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ARCH5023/ARCH4333 -
ARCHITECTURAL STRUCTURES I
][ STEEL
STRUCTURES ][
THE UNIVERSITY OF OKLAHOMA
- COLLEGE OF ARCHITECTURE
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LOADS
Dead Loads
weight of building materials
roof
floor
ceiling
beams
walls
columns
Live Loads
Roof Loads
snow
Floor Loads
occupancy
Lateral Loads
wind loads
earthquake
DEAD LOADS
weight of building construction
roofs
ceilings
floors
walls
weight of the structural element itself
permanent load
loads due to gravity
result in a downward vertical force
Weight of Building Construction
P&A 557 - 558
ROOF
Given: Asked:
18 ga steel deck painted 2.6 psf
Fiberglas insulation bats 0.5 psf
mtl. purlins estimate 1.0 psf
steel joists estimate 1.5 psf
suspended steel channels 1.0 psf
dry wall gypsum board 1/2" 2.5 psf
suspended lighting and 3.0 psf
air distribution systems average ________
Total Roof Dead Load 12.1 psf
========
Live Load
all non-permanent loadings
Roof Loads
========
construction/maintenance loads
snow
PA p.562 table 22.2 - UBC
lack of total surface loading
ground snow load - BOCA
roof pitch
wind pressure
uplift
water
ponding= water causes deflection
flat roof = 1/4" / ft min.
Floor Loads
=========
occupancy dependent
humans
furniture
equipment
stored materials
moveable partitions 15-20 psf
PA 563 Table 22.3
Residential 40 psf
Office Buildings 50 psf
Res. Balconies 60 psf
Libary 125 psf
Lateral Loads:
==========
Wind
====
Basic Wind Speed
Exposure
Wind Stagnation Pressure qs
Design Wind Pressure p
Design Methods
Method 1 Normal Force Method
Method 2 Projected Area Method
Uplift
Overturning Moment
Drift
LOAD REDUCTION FACTORS
When structural framing members support large area, most codes allow some reduction in the total live
load to be used for design.
ROOF LOADS
===========
These reductions, in the case of roof loads are incorporated into the data in Table 22.2. Page 562 P&A
Flat roof
4:12
12:12
FLOOR LOADS
============
The following example shows the method given in the UBC for determining the reduction permitted for beams, trusses, or columns that support large floor areas.
EXAMPLE - PROBLEM
LOAD REDUCTION FACTOR
A wide flange is supporting a floor system
with a dead load of 20 psf and a
residential floor live load of 40 psf.
The floor area supported by the structural member
is 25 ft by 30 ft, a total of 750 sqft.
A. GIVEN
A = area of floor supported by a member
in sqft
A = 750 sqft
D = unit dead load/sq ft of supported area
D = 20 psf
L = unit live load/sq ft of supported area
L = 40 psf
R = reduction in percent
A = area of floor supported by a member
in sq.ft.
D = unit dead load/sq ft of supported area
L = unit live load/sq ft of supported area
B. GRAPH
30 ft L=40psf
--------------------------
/ /
/ /
/_________________/ 25 ft
][ D=20 psf
C. ASKED
Load Reduction Factor
R = reduction in percent [%]
D.CALCULATIONS
SOLUTION
Step by step solution
# indicates the steps
#1. No reduction allowed for:
assembly theater buildings
live loads greater than 100 psf
#2. Reduction allowed according to:
P&A 564
#3. R = 0.08 (A[sqft] - 150[sqft])
#4. R = 0.08 (750 [sqft] - 150 [sqft])
#5. R = 0.08 (600 [sqft])
#6. R = 48%
but
#7. R <40%
Reduction shall not exceed:
#8. 40% for horizontal members
#9. 40% for vertical members receiving
load from 1 level only
#10. 60% for other vertical members
nor R as determined by the
following formula
#11. R = 23.1 ( 1 + D[psf]/L[psf])
#12. R = 23.1 ( 1 + D[psf]/L[psf])
#13. R = 23.1 ( 1 + 20[psf]/40[psf])
#14. R = 23.1 ( 1 + 0.5 )
#15. R = 23.1 * 1.5
#16. R = 34.65 % governs
==========
#17. R = 48% does not govern
#18. R = 40% does not govern
E. ANSWER
Please clearly indicate your answer and underline it!
#19. R = reduction in percent
The total live load to be used for design
of a member may be reduced by %.
R = 34.65 %
==========
34 % would also be an acceptable answer
35 % would be the wrong answer
you have to round down,
otherwise your safety factor is too great!
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