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N-W.F.P. University of Engineering and Technology Peshawar Lecture 03: Design Loads By: Prof Dr. Akhtar Naeem Khan [email protected] 1 Topics to be Addressed Types of loads Wind Load Earthquake Load Load Combinations CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 2 Feeling Responsibility CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 3 Types of Loads Determination of loads for which a given structure may be designed for is a difficult problem. Questions to be Answered: • What loads may structure be called upon during its lifetime? • In what combinations these loads occur? • The probability that a specific live load be exceeded at some time during lifetime of structure? CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 4 Types of Loads Design load should be rational such that considering 150mph wind load for a tower is reasonable but not the load of a tank on top of the tower. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 5 Types of Loads Three broad categories: 1. Dead load 2. Live load 3. Environmental load CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 6 Types of Loads 1. Dead load Dead Loads consist of the weight of all materials and fixed equipment incorporated into the building or other structure. (UBC Section 1602) Weight of structure Weight of permanent machinery etc. Dead loads can be reasonably estimated if the member dimensions and material densities are known. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 7 Types of Loads 2. Live load: Live loads are those loads produced by the use and occupancy of the building or other structure and do not include dead load, construction load, or environmental loads. Weight of people, furniture, machinery, goods in building. Weight of traffic on bridge CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 8 Types of Loads 2. Live load: • Buildings serve such diverse purposes that it is extremely difficult to estimate suitable design loads. • Different building codes specify live load requirements. • Uniform Building Code (UBC) • Southern Standard Building Code • BOCA National Building Code CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 9 Types of Loads 2. Live load: (UBC Table 16-A) Live loads for various occupancies Occupancy Live load,psf Residential 40 Libraries(reading room) 60 Mercantile 75-125 Heavy manufacturing 125-150 Light storage 120-125 Heavy storage 250 minimum CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 10 Types of Loads 2. Live load: The 40psf L.L specified by code for Residential Buildings is too Conservative to account for the uncertainties in structural actions Such as impact, fatigue, temp. effects etc. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 11 Types of loads 3. Environmental Loads Environmental loads include wind load, snow load, rain load, earthquake load, and flood load. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 12 Live load reduction The Uniform building code and BOCA National building code permit reduction in basic design live load on any member supporting more than 150ft2 R = r(A-150) Or Where CE-409: Lecture 03 R = 23.1(1+D/L) R = reduction, percent r = rate of reduction = 0.08% for floors A = area supported by floor or member D = dead load, psf L = basic live load,psf Prof. Dr Akhtar Naeem Khan 13 Wind load Bernoulli’s equation for stream flow is used to determine local pressure at stagnation point, considering air to be non-viscous & incompressible. q = (ρv2/2) q: pressure ρ: mass density of air v: velocity • This pressure is called velocity pressure, dynamic pressure, stagnation pressure. • This equation is based on steady flow. • It does not account for dynamic effects of gusts or dynamic response of body. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 14 Wind load Resultant wind pressure on body depends upon pattern of flow around it. Pressure vary from point to point on surface, which depends on shape & size of body. Resultant wind pressure is expressed as: PD = CDA(ρv2/2) PL= CLA(ρv2/2) CD : Drag coefficient CL : Lift coefficient CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 15 Wind load • For buildings bridges and the like pressure is expressed in terms of Shape Factor CS (pressure coefficient) P = CSq = CS(ρv2/2) •Air at 15C weighs 0.0765pcf P=0.00256CSV2 CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan V: mph 16 Wind load Measured wind velocities are averages of fluctuating velocities encountered during a finite time. In US average of velocities recorded during the time it takes a horizontal column of air 1 mile long to pass a fixed point. Fastest mile is highest velocity in 1 day. Annual extreme mile is the largest of the daily maximums. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 17 Wind load Wind pressure to be used in design should be based on a wind velocity having a specific mean recurrence interval. The flow of air close to ground is slowed by surface roughness, which depends on density, size and height of buildings, trees, vegetation etc. Velocity at 33ft (UBC: Sec 1616) above ground is used as the basic values for design purpose. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 18 Wind load CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 19 Wind load Shape factor varies considerably with proportion of structure & horizontal angle of incidence of the wind. • CS for windward face of flat roofed rectangular building is 0.9 • CS for negative pressure on rear face varies from -0.3 to -0.6 • For such building resultant pressure be determined by shape factor 1.2 to 1.5 • Commonly used is 1.3 • CS for Side walls -0.4 to –0.8 • CS for roof –0.5 to –0.8 CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 20 Wind load Wind forces on trussed structures e.g. bridges, transmission towers, beam bridges, girder bridges etc. difficult to assess because of leeward parts of structure. Recommended coefficients for walls of buildings, gabled roofs, arched roofs, roofs over unenclosed structures(stadium), chimneys, tanks, signs, transmission towers etc. are given in ASCE 7-02 Wind pressures specified by building codes include allowance for gust and shape factors. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 21 Wind load • • • • CE-409: Lecture 03 Pressure acts on the windward face of the building Suction acts on the leeward face of the building Suction acts on the sides of the building so a person standing in The window may be thrown outside Suction acts on the floor so that GI sheet floors are blown away During strong wind storms Prof. Dr. Akhtar Naeem Khan 22 Wind load The revolving restaurant supported by a concrete column will Experience suction which will cause tension in the column and as Concrete is weak in tension so it may crack. As a result the lateral Wind load may collapse the restaurant. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 23 Wind load AASHTO specification for Bridge Truss The pressure face is taken as a solid without openings and suction on the leeward face is neglected (its still quiet Conservative) CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 24 Wind Pressure UBC 97 Design Wind Pressure: P CeCqqsIw UBC (20-1) Ce: combined height, exposure and gust factor (Table 16-G) Cq (or Cs): Pressure coefficient for the structure or portion of structure under consideration (Table 16-H) qs : wind stagnation pressure at the standard height of 33ft (Table 16-F) Iw: importance factor (Table 16-k) CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 25 Wind Load Example Example: Calculate the wind pressure exerted by a wind blowing at 100mph on the civil engineering department old building. Sol: According the formula given above: P=0.00256CSV2 V: mph For windward face: Cs = .8 inward (UBC97 Table 16-H) For Leeward face: Cs = .5 outward (UBC97 Table 16-H) CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 26 Wind Load Example P=0.00256CSV2 Pwindward = 20.48 psf Pleeward = 12.80 psf Ptotal CE-409: Lecture 03 = 33.28 psf Prof. Dr. Akhtar Naeem Khan 27 Wind Load Example Alternate Method: P CeCqqsIw UBC (20-1) Ce = 0.76 ( For 30ft height & Exposure B, Table 16-G) Cq = 0.8 ( For windward wall, Table 16-H) = 0.5 ( For leeward wall, Table 16-H) qs = 25.6 psf (For 100mph velocity, Table 16-F) Iw = 1.0 (According to occupancy category, Table16-K) CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 28 Wind Load Example P CeCqqsIw Pwindward = 15.56 psf Pleeward = 9.73 psf Ptotal CE-409: Lecture 03 = 25.29 psf Prof. Dr. Akhtar Naeem Khan 29 Wind Load Example CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 30 Earthquake Load Earthquake Waves Earthquake loads are necessary to consider in earthquake prone regions. Earthquake waves are of two types: Body waves Surface waves CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 31 Earthquake Load Earthquake Waves • Body waves consists of P-waves & S-waves •These waves cause the ground beneath the structure to move back and forth and impart accelerations into the base of structure. •Period and intensity of these acceleration pulses change rapidly & their magnitude vary from small values to more than that of gravity. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 32 Earthquake Load Earthquake Waves CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 33 Earthquake Load Earthquake Waves Body waves reach the buildings first, followed by the more Dangerous surface waves CE-409: Lecture 03 A linear increase in magnitude of EQ causes approximately cubic increase in the corresponding amount of energy released Prof. Dr. Akhtar Naeem Khan 34 Earthquake Load Earthquake Waves Shallow EQ of depth, say, 15-20km are far more dangerous than deep EQ of depth, say, 150-200km. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 35 Earthquake Load Factors effecting earthquake response of structures Structure response to an earthquake primarily depends upon: • Mass • stiffness • natural period of vibration • damping characteristics of structure • location from epicenter • topography & geological formation. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 36 Earthquake Load Factors effecting earthquake response of structures CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 37 Earthquake Load Response Modification Factor CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 38 Earthquake Load Response Modification Factor CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 39 Earthquake Load Natural Time period of structures EQ generally have short periods which may match the natural period of the low rise buildings, say 10 to 20 stories which causes resonance results in serious damages. The possibility of resonance for high rise buildings is low due to longer time periods. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 40 Earthquake Load UBC 97 Static Lateral force procedure: Limitations 1629.8.3: The static lateral force procedure of Section1630 may be used for the following structures: 1. All structures, regular or irregular, in Seismic Zone 1 and in Occupancy Categories 4 and 5 in Seismic Zone 2. 2. Regular structures under 240 feet in height with lateral force resistance provided by systems listed in Table 16-N, except where Section 1629.8.4, Item 4, applies. 3. Irregular structures not more than five stories or 65 feet in height CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 41 Earthquake Load UBC 97 Static Lateral force procedure: Limitations 1629.8.3: The static lateral force procedure of Section1630 may be used for the following structures: 4. Structures having a flexible upper portion supported on a rigid lower portion where both portions of the structure considered separately can be classified as being regular, the average story stiffness of the lower portion is at least 10 times the average story stiffness of the upper portion and the period of the entire structure is not greater than 1.1 times the period of the upper portion considered as a separate structure fixed at the base. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 42 Earthquake Load UBC 97 1630.2.1 Design base shear. The total design base shear in a given direction shall be determined from the following formula: V = (Cv I/R T) * W (30-4) The total design base shear need not exceed the following: V = (2.5 Ca I/R) * W (30-5) The total design base shear shall not be less than the following: V = (0.11 Ca I) W (30-6) In addition, for Seismic Zone 4, the total base shear shall also not be less than the following: V = (0.8 ZNv I/R) * W (30-7) CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 43 Earthquake Load UBC 97 1630.2.1 Design base shear. For Seismic Zones 1, 2A, 2B, and 3: V = (0.11 Ca I) W ≤ V = (Cv I/R T) * W ≤ V = (2.5 Ca I/R) * W For Seismic Zone 4: V = (0.11 Ca I) W ≤ V = (Cv I/R T) * W ≤ V = (2.5 Ca I/R) * W V = (0.8 ZNv I/R) * W CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 44 Earthquake Load UBC 97 1630.2.1 Design base shear. V = total base shear *Ca & CV = seismic dynamic response spectrum values. (table 16-Q & table 16-R) Z = seismic zone factor. (Table 16.I) Nv & Na = near source factors that are applicable in only seismic zone 4. (Table 16-T & Table 16-S) *Depends on Seismic zone and soil profile CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 45 Earthquake Load UBC 97 Soil profiles CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 46 Earthquake Load UBC 97 Seismic Zone BCP 07 CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 47 Earthquake Load UBC 97 1630.2.1 Design base shear. I = Importance factor (Table 16-K) W = Total seismic dead load R = Response factor depends on type of structural system (Table 16-N) T =Elastic fundamental period of vibration. T = Ct hn¾ Ct = 0.035 for steel moment resisting frame CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 48 Earthquake Load UBC 97 1630.2.1 Design base shear. (0.11 Ca I): this coefficient is also independent of the period of vibration. It is a lower bound value, keeping V at some minimum value. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 49 Earthquake Load UBC 97 1630.2.1 Design base shear. (Cv I / R T): acceleration factor (also known as a seismic base shear coefficient). This coefficient will govern V for buildings with medium to long fundamental period of vibrations. The forces in these buildings are induced by the velocity component of the bedrock motion. Hence the "v" subscript. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 50 Earthquake Load UBC 97 1630.2.1 Design base shear. (2.5 Ca I/R): this coefficient is independent of the period of vibration. It will govern V for buildings with short fundamental periods of vibrations, like the buildings being studied in this class. The forces in these stiff buildings are generated by the acceleration component of the bedrock motion. Hence the "a” subscript. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 51 Earthquake Load UBC 97 1630.2.1 Design base shear. (0.82 N v I / R): this lower bound coefficient is only applicable to structures located in seismic zone 4 and within 9.3 miles (15 km) of a known seismic fault. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 52 Earthquake Load UBC 97 Typical Base shear coefficient for Masonry Typical base shear coefficient values for a regular, singlestory masonry building not located near a fault. In addition, we conservatively assumed that a geotechnical site investigation was not completed. Because this type of building is so stiff, the (2.5 Ca I / R) coefficient governs V. Zone 1 2a 2b 3 4 CE-409: Lecture 03 Coefficient V = .067W V = .122W V = .156W V = .200W V = .244W Prof. Dr. Akhtar Naeem Khan 53 Earthquake Load UBC 97 Vertical distribution: Total force shall be distributed over height in the following manner: V=Ft + Fx • Concentrated force Ft at top shall determined by: .07TV ≤ .25V Ft = 0 if T ≤ .7 sec. • Force Fx at each level including level n: CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 54 Earthquake Load UBC 97 Vertical distribution: The single story building is a special case. In most cases, T ≤ .7 and Ft then is taken as zero. From equation 30-15: CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 55 Base Shear Example Calculate and distribute the base shear for a five story residential steel building 50 ft high, located at Peshawar. Assuming SD soil profile. V = (0.11 Ca I) W ≤ V = (Cv I/R T) * W ≤ V = (2.5 Ca I/R) * W CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 56 Base Shear Example Z Ca CV I R = 0.2 = 0.28 = 0.40 =1 = 4.5 Table 16.I Table 16-Q Table 16-R Table 16-K Table 16-N T = Ct hn¾ = (.035)(50)¾ = 0.66 sec CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 57 Base Shear Example V = (0.11 Ca I) W ≤ V = (Cv I/R T) * W ≤ V = (2.5 Ca I/R) * W (0.11 Ca I) = 0.0308 (Cv I/R T) = 0.1347 (2.5 Ca I/R) = 0.1556 Therefore base shear is equal to: V = 0.1347W CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 58 Earthquake Load UBC 97 Total force shall be distributed over height in the following manner: W5 W4 W3 W2 W1 CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 59 Base Shear Example As T < 0.7 therefore Ft = 0 W2 = W3 = W5 = 14 kips W1 = W4 = 20 kips W = 14 x 3 + 20 x 2 = 82 kips V = 0.1347 x 82 = 11.04 kips CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 60 Base Shear Example i. Story level 1: F1 W h = 20 x 10 = 200 k-ft x x ∑ W h = 20 x 10 + 14 x 20 + 14 x 30 + 20 x 40 + 14 x 50 = 2400 k-ft i i F1 = 0.92 kips CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 61 Base Shear Example i. Story level 2: F2 W h = 14 x 20 = 280 k-ft x x ∑ W h = 20 x 10 + 14 x 20 + 14 x 30 + 20 x 40 + 14 x 50 = 2400 k-ft i i F2 = 1.29 kips CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 62 Base Shear Example i. Story level 3: F3 W h = 14 x 30 = 420 k-ft x x ∑ W h = 20 x 10 + 14 x 20 + 14 x 30 + 20 x 40 + 14 x 50 = 2400 k-ft i i F3 = 1.93 kips CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 63 Base Shear Example i. Story level 4: F4 W h = 20 x 40 = 800 k-ft x x ∑ W h = 20 x 10 + 14 x 20 + 14 x 30 + 20 x 40 + 14 x 50 = 2400 k-ft i i F4 = 3.68 kips CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 64 Base Shear Example i. Story level 5: F5 W h = 14 x 50 = 700 k-ft x x ∑ W h = 20 x 10 + 14 x 20 + 14 x 30 + 20 x 40 + 14 x 50 = 2400 k-ft i i F5 = 3.22 kips CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 65 Mean Return Period The average Time Period (in years) based on geological and historical records in which there is a good statistical probability that an earthquake of a certain magnitude or a hurricane will recur is called Mean Return Period or Recurrence Interval R. Probability of Exceedence of the event in any one year is the inverse of the Mean Return Period = 1/R Probability that an event will be exceeded at least once in the n years is Pn= 1-( 1-1/R)n CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 66 Mean Return Period Considering 150mph with a return period of, say, 100years is Reasonable as compared to 500mph with a return period of, say, 1000 years. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 67 Mean Return Period Example:- A structure expected to have a life of 50 years built in locality where mean recurrence interval of an windstorm of 150mph is 95 yrs. The probability that structure will encounter an windstorm exceeding 150mph during its life is? P50=1-( 1-1/95)50 =1- 0.589 = 0.41 or 41% There is 41 percent chances that the structure will be exposed to a windstorm exceeding 150mph. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 68 Mean Return Period Example:- A structure expected to have a life of 50 years built in locality where mean recurrence interval of an earthquake of 0.4g is 95 yrs. The probability that structure will encounter an earthquake exceeding 0.4g during its life is? P50=1-( 1-1/95)50 =1- 0.589 = 0.41 or 41% There is 41 percent chances that the structure will be exposed to an earthquake exceeding 0.4g CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 69 Mean Return Period Uniform Building Code specifies that the earthquake for which a building has to be designed should correspond to an earthquake with a return period of 475 years. Assuming that a building has service life of 50 years. The probability that it will experience and earthquake of mean return period 475 in its design life would be: P50=1 - ( 1 - 1/475)50 =1- 0.90 = 0.01 or 10% CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 70 Impact Load Spring Example It is customary to express Impact load as percentage of static force. Effect of impact load is taken into account in calculation of loads. If impact is 25 %, Live load is multiplied by 1.25 According to AISC live load on hangers supporting floor and balcony construction should be increased by one-third for impact. CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 71 Load Combinations ASD Load combinations 1. 1.0D + 1.0L 2. 0.75D + 0.75L + 0.75W 3. 0.75D + 0.75L + 0.75E D = dead load L = Live load W = Wind load E = Earthquake load CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 72 Load Combinations ASD Load combinations You can use following load combinations with the parameter ALSTRINC (Allowable Strength Increase) to account for the 1/3 allowable increase for the wind and seismic load 1. 1.0D + 1.0L 2. 1.0D + 1.0L + 1.0W 3. 1.0D + 1.0L + 1.0E CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 73 Load Combinations LRFD Load Combinations 1. 1.4D 2. 1.2D + 1.6L + 0.5(Lr or S or R) 3. 1.2D + 1.6(Lr or S or R) + (0.5L or 0.8W) 4. 1.2D +1.3W + 0.5L + 0.5(Lr or S or R) 5. 1.2D ± 1.0E + 0.5L + 0.2S 6. 0.9D ± (1.3W or 1.0E D = Dead load Lr = L = Live load Roof Live Load S = Snow Load R = Rain Water or Ice CE-409: Lecture 03 W = Wind load E = Earthquake load Prof. Dr Akhtar Naeem Khan 74 Load Combinations LRFD Load Combinations Why only Dead load in equation (1) ? There may be a significant live load on a structure during construction. Moreover, the structure may have not reached its full 28 days strength as further construction is usually carried out . CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 75 Load Combinations LRFD Load Combinations Example: increase in dead load on the ground floor due bricks lying on the roof for the construction of the first floor CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 76 Load Combinations LRFD Load Combinations Why negative sign in equation (6) ? It accounts for the stability of structures due to lateral loadings. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 77 Load Combinations LRFD Load Combinations The stabilizing effect of gravity is reduced and the destabilizing effect of lateral load due to wind or earthquake is increased to have the worse situation CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 78 Load Combinations Example: Roof beams W16X31, spaced 7ft-0in center-to-center, support a superimposed dead load of 40 psf. Code specified roof loads are 30 psf downward (due to roof live load, snow, or rain) and 20 psf upward or downward (due to wind). Determine the critical loading for LRFD. D = 31 plf + 40 psf X 7.0 ft = 311 plf L=0 (Lr or S or R) = 30 psf X 7.0 ft = 210 plf W = 20 psf X 7.0 ft = 140 plf E=0 CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 79 Load Combinations 1) 1.4D 1.4(311 plf) = 435 plf 2) 1.2D + 1.6L + 0.5(Lr or S or R) 1.2(311 plf) + 0 + 0.5(210 plf) = 478 plf 3) 1.2D + 1.6(Lr or S or R) + (0.5L or 0.8W) 1.2(311 plf) + 1.6 (210 plf) +0.8(140 plf) = 821 plf 4) 1.2D + 1.3W + 0.5L + 0.5(Lr or S or R) 1.2(311 plf) + 1.3(140 plf) + 0 +0.5(210 plf) = 660 plf CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 80 Load Combinations 5) 1.2D ± 1.0E + 0.5L + 0.2S 1.2(311 plf) + 0 + 0 + 0.2(210 plf) = 415 plf 6) 0.9D ± (1.3W or 1.0E) a) 0.9 (311 plf) + 1.3 (140 plf) = 462 plf b) 0.9(311 plf) - 1.3(140 plf) = 98 plf The critical factored load combination for design is the third, with a total factored load of 821 plf. CE-409: Lecture 03 Prof. Dr. Akhtar Naeem Khan 81 Thank You! CE-409: Lecture 03 Prof. Dr Akhtar Naeem Khan 82