# What Is the Maximum Load a Concrete-Filled Cast Iron Pipe Can Support?

• anthonyk2013
In summary: You were mixing up Newtons and pascals.In summary, the conversation is about finding the maximum permissible value for the load W on a cast iron pipe filled with concrete used as a column. The compressive stress in the concrete is limited to 5 KN/m2 and the E of concrete is one tenth of that of cast iron. The summary includes incorrect calculations of the area and units, leading to an incorrect final result of 934 tons instead of the correct answer of 79 tons. Remembering the definition of a pascal (1 Pa = 1 N / m^2) and using the correct units is important in calculations.
anthonyk2013
A cast iron pipe is filled with concrete and used as a column to support a load W.If the outside diameter of the pipe is 200mm and the inside diameter is 150mm, what is the maximum permissible value for W if the compressive stress in the concrete is limited to 5 KN/m2.
Take E for concrete as one tenth that of cast iron.

σc=5KN/m2
Area of concrete is 3.14*1502/4=17671*10-3mm2
Area of pipe is 3.14*2002/4=31415-17671=13744*10-3mm2

σp/Epc/Ec

σpc*.1N E of concrete one tenth that of cast iron=.1

σpc*.1N

σp=5*106N*.1N

σp=.5*106N

F=Fc+Fp

F=σp*Apc*Ac

F=(.5*106N*17771*10-3mm)+(5*106N*13744*10-3MM2)

F=8835.5*103N+68720*103N

F=77555.5N*106

F=77555.5N/9.81=7905kg=7.9tons

The answer in my notes is 79tons, where have I gone wrong with my decimal point

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It's really hard to follow your calculations because you don't indicate the units of your results. You must get into the habit of writing the units along with the magnitudes of your results.

SteamKing said:
It's really hard to follow your calculations because you don't indicate the units of your results. You must get into the habit of writing the units along with the magnitudes of your results.

The units of my result are in tons which is what is asked for in the question and are indicated Do you see something else I m missing?

Yeah, but what about all the calculations it took to get you to the final result? Units are important there, too.

SteamKing said:
Yeah, but what about all the calculations it took to get you to the final result? Units are important there, too.

Just so I understand properly would you expect units in this part of the solution

F=(.5*106*17771*10-3)+(5*106*13744*10-3)

Yup. And BTW, your area calculations are wrong. 1 m^2 = 10^6 mm^2

Any help would be appreciated

anthonyk2013 said:
A cast iron pipe is filled with concrete and used as a column to support a load W.If the outside diameter of the pipe is 200mm and the inside diameter is 150mm, what is the maximum permissible value for W if the compressive stress in the concrete is limited to 5 KN/m2.
Take E for concrete as one tenth that of cast iron.

σc=5KN/m2
Area of concrete is 3.14*1502/4=17671*10-3mm2
Area of pipe is 3.14*2002/4=31415-17671=13744*10-3mm2

Incorrect area calculations. If your dimensions are in mm, then area will be in mm2. 1 m2 = 106 mm2

σp/Epc/Ec

σpc*.1N E of concrete one tenth that of cast iron=.1

σpc*.1N

σp=5*106N*.1N

σp=.5*106N

The units of stress are pascals, not Newtons.

F=Fc+Fp

F=σp*Apc*Ac

F=(.5*106N*17771*10-3mm)+(5*106N*13744*10-3MM2)

Incorrect areas used here.

F=8835.5*103N+68720*103N

F=77555.5N*106

F=77555.5N/9.81=7905kg=7.9tons

The answer in my notes is 79tons, where have I gone wrong with my decimal point

σc=5KN/m2
Area of concrete is 3.14*1502/4=17671mm2
Area of pipe is 3.14*2002/4=31415-17671=13744mm2

σp/Epc/Ec

σpc*.1pa E of concrete one tenth that of cast iron=.1

σpc*.1pa

σp=5*106pa*.1pa

σp=.5*106pa

F=Fc+Fp

F=σp*Apc*Ac

F=(.5*106pa*13744mm2)+(5*106*17671mm2)

F=(6872*106)+(88355*106)

F=95227*1012*9.81=934176.87kg/1000=934tons

Seem to have gone a long way from the correct answer of 79tons

You've got to remember what the definition of a pascal is. 1 Pa = 1 N / m^2

Check your force calculations. This is why units are important.

## 1. How do you find the load in a compound bar?

To find the load in a compound bar, you can use the formula F = EAΔL/L, where F is the load, E is the modulus of elasticity, A is the cross-sectional area of the bar, ΔL is the change in length, and L is the original length of the bar. This formula takes into account the different materials and dimensions of each section of the compound bar.

## 2. What is a compound bar?

A compound bar is a structure made up of two or more bars of different materials or dimensions that are connected together. This allows for the bar to have different properties and characteristics depending on the materials used and their arrangement.

## 3. Why is it important to find the load in a compound bar?

Finding the load in a compound bar is important because it helps determine the structural integrity and strength of the bar. It also allows for the proper selection of materials and dimensions to ensure the bar can withstand the expected load without failure.

## 4. What are some factors that can affect the load in a compound bar?

The load in a compound bar can be affected by various factors such as the materials used, their arrangement, the dimensions of each section, and the applied force or load. Temperature changes can also cause changes in the load due to thermal expansion or contraction of the different materials.

## 5. Can the load in a compound bar be greater than the sum of its individual parts?

Yes, the load in a compound bar can be greater than the sum of its individual parts. This is due to the phenomenon of stress concentration, where the stress is concentrated at the interface between the different materials. This can result in a higher load on the bar than if it were made of a single material.

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