Reactions of a redundant frame structure using Unit Load Method

[spggodd]

Homework Statement

Hi, I am having trouble with the following problem and I can't seem to find any examples:

I am trying to determine reactions at points A and D of the redundant structure below using the unit load method, there is a 10kN load at the top left point (B), all lengths are 1m. Apologies for the poor diagram, please ask if you need clarification.

Edit: Diagram through text didn't work, uploading a picture now..

Homework Equations

Equilibrium Equations?

The Attempt at a Solution

I started by working out forces in the x direction.

10000+Dx+Ax =0

Then in the y direction:

Ay + Dy = 0

And then moments around D:

10000 x 1 + Ay

This then gives a series of un-solvable equations.
I think I may be ok when getting to the actual unit load method but I can't convince myself that I'm right for the equilibrium equations.

Can you help out?

Thanks
Steve

 

[spggodd]

Bump?
Can anyone please help with this?

 

[PhanthomJay]

Homework Statement

Hi, I am having trouble with the following problem and I can't seem to find any examples:

I am trying to determine reactions at points A and D of the redundant structure below using the unit load method, there is a 10kN load at the top left point (B), all lengths are 1m. Apologies for the poor diagram, please ask if you need clarification.

Edit: Diagram through text didn't work, uploading a picture now..

Homework Equations

Equilibrium Equations?

The Attempt at a Solution

I started by working out forces in the x direction.

10000+Dx+Ax =0

good

Then in the y direction:

Ay + Dy = 0

yes.

And then moments around D:

10000 x 1 + Ay

there is a fixed end moment A , M_Az, that you must include on this equation.

This then gives a series of un-solvable equations.
I think I may be ok when getting to the actual unit load method but I can't convince myself that I'm right for the equilibrium equations.

Can you help out?

Thanks
Steve

you have 3 equilibrium equations with 5 unknowns, so the frame is statically indeterminate to the second degree. You must remove the support at D, calculate the deflection of D in the x and y directions, and apply the unit loads at D in the x direction and then in the y direction. What have you tried beyond writing the equilibrium equations?

 

[spggodd]

Ok I have got to the point where I have

Forces in X = 10000 - Ax - Dx = 0
Forces in Y = Ay + Dy = 0
Moments about D = 10000 x 1 - Ma + Ay = 0

Which gives me: Ax, Dx, Ay, Dy and Ma as my unknowns with 3 equations to solve.

-------------------------------

Removing supports at D and doing the same again gives:

Forces in X = 10000 - Ax = 0 -------> Ax = 10000 N
Forces in Y = Ay = 0 -------> Ay = 0 N
Moments about A = 10000 x 1 - Ma -------> Ma = 10000 Nm

Where next?
Is there any reference sources you can provide so I can have a look at any example problems?

I need to work out deflection in the x and y direction as you've mentioned above, both individually but how do you go about this. I assume you need to create 2 normal bending moment equations?

 

[PhanthomJay]

Ok I have got to the point where I have

Forces in X = 10000 - Ax - Dx = 0
Forces in Y = Ay + Dy = 0
Moments about D = 10000 x 1 - Ma + Ay = 0

Which gives me: Ax, Dx, Ay, Dy and Ma as my unknowns with 3 equations to solve.

you need 5 equations, three from the static equilibrium equations and the other 2 from deflection compatability equations

-------------------------------

Removing supports at D and doing the same again gives:

Forces in X = 10000 - Ax = 0 -------> Ax = 10000 N

pointing left or right?

Forces in Y = Ay = 0 -------> Ay = 0 N

yes

Moments about A = 10000 x 1 - Ma -------> Ma = 10000 Nm

cw or ccw?

Where next?

Now calculate the deflection of joint D for that load case. You will need to know this, because ultimately, when you restore the support reactions at D, deflection is 0 at D.

Is there any reference sources you can provide so I can have a look at any example problems?

http://www.public.iastate.edu/~fanous/ce332/force/cdframefp.html

I need to work out deflection in the x and y direction as you've mentioned above, both individually but how do you go about this. I assume you need to create 2 normal bending moment equations?

Very tedious process, I am afraid. I'm glad I don't do these anymore. Yet, it is imperative that you understand the concept of how and why this works. Otherwise, you can shove this stuff into the computer and have no idea of the accuracy of the results. Very dangerous.

 

[spggodd]

Hi PhantonJay, I think I may have got somewhere with this last night..

Replacing support D with two redundant reactions R₁ and R₂ gives the compatibility equations of:

ΔD_y+a₁₁R₁+a₁₂R₂=0
ΔD_x+a₂₁R₁+a₂₂R₂=0

Using the unit load method to calculate the displacements and flexibility coefficients using the equations:

ΔD_y=∑∫_L((M₀M_1y)/EI)dz

ΔD_x=∑∫_L((M₀M_1x)/EI)dz

a₁₁=∑∫_L((M_1y²)/EI)dz

a₂₂=∑∫_L((M_1x²)/EI)dz

a₁₂=a₂₁=∑∫_L((M_1yM_1x)/EI)dz

The expressions for the bending moments of the frame are:

In DC:

M₀=0
M_1y=0
M_1x=-1z₁

In CB:

M₀=0
M_1y=-1z₂
M_1x=-1

In BA:

M₀=10Z₃
M_1y=-1
M_1x=-1(1-z₃)

Substituting these values into the equations above and solving gives:

Δ_dy=1/EI(∫¹₀(-10z₃)dz₃=-5/EI

Δ_dx1/EI(∫¹₀(-(10z₃)(1-z₃))dz₃=-1.666/EI

a₁₁=1/EI(∫¹₀(-z₂)²)dz₂+∫¹₀((-1)²)dz₃)=1.333/EI

a₂₂=1/EI(∫¹₀(-1z₁)²dz₁+∫¹₀((-1)²)dz₂+∫¹₀(-1(1-z₃))²dz₃)=1.666/EI

a₁₂=a₂₁=1/EI(∫¹₀(-1z₂)(-1)dz₂+∫¹₀(-1)(-1(1-z₃))dz₃)=1/EI

Substituting these values into the compatibility equations gives:

-5/EI+(1.333/EI)R₁+(1/EI)R₂=0

And

-1.666/EI+(1/EI)R₁+(1.666/EU)R₂=0

Solving for R₁ and R₂ gives:

R₁=5.459 kN
R₂=-2.277 kN


-------------------------------

I'm convinced this is the right method however I am not overly confident with my expressions for the bending moments in each of the frame members?

Can I use another simpler method to check my answer etc?

 

[nvn]

spggodd: I currently think we cannot check your final answer until you post the E, I, and A value you used, where A = cross-sectional area of each member. Please post the E, I, and A values you assumed for your R1 and R2 answers.

 

[spggodd]

Hi NVN,

To get rid of EI I just multiplied both sides by EI, this gets rid of the term right?

Also, where do I need to use A?Sent from my iPhone using Physics Forums

 

[nvn]

spggodd: I retract my statement in post 7 now. Excellent work. Your R1 and R2 answers in post 6 are correct. The correct answers are R1 = -2.273 kN, and R2 = 5.455 kN, where R1 is the horizontal reaction force at point D.

 

[spggodd]

Ok, no worries.

Do you agree with the rest of my calculations though?Sent from my iPhone using Physics Forums

 

[spggodd]

NVN, thanks for your response.
I'm glad I've finally seemed to crack this! Sent from my iPhone using Physics Forums

 

[spggodd]

I've been going over this and I'm struggling to grasp how I got the bending moment equations for each member.

Also, my lecturer seems to specifically use a "System 0" and "System 1"... Etc.

The method in using seems to kind of match some of the examples, such that it uses the Stationary of Complementary Strain Energy equations and so forth.

I think I have the correct answers but I want to understand what I'm doing a bit better.

Is someone able to shed any light, or give a more detailed description of the process?

Thanks