# Statically Indeterminate Stepped Bar Problem

• noboost4you
In summary, you are trying to find the stresses in a bar section and you have determined that the stresses are P/A.
noboost4you
I've been having difficulty with this problem for the past 5 hours. I am understanding some part, but I can't get the final answer.

http://img223.imageshack.us/img223/1103/1015zz1.th.jpg

This is what I got so far:

EFx = Ra + Rb - P = 0
Ra + Rb = P

Sab = Sac + Scb = [(-Ra)(a)/AE] + [(Rb)(b)/AE] = 0
-Ra(a) + Rb(b) = 0
-Ra(a) + (P - Ra)b = 0
Ra = (b/a+b)*P
Rb = (a/a+b)*P

Ra = (8in/18in)*100kips = 44.44 kips
Rb = (10in/18in)*100kips = 55.56 kips

I found the reactions but I cannot find the stress of each segment. The fact that the problem isn't giving me the modulus of elasticity is throwing me off.

I know stress = P/A where P = force and A = cross-sectional area

I figured then stress for segment AC = 44.44kips/6in^2 but it doesn't equal the answer in the back of the book. I also figured that would be the same for segment CB but it's not.

Thanks

Last edited by a moderator:
noboost4you said:
I've been having difficulty with this problem for the past 5 hours. I am understanding some part, but I can't get the final answer.

http://img223.imageshack.us/img223/1103/1015zz1.th.jpg

This is what I got so far:

EFx = Ra + Rb - P = 0
Ra + Rb = P

Sab = Sac + Scb = [(-Ra)(a)/AE] + [(Rb)(b)/AE] = 0
-Ra(a) + Rb(b) = 0
-Ra(a) + (P - Ra)b = 0
Ra = (b/a+b)*P
Rb = (a/a+b)*P

Ra = (8in/18in)*100kips = 44.44 kips
Rb = (10in/18in)*100kips = 55.56 kips

I found the reactions but I cannot find the stress of each segment. The fact that the problem isn't giving me the modulus of elasticity is throwing me off.

I know stress = P/A where P = force and A = cross-sectional area

I figured then stress for segment AC = 44.44kips/6in^2 but it doesn't equal the answer in the back of the book. I also figured that would be the same for segment CB but it's not.

Thanks
You don't need to know the modulus of elasticity since it is given that both materials have the same modulus.
It sure looks like the fat piece will be in compression and the skinny piece will be in tension. To get the force in each, you must take note that the deflections of each must be the same. The fat piece compresses by the same amount that the skinny piece elongates. What is the formula for axial deformation? Apply that to solve for the ratio of the loads in each piece, and thence the load in each; then stress is just P/A for each piece.
EDIT: Oh, sorry, you seem to be on the right track, however, you assumed that A is the same for each section. It is not. Plug in the values of A for each and I think you've got it.

Last edited by a moderator:
I didn't assume the cross-sectional areas were the same. The fat piece is 6 in^2 and the skinnier piece is 3 in^2.

In my thinking:
Stress of AC should = 44.44 kips/6 in^2 = 7.41
and
Stress of BC should = 55.56 kips/3 in^2 = 18.52

However those are not the answers. In the back of the book, AC is 10.xx and BC is 12.xx

Any ideas?

Thanks

noboost4you said:
I didn't assume the cross-sectional areas were the same. The fat piece is 6 in^2 and the skinnier piece is 3 in^2.

In my thinking:
Stress of AC should = 44.44 kips/6 in^2 = 7.41
and
Stress of BC should = 55.56 kips/3 in^2 = 18.52

However those are not the answers. In the back of the book, AC is 10.xx and BC is 12.xx

Any ideas?

Thanks
Go back to your equation with Sab etc you will see that you canceled out the AE term; this is where you assumed the x-section was the same...you can cancel out the E but not the the A's, they are different...

Ah excellent! Thank you very much. The equations for Ra and Rb were a little more complicated but I got the answers the back of the book has.

I'm surprised I didn't see that I was cancelling out both A's when they clearly aren't equal to each other.

Thank you very much. This forum has always been helpful.

Once you've found your reactions from the applications of Statics and your knowledge of Deformations, the determination of the stresses in a bar section reduces to

P/A

where P is the resultant force acting on each section. For section AC, P = FBx - 100. Simlarly for the other section.

You have the hard part figured out. Little details always hold you back.

## 1. What is the definition of a "Statically Indeterminate Stepped Bar Problem"?

A statically indeterminate stepped bar problem refers to a structural engineering problem in which the number of unknown forces or reactions is greater than the number of equations of static equilibrium. This means that the structure cannot be solved using traditional methods and requires advanced techniques such as the flexibility method or the stiffness method.

## 2. What are the main challenges in solving a "Statically Indeterminate Stepped Bar Problem"?

One of the main challenges in solving a statically indeterminate stepped bar problem is determining the appropriate number of unknowns and equations to use in the solution. This can be difficult as it requires a thorough understanding of the structure and its behavior. Additionally, the solution may involve complex calculations and iterations, making it time-consuming and prone to errors.

## 3. How does the flexibility method help in solving a "Statically Indeterminate Stepped Bar Problem"?

The flexibility method, also known as the force method, is a technique used to solve statically indeterminate problems by considering the flexibility or deformability of the structure. It involves determining the deflections of the structure due to each load and then using these deflections to calculate the unknown forces or reactions. This method is particularly useful for solving problems with multiple unknowns and complex geometries.

## 4. What is the difference between a "Statically Indeterminate Stepped Bar Problem" and a "Statically Determinate Stepped Bar Problem"?

A statically determinate stepped bar problem is one in which the number of unknown forces or reactions is equal to the number of equations of static equilibrium, making it solvable using traditional methods. On the other hand, a statically indeterminate stepped bar problem has more unknowns than equations, making it impossible to solve using traditional methods and requiring advanced techniques.

## 5. How does the stiffness method differ from the flexibility method in solving a "Statically Indeterminate Stepped Bar Problem"?

The stiffness method, also known as the displacement method, is another technique used to solve statically indeterminate problems. Unlike the flexibility method, which considers the flexibility of the structure, the stiffness method considers the stiffness or rigidity of the structure. It involves determining the stiffness of each member and using these stiffness values to calculate the unknown forces or reactions. This method is particularly useful for solving problems with redundant supports or members.

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