Design a tension member w/ cross section

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 safety factors of 2 against yield and 3 against static fracture. The calculated design stress is 429 MPa, derived from yield and static fracture considerations. The force exerted on the member is calculated to be 5.641 MN, leading to an area of 13,148.6014 mm². Participants emphasize using the lowest allowable stress rather than an average for safety-critical applications.

PREREQUISITES
  • Understanding of tensile strength and safety factors in structural engineering.
  • Familiarity with stress calculations, specifically using the formula [sigma] = F / Ao.
  • Knowledge of material properties, including ultimate tensile strength (UTS) and yield strength.
  • Basic grasp of unit conversions, particularly between Newtons, megapascals (MPa), and square millimeters (mm²).
NEXT STEPS
  • Research the implications of safety factors in structural design.
  • Learn about material selection criteria for tension members in engineering applications.
  • Study the calculation of cross-sectional areas for various shapes under tensile loads.
  • Examine case studies on tension member failures to understand the importance of using conservative design stresses.
USEFUL FOR

Structural engineers, mechanical engineers, and students involved in designing tension members or studying material strength and safety factors in engineering applications.

General_Sax
Messages
445
Reaction score
0
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.
 
Physics news on Phys.org
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!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 21 ·
Replies
21
Views
24K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
1
Views
3K
Replies
6
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K