# Stiffness Matrix method

1. May 15, 2014

### MMCS

This problem has alot of calculations from the beginning so i have skipped to the part i am stuck with and tried to include relevant information, apologies if i have missed anything. Assume all working is correct as i was given the answers.

After carrying out the stiffness matrix method I am trying to construct a bending moment diagram using member equilibrium, see attached.

The specific problem is member 1-5

working

member 1-4

bending moment at 1 (given) 123.33Kn
vertical force 57.44Kn
length 2.5

therefore

123.33 - 57.44Kn*2.5 = -20.27 (correct)

Member 4 -5
-20.67 + 160KN(moment from pre-eliminated beam) = 139.63Kn (correct)

however i can not get the moment of -123.98 at point 5

attempt member 4-5
vertical force relative to beam = 80 (sin(36.87)
moment = 139.63Kn
139.63Kn - 80 sin(36.87)*2.5 = 19.629 (incorrect)

#### Attached Files:

• ###### stiffness.jpg
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2. May 17, 2014

### paisiello2

It's difficult to follow what you doing here but is there not a shear force at point 4 in addition to the 80?

3. May 17, 2014

### MMCS

Yes I appreciate it is fairly difficult to follow. Essentially I have got the bending moments on one side using stiffness matrix method and the used these values and member equilibrium for the rest of the beam. There is a shear force yes, I have tried to account for that, that's why I have used 80sin(36.87) as this would give me the vertical component of the force (vertical relative to the member)

4. May 17, 2014

### paisiello2

The shear force is the internal shear in addition to the 80 external force.

5. May 17, 2014

### MMCS

Would that have to be included in the bending moment diagram? If so, how is the force calculated? And would it be used in the same way as a vertical force e.g F x distance

6. May 17, 2014

### paisiello2

1) Yes.

2) Same way you got the bending moment at point 4.

3) Yes.

7. May 18, 2014

### MMCS

I got the moment by using (UDL x L) x L/2 ?

8. May 18, 2014

### paisiello2

But you added another term to it as well.

9. May 18, 2014

### MMCS

Yes I added the moment from point one. For point 5 this is the method I have tried. I don't follow what you are suggesting?

10. May 18, 2014

### paisiello2

OK, we are trying to get the moment at point 5, correct? So you need to take the moment at point 4 plus the shear at point 4 multiplied by L = 2.92m.

The shear at point 4 is 80kN*sin(37.9°) plus the shear from member 4-1.

11. May 18, 2014

### MMCS

Attempt

139.63-((80sin(37.9))+57.44)*25 = -126.83

This is pretty close to -123.98 but still off,