# Force/Stress Along a Beam

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1. Sep 27, 2015

### Gwozdzilla

1. The problem statement, all variables and given/known data

Two beams of different cross sectional areas are fused together to create a single beam, as shown in the figure above. The following relationships apply: A1 = 2A2, L2 = 2L1, E1 = 3E2. A tensile force F is applied at the ends of the bar, as shown. What is the ratio of the stress in beam 1, σ1, to the stress in beam 2, σ2?

2. Relevant equations
stress = σ = F/A0
strain = ε = ΔL/L0
σ = Eε

3. The attempt at a solution
σ1 = F/A1 = F/(2A2)
σ2 = F/A2

σ12 = 1/2

I think that my answer is probably wrong because I don't think that the force being applied to the beam should just be equal to F. Is the force at the location of fusion different from F that's being applied at both ends? How would I go about calculating this force? Would this make the F in the formula for stress an average of F and the stress at the fusion point?

2. Sep 27, 2015

### Staff: Mentor

Your answer is correct. In a real situation, the stresses in the vicinity of the fusion cross section would vary a lot with spatial position within the beam (you were very perceptive in recognizing this). But, at distances of only one or two cross section diagonal lengths away on either side of the fusion cross section, the stresses would become uniform and purely axial. So this is the state of stress that would prevail throughout most of the two pieces.

Chet

3. Sep 27, 2015

### paisiello2

I think they are talking about average stresses here. There will be some stress concentrations near the fusion point that exceed the average but these are localized points and not the average across the cross section.

4. Sep 27, 2015

### Gwozdzilla

So when a tensile force is applied to a beam, this force is generally uniform throughout its cross section and does not vary with the length or mass of the beam?

5. Sep 27, 2015

### SteamKing

Staff Emeritus
No. You can take a cut thru the cross section of the beam at any point along its length and make a free body diagram out of the pieces. The forces which are required to keep the beam in equilibrium can then be used to calculate the axial stress.

6. Sep 27, 2015

### Gwozdzilla

How would axial stress be calculated? How would this differ from the stress used in the problem?

7. Sep 27, 2015

### Staff: Mentor

It would be as you already calculated it. It would be the force F divided by the local cross sectional area.

8. Sep 27, 2015

### Gwozdzilla

So if the force is different at each cross sectional area due to the length and mass of the beam, then why isn't the force for the beam with area A1 different from the force for the beam with area A2? Shouldn't this mean that the force providing the tensile stress isn't equal to F?

9. Sep 27, 2015

### Staff: Mentor

Why should the length and mass affect the tensile force?

10. Sep 27, 2015

### Gwozdzilla

I suppose the tensile force is only being applied in the x direction, whereas forces caused by the mass of the beam would be in the y direction, causing them to not affect each other. But I know I've seen other beam problems before where the beam is fixed to a wall and an object is placed on it, and the distance of the object from the wall affects the force, but it's not affecting a tensile force in that situation, it's affecting the force in the y direction.

11. Sep 28, 2015

### SteamKing

Staff Emeritus
A cantilever beam (that is, a beam with one end fixed and one end free) behaves a little differently when it is loaded laterally as opposed to a beam which is loaded only in the axial direction.

A cantilever beam with a lateral load has a bending moment created, and due to curvature of the beam with a bending moment, a completely different stress pattern is created as opposed to a beam which has only an axial load applied.

For the beam undergoing bending, σ ≠ F / A. For the beam undergoing bending, one portion of the cross section will have a tensile stress and the rest of the same section will have a compressive stress.

In the case of bending, the net axial force on the beam section is zero, in order for the beam to remain in equilibrium.

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