# How can I get the force from the bending stress

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1. Jan 18, 2016

### atlas_2010

Hello all,
I am a new member of PF.I have a basic question about the bending stress, I remember somewhere I saw this part, but can't find out it. Could you give me a answer?
Given:
1, M-moment applied on the section.
2, I-Section modules
3. y-distance from shear center
Solution
F,how much is the force at point y?

2. Jan 18, 2016

### SteamKing

Staff Emeritus
In general, the location of the shear center is separate from the location of the centroid of the section, which is where the neutral axis in bending is located. The location of the shear center is that point at which a shearing force can be applied to a section which does not create a torsional moment on that section.

In some beams, the location of the maximum bending moment is also the location where the shear force on the beam is zero. In other beams, especially those which have fixed supports, there can be large bending moments and shear forces applied simultaneously.

At the location of the neutral axis, the bending stress, σ = 0 by definition.

The formula for bending stress is

$$σ = \frac{M ⋅ y} { I }$$

where

σ = bending stress
M = bending moment
I = second moment of area (not the section modulus, which is a different property)
y = the distance from the neutral axis of the section to the point where the bending stress is calculated.

In the section below:

the bending moment Mx = 150 N-m and the centroid of the section $\bar y$ is located as shown. In this particular section, $\bar y = 66.40$ mm.
$\bar y$ would also be the distance used in the bending stress formula above to calculate the bending stress at the bottom edge of the web.

(Note: The second moment of area for this section is I = 1,202,840 mm4)

The bending stress at the bottom of the flange is

$$σ = \frac {M ⋅ y}{I}$$

$$σ = \frac{(150 N⋅m) ⋅ 1000\, mm / m ⋅ 66.40\, mm}{1,202,840\, mm^4}$$

σ = 8.28 N/mm2 = 8.28 MPa compression

The bending stresses will be distributed as in this diagram:

3. Jan 18, 2016

### atlas_2010

Hello Steamking,
I am much more want to know hou much is the force rather than the stress, the moment can be equal to a couple of force, and the stress is caused by the force?
Based on the definition, we can get the following result
F=σdA, A=σdy, σ=ky
If I were wright, what's the solution of this equation?

4. Jan 18, 2016

### SteamKing

Staff Emeritus
If you calculated the total force due to the bending moment above and below the neutral axis, F = σ ⋅ dA, that total force would equal zero, since the beam is in equilibrium. You have one part of the beam cross section in compression, and the other in tension, so the net force on the cross section is zero.

However, the moment of these forces about the neutral axis, M = σ ⋅ y ⋅ dA, must be equal to the bending moment at at that location of the beam.

5. Jan 18, 2016

### atlas_2010

I will take compression area instead of total area, the integral area is from shear center to the section edge,but the result is resultant force. still can't get the exactly force at any point. Can't get a link between y and F, something like y and σ.

6. Jan 18, 2016

### SteamKing

Staff Emeritus
The shear center doesn't play a role in calculating bending stress, but the location of the neutral axis does. (The neutral axis and the shear center are two different things).

If you want to calculate the force over a particular portion of the cross section, dF = σ ⋅ dA, and σ = M ⋅ y / I. For a particular distance y from the neutral axis, the bending stress is constant all along that line, just like along the neutral axis, σ = 0. The bending stress σ is proportional to y and the constant of proportionality is the quantity (M / I).

In the section below:

If you wanted to calculate the force acting on the web below the neutral axis, you would proceed thus:
dF = σ ⋅ dA and
σ = M ⋅ y / I

From the previous calculations,
$\bar y = 66.40$ mm above the bottom edge of the web

The web has a constant thickness of 8 mm, so dA = 8 ⋅ dy, where dy is the height of a tiny slice of the web.

Then,

dF = (M / I) ⋅ y ⋅ 8 ⋅ dy = 8 ⋅ (M / I) ⋅ y ⋅ dy

To find F, we must integrate both sides of the equation for dF above:

$$\int\, dF = F = \int_0^{66.40} 8 ⋅ (M / I) ⋅ y ⋅ dy$$

$$F = 8 ⋅ (M / I) \int_0^{66.40} y ⋅ dy$$

$$F = 8 ⋅ (M / I) \frac{y^2}{2} \vert_0^{66.40}$$

$$F = 8\,mm ⋅ \frac{150,000\, N⋅mm}{1,202,840\,mm^4} ⋅ \frac{66.40^2\, mm^2}{2}$$

$F = 2,199.30\,N$ (in compression)

You can also compute the average bending stress acting over an area, and the force F = σ(avg) ⋅ Area.

7. Jan 19, 2016

### atlas_2010

Hello Steamking
Thank you for your correction and help.
Based on this conclustion , I can decide the weld size(connect web and flange)
Thanks again.

8. Jan 20, 2016

### SteamKing

Staff Emeritus
If you are doing weld design, you may need to consider the effects of shear, especially if it is combined with bending. The joint between a flange and a web is a point where shear stresses take a large jump, and this should not be neglected.

Here is a comparison of bending stress and shear stress for a typical T-section:

Bending Stress distribution

Shear Stress distribution
Just as the location of the centroid of a cross section is important in evaluating bending stress, the centroid location is often where the maximum shearing stresses are found as well.

9. Jan 24, 2016

### atlas_2010

Thank you very much for your kindly suggestion, we also decide the weld by rules, which just given a factor depond on different type and location weld,as usual we take the weld leg as 0.7*thinner platethickness, I want to know clear about this, so posted this topic,now I am clear, thanks for your help.

Steamking, I have some queation about the weld group check, the link is following, could you have a look when you are free, thanks very much.