CH. 2 LOADS ON BUILDINGS

CH. 2 LOADS ON BUILDINGS GRAVITY LOADS Dead loads Vertical loads due to weight of building and any permanent equipment Dead loads of structural elements cannot be readily determined b/c weight depends on size which in turn depends on weight to be supported initially weight must be assumed to make a preliminary calculation, then actual weight can be used for checking the calculation Easily calculated from published lists of material weights in reference sources IBC requires floors in office buildings and others with live loads of 80psf or less where partition locations are subject to change be designed to support a minimum partition load of 20psf & is considered part of the live load Example 2.1: find the uniform load on a typical interior beam supporting the floor shown - From Table 2.1 determine weight per square foot of materials comprising the floor. Since concrete is on fluted steel deck, take average thickness of 5 total weight is: Quarry tile Concrete Steel deck Suspended ceiling Total 5.8psf 62.5psf (5 x 12.5psf) 2.5psf 1.0psf 71.8psf - The beam supports a portion fot he floor half the distance of the beam spacing on either side or 8. 8 x 17.8psf = 574plf. - To determine total load multiply by the length of beam - In practice 71.8 would be rounded to nearest whole number so weight in this example is 72psf and the load would be 576plf Live Loads Nomenclatures - L O unreduced floor live load - L floor live load - K LL live load element factor - R allowable reduction of floor live load - r rate of reduction of live load - A area of floor or roof - A T tributary floor area supported by a structural member Loads imposed on a building by use and occupancy and are considered movable or temporary - People, furniture and movable equipment Wind, earthquake or snow loads are not considered live loads but transient loads b/c so variable and are determined by consulting codes Live loads are established by code for different occupancies Roofs: must be designed for uniform live loads or show loads via American Society of Civil Engineers publication 7-02 Code requires floors be designed to support concentrated loads if the specified load on an otherwise unloaded floor would produce stresses greater than those caused by the uniform load

- Concentrated load is assumed to be located on any space 30 square IBC allow for live load to be reduced in most cases but ay not b e reduced for any public assembly occupancy with: - Live load less than or equal to 100psf - For any member supporting one floor of parking garage - With live load exceeding 100psf L O = uniform live load from Table 2.2 K LL = given in Table 2.3 L = L O (.25 + 15/ K LL A T ) - Final value of L must not be: - Less than.5l O for members supporting one floor - Less than.4 L O for members supporting more than one floor - L must not be less than.8 L O for members supporting more than one floor with live load exceeding 100psf or in parking garages IBC allows for alternate method to calculate live load reduction to members supporting more than 150ft 2 - May not be used for public assembly occupancy or live loads exceeding 100psf r = rate of reduction & is equal to.08 A = tributary area R = r(a 150) - Limitations: - Reduction cannot exceed 40% for horizontal members - Reduction cannot exceed 60% for vertical members - Reduction cannot exceed the percentage determined by the formula: R = 23.1(1+ D/L O ) Example 2.2: What live load should be used to design an interior beam that supports 225ft 2 of office space, a live load of 50psf, and a dead load of 72psf? Calculate and compare both code methods - Since the live load is less than 100psf and the office is not a public assembly space ro a parking garage, a reduction is permitted by either method. For each, first determine the reduction and then check against the limitations Since an interior beam, the value of K LL from table 2.3 is 2 L = L O (.25 + 15/ K LL A T ) L = 50[.25 + 15/ (2)(225)] L = 47.86psf The limiting load is:.5l O (.5)(50) = 25psf Value is less than the calculated value use 47.9psf - The alternate reduction yields a similar result: R = r(a 150) R = (.08)(225 150) R = 6%

The limiting reduction is 40% for horizontal members and it cannot exceed the percentage determined by the following formula: R = 23.1(1+ D/L O ) R = (23.1)(1 + 72/50) R = 56.36% Of the three values, 6% is the least so the reduced live load will be: 50 (.06)(50) Reduced live load = 47psf Minimum roof live loads: prescribed by code and act on a horizontal projected area Nomenclature - A t Tributary roof area supported by structural member - R 1 Roof area reduction factor - R 2 Roof slope reduction factor - L r roof live load - F roof slope Roof live load, L r is given by the following: L r = 20R 1 R 2 - R 1 is based on the tributary area A t - If A t does not exceed 200ft 2, R 1 is 1.0 - If A t exceeds 600ft 2, R 1 is.6 - For values of A t between these two limits the following equation is used: R 1 = 1.2 -.001A t - The value of R 2 is based on the roof slope F in inches of rise per foot of run. - If F does not exceed 4 /ft, R 2 is 1.0. - If F exceeds 12 /ft, R 2 is.6. - For values of F between these two limits the following equation is used: R 2 = 1.2 -.05F Example 2.3: What roof live load should be used to design a column that supports 450ft 2 of a roof with a slope of 5 of rise per foot of run? Calculate the reduction for tributary area and slop and then find the design roof live load R 1 = 1.2 -.001A t R 1 = 1.2 -.001(450) R 1 =.75 R 2 = 1.2 -.05F R 2 = 1.2 -.05(5) R 2 =.95 L r = 20R 1 R 2 L r = 20(.75)(.95) L r = 14.25psf Since the calculated roof live load is greater than the minimum value of 12, use 14.3psf b/c value is heavier

Load Combinations When calculating load, all sources will not act at full values at once - Snow load will not be present when full wind load exist b/c wind will blow snow off building IBC recognizes this and requires several combinations of loads be calculated to rind the most critical case Basic load combinations per IBC Section 1605.2.1 using strength design or load and resistance factor design are: LATERAL LOADS Wind Wind loading: pressures, direction and timing are constantly changing For purpose of calculation however wind is considered a static force Factors: - Velocity: pressure on building varies as the square o the velocity p psf =.00256v 2 mph - Height of wind above ground - Surroundings: other buildings, trees and topography - Size, shape and surface texture Positive pressure on windward side Negative pressure (suction) on leeward side & roof Local areas where pressure is greater such as corners, overhangs & parapets Building drift: distance a building moves in wind Tall buildings = several feet Max drift = 1/500 height of building Earthquake Dynamic structural analysis: Tall buildings with complex shapes or unusual conditions a computer model is used to study what forces are developed - Most cases building codes allow a static analysis of the loads Static analysis method: total horizontal shear at base is calculated according to standard formula - Total lateral force is distributed to various floors MISCELLANEOUS LOADS Dynamic Loads Dynamic load: when load is applied suddenly or changes rapidly - Dynamic load is treated as a static load multiplied by an impact factor Impact load: when force is only applied suddenly - Automobiles in parking garage - Elevators in shaft - Helicopter IBC specifies minimum requirements for these type of loads Resonant load: rhythmic application of a force to a structure with same fundamental period as structure

Fundamental period: time it takes structure to complete one full oscillation such as complete swing from side to side in a tall building or up-and-down bounce of a floor - Usually small but slowly build over time as load repeatedly amplifies the motion of the structure - Can effect entire structure as repeated gusts of wind on a tall building - Common problem is vibrating machines where the floor has same period as the machines vibrations - Resilient pads or springs to alleviate - Stiffening floor to change its fundamental period - Tuned dynamic damper: dampen effects of wind sway that is a very heavy mass attached to the sides of a building with springs of the same period as the building and when building oscillates in one direction the spring mounted mass moves in opposite direction counteracting the action of the wind - Can aide in the reduction of costly wind bracing normally required Temperature-Induced Loads Coefficient of expansion: measured in inches per inch per degree Fahrenheit - Low coefficient: wood - High coefficient: plastics If material is retained so it cannot move and is subjected to temperature change, a load is introduced on the material Failing to account for temperature induced loads causes other failures - Tight fitting glass breaking when metal frame contracts - Masonry wall cracking when expansion joints are not provided Soil Loads Retaining wall required to resist lateral pressure of retained material - In addition to any vertical loads near the top of the wall or other lateral loads, retaining walls must resist sliding by at least 1.5 times the lateral force and resist overturning by at least 1.5 times the overturning moment To calculate pressure at the bottom of the wall (p) multiply design lateral soil pressure (q) by the depth of the wall (h) to get pounds per square foot Pressure varies uniformly from zero at top of wall (no earth retained) to maximum at the bottom total horizontal load per linear foot acting on the wall is found by calculating the area of the triangular distribution or the maximum earth pressure at the bottom time the height divided by two P = p(h/2) Example 2.4: What is the total horizontal load exerted on a retaining wall 8 high? Te wall is retaining free drained silty sand. From table 2.4 and active pressure of 45psf may be used for the silty sand The pressure at the bottom of the wall is (8)(45) = 360psf total horizontal load is (360/2)(8) = 1440psf

Water Load developed from water is equal to the unit weight of the fluid in pounds per cubic foot multiplied by its depth Water weighs approx 62lbf/ft 3