Solid Mechanics Stiffness Question

Click For Summary
SUMMARY

The discussion centers on calculating the number of steel rods required to achieve a minimum axial stiffness of 2000 kN/mm for a concrete post measuring 0.5m in height and 100mm in diameter, reinforced with 10mm diameter steel rods. The formula derived for the number of rods is n = (4D²)/(5d²), which results in an incorrect value of 80. Participants highlight the need for clarity regarding the units of stiffness (k) and the forces involved in the equation k = F/(EA), indicating potential errors in the initial calculations.

PREREQUISITES
  • Understanding of axial stiffness in solid mechanics
  • Familiarity with the properties of concrete and steel materials
  • Knowledge of basic structural engineering equations
  • Proficiency in unit analysis for stiffness calculations
NEXT STEPS
  • Review the derivation of axial stiffness equations in solid mechanics
  • Investigate the properties of concrete and steel to understand their contributions to stiffness
  • Learn about strain compatibility and its application in composite materials
  • Explore unit analysis techniques to ensure dimensional consistency in engineering calculations
USEFUL FOR

Students in civil engineering, structural engineers, and professionals involved in material design and analysis will benefit from this discussion.

adlinkie1
Messages
2
Reaction score
0

Homework Statement



A 0.5m high and 100mm diameter post is to be made from concrete reinforced by 10mm diameter steel rods. How many steel rods should be used if the axial stiffness of the post is to be at least 2000kN/mm?

Homework Equations



n = number of steel rods
D = diamater of the post
d = Diameter of the steel rods

A Concrete= [tex]\stackrel{piD 2}{4}[/tex] - n [tex]\stackrel{pid 2}{4}[/tex]
A Steel = n [tex]\stackrel{pid 2}{4}[/tex]k (Stiffness) = F/(EA)

The Attempt at a Solution



Strain compatibility is assumed
[tex]\frac{E SteelA Steel}{L}[/tex] = [tex]\frac{E ConcreteA Concreete}{L}[/tex]

Substitute in Areas, and cancel where possible, I am left with n = [tex]\frac{4D<sup>2</sup>}{5d<sup>2</sup>}[/tex] which = 80, which is not right but i am struggling to see what I have done wrong.
 
Physics news on Phys.org
In your equ k=F/EA, what are the units of k on the LHS and what are the units of F/EA on the RHS? Isn't there something missing?
 

Similar threads

Replies
20
Views
14K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
12K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 26 ·
Replies
26
Views
9K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K