Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations SDETERS on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Fixed connections formulas 1

Status
Not open for further replies.
Replies continue below

Recommended for you

By symmetry the BM's at supports and midspans are just going to be PL/8 if the left, middle and right supports are encastre. Your symmetrical half model is basically just a simple fixed ended beam then to calculate your reactions and shears and moments (assuming of course the LHS and RHS loading conditions are exactly the same).

Of course, dont forget that the middle support will take "P" kN of load, whereas the right and left supports will take "P/2" kN of load.

 
Can anyone show how exactly is the 5P/16 and 11P/8 derived in the original 3 supports pinned configuratuon in the original attachment in the first message? Thank you.
 
one solution method:

Remove R2 (R2 is located at the centerline)

Deflection at R2 for the simple condition:
Pa/24EI * (3*l^2 - 4*a^2)
where: a = L/4
Pl/96EI * (3*l^2 - l^2/4)
Pl/96EI * (12*l^2/4 - l^2/4)
Pl/96EI * (11*l^2/4)
11*Pl^3/384EI

Too restore that deflection consider a single point load at R2:
R2 l^3 / 48*EI

set those two equations equal to each other and solve for R2
R2 = 11*P/8

from here it is simple statics to determine R1 and R3
Sum Fy = 0 -> R1 + R2 + R3 = 2P
Sum M1 = 0 -> R2*l/2 + R3 l - P*l/4 - P*3l/4 = 0 -> R3 = P - R2/2 = P - 11*P/16 = 16*P /16 - 11*P/16 = 5P/16

R1 + 11*P/8 + 5*P/16 = 2P
R1 + 27*P/16 = 2P
R1 = 2P - 27*P/16
R1 = 32*P/16 - 27*P/16
R1 = 5P/16



Open Source Structural Applications:
 
From symmetry, the slope at R2 must be zero. This is the same as a fixed end, so you could refer to Diagram 13 of AISC "Beam Diagrams and Formulas" for all the reactions, shears, moments and deflections for a concentrated load at midspan with one end fixed and one end supported.

If you think this is cheating, it is not difficult to determine all of these quantities from first principles.

BA
 
 https://files.engineering.com/getfile.aspx?folder=2eab48f3-d1f8-4b5d-896d-00202cbb18be&file=BeamDiagrams&Formulas.pdf
In column-beam joints.. do you model it as pinned or fixed? But how can you make it fixed.. because when the moment of the column can affect the moment of the beam.. it is not really "fixed". How do you make a column-beam joint such that no moment can be transfered and it is really fixed?
 
Quence said:
In column-beam joints.. do you model it as pinned or fixed?

If the joint is pinned, none of the members can have any moment at that joint. If the joint is fixed, none of the members can have any rotation at that joint. If the joint is free to rotate, then all members meeting at that joint will rotate by the same amount.

Quence said:
But how can you make it fixed.. because when the moment of the column can affect the moment of the beam.. it is not really "fixed".

If a joint is fixed, then the moment of one member cannot affect the moment of any other. Typically, in a rigid frame, joints are free to rotate and translate. Any displacement of the joint can affect the axial force, shear and bending moment of all members meeting at that joint.

Quence said:
How do you make a column-beam joint such that no moment can be transfered and it is really fixed?

Whether the joint is pinned or fixed, no moment can be transferred from one member to another; but if you want it to be really fixed, then you prevent translation and rotation of the joint.

BA
 
In real life structures where the joints are neither pinned not fixed.. what do you called them then? Semi fixed? And what formulas or modelling should you use then to compute since you cant use the formulas of pinned or fixed?
 
In that case, you can add a virtual member with a stiffness appropriate to the situation.

BA
 
How common is this method? Or does one just use approximate pinned or fixed model method?
 
See:

for a spreadsheet that will return fixed end moments and reactions for any specified loading and end conditions.

The latest version can be downloaded from:

macaulay2-02.jpg


Doug Jenkins
Interactive Design Services
 
What kind of analysis does etabs use in the computations of the reactions in the column beam joints.. is it pinned, fixed or does it use virtual member method or others? Anyone got a clue?
 
Quence said:
What kind of analysis does etabs use in the computations of the reactions in the column beam joints.. is it pinned, fixed or does it use virtual member method or others? Anyone got a clue?
I don't understand your question.
Any joint can be pinned, fixed, restrained or free. When using a frame program to analyze a structure, the user decides the category of every joint.


BA
 
In etabs.. either the joints are moment connected or released.. so I guess the choice is either pinned or moment connected.. which is not fixed.. so there is no fixed option in Etabs or other software.. I guess the program is using virtual member with stiffness appropriate to the situation? Anyone got a clue what method of computation it and other program uses? I think lesser program doesn't use virtual member with stiffness analysis?
 
The choice is not restricted to pinned or moment connected. "Tutorial for Beginners" for etabs is very hard to follow because the announcer speaks much too fast. It is really one of the worst presentations I have ever seen, but at about minute 8:45, a selection of joint properties is made. A roller is selected instead of a hinge.

Under the "Assign" heading, you can select a hinge, roller, fixed joint and in fact any combination of joint restraints that you want. Check it out here.


BA
 
If restraints against rotation is not checkes and moment connection is used..so the program indeed use virtual members or more accurate methods of computing for reactions?
 
No! Virtual (or hypothetical) members could be introduced by the user to represent a condition of partial constraint. If a column bears on a footing or pile with a full moment connection, the joint at the base of the column is neither fixed nor pinned because the foundation can rotate within the soil mass. It is actually somewhere between fixed and pinned. The user can introduce a fictitious spring at that joint to account for the restraint of the soil around the foundation.

In most structures, typical joints have no restraints applied. They simply rotate and translate as dictated by the applied loads and continuity of members. That is the default condition. The exception is usually the base of columns which may be hinged, fixed or somewhere in between. In some cases, the designer may designate a particular joint to be a hinge or other type of support for reasons of his own.

Structural programs such as etabs should be used only by experienced engineers who have a clear understanding of the principles involved and are capable of doing a sanity check on the output.

BA
 
Lets focus on column-beam joints. We know it is neither pinned nor fixed.. yet the program can use moment connection and output reactions.. so I was asking that since it is neither pinned nor fixed.. could it be using virtual member method as default in the analysis of the reactions? If not.. then what formulas does it use? the pinned formulas? I'm talking about column-beam joints only now. Thank you.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor