CHAPTER 1
TYPES OF
STRUCTURE &
LOADS
EAT 251 STRUCTURAL
THEORY
Introduction
●
Structural analysis is the analysis
of a given structure subject to
some given loads in order to
predict the response of the
structure system.
● The structural analysis is the
application of solid mechanics to
predict the response such as
forces and displacement of a
given structure which are
subjected to a specified load.
Types of Structures
 Frame structure – are composed of beams and columns that




are connected either pin or fixed
Truss structure – consist of slender elements can resist axial
tensile or compressive forces
Surface structure – made from a material having a very
small thickness compare to its other dimensions
Arch structure – support their loads in compression
Cables – support their loads in tension
Types of Structures
STEEL
STRUCTURES
TRUSS
SYSTEM
CABLE
ARCH
Classification of Structures
Structural elements can be
tie rod, beams, columns
Types of Loads
 Loads are classified into 2 parts :
a) Dead loads (Gk)
-Self-weight of the structure components
- weight of objects that are permanently attached to the structure
b) Imposed loads and forces (Qk)
-Live, wind, snow, lateral loads and others.
- Weight of temporary objects such as moving vehicle,
natural forces
Types of Loads
 Wind load (Wk)
- Occur when structures block the flow of wind
- Cause by the conversion of wind’s kinetic energy into potential
energy of pressure and known as wind load
 Snow Load
 Earthquake Load
 Other Natural Loads (temperature change, blast effect and
foundation settlement
LOAD COMBINATIONS
Dead Load, Gk
(1.4 x dead load) = 1.4Gk
Dead, Gk and wind Load, Wk
(0.9 x dead load) + (1.4 x wind load) = 0.9Gk + 1.4 Wk
Dead, imposed and snow load
(1.2 x dead load + (1.6 x live load) + (0.5 x snow load) = (1.2Gk
+ 1.6Qk + 0.5Wk)
Dead load and earthquake load
(0.9 x dead load) + (1.0 x earthquake load)
Data for Loading Calculation
Reduced Live Load


For some types of buildings having very large floor areas, many codes
will allow a reduction in the uniform live load for a floor, since it is
unlikely that the prescribed live load will occur simultaneously
throughout the entire structure at any one time.
The reduction live load can be made if influence area, KLLAt > 37.2m2
(proposed by ASCE-7).
Example 1.1
The floor beam in Figure Eg. 1.1 is used
to support the 1.8m width of a
lightweight plain concrete slab having a
thickness of 100mm. The slab serves as a
portion of the ceiling for the floor below
and therefore its bottom is coated with
plaster. Furthermore, a 3m high, 300mm
thick lightweight solid concrete block wall
is directly over the top flange of the
beam. Determine the loading on the
beam measured per meter of length for
the length of the beam.
Figure Eg.1.1
Solution 1.1
Example 1.2
A 2-storey office building shown in the
Figure Eg. 1.2 has interior columns
that are spaced 6.6m apart in two
perpendicular directions. If the (flat)
roof loading is 0.96kN/m2, determine
the reduced live load supported by a
typical interior column located at
ground level.
.
Figure Eg. 1.2
Solution 1.2
Wind Loads
 When the speed of the wind is very high, it can cause massive
damage to structure. To understand the effect of horizontal wind
blowing over and around of building, figure below can be
considered.
Wind Loads
● Wind pressure is the
force per square
metre on the building
● Wind pressure for
building :a) External pressure
b) Internal pressure
Surface
L/B
Cp
Use with
Wind
Windward
Leeward
Windward wall
All values
0.8
qz
direction
angle ϴ
angle
Leeward
0–1
-0.5
qh
wall
2
-0.3
≥4
-0.2
All values
-0.7
Side Walls
Wall pressure coefficients, Cp
***refer page 36 in text book
qh
h/L
10°
ϴ = 10°
Normal to
≤0.25
-0.7
-0.3
ridge
0.5
-0.9
-0.5
> 1.0
-1.3
-0.7
Maximum negative roof pressure
coefficients, Cp for use with qh
***refer Fig 1-13 page 37 in text book
P = qGC
Where;
q = basic pressure at the height of 10m
Where;
P = wind pressure
G = gust factor
(0.85)
C = shape factor
q = 0.613KzKztKdKeV2 (q in N/m2 and V in m/s)
V = velocity of a 3 seconds gust of wind measured 10m above the
ground during a 50-year recurrence period. Normal basic wind load is
47m/s (for storage building) unless is stated.
Kz =The velocity pressure exposure coefficient, which is a function of
height and depends upon the ground terrain. Table 1-5 lists values for
a structure which is located in open terrain with scattered low-lying
obstructions.
Kzt = a factor that accounts for wind speed increases due to hills and
escarpments, for flat ground Kzt = 1
Kd = a factor that accounts for the direction of the wind. It is used only
when the structure is subjected to combinations of loads. For wind
acting alone, Kd = 1
Ke = a ground elevation factor, for conservative design use Ke = 1.0
Coefficients
P = qGCp
q = 0.613KzKztKdV2I (q in N/m2 and V in m/s)
Where;
Cp = a wall or roof pressure coefficient
qz = windward wall
qh = Leeward wall
P = qGCp
q = 0.613KzKztKdKeV2 (q in N/m2 and V in m/s)
Where;
G = 0.85
Kzt = 1 for flat area
Elevation
Plan
Cp refer Figure 1-13 in
textbook
▪ Internal Pressure for Enclosed Building
0.85qh
0.85qh
▪ Internal Pressure for Partially Open Building
0.55qh
0.55qh
Total Wind Pressure on the Main Wind-Resisting
Frame System for Enclosed Building
qhGC
p
qhGC
qhGC
p
p
qhGC
qzGC
p
p
0.85qh
0.85qh
Ph
Ph
Ph
Ph
Pz
Ph
Pz
Ph
Design wind pressure for enclosed buildings
P = qGCp-qh(GCpi)
(Eq. 1-3)
qz = 0.613KzKztV2 (q in N/m2 and V in m/s)
Where;
GCpi = ±0.18 for enclosed building
GCpi = ±0.55 for partially enclosed building
GCpi = 0 for open building
q
= qz for windward wall at height z (refer Eq. 1-2)
= qh for leeward walls, side walls and roof;
where z = h (mean height of the roof)
Cp = a wall or roof pressure coefficient determine from table. –ve value indicate
pressures acting away from the surface
Example 1.3
The enclosed building as shown in Figure
Eg. 1.3, is used for storage purposes and
is located outside of Chicago, Illinois on
open flat terrain. When the wind is
directed as shown, determine the design
wind pressure acting on the roof and
sides of the building using the ASCE 7-10
Specifications.
Figure Eg.1.3
Solution:Basic wind speed, V = 47m/s
For flat terrain, Kzt = 1.0
Consider only wind load, Kd = 1.0
Using eq. 1-2 ;
qz = 0.613KzKztKeV2 (q in N/m2 and V in m/s)
= 0.613 x Kz x 1.0 x 1.0 x (47)2
= 1354Kz
From Figure 1-13 (a), h’ = 22.5 tan 10° = 3.97m
Thus, the mean height of the roof is h = 7.5 + 3.97/2 = 9.48m
By using interpolation method of pressure profile values listed in Table 1.5
z = h = 9.48m
Z (m)
Kz/z
0-4.6
0.85
6.1
0.90
7.6
0.94
9.1
0.98
12.2
1.04
15.2
1.09
Kz = 0.987m
qh = 1354Kz
= 1354 (0.987)
= 1337N/m2
Using Eq 1-3, the pressure loading can be determine as follows:-
= q (0.85) Cp – 1337 ( ± 0.18)
= 0.85qCp ± 241
Windward wall
Since the pressure varies with height, qzGCp need to be considered. From table in Page
36, for all values L/B, Cp = 0.8 Using eq. 1-3;
P(0-4.6) = 542 N/m2
P(6.1) = 588 N/m2
P(7.6) = 625 N/m2
Leeward wall (refer table in page 36)
L/B = (2 x 22.5)/45 =1, Cp = -0.5 and use q=qh.
Using eq. 1-3, P = -809 N/m2
Side walls (refer table in page 36)
For all values of L/B, Cp = -0.7 and q=qh.
Using eq. 1-3, P = -1037 N/m2
Windward Roof (refer table in page 37)
h/L = 9.48/(2x22.5) = 0.211 < 0.25, Cp = -0.7 and q = qh.
Using eq. 1-3, P = -1037 N/m2
Leeward Roof (refer table in page 37)
h/L < 0.25, Cp = -0.3 and q = qh.
Using eq. 1-3, P = -582 N/m2
Other Types of Loads
EARTHQUAKE
HYDROSTATIC
SOIL
PRESSURE
SNOW
Do you have any questions?
ATTENDANCE
https://tinyurl.com/yxr5nglc
You can voice out your
questions during next class or
in the google classroom
platform.