Calculating size of rectangle cross section

Click For Summary

Discussion Overview

The discussion revolves around calculating the size of a rectangular cross section for a medium carbon steel rod subjected to a specific tension. The context includes homework-related problem-solving involving material properties and mechanical calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Post 1 presents a calculation for the cross-sectional area using the formula A=P/Ok, with initial values provided for tension, thickness, and ultimate tensile strength (UTS).
  • Post 2 critiques the initial calculations, pointing out errors in notation and suggesting a re-evaluation of the math.
  • Post 3 attempts to recalculate the area, arriving at a new size for the cross section but still expressing uncertainty about the correctness of the dimensions.
  • Post 4 advises against taking the square root in the calculations, indicating a potential misunderstanding of the problem.
  • Post 5 questions whether the cross section can be stated as 15 mm x 10.4 mm based on the previous calculations.
  • Post 6 emphasizes the importance of maintaining significant digits in calculations and notes a rounding error in the final answer.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correctness of the calculations and the methodology used. There is no consensus on the final dimensions of the cross section, and uncertainty remains about the proper approach to the problem.

Contextual Notes

Limitations include potential misunderstandings of the mathematical steps involved, the significance of significant digits, and the implications of rounding errors. The discussion does not resolve these issues.

rad10k
Messages
60
Reaction score
0

Homework Statement



A cmedium carbon steel rod has a rectangular cross section and must cope with a tension of 29 430 Newtons . The thickness of the rod is 15 mm calculate the siz of the cross section?

Homework Equations



A=P/Ok


3. The Attempt at a Solution [/b

O = UTS = 750 Mpa k = 0.25 P = 29 430 N t = 15 mm

A= P/Ok

29 430 / 200 = 187.50 mm^2

187.5 / t = 187.5 / 15 = 12.5 sqrt = 3.5355

Size of cross section = 12.5 mm x 3.5 mm

I believe I have the area correct but not sure if the sides are correct ?
 
Physics news on Phys.org
rad10k: Your answer is currently incorrect. Check your math. MPa should be spelled MPa, not Mpa. The name Newton is the name of a man, whereas Newton (N) is a unit of force. Also, place a comma between separate equations on the same line, for clarity. Try again.
 
ok thanks

A=P/Ok

750 * 0.25 = 187.5, 29 430 / 187.5 = 156.96

Area = 156.96 mm^2 from this;

A/t sqrt = 156.96 / 15 = 10.464, sqrt = 3.2348

size of cross sectional area is 10.4 mm x 3.2 mm
 
rad10k: You should not perform sqrt.
 
so the cross section is 15 mm x 10.4 mm ?
 
rad10k: Your answer is almost correct. Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits.

Even if you round the final answer to three significant digits, you currently rounded it wrong.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
30K
Replies
48
Views
19K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
19
Views
34K
Replies
1
Views
5K