Bridge Construction Manual - Falsework and Forms

November 1, 2005 BRIDGE CONSTRUCTION MANUAL 5-393.200

FALSEWORK AND FORMS 5-393.200

(Note: This section uses English units only)

5-393.201 INTRODUCTION The finished appearance of a concrete structure is dependent to a large extent on the forms and falsework used in the concrete construction. Concrete lines and surfaces will usually reflect the degree of care and skill used in the design and construction of the forms and falsework. Unsatisfactory concrete construction due to faulty form and falsework design or construction is very difficult to correct after the concrete is cast. In an extreme case, the result of a faulty design or careless workmanship on the forms and falsework could lead to their collapse during construction operations, with serious consequences. The Department, by its Specifications, has placed certain restrictions on the design and construction of forms and falsework. These restrictions do not prohibit the Contractor from exercising ingenuity in the construction of form details and the economical selection of materials. The Specifications do provide guidelines for forms and falsework that should be safe and that will result in satisfactory concrete lines and surfaces. A. Definitions and Nomenclature Forms are those members (usually vertical) that are required to maintain plastic concrete in its desired shape until it has set up. Forms resist the fluid pressure of the plastic concrete, the additional fluid pressure generated by mechanical vibration of the concrete and the impact of placing the concrete in the forms. Forms may be stripped when the concrete is set up and its fluid pressure is reduced to zero. Falsework is the supporting framework required to maintain a concrete unit in the desired position (when it cannot be supported directly on the ground, as a footing or on previously cast concrete) until the concrete is strong enough to carry its own dead weight. Falsework must be strong enough to carry its own dead weight, the dead weight of the forms and concrete and the live load of construction crews and their equipment. Sheathing used on the underside of a concrete unit, such as sheathing under deck slabs or pier caps, serves the dual purpose of a form and a falsework member but, for design, is classified as a falsework member, as the dead weight of the concrete plus live load is used in the design loading rather than the fluid concrete pressure. Common terminology for members of a falsework pile bent and for a typical form system are shown on Figure A 5-393.201. In addition, the following nomenclature will be used in this manual when referring to form lumber: Side = wider face }regardless of how the piece is positioned Edge = narrower face}

Dressed or surfaced lumber - lumber which, to attain smoothness has been planed on a planing machine on either one side (S1S), two sides (S2S), one edge (S1E), two edges (S2E) or a combination thereof (including complete planing S4S) Rough timber - lumber which has not been dressed but has been sawn to the extent of showing saw marks on the full width of each of the 4 faces, also described as “full sawn lumber.” Sized lumber - lumber, either rough or dressed, which is prepared in standard dimension increments, with a plus and minus tolerance assigned to each size. Nominal dimensions - the cross-section dimensions of the piece in inches as a full sawn piece (dimension prior to surfacing). B. Falsework Requirements Usually the need for falsework is self-evident; however, bridge plans often require the construction of some concrete item at or below ground level which has no footing. Typical of these are: 1. cantilevered wingwalls on the abutments, 2. tie beams between concrete walls or between footings, 3. high abutment side walls (or curtain walls) as on box type

abutments. Judgment must then be exercised as to whether the existing soil can adequately support the weight of the concrete without settlement, or if falsework support is necessary. (Note, the falsework referred to is in addition to that which is used to hold sheathing to shape the lower concrete surface, as on the bottom of most abutment wingwalls.) Wingwalls and tie beams can normally be formed directly on the underlying soil, providing that the soil is stable and well compacted. (Beware of plastic soils that will become unstable with rain.) However, the higher abutment sidewalls will usually require falsework and falsework piling when the underlying material is not rock. Specific falsework requirements are given in Specifications 2401.3B4. For certain structures, the special provisions will contain requirements for falsework plans and construction. These requirements are in addition to those contained in construction specifications and this Manual. Forms must be used on all vertical or battered surfaces except the portions of footings that extend into solid rock. Casting concrete against an earth face will not be permitted (see Specifications 2401.3B). One exception to the above is that

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5-393.201 BRIDGE CONSTRUCTION MANUAL November 1, 2005

the 1 inch radius required at the front face of sidewalks can normally be shaped with a 1 inch radius edger. However, the 3 inch radius required at the front face of 6 inch high curbs must normally be shaped by a form built to the specified radius since hand shaping of this radius is seldom performed satisfactorily. C. Contractor Responsibilities The Contractor is responsible for the design of all falsework and forms and shall submit detailed plans of the proposed falsework and form construction to the Engineer on request (see Specifications 1502 and 2401.3B). In accordance with these Specifications, all working drawings are subject to review and approval or rejection by the Engineer prior to performance of the work involved. When trying new or untried devices or systems for falsework and/or forms and when using materials with unknown strength properties, it is the Contractor’s responsibility to verify to the Engineer’s satisfaction the strength and safety of the device or system and the workability of the device or system as related to the desired end product. This verification has in the past been provided in the form of (a) full scale field tests, (b) tests by a reputable testing laboratory, (c) certified design calculations, (d) manufacturers literature, or a combination of the above items. The Contractor is responsible for constructing the falsework and forms in reasonably close conformity with the approved falsework plans. D. Engineer's (Inspector's) Responsibilities When falsework and/or form plans are requested by the Engineer or required by the Contract, these plans should be reviewed for acceptability as to strength, method of construction, safety, potential problems, and ability to produce the desired finished product. Approval to use such plans should be noted as being approved as to type of construction and should also bear a note that such acceptance is conditional to making changes which the Engineer has noted thereon. When evaluating a new or untried device or system, approval (if given) should be given only on a performance basis. Such approval of plans does not relieve the Contractor of responsibility for results obtained by use of the plans (see Specifications 1502 and Specifications 2401.3B). For certain types of structures, a review by the Contractor's engineer is required prior to acceptance of the completed falsework. The inspector should be present during this review and no use of the falsework should be permitted until this engineer has completed the review and authorized use (normally a written certification is provided). Check the material that will be used for forms and falsework for agreement with sizes, shapes and qualities shown on the Contractor’s plans. If not in conformance with approved plans, the material must be evaluated as to its ability to

function as intended. See Section 5-393.202 for more information. A continuing inspection should be made during placement of form and falsework members to assure conformance with approved plans (if used), to assure structural soundness and accuracy, and to minimize the need for last minute corrections. This inspection is discussed in more detail in Section 5-393.208. Concrete pours are to be made in accordance with approved pour sequences. Where approval of pour sequences is not required, pours should be with the form or falsework design and should provide balanced loading to the extent possible. A follow-up inspection during and after concrete placement should be made to assure that the forms and/or falsework function as intended with regard to deflections, tolerances, etc.





November 1, 2005 BRIDGE CONSTRUCTION MANUAL Figure A 5-393.201

CROSS SECTION OF VERTICAL FORMS

TYPICAL FALSEWORK PILE BENT

Bridging Sheathing StringersPile cap Pile cut off

Drift bolts Pile capsplice point

Corbel

Swaybraces

Posts or piles

Bolt, nutand washers

Concrete strike-off elevation Upper plate

Strongback

Sheathing or form liningbacked with sheathing

Wood spreader

Form bolt(tie bolt)(tie rod)

Chamfer strip

Tie plate (washer)

Stud

Waler

Tie cone

Lower plate

5-393.201 (2) BRIDGE CONSTRUCTION MANUAL November 1, 2005

Symbols and Units The following symbols and abbreviations will apply to forms and falsework:

psi = pounds per square inch psf = pounds per square foot pcf = pounds per cubic foot

Symbol Description Units

A

B

h

t

w

d

D

ℓ

E

I

c

S

r

P

p

R

R1

T

v

V

V1

M

f

H

∆

area

width of beam

depth of beam

thickness of web for steel member

uniform load per foot of length

least dimension of a column

diameter

length (center to center of supports for beam spans; unsupported length for columns)

modulus of elasticity

moment on inertia

distance from neutral axis to extreme fiber

section modulus

least radius of gyration (for steel columns)

concentrated load

lateral pressure of concrete

reaction at beam support

rate or pour for concrete

concrete temperature in Fahrenheit at time of placement

shear stress

shearing force

the vertical force causing horizontal shear in a timber beam

external bending moment

stress in member

maximum horizontal shear stress in timber beams

deflection

in.2

inches

inches

inches

lb/lin.ft

inches

inches

inches

psi

in.4

inches

in.3

inches

lb

psf

lb

ft/hour

°F

psi

lb

lb

inch lb

psi

psi

inches

November 1, 2005 BRIDGE CONSTRUCTION MANUAL 5-393.202 (1)

5-393.202 FORM AND FALSEWORK MATERIALS Form and falsework materials described below are listed with either an allowable maximum working stress or a basis for determining safe load. Working stresses, when shown, are based on use of sound material for temporary construction. In general, used material is permitted, provided it is in good condition. The material requirements for falsework piling are stated in Specifications 2401.3B. Maximum allowable pile loads are as follows:

Size of Dia. At Cut-Off (inches)

Timber (tons) Steel Friction (tons)

Steel Point Bearing (tons)

8 10 12 14 16

Butts smaller than 8 in. are not permitted

16 20 24 28 32

16 20 24 28

9000 lb per sqare inch of point area (or least cross-sectional area of the pile)

General requirements for lumber for falsework and forms are specified in Specifications 2401.3B1c. In addition to these general requirements, it is specifically recommended that material used for studs and walers be sized and dressed to at least S2E to provide for true concrete lines. Lumber that must withstand stress should be checked for conformance with the appropriate allowable stresses shown in the table of “Allowable Working Stresses for Design of Wood Forms and Falsework” in this section. The following notes apply to use of this table: 1. NEW LUMBER -Each piece of graded lumber is stamped. On new material, information as to timber species and grade or

stress rating can be obtained from this stamp for use with the allowable stress table in this section of the manual. 2. USED LUMBER -In the event the mill stamp is missing or eradicated, the species and grade or stress rating must either be

determined by visual examination or judgement or an assumed identification must be applied. In case of uncertainty, assume Norway Pine common structural grade to be on the safe side.

3. Regardless of whether new or used lumber will be used, a visual check should be made of stressed members with the

following consideration in mind: Any reduction in section in or near the middle 1/3 of the length of a beam reduces the capacity to resist bending. Such reduction in section could be a damaged area, large knots, notches, or holes in the upper or lower 1/3 of the section. If such pieces are used for beams, only the sound portion of the section can be considered as effective for calculating stresses. Notches or reduction in beam depth near the support point will reduce the beam’s capacity to resist horizontal shear. Special calculations are necessary to determine the horizontal shear stress when such pieces are used (see Section 5-393.204). When forms or falsework are constructed of used material which is judged to be not equal in strength to sound material, the allowable stresses in the table should be reduced by an appropriate amount. The allowable stresses and modulus of elasticity (E) values listed in the chart below are in accordance with the recommendations of AASHTO Standard Specifications for Highway Bridges. A 25% stress increase has been included in these values (except E values) in view of the anticipated short time loading. Stresses for species or grades not listed below should be obtained from the Office of Bridges and Structures and conform to AASHTO Specifications.




ALLOWABLE WORKING STRESSES FOR DESIGN OF WOOD FORMS AND FALSEWORK

Max. Allowable Fiber Stress, psi

Species and Commercial grade

Size Classification Bending

Horizontal Shear

Side Bearing

End* Bearing

Modulus of Elasticity, E, psi

Douglas Fir - larch, No. 1 2 to 4 in. thick 1875 120 480 1562* 1,800,000

Southern Pine, No. 1 No. 2

≤ 4 in. thick 1812 1500

112 112

506 506

1562* 1250*

1,700,000 1,600,000

*The strength of a wood column is dependent on its ℓ/d ratio, which must not exceed 50. The allowable stress in a wood column will be either 1. f = the allowable end bearing stress listed in this section for the wood species used

2. f = 0.30 E

d

d = dimension of least side of columnl⎛

⎝⎜⎞⎠⎟

2

Use whichever allowable stress value is smallest. The maximum allowable compression stress for Douglas Fir columns and Norway Pine columns (as determined by the above criteria) may be obtained from the graph below.

. For convenience in making calculations involving dimension lumber, a tabulation of standard lumber sizes and their

Douglas Fir-Larch1500

1000

500

0

Allowablecompressivestress forcolumns, psi

1562 psi

No. 1 Grade

No. 1 Grade

Red Pine

1030 psi

5 10 15 20 25 30 35 40R/d

d

R

3

respective properties has been included in Figure A 5-393.202.


General requirements for plywood sheathing are specified in Specifications 2401.3B. The plywood sheathing most commonly used is Douglas Fir Associations Exterior “Plyform,” which is available in two strength varieties known as Class I and Class II. Both Classes are fabricated using exterior glue and have sanded grade B face plies. New panels of plyform can be identified by the following trademark stamped on the panels: In considering the bending strength, shearing strength, or deflection of a panel, only those plies have their grain perpendicular to the supporting joist or stud are assumed to be stressed. The safe span length is therefore dependent not only on whether Class I or Class II Plyform is used but also on whether the grain of the face plies runs across supports (perpendicular to the joist or stud), or parallel to supports (parallel to the joists or stud).

Plywood used the strong way(Face ply grains are perpendicular to supports)

B - B P L Y F O R M

B - B P L Y F O R M

CLASS IEXTERIOR

CLASS IIEXTERIOR

PS 1-66 000

PS 1-66 000

The data in Figure B 5-393.202 may be used for quickly determining the safe spacing of studs or joists using Plyform Class I or Plyform Class II under two different loading conditions. These charts are recommended by the American Plywood Association. The applicable allowable stress values for bending and horizontal shear are shown for each situation. When the brand name or grade stamp is visible on the plywood being used, the requirements of Specifications 2401.3B can be quickly verified. When no grade stamp is visible, it is the Contractor’s responsibility to verify to the satisfaction of the inspector that concrete form grade plywood has been furnished. When it is determined that form grade plywood has been furnished but the specific Class of plywood is unknown, the following limiting stress values will apply: Maximum allowable bending stress = 1,500 psi Maximum allowable shear stress = 70 psi Modules of Elasticity = 1,600,000 psi Maximum allowable bearing stress at right angle to the plane of the plywood = 285 psi

Plywood used the weak way(Face ply grains are parallel to supports)




Plywood section properties, which will be necessary for checking stresses when not using the plyform charts, are tabulated below:

EFFECTIVE SECTION PROPERTIES FOR PLYWOOD (12-IN. WIDTHS)*

12-in. width, used with face grain perpendicular to supports

12-in. width used with face grain parallel to supports

Sanded plywood,

net thickness,

inc.

Number of plies

Effective thickness for shear

all grades, using

exterior glue

Area for tension and compression

(in.2)

Moment of inertia

I (in.4)

Effective section

modulus S (in.3)

Rolling shear

constant I/Q (in.)

Area for tension and compression

(in.2)

Moment of inertia

I (in.4)

Effective section

modulus S (in.3)

Rolling shear constant I/Q (in.)

1/4 3 0.241 1.680 0.013 0.091 0.179 0.600 0.001 0.016 -

3/8 3 0.305 1.680 0.040 0.181 0.309 1.050 0.004 0.044 -

1/2 5 0.450 2.400 0.080 0.271 0.436 1.200 0.016 0.096 0.215

5/8 5 0.508 2.407 0.133 0.360 0.557 1.457 0.040 0.178 0.315

3/4 5 0.567 2.778 0.201 0.456 0.687 2.200 0.088 0.305 0.393

7/8 7 0.711 2.837 0.301 0.585 0.704 2.893 0.145 0.413 0.531

1 7 0.769 3.600 0.431 0.733 0.763 3.323 0.234 0.568 0.632

1 1/8 7 0.825 3.829 0.566 0.855 0.849 3.307 0.334 0.702 0.748 Information from "Plywood Design Specification", American Plywood Association, Tacoma, Washington. Use listed S values in bending calculations, and use I only in deflection calculations.


The reuse of plywood sheathing will be dependent on its condition with respect to damage due to prior use, amount of permanent set from prior use, amount of face ply separation, and the nature of the concrete surface being formed (exposed or not exposed, etc.). Plywood that is no longer suitable for its intended purpose must be rejected. Form lining requirements both as to material and usage, are specified in detail in Specifications 2401.3B. Forms incorporating form lining backed by sheathing will be used rarely except in the case of architectural treatment of concrete surfaces. In situations like this, special form lining material requirements will usually be specified in the contract. When angles, channels, wide flange beams, H-piles or other rolled shapes are used in critical portions of the falsework, the section should be identified by making measurements of the depth, width and thickness. These dimensions can then be used to identify the member further by referring to the AISC Steel Construction Manual, where all standard rolled sections are listed along with their dimensions, weights and the necessary design properties. Since this material cannot be visually identified as to grade of steel, the following allowable stress limits should be assumed, unless the Contractor furnishes satisfactory assurance that the steel is of a higher grade. Rolled Steel Shapes (Assume ASTM A36 Steel) Maximum allowable bending stress 25000 psi *Maximum allowable compression stress (columns) =

16,980 - 0.53 x KLr

⎛⎝⎜

⎞⎠⎟

2

L = unsupported length K = 1.0 for pinned ends r = governing radius of gyration

* Lr

must not exceed 120.

The values listed above will be sufficient for checking most falsework problems involving rolled steel members. Any additional design considerations (as for steel falsework trusses and other special cases) should conform to the provisions of AASHTO Standard Specifications for Highway Bridges, as required in Specifications 2401.3B. When used material is to be incorporated into the work, the extent of damage (caused by previous usage) and corrosion should be evaluated. If corrosion is determined to have reduced the net thickness of a section, it is allowed to use the section properties of a rolled shape in the AISC manual with thickness dimensions compared to those of the intact material. Additional requirements for structural shapes are given in Specifications 2401.3B.

The increasing use of special devices, (made of material other than wood) for forms and falsework has, in general, resulted in a speed-up of work as well as improved quality of work. However, there is usually a degree of uncertainty about each new device until it is proven in use. A partial listing of devices which have been used both successfully, and in some instances unsuccessfully, is as follows: 1. Wall Form Panels The form panels referred to are the mass produced brand

name form sections (constructed either of steel or steel and wood) which are produced in small segments so as to be adaptable to a variety of concrete shapes and a variety of types of construction. Past experience with certain brands of these form panels resulted in the recommendation that form panel construction should not be permitted for concrete exposed to view. The reason for dissatisfaction on the work referred to was as follows:

a. Objectionable offsets existed at abutting panel edges. b. There were an excessive number of joints. (The

frequency of panel joints should generally be no greater than in conventional plywood-form construction.)

c. After being reused a number of times, permanent set

(permanent deflection) in the panels became excessive.

d. Adequate provisions were not made for overall

alignment of the form work nor for providing mortar-tight joints.

Only a form panel system which adequately overcomes

these objections with respect to appearances can be considered for use on concrete surfaces exposed to view.

Design of the forms, with respect to size and spacing of

members, is normally furnished by the manufacturer either as part of the advertising literature or as a special design for the job along with a safe rate of pour for concrete in the form system. These should be carefully adhered to.

2. Circular Column Forms Specific requirements for circular column forms are stated

in Specification 2401.3B. Such forms have been fabricated of steel, fiberglass and paper or other fibers and all have been used with varying degrees of success.

Since some circular forms can be damaged through

mishandling or improper storage, it is necessary to check the roundness and smoothness when making a judgment as to acceptability of each individual form. The form diameter on any axis should not be more than 1/2 inch under the specified diameter. This requirement is to assure proper cover on the column reinforcement. (Dents in paper tubes are normally not so critical since concrete






pressures during filling will round out the tubes. Flattened or elliptical tubes will not be perfectly round when filled with concrete).

Reusable steel forms are susceptible to damage in the

form of small dents and kinks. These result in unsightly dimples on the concrete surface. Repair of such forms should normally be requested prior to permitting their use. In addition, abutting panels should be adjusted so as to eliminate offsets at panel joints. If they have been overpoured in the past, the face panels may be stretched. Straight edge for acceptability.

Due to the possibility of very fast rates of concrete

placement in column forms, the pressure at the bottom of the form can be extremely high. Fasteners for the vertical form-joint on segmental forms (such as on steel or fiberglass column forms) can readily be checked for ability to withstand these pressures. (NOTE: These forms usually provide for a variable number of bolts or pins in this joint.)

Since circular paper or fibre forms are commercially

mass-produced in several strength grades, the adequacy of their design for a specific case will normally be determined by checking the manufacturer’s literature. Note carefully whether this literature lists a safe loading or a failure loading (or bursting pressure). When only the bursting pressure is given, a safety factor must be applied to determine a safe load. Normally a safety factor of 2 will be adequate.

If paper tubes have become wet prior to use, they should

be inspected for weak areas in advance of concrete placement. Paper tubes should also be checked to assure that no conspicuous seam ridges are present on the inside surface since these cause objectionable spiral ridges on the finished concrete surface.

3. Friction Collars For Pier Caps Friction collars for pier caps are steel devices which are

clamped around the top of circular concrete columns to support the pier cap falsework and pier cap concrete. Serious failures have resulted because of inattention to the placement of these collars. Since the entire falsework in this case is dependent on the stability of the collar, the tightening of the collars must be properly performed. The collars must be level to assure full bearing on the concrete. Manufacturers literature should be used to determine the necessary bolt tension. In addition, the total applied vertical load must not exceed the safe load specified in the manufacturer’s literature.

4. Slab Falsework - Interior Bays Several types of slab falsework other than the all-wood

type which have been successfully used by contractors are as follows:

a. Adjustable Steel Posts (See Figure C and D 5-393.202, types 4 and 5.)

This system basically replaces the wood legs of the

wooden "horse" system with adjustable steel posts. These posts are normally supported on wood joists spanning between the bottom flanges of adjacent beams. The strength of the system will normally be controlled by the wood members in the system.

b. Steel Hangers (See Figure C and D 5-393.202, types

2, 3 and 6.) This is basically a hardware item which is laid

transversely across the top flange of the beam to receive a vertical bolt on either side of the flange. The bolt in turn supports the main falsework member.

Balanced loading of the two sides of the hanger

(simultaneous loading of both bolts) is important in the early stage of falsework construction unless measures are taken to hold the hangers in place. Welding of these hangers to the shear connectors is a permissible method of hold down; however, welding to the beam flange is not permitted.

Safe working loads for steel hangers are listed in the

manufacturer’s literature. c. Steel Bar Joints (See Figure D 5-393.202, Type 7.) This is a steel falsework beam which can be adjusted

to a variety of lengths. Load capacity, allowable spacing and deflection data are available from the manufacturer’s literature which should be used for checking the system.

Such steel bar joists have been used as joists to

support longitudinal falsework stringers and also used at closer spacing with the sheathing placed directly on them. In the event the latter system is used, no wood nailer is available to hold down the sheathing and a system of wire ties or some other approved method of hold down is necessary.

Precautions must be taken to allow for residual

camber in this type of falsework system. The amount of residual camber anticipated after placement of the concrete should be determined (by field tests if necessary) and adequate allowance made in setting stool heights to obtain the specified slab thickness.

d. Corrugated Steel Forms Corrugated Steel Forms are commercially mass-

produced corrugated sheet metal forms for the bottom of the slab which require no additional supporting falsework.

Each unit spans transversely from beam to beam on

the bridge and acts in the capacity of a complete structural entity of falsework and sheathing. These


units are galvanized and are normally intended to remain in place at completion of the work. Safe loads and deflections for each size of member are available in the manufacturer’s literature.

Under the current policy, this type of falsework is not

permitted unless specifically indicated in the Special Provisions.

5. Slab Overhang Falsework Several types of slab overhang falsework (other than the

all-wood type) that have been successfully used by contractors are steel hangers, which have been previously discussed, and Steel Overhang Brackets.

Typical application of steel overhang brackets is shown in

Figures C and D 5-393.202, Types 3, 4, 5 and 6. Details and design data pertaining to two commonly used

overhang brackets (Capitol and Superior) are given in Figures E and F 5-393.202. It is intended that spacing and deflection of these brackets be determined by these details as furnished by the manufacturer. However, several precautions must be observed as described below.

Information for the Capitol brackets states that the

brackets should be spaced at 6'0" centers. However, experience has shown that the 6'0" spacing must be reduced under certain conditions. For example, when the strike-off rails are placed on top of the coping forms or when a very wide slab overhang is specified in the plans, a much higher load is applied to each bracket unless this spacing is reduced.

When installing Capitol brackets, the 2" x 4" member

placed in the top horizontal member of the bracket must be firmly seated and the hanger “chain” must be tight. Poorly aligned concrete surfaces have resulted when seating occurred during concrete placement.

The influence lines in Figure F 5-393.202 for checking

the Superior brackets may be used with a variety of loading conditions. The actual load in the critical members can be determined by use of this chart and checked against the safe working loads shown on the Figure.

A wood filler block is required when using these brackets

on prestressed concrete girders. This filler must be varied as necessary to provide a bearing surface on the beam at the end of the top horizontal member and at the end of the diagonal member. The filler should not bear on the vertical member of the Superior bracket.

The deflection graphs given for each of these brackets

should be used only as a guide since the graphs apply only to the specific loadings pictured on the manufacturer’s details.

For either bracket, when unusual loading conditions are encountered, a full scale field test is recommended. An overload should be applied to assure that there is a safety factor.

Since cantilever brackets tend to rotate the fascia beam

(push the bottom flange inward), special bracing precautions, as specified in Specification 2401.3B4, are occasionally necessary. For beams depths of 24 in. or less, the difficulty of obtaining good concrete lines increases when this type of overhang falsework is used and serious consideration should be given to the use of needle beams as shown in Type 1, Figure C 5-393.202.

6. Tubular Steel Scaffolding The basic components of Tubular Steel Scaffolding

Shoring as shown by the following picture are end frames of various designs and dimensions which are assembled with diagonal bracing and lock clamps. Vertical adjustments are made by adjustable jacks either at the bottom or top of the frames. Frames are normally fitted either with flat top plates or U-heads for supporting the falsework and forms.

These towers are rated by the load carrying capacity of

either one leg or of one frame (two legs). The manufacturer’s rated capacity should not be exceeded.

Adequate rigid bracing involving several units of steel

shoring should be provided. Full bearing for the base plates should be provided, such as being set in fresh mortar pads when resting on rock-like formations. Mudsills placed on yielding earth should not be permitted for supports.

U-Head or other framing

Cross braces

End framesjoined here

Adjustable jacksEnd frames

Lock forbraces

7. Void Tubes For Voided Slab Spans Void tubes for voided slab spans are similar to the fibre

tubes used for column forms except that galvanized steel tubes are also permitted. The circumferential crushing pressure and straight crushing pressure of these tubes will



normally be listed in the manufacturer’s literature. When checking stresses, it is necessary to determine if the manufacturer has listed a safe pressure or a failure pressure.

Since stress in the void tube is very high at the tie-down

straps, a careful visual inspection is necessary at this location. Wetting of paper tubes can result in isolated weak spots where the waterproof coating has been scratched or damaged and the water has penetrated into the paper or fibre layers. Such pieces should be rejected unless they can be satisfactorily reinforced.

Void tubes must be mortar tight. When several lengths of

tube are necessary to make up the length of void shown in the Plans, each segment of tube should have sealed ends. Butting tube ends together and taping around the perimeter of the joint will normally not be acceptable since deformation of one of the joined tubes during concrete placement would likely rupture a taped splice.

Information pertaining to nails and spikes are located on Figure H and I 5-393.202. General requirements governing bolts or form ties are given in Specifications 2401.3B. One specific provision is that a major portion of the device must remain permanently in the concrete. Several types of commercially available form ties meeting this description are shown below. Normally the manufacturer’s literature will list the safe load that may be applied. However, when the load capacity is questionable or unknown, laboratory tests will be necessary to determine the safe load.

In this event, the safe load may be set by determining the cross-sectional area of the member and the yield point of the steel by tension test in the laboratory. The applied load should not exceed 70% of the yield strength of the device. NOTE: the yield point of the steel (psi) is not the same as the yield strength of a particular bar. On portions of the structure exposed to view, form bolts must be so designed that all metal can be removed to a depth of not less than 1 inch from the concrete surface. Tie wires may be used only in locations where they will not extend through surfaces exposed to view in the finished work. The hardware used to secure form bolts against the forms is usually reusable. This hardware is normally designed to be stronger than the portion of the device that remains in the concrete and, therefore, will not be the limiting strength factor in the form tie. Crimp ties or snap ties are wire form ties with a notch or reduced cross-section at the point of break-back. These ties are not reusable. After the concrete is set, the portion of the wire which extends outside of the concrete surface is twisted off and removed. A washer is sometimes welded to the wire at the face of the form to act as a form spreader. On concrete surfaces exposed to view, a cone should be used in place of the washer since satisfactory patching of the shallow depression left by the washer is very difficult. Because these ties do not always break off at the intended point, but sometimes break instead at the face of the concrete, plus the fact that they do not provide a rigid member for support of the workcrew, they are not recommended for use on heavy construction. Their use is primarily restricted to light work such as box culverts, etc.

CONE NUTS AND INSIDE THREADED RODS

PLAIN TIE WITH SHE-BOLT DISCONNECTING ENDS

Nut washer orother locking unit

May have hole fornailing to stud

CRIMPED TIE WITH DISCONNECTING ENDS

Wedge shaped holder

COIL TYPE TIE WITH CONE SPREADERWasher

Coil bolt



NOMINALSIZE

PROPERTIES OF DRESSED SIZES(S4S)

DESIGN PROPERTIES FOR AMERICAN STANDARD LUMBER SIZES

MOMENT OF INERTIA SECTION MODULUS

I = inches4bh3

12 S = inches3

6bh2

load

hb

b(in.)

b(in.)

h(in.)

h(in.)

Area = b x h

Weightlb / lin. ft. S4S S4SFull Sawn Full Sawn

11 1/41 1/2

2468101214468101214468101214166810121416681012141668101214166810121416

11 1/4

1 1/2

2 1/2

3 1/2

5 1/2

7 1/4

9 1/4

11 1/4

3/41

1 1/41 1/23 1/25 1/27 1/49 1/411 1/413 1/43 1/25 1/27 1/49 1/411 1/413 1/43 1/25 1/27 1/49 1/411 1/413 1/415 1/45 1/27 1/49 1/411 1/413 1/415 1/45 1/27 1/49 1/411 1/413 1/415 1/45 1/27 1/49 1/411 1/413 1/415 1/45 1/27 1/49 1/411 1/413 1/415 1/4

8.4411.2514.0616.885.258.25

10.8813.8816.8819.888.75

13.7518.1323.1328.1333.1312.2519.2525.3832.3839.3846.3853.3830.2539.8850.8861.8872.8883.8839.8852.5667.0681.5696.06110.5650.8867.0685.56104.06133.56141.0661.8881.56104.06126.56149.06171.56

2.33.13.94.71.52.33.03.94.75.52.43.85.06.47.89.23.45.37.09.010.912.914.88.411.114.117.220.223.311.114.618.622.726.730.714.118.623.828.934.039.217.222.728.935.241.447.7

0.400.941.833.165.36

20.8047.6398.93177.98290.788.93

34.6679.39164.89296.63484.6212.5148.53111.15230.84415.28678.481034.4276.26174.66362.75652.591066.181625.51100.52230.23478.17860.231405.412142.72128.25293.75610.081097.531793.112733.82155.98357.26741.991334.842180.823324.92

1.001.953.388.0010.6736.0085.33166.67288.00457.3316.0054.00128.00250.00432.00686.0021.3372.00170.67333.33576.00914.67

1365.33108.00256.00500.00864.00

1372.002048.00144.00341.33666.671152.001829.332730.67180.00426.67833.33

1440.002286.673413.33216.00512.00

1000.001728.002744.004096.00

1.051.882.934.223.067.56

13.1421.3931.6443.895.10

12.6021.9035.6552.7373.157.15

17.6530.6649.9173.83102.41135.6627.7348.1878.43116.02160.93213.1836.5563.51103.39152.93212.14281.0146.6481.03131.91195.12270.66358.5356.7298.55160.43237.30329.18436.05

2.003.134.508.005.3312.0021.3333.3348.0065.338.0018.0032.0050.0072.0098.0010.6724.0042.6766.6796.00

130.67170.6736.0064.00

100.00144.00196.00256.0048.0085.33

133.33192.00261.33341.3360.00

106.67166.67240.00326.67426.6772.00

128.00200.00288.00392.00512.00

Figure B 5-393.202 BRIDGE CONSTRUCTION MANUAL November 1, 2005

PLYWOOD SHEATHING FOR CONCRETE FORMS

PLYFORM CLASS IFACE GRAIN ACROSS SUPPORTS FACE GRAIN PARALLEL TO SUPPORTS

PLYFORM CLASS IIFACE GRAIN PARALLEL TO SUPPORTSFACE GRAIN ACROSS SUPPORTS

1400

14001400

1400

1200

12001200

1200

1000

10001000

1000

800

800800

800

600

600600

600

400

400400

400

200

200200

200

0

0 0

00

0 0

08

8 8

812

12 12

1216

16 16

1620

20 20

2024

24 24

2432

32

1/2

1/2 1/2

1/25/8

5/8 5/8

5/83/4

3/4 3/4

3/47/8

7/8 7/8

7/81

1 1

11-1/8

1-1/8 1-1/8

1-1/8

Panels continuous across two or morespans

Panels continuous across two or morespans

Panels continuousacross two or morespans

Panels continuousacross two or morespans

Flexure stress of 2000 psi

Flexure stress of 1500 psi Flexure stressof 1500 psi

Flexure stressof 1700 psi

Shear stress of 75 psi

Shear stress of 70 psi Shear stressof 70 psi

Shear stressof 70 psi

Average panel deflection of 1/270thof span

Average panel deflection of 1/270thof span Average panel

deflection of1/270th of span

Average paneldeflection of1/270th of span

Stud or joist spacing,inches. (center to center)




CO

NC

RE

TE

PRE

SSU

RE

, psf

CO

NC

RE

TE

PR

ESS

UR

E, p

sf

CO

NC

RET

E P

RE

SSU

RE

, psf

CO

NC

RE

TE

PRE

SSU

RE

, psf

November 1, 2005 BRIDGE CONSTRUCTION MANUAL Figure C 5-393.202

Filler on top of each joist Sheathing

Sheathing

Stringers

JoistJoist

Posts undereach joist

Walk RunnerWedges

Cleat (preferably on both sides of each post)

Supporting bolts for needle beam

Posts under each joist

Needle beam

TYPICAL SLAB FALSEWORK DETAILS

TYPE 1

TYPE 2

TYPE 3

TYPE 4

Hanger Stringers

Stringers

Joist

Wood overhangbracket withhanger support

Add struts as necessary

Add struts as necessary

Hanger

Borg hangerBolt thru web

Bolt thru web

Adjustable steel posts

Add wedges when struttingof fascia is necessary (both ends)

Steel overhangbracket withhanger support

Steel overhangbracket supportedby bolt thrubeam web

Figure D 5-393.202 BRIDGE CONSTRUCTION MANUAL November 1, 2005

TYPE 7

TYPE 6

TYPE 5

Sheathing

Sheathing

Sheathing

Sheathing

Stringer

Stringer

Sheathing

Adjustable steel postsBolt anchorage

Joist

Joist

Joist

Support for steel postsSteel overhangbracket supportedby bolt inbeam web

Steel overhangbracket withhanger support

Overhang falseworksimilar to type 5 or 6

Hanger

Borg hanger

Wood filleras necesary

Hanger

November 1, 2005 BRIDGE CONSTRUCTION MANUAL Figure E 5-393.202

9"

For light beamsuse a drive fitbrace as shown Adjustment 1

2 x 4

Adjustment 2

See anchor detail

11" for 36 " girder13" for 42" girder15-1/2" for 54" girder

2 x 6 for 36 " girder2 x 8 for 42" girder

EOH

7500

6000

4500

DOH

Precastform

AOH

COH22"

22"

1"

Adjustable. Use max.that beam permits.

60" or 72"

6-1/2" adjustable

1"

ITEM COH Plastic Cone

GENERAL DIMENSIONSANCHOR DETAIL

ITEM BOH 3/4 x 3 Stud TappedReusable

ITEM DOH 1/4 x 2 Cap ScrewReusable

BOH

Beam PrecastingLo

ad -

Lbs.

3000

1500

0 1/2 1 1-1/2Deflection Inches

LOAD DEFLECTION CURVE

4'-0"

18"

Deflection taken asshown above

CAPITAL OVERHANG BRACKETPat. Applied for

CAPITAL ENGINEERING CO.

ITEM EOH 3/4 x 3 BoltITEM AOH Special Anchor

9600 Capacity @ 2000 psi concrete

VerticalDistance

VerticalDistance

30"

18"

Figure F 5-393.202 BRIDGE CONSTRUCTION MANUAL November 1, 2005

DEFLECTION GRAPHS FOR

ADJUSTABLE BRIDGE OVERHANG BRACKET(As produced by Superior Concrete Accessories Inc.)

SECTION A-ADuctileCoil wing nut

54"

SlotA

A

A

14" max.

Coil rod

23"

* LOAD is the total weight of concrete and forms applied on the bracket by the stringers

9000 Failure: Bucklingof Diagonal Member

71006000

4500

3000

1500

7500

* L

oad

(lbs)

00 .25 .50 .75 1.00 1.25

Deflection (inches)

3 Point loadingWF connection

9000

47506000

4500

3000

1500

7500

* Lo

ad (l

bs)

00 .25 .50 .75 1.00 1.25

Deflection (inches)

3 Point loading concrete beam connection

Failure: Bucklingof Diagonal Member

4 Bracket support points against beam

Filler block

23"

Coil rod

A

ASlot

14" max.Ductile coilwing nut

54"

November 1, 2005 BRIDGE CONSTRUCTION MANUAL Figure G 5-393.202

EXA

MPLE

Determ

ine load on 45 coil rod due to joist loads shown above.

Joist "A" = (200 lb) (2.2) = 440

Joist "B" = (500 lb) (1.85) = 925

Joist "C" = (400 lb) (1.75) = 700

Total load on rod 2065 lb(Safe w

orking load of rod 9000 lb)

Loads on the diagonal mem

ber are determined in a sim

ilarm

anner.A

rea of diagonal mem

ber: 0.44 in.. Allow

able load (lb) ondiagonal m

ember 4733 lbs. Com

pare this allowable w

ith actualload that is obtained from

influence chart.

2

NO

TE: Pres-steel hangermustalso

havea

safew

orkingload

of9000lbs.

NO

TE:Foruse

with

Superiorbracketsonly.

USE

OF

INFL

UE

NC

EL

INE

SFO

RO

VER

HA

NG

BR

AC

KE

T

These influence curves indicate the effect a unit joist load, at any point alongthe horizontal m

ember, has on other m

embers of the bracket. Loads are

cumulative depending upon the num

ber of joists that are used. Note that the

influence factor (vertical axis) has two unit designations, one for the vertical

mem

ber and one for the coil rod.

Joistspacing

6"12"

12"

Joistloads400

lbs.500lbs.

200lbs."A

""B

""C

"

Out board end

of bracketJO

IST LOC

ATION

S (Inches)

INFLUENCE FACTOR45 Coil rod and diagonal member: lbs/lb of applied load 5.0

4.0

3.0

2.0

1.0

5448

4236

3024

1812

60

ABC45 Coil Rod

Diagonal Member

Figure H 5-393.202 BRIDGE CONSTRUCTION MANUAL November 1, 2005

SAFE LATERAL (SHEAR) LOADS ON NAILS AND SPIKES Driven into the side grain of seasoned wood. Load applied in any lateral direction.

*SAFE LATERAL LOAD ON EACH, IN LBS

At penetration in diameters noted for each group, into piece holding the point. SIZE

PENNY WEIGHT

LENGTH (Note 1) INCHES

DIAMETER “D”

IN INCHES GROUP I

10 x D GROUP II

11 x D GROUP III

13 x D GROUP IV

14 x D COMMON NAILS (Flat Head, Diamond Point)

Note (1) Length from underside of head to tip of point. 6d 2 0.113 104 84 68 54 8d 2 1/2 0.131 129 104 86 68

10d 3 0.148 154 126 102 82 12d 3 1/4 0.148 154 126 102 82 16d 3 1/2 0.162 176 142 118 93

SPIKES (Countersunk Head Diamond Point) Note (1) Length overall

10d 3 0.192 228 186 151 121 12d 3 1/4 0.192 228 186 151 121 16d 3 1/2 0.207 254 206 168 134

DUPLEX HEAD NAILS (Heavy Double Head, Diamond Point) Note (1) Length from underside of head to tip of point.

6d 1 3/4 0.113 104 84 68 54 8d 2 1/4 0.131 129 104 86 68

10d 2 3/4 0.148 154 126 102 82 16d 3 0.162 176 142 118 93 20d 3 1/2 0.192 228 186 151 121 30d 4 0.207 254 206 168 134

SMOOTH BOX NAILS (Large Flat Head, Diamond Point) Note (1) Length from underside of head to tip of point.

6d 2 0.099 84 68 56 44 7d 2 1/4 0.099 84 68 56 44 8d 2 1/2 0.113 104 84 68 54

10d 3 0.138 136 101 83 67

COOLERS (Flat Head, Diamond Point). SINKERS (Flat Countersunk Head, Diamond Point) as per BOX NAILS except length overall is 1/8” less than shown. When the penetration in nail or spike diameters is less than that shown in the above heading for each timber species group, but at least equal to 1/3 of that shown in the heading, the safe load may be determined by straight line interpolation between zero and the load tabulated above. For example, driven only 7 diameters into Group II timber species, the safe load would be only 7 / 11 of the tabulated load. Driven less than 1/3 that shown in the above heading, the nail or spike should not be considered as capable of carrying any lateral load. For example, for Group II timber species, the minimum penetration should be 11 / 3 or 3 2/3 diameters, at when penetration the safe load would be 1/3 of the tabulated load. End distance, side distance and spacing to be such that unusual splitting of the wood is avoided: pieces to be in close contact. Stagger nails, with fairly uniform spacing, along contact surface. Basic Formulas: Safe Load = 1.33 x K x D 3/2 Group I : K = 2040, Group II : K = 1650 Group III : K = 1350, Group IV : K = 1080 Formulas may be used when nails other than diameters listed are used.

* Based on values given in Nat’l. Design Spec., “Stress Grade Lbr. and It’s Fastenings”, recommended by the Nat’l. Lbr. Mfg. Assoc. latest Edition. The values given in this publication, which are for normal load duration of 10 years have been increased by 1/3 due to short duration of static load on falsework and high factor of safety against failure in the values.

TIMBER SPECIES GROUPS GROUP I Ash, Elm, Maple, Oak GROUP II Douglas Fir, Larch, Southern Pine GROUP III Hemlock, Red Pine GROUP IV Cedar, White & Balsam Fir, White Sugar Ponderosa and Lodgepole Pines, Cottonwood, Spruce, Yellow Poplar

November 1, 2005 BRIDGE CONSTRUCTION MANUAL Figure I 5-393.202

NAILS

The following diagram may be used as an aid in size identification of nails used in formwork.

60d 50d 40d 30d 20d 6d 5d 4d 16d 12d 10d 9d 8d 7d 3d 2d COMMON NAILS - Flat Head Diamond Point STOCK ITEMS Diamond Approx.Length GaugeSize Head No. No. to LbIn.

11/64 2d 1 15 84713/64 3d 1 1/4 14 543

1/4 4d 1 1/2 12 1/2 294 1/4 5d 1 3/4 12 1/2 25417/64 6d 2 11 1/2 16717/64 7d 2 1/4 11 1/2 150 9/32 8d 2 1/2 10 1/4 101

9/32 9d 2 3/4 10 1/4 925/16 10d 3 9 665/16 12d 3 1/4 9 61 11/32 16d 3 1/2 8 47

13/32 20d 4 6 297/16 30d 4 1/2 5 22

15/32 40d 5 4 17 1/2 50d 5 1/2 3 13 17/32 60d 6 2 10

Length from underside of head to tip of point.

Safe loads for nails shall conform to the values listed in Fig. H 5-393.202.


5-393.203 DEFLECTIONS AND ALIGNMENT Deflection will occur in any form or falsework member in which beam action is involved regardless of the design used or the material of which the forms or falsework are constructed. The surface and lines of the concrete being formed will reflect these deflections. Such deflections may detract from the appearance of lines or surfaces which are exposed to view. For this reason, a limit is placed on the amount of anticipated deflection of the form and falsework members which will be used for concrete exposed to view. The anticipated deflection of the members involved should be computed and checked against the allowable deflection described later in this section prior to approving a form or falsework system. The dead loads and concrete pressures used to compute deflections are the same as those used to check stresses in the member except that live load will normally not be included in the deflection loading. When it is anticipated that the allowable deflection will be exceeded, the size or spacing of the members must be modified. In lieu of such modification of the members, in certain cases it is possible to compensate for these deflections during construction of the forms or falsework by use of profile strips or wedging to induce reverse deflection equal in amount to the computed deflection. Certain restrictions are placed on this practice of compensating for deflections which are discussed below. The following criteria will govern acceptance or rejection of the Contractor’s falsework details with regard to deflection of structures that are exposed to view: 1. On concrete surfaces exposed to view the computed

deflection of any member shall not exceed 1/270 of its span or 1/4 inch, whichever is least, unless adequate provision is made to compensate for the deflection as was discussed above. (The 1/270 criteria will be applicable for spans up to 67 inches.)

2. Between fascia beams, the falsework supporting the deck

slab will not be limited by the foregoing. In this area, a limiting cumulative deflection (deflection of sheathing plus deflection of stringers plus deflection of joists, etc.) of ½ inch should be applied. This limit is to avoid excessive addition of dead weight to the superstructure.

3. At locations of transverse construction joints in the

roadway slab, the falsework supporting the bulkhead must be sufficiently strong to reduce the computed bulkhead deflection to not more than 1/16 inch.

4. Deflection of slab overhang falsework must normally be

compensated for by wedging or raising the edge of overhang falsework by an amount equal to that of the anticipated deflection. The anticipated cumulative

deflection of the overhang falsework must not exceed 1/2 inch even though compensated for. When the main overhang falsework support members (the overhang bracket, needle beam or equivalent) are spaced at less than 48 inches, the anticipated deflection of these main members must not exceed S/100, where S = member spacing in inches.

The following criteria will govern acceptance or rejection of the Contractor’s form details with regard to deflection of structures that are exposed to view: 1. Deflection in any form member which acts as a beam

should not exceed 1/270 of its span or 1/8 inch, whichever is least.

2. Concrete diaphragms for prestressed concrete girder spans

will normally not, for this purpose, be considered to be exposed to view and their forms will not be subject to deflection controls.

3. Except when used as pedestrian underpasses, the walls of

box culverts will not be classified as concrete exposed to view. Head walls and wings of box culverts, or the projecting ends of interior walls of multiple opening box culverts, will be classified as concrete exposed to view.

Forms for concrete surfaces that will be exposed to view must be so aligned and of sufficient stiffness that irregularities in any 10 foot length on the finished concrete surface will not exceed ¼ inch. When the Plans indicate that a bridge (or portions thereof) is to be constructed to a horizontal curve, the forms for edges of slab, curbs, copings, medians and railings must be constructed to their proper degree of curvature within a tolerance of 1/8 inch in 10 feet. Midordinates of 1/8 inch or more in 10 feet will occur with a degree of curvature of about 5Eor more. For a degree of curvature of less than 5E, concrete forms may be constructed on short chords along the intended curve line. It is intended that forms which can easily be placed to a scribed line on the falsework or on previously placed concrete, will be placed on the specified curved alignment. This would include forms for the edges of slabs, curbs and medians. Forms for curved railings, however, cannot always be aligned this easily. It is the intent of the Specification tolerance to permit rail construction on chords between railposts when the curvature is small enough (5E or less) so that a chord construction would not be visually objectionable. No offsets should exist at abutting joints of sheathing or at abutting form panels. The variation from plumb or from the specified batter in the lines and surfaces of columns, piers and walls should not exceed ¼ inch per 10 feet of height and, in any event, shall not exceed ½ inch.


5-393.204 FORMULAS AND STANDARD LOADS Practically all falsework members act either as columns or as beams. The internal stresses and the deflections in these members due to the weight of the various construction loads (weight of forms and falsework, weight of concrete, weight of equipment and workcrew, etc.) can be determined by standard methods of Engineering mechanics and compared to the allowable stresses listed in Section 5-393.202 and the allowable deflections listed in Section 5-393.203 to aid in determining the acceptability of a proposed falsework system. Typical examples of the calculations necessary for checking the falsework systems are given in Sections 5-393.206 to 5-393.209. The following formulas will apply to falsework and form analysis:

1. Flexure formula (bending stress) f = MS

2. Direct stress formula

a. Direct tension stress f = PA

b. End bearing f = PA

The allowable side bearing stress given in Section 5-

393.202 applies to bearing surfaces 6 inches or more in length. For bearing surfaces less than 6 inches in length and located 3 inches or more from the end of the timber, the allowable working stress may be increased by a factor of

l

l

+ 3 / 8

where R is the length in bearing. (For circular bearing areas, R= diameter.) Figure A 5-393.204 lists safe side bearing loads for several common form and falsework details as derived from the above formula.

3. Shearing stress formulas

a. Shear stress in steel members v = Vth

b. Shear stress in plywood, known as rolling shear

( )v = V

I / Q b′

NOTE: I/Q is known as the “rolling shear constant” and is tabulated in Section 5-393.202 for the various thicknesses.

c. Horizontal shear stress in timber beams H = 1.5 Vbh′

To simplify the calculation for determining VN, use the appropriate beam formula for determining shear but insert the

term (R-2h) in place of R. This applies to both continuous and simple spans.

h hh

V' based on loads in this area only

Concrete pressure on forms have formulas that are recommended by ACI Committee 347. The formulas apply to standard concrete weighing 150 pcf having a maximum slump of 4 inches and internally vibrated during placement. For forms with rate of concrete placement not exceeding 7 feet per hour, and for all column pours, the maximum lateral pressure at a given point shall be equal to:

a. p = 150 + 9000R

T

1

or b. p = 150h or c. p = 3000 psf Use whichever is least. Note: R1 = rate of concrete placement per hour T = temperature in degrees Fahrenheit The forms with a rate of concrete placement exceeding 7 feet per hour, the maximum lateral pressure at a given point shall be equal to:

a. P = 150 + 43400

T +

2800RT

1

or b. p = 150h or c. p = 2000 psf Use whichever is least. In the event that the above conditions do not apply (for example, if a greater than 4 inch slump is to be used), or if other factors are involved such as the proposed use of a retarder, the actual concrete pressure will increase and appropriate adjustment must be made in the calculated pressures. Vibration and depth of placement in layers should conform to Specifications if these formulas are to be applicable.

Figure A 5-393.204 BRIDGE CONSTRUCTION MANUAL November 1, 2005

NTACT AREAS AND ALLOWABLE STRESSEASE FACTORS FOR STUDS AND WALERS

NTACT AREAS AND ALLOWABLE STRESSEASE FACTORS FOR WALERS AND TIE PLATES

EARING AREA IN SQ. IN. BETWEEN CAPS AND PILES OF VARIOUS SIZES

UDDTH

WIDTH OF ONE WALER1 1/2

4.501.256.501.19

7.501.159.001.13

10.501.11

12.001.09

1 1/2

2

2 1/2

3

3 1/2

4

2

6.001.198.001.19

10.001.15

12.001.13

14.001.11

16.001.09

2 1/2

7.501.15

10.001.15

12.501.15

15.001.13

17.501.11

20.001.09

3

9.001.13

12.001.13

15.001.13

18.001.13

21.001.11

24.001.09

3 1/2

10.501.11

14.001.11

17.501.11

21.001.11

24.501.11

28.001.09

4

12.001.0916.001.09

20.001.09

24.001.09

28.001.0932.001.09

(1)(2)(1)(2)

(1)(2)(1)(2)

(1)(2)(1)(2)

4"5"

4"3"5"

4"4"

6"4"4"

5"5"

4"4"

6"4"4"

1413 1/2

1312 1/2

1211 1/2

1110 1/2

109 1/2

98 1/2

8

L

3/4"3/4"3/4"3/4"3/4"3/4"3/4"3/4"3/4"3/4"1"1"1"1"1"1"1"

Y

9.7512.5010.319.0015.0015.7517.2518.0018.7520.2513.7515.0015.7517.2518.0018.7520.25

CONTACT AREASQ. IN.

1.121.081.101.131.081.071.001.001.001.001.081.081.071.001.001.001.00

81.478.275.172.068.965.762.659.456.252.949.746.443.0

99.795.891.887.883.779.775.671.467.262.958.654.049.3

105.6101.397.092.788.383.979.575.070.465.760.855.850.3

121.9116.7111.4106.0100.695.189.483.677.570.963.656.7

127.0121.4115.7109.9104.198.191.985.578.5

140.4133.6126.6119.4111.9103.995.086.6

144.2136.9129.4121.5113.1

152.7143.1132.7122.7

153.9

ALLOWABLE STRESSINCREASE FACTOR

NOTE: Bearing area at rt. end ofeach line = area of pile at cut-off ofdiameter shown at left. Use whencap width equals or exceeds pile diameter.

3 3/4"3 1/4"3 1/2"3 3/4"3 3/4"3 3/4"3 3/4"3 3/4"3 3/4"3 3/4"3 3/4"

4"4"4"4"4"4"

or more

X

*******

*******

Dia

. of P

ile a

t Cut

-off

in In

ches

6

(Piles assumed circular)ACTUAL WIDTH OF CAP IN INCHES7 1/2 8 9 1/2 10 11 1/2 12 13 1/2 14

CO INCR ST WI

COINCR

3 1/ 3 3/ 5 1/ 5 3/ 6 1/6 3/ 5 1/ 5 3/ 6 1/6 3/ B

NOTE: Allowable stress increase factor = where L is the L length of bearing along grain. Use only if stud is less than 6" and not nearer than 3" from the end of a member

L+3/8

(1) = CONTACT AREA SQ. IN.(2) = ALLOWABLE STRESS INCREASE FACTOR

3/4"

Min

.

Wal

erw

idth

Bearing AreaShown Hatched

Studwidth

Assume waler@ 2" nominal width

Tie plate

L

X1 1/2 1 1/2

X + 3"

Y

Capwidth

Bearingarea shownhatched

W/2

Pilediameter

sin 0 = R

W/2Area = (W) (h)

+ (M) (R)2090( )


When the ends of a beam are notched, as shown below, the horizontal shear stress should be determined by use of theaccompanying formula. As shown by the formula, notching beams adds proportionately to the horizontal shear stress andshould be avoided.

Column deflection formula (elastic shortening)

Beam Formulas - Reactions, Moments, Shears and Deflections.

a. Simple span - uniformly loaded

b. 2 continuous spans - uniformly loaded

c. 3 or more spans - uniformly loaded

d. Cantilever beam - uniformly loaded

V = (L - 2h )1e

W2H = 1.5 V1

b(h )e

hhe

heh

SupportBeam

h

he

A more complete listing of beam formulas for use with point loads or other load variations may be found in the AISC Steel Con-struction Manual.

= PLAE

w

w

w

w

1

1

1

1

2

2

3

1

3

2

2

1

1

1

2

21

1

2

2

2

3 3

3

2

3

L

L

L

L

L

L L

max. R max. M max. V max.

wL wL2 2

wL2

85wL4

wL4

0.0069 wL4

wL4

384 EI

185 EI

EI

8 EI

R =

R = 0.4 wL

R = wL

R =

R = 1.1 wL

M =

M =

M = 0.08 wL2

M =

M = 0.10 wL2

V =

V = 0.4 wL

V = wL

V =

V = 0.6 wL

=

=

=

3wL

5wL

wL2

wL2

wL2

3wL

5wL

8

8

14.2

2

8

8

4

x


The following loads and unit weights will be used in falsework analysis: lumber - 40 pcf steel - 490 pcf plain concrete and reinforced concrete - 150 pcf construction live load* - 50 psf on the upper concrete surface

*This live load is considered a minimum and must be increased where known concentrated loads may produce higher live loads on the member. In addition, when a falsework platform extends outside of the concrete surface to provide working room, as for most pier cap construction, the 50 psf live load should be applied to such walk space as well as to the upper concrete surface. 5-393.205 FALSEWORK DETAILS AND ANALYSIS The inspector must be satisfied that the Contractor’s falsework plan or scheme is in conformance with the Specifications. A common way to do this is to compute the maximum deflection and maximum stresses (bending, bearing, shear, etc) based on plans of the proposed falsework and the assumed loading conditions. These computed stresses and deflections are then compared with the allowable values. If the computed stress and deflection is less than or equal to the allowable value, the member qualifies for use. The examples of falsework investigation given in this section are presented as being representative of commonly used methods. It is not intended that the Contractor’s methods be restricted to the details shown here. However, when it appears that unsafe or improper methods are being used, these details may be suggested as a guide. The American Concrete Institute recommends that three basic simplifications be used in checking forms and falsework. These are as follows: 1. Beams such as joists, studs and walers should generally be assumed to be uniformly loaded with the exception of that when

only one or two point loads occur in a span then the assumption of uniform load should not be used. 2. Beams supported over three or more spans are regarded as continuous and the appropriate continuous beam formulas should

be used. 3. For beams continuous over two spans, design values for simple spans may be safely used except for reaction loads. In the event that the results of the preliminary investigation of the falsework appear marginal, more exact methods should be used. For very large members (falsework pier caps, etc.) more exact methods are recommended. Formulas used in the following computations are found in Section 5-393.204. Allowable stresses are listed in Section 5-393.202 and allowable deflections are listed in Section 5-393.203. Particular attention should be given to writing the unit (inches, feet, pounds, etc.) with each number used in the calculations to assure correct answers. 5-393.206 PIER CAP FALSEWORK EXAMPLE A check of the pier cap falsework details shown in Figure A 5-393.206 would require the following investigation: (NOTE: items above plyform for bottom of pier cap in this figure will be checked in Section 5-393.206.) 1. Plyform for Bottom of Pier Cap a. bending stress b. rolling shear stress c. deflection


A

A

2'-10"

3/4" Plyform, Class I2" x 4" @ 16" O.C.

1/2" dia. coil bolt with 4"x5"washer spaced @ 4'-0"

2 - 2" x 6" (walers)

4' x 8' x 3/4" Plyform, Class I

2" - 6" x 10' Long @ 10" O.C.HP 12x53 x 40' long

Friction Collar

3'-10"

SECTION A-A

PIER ELEVATION

All lumber to be used material (species unknown)

2" x 4" Brace @4' O.C. both sides

3'-8

"

10"

2'-2

"8"

2'-8" dia.

37'-6"

6'-6" 24'-6" 6'-6"

EXAMPLE: PIER CAP FORMS AND FALSEWORK


2. Joists a. bending stress b. horizontal shear stress c. bearing stress d deflection 3. Main Support Beams (HP12x53) a. bending stress b. shear stress c. deflection 4. Friction Collar a. Check against manufacturer’s safe carrying capacity. b. Tighten collar bolts to correct torque. The necessary calculations corresponding to the above example items are as follows: 1. Plyform for bottom of pier cap Determine applied uniform load, w: concrete 3.67 ft x 150 lb/ft3 = 550 lb/ft2

w

10"

2"x6"Joist3/4"

Plyform plywood .06 ft x 40 lb/ft3 = 3 lb/ft2 live load = 50 lb/ft2 Total load w = 603 lb/ft2 From the chart for Class I Plyform in Figure B 5-393.202 for face grain parallel to supports, for a concrete pressure of 603

psf, the maximum allowable stud spacing would be about 11 inches. Therefore, the proposed 10 inch spacing is safe. (Note that the 8 foot panels must be parallel to the joists in this detail and, therefore, the grain of face plies will be parallel to the joist.)

In lieu of using the chart (as when the class of plywood is unknown), the following calculations would be necessary: (Assume 3 span continuous design conditions.)

a. Bending stress f = MS

M = 0.10 wL2 = 0.1 x 603 lb/ft x (10 in.)2 x 1 ft/12 in.

= 503 in. lb S = 0.305 in.3 [from Section 5-393.202]

F = 503 in. lb0.305 in.3

=1649 lb/in.2

This stress is higher than the allowable stress of 1500 psi (which would apply when the class of plywood is unknown).

Therefore, care must be taken in determining type of plyform used. NOTE: the allowable stress of 1700 psi can only be used when it has been determined that a concrete form grade of plyform Class I is being used.

b. Rolling shear stress v = ( )V

I / Q b V = 0.6 wL

= 0.6 x 603 lb/ft x .83 ft = 300 lb


From Section 5-393.202, IQ

= 0.393 in. b = 12 in. v = 300 lb12 in. x 0.393 in.

= 64 lb/in.2

This is acceptable since it is less than the allowable stress of 70 psi.

c. Deflection - ∆ = 0069 w L

EI

4 w = (603 lb/ft) - (50 lb/ft live load) = 553 lb/ft, L = 10 in.

NOTE: live load is not to be included in deflection computations. From Section 5-393.202 E = 1,600,000 lb/in.2 From Section 5-393.202 I = 0.088 in.4 ∆ =

⎛⎝⎜

⎞⎠⎟ =

0 00691 600 000 0 088

10 023

., , .

.x 553 lb / ft x (10 in.)

lb / in. in.x

ft.12 in.

in.4

2 4

This span is less than 67 inches long; therefore, the allowable deflection = 1/270 x 10 in. = 0.037 in. Since actual deflection (0.023 in.) is less than allowable, (0.037 in.) the sheathing is acceptable. 2. Joists Determine applied uniform load due to weight of forms per linear foot along cap: plywood = 16 ft x 1 ft x .062 ft x 40 lb/ft3 = 39.7 lb 39.7 lb

2"x6" joist

Plyform

Width of applied load10"

w

studs = 3.67 ft x 2 x 12 in.16 in.

x 1.5 lb/ft = 8.3 lb 8.3 lb

plates = 4 x 1 ft x 1.5 lb/ft = 6.0 lb 6.0 lb walers = 8 x 1 ft x 2.5 lb/ft = 20.0 lb 20.0 lb . Total 74.0 lb/lf of cap 74.0 lb/lf of cap forms = 74 lb / lf of cap x

10 in.12 in.

⎛⎝⎜

⎞⎠⎟

= 62 lb/joist 55.5 lb/joist

concrete = 2.83 ft x 3.67 ft x 150 lb / ft x

10 in.12 in.

3 ⎛⎝⎜

⎞⎠⎟

= 1298 lb/joist 1168.4 lb/joist

Total 1360 lb/joist 1224 lb/joist (10 in. Spacing) (9 in. Spacing)


This weight is spread over a length of 3.0 feet for all practical purposes; therefore, the corresponding uniform load on the joists is:

form + concrete = 1360 lb / joist3 ft

1224 lb / joist3 ft

= 453.3 lb/lf of joist 408 lb/lf of joist

live load = 50 lbft

x 10 in.12 in.2 50 lb

ft x

9 in.12 in.2

= 41.7 lb/lf of joist 37.5 lb/lf of joist weight of joist = 2.3 lb/lf 2.3 lb/lf Total uniform w = 497 lb/lf = 448 lb/lf (10 in. Spacing) (9 in. Spacing)


2"

3'-0"Uniform load

2" x 6"Joist

L = 3'-2"3'-10"

NOTE: for beams with a very wide bearing area (such as the 12 inches wide beam flange in this example), it is reasonable to assume the span begins about 2 inches back from the edge of the support. For example span length would then be = (3'-10") - 8" = 3' - 2". To simplify calculations, assume the load w is for full length of the 3'-2" span. This will result in only an insignificant stress increase. Maximum bending stress in this example occurs with no load on the cantilevers.

M = ( )wL =

497 lb / ft x 3.17 ft2

8 8

2 = 624 ft lb

S = 7.56 in.3 (For the member sizes used here, all lumber will be S4S.) f =

624 ft lb7.56 in.

x 12 in.

ft990 lb / in.3

2⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ =

1

Allowable bending stress (assuming the proposed form lumber is used material with no visible grade stamp, the allowable stress for Red Pine will be used) = 1375 psi. The member is acceptable with regard to bending stress.

b. Horizontal Shear Stress H = 1.5 Vbh

1 V1 = ( )w (L - 2h)

2 =

497 lb / ft 3.17 ft - 2 x 5.5 in./12 in./ft2

= 560 lb

b = 1 ½ in. h = 5 ½ in. H = 1.5 x 560 lb in. x 5.5 in.15.

= 102 lb/in.2

This is more than the allowable horizontal shear stress of 88 psi for Red Pine and is, therefore, not acceptable. Thus,

reduce the 2" x 6" joist spacing from 10 inches to 9 inches. (9/10) x 102 = 92 psi about 88 psi and, therefore, acceptable. Bending stress will reduce also, so it is also acceptable.

c. Bearing stress in joist on the HP12x53 beam Determine total load of joist

Form Lumber = 74 x 9

12 = 55.5 lb / joist⎛

⎝⎜⎞⎠⎟


Concrete = 2.83 ft x 3.67 ft x 150 lb / ft x 9 in.

12 in. = 1168.4 lb / joist3 ⎛

⎝⎜⎞⎠⎟

Live Load = 50 lb / ft x 8 ft x

9 in.12 in.

= 300.0 lb / joist2 ⎛⎝⎜

⎞⎠⎟

Weight of Joist = 2.3 lb/ft x 10 ft = 23.0 lb Total = 1547 lb/joist

The bearing weight at each support = 1547 lb2

= 774 lb

8'-0"

Live load Dead load + live load

2"x6" joist x 10'-0" long

861 lb 861 lb

Bearing stress f = PA

P = 774 lb A = 1 1/2 in. x 12 in. = 18.0 in.2

f = 77418

= 43 lb/in2

This is less than the allowable side bearing stress of 350 psi (for Red Pine). d. Deflection of joist Assume similar loading condition to that which causes maximum bending stress. w = 448 lb/lf (for 9 inch spacing)

) = 5wL384 EI

4 w = (448 lb/lf) - (37.5 lb/ft live load) = 410 lb/ft

L = 38 in. E = 1,300,000 psi (Red Pine) I = 22.53 in.4 ∆ =

⎛⎝⎜

⎞⎠⎟

5 x 410 lb / ft x (38 in.)384 x 1,300,000 lb / in. x 22.53 in.

x 1 ft.

in. = 0.032 in.

4

2 4 12

The allowable deflection = 1270

x 38 in. = 0.141 in.

Since actual deflection is less than allowable, the member is acceptable. 3. Main Support Beams (HP12x53) Loads will be as determined for bearing stress in 2 c) above except that the live load can reasonably be reduced to 50 psf on only the horizontal concrete surface area for this member. Determine dead load on each joist which bears on the two HP 12x53 beams.

Form Lumber = 55.5 lb/joist

Concrete = 1168.4 lb/joist

Weight of Joist = 23 lb

Total Applied Dead Load = 1247 lb/joist


Convert to uniform load on each 12HPx53 beam.

Dead Load =

1247 lb / joist2

x 12 in.9 in.

= 831 lb / ft

Live Load = 50 lb / ft x 2.83 ft

2 = 71 lb / ft

Weight of Beam w ==

53 lb / ft955 lb / ft

2

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟


Bending stress must be checked at locations Î and Ï. There are no available formulas to determine these moments

directly. Therefore, moments will be determined by combining two known loading conditions in the AISC Steel Construction Manual as follows:

2 1

6'-6"6'-6"

6'-6" 6'-6"

24'-6"

24'-6"

Friction collar supports

HP12x53

w = 955 lb/ft

w = 955 lb/ft w = 955 lb/ft w = 955 lb/ft

This results in the following bending moment diagrams:

2

11

2

2017

4 ft

lb71

655

ft lb

5148

1 ft

lb

( )M =

wL2

= 955 lb / ft x 6.5 ft

= 20,174 ft lb1

2 2

2

( )M =

wL8

= 955 lb / ft x 24.5 ft

8 = 71,655 ft lb2

2 2


For an HP12x53, S = 66.8 in.3 (from AISC Manual). Use moments from summarized diagram at right above.

At location Î, f1 = ( )20,174 ft lb 12 in./1 ft66.8 in.3

= 3624 lb/in.2

At location Ï, f2 = ( )51,481 ft lb 12 in./1 ft66.8 in.3

= 9248 lb/in.2

Assuming that the steel pile material would likely be ASTM A 36 grade, the allowable temporary bending stress is

25,000 psi. Therefore, this member qualified in bending. 955 lb/ft

1 2

21

6'-6" 6'-6"24'-6"

b. Shear Stress in HP12x53 v = Vth

Load Diagram at location Î V1 = 6.5 ft x 955 lb/ft = 6208 lb Shear Diagram at location Ï v2 = 955 lb/ft x = 11700 lb From AISC Manual, for HP12x53, web thickness =0.435 in. height = 11.78 in.

At the point of maximum shear, v = 11700 lb0.435 in. x 11.78 in.

= 2283 lb/in.2

This is less than the allowable temporary shear stress of 15000 psi. c. Deflection of HP12x53 The loading diagram will be as shown above for shear except that live load will not be included for deflection

computations. Therefore, w = (955 lb/ft) - (71 lb/ft live load) = 884 lb/ft. Deflections must be determined at points Î and Ï. Since there are no readily available formulas for determining these

deflections directly, this loading situation may be duplicated by combining two of the available loading diagrams in the AISC Steel Construction Manual as follows:

1

2

1

2

1

2

1

2

w = 884 lb/ft


Deflection at point Ï (midspan) is determined by the following formulas from the AISC manual for the loading diagrams shown above.

( ) ( )∆ 2 2 =

wx24EIL

L L x + Lx 2a L 2a x - wa x12EIL

L x4 2 2 3 2 2 2 22

2 2− − +⎡⎣⎢

⎤⎦⎥

−⎡

⎣⎢

⎤

⎦⎥

w = 884 lb/ft L = 24.5 ft x = ½ x 24.5 ft = 12.25 ft a = 6.5 ft E = 29,000,000 psi I = 393 in.4 (from AISC manual) ∆ 2

224

= 884 lb / ft x 12.25 ft x (24.5 - (2 x 24.5 x 12.25 (24.5 x 12.25 x 6.5 x 24.5 + (2 x 6.5 x 12.25

x 29,000,000 psi x 393 in. x 24.5 ft

4 2 2 3 2 2 2

4

) ) ( )+ −⎡

⎣⎢

3 ))

( )x

12 in.1 ft

- 884 lb / ft x (6.5) ft) x 12.25 ft

x 29,000,000 psi x 393 in. x 24.5 ftx 24.5 - 12.25 ft x

12 in.1 ft

2

42 2 2⎛

⎝⎜⎞⎠⎟

⎤

⎦⎥

⎛

⎝⎜

⎞

⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

3 3

12)

)2 = 0.523 in. - 0.106 in. = 0.416 in. The maximum allowable deflection in this member will be 1/4 inch. See Section 5-393.203 for further details. Since the

allowable deflection at this point is exceeded, the member must either be increased in size or wedges must be placed to compensate for this deflection. (For example, at midspan 0.416 of wedging is necessary)

The deflection at point â may be determined with sufficient accuracy by use of the following loading condition from the

AISC Manual:

w = 884 lb/ft

6'-6"24'-6"

∆ 1

3

24 =

884 lb / ft x 6.5 ft x (4 x 6.5 x 24.5 - 24.5 + 6 x 6.5 x 6.5 - 4 x 6.5 x 6.5 6.5 ft x 29,000,000 psi x 393 in.

x 12 in.1 ft

= - 0.3542 3 2 2 3 3

4

+ ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

)

Since this exceeds the allowable deflection of 1/4", compensation (by wedging or other) must be made in the falsework

construction, in order to obtain true lines.

November 1, 2005 BRIDGE CONSTRUCTION MANUAL 5-393-207 (1)

HP 12x53 in deflected position(without deflection compensation)

0.35

4"

6'-6" 24'-6" 6'-6"C Columns

and friction collarsL

0.41

6"

NOTE: The minus sign indicates an upward deflection of the ends of the HP 12x53 as indicated in the diagram.

4. Friction Collar The reaction on each side of the friction collar may be determined from the shear diagram in part b. F = 6208 lb + 11700 lb = 17908 lb Total load on each friction collar = 17908 lb x 2 beams = 35816 lb NOTE: The collar bolts shall be torqued adequately to develop friction collar load. This load should be checked against the

allowable load listed in the manufacturer’s literature for the proposed friction collars. 5-393.207 ROADWAY SLAB FALSEWORK EXAMPLE Assume the Contractor has proposed the slab falsework details shown in Figure A 5-393.207. Assume also that rails for the strike off machine will be placed on the fascia beams. The following investigations will then be necessary to determine the acceptability of the proposed method:

Interior Bays Slab Overhang Falsework

1. Plywood Sheathing 1. Plywood sheathing

a. Bending stress a. Bending stress

b. Rolling shear stress b. Rolling shear stress

c. Deflection c. Deflection

2. Stringers 2. Stringers


b. Horizontal shear stress b. Horizontal shear stress

c. Bearing stress c. Bearing stress

d. Deflection d. Deflection

3. Joists (double 2" x 12" member) 3. Steel overhang bracket

a. Bending stress a. Safe load

b. Horizontal shear stress b. Deflection

c. Bearing stress on washer 4. Hanger

d. Deflection a. Direct tension on bolt

4. Hanger b. Capacity of hanger

a. Direct tension on bolt

b. Capacity of hanger


A

A

EXAMPLE: ROADWAY SLAB FALSEWORK

CROSS SECTION OF DECK FALSEWORK

SECTION A-AAll lumber to be Douglas Fir, No. 1.

Hangers @ 5'-0" O.C. 8" Min. conc. slab

8" Min. conc. slab

3/4" Plyform, Class I

3/4" Plyform, Class I

1/2" Dia. coil bolt (typ.)

Double 2" x 10" @ 5'-0" O.C.

3'-10" 8'-2" (typical)

2" x 4" @12" O.C.

5'-0" 5'-0"

2" x 10"1/2" Dia. coil boltwith 3" x 4" washer

2" x 4" @ 12" O.C.

2" x 4" @ 1'-0" O.C.

Figure A 5-393.207 INTERIOR BAYS 1. Plywood Sheathing Determine applied load w. Concrete 0.67 ft x 1 ft x 150 lb/ft3 = 100.0 lb/ft Plywood 0.06 ft x 1 ft x 40 lb/ft3 = 2.5 lb/ft Live Load = 50.0 lb/ft w = 152.5 lb/ft a., b., and c. - Bending, Rolling Shear and Deflection are all automatically checked when using Figure B 5-393.202.

Assume face grains will be placed the weak way, (face grain parallel to supports). According to this Figure, 12 inch spacing on ¾ inch plyform Class I will safely support about 550 psf; therefore, the sheathing is acceptable.

Superior overhangcket @ 5'-0" O.C.bra

w

12"2" x 4"Stringers@ 12" O.C.

3/4" Plyform


2. Stringers 12" w = 152.5 lb/ft

2" x 4" Stringers

Determine applied load w per foot of stringer. Concrete, plywood and live load = 152.5 lb/ft Weight of member (2 x 4) = 1.5 lb/ft w = 154.0 lb/ft a. Bending Stress A 2 or 3 span continuous situation will very likely apply for the spans given (5'-0"). Assume 2 span continuous for the

design check. w = 155 lb/ft

5'-0" 5'-0"

2" x 4" Stringer

Waler spacing

M = ( )wL =

154 lb / ft x 5 ft2

8 8

2 = 481.2 ft lb

S = 3.06 in.3

F = MS

b3.06 in.

x 12 in.1 ft

= 1887 psi3= ⎛⎝⎜

⎞⎠⎟

481.2 ft l

Since the allowable bending stress in Douglas Fir is 1875 psi, this member is acceptable in bending with slight

overstress.


1⎛

⎝⎜

⎞

⎠⎟

Assuming the stringers are continuous over two spans, the maximum shear occurs at the center support and is equal to V

= 5/8 wR. To convert this to the applicable horizontal shear, use L = (L-2h) as follows:

V = 5 / 8 w (L - 2h) = 5 / 8 x 154 lb / ft 5 ft - 2 x 3.5 in. x 1 ft12 in.

= 425 lb1 ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜

⎞

⎠⎟

b = 1 ½ in.

h = 3 1/ 2 in. H = 1.5 x 426 lb1.5 in. x 3.5 in.

= 121 psi⎛⎝⎜

⎞⎠⎟

The allowable horizontal shear for Douglas Fir is 120 psi; therefore, the member is acceptable with regard to horizontal

shear with a slight overstress.


c. Bearing Stress - f = P/A For a two span continuous stringer, the maximum P will be at the center reaction point.

P = R2 = 5wL4

1 1/2"1 1/2"

2" x 4" Stringer

1 5/

8"

2 - 2" x 12"Bearingsurface(shaded)

5 x 154 lb / ft x 5 ft4

= 962.5 lb

A = 2 (1 ½ in.) (1 ½ in.) = 4.5 in.2

f = 962.5 lb4.5 in.

= 2 214 psi

The temporary allowable side bearing stress for Douglas Fir is 480 psi; therefore, member is acceptable.

d. Deflection of Stringers ) = wL185 EI

4

Since deflection is to be based on dead load only, the value for w will be: w = 154 lb/ft - 50 lb/ft = 104 lb/ft L = 5 ft E = 1,760,000 lb/in.2 I = 5.36 in.4

∆ =

⎛⎝⎜

⎞⎠⎟

104

185

3

lb / ft x 5 ft x 12 in.1 ft

x 1,800,000 lb / in. x 5.36 in. = 0.063 in.

4

2 4

The surface being formed is not exposed to view and is, therefore, not subject to the normal deflection limitation.

However, this value will be used later to determine the cumulative deflection of the falsework. 3. Joists (double 2 x 10 member) Dead load is applied to this member through eight 2 x 4 stringers. As a general rule, when the concentrated loads are applied

through 3 or more crossing members, the assumption of uniform loading may be used.


7'-4" (Concrete carried by joist)

8"

8'-2" O.C.

Double 2" x 10" (joist)

Determine uniform load on the joist:

Concrete 812

in. x 1 ft in.

x 1 ft x 150 lb / ft x 5 ft = 500.0 lb3⎛⎝⎜

⎞⎠⎟

Plywood 0.06 ft x 1 ft x 40 lb/ft3 x 5 ft = 12.5 lb

Stringers 8 x 1.5 lb / 1f x 5 ft x 17.33

= 8.2 lb⎛⎝⎜

⎞⎠⎟

Double 2 x 10 2 x 3.9 lb/lf = 7.8 lb Live Load 50 lb/ft2 x 5 ft x 1 ft = 250.0 lb Total w = 778.5 lb/lf w = 778.5 lb/ft

7'-4"

2 - 2" x 10"

Assumed design condition

a. Bending Stress - f = MS

M = ( )wL8

= 778.5 lb / ft x 7.333 ft.2 2

8 = 5233 ft lb

for two (2) 2 x 10's, S4S, S = 2 x 21.39 in.3 = 42.78 in.3

f = 5233 ft lb x 12 in.

1 ft4278 in.

= 1468 psi3

⎛⎝⎜

⎞⎠⎟


The allowable bending stress for Douglas Fir is 1875 psi; therefore, the member is acceptable in bending.


1

V = w (L - 2h)2

= 778.5 lb / ft x 7.333 ft - 2 x 9.5 in. x 1 ft12 in.

/ 2 = 2238 lb1 ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

Half of this amount is carried on each 2 x 10 or V1 = 1119 lb per 2 x 10. b = 1 ½ in. h = 9 ½ in.

H = 1.5 x 1119 lb1.5 in. x 9.5 in.

= 118 psi

This is less than the allowable stress of 120 psi and is acceptable. c. Bearing Stress on Washer f = P/A

4"

1 1/2"1 1/2"

3/4"

3"

P = 778.5 lb / ft x 7.33 ft2

= 2853 lb

With a 3" x 4" washer placed as shown, and assuming a 3/4 inch space is used between the 2" x 10" members, A = 3 in. x 3 in. = 9.0 in.2

f = 2853 lb9 in.2

= 317 psi

The allowable side bearing stress for Douglas Fir is 480 psi. A stress increase factor of 1.13 is permitted in accordance

with Figure A 5-393.204 resulting in a total allowable stress of 480 x 1.13 - 542 psi which is considerably more than the actual stress.

d. Deflection ) = 5wL384 EI

4 w = 778.5 lb/ft - 250 lb/ft = 528.5 lb/ft

= 7' - 4" E = 1,760,000 psi For 2-2 x 10's I = 2 x 98.93 in4 = 197.86 in.4

∆ = 5 x 528.5 lb / ft x (7.333 ft) x 12 in.

1 ft384 x 1,800,000 lb / in. x 197.86 in.

= 0.097

4

2 4

⎛⎝⎜

⎞⎠⎟

3


The cumulative deflection of the falsework in the interior bays is limited to about 1/2 inch (see Section 5-393.203). It can be seen that the cumulative deflection of stringers (0.063 in.) plus joists (0.097 in.) will be only 0.160 inches (approximately 3/16 inch) and is, therefore, acceptable.

4. Hanger Rods The load on each hanger rod will be equal to the bearing load on the plate washers, or 2853 pounds. The ½ inch diameter coil bolts for the hangers are manufactured in various strengths such as 6000 pound capacity, 9000

pound capacity, etc.. When required, the Contractor should furnish evidence of the safe capacity of the proposed coil bolts. In addition to checking the coil bolt, the hanger must be checked for rated capacity. Most hangers are rated for the load

carrying capacity of the entire hanger. The load on either side should not exceed one-half of this value. SLAB OVERHANG FALSEWORK 1. Plywood Sheathing Maximum stress in the sheathing will occur adjacent to the beam, at the point where concrete depth is a maximum. 8" 3'-10"

8"

8 3/

4"

9 1/

2"

10 1

/4"

3"8"

2" 1'-0" 1'-0" 1'-0" 1'-0" 4"54" (Top member of superior bracket)

* Stool height is an estimated valuefor computation purposes only.

*

Assume the concrete stool plus flange thickness at the maximum depth will be 3 inches. Where this value is known to be

greater, use the known maximum value. Determine uniform dead load on the sheathing based on this maximum thickness:

Concrete 11 in x 1 ft12 in.

x 1 ft x 150 lb / ft = 137.5 lb / ft3⎛⎝⎜

⎞⎠⎟

Plywood 0.06 ft x 1 ft x 40 lb/ft3 = 2.5 lb/ft Live Load = 50 lb/ft Total w = 190.0 lb/ft


From Figure B 5-393.202, the safe carrying load for Class I plywood placed weak way with supports at 12 inches is 550 psf. Therefore, the sheathing is acceptable.

2. Stringers The second stringer from the right will be the controlling stringer for design. (The stringer on the right carries only about

one-half as much load.) The average slab thickness at this controlling stringer can be determined by calculation or by scaling the drawing. In this case, an average thickness of 10 ¼ inches was scaled. The uniform load on this stringer will be:

1'-0"

2" x 4" Stringer

10 1

/4"

(Ave

.)

Concrete 10.25 in. x 1 ft12 in.

x 1 ft x 150 lb / ft = 128.1 lb / ft3⎛⎝⎜

⎞⎠⎟

Plywood .06 ft x 1 ft x 40 lb/ft3 = 2.5 lb/ft Stringer = 1.5 lb/ft Live Load = 50 lb/ft Total w = 182.1 lb/ft The uniform load on interior stringers was 154 lb/ft. Since stringers on the overhang have the same span length as the stringers on the interior bays, their stresses may quickly be checked by ratios as follows:

a. Bending Stress F = 182.1 lb / ft lb / ft154

x 1887 psi = 2231 psi

This is 19% over the allowable bending stress of 1875 psi for Douglas Fir. Therefore, this member should have its

spacings reduced 20% or down to 10 inch spacing.

Concrete 10.40 in. x 1 ftin.

x 10 in. x 1 ft12 in.

x 150 lb / ft = 108.3 lb / ft3

12⎛

⎝⎜

⎞

⎠⎟

⎛⎝⎜

⎞⎠⎟

Plywood 0.06 ft x 0.833 ft x 40 lb/ft3 = 2.1 lb/ft Stringer = 1.5 lb/ft Live Load 50 lb/ft2 x (0.833 ft) = 41.7 lb/ft Total w = 153.6 lb/ft 10"

2" x 4" Stringer

10 1

/4"

(Ave

.)

Repeat a. and b. steps as follows: The uniform load on interior stringers was 154 lb/lf. Since stringers on the overhang have the same span length as the

stringers on the interior bays, their stresses may quickly be check by ratios as follows:


a. Bearing Stress f = 153.6 lb / ft154 lb / ft

x 1887 psi = 1882 psi

This is about the allowable bending stress of 1875 psi for Douglas Fir; therefore, this member is acceptable in bending.

b. Horizontal Shear Stress H = 153.6 lb / ft154 lb / ft

x 121 psi = 120.7 psi

This is about the same as the allowable stress of 120 psi and is, therefore, acceptable. c. Bearing Stress f = P A Using the Superior bracket as recommended by the manufacturer with a slotted 2" x 6" top bearing surface, the bearing

area is:

END VIEW

PLAN VIEW

2" x 4" Stringer

2" x 6" withcenter slot

Contact surfaces(shaded)

1-1/2" Wide slot

1-5/

8"

5 1/2"

A = 4 in. x 1 1/2 in. = 6.0 in.2

P = 153.6 lb / ft154 lb / ft

x 962.5 lb = 960 lb

f = 960 lb6.0 in.2

= 160 psi

The allowable stress increase factor need not be figured since this stress is much less than the allowable stress of 490 psi.

d. Deflection of Stringers = wL185EI

4

The uniform load is the only factor which differs from the calculation for deflection of the interior stringers. For this

member, w = 153.6 lb/ft - 41.7 lb/ft (live load) = 111.9 lb/ft. Deflection of the overhang stringers can be determined by using a ratio of the uniform loads.

) = 111.9 lb / ft104 lb / ft

x 0.063 in. = 0.068 in.

Since this surface is considered to be exposed to view and the span length is less than 67 inches, the maximum allowable

deflection will be:

L / 270 = 5 ft x 12 in.

1 ft270

= 0.222 in.

⎛⎝⎜

⎞⎠⎟

The actual deflection is less than this; therefore, the member is acceptable.


3. Steel Overhang Brackets Superior brackets may be checked using the influence lines in Figure F 5-393.202. To use this chart, the load on individual

stringers must be determined and the distance from the outboard end of the bracket to each stringer must be determined. A calculation summary of the loads and distances are on the next page.

8"

8.15

"

9.65

"

8.90

"

10.4

0"

11"

B C D E F

SLAB DEPTHS AT STRINGER LOCATIONS

A B C D E F

2"10"

20"30"

40"50"

Spacing of brackets = 5'-0"

Use 8.30' because most of concrete section is to the right of "B".

Stool height is an estimated 3" for computation purposes only.

*

**

*

* *


Plywood load = 34 in.

x 1 ft12 in.

x (10 in.) x 1 ft12 in.

x (40) = 2.1 lbs / linear ft⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

Stringer load = 1.2 lbs/linear ft

Live load = (50 lbs / ft x 10 in.12 in.

= 1.2 lbs / linear ft2 ) ⎛⎝⎜

⎞⎠⎟

Concrete loads:

P = 8.075 in. x 1 ft12 in.

x (2 in.) x 1 ft.12 in.

x 1 in.10 in.

x 150 lbs / ft = 1.7 lbs / linear ftA3⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

P = 8.30 in. x 1 ft

12 in. x (5 in.) x

1 ft12 in.

+ 8.075 in. x 1 ft

12 in. x (2 in.) x

1 ft12 in.

x 9 in.

10 in. x (150 lbs / ft ) = 58.4 lbs / linear ftB

3⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

P = 8.90 in. x 1 ft12 in.

x (10 in.) x 1 ft12 in.

x 150 lbs / ft = 92.7 lbs / linear ftC3⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

P = 9.65 in. x 1 ft12 in.

x (10 in.) x 1 ft12 in.

x 150 lbs / ft = 100.5 lbs / linear ftD3⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

P = 10.4 in. x 1 ft in.

x (9 in. ) x 1 ft12 in.

x 150 lbs / ft = 97.5 lbs / linear ftE3

12⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

P = 10.7 in. x 1 ft12 in.

x 8 in.2

x 1 ft12 in.

x 150 lbs / ft = 44.6 lbs / linear ftF3⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

TOTAL =395.4 lbs/linear ft

LIVE LOADS BRACKET LOADS = 5/4 wL

w = dead load plus live load

PA = 25 lbs/linear ft PA = 188 lbs

PB = 41.7 lbs/linear ft PB = 646 lbs

PC = 41.7 lbs/linear ft PC = 861 lbs

PD = 41.7 lbs/linear ft PD = 909 lbs

PE = 37.5 lbs/linear ft PE = 865 lbs

PF = 16.7 lbs/linear ft PF = 403 lbs

Examples: Bracket load PA = 5/4 x (1.7 + 2.1 + 1.2 + 25) x (5.0') = 188 lbs

PB = 5/4 x (58.4 + 2.1 + 1.2 + 41.7) x (5.0') = 646 lbs


Stringer No. Load On 45ECoil Rod Load on Diagonal Member

A

B

C

D

E

F

188 lb x 2.7 = 508 lb

646 lb x 1.3 = 840 lb

861 lb x 1.95 = 1679 lb

909 lb x 1.7 = 1545 lb

865 lb x 1.58 = 1367 lb

403 lb x 1.45 = 584 lb

Total Load = 6523 lb

188 lb x 2.7 = 508 lb

646 lb x 1.9 = 1227 lb

861 lb x 1.3 -= 1119 lb

909 lb x 0.80 = 727 lb

865 lb x 0.40 = 346 lb

403 lb x 0.10 = 40 lb

Total Load = 3967 lb

Manufacturer’s Allowable Load = 9000 lb

Manufacturer’s Allowable Load = 4733 lb*

* This load is only for overhang brackets on steel beams. Since the applied loads are less than the allowable load, the coil rod and diagonal are acceptable with regard to strength.

However, other bracket components such as the hanger assembly must also be checked for strength requirements as per manufacturer’s allowable loads.

b. Deflection of Overhang Bracket The manufacturer’s literature indicates that the deflection is determined by summarizing the total vertical weight on the

bracket. Only the weight of concrete need be applied since deflection due to dead weight of the falsework may be allowed for prior to concrete placement.

Total weight of concrete = 8 in. + 11 in.2

x 3.33 ft x 5 ft x 150 lb / ft x 1 ft12 in.

= 1977 lb⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

Using Figure F 5-393.202 as a guide, the deflection resulting from a load of 1977 lb would be about 3/16 inch. The

cumulative deflection of the overhang falsework may now be summarized. Deflection of sheathing negligible Deflection of stringers 0.068 in. Deflection of brackets 0.190 in. Seating of wood members (2 x 1/16")* 0.120 in. Total Deflection at center of stringer span = 0.378 in. *Abutting faces of wood members are assumed to crush about 1/16 inch when heavy load is applied. This value will be

less for tightly constructed falsework. In addition, wood fillers against the web as used on prestressed concrete girders must be uniformly fitted and seated to prevent uneven overhang deflections.

The falsework along the edge of coping should, therefore, be set about 3/8 in. above final grade to compensate for the

anticipated deflection.


4. Hanger a. The bolt on this hanger is actually the 45E coil rod which was checked in Item 3. above. Note that the manufacturer

specifies a 9000 lb capacity coil bolt. b. Hangers are normally rated for vertical load carrying capacity. The vertical component on this hanger can be determined

as follows: P = 0.707 x 6523 lb = 4612 lb

6523

45

P

This value should not exceed 1/2 of the safe working load for the total hanger. Preferably, the manufacturer should

furnish information as to the safe load along the 45E angle for the overhang hangers. Note: The safe working loads ascribed to these hangers only applies when the device has full bearing contact on the top flange of the beam and when the hanger bolts are flush with the edge of the beam flange.

5-393.208 SLAB SPAN FALSEWORK Assume the Contractor has proposed the falsework scheme shown in Figure A 5-393.208. In addition, assume they have stated that a strike-off machine weighing 8000 pounds will be used and strike-off rails will be located as shown in the figure (outside berm). The following stress investigation would be necessary.

1. Sheathing 4. Pile Cap


b. Rolling shear stress b. Horizontal shear stress

c. Deflection c. Bearing stress

d. Deflection

2. Joist (2 x 6)

a. Bending stress 5. Pile - total reaction

b. Horizontal shear stress

c. Bearing stress 6. Strike-off machine support system

d. Deflection a. Bending stress


3. Beams (6 x 14) c. Bearing stress

a. Bending stress d. Deflection


c. Bearing stress

d. Deflection


EXAMPLE: FALSEWORK FOR SLAB SPAN

CROSS SECTION OF SLAB FALSEWORK

SECTION A-AAll lumber to be Douglas Fir, Construction Grade

Concrete slab

Concrete slab

3/4" Plyform sheathing

C Supporting rail forstrike-off machineL

2" x 6" @ 1'-0" O.C.

2" x 6" @ 1'-0" O.C.

Drift pins

2" x 12" Bracing

2" x 12" Bracing

6" x 14" Beams@ 5'-0" O.C.

12" x 12" Pile cap

12" x 12" cap

Bolts

Timber pilesbutt dia. 12"

10'-0" 10'-0" 10'-0"

10'-0"10'-0"10'-0"

1'-9

"

Timber piles

6" x 14" @ 5'-0" O.C.

A

A


Calculations are as follows: 1. 3/4 inch Sheathing Determine applied uniform load

Concrete 1.75 ft x 1 ft x 150 lbft3 = 262.5 lb/ft

Plywood 0 .06 ft x 1 ft x 40 lbft3 = 2.5 lb/ft

Live Load = 50.0 lb/ft Total w = 315.0 lb/ft Figure B 5-393.202 indicates that even the lowest grade plyform (Class II) placed in the weak direction will safely support

about 500 psf; therefore, the sheathing is acceptable. 2. Joist (2 x 6) Since these members are spaced at 1'-0", the applied uniform load = 315 lb/ft + 2.3 lb/ft (weight of joist) = 317.3 lb/ft.

Assume two span continuous design with L = 5 feet. a. Bending Stress

f = MS

M = wL8

2 = ( )317.3 lb / ft x 5 ft 2

8 = 991.5 ft/lb

S = 7.56 in.3

f = 991.5 ft / lb 12 in.

1 ft7.56

= 1574 psi

⎛⎝⎜

⎞⎠⎟

Since this is less than the 1875 psi allowable stress, it is acceptable.


1

At the center support, V = 5 / 8 w (L - 2h) = 5 / 8 x 317.3 lb / ft x 5 ft - 2 x 5.5 in. x 1 ft12 in.

= 809.8 lbs1 ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜

⎞

⎠⎟

b = 1 1/2 in. h = 5 1/2 in

H = 1.5 809.8 lb1.5 in. x 5.5 in.

= 147.2 psi

Allowable horizontal shear is 120 psi; therefore, this is not acceptable, so reduce joist spacing to 10 inch.

w = 264.8 V = 5 / 8 (264.8) x 5 ft - 2 x 5.5 in. x 1 ft12 in.

= 675.8 lbs1 ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜

⎞

⎠⎟

H = 1.5 (675.8)1.5 in. x 5.5 in.

= 122.9 psi 120 psi allowable.≈

Also, bending is OK with the reduced spacing.


c. Bearing Stress f = PA

2" x 6" Joist

6" x 14" Beam

PLAN VIEW

Bearingsurface For two span continuous, maximum P is at the

center support.

P = R2 = 5wL4

= 5 x 264.8 lb / ft x 5 ft

4 = 1655 lbs

A = 1 1/2 in. x 5.5 in. = 8.25 in.2 NOTE: The 6 x 14 beams will normally be rough cut and, therefore, will have the full 6 x 14 dimensions and the area will be 1 1/2 in x 6.0 in. = 9.0 in.2.

f = 1655 lb8.25 in.2

= 201 psi

This is much less than the 480 psi allowable and is, therefore, acceptable.

d. Deflection of 2 x 6 joist ) = wL185EI

4

w = 264.8 lb/ft - 41.7 lb/ft (live load) = 223.1 lb/ft L = 5 feet

E = 1,800,000 lb/in.2 ∆ = 223.1 lb / ft x (5 ft) x 12 in.

1 ft x 1,800,000 lb / in. x 20.80 in.

= 0.035 in.

4

2 4

⎛⎝⎜

⎞⎠⎟

3

185

I = 20.80 in.4

The limiting deflection is L270

= 5 ft x 12 in.1 ft

= 0.22 in. (as specified in Section 5- 393.202).⎛⎝⎜

⎞⎠⎟

Since .035 inches is less than the allowable, the member is acceptable) 3. Beams (6 x 14) Assume the Contractor has stated that these beams will be furnished in 22 foot lengths. Two span continuous design will

then apply. Determine applied, uniform load: Live load, concrete, sheathing and joists = 264.8 lb/ft2 x 5 ft = 1324.0 lb/ft

Weight of 6 x 14 member (rough cut) = 6 in. x 14 in. x 1 ft144 in.

x 40 lb / ft = 23.3 lb / ft2

23⎛

⎝⎜

⎞

⎠⎟

w = 1347.3 lb/ft NOTE: It can be assumed that the ends of joists will be staggered so that the critical load determined in 2c. above will not

occur on any one beam.


M = ( )wL8

= 1347.3 lb / ft x 10 ft2 2

8 = 16841.3 ft lb

S = 196.0 in.3 (for rough cut 6 x 14)


f = 16841.3 x 12 in.

1 ft196.0 in.

= 1031 psi3

⎛⎝⎜

⎞⎠⎟

The allowable bending stress is 1875 psi; therefore, this is acceptable.

b. Horizontal shear stress H = 1.5 Vbh

1

( )

For 2 span continuous, V = 5w (L - 2h)

8 =

5 x 1347.3 lb / ft x 10 ft - 2 x 14 in. x 1 ft

12 in.

8 = 6456 lb1

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

b = 6 in. h = 14 in. H = 1.5 x 6456 lb6 in. x 14 in.

= 115 psi

This is less than the allowable stress of 120 psi and is, therefore, acceptable.

c. Side bearing stress f = PA

For 2 span continuous, the maximum P will be over the center support.

P = R2 = 5wL4

= 5 x 1347.3 lb / ft x 10 ft

4 = 16841 lb

A = 6 in. x 14 in. = 84 in.2 f = 16841 lb84 in.2

= 200 psi

Allowable side bearing stress is 480 psi; therefore, this is acceptable.

d. Deflection ) = wL EI

4

185

w = 1347.3 lb/ft - 41.67 lb/ft2 x 5 ft (live load) = 1139.0 lb ft L = 120 in. E = 1,800,000 lb/in.2 I = 1372 in.4 (rough cut)

∆ = 1139.0 lb / ft x (10 ft) x 12 in.

1 ft185 x 1,800,000 lb / in. x 1372 in.

= 0.043 in.

4

2 4

⎛⎝⎜

⎞⎠⎟

3

This is less than the allowable deflection of 1/4 inch for the member but must also be checked later as part of the

cumulative deflection. 4. Pile cap (12 x 12) The reaction of the 6 x 14 beams on the pile cap will be as follows: NOTE: A simple span reaction will be used since the higher reaction R2 determined in Step 3 c) above will occur at random

locations rather than all on one pier cap. This simplification is also in agreement with ACI recommendations. Live load, concrete, sheathing, joist and beam = 1347.3 lb/lf of beam Load on pile cap = 1374.3 lb/ft x 10 ft = 1347.3 lb per beam


Assume the Contractor has stated that pile caps will be furnished in 20 foot lengths. Two span continuous design will apply. The following loading diagram will be typical of each two span segment:

1 2

13473 lb 13473 lb 13473 lb 13473 lb 13473 lbw = 40 lb/ft (due toweight of pile cap)

12" x 12" Pile cap

10'-0" Pile spacing10'-0"

a. Bending Stress in Pile Cap Maximum bending stresses should be checked at points â and ã. To determine the bending moments in the cap, use the applicable load diagrams from the AISC Manual. For this

example, there is no identical loading diagram in the manual, but it is possible to obtain the moments by summarizing diagrams for each of the individual loads as follows:

1

2

1

2

1

2 w = 40 lb/ft13473 lb 13473 lb Note: The load from the 6” x 14” beams directly over the piles are not shown since they do not cause bending in the pile

cap.

Mâ ( ) =

1364

PL - 1 / 2 3

32 PL +

wL14.2

= 1364

x 13473 lb x 10 ft - 3

64 x 13473 lb x 10 ft +

40 lb / ft x 10 ft = 21333 ft lb

2 2

14 2.

Mã( )

= 2x 3

32 PL +

wL8

= 2 x 3

32 x 13473 lb x 10 ft +

40 lb / ft x 10 ft = 25762 ft lb

2 2

8

f = MS

for 12" x 12" rough cut, S = 288 in.3

f = 25762 ft lb x

12 in.1 ft.

288 in. = 1073 psi3

This is less than the allowable stress of 1875 psi and is therefore acceptable.


b. Horizontal shear stress in pile cap H = 1.5 V

bh

1

This stress will be maximum over the center support. Summarize the shear formulas for the three diagrams used to

determine bending moments.

V = 1932

P + 12

3

32 P +

5w (L - 2h)8

= 20.532

x 13473 lb + 5 x 40 lb / ft 10 ft - 2 x 12 in. x

1 ft12 in.

= 8831 lb1 ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

8

H = 1.5 x 8831 lb144 in.2

= 92 psi


c. Bearing stress of pile cap on pile. f = PA

The maximum P will be over the center support. Use the applicable formulas for reactions for the load diagrams used to

determine bending moments.

P = R2 = 2 x 1116

P + (Reaction from beam directly over the pile) + 5wL4

= 2 x 1116

x 13473 lb + 13473 lb + 5 x 40 lb / ft x 10 ft4

= 32498 lb

Assuming 12 inch diameter piles under the 12 x 12 cap, the contact area would be: A = Br2 = 3.14 x 62 = 113.1 in.2 [This value may also be determined by Figure A 5-393.204]

f = 32498113.1 in.2

= 287 psi

This is less than the allowable side bearing stress of 480 psi on the 12 x 12; therefore, it is acceptable. d. Deflection of pile cap The exact deflection of the pile cap cannot be readily determined since a formula to cover this load situation is not

available in the AISC Manual. However, formulas are available to determine an approximate value of the deflection, assuming a simple span loading condition as shown below: (NOTE: This deflection will be slightly greater than the actual deflection of the two span continuous pier cap.)

13473 lb minus live load

10'-0"

40 lb/ft


At â, ) = PL 3EI

+ 5wL384EI

4

48

P = Total reaction minus live load = 13473 lb - (50 lb/ft2 x 5 ft x 10 ft) = 10973 lb L = 10 ft E = 1,800,000 lb/in.2 I = 1728 in.4 (full sawn) w = 40 lb/ft

)Î = 10973

48 384

3 3 lb x (10 ft) x

12 in.1 ft

x 1,800,000 lb / in. x 1728 in. +

5 x 40 lb / ft x (10 ft) x 12 in.1 ft

x 1,800,000 lb / in x 1728 in. = 0.130 in.

3

2 4

4

2 4

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

The maximum cumulative deflection of the joists, beams and pile caps will be as follows:

. Joists 0.035 in. Beams 0.043 in. Pile Cap 0.130 in. (conservative value) Total 0.208 in. It can be concluded that deflections will approach a value of 1/4 inch at points of maximum deflection. Each of the

individual members (joists, beam and pile cap) are within the limiting deflection value of 1/4 inch and the cumulative deflection is also close enough to this value to be acceptable.

5. Pile Load The maximum pile load will be as shown in 4c above. P = 32498 lb = 16.25 tons. This pile load is not an average pile load

but rather is based on the assumption of two span continuous action of the pile caps.

The average load per pile is as follows (assume each pile supports a 10 foot square area above it since piles are spaced at 10 feet in both directions):

Sheathing, concrete and live load 315.0 lb/ft2 x 10 ft x 10 ft = 31500 lb Joists 12 x 10 ft x 2.3 lb/ft = 276 lb Beams 2 x 10 ft x 23.3 lb/ft = 467 lb Pile cap 1 x 10 ft x 40 lb/ft = 400 lb Total = 32643 lb = 16.32 tons

The chart of page 5-393.202 indicates that piles having 12 inch butts may be used for loads of up to 18 tons and that piles having 14 inch butts may be used for loads of up to 21 tons. In consideration of the relative values of the maximum pile load and the average pile load shown above, it would be reasonable to permit the use of piles having 12 inch butts for the falsework in this example. Had the maximum pile load been significantly larger, some falsework revision would have been necessary.


6. Strike-off Machine Support Assume the Contractor (for this example) has provided information regarding the strike-off machine which indicates a total

weight of 8000 pounds. Assume also that the machine wheel base is 5' 0" and that posts for the strike-off rail are spaced at 5'0". The maximum loads on the 6 x 14 beams supporting the strike-off machine can then be determined.

CROSS SECTION OF FALSEWORKNEAR EDGE OF SLAB

Falsework piles

Pile cap 1 2 3

1 2 3Load on beam= 1618 lb/ft

Load on beam= 1415 lb/ft

Load on beam= 202 lb/ft

Conc. slab

2 x 6 Joist

Wt. of strike-off machine 8000 lb

6 x 14 Beams@ 5'-0" O.C.

Beam Î will support the full design loads determined in part 2. of this section. With the edge of slab ending midway

between beams Ï and Ð and, assuming the joists are simple spans, it can be shown that beam Ï will carry about 7/8 of the load carried by beam Î, and beam Ð will carry about 1/8 of the load carried by beam Î (plus the weight of the strike-off machine). This will result in the following loading diagram for beam Ð:

ASSUMED CRITICAL LOAD CONDITION FOR BEAM 3

ELEVATION OF BEAM 3

w = 202 lb/ft

10'-0"10'-0"

2000 lb2000 lb

5'-0"2'-6" 2'-6" Falsework pile

12" x 12" Cap

6" x 14" Beam

Rail support posts

Rail for strike-off machine

Strike-off machine


The position of the strike-off machine shown in the load diagram will result in the maximum bending stress and maximum deflection of the 6 x 14 beam. Note that the rail support posts are placed in locations which will have approximately equal deflections. This is preferable to placing one post over the non-deflecting pier cap and having the remaining posts fall at mid-span where deflection is greatest.

The strike-off machine will not appreciably affect the falsework joists since the rail supports fall directly over the 6 x 14

beams. In addition, the strike-off machine will not cause bending, deflection or horizontal shear in the pile cap, since the supporting beams fall directly over the outside row of piles. Therefore, only the 6 x 14 beam (beam Ð) will be investigated. To simplify calculations, this will be assumed to be a simple span rather than two span continuous. (Use the left half of the load diagram shown above.)

a. Bending stress in 6 x 14 beam f = MS

M = wL8

2+ Pa (formulas from AISC Manual)

( )202 lb / ft x 10 ft 2

8+ 2000 lb x 2.5 ft = 7525 ft/lb

for rough cut 6 x 14, S = 196.0 in.3

f = 7525 ft / lb x

12 in.1 ft

in. = 461 psi3

⎛⎝⎜

⎞⎠⎟

196 0.

This is less than the allowable bending stress of 1875 psi and is, therefore, acceptable.

b. Horizontal shear stress in 6 x 14 beam H = 1.5 Vbh

1

V = 2000 lb +

202 lb / ft x 10 ft - 2 x 14 in. x 1 ft

12 in.

= 2774 lbs1

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

2

H = 1.5 x 2774 lb = 49.5 psi 6 in. x 14 in. This is less than the allowable stress of 120 psi and is, therefore, acceptable.

c. Bearing stress on 6 x 14 beam f = PA

This critical bearing load would occur with the strike-off machine centered over a pile cap. The following load diagram

would apply: 2000 lb 2000 lb

2'-6"10'-0"

2'-6"C pile bents10'-0"

202 lb/ft

L

LOADING CONDITION FOR HORIZONTAL SHEAR IN BEAM 3

R R1 2


Assume two simple spans:

P = R = 2 x PaL

+ 2 x wL22

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

a = 10 ft less 2' 6" = 7' 6" (First portion of formula from AISC Manual)

= 2 x 2000 lb x 7.5 ft

ft + 201 lb / ft x 10 ft

10⎛⎝⎜

⎞⎠⎟

= 5010 lb (Area of 6 x 14 beam on pile cap = 6 in. x 12 in. = 72 in.2)

f = 5010 lb72 in.2

= 69.6 psi

This is less than the allowable side bearing stress of 480 psi and is, therefore, acceptable. d. Deflection of 6 x 14 beam under strike-off machine Assume simple span with loading as for maximum bending stress.

) = 5wL EI

+ Pa

24 EI

4

384(3L2 - 4a2) (formulas from AISC Manual)

w = 202 lb/ft L = 10 ft E = 1,800,000 psi I = 1372 in.4 a = 2.5 ft P = 2000 lb

= 5 x 202 lb / ft x (10 ft) x

12 in.1 ft

x 1,800,000 lb / in. x 1372 in. =

2000 lb x 2.5 ft x 12 in.1 ft

24 x 1,800,000 lb / in. x 1372 in.

4

2 4 2 4

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

3 3

384

x [3 x (10 ft)2 - 4 (2.5 ft)2 ] ) = 0.58 in.

This represents a deflection of approximately 1/16 inch and could be ignored. However a provision should be made for seating of wood members (about 1/16 inch per wood interface) when setting strike-off rail to grade.


5-393.209 NEEDLE BEAM EXAMPLE Assume that a Contractor is submitting slab falsework plans for a bridge which has shallow steel beams. Due to the difficulty of preventing rotation of the fascia beam which would occur with a cantilevered overhang bracket, they have proposed a scheme which includes the needle beam falsework shown in Figure A 5-393.209. Assume the strike-off machine will be run on the fascia beams. The following stressed items must be investigated: 1. Sheathing a. Bending stress b. Rolling shear stress c. Deflection 2. Joist (2 x 4) a. Bending stress b. Horizontal shear stress c. Bearing stress on runner d. Deflection 3. Runner (4 x 4) a. Bending stress b. Horizontal shear stress c. Bearing stress on post d. Deflection 4. Post (2 x 4) end bearing stress-column stress 5. Needle beams a. Bending stress b. Horizontal shear stress c. Bearing stress on plate washer d. Deflection 6. Supporting bolt a. Tension Calculations will be based on the assumed loading condition shown below:

6"Conc. supported oninside runner = 1'-4"

Conc. supported onoutside runner = 2'-0"

11" 9.8"

8"

1'-7" 1'-7"

3" 3'-2" 5" C of 4 x 4 runnersL


SECTION A-A

EXAMPLE: NEEDLE BEAM OVERHANG FALSEWORK

First. Interiorbeam

W 24x1043/4" Dia. bolt4" x 5" Washer

Filler plate

Needle beam

4x4 Runner2" x 4" Post @5'-0" O.C.

Fascia beam

Supporting bolt

4x4 Runner

3/4" PlyformClass II

3'-10"8'-0"

6"

3" 3'-2"

5'-3"

2" x 4" Post @5'-0" O.C.

2" x 6" Plate1" x 4" Braces

Wedges

9"

2" x 4" Joists @ 16" O.C. Sheathing Conc. slab

2" x 4" Posts4" x 4" Runner

Wedges W24x104Bracing(as necessary)

5'-0" 5'-0"

A

A

2 - 2" x 10" @ 5'-0" O.C.

2" x 4" Joist @ 16" O.C.

All material to be Douglas Fir, Construction Grade


1. Sheathing Sheathing is supported on joists spaced at 16 inches. The maximum load on the sheathing will be near the beam flange with

an estimated concrete depth of 11 inches.

Determine the uniform applied load:

Concrete 150 lb / ft x 1 ft x 11 in. x 1 ft

12 in. = 137.5 lb / ft3 ⎛

⎝⎜⎞⎠⎟

Sheathing 40 lb/ft3 x 1 ft x 0.06 ft = 2.5 lb/ft Live Load = 50 lb/ft w = 190 lb/ft Refer to Figure B 5-393.152: For 3/4 inch thick Class II Plyform placed the strong way (which is most likely here) the safe load for 16 inch spacing is

about 240 psf. The sheathing is therefore acceptable. 2. Joists (2 x 4) Check the joists using the average slab thickness of 9.8 inches. Determine uniform applied load:

Concrete 9 8. in. x 1 ft

12 in. x 16 in. x

1 ft12 in.

x 150 lb / ft = 163.3 lb / ft3⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

Live Load 1.33 ft x 50 lb/ft2 = 66.7 lb/ft Sheathing 40 lb/ft3 x 1.33 ft x .06 ft = 3.3 lb/ft Joist = 1.5 lb/ft w = 234.8 lb/ft


Check joists as a simple span.

M = wL8

2 L = 3' 2"

M = ( )234.8 lb / ft x 3.167 ft 2

8 = 294.4 ft/lb

S = 3.06 in.3

f = 294.4 ft / lb x

12 in.1 ft

in. = 1155 psi3

⎛⎝⎜

⎞⎠⎟

306.




H = 1.5 V

bh

1

V = w (L - 2h)

2 =

234.8 lb / ft x 3.167 ft - 2 x 3.5 in. x 1 ft

12 in.2

= 303.3 lbs1

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

b = 1.50 in. h = 3.50 in.

( )H =

1.5 x 303.3 lbs1.5 in. x 3.5 in.

= 86.7 psi


c. Bearing stress on runner f = PA

Determine the reaction on the outer runner by assuming that the outer 2' 0" of slab concrete is supported on this runner as indicated on the previous sketch.

Concrete = 9.9 in. + 8 in. x

1 ft12 in.

x 2 ft x 150 lb / ft = 222.5 lb / f32

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

Sheathing = 5 ft x .06 ft x 40 lb/ft3 = 12.5 lb/ft (for falsework and for edge of slab form) 2 x 4 = 7 lf/ft x 1.5 lb/ft = 10.5 lb/ft (for falsework joist and edge of slab form) Runner (4 x 4) = 3.4 lb/ft Live Load = 50 lb/ft2 x 2 ft = 100 lb/ft w = 348.9 lb/ft Since joists are spaced at 16 inches or 1.333 ft, the reaction per joist is: P = 1.333 ft x 348.9 lb/ft = 465.2 lb A = 1.5 in. x 3.5 in. = 5.25 in.2

f = 464.4 lbs5.25 in.

= 88.6 psi2

This is much less than the allowable stress of 480 psi and is, therefore acceptable.

d. Deflection of joist ∆ = 5wL384 EI

4

Use the same loading criteria for deflection as was used for determining bending stress except that live load is deleted

from the uniform load. w = 234.8 - 66.7 = 168.1 lb/ft L = 3' 2"


E = 1,800,000 lb/in.2 I = 5.36 in.4

( )

∆ = 5 x 168.1 lb / ft 3.167 ft) x

12 in.1 ft

384 x 1,800,000 lb / in. x 5.36 in. = 0.039 in.

4

2 4

⎛⎝⎜

⎞⎠⎟

3

Since this is a surface that is exposed to view, the allowable deflection is:

L270

= 3,167 ft x

12 in.1 ft

270 = 0.141 in.

⎛⎝⎜

⎞⎠⎟

3. Runners (4 x 4) Approximately 3 joists will bear on each runner span; therefore, assume the joists produce a uniform load on the runners.

This uniform load has been determined in part c) above. w = 348.9 lb/ft

a. Bending stress in runners f = MS

Assume the runners will be furnished in lengths of two spans or more. In following the recommended ACI design

simplifications, simple span design will be used. L = 5 ft

( )M =

wL8

= 348.9 lb / ft x 5 ft

= 1090.3 ft lb2 2

8

S = 7.15 in.3

f = 1090.3 ft lb x

12 in.1 ft

7.15 in. = 1830 psi3

⎛⎝⎜

⎞⎠⎟

Since this is less than the allowable stress of 1875 psi, the member is acceptable in bending.

b. Horizontal shear stress in runners H = 1.5 V

bh

1

( )V =

w L - 2h =

348.9 lb / ft x 5 ft - 2 x 3.5 in. x 1 ft

12 in.2

= 770.5 lbs12

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

b = 3.5 in. h = 3.5 in.

H = 1.5 x 770.5 lbs3.5 in. x 3.5 in.

= 94.3 psi



c. Bearing stress on 2 x 4 post f = PA

P = 348.9 lb/ft x 5 ft = 1744 lbs

f = 1744 lb

3.5 in. x 1.5 in. = 332 psi

This is less than the allowable side bearing stress of 480 psi and is, therefore, acceptable d. Deflection of runner

∆ = 5wL384 EI

4

w = 348.9 lb/ft - 100 lb/ft (live load) = 248.9 lb/ft L = 5 ft E = 1,800,000 lb/in.2 I = 12.51 in.4

( )

∆ = 5 x 248.4 lb / ft x 5 ft x

12 in.1 ft

x 1,800,000 lb / in. x 12.51 in. = 0.155 in.2 4

43

384

⎛⎝⎜

⎞⎠⎟

Since this concrete surface will be exposed to view, the allowable deflection is:

L270

= 5 ft x

12 in.1 ft

270 = 0.222 in.

⎛⎝⎜

⎞⎠⎟

The actual deflection is less than the allowable deflection; therefore, the member is acceptable. 4. 2x4 Post The total load and the resulting bearing stress on this post was determined in part c. above (f = 332 psi). By measurement on

the falsework plan, the post height is determined to be about 15 inches. The L/d ratio can then be determined:

Ld

= 15 in.1.5 in.

= 10.0

The allowable column stress will be 1562 psi as determined by the graph in Section 5-393.202. Actual stress (332 psi) less

than allowable stress; therefore, the column is acceptable. 5. Needle Beam Assume each needle beam supports 5 feet of falsework. (Although the runners are continuous members, they are quite

flexible; therefore, simple span reactions can be safely used to determine the applied load on the needle beam.)


The loading diagram for the needle beam is as follows:

First interiorbeam

Fascia beamSupporting bolt

Runner Walkway

Weight of member

Needle beam

8'-6" 2.92'

3.00'

5.25'

6.00'

Determine loads on needle beam: Concrete, live load, sheathing, joists, runner = 348.9 lbs/ft x 5 ft = 1744.5 lb Post = 1.5 lb/ft x 1.3 ft = 2.0 lb 2 x 6 plate = 2.3 lb/ft x 5 ft = 11.5 lb Total reaction at runner = 1758.0 lb 2 x 8 walk plank = 3.0 lb/ft x 5 ft = 15.0 lb Live load on walkway = 5 ft. x 50 lb/ft = 250.0 lb Total reaction at walkway = 265.0 lb Weight of cantilevered needle beam = 3.9 lb/ft x 2 x 6 ft = 46.8 lb

a. Bending stress in needle beam f = MS

Maximum bending moment will be at the supporting bolt. The bending moment is determined as follows:

Reaction x distance = Moment

1758 lb x 2.92 ft = 5133.4 ft lb

265 lb x 5.25 ft = 1391.3 ft lb

46.8 lb x 3.00 ft = 140.4 ft lb

2069.8 lb Total = 6665.1 ft lb

for two 2 x 10's S = 21.39 in.3 x 2 = 42.78 in.3

f = 6665.1 ft lb x

12 in.1 ft = 1869.6 psi

⎛⎝⎜

⎞⎠⎟

42 78.

This is less than the allowable stress of 1875 psi; therefore, the member is acceptable in bending.


b. Horizontal shear stress in needle beam H = 1.5 V

bh

1

The shear in this member can most easily be visualized by drawing a shear diagram. To do this, the reaction (P) at the

first interior beam must be determined.

P = ML

= 6665.1 ft / lb

8.50 ft = 784.1 lb

C First interior beam C Supporting bolt

on fascia beamL

L

L

L

784.

1 lb

Zero shear

784.1 lb

265

lb

C Walkway1758

lb

C Runner

2069.8 lb

SHEAR DIAGRAM

The maximum shear (V1) will be about 2069.8 pounds. (Since there is no significantly large uniform load, the shear is not

noticeably reduced by using the shear at a distance h from the support.) b = 2 x 1.50 in. = 3.00 in. h = 9.25 in.

H = 1.5 x 2069.8 lb

3.00 in. x 9.25 in. = 111.9 psi



c. Bearing stress on plate washer f = PA

The bearing reaction, as determined from the shear diagram above, will be: P = 7841 lb + 2069.8 lb = 2853.9 lb The area of contact for a 4 x 5 washer, as determined from Figure A 5-393.204 is: A = 15.0 in.

f = 2853.9 lb.15.0 in.

= 190.3 psi

This is less than the allowable stress of 480 psi and is, therefore, acceptable. d. Deflection of needle beam The needle beam can be set to plan elevation after the deflection due to weight of the members has occurred. Therefore,

the calculations for deflection must only determine the additional amount of deflection due to the weight of concrete applied through the runner.

The uniform weight of concrete on the outside runner has already been determined to be 222.5 pounds per foot. The

concrete load on each needle beam will be: P = 222.5 lb/ft x 5 ft = 1112.5 lb

1112.5 lb

Supporting bolt

Needle beam

Beam indeflectedposition

8.50' 2.92'

1112.5 lbR2

R1

R1 x 8.50 ft = 1112.5 lb x 2.9167 ft R1 = 381.7 lb R2 = 1112.5 l + 381.7 lb = 1494.2 lb The formula for determining this deflection, which can be found in the AISC Manual, is as follows:

∆ = Pa3EI

(L + a)2

P = 1112.5 lb L = 8.5 ft a = 2.9167 ft for two 2 x 10s, I = 2 x 98.93 in.2 = 197.86 in.4

November 1, 2005 BRIDGE CONSTRUCTION MANUAL 5-393.210

( )

∆ = 112.5 lb x (8.5 ft + 2.9167 ft) x

12 in.1 ft

x 2.9167 ft x 12 in.1 ft

x 1,800,000 lb / in. x 197.86 in. = 0.175 in.2 4

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

22

3

The falsework should be set 0.175 inches high at the outer runner to compensate for deflection of the needle beam. In

addition, five or more wood to wood surfaces should exist in the support falsework, all of which will tend to seat (deflection downward), when the concrete load is applied. A commonly used practice is to set the falsework “high” by 1/16 inch per interface or 5/16 inch in this example. The net height adjustment to the outer runner would then be:

0.175 in. + 0.31 in. = 0.485 in. (above plan elevation) 5-393.210 COLUMN EXAMPLES 1. Wood column 2. Steel column 1. Wood Column A Douglas Fir 6 x 8 S4S member will be used as a falsework column to support a load of 16,000 pounds. The unsupported

length of the column is 14 feet. To determine if this member is acceptable with regard to calculated stresses, the following computations are necessary:

The actual stress in the member = f = PA

= 16000 lb

5.5 in. x 7.25 in. = 401.3 psi

Allowable stress is dependent on the L/d ratio.

L / d = 14 ft

5.5 in. x

12 in.1 ft

= 30.55

From the graph in Section 5-393.202, the allowable stress for a Douglas Fir column with L/d ration of 30.55 is 578 psi.

Since the actual stress (401.3 psi) is less than the allowable stress, the column is acceptable. 2. Steel Column A length of new HP10x42 piling will be used as a falsework column to support a load of 40,000 pounds. The unsupported

length of the column is 16 feet. The following calculations are necessary to determine acceptability of this column.

The actual stress in the member will be f = PA

= Area of HP 10 x 42 = 12.4 in.2

= 40,000 lb

in. = 3226 psi212 4.

(NOTE: Areas of steel members are found in the AISC Manual.) The allowable stress is determined by the appropriate formula from Section 5-393.202.

allowable f = 16980 - 0.53 x kLr

⎛⎝⎜

⎞⎠⎟

2

L = 16 ft x 12 in.1 ft

= 192 in.⎛⎝⎜

⎞⎠⎟


from the AISC Manual, r = 2.41 (use the smallest r value)

allowable f = 16,980 - 0.53 x 192 in.2.41 in.

= 13616 psi⎛⎝⎜

⎞⎠⎟

2

The member will obviously qualify for use with regard to stress. 5-393.211 JOIST AND STRINGER TABLES The following bridge deck falsework Joist and Stringer Tables can be used as an aid to checking slab falsework. These tables show the maximum allowable spans and dead load deflections for joists and stringers for commonly used allowable bending and shear stresses. The applied dead load and live load per square foot and allowable lumber stresses must be known to obtain the allowable span lengths from the tables. See the following example illustrating the use of the tables.

5" 5"8'-8"9'-6"

Double 2" x 10" joists @ 4'-0" O.C.

8 1/

4"m

in.

9" 4 sps. @ 2'-0" (2" x 6" stringers) 9"

FALSEWORK EXAMPLE

1. Stringers Determine applied load per square foot:

Concrete 8.25 in. x 1 ft

12 in. x 150 lb / ft = 103.1 lb / sf3⎛

⎝⎜⎞⎠⎟

Plywood 0.75 in. x 1 ft

12 in. x 40 lb / ft = 2.5 lb / sf3⎛

⎝⎜⎞⎠⎟

2" x 6" Stringer 2.3 lb / ft x 1

2 ft (Spacing) = 1.2 lb / sf

Live Load = 50.0 lb/sf Dead Load + Live Load = 156.8 lb/sf Read the allowable span from the stringer table using the 160 psf load. The maximum allowable span in the column for

lumber having an allowable bending stress of 1875 psi and an allowable shear of 120 psi, 4.22 feet which is greater than the 4' 0" joist spacing and, therefore, the stringer design meets shear and bending requirements. The deflection may be estimated from the table value as 0.017 inches. The stringer bearing stress should be checked as indicated in previous examples in the Bridge Construction Manual.

2. Joists Determine applied load per square foot: Stringer - Dead Load + Live Load = 156.9 lb/square foot


Double 2 x 10 - 2 x 4.3 lb / ft x 1

4 ft (spacing) = 2.2 lb / square foot

Total = 159.1 lb/square foot Read the allowable span from joist table using 160 psf load. The maximum allowable span in the column for lumber, having an allowable bending stress of 1875 psi and an allowable shear of 120 psi, is 8.48 feet which is approximately the 8'8" joist span and therefore the stringer design meets shear and bending requirements. The deflections will be slightly greater than the 0.144 in. shown in the table. The bearing stress in the lumber and the stress in the hanger hardware should be checked as indicated elsewhere in the Bridge Construction Manual. If lumber is used which has allowable stresses differing from those used in the tables, and the tables indicate that the design is questionable, exact formulas which are shown in the Bridge Construction Manual should be used to check the falsework design.

STRINGER TABLE (for two continuous spans)

M = wL

8

2 H =

1.5 Vbh

1 V =

5 w (L - 2h)8

, ∆ = wL

185 EI

4

Allowable Stress 1875 psi -bending

120 psi- shear E = 1,800,000


94 psi - shear E = 1,400,000

Nominal Stringer Size (actual size)

inches

Ctr. to Ctr. Spacing of Stringers

DL+LL lb/ft2

Allowable Span in ft

DL ) in inches


DL ) In inches

2 x 4 (1 ½ x 3 ½ )

12" 12" 12" 12"

150 160 170 180

5.05 4.78 4.54 4.32

0.063 0.056 0.049 0.044

4.09 3.87 3.68 3.51

0.035 0.031 0.027 0.025

2 x 6 (1 ½ x 5 ½ )

12" 12" 12" 12"

150 160 170 180

7.94 7.52 7.13 6.78

0.099 0.088 0.077 0.068

6.43 6.09 5.78 5.51

0.055 0.048 0.043 0.039

2 x 4 (1 ½ x 3 ½ )

16" 16" 16" 16"

150 160 170 180

3.94 3.73 3.55 3.38

0.031 0.028 0.025 0.022

3.22 3.05 2.91 2.78

0.018 0.016 0.014 0.013

2 x 6 (1 ½ x 5 ½ )

16" 16" 16" 16"

150 160 170 180

6.20 5.87 5.58 5.32

0.049 0.043 0.039 0.035

5.05 4.79 4.57 4.36

0.028 0.025 0.022 0.020

2 x 4 (1 ½ x 3 ½ )

24" 24" 24" 24"

150 160 170 180

2.82 2.68 2.56 2.45

0.012 0.011 0.010 0.009

2.34 2.23 2.13 2.05

0.007 0.007 0.006 0.006

2 x 6 (1 ½ x 5 ½ )

24" 24" 24" 24"

150 160 170 180

4.44 4.22 4.02 3.85

0.019 0.017 0.016 0.014

3.67 3.50 3.35 3.21

0.012 0.011 0.010 0.009


JOIST TABLE (for a simple span)

∆ = 5 w L

384 EIDL

4 M =

w L8

2 H =

1.5 Vbh

1 V =

w (L - 2h)2

,


120 psi- shear E = 1,800,000


94 psi - shear E = 1,400,000


inches


DL+LL lb/ft2


DL ) in inches


DL ) In inches

2 x 8 (1 ½ x 7 ¼ )

2'-0" 2'-0" 2'-0" 2'-0"

150 160 170 180

7.01 6.65 6.33 6.04

0.127 0.113 0.101 0.091

5.75 5.47 5.22 4.99

0.074 0.066 0.060 0.055

2 x 10 (1 ½ x 9 ¼ )

2'-0" 2'-0" 2'-0" 2'-0"

150 160 170 180

8.94 8.48 8.07 7.71

0.162 0.144 0.129 0.116

7.34 6.98 6.66 6.37

0.094 0.085 0.077 0.070

Double 2 x 8

(1 ½ x 7 ¼ )

2'-0" 2'-0" 2'-0" 2'-0"

150 160 170 180

10.46 10.13 9.83 9.55

0.315 0.304 0.294 0.284

8.75 8.48 8.22 7.99

0.200 0.192 0.185 0.179

Double 2 x 10

(1 ½ x 9 ¼ )

2'-0" 2'-0" 2'-0" 2'-0"

150 160 170 180

13.35 12.93 12.54 12.19

0.402 0.388 0.375 0.362

11.17 10.81 10.49 10.20

0.253 0.244 0.236 0.229

2 x 8 (1 ½ x 7 ¼ )

3'-0" 3'-0" 3'-0" 3'-0"

150 160 170 180

5.08 4.83 4.62 4.43

0.052 0.047 0.043 0.039

4.24 4.05 3.88 3.73

0.033 0.030 0.028 0.026

2 x 10 (1 ½ x 9 ¼ )

3'-0" 3'-0" 3'-0" 3'-0"

150 160 170 180

6.48 6.17 5.89 5.65

0.067 0.060 0.055 0.050

5.41 5.16 4.95 4.76

0.042 0.038 0.035 0.033

Double 2 x 8

(1 ½ x 7 ¼ )

3'-0" 3'-0" 3'-0" 3'-0"

150 160 170 180

8.54 8.27 8.03 7.65

0.210 0.203 0.197 0.176

7.15 6.89 6.55 6.26

0.132 0.125 0.112 0.101

Double 2 x 10

(1 ½ x 9 ¼ )

3'-0" 3'-0" 3'-0" 3'-0"

150 160 170 180

10.90 10.56 10.24 9.76

0.268 0.259 0.251 0.224

9.12 8.79 8.36 7.98

0.169 0.160 0.143 0.128

2 x 8 (1 ½ x 7 ¼ )

4'-0" 4'-0" 4'-0" 4'-0"

150 160 170 180

4.11 3.93 3.77 3.63

0.030 0.028 0.025 0.024

3.48 3.34 3.21 3.10

0.020 0.018 0.017 0.01

2 x 8 (1 ½ x 9 ¼ )

4'-0" 4'-0"

150 160

5.24 5.01

0.038 0.035

4.44 4.26

0.025 0.024



120 psi- shear E = 1,800,000


94 psi - shear E = 1,400,000


inches


DL+LL lb/ft2


DL ) in inches


DL ) In inches

2 x 10 (1 ½ x 9 ¼ )

4'-0" 4'-0"

170 180

4.81 4.63

0.032 0.030

4.10 3.96

0.022 0.021

Double 2 x 8

(1 ½ x 7 ¼ )

4'-0" 4'-0" 4'-0" 4'-0"

150 160 170 180

7.01 6.65 6.33 6.04

0.127 0.113 0.101 0.091

5.75 5.47 5.22 4.99

0.074 0.066 0.060 0.055

Double 2 x 10

(1 ½ x 9 ¼ )

4'-0" 4'-0" 4'-0" 4'-0"

150 160 170 180

8.94 8.48 8.07 7.71

0.162 0.144 0.129 0.116

7.34 6.98 6.66 6.37

0.094 0.085 0.077 0.070

2 x 8 (1 ½ x 7 ¼ )

5'-0" 5'-0" 5'-0" 5'-0"

150 160 170 180

3.53 3.38 3.26 3.14

0.020 0.019 0.018 0.017

3.03 2.91 2.81 2.72

0.014 0.013 0.013 0.012

2 x 10 (1 ½ x 9 ¼ )

5'-0" 5'-0" 5'-0" 5'-0"

150 160 170 180

4.50 4.32 4.15 4.01

0.026 0.024 0.023 0.021

3.86 3.72 3.59 3.47

0.018 0.017 0.016 0.015

Double 2 x 8

(1 ½ x 7 ¼ )

5'-0" 5'-0" 5'-0" 5'-0"

150 160 170 180

5.85 5.56 5.30 5.08

0.077 0.069 0.062 0.057

4.84 4.62 4.42 4.24

0.046 0.042 0.039 0.035

Double 2 x 10

(1 ½ x 9 ¼ )

5'-0" 5'-0" 5'-0" 5'-0"

150 160 170 180

7.46 7.09 6.77 6.48

0.098 0.088 0.079 0.072

6.18 5.89 5.63 5.41

0.059 0.054 0.049 0.045

2 x 8 (1 ½ x 7 ¼ )

6'-0" 6'-0" 6'-0" 6'-0"

150 160 170 180

3.14 3.02 2.91 2.82

0.015 0.014 0.014 0.013

2.72 2.63 2.54 2.47

0.011 0.011 0.010 0.010

2 x 10 (1 ½ x 9 ¼ )

6'-0" 6'-0" 6'-0" 6'-0"

150 160 170 180

4.01 3.85 3.72 3.60

0.020 0.018 0.017 0.017

3.47 3.35 3.25 3.15

0.014 0.014 0.013 0.013

Double 2 x 8

(1 ½ x 7 ¼)

6'-0" 6'-0" 6'-0" 6'-0"

150 160 170 180

5.08 4.83 4.62 4.43

0.052 0.047 0.043 0.039

4.24 4.05 3.88 3.73

0.033 0.030 0.028 0.026

Double 2 x 10

(1 ½ x 9 ¼)

6'-0" 6'-0" 6'-0" 6'-0"

150 160 170 180

6.48 6.17 5.89 5.65

0.067 0.060 0.055 0.050

5.41 5.16 4.95 4.76

0.042 0.038 0.035 0.033


5-393.212 FORM DETAILS All of the comments in Section 5-393.205 regarding Falsework Details and Analysis apply equally to this Section on Form Details and Analysis. In addition to the formulas used in the falsework examples, the formula for lateral concrete pressure (see Section 5-393.204) will be used here for the examples concerning form plans. The pressure determined by these formulas is based on anticipated concrete temperatures and an anticipated rate of concrete placement. The actual value of these two items must be determined during the operation of concrete placement. Any deviation from the anticipated values used for checking stresses must be evaluated to assure that unsafe (over stressed) conditions will not result. 5-393.213 PIER CAP FORM EXAMPLE Assume the Contractor has proposed that pier cap form plans shown in Figure A 5-393.206. Assume all lumber will be Douglas Fir, No. 1. The members which require stress investigation are as follows: (NOTE: Items defined as falsework are checked in Section 5-393.205.) 1. Sheathing a. Bending stress b. Rolling Shear Stress c. Deflection 2. Studs a. Bending stress b. Horizontal shear stress c. Bearing stress on walers d. Deflection 3. Walers a. Bending stress b. Horizontal shear stress c. Bearing on tie plates d. Deflection 4. Tie Rods a. Tension stress or manufacturers safe load Calculations would be as follows: The lateral concrete pressure is the only load applied on the forms. About 17 cubic yards of concrete are required for the pier cap. Assume the Contractor anticipates placing this concrete in a thirty minute period. The rate of placement would then be: 3.67 feet or 7.33 feet 30 minutes 1 hour The formula for rate of pour exceeding 7 feet/hour would then apply. The three conditions for determining this pressure are as follows: (Assume T = 70ºF) T = 70º and R1 = 7.33 feet/hour 1. P = 150 + 43400 + 2800R1 T T = 150 + 43400 + 2800 x 7.33 70 70 = 1063.2 psf (maximum pressure at any depth) 2. P = 150h = 150 x 3.667 feet = 550 psf (at bottom of cap forms)


3. Neither of the above can exceed 2000 psf. The following sketch illustrates the concrete pressure as determined above.

p = 0

p = 550 psf

p = 1060

h = 3.67 ft(Bottom of pier cap)

h = 0(Top of concrete)

Only the cross-hatched portion of the pressure diagram applies in this example. 1. Sheathing

When a triangular shaped pressure diagram is involved, check sheathing for the maximum pressure. In this case, check the sheathing for a pressure of 550 psf on a stud spacing of 16 inches. The sheathing material is 7/8 inch plyform, Class I. The chart for face grain across (see Figure B 5-393.202) indicates that 7/8 inch plyform with 16 inch stud spacing can safely carry just 550 psf. It must be verified later that the Contractor actually places the plyform the “strong way.”

2. Studs

The studs in this example should be checked as a simple span. In the following sketches, the stud will be shown horizontal to more clearly illustrate its beam action.

550 psf

450 psf

287.5 psf @ mid span

125 psf2 x 4 stud1'-1" 1'-1"

2'-2"8" 10"

C WalersL

a. Bending stress

The pressure at mid span (287.5 psf) may be used as a uniform load for computing bending moments. The results will be slightly more conservative than would result from use of the actual loading.

The pressure of 287.5 psf must be converted to a load per linear foot on the studs with a 16 inch spacing.

w = 287.5 lb/ft2 x 16 in. x 1 ft = 383.3 lb/ft 12 in.


M = wl2 = 383.3 lb/ft x (2.167 ft)2 = 225 ft lb 8 8 S = 3.06 in.3 225 ft/lb 12 in. f = M = 1 ft = 882.3 psi S 3.06 in.3

This is less than the allowable bending stress of 1875 psi for Douglas Fir and is, therefore, acceptable.


This should be checked by assuming that the load at the left support (450 psf) extends uniformly across the simple span. Results will be slightly more conservative than would result from the use of the actual loading.

First convert the load to a uniform load for studs at 16 inch spacing.

w = 450 lb/ft2 x (16 in.) x 1 ft = 600 lb/ft 12 in. 600 lb/ft x 2.167 ft - 2 x 3.5 in. x 1 ft V1 = w (L-2h) = 12 in. = 475 lb 2 2 H = 1.5 V1 = 1.5 x 475 lb = 135.7 psi bh 1.5 in. x 3.5 in. This is more than the allowable stress of 120 psi and is not acceptable. Therefore, reduce the stud spacing to 12", then H = 101.8 psi, which is less than the allowable value of 120 psi.

29.5"14.5"

1009.3

8" 26" 10"

3.67' = 44"

26"

19.3"6.7"

R1 R2

1009.3

c. Bearing stress of studs on walers

The maximum reaction will be at the lower waler. Actual reactions at each waler can be determined as follows: The total weight of the pressure block on each stud is: = (550 lb/ft x 3.67 ft x 1/2) x 12 in. x 1 ft = 1009.3 lb 12 in. 1009.3 lb x 19.3 in. = R1 x 26 in. R1 = 749.2 lb R2 = 1009.3 - 749.2 = 260.1 lb Bearing Stress f = P A P = 749.2 lb A = 4.50 in.2 [from Figure A 5-393.204] f = 749.2 lb = 166.5 psi 4.50 in.2 This is less than the allowable stress of 480 psi and is, therefore, acceptable.


d. Deflection of studs - The load condition used for determining maximum bending stress will be used for determining deflection.

∆ = 5wL4 384EI w = 287.5 lb/ft L = 2.167 ft E = 1,800,000 lb/in.2 I = 5.36 in.4 5 x 287.5 lb/ft x (2.167 ft)4 x 12 in. 3 ∆= ft = 0.015 in. 384 x 1,800,000 lb/in.2 x 5.36 in.4 2.167 ft x 12 in. Allowable deflection = 1 ft = 0.096 in. 270 The studs are, therefore, acceptable with regard to deflection. 3. Walers

The bottom waler will be checked since the higher stud reaction was found to exist at this location. A condition of uniform loading may be assumed to exist since three studs bear on each waler span (between tie rods).

w = 749.2 lb/ft

a. Bending stress in waler f = M S The waler span length is equal to the tie rod spacing (4 feet).

This member will be continuous over two or more spans. In keeping with the recommended simplifications, the assumption of simple spans may be used here.

M = wL2 749.2 lb/ft x (4 ft)2 = 1498.4 ft/lb 8 8 For two 2 x 6s S = 2 x 7.56 in.3 = 15.12 in.3 1498.4 ft/lb x 12 in. f = 1 ft = 1189.2 psi 15.12 in.3 This is less than the allowable bending stress of 1875 psi and, therefore, is acceptable.

b. Horizontal shear stress in walers H = 1.5V1 bh 749.2 lb/ft x 4 ft - 2 x 5 1/2 in. x 1 ft V1 = w(L - 2h) = 12 in. = 1155 lb 2 2


H = 1.5 x 1155 lb = 105.0 psi 2 x 1.5 in. x 5.5 in. This is less than the allowable horizontal shear stress of 120 psi and, therefore, is acceptable.

c. Bearing stress of waler on plate washer f = P A P = 749.2 lb/ft x 4 ft = 2996.8 lb From Figure A 5-393.204, for a 4" x 5" plate washer: A = 15.0 in.2 f = 2996.8 lb = 199.8 psi 15 in.2

This is less than the allowable side bearing stress of 480 psi and is, therefore, acceptable.

d. Deflection of waler Using the assumption of simple span: ∆= 5wL4 384EI w = 749.2 lb/ft L = 4 ft E = 1,800,000 lb/in.2 For two 2 x 6's I = 20.80 in.4 x 2 = 41.60 in.4 5 x 749.2 lb/ft x (4 ft)4 x 12 in. 3 ∆= 1 ft = 0.058 in. 384 x 1,800,000 lb/in.2 x 41.60 in.4

This surface is exposed to view. The allowable deflection of the span will be 1/8 inch since the L/270 value for this span is greater than 1/8 inch.

4 ft x 12 in. (NOTE: L = 1 ft = 0.178 in.) 270 270 Actual deflection is less than allowable deflection; therefore, the member is acceptable. 4. Tie Rods f = P A

As determined in part 3 c. above, P = 2996.8 pounds

The form plan indicates that 1/2 inch diameter coil bolts (and coil ties) will be used as form ties. The manufacturer’s literature must indicate a load capacity of at least 2996.8 pounds for both the coil bolt and the coil tie.


5-393.214 ABUTMENT WALL FORM EXAMPLE A check for the abutment forms shown in Figure A 5-393.214 would require the following investigations. The anticipated rates of concrete placement are indicated on the figure. 1. Sheathing a. Bending stress b. Rolling shear stress c. Deflection 2. Studs a. Bending stress b. Horizontal shear stress c. Bearing stress on walers d. Deflection 3. Walers a. Bending stress b. Horizontal shear stress c. Bearing on tie plate d. Deflection 4. Tie Rods a. Tension stress or manufacturer’s safe loading The stress investigation listed above will be necessary for both the main wall forms and the parapet forms. Calculations for the main wall forms are as follows: First determine the amount of pressure on the forms. The Contractor has indicated a proposed rate of pour of 3 feet per hour in this example. Assuming this concrete will be placed in mid-July, an anticipated temperature of 75º may be used. The three pressure criteria for rates of pour less than 7 feet per hour are as follows: a. p = 150 + 9000R1 T R1 = 3 ft per hour T = 75º = 150 + 9000 x 3 75 = 510 psf (maximum pressure at any depth) b. p = 150h - (150 x 18 feet) = 2700 psf (this will not govern) This formula will only apply to the upper portion of the pressure diagram on a high pour such as this: 150 h = 510 psf. h = 510 psf = 3.4 feet 150 psf c. Neither of the above can exceed 3000 psf. The resulting pressure diagram is shown on Figure A 5-393.214.


150h

3'-4

"

Strongbacksas necessary

7 Sp

aces

@ 2

'-6"

510 psf

Concrete Pressure DiagramEstimated conc. temp. = 75Rate of pour = 3 ft/hr

Concrete Pressure DiagramEstimated conc. temp. = 70Rate of pour = 4 ft/hr

MAIN WALL FORMS

18'-6

"

1/2" Dia. coil bolt,coil ties, and3 x 4 flat washersspaced @ 3'-0"

Double 2 x 6 waler

2 x 6 Studs@ 1'-0" O.C.

3/4" PlyformClass I

2 x 4 Studs@ 1'-0" O.C.

664 psf

150h

4.43

"

PARAPET FORMS

All lumber to be Douglas Fir, Construction grade.

6"4

Spac

es @

1'-1

0"

3/4" PlyformClass I 2 x 4 Studs

@ 1'-0" O.C.

1/2" Coil tieswith 3 x 4washers spaced@ 2'-6"

Double 2 x 6walers

8'-6

"

Example Abutment Forms

6"


1. Sheathing

The sheathing must support a pressures of 510 psf over a stud spacing of one foot. Assume the Class I Plyform will be placed the "strong" way (face grain across supports). From Figure B 5-393.202 for 3/4" Plyform, it can be seen that the safe load is about 760 psf. The sheathing is, therefore, acceptable.

2. Studs

The 2 x 6 studs are spaced at one foot with a uniform load of 510 psf. Span length is 2' 6". Assume these are continuous for more than three spans.

a. Bending stress in studs f = M w = 510 psf

S For 3 span continuous, M = 0.1wL2 = 0.1 x 510 lb/ft (2.5 ft)2 = 318.8 ft lb For one 2 x 6, S = 7.56 in.3 318.8 ft/lb x 12 in. f = 1 ft = 506 psi 7.56 in.3 This is less than the allowable stress of 1875 psi and, therefore is acceptable.

b. Horizontal shear stress in studs H = 1.5V1 bh For 3 span continuous, V1 = 0.6 w (L-2h) = 0.6 x 510 lb/ft x 2.5 ft - 2 x 5.5 in. x 1 ft = 484.5 lb 12 in. H = 1.5 x (484.5 lb) = 88.1 psi 1.5 in. x (5.5 in.) This is less than the allowable stress of 120 psi and is, therefore is acceptable.

c. Bearing stress of stud on waler f = P A For 3 span continuous, P = 1.1wL = 1.1 x 510 lb/ft x (2.5 ft) = 1402.5 lb A = 1.50 in. x 1.50 in. x 2 = 4.5 in.2 f = 1402.5 = 311.7 psi 4.5 This is less than the allowable stress of 480 psi and, therefore is acceptable.

d. Deflection of studs For 3 span continuous, ∆= 0.0069 wL4 EI w = 510 lb/ft L = 2.5 ft E = 1,800,000 lb/in2 I = 20.80 in.4


0.0069 x 510 lb/ft x (2.5 ft)4 x 12 in. 3 ∆= 1 ft = 0.006 in. 1,800,000 lb/in.2 x 20.80 in.4 2.5 ft x 12 in. The allowable deflection is L = 1 ft = 0.111 in. 270 270

Since actual deflection is less than allowable, the studs are acceptable. However, cumulative deflection of sheathing plus studs plus walers must not exceed 1/8 inch to meet the alignment and stiffness criteria of Section 5-393.203.

3. Walers (double 2 x 8 member)

Tie rods are spaced 3' 0". Assume walers will be continuous for three spans or more and use the three span continuous formulas. Since studs are spaced at 12 inches, there are at least 3 studs in each waler span and a condition of uniform load may be assumed on the walers.

Uniform load w = 510 lb/ft2 x 2.5 ft = 1275 lb/ft of waler

a. Bending stress of waler f = M

S M = 0.10 wL2 = 0.1 x 1275 lb/ft x (3.0 ft)2 = 1147.5 ft/lb For two 2 x 8's S = 13.14 in.3 x 2 = 26.28 in.3 1147.5 ft/lb x 12 in. f = 1 ft = 524 psi 26.28 in.3 This is less than the allowable stress of 1875 psi and is, therefore, acceptable.

b. Horizontal shear stress in waler H = 1.5 V1 bh V1 = 0.6 w (L-2h) = 0.6 x 1275 lb/ft x 3.0 ft - 2 x 7.25 in. 1 ft = 1370.6 lb 12 in. H = 1.5 (1370.6 lb) = 94.5 psi 2 (1.5 in.) (7.25 in.) This is less than the allowable horizontal shear stress of 120 psi and is, therefore, acceptable.

c. Bearing on tie plate f = P A P = R2 = 1.1 wL = 1.1 x 1275 lb/ft x (3.0 ft) = 4207.5 lb From Figure A 5-393.204, for a 3 x 4 flat washer with a 3/4 in. Spacer between the waler members, A = 9.0 in.2. f = 4207.5 lb = 467.5 psi 9 in.2 Allowable stress = 480 psi x 1.13 (stress increase factor) = 542.4 psi Since actual stress is less than allowable stress, the member is acceptable.


4. Tie Rods - Tension Stress

The tension load as shown in part 3 c. above is 4207.5 pounds. The form details indicate that 1/2 inch diameter coil bolt and coil tie will be used. The manufacturer’s literature must be checked to determine that these bolts and ties will safely carry the 4207.5 pound load.

Calculations for checking the parapet forms (see Figure A 5-393.214) are as follows:

First determine the amount of pressure on the forms. The form plan indicates a proposed rate of concrete placement of 4 feet per hour. Assuming concrete placement will be late in August, a concrete temperature of 70º may reasonably be used. The three criteria for determining form pressure with rates of pour less than 7 feet per hour are as follows:

a. p = 150 + 9000R1 R1 = 4 ft per hour T = 70º

T = 150 + 9000 x 4 70 = 664 psf (maximum pressure at any depth)

b. p = 150 h = 150 x 8.5 ft = 1275 psf (this will not govern)

Form pressure will be determined by this formula above the point where pressure is 664 psf.

150 h = 664 psf h = 664 = 4.43 ft 150

c. Neither of the above can exceed 3000 psf.

The resulting concrete pressure diagram is shown on Figure A 5-393.214. The actual stress calculations for the parapet forms will be similar to those for the main wall forms and, therefore, will not be repeated in this example. However, it would be necessary to perform these calculations since the concrete pressure and member spacings differ from those of the main wall forms.

Bridge Construction Manual - Falsework and Forms

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