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lectures of loads

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    lectures of loads

    [LEFT][LEFT]Loads


    The intent of this module is to provide you with the skills to estimate basic building loads according to the Uniform Building Code. The load information and examples are tailored to address the requirements of our single-story, masonry-wall, wood-roof building.

    This deep module has been subdivided into three secondary modules called - gravity loads, lateral loads, and load combinations - each of which are further divided into topics and even sub-topics. Directions for using each of the secondary modules are provided on the corresponding top pages.

    The organizational structure of this loads module is given below.
    Introduction
    Basic Structural System Behavior
    1. Box system  code designation as a bearing wall system (also known as shearwall systems or dual function walls)
    • Stable
    • Often very symmetrical
    • Commonly used in low-rise construction.
    • The dual function walls transfer both vertical and horizontal loads to the ground.
    2. Vertical loads are transferred to the ground using a post and beam mechanism.
    • The basic vertical load path is given by the following pictures 1  2  3.
    3. Horizontal loads are transferred to the ground through shearwalls.
    • The hard problem in structural engineering is to economically design for lateral loads due to:
    o Wind
    o Earthquakes
    o Blasts
    • Bearing wall structures make use of specially designed and carefully connected horizontal diaphragms and vertical shearwalls to ensure lateral stability, as shown below.
    Gravity Loads

    Dead Loads, D

    1. Dead loads are long-term stationary forces that include the self weight of the structure and the weight of any permanent equipment.
    2. How much do various building materials weigh?
    • For example see Appendices A and B in Design and Wood Structures by D. Breyer.
    • These material weights are given in lb/ft2 (also written as psf).
    • How do you use this information?
    3. Consider the following flat roof:
    A flat roof is any roof with a slope less than:
    a. Compute the dead load carried by a typical 2 X 10 beam and draw its proper load diagram.
    i. Consider the weight of the basic roof.
    • The basic roofing loads are uniformly applied over the entire roof:
    Material psf
    3 ply with gravel 5.5
    5/8" plywood 5/8 (3.0) = 1.9
    4" loose insulation 4 (.5) = 2.0
     9.4 psf

    ii. How much of this uniform load does the typical 2 X 10 carry?
    • It is a function of the member's tributary area.
    • The area assumed to load any given member is called tributary area (t.a.)
    o t.a.= span X tributary width
    o Where tributary width is often equal to the on center, (o.c.) spacing.

    • For a typical 2 X 10, the tributary area = 20 X 8 = 160 ft2
    • For an exterior 2 X 10, tributary area = 20 X 4 = 80 ft2
    • Model the 2 X 10 with basic roof load as:

    with  = uniformly distributed line load = unit psf load X tributary width.
    For this example then,  = 9.4 lb/ft2 (8ft) = 75.2 lb/ft.

    iii. But is this the only dead load carried by this beam?
    • Need to also consider the 2X10's self weight and the weight of the 2 X 4 members framing into the beam.
    • There are a number of ways to calculate a member's self weight, but all methods:
    o Are a function of wood species, moisture *******, and dimensions.
    o And, you must be careful to properly track units.
    • For this example problem, self-weight can be found as:
    o Assume that the 2X10's are dry Douglas Fir-Larch, (DF-L), members with a specific weight of  35 lb/ft3.
    o 2" nominal  1.5" actual.
    o 10" nominal  9.25" actual.

    • Calculate the dead weight of 2 X 4 at 2' o.c. spanning into the supporting 2X10 members.
    o Assume DF-L again
    o One 2 X 4 member weighs:
    • 4" nominal  3.5" actual.

    o Each 2X4 spans 8' with a 2X10 supporting each end of the 2X4:
     1/2 its weight is supported by each 2X10 or
     1/2 (1.28 lb/ft) (8ft) = 5.12 lbs.
    o Each typical 2X10 supports 9-2X4's per side:
     the total load on the 2 X 10 is 2 (9) (5.12 lb) = 92.16 lbs.
     distribute this over 20' span yielding 92.1 lbs / 20 ft = 4.61 lb/ft.
    But there is an easier way to get the 2X4 dead weight effect on a typical 2X10 member:

    i. Therefore the total dead weight
    supported by a typical 2X10 is the summation of the basic roof (and ceiling, if directly attached to underside of roof members), self-weight, and self-weight of any supporting members.

    Gravity Loads
    Live Loads, L or Lr

    • Live Loads, L:
    o Short duration forces which change in location and magnitude.
    o Include people and furniture.
    o Based upon intended use of building = occupancy.
    o Tabulated in UBC Table 16-A:
    • floor uniform loads are a function of use/occupancy.
    • for examples, the uniform floor load for:
    o residential = 40 psf
    o offices = 50 psf
    o light storage = 125 psf
    • Roof Live Loads, Lr:
    o UBC recognizes that roofs carry lower loads than floors, since roofs are generally not occupied.
    o UBC specified roof loads account for miscellaneous loads like roofing, equipment, servicing.
    o Obtain minimum unit roof loads from 1994 or 1997 UBC Table 16-C:
    • These live loads are assumed to act vertically over the area projected onto a horizontal plane.
    • They are a function of roof slope and tributary load area with smaller unit loads for steeper members and large tributary areas:
    o Flat roofs: higher probability that high unit live loads could occur.
    o Tributary area of member under consideration:
    • "If a member has a small t.a., it is likely that a fairly high unit live load could be imposed over that entire small surface area... It is less likely that a large t.a. will be uniformly loaded by the same high unit load considered for a small t.a. member." (Design of Wood Structures by D. Breyer)
    • Procedure: (see Breyer, Example 2.3)
    1. Calculate roof slope (generally the same as the member slope)
    2. Calculate t.a. of member.
    3. Choose Method 1 or Method 2 to obtain unit load.
    o Method 1 is straightforward, but incremental.
    o Method 2 provides a continuous range of loads, but requires the checking of two equations which can be found in '94 UBC 1606 or '97 UBC 1607.5.
    4. Apply the unit load to the horizontal plane. See upcoming example under snow loads for an explanation of horizontal plane.
    Snow Loads, S
    Introduction

    • Snow loads are often established by the local building official as they can vary greatly over relatively small geographic areas.
    • Appendix, Chapter 16 of the UBC covers snow loads in much more detail than '94 UBC 1605.4 or '97 UBC 1614 sections.
    o This appendix, which is not addressed here, provides detailed information to calculate:
    • Roof snow load as a function of ground snow load, building exposure, and importance.
    • Unbalanced snow loads.
    • Drift potential.
    • Snow loads, full uniform or unbalanced, should be considered in place of roof live loads when their effect results in larger members:
    o Because various loads may act on the structure (in this case, the roof) simultaneously, you need to consider combinations of loads which are discussed in further detail in the Load Combinations sub-module.
    o One of the five combinations of concern here, however, is:
    • D + L + (Lr or S) ('94 UBC 1603.6 or '97 UBC 1612.3.1).
    • For roofs with S > Lr, this load combination simply becomes D + S.
    • Design snow load = f (roof slope):
    o If basic roof snow load is greater than 20 psf and if roof slope  20, then Ra = S/40 - 1/2, where Ra = reduction in S in psf per degree > 20.
    • Flagstaff's basic S = 35 psf.
    • Snow load is given along the horizontal plane.
    • Other facts:
    o 1" newly fallen snow  .5 psf.
    o 1" packed snow  1 psf.

    Snow Loads, S
    Example
    Example Problem:
    This problem has been adapted and modified from Structural Design in Wood by J. Stalnaker and E. Harris. It demonstrates the calculation of roof live load, snow load, dead load, summation of loads, and corresponding load, shear, and moment diagrams.

    Problem Statement:
    Determine the load, V and M diagrams for a typical 6 X 12 rafter using both the sloping roof method and the horizontal plane method. The basic roof S is given as 35 psf.
    Solution:
    1. Roof dead load determination: The following roof dead loads were assumed.
    Roofing 2.5 psf
    Decking 6.6 psf
    Insulation 2.0 psf
    Rafter weight 15.4 lbs/ft (1/10) = 1.5 psf
     12.6 psf =D applied along the slope.
    2. Roof live load for typical rafter:
    o Determine tributary width on the horizontal plane: t.a. = 24(10) = 240 ft2
    o Roof slope: tan  = 1/2,  = 26.5 or 1/2 = 6/12  6:12
    o Therefore according to UBC Table 16-C, basic Lr = 14 psf.

    3. Snow load:
    o Since  > 20 and S > 20 psf, can reduce basic load as follows:
    • Rs = 35/40 - 1/2 = .375 psf/degree > 20
    • S = 35 - .375(26.5 - 20) = 32.6 psf

    4. Load combinations:
    o See '94 UBC 1603.6, or case 1 of the '97 UBC 1612.3.2. This equation reduces to D + S since S > Lr

    5. Summing loads wrt rafter:

    a. To add these loads together, they must be consistent - either along the slope or along the horizontal plane. Refer to adding loads for additional information.
    b. The resulting addition of loads with respect to the two different planes are given in the following two load diagrams.

    6. Once the loading diagrams have been constructed, it is possible to determine the V and M diagrams for both methods.
    . Find support reactions:

    a. Calculate and construct the V and M diagrams. Compare the results.

    b. The horizontal plane method, which is the one most commonly used, is an approximation of the more accurate sloping beam model. It yields conservative shear values. For the bending moment analysis, however, the max -M and +M, are almost the same as that found from the sloping beam method. Not obvious in the horizontal plane method, however, is the thrust force acting on the exterior walls due to the component of the load parallel to the roof. Care must be taken to not forget this thrust effect on the vertical support walls.
    Zone Coefficient
    1 V = .067W
    2a V = .122W
    2b V = .156W
    3 V = .200W
    4 V = .244W

    Wind Loads
    Wind Pressure

    The design of the structure's primary lateral force system, as well as its elements and components, shall be based upon wind pressures determined from the following equation:
    P = Ce Cq qs I
    where
    P = design wind pressure in psf.

    a. Qs = wind stagnation pressure in psf at a height of 33' above the ground.
    • This wind force is a function of wind velocity:
    o Qs = .00256 V2
    Wind Stagnation Pressure
    Wind stagnation pressure is based upon Bernoulli's equation and its a function of the wind velocity squared and the mass density of air. The UBC equation assumes a standard air density of 0.0765 pcf and does not account for density changes as the result of altitude, season of the year, weather, and latitude.

    where V = windspeed in mph found from UBC Figure 16-1.
    • The applicable wind speed map can be found as Figure 2-1 in the article Nature of Wind
    o Can also find Qs using UBC Table 16-F.
    o The minimum design speed is taken as 70 mph.
    • Refer to Wind Speeds for information relating windspeeds in mph to observations.
    o V is based upon the 50-year mean recurrence interval for the fastest-mile wind speeds.
    50-Year Mean Recurrence Interval
    This means that the probability of experiencing a basic wind speed faster than the indicated value in any one-year period is one in fifty of 2%.
    Fastest-Mile Wind Speeds

    In the U.S. wind speeds are measured by recording the time a constant number of turns of the anemometer occur equivalent to one mile of wind passing by.
    Other countries use different windspeed measuring methods. This results in different reported values for similar intensity winds.
    b. Iw = Importance factor taken from UBC Table 16 - K.
    • Essential facilities and hazardous facilities are designed to withstand higher forces by setting Iw = 1.15, which is approximately equivalent to a 100 year recurrence interval.
    • Otherwise, Iw = 1.0

    c. CE = combined height, exposure, gust coefficient found in UBC Table 16 - G.
    • Wind pressure increases with building height:
    o For the windward wall, UBC uses a stepped pressure diagram as f(height).
    o For the leeward wall, pressure taken as constant over the full wall height and is calculated using the mean roof height for CE determination.

    • Turbulence (gusting) = f(terrain), is quantified in terms of an exposure rating:
    GustingThis component accounts for the effects of air turbulence and dynamic building behavior. The underlying gust response assumptions are only appropriate for structures that have stiffness, mass, and damping characteristics similar to ordinary buildings with a period of vibration less than one second. More flexible structures may experience wind effects that are substantially greater than that predicted by UBC; particularly with respect to the across-wind response.

    Exposure Effects
    Ground surface irregularities and flow obstructions such as trees and other buildings reduce wind forces. The UBC broadly categorizes these terrain influences through the use of exposure ratings.

    Although a building site may have different exposure in different directions, the most severe exposure governs for all wind load calculations.
    o B - urban, suburban, closely spaced obstruction the size of single - family residences.
    o C - open country and grasslands that are generally flat.
    o D - unobstructed flat terrain facing a large body of water extending inland 1/4 mile or ten times building height, whichever is greater.
    o UBC recognizes only 3 of the 4 possible exposure classes.
    • It has not adopted Exposure A - building sites in large city centers.

    d. Cq = Pressure coefficient.
    • Found in UBC table 16 - H.
    • Cq = f (structure or part of, analysis method, openings)
    • Structure or part of:
    o Cq for primary LFRS.
    LFRS
    The lateral force resisting system (LFRS) is that part of the structural system designed to resist lateral loads. For bearing wall type building considered here, the horizontal diagrams and the shear walls make up the LFRS.
    • Considering the whole structure as it resists lateral forces.
    • Applied to horizontal diaphragms and shearwalls.
    o Cq for elements and components:
    • Accounts for higher pressures that occur locally.
    • Taken at locations away from discontinuities, and
    • At discontinuities.
    • Primary frame analysis methods:
    o Method 1: normal force method


    • A more accurate description of wind forces.
    • Windward and leeward forces acting normal to all exterior surfaces simultaneously.
    o Method 2: the projected area method.
    • Carry over from earlier codes.
    • Simpler than method 1.
    • Generally more conservative.
    • Cannot be used for gabled rigid frames or structures greater than 200' in height.
    • Simultaneous application of horizontal pressures on the vertical projected area and vertical pressure on the horizontal projected area.
    • Openings:
    o Open (known as partially enclosed in newer codes like the '94 and '97 UBC's) structures have higher outward pressures than enclosed or unenclosed structures.
    o Doors and windows are considered openings unless protected.
    o Account for the effect of openings in partially enclosed buildings by using a bigger Cq coefficient according to footnote 1 Table 16 - H.
    Footnote 1, Table 16-H
    1For one story or the top story of multistory partially enclosed structures, an additional value of 0.5 shall be added to the outward Cq. The most critical combination shall be used for design. For definition of open structures, see section 1613.
    • This effect could be simply thought of as what happens to the surface of a balloon as it is being blown up.
    • For primary LFRS, method 1: increase Cq for roofs according to footnote 1 of UBC 16-H.
    o For elements and components not at discontinuity: select Cq according to the element under consideration, and if building is unenclosed, enclosed, or partially enclosed.
    o Definitions, according to '97 UBC 1616, for enclosed, partially enclosed, and unenclosed:
    • Aoi = area of opening on projected side i.
    • Ai = total projected wall area of side i.
    • Aoj = area of opening on projected side j.
    • If Aoi / AI  0.85 for all sides, then structure is unenclosed.
    • If Aoi / AI > 0.15 for side i and if
     Aoj < .5 Aoi for all other sides j, then structure is partially enclosed.
    • All other cases, consider the building as enclosed.



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