Design a tension member w/ cross section

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The discussion focuses on designing a tension member, specifically a square rod, to support a mass of 575,000 kg with specified safety factors against yield and static fracture. The proposed design stress is calculated as the average of two values derived from yield and fracture considerations, resulting in approximately 429 MPa. However, there is confusion regarding the source of the initial stress values and the correct approach for determining the design stress, with advice given to use the lowest allowable stress instead of the average. Additionally, the calculations for the cross-sectional area and dimensions are clarified, emphasizing the importance of unit consistency. The thread highlights the need for accurate understanding of stress calculations in structural design.
General_Sax
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design a tension member w/ cross section...

Homework Statement



You are to design a tension member (square rod) with a cross section to carry a mass of 575 000 kg with a safety factor of 2 against yield and a safety factor of 3 against static fracture (overload).

a) What is the design stree that you will use for the application? Why?

b) What are the appropriate bar dimensions for use?


Homework Equations



FS = UTS / R

[sigma] = F / Ao

Ao = L * w

The Attempt at a Solution



FS = 1279MPa / 3 = 426.33333... MPa ~ 426 MPa
that's the design stress considering static fracture

FS = 862 / 2 = 431 MPa
that's the design stress consider yield stress/strength

So, I'm considering using a design stress that is the average of these two values, which would be: 429 MPa

I'm just confused, because we haven't done anything like this in class.

Now for part b)

[sigma] = F / l*w

F = 575 * 103(kg) * 9.81(m/s2)
F = 5.641 MN (mega Newtons)

429 = 5.641 MN / Ao[/SUP]

Ao = 13148.6014 MM (mega meters? I'm not sure about the units on this one)

l = w = 6574 MM

Please help and thank you.
 
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General_Sax: First, where did 1279 MPa, and 862 MPa, come from? I do not see that in the problem statement. Secondly, MN/MPa = MN/(MN/m^2) = m^2. Or N/MPa = mm^2. Third, use the lowest allowable stress, not the average.
 


Thanks for the help!
 

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