# Lecture 3 – Design Loads

```N-W.F.P. University of Engineering and
Technology Peshawar
By: Prof Dr. Akhtar Naeem Khan
chairciv@nwfpuet.edu.pk
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Prof. Dr Akhtar Naeem Khan
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Feeling Responsibility
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 Determination of loads for which a given
structure may be designed for is a
difficult problem.
• What loads may structure be called upon during its
• In what combinations these loads occur?
• The probability that a specific live load be
exceeded at some time during lifetime of
structure?
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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.
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materials and fixed equipment
incorporated into the building or other
structure. (UBC Section 1602)



Weight of structure
Weight of permanent machinery etc.
the member dimensions and material
densities are known.
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use and occupancy of the building or other
 Weight of people, furniture, machinery,
goods in building.
 Weight of traffic on bridge
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• Buildings serve such diverse purposes that it is
extremely difficult to estimate suitable design
• Different building codes specify live load
requirements.
• Uniform Building Code (UBC)
• Southern Standard Building Code
• BOCA National Building Code
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2. Live load: (UBC Table 16-A)
Occupancy
Residential
40
60
Mercantile
75-125
Heavy manufacturing
125-150
Light storage
120-125
Heavy storage
250 minimum
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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.
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 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
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R = 23.1(1+D/L)
R = reduction, percent
r
= rate of reduction = 0.08% for floors
A = area supported by floor or member
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 Bernoulli’s equation for stream flow is used to
determine local pressure at stagnation point,
considering air to be non-viscous &amp;
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.
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 Resultant wind pressure on body depends upon
pattern of flow around it.
 Pressure vary from point to point on surface,
which depends on shape &amp; size of body.
 Resultant wind pressure is expressed as:
PD = CDA(ρv2/2)
PL= CLA(ρv2/2)
CD : Drag coefficient
CL : Lift coefficient
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• 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
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V: mph
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 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.
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 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.
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 Shape factor varies considerably with proportion
of structure &amp; 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
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 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
transmission towers etc. are given in ASCE 7-02
 Wind pressures specified by building codes include
allowance for gust and shape factors.
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•
•
•
•
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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
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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.
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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)
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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)
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 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)
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P=0.00256CSV2
 Pwindward = 20.48 psf
 Pleeward = 12.80 psf
 Ptotal
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= 33.28 psf
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 Alternate Method:
P  CeCqqsIw
UBC (20-1)
 Ce = 0.76 ( For 30ft height &amp; 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)
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P  CeCqqsIw
 Pwindward = 15.56 psf
 Pleeward = 9.73 psf
 Ptotal
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= 25.29 psf
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Earthquake Waves
 Earthquake loads are necessary to
consider in earthquake prone regions.
 Earthquake waves are of two types:

Body waves

Surface waves
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Earthquake Waves
• Body waves consists of P-waves &amp; 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 &amp; their magnitude vary from small
values to more than that of gravity.
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Earthquake Waves
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Earthquake Waves
Body waves reach the
buildings first, followed by
the more Dangerous
surface waves
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A linear increase in magnitude
of EQ causes approximately cubic
increase in the corresponding
amount of energy released
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Earthquake Waves
Shallow EQ of depth, say, 15-20km are far more dangerous than
deep EQ of depth, say, 150-200km.
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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 &amp; geological formation.
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Factors effecting earthquake response of structures
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Response Modification Factor
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Response Modification Factor
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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.
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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
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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.
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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)
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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
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1630.2.1 Design base shear.
V = total base shear
*Ca &amp; CV = seismic dynamic response spectrum values.
(table 16-Q &amp; table 16-R)
Z = seismic zone factor. (Table 16.I)
Nv &amp; Na = near source factors that are applicable in only
seismic zone 4. (Table 16-T &amp; Table 16-S)
*Depends on Seismic zone and soil profile
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Soil profiles
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Seismic Zone BCP 07
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1630.2.1 Design base shear.
I = Importance factor (Table 16-K)
R = Response factor depends on type of structural system
(Table 16-N)
T =Elastic fundamental period of vibration.
T = Ct hn&frac34;
Ct = 0.035 for steel moment resisting frame
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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.
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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 &quot;v&quot;
subscript.
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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 &quot;a”
subscript.
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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.
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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
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Coefficient
V = .067W
V = .122W
V = .156W
V = .200W
V = .244W
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 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:
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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:
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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
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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 hn&frac34; = (.035)(50)&frac34; = 0.66 sec
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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
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 Total force shall be distributed over height in the
following manner:
W5
W4
W3
W2
W1
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Base Shear Example
As T &lt; 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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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
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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%
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 Spring Example
 It is customary to express Impact load as
percentage of static force.
 Effect of impact load is taken into account in
 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.
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1. 1.0D + 1.0L
2. 0.75D + 0.75L + 0.75W
3. 0.75D + 0.75L + 0.75E
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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
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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 &plusmn; 1.0E + 0.5L + 0.2S
6. 0.9D &plusmn; (1.3W or 1.0E
D
Lr =
S
R
= Rain Water or Ice
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 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 .
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lying on the roof for the construction of the first floor
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 It accounts for the stability of structures due
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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
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Example: Roof beams W16X31, spaced 7ft-0in center-to-center,
loads are 30 psf downward (due to roof live load, snow, or rain)
and 20 psf upward or downward (due to wind). Determine the
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
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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
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5) 1.2D &plusmn; 1.0E + 0.5L + 0.2S
1.2(311 plf) + 0 + 0 + 0.2(210 plf) = 415 plf
6) 0.9D &plusmn; (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.
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Thank You!
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