# Loads on 2 fixed supports

1. Mar 15, 2017

### dawnchang888

1. The problem statement, all variables and given/known data
Find the loads on A and B due to the force F at C

2. Relevant equations
T=Fd
M=Fd

3. The attempt at a solution
see image. Im lost at getting the reactions with respect to Y.

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2. Mar 15, 2017

### PhanthomJay

There are several issues here in your workings.
-You show what appears to be fixed end moments at the rigid supports. Do such moments exist?
- you have signage errors in your calculation for the horizontal forces at the supports. The numerical value of 800 N at each is correct, but in which direction do they act?
- to determine how the 400 N vertically applied load splits amongst the two supports, take advantage of the fact that the rigid supports cannot move. What then must be the force in Member AB if AB cannot deform?

3. Mar 16, 2017

### CivilSigma

I would say yes because they are fixed and not allowed to rotate. So there should be internal moments at A and B.

Also, I attempted a go at the solution, and I get 6 unknowns (Ma , Mb ,Ay, By, Ax, Bx) in 5 equations (Moment about A, B and C=0, Sum Fy=0 and Sum Fx=0) and that system of equations is inconsistent.

The structure has static indeterminacy of SI=6 unknowns - 3(2 members) = 0 , so we should be able to solve it using statics only. Are you suggesting we use displacement equations?

4. Mar 16, 2017

### PhanthomJay

This is all incorrect. Suppose you had a horizontal beam of negligible mass that was fixed to a wall at one end and free at the other (a cantilever beam) and you applied an axial horizontal load at the free end directed along the axis of the beam away from the support, would there be a fixed end moment at the wall under this loading?

5. Mar 16, 2017

### CivilSigma

No, there wouldn't be any moments in that case.

But in this scenario, beam BC and AC will experience some sort of bending due to the applied force causing internal moments at fixed support A and B, correct?

6. Mar 16, 2017

### PhanthomJay

well, there will be some very small internal secondary
Moments due to the restraint against rotation at the fixed end as thise members deflect downward under the load, but these moments can be neglected as they don't amount to much. Therefore you can assume the support joints Aand B as pinned supports that cannot move vertically or horizontally.

7. Mar 16, 2017

### CivilSigma

They may be negligible , but what's the proof?

How would you calculate them?

8. Mar 16, 2017

### PhanthomJay

It would be difficult by hand calc because if the high degree of indeterminancy and you would need to know the stiffness of the members (beam properties E, I and A must be known) or else you can approximate it by determining the deflection of joint C as if the supports were pinned using virtual work method then back calc the moment base on that deflection , or shove it into a computer using a non linear analysis with correctly input loads and geometry and joint fixity, and you will see how small these moments are. When I was at the university x years ago before computers, we built a simple truss with pinned joints and determined stresses using strain gages, then we fixed the joints with gusset plates and multiple bolts which is common in practice and which made the joints more rigid, then we applied the same loading and measured stresses with strain gages, and there was very little change in the results. So I guess the proof is in the pudding if you will.

So the bigger question is what are the vertical reactions at A and B as asked by the OP who appears at this point to have gone into hiding ....you can find them without actually doing an indeterminate analysis by using the hint I gave earlier.

9. Mar 17, 2017

### CivilSigma

The structure is in-determinant to the third degree, correct?
$$SI = 3m +r - (3j+e_c)=3(2)+6 - (3(3)+0)=3$$

(I previously calculated that it was 0 - which made me believe we could solve for the moments using statics only)

Last edited: Mar 17, 2017
10. Mar 17, 2017

### PhanthomJay

I grossly dislike these indeterminancy equations because you must be very careful how you use them, and they are often not correct anyway. The number of truss members don't enter into this. The structure is externally statically indeterminate to the first degree, because you have 3 equilibrium equations (sum of x forces = 0, sum of y forces = 0, and sum of moments about any point = 0), and 4 unknowns (Ax, Ay, Bx, and By) and as I mentioned we are ignoring any small fixed end secondary moments so his may be considered a pure truss with 2 pin supports, and axial forces only in the members. So in order to find all 4 reaction forces, you need one more equation, either using indeterminate virtual work analysis, or by inspection of any force in member AB.

11. Mar 17, 2017

### CivilSigma

Ok! Here is my attempt at the solution. I treated the frame as a truss ( originally I interpreted the problem with only two members AC and BC ).

I did simple truss analysis using your hint:
which implies that: $$\delta = \frac{NL}{AE}=0 \to N=0$$

So starting at node C, I calcualted:
$$F_{AC}= 400 \cdot \frac{0.2}{ 0.2 \cdot \sqrt{5} } = 179 N \, (C)$$
$$F_{BC}= 179 \cdot \frac{0.4}{ 0.2 \cdot \sqrt{5} } = 160 N\, (T)$$

Moving to Node A , and known F_{AB}=0, I calculated the reaction forces:

$$F_{A_x} = 178.9 \cdot \frac{0.4}{0.2\sqrt{5}} = 160 N (Left)$$
$$F_{A_y} = 178.9 \cdot \frac{0.2}{0.2 \sqrt{5}} = 80 N (Up)$$

Now analyzing the entire truss:

$$\Sigma F_x=0 \to F_{B_x} = 160 N (Right)$$
$$\Sigma F_y=0 \to F_{A_y}+F_{B_y}=400 \to F_{B_y} = 240 N (Up)$$

is this correct?

12. Mar 17, 2017

### PhanthomJay

You are messing up your calcs for the forces big time.

Firstly, if you assume that member AB is not there, then you get the same result as if AB was there, that is, F_AB = 0.

Now to solve for the vertical reactions at A and B, your 4th equation to solve for the unknown reactions comes from the fact that truss members are so-called 2-force members, that is, they take axial loads only and the resultant reaction forces must line up with the axial axis of the member, resulting in no vertical reaction force at joint A.
But otherwise, your calcs for determining member forces are way off, when you look at joint C for example using method of joints, you must have a 400 N vert component in BC , and an 800 N horizontal component in BC, thus F_BC = 400*(sq root of 5) = 894 N Tension, and F_AC = 800 N compression. Continue.

13. Mar 22, 2017

### CivilSigma

Sorry for the late reply, I fixed the calculations:

F_AC= 894 N (T)
F_BC= 800 N (C)
F_AB = 0

A_x=800 (Left)
B_x = 800 (Right)

A_y = 400 (up)
B_y = 0

But I still have to admit, it feels really weird that we are ignoring the moments at the fixed supports (to a point where it bothers me) because I have never done an exercise like this where I treated a fixed support as a pin support. However, I do understand the need to do so in order to solve the problem which lacks member properties.

Also, the trick to realizing that member AB can't deform and thus has internal force of 0 N would have never crossed my mind, good to know !

14. Mar 22, 2017

### PhanthomJay

Yes, this is good now. This is essentially a truss that takes axial loads only, with no moments or shears in the members except for insignificant moments and shears due to secondary stresses from small member deflections that often as in this case can be ignored.