FIELD FABRICATION OF STRUCTURAL COMPONENTS (MIXTURE AND COMPONENTS) BASIC INFORMATION
What Are Field Fabrication Of Structural Components?
Structural components that are fabricated on site by trades people constitute the greatest risk for a catastrophic failure. This is due to the fact that control of putting parts together in the field is not done with the same diligence and controlled environment as a factory-made component.
Thus, great care must be taken to ensure that proper testing is performed so that a failure will not occur. The erection of a concrete structure is an excellent example where the use of a mixed type material must have adequate testing.
Concrete is a very viable construction material if placed according to the standards established by the organizations. However, due to the complexity of mixing the ingredients at the plant and transporting it to the site, placing the concrete at the site requires numerous controls to obtain an excellent final product.
The testing of concrete should include:
1. A trial concrete mix approved by the owner’s engineer
2. Proper mixing procedures at the concrete plant
3. Timing for the transportation of the concrete mix
4. Designed and properly installed form work and shoring so that they will not collapse or deflect
5. Temperature monitoring of the concrete at the site (to make sure that flash setting will not occur)
6. Ambient temperature monitoring (too hot for flash setting and too cold for freezing)
7. Slump test to confirm water/cement ratio of the concrete
8. Supervision for concrete vibration and dropping height for the actual placement of the concrete
9. Monitoring the thickness of a concrete slab
10. Assurance that all the concrete encapsulates the reinforcing bars, especially when
pouring columns
11. Placement of a sample of the concrete into concrete cylinders to determine the compressive strength of the concrete at 7, 14, and 28 days (via testing in the laboratory). This will be accomplished for design strength conformance and to know when the forms can be stripped
12. Checking the number and location of the reinforcing bars required for the pour
13. Proper curing of the concrete
14. Assurance that reinforcing bars are properly lapped
15. Assurance that all exterior exposed concrete is covered by 3 inches of concrete
(2 inches for interior concrete) over the reinforcing steel
Even though steel sections are fabricated in a controlled environment at a plant, the steel members must be connected in the field by iron workers with bolts and/or welding.
Thus, stringent testing is also required for a steel structure. Some of the tests that would have to be considered when erecting steel are the following:
1. Proper bolts are being utilized.
2. Required tightening (torque) of the bolts needs to be accomplished by code standards.
3. Steel sections as indicated on the approved shop drawings are in fact being installed.
4. Welds have to be checked for proper thickness and continuity.
5. All welders have to be certified.
6. Shear stud connectors have to be attached to the steel with proper spacing and welds.
7. The steel has to be fireproofed with approved material that will have proper thickness, adhesion, and density.
8. All columns are perfectly aligned (plumbed).
9. Correct steel is being used (i.e., A36).
10. Proper steel camber has been placed on the steel as specified by the consultants.
11. Splice plates must be of the approved thickness.
12. Inspection at the fabricator’s shop would be helpful for checking beam camber and obtaining coupons.
Saturday, April 7, 2012
METHODS OF JOINT TRUSS MEMBERS ANALYSIS BASICS AND CIVIL ENGINEERING TUTORIALS
METHODS OF JOINT TRUSS MEMBERS ANALYSIS BASIC INFORMATION
How To Do The Methods Of Joint Truss Members Analysis?
This method for finding the forces in the members of a truss consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint. The method therefore deals with the equilibrium of concurrent forces, and only two independent equilibrium equations are involved.
We begin the analysis with any joint where at least one known load exists and where not more than two unknown forces are present. The solution may be started with the pin at the left end. Its free-body diagram is shown in Fig. 4/7.
With the joints indicated by letters, we usually designate the force in each member by the two letters defining the ends of the member. The proper directions of the forces should be evident by inspection for this simple case.
The free-body diagrams of portions of members AF and AB are also shown to clearly indicate the mechanism of the action and reaction. The member AB actually makes contact on the left side of the pin, although the force AB is drawn from the right side and is shown acting away from the pin.
Thus, if we consistently draw the force arrows on the same side of the pin as the member, then tension (such as AB) will always be indicated by an arrow away from the pin, and compression (such as AF) will always be indicated by an arrow toward the pin.
The magnitude of AF is obtained from the equation ΣFy = 0 and AB is then found from ΣFx = 0. Joint F may be analyzed next, since it now contains only two unknowns, EF and BF. Proceeding to the next joint having no more than two unknowns, we subsequently analyze joints B, C, E, and D in that order.
Figure 4/8 shows the free-body diagram of each joint and its corresponding force polygon, which represents graphically the two equilibrium conditions ΣFx = 0 and ΣFy = 0. The numbers indicate the order in which the joints are analyzed.
We note that, when joint D is finally reached, the computed reaction R2 must be in equilibrium with the forces in members CD and ED, which were determined previously from the two neighboring joints. This requirement provides a check on the correctness of our work.
Note that isolation of joint C shows that the force in CE is zero when the equation ΣFy = 0 is applied. The force in this member would not be zero, of course, if an external vertical load were applied at C.
It is often convenient to indicate the tension T and compression C of the various members directly on the original truss diagram by drawing arrows away from the pins for tension and toward the pins for compression.
This designation is illustrated at the bottom of Fig. 4/8. Sometimes we cannot initially assign the correct direction of one or both of the unknown forces acting on a given pin. If so, we may make an arbitrary assignment. A negative computed force value indicates that the initially assumed direction is incorrect.
How To Do The Methods Of Joint Truss Members Analysis?
This method for finding the forces in the members of a truss consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint. The method therefore deals with the equilibrium of concurrent forces, and only two independent equilibrium equations are involved.
We begin the analysis with any joint where at least one known load exists and where not more than two unknown forces are present. The solution may be started with the pin at the left end. Its free-body diagram is shown in Fig. 4/7.
With the joints indicated by letters, we usually designate the force in each member by the two letters defining the ends of the member. The proper directions of the forces should be evident by inspection for this simple case.
The free-body diagrams of portions of members AF and AB are also shown to clearly indicate the mechanism of the action and reaction. The member AB actually makes contact on the left side of the pin, although the force AB is drawn from the right side and is shown acting away from the pin.
Thus, if we consistently draw the force arrows on the same side of the pin as the member, then tension (such as AB) will always be indicated by an arrow away from the pin, and compression (such as AF) will always be indicated by an arrow toward the pin.
The magnitude of AF is obtained from the equation ΣFy = 0 and AB is then found from ΣFx = 0. Joint F may be analyzed next, since it now contains only two unknowns, EF and BF. Proceeding to the next joint having no more than two unknowns, we subsequently analyze joints B, C, E, and D in that order.
Figure 4/8 shows the free-body diagram of each joint and its corresponding force polygon, which represents graphically the two equilibrium conditions ΣFx = 0 and ΣFy = 0. The numbers indicate the order in which the joints are analyzed.
We note that, when joint D is finally reached, the computed reaction R2 must be in equilibrium with the forces in members CD and ED, which were determined previously from the two neighboring joints. This requirement provides a check on the correctness of our work.
Note that isolation of joint C shows that the force in CE is zero when the equation ΣFy = 0 is applied. The force in this member would not be zero, of course, if an external vertical load were applied at C.
It is often convenient to indicate the tension T and compression C of the various members directly on the original truss diagram by drawing arrows away from the pins for tension and toward the pins for compression.
This designation is illustrated at the bottom of Fig. 4/8. Sometimes we cannot initially assign the correct direction of one or both of the unknown forces acting on a given pin. If so, we may make an arbitrary assignment. A negative computed force value indicates that the initially assumed direction is incorrect.
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