Iron bar bending force

  • Thread starter cooper1
  • Start date
  • #1
1
0

Homework Statement



Given an iron bar, of round cross-section, fixed by its extremities
(bar measures 152 mm in length and 4.8 mm in diameter),
a weight of 88.5 kgs is hung from its middle, bending it at least
90 degrees.

How much weight/force would be needed to bend another
bar of the same material and size, but of hexagonal or square
section?


Homework Equations



Torque = Force * R

Sin(90°) * R


The Attempt at a Solution



Haven't got any idea. There should
be something lacking in the text or
more probably some calculus skills
given as knows and hence omitted.

Hope you could clear up this question,
thank you in advance
 

Answers and Replies

  • #2
207
1
I'm not sure, but I think that you need a second moment of inertia over here, which is defined as

[tex]I = \int r^2\, dA[/tex]

where the are is perpendicular to the axis of bending. The bigger this moment of inertia is, the harder it is to bend or deform the beam, so you just use proportion to find what you need.
 

Related Threads on Iron bar bending force

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
23K
Replies
1
Views
9K
Replies
6
Views
2K
Replies
7
Views
2K
Replies
3
Views
3K
  • Last Post
Replies
1
Views
828
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
0
Views
1K
Top