## Monday, December 23, 2013

### TRUSS EXAMPLE

Some trusses have bars with zero force under certain loads.  The example here has zero force in bars HG, LM, and PG under the given  load.  Under asymmetrical loads these bars would not be zero and, therefore, cannot be eliminated.  Bars with zero force have  vectors of zero length in the equilibrium polygon and, therefore, have both letters at the  same location.

Tension and compression in truss bars can be visually verified by deformed shape (4),  exaggerated for clarity.  Bars in tension will elongate; bars in compression will shorten.

In the truss illustrated the top chord is in compression; the bottom chord is in tension;  inward sloping diagonal bars in tension; outward sloping diagonal bars in compression.

Since diagonal bars are the longest and, therefore, more likely subject to buckling, they  are best oriented as tension bars.

1 Truss diagram
2 Force polygon
3  Tabulated bar forces (+ implies tension, - compression)
4  Deformed truss to visualize tension and compression bars
A  Bar elongation causes tension
B  Bar shortening causes compression

## Tuesday, December 17, 2013

### STRUCTURES - TRUSS ANALYSIS

Graphic truss analysis (Bow’s Notation) is a method to find bar forces using graphic vectors as in the following steps:

A  Draw a truss scaled as large as possible (1) and compute the reactions as for beams (by moment method for asymmetrical trusses).

B  Letter the spaces between loads, reactions, and truss bars.  Name bars by adjacent letters: bar BH between B and H, etc.

C  Draw a force polygon for external loads  and reactions in a force scale, such as  1”=10 pounds (2).  Use a large scale for accuracy.  A closed polygon with head-to-tail arrows implies equilibrium.  Offset the reactions to the right for clarity.

Draw polygons for each joint to find forces in connected bars.  Closed polygons
vector parallel to bar BH   through B in the polygon.  H is along BH.  Draw a vector
parallel to bar HG through G to find H at intersection BH-HG.

E  Measure the bar forces as vector length in the polygon.

F  Find bar tension and compression.  Start with direction of   load  AB  and  follow  polygon ABHGA with head-to-tail arrows.  Transpose arrows to respective bars in  the truss next to the joint.  Arrows pushing toward the joint are in compression; arrows pulling away are in tension.  Since the arrows reverse for adjacent joints,  draw them only on the truss but not on the polygon.

G  Draw equilibrium arrows on opposite bar ends; then proceed to the next joint with  two unknown bar forces or less (3).  Draw polygons for all joints (4), starting with  known loads or bars (for symmetrical  trusses half analysis is needed).

1 Truss diagram
2  Force polygon for loads, reactions, and the first joint polygon
3  Truss with completed tension and compression arrows
4  Completed force polygon for left half of truss
5  Tabulated bar forces (- implies compression) STRUCTURES - TRUSS ANALYSIS

## Monday, December 9, 2013

### STRUCTURES VECTOR ANALYSIS

First used by Leonardo da Vinci, graphic vector analysis is a powerful method to analyze  and visualize the flow of forces through a structure.  However, the use of this method is restricted to statically determinate systems.  In addition to forces, vectors may represent  displacement, velocity, etc.  Though only two-dimensional forces are described here, vectors may represent forces in three-dimensional space as well.  Vectors are defined by  magnitude, line of action, and direction, represented by a straight line with an arrow and  defined as follows:

Magnitude is the vector length in a force scale, like 1” =10 k or 1 cm=50 kN
Line of Action is the vector slope and location in space
Direction is defined by an arrow pointing in the direction of action

1  Two force vectors P1 and P2 acting on a body pull in a certain direction.  The resultant R is a force with the same results as P1 and P2 combined, pulling in the  same general direction.  The resultant is found by drawing a force parallelogram [A]  or a force triangle [B].  Lines in the vector triangle must be parallel to corresponding  lines in the  vector plan [A].  The line of action of the resultant is at the intersection  of P1 / P2 in the vector plan [A].  Since most structures must be at rest it is more  useful to find the  equilibriant E that puts a set of forces in equilibrium [C].  The  equilibriant is equal in magnitude but opposite in direction to the resultant.  The  equilibriant closes a force triangle with all vectors connected head-to-tail.  The line  of action of the equilibriant is also at the intersection of P1/P2 in the vector plan [A].

2  The equilibriant of three forces [D] is found, combining interim resultant R1-2 of  forces P1 and P2 with P3 [E].  This process may be repeated for any number of  forces.  The interim resultants help to clarify the process but are not required [F].  The line of action of the equilibriant  is located at the intersection of all forces in the  vector plan [D].  Finding the equilibriant for any number of forces may be stated as  follows:

The equilibriant closes a force polygon with all forces connected head-to-tail,  and puts them in equilibrium in the force plan.

3  The equilibriant of forces without a common cross-point [G] is found in stages:   First the interim resultant R1-2 of P1 and P2 is found [H] and located at the  intersection of P1/P2 in the vector plan [G].  P3 is then combined with R1-2 to find  the equilibriant for all three forces, located at the intersection of  R1-2 with P3 in the  vector plan.  The process is repeated for any number of forces.

## Monday, December 2, 2013

### STRUCTURES - BEAM REACTIONS

To find reactions for asymmetrical beams:

•  Draw an abstract beam diagram to illustrate computations
•  Use Σ M = 0 at one support to find reaction at other support
•  Verify results for vertical equilibrium

1 Floor framing
2  Abstract beam diagram

Support reactions:

Alternate method (use uniform load directly)

1  Simple beam with point loads

2  Beam with overhang and point loads