rein shear wall


Graphic Vector AnalysisFirst used by Leonardo da Vinci, graphic vector analysis is apowerful method to analyze and visualize the flow of forces. Graphic vectors are only for statically determinate systems.Only 2-D vectors are described but 3-D vectors are possible Vectors have magnitude, line of action, direction, presented by straight lines with an arrow:

Direction: arrow pointing in direction of force

Magnitude: vector length at a scale, like 1” = 1 k

Line of Action: slope and location in space


Resultant and Equilibriant1 Two force vectors P1 & P2 pull a body in a certain direction.

The resultant R has the same results as P1 & P2 combined. The resultant is found by drawing a force parallelogram [A] or a force triangle [B]. Lines in the force triangle are parallel to corresponding linesin the vector plan [A]. The resultant R intersects with P1/P2.Since structures must be at rest we can find the equilibriant Ethat puts forces in equilibrium [C]. Equilibriant and Resultant have equal magnitude but oppositedirection. The equilibriant is also at the P1/P2 intersection.The equilibriant closes a polygon with head-to-tail vectors.

This process may be repeated for any number of forces.Interim resultants clarify the process but are not required [F].The line of action of the equilibriant is located at theintersection of all forces in the vector plan [D]. The equilibriant method may be stated as follows: The equilibriant puts any number of forces in equilibrium byclosing the force polygon with all vectors joined head-to-tail

3 The equilibriant of forces without a common cross-point [G] isfound in stages: First the interim resultant R1-2 of P1 and P2is found [H] and located at the P1/P2 intersection in plan [G].

2 The equilibriant of three forces [D] is found, combining aninterim resultant R1-2 of forces P1 & P2 with P3 [E].

Next P3 is combined with R1-2 to find the equilibriant forall forces, located at the intersection of R1-2 with P3.This process too can be repeated for any number of forces.


Component VectorsComponent vectors have the same effect on a body as theVectors for which they are components . Thus componentsrelate to a vector as two vectors relate to an equilibriant.

1. The component forces C1 and C2 of two rods supportingload P are found by force triangle [B] with corresponding lines parallel to those in the vector plan [A].

2. Forces in more than two rods supporting a load P areindeterminate since they have infinite solutions [C] andcannot be found graphically [D]. The solution requiresstiffness (cross section, slope, and E-modulus). Hence:Only 2 components can be found by graphic method.

3. Graphic vector example: Forces are drawn on a forceplan [E]. Magnitude is written on each force or drawn ata force scale. A force polygon [F] is drawn at a force scale. For accuracy the force scale should be as large as space permits. The line of action of the equilibriant (or resultant) is transposed to the force plan at the intersection of all force vectors [E]


Numeric Vector AnalysisNumeric vector analysis is more accurate than graphic vectoranalysis but more cumbersome, as demonstrated below.P1 = 4 kP2 = 6 kA = 45°B = 26.57°

Find resultant R for vectors P1 and P2

P1 components:P1x = P1 cos (A) = 4 (-0.707) P1x = - 2.83P1y = P1 sin (A) = 4 (0.707) P1y = + 2.83

P2 components:P2x = P2 cos (B) = 6 (0.894) P2x = +5.37P2y = P2 sin (B) = 6 (0.447) P2y = +2.68

Resultant components:Rx = ΣPx = -2.83 + 5.37 Rx = +2.54Ry = ΣPy = +2.83 + 2.68 Ry = +5.51

Resultant R:R = (Rx2 + Ry2)1/2 = (2.542 + 5.512)1/2R = + 6.06 kResultant slope C:C = Atan ( Ry / Rx) = Atan (5.51 / 2.54) C = 65.25°


R = 6

Which method is faster ? Graphic Which visualizes better ? Graphic

Fall in love with graphic vectors

Graphic Vector Analysis


Arch and cable forces1 Arch forces are found by graphic method:

Draw arch of span A-C and rise D-E. Draw support tangents A-B and C-B. For quadratic parabola (uniform load) B-E = D-E.Compute vertical reactions:V = wL/2 (L = arch span, w = uniform load)Draw vector triangle at C, starting with reaction V.Complete triangle with vectors parallel to F and H.F is the arch force at C and H the horizontal reaction at C.Vector lengths represent vector magnitudes at force scale.Symmetric loads and supports have equal forces at both ends.

2 Cable forces are found in similar manner:Draw cable of span A-C, sag D-E and unequal support.For quadratic parabola (uniform load) D-E = B-E.Compute and draw vector of total vertical load W = wL (L = cable span, w = uniform load)Draw vector parallel to A-B, representing cable force at ADraw vector parallel to B-C, representing cable force at CDraw reaction H and V1 at A. Draw reaction H and V2 at C.Measure magnitude of all forces and reactions at the same force scale as the initial vector W.

rein shear wall

Documents

Transcript of rein shear wall