What is the Correct Calculation for Stress in Steel and Concrete?

AI Thread Summary
The discussion centers on calculating stress in steel and concrete, where the stress in steel is 12.5 times that of concrete. The maximum allowable stresses are 175 MPa for steel and 20 MPa for concrete, leading to a conflict in stress calculations. The user initially calculated the area of steel as 0.52 mm², which was deemed impractical, prompting a reevaluation that resulted in an area of 3.174 mm². However, confusion persists regarding unit declarations and the application of loads on the combined areas of steel and concrete. Clarification on these calculations and assumptions is necessary for accurate results.
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Homework Statement


here's the question , Part of solution was given , but not the final ans . I have tried to do , but i need someone to verify my ans .

Homework Equations

The Attempt at a Solution


Since , stress of steel is 12.5 times of stress of concrete . The max allowable stress in steel is 175MPa , while for the concrete is 20MPa . The author choose stress of steel = 175MPa , because by doing so , the stress in concrete is 14MPa(<20MPa) ,
But if the author choose stress of concrete = 14MPA , then the stress of steel would be = 12.5(14) = 250MPa , which is (>175MPa)

So , in 2nd part , my working = (17.5x10^6)(Area of steel) + (14x10^6)(10 Area of steel ) = 1000x10^3

so , Area of steel = 0.52(mm^2) , is my ans and concept correct ?
 

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or there's something wrong with my calculation ?
 
Anyone can response?
 
anyone ?
 
Or each of the steel and concrete received forces of 1000kN ?
 
The second part doesn't look right.

Note the area of the concrete, a2, is 10 times the area of the steel. What is the area of the steel?

What does one assume about load P on the combined area?
 
Your answer of 0.52 mm2 seems improbable, if not impractical, and so, without looking at your working, I am inclined to try a different approach. The usual problem here is often the units; so stick to, say N and mm. Try a concrete area of say 50 mm x 50 mm. At 14 MPa (ie N/mm2) , its resistance is? The steel would have a thickness of 250/200 = 1.25 mm and this does seem practical. So what is the steel's resistance? When you add these two resistances together, how does this compare with the applied load? So how far are you from the problem now you have a better feel for the quantities involved? - perhaps this is not the help you were expecting, but worth trying?
 
Astronuc said:
The second part doesn't look right.

Note the area of the concrete, a2, is 10 times the area of the steel. What is the area of the steel?

What does one assume about load P on the combined area?
sorry , i made a typo , it should be (175x10^6)(Area of steel) + (14x10^6)(10 Area of steel ) = 1000x10^3
Area of steel = 3.174(mm^2) , is it correct now ?
 
Astronuc said:
What does one assume about load P on the combined area?
total force = pressure act on steel (area of steel) + pressure on concrete (area of concrete)
= pressure act on steel (area of steel) + pressure on concrete (10 area of steel)
 
  • #10
Still looks funny. I am struggling to visualise 3.174 mm2. You haven't declared your units. Can you explain the 10^6 and the 10^3 in your corrected equation?
 
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  • #11
pongo38 said:
Still looks funny. I am struggling to visualise 3.174 mm2. You haven't declared your units. Can you explain the 10^6 and the 10^3 in your corrected equation?
10^6 = MPa , 10^3 = kN , are they wrong ?
 
  • #12
are they wrong ?
 
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