540550

and when i add those up i get 18733250. this is a far cry from the 362 i started out with, even if there are problems with the decimal.

can someone let me know where i went wrong?

steve321 said:
b. σconcrete(8)(2850)+σconcrete(59,650)=362kN
σconcrete = 227.76

227.76 what? Work this through carefully with the units and you'll see where you went off track.

## 1. What is simple axial stress?

Simple axial stress is a type of mechanical stress that occurs when a force is applied along the axis of an object, causing it to elongate or compress. It is also known as tensile or compressive stress.

## 2. How is simple axial stress calculated?

To calculate simple axial stress, you need to divide the force applied by the cross-sectional area of the object. The formula for simple axial stress is stress = force/area.

## 3. What is the difference between tensile and compressive stress?

Tensile stress occurs when a force is applied to stretch an object, while compressive stress occurs when a force is applied to compress or squeeze an object. They both fall under the category of simple axial stress, with the only difference being the direction of the applied force.

## 4. What are some examples of simple axial stress in everyday life?

Some examples of simple axial stress in everyday life include stretching a rubber band, compressing a spring, or pulling on a rope. It can also occur in structural elements such as bridges, buildings, and machinery.

## 5. How can simple axial stress be managed or reduced?

To manage or reduce simple axial stress, engineers may use strategies such as increasing the cross-sectional area of an object, using stronger materials, or adding additional support structures. Proper design and maintenance can also help prevent excessive stress on objects.

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