You are to design a tension member

Click For Summary
SUMMARY

The discussion focuses on designing a tension member, specifically a square rod, to support a mass of 575,000 kg with a safety factor of 2 against yield and 3 against static fracture. The calculated design stresses are 426 MPa for static fracture and 431 MPa for yield strength, leading to an average design stress of 429 MPa. The force exerted by the mass is calculated as 5.641 MN, and the required cross-sectional area is determined to be approximately 13,148.6 mm², with dimensions of 6,574 mm for both length and width.

PREREQUISITES
  • Understanding of tensile stress and safety factors in structural engineering
  • Familiarity with material properties, specifically Ultimate Tensile Strength (UTS)
  • Knowledge of basic mechanics, including force calculations and unit conversions
  • Experience with cross-sectional area calculations for structural members
NEXT STEPS
  • Research material selection criteria for tension members in structural applications
  • Learn about the implications of safety factors in engineering design
  • Study the calculation of cross-sectional areas for different shapes and materials
  • Explore advanced topics in structural analysis, including load distribution and failure modes
USEFUL FOR

Structural engineers, mechanical engineers, and students involved in design and analysis of tension members in engineering applications.

General_Sax
Messages
445
Reaction score
0
You are to design a tension member...

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?


=============================================================

FS = UTS / R

[sigma] = F / Ao

Ao = L * w

======================================================

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

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; ie, does this look right to you?
 
Engineering news on Phys.org


Of the three stresses (426, 429 & 431) you have calculated only one gives a safety factor of 3 against fracture.

Which one is it?

This is the only one which satisfies both the safety factors of at least 2 and at least 3.

I'm sure if you check the units of (426, 429 & 431) in part (b) you will be able to calculate a more sensible area.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
6
Views
5K