What is the correct stress in the concrete column?

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In summary, the conversation discusses a reinforced concrete column carrying a load of 20t with an overall cross section area of 0.1m^2, with 3% of the area being accounted for by steel bars. The stress in the concrete is found using the formula σs=σc*200/20 and the length of the column is 4m. The output is 1.54MN/m^2, which is equivalent to 1.54x10^6 N/m^2. There may be an error in standard form in the notes as the calculated answer is 1,543,000 N/m^2.
  • #1
anthonyk2013
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A concrete column is reinforced with steel bars and carries a load of 20t. The overall cross section area of the column is 0.1m^2 and the steel bars accounts for 3% of this area.
Find the stress in the concrete
If the length is 4m< how much does it shorten.

E of steel 200GN/m^2
E of concrete 20 GN/m^2

Load=20*1000=20000*9.81=196200 or 19.6*10^4

Area of concrete =.1*97%/100=.097m^2
Area of steel=.1*3%/100=.003m^2
Area of concrete=.1*97%/100=.097m^2
steel=s
concrete=c

σs/Es=σc/Ec→σs=σc*Es/Ec→σs=σc*200/20→σs=10*σc

F=σs*As+σc*Ac
19.6*10^4=(10*σc*.003)+(σc*.097)
19.6*10^4=σc(10*.003+.097)
19.6*10^4=σc(.127)

19.6/.127=154.3*10^4
σc=154.3*10^4

My question is, the answer I have in my notes is 1.54MN/m^2. Why is my decimal point wrong?
 
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  • #2
I'm not sure if it isn't. 1.54MN is the same as 1.54 x 10^6. 154.3 x 10^4 is the same as 1.54x10^6
 
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  • #3
Fineline00 said:
I'm not sure if it isn't. 1.54MN is the same as 1.54 x 10^6. 154.3 x 10^4 is the same as 1.54x10^6

Would be able to break that down more, still don't understand. Thanks
 
  • #4
I just want to stress I'm not so sure on the physics but I feel it is just a error in standard form

Your notes give you the answer 1.54MN/m2, which is 1.54 Mega nm2 which is 1.54x106 N/m2 or 1,540,000 n/m2

Your answer is 154.3x104 N/m2which is 154.3x10,000 which is 1,543,000 N/m2
 
  • #5
Fineline00 said:
I just want to stress I'm not so sure on the physics but I feel it is just a error in standard form

Your notes give you the answer 1.54MN/m2, which is 1.54 Mega nm2 which is 1.54x106 N/m2 or 1,540,000 n/m2

Your answer is 154.3x104 N/m2which is 154.3x10,000 which is 1,543,000 N/m2

Thanks
 
  • #6
Appreciate if someone could check the physics work in my problem.
 
  • #7
anthonyk2013 said:
Appreciate if someone could check the physics work in my problem.

Anyone thanks
 

1. What is a compound bar question?

A compound bar question is a type of problem in which two or more bars are connected together, with different weights or forces acting on each bar. The goal is to find the resulting force or weight of the entire system.

2. How do you solve a compound bar question?

To solve a compound bar question, you need to use the principle of moments or the law of equilibrium. This involves balancing the clockwise and counterclockwise moments around a pivot point to find the unknown force or weight.

3. What are the key concepts to keep in mind when solving a compound bar question?

The key concepts to remember in solving a compound bar question are the principle of moments, the law of equilibrium, and the relationship between force, weight, and distance from the pivot point. It is also important to correctly identify all the forces and weights acting on the bars.

4. Can you provide an example of a compound bar question and its solution?

Sure, let's say we have a compound bar with two bars connected at a pivot point. The first bar has a weight of 10kg and is 2m long, while the second bar has a weight of 5kg and is 3m long. If the pivot point is located 1m from the first bar's end, the resulting force on the system would be calculated as follows: (10kg x 2m) + (5kg x 3m) = 35kgm. Since the pivot point is 1m from the first bar's end, the resulting force would be 35kgm/1m = 35kg.

5. Are there any tips for solving compound bar questions more efficiently?

One tip for solving compound bar questions more efficiently is to draw a clear and accurate diagram of the system, labeling all the forces and distances. It is also helpful to set up an equation with the unknown force or weight on one side and all the other forces and weights on the other side. Additionally, double-checking your calculations and using units consistently can help avoid mistakes and make the solving process smoother.

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