Architectural Structures

Vector components

Vector components have the same effect on a body as the initial vector. Thus components relate to a vector as two vectors relate to a resultant or equilibriant.

1The component forces C1 and C2 in two cables supporting a load P are found by drawing a force triangle [B] with corresponding lines parallel to the those in the vector plan [A].

2Forces in more than two cables supporting a load P are indeterminate [C] and cannot be found by graphic vector method since there are infinite number of solutions [D]. A problem with more than two unknown force components requires consideration of relative cable stiffness (cross-section area, length, and stiffness). Hence we can state:

Only two components can be found by graphic vector method

3This example demonstrates graphic vector analysis: Forces are drawn on a vector plan with line of action and direction [E]. The magnitude may be written on each vector or the vector may be drawn at a force scale. A force polygon [F] is drawn next at a force scale, such as 1” = 1k. For good accuracy, the force scale should be as large as space permits. The line of action of the equilibriant (or resultant) is then transposed into the vector plan at the intersection of all force vectors [E].

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Truss Analysis

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

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

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

CDraw 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.

DDraw polygons for each joint to find forces in connected bars. Closed polygons with head-to-tail arrows are in equilibrium. Start with left joint ABHG. Draw a 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.

EMeasure the bar forces as vector length in the polygon.

GDraw 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).

1Truss diagram

2Force polygon for loads, reactions, and the first joint polygon

3Truss with completed tension and compression arrows

4Completed force polygon for left half of truss

5Tabulated bar forces (- implies compression)

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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.

1Truss diagram

2Force polygon

3Tabulated bar forces (+ implies tension, - compression)

4Deformed truss to visualize tension and compression bars

ABar elongation causes tension

BBar shortening causes compression

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Funicular

Graphic vector are powerful means to design funicular structures, like arches and suspension roofs; providing both form and forces under uniform and random loads.

Arch

Assume:

Arch span L = 150, arch spacing e = 20’ DL = 14 psf

LL = 16 psf Σ = 30 psf

1Arch structure

2Parabolic arch by graphic method Process:

Draw AB and AC (tangents of arch at supports) Divide tangents AB and AC into equal segments

Lines connecting AB to AC define parabolic arch envelope

3Arch profile Process:

Define desired arch rise D (usually D = L/5) Define point A at 2D above supports

AB and AC are tangents of parabolic arch at supports Compute vertical reactions R = w L /2

4Equilibrium vector polygon at supports (force scale: 1” = 50 k) Process:

Draw vertical vector (reaction R)

Complete vector polygon (diagonal vector parallel to tangent) Measure vectors (H = horizontal reaction, F = max. arch force)

Note:

The arch force varies from minimum at crown (equal to horizontal reaction), gradually increasing with arch slope, to maximum at the supports.

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Suspension roof

Assume:

Span L = 300, cable spacing e = 10’, sag f = 30’, height difference h = 50’ DL = 14 psf

LL = 16 psf Σ = 30 psf

1Cable roof structure

2Parabolic cable by graphic method Process:

Draw AB and AC (tangents of cable at supports) Divide tangents AB and AC into equal segments

Lines connecting AB to AC define parabolic cable envelop

3Cable profile Process:

Define desired cable sag f (usually f = L/10) Define point A at 2f below midpoint of line BC

AB and AC are tangents of parabolic cable at supports Compute total load W = w L

4Equilibrium vector polygon at supports (force scale: 1” = 50 k) Process:

Draw vertical vector (total load W) Draw equilibrium polygon W-Tl-Tr

Draw equilibrium polygons at left support Tl-H-Rl

Draw equilibrium polygons at right support Tr-Rr-H

Measure vectors H, Rl, Rr at force scale

Note:

This powerful method finds five unknowns: H, Rl, Rr. Tl. Tr

The maximum cable force is at the highest support

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Axial stress

Axial stress acts in the axis of members, such as columns. Axial tension is common in rods and cables; wile axial compression is common in walls and columns. The following examples illustrates axial design and analysis. Analysis determines if an element is ok; design defines the required size. The equation, fa = P/A, is used for analysis. The equation A = P/Fa, is used for design. Allowable stress, Fa, includes a factor of safety.

1Crane cable design

Assume: P = 12 k, Fa = 70 ksi Find required cable size

Metallic cross section Am (cables are about 60% metallic)

3Post/footing analysis

Assume: P = 12,000 #, 3’x3’x2’ footing at 150 pcf, 4x4 post (3.5”x3.5” actual) Allowable post stress Fa = 1000 psi, allowable soil pressure Fs = 2000 psf Post stress

P/A = 12,000 # / (3.5”x 3.5”) Soil pressure

fs = P/A = (12,000 # + 3’x3’x2’x150 pcf)/ (3’x3’)

4Slab/wall/footing, analyze a 1’ wide strip

Assume: allowable wall stress Fa = 360 psi; allowable soil pressure Fs=1500 psf Concrete slab, t =8” thick, L = 20’ span

CMU wall, h =10’, DL = 80 psf, t = 8” nominal (7 5/8” = 7.625” actual)

Shear stress

Shear stress occurs in many situations, including the following examples, but also in conjunction with bending, described in the next chapter on bending. Shear stress develops as a resistance to sliding of adjacent parts or fibers, as shown on the following examples. Depending on the number of shear planes (the joining surface [A] of connected elements) shear is defined as single shear or double shear.

AShear plane

BShear crack

1Single shear

Assume: P = 3 k = 3000 #, 2”x4” wood bars with ½” bolt of Fv = 20 ksi Shear area (bolt cross section)

2Check end block (A)

Assume: Block length 6”, wood Fv = 95 psi, all other as above

3Double shear

Assume: P = 22 k, 2 5/8” bolts of Fv = 20 ksi

4Double shear, glued

Assume: P = 6000 #, Wood bars, Fv = 95 psi

5Twin beam double shear

Assume: P = R = 12 k, 2 ½“ bolts, Fv = 20 ksi

6Shear wall

Assume: P = 20 k, 8” CMU wall, t = 7.625”, L = 8’, Fv = 30 psi