# Section Modulus, major & minor axis

• zaurus
In summary, the conversation discusses calculating section modulus and stress in a fillet weld for a cantilever beam. The section modulus equations for the major and minor axes are bh^2/6 and hb^2/6, respectively. The conversation also mentions using a full penetration double bevel butt weld or designing fillet welds to take shear and bending stresses. The process for calculating the stress in a fillet weld involves finding the unit area, calculating the moment of inertia, and plugging the values into various equations. However, it is important to also consider the allowable weld shear stress and compare the resultant stress to the yield of the material.
zaurus

## Homework Statement

I have attached a figure showing beam bending around 2 axis. I need to calculate section modulus Sx and Sy but seem to be getting the major and minor axis confused. I guess the problem comes in when I go to select which term is squared in my equation below.

## Homework Equations

S = 1/6 * b * h^2 for rectangular cross section

## The Attempt at a Solution

Sx = (70*750^2)/6 = 6.56E+6

Sy = (750*70^2)/6 = 6.13E+5

If this is correct, could you please help clarify why b and h are what they are in the section modulus equation for 1) major axis and 2) minor axis.

Thanks.

#### Attachments

• bending.JPG
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Yes, you have it correct. The major axis (X-X) bending section modulus is bh^2/6, and the minor (Y-Y) axis modulus is hb^2/6. What you have is a cantilever beam, 70 units x 750 units in cross section, and 320 units in length. The Fy force creates bending moments about the X axis, called the major axis because the section modulus is greater about that axis, and the Fx force creates bending moments about the (minor) Y axis. Visualize that under the Fy force, the beam is strong (large S_x) because it is 750 units deep, whereas under the Fx force, the beam is weak (small S-y), because it is only 70 units deep. Did I answer your question?

Yes, thank you for your help. I don't know if you would also be able to help with this question, but how would you go about calculating stress in a fillet weld at the base of the cantilever due to this bending?

I found some information online about finding where you find the throat of the weld by width / sqrt(2) but then how do you use this with the bending (moment)? Do you use a moment diagram / shear diagram to take max shear and divide by weld throat * weld length?

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You should probably go for a full penetration double bevel butt (tee) weld using the appropriate electrode and be done with it, but if you want practice in designing fillet welds to take shear and bending stresses, here's a site you can check: http://www.roymech.co.uk/Useful_Tables/Form/Weld_strength.html
I used to design welds of all types 25 years ago, but I've since moved on to just specifying the loadings on the welds and have the designers do the calcs, so I've gotten away from the specifics of weld design (I've earned it!).

Thanks PhanthomJay, this is exactly what I am looking for.

Just to clear up what I read on the site,

basically I take .707*weld leg and multiply it by the weld length to get a unit Area

then calculate the moment of inertia for the weld as a line: 1/12 L^3

Tbending = M.y/I u
Tshear = P /A
Tresultant = Sqrt (τ b2 + τ s2 )

then plug into the above equations, and Presto! I can compare Tresultant to yield of material.

zaurus said:
Just to clear up what I read on the site,

basically I take .707*weld leg and multiply it by the weld length to get a unit Area

then calculate the moment of inertia for the weld as a line: 1/12 L^3

Tbending = M.y/I u
Tshear = P /A
Tresultant = Sqrt (τ b2 + τ s2 )

then plug into the above equations, and Presto! I can compare Tresultant to yield of material.
I don't think that is quite correct; once you get T_resultant, that is the stress for a 1 inch weld; then you must divide T_resultant by the allowable weld shear stress, to get the required weld thickness. Note also that the allowable filet weld stresses are the shear allowables, I think it's 0.3*(nominal tensile strength of weld metal), but not to exceed 0.4*(F_yield) of base metal.

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## 1. What is the definition of section modulus?

The section modulus is a geometric property of a cross-sectional shape that represents the resistance to bending. It measures the shape's ability to resist bending forces and is used to determine the strength and stiffness of a structural member.

## 2. What is the difference between major axis and minor axis in section modulus?

The major axis is the axis that passes through the centroid of the cross-section and is perpendicular to the direction of the applied force. The minor axis is the axis perpendicular to the major axis. The section modulus differs for the major and minor axis as the shape's geometry and resistance to bending forces vary along each axis.

## 3. How is section modulus calculated?

The section modulus is calculated by dividing the moment of inertia of the cross-section by the distance from the neutral axis to the outermost point of the section. This value represents the distribution of the shape's cross-sectional area, which determines its resistance to bending.

## 4. What are the units of section modulus?

The units of section modulus are typically given in cubic inches (in3) or cubic centimeters (cm3) for smaller cross-sections, and cubic feet (ft3) or cubic meters (m3) for larger cross-sections. However, the units can vary depending on the units used for the moment of inertia and the distance measurement.

## 5. How is section modulus used in structural design?

Section modulus is an important factor in structural design as it helps engineers determine the appropriate size and shape of structural members to resist bending forces. It is also used to calculate the maximum allowable stress in a member and to compare different cross-sectional shapes to determine the most efficient design for a given application.

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