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Chapter4 Thesis Proposed 3-Storey MixedUse Building

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CHAPTER IV
ANALYSIS AND DESIGN
4.1 PRESENTATION OF DATA
4.1.1 Introduction
This chapter includes the determination and computation of loadings, structural analysis,
and designing of structural members.
Structural analysis is the study of the consequences of forces operating on structures and
their components. Bridges, buildings, towers and other vertical structures are structures that are
entitled to this kind of analysis include all that can resist structural loads. In order to derive
values for structure deformation, internal forces, stresses, response at supports, acceleration, and
stability, this analysis incorporates fields such as applied mechanics, materials science, and
applied mathematics. Finally, the outcome of the analysis shall determine whether the structure
is capable for use which saves various physical testing. Moreover, structural analysis is critical in
structural design. The primary method used in the analysis is the Moment Distribution Method
(MDM).
Designing of structural members refers to the selection of materials and member
properties such as its type, size, and composition to safely and controllably support the loads
applied to the structure. This chapter includes all the design of structural members such as slabs,
ramps, stairs, beams, columns, and footing.
General Classification and Category
Occupancy Category ÅÍÍ
IV – Standard Occupancy
Type of Construction
Type V – Reinforced Concrete Special Moment
Resisting Framework
Reference for Architectural Standards
National Building Code of the Philippines (NBCP)
Reference for Loading
National Structural Code of the Philippines (NSCP)
Structural Analysis Method
Moment Distribution Method
Assumptions
o The frames are fixed at the supports.
o The frame is unbraced from lateral loads due to the presence of openings in walls.
o Each frame carries its respective tributary area.
o The design of other structural members on the same level follow the same design of
critical members.
o Each footing carries the greatest load combination of the bottom column in the whole
structure.
o The columns are unbraced and most of them require moment magnification
o The uplift pressure of wind reduced the effect of the gravity loads and is therefore
negated.
4.1.2 Framing Plans and Properties
Figure: Framing Plan – Top View
Figure: Longitudinal Frame Elevation with Assumed Dimensions
Figure: Transverse Frame Elevation with Assumed Dimensions
4.1.3 Structural Elements
4.1.3.1 Slabs
Ground Level: 250-mm Concrete Slab
2nd Floor Level: 200-mm Concrete Slab
3rd Floor Level: 200-mm Concrete Slab
Roof Deck: 200-mm Concrete Slab
4.1.3.2 Beams (B1 and B2)
Dimensions: 400 π‘šπ‘š π‘₯ 600 π‘šπ‘š (Assumption) π‘œπ‘Ÿ
0.40 π‘š π‘₯ 0.60 π‘š
Uniformly Distributed Dead Load of the Beams/Girders (π‘€π΅π‘’π‘Žπ‘š ),
π‘€π΅π‘’π‘Žπ‘š = 23.6 π‘˜π‘/π‘š3 (0.40 π‘š)(0.60 π‘š)
π‘€π΅π‘’π‘Žπ‘š = 5.664 π‘˜π‘/π‘š
4.1.3.3 Columns (C1)
Dimensions: 800 π‘šπ‘š π‘₯ 800 π‘šπ‘š (Assumption) π‘œπ‘Ÿ
0.80 π‘š π‘₯ 0.80 π‘š
4.1.4 Determination of Applied Loadings
Tributary Areas of the Frames
The tributary areas are the surface areas that cause distribution of the loadings such as
load pressures on the structural elements. The structure has two (2) types of tributary areas:
horizontal and vertical. The horizontal tributary areas are surface areas on the slabs that support
the applied vertical loads such as dead loads, live loads, and roof live loads. Whereas vertical
tributary areas are the projection on the wall surface areas that support the applied horizontal or
lateral loads such as the wind load (wind pressure).
(1) Horizontal Tributary Area
Figure: Horizontal Tributary Areas of the Longitudinal Exterior and Interior Frames:
Frame 1 (Left), Frame 2 (Right)
Figure: Horizontal Tributary Areas of the Transverse Exterior and Interior Frames:
Frame 3 (Left), Frame 4 (Right)
(2) Vertical Tributary Area
Figure: Vertical Tributary Areas of the Longitudinal Exterior and Interior Frames Projected on
the Transverse Frame: Frame 1 (Left), Frame 2 (Right)
Figure: Vertical Tributary Area of the Transverse Exterior Frame
Projected on the Longitudinal Frame: Frame 3
Figure: Vertical Tributary Area of the Transverse Interior Frame
Projected on the Longitudinal Frame: Frame 4
4.1.4.1 Dead Loads
The dead load contains the weight of all materials of construction incorporated into the
building or other structure, including but not limited to walls, floors, roofs, ceilings, stairways,
built-in partitions, finishes, cladding and other similarly incorporated architectural and structural
items, and fixed service equipment, including the weight of cranes. The dead loads of the
building are acquired from the Chapter two of the NSCP 2015.
4.1.4.1.a Roof Deck (RD)
From Table 204-1 Minimum Densities for Design Loads from Materials
200-mm Concrete Roof Deck
Concrete, Reinforced
Stone, including gravel = 23.6 π‘˜π‘/π‘š3
Masonry, Medium Weight Units = 19.6 π‘˜π‘/π‘š3
From Table 204.2 Minimum Design Dead Loads (kPa)
Ceilings:
Acoustical Fiber Board = 0.05 π‘˜π‘ƒπ‘Ž
Mechanical Duct Allowance = 0.20 π‘˜π‘ƒπ‘Ž
Suspended Steel Channel System = 0.10 π‘˜π‘ƒπ‘Ž
Coverings (Roof and Wall):
Rigid Insulation, 13mm = 0.04 π‘˜π‘ƒπ‘Ž
Skylight, metal frame, 10mm Wire Glass = 0.38 π‘˜π‘ƒπ‘Ž
Floor Fill:
Stone Concrete per mm (30-mm thickness) = 0.023 π‘˜π‘ƒπ‘Ž/π‘šπ‘š
Therefore, the Total Uniform Dead Load on the Roof Deck (π‘žπ·π‘…π· ),
π‘žπ·π‘…π· = Ceilings + Coverings + Concrete Roof Deck Slab
= 0.05 π‘˜π‘ƒπ‘Ž + 0.20 π‘˜π‘ƒπ‘Ž + 0.10 π‘˜π‘ƒπ‘Ž + 0.04 π‘˜π‘ƒπ‘Ž + 0.38 π‘˜π‘ƒπ‘Ž
+ 0.023 π‘˜π‘ƒπ‘Ž/π‘šπ‘š (30 π‘šπ‘š) + 23.6 π‘˜π‘/π‘š3 (0.2 π‘š)
π‘žπ·π‘…π· = 6.18 π‘˜π‘ƒπ‘Ž
4.1.4.1.b 3rd Floor Level (3L)
From Table 204-1 Minimum Densities for Design Loads from Materials
200-mm Concrete Slab
Concrete, Reinforced
Stone, including gravel = 23.6 π‘˜π‘/π‘š3
Masonry, Medium Weight Units = 19.6 π‘˜π‘/π‘š3
From Table 204.2 Minimum Design Dead Loads (kPa)
Ceilings:
Acoustical Fiber Board = 0.05 π‘˜π‘ƒπ‘Ž
Mechanical Duct Allowance = 0.20 π‘˜π‘ƒπ‘Ž
Suspended Steel Channel System = 0.10 π‘˜π‘ƒπ‘Ž
Coverings (Roof and Wall):
Floor Fill:
Floor and Floor Finishes:
Ceramic Tile on 25 mm Mortar Bed = 1.10 π‘˜π‘ƒπ‘Ž
Frame Partitions:
Wood or steel studs, 13mm gypsum board each side = 0.38 π‘˜π‘ƒπ‘Ž
Frame Walls:
Windows, glass, frame and sash = 0.38 π‘˜π‘ƒπ‘Ž
Therefore, the Total Uniform Dead Load on the 3rd Floor Level (π‘žπ·3𝐿 )
π‘žπ·3𝐿 = Ceilings + Coverings + Floor Fill + Floor and Floor Finishes
+Frame Partitions + Frame Walls + Concrete Slab
= 0.05 π‘˜π‘ƒπ‘Ž + 0.20 π‘˜π‘ƒπ‘Ž + 0.10 π‘˜π‘ƒπ‘Ž + 1.10 π‘˜π‘ƒπ‘Ž + 0.38 π‘˜π‘ƒπ‘Ž + 0.38 π‘˜π‘ƒπ‘Ž
+ 23.6 π‘˜π‘/π‘š3 (0.2 π‘š)
π‘žπ·3𝐿 = 6.93 π‘˜π‘ƒπ‘Ž
Concrete Masonry Units:
Exterior CHB Wall: Full Grout with Plaster on Both Sides (150-mm thickness),
π‘žπΆπ»π΅πΈπ‘₯𝑑 = 2.82 π‘˜π‘ƒπ‘Ž
Interior CHB Wall: Full Grout with Plaster on Both Sides (100-mm thickness),
π‘žπΆπ»π΅πΌπ‘›π‘‘ = 2.69 π‘˜π‘ƒπ‘Ž
Uniformly Distributed Dead Load of CHB on the 3rd Floor Level (𝑀𝐢𝐻𝐡3𝐿
𝐸π‘₯𝑑
Wall height, h = 3.5 π‘š
Exterior Side:
𝑀𝐢𝐻𝐡3𝐿
= π‘žπΆπ»π΅πΈπ‘₯𝑑 (β„Ž) = 2.82 π‘˜π‘ƒπ‘Ž (3.5 π‘š)
𝑀𝐢𝐻𝐡3𝐿
= 9.87 π‘˜π‘/π‘š
𝐸π‘₯𝑑
𝐸π‘₯𝑑
Interior Side:
𝑀𝐢𝐻𝐡3𝐿
= π‘žπΆπ»π΅πΌπ‘›π‘‘ (β„Ž) = 2.69 π‘˜π‘ƒπ‘Ž (3.5 π‘š)
𝑀𝐢𝐻𝐡3𝐿
= 9.415 π‘˜π‘/π‘š
𝐼𝑛𝑑
𝐼𝑛𝑑
4.1.4.1.c 2nd Floor Level (2L)
),
From Table 204-1 Minimum Densities for Design Loads from Materials
200-mm Concrete Slab
Concrete, Reinforced
Stone, including gravel = 23.6 π‘˜π‘/π‘š3
Masonry, Medium Weight Units = 19.6 π‘˜π‘/π‘š3
From Table 204.2 Minimum Design Dead Loads (kPa)
Ceilings:
Acoustical Fiber Board = 0.05 π‘˜π‘ƒπ‘Ž
Mechanical Duct Allowance = 0.20 π‘˜π‘ƒπ‘Ž
Suspended Steel Channel System = 0.10 π‘˜π‘ƒπ‘Ž
Coverings (Roof and Wall):
Floor Fill:
Floor and Floor Finishes:
Ceramic Tile on 25 mm Mortar Bed = 1.10 π‘˜π‘ƒπ‘Ž
Frame Partitions:
Wood or steel studs, 13mm gypsum board each side = 0.38 π‘˜π‘ƒπ‘Ž
Movable Partitions = 0.24 π‘˜π‘ƒπ‘Ž
Additional Partition Loads (Section 204.3) = 1 π‘˜π‘ƒπ‘Ž
Frame Walls:
Windows, glass, frame and sash = 0.38 π‘˜π‘ƒπ‘Ž
Therefore, the Total Uniform Dead Load on the 2nd Floor Level (π‘žπ·2𝐿 ),
π‘žπ·2𝐿 = Ceilings + Coverings + Floor Fill + Floor and Floor Finishes
+Frame Partitions + Frame Walls + Concrete Slab
= 0.05 π‘˜π‘ƒπ‘Ž + 0.20 π‘˜π‘ƒπ‘Ž + 0.10 π‘˜π‘ƒπ‘Ž + 1.10 π‘˜π‘ƒπ‘Ž + 0.38 π‘˜π‘ƒπ‘Ž
+0.24 π‘˜π‘ƒπ‘Ž + 1 π‘˜π‘ƒπ‘Ž + 0.38 π‘˜π‘ƒπ‘Ž + 23.6 π‘˜π‘/π‘š3 (0.2 π‘š)
π‘žπ·2𝐿 = 8.17 π‘˜π‘ƒπ‘Ž
Concrete Masonry Units:
Exterior CHB Wall: Full Grout with Plaster on Both Sides (150-mm thickness),
π‘žπΆπ»π΅πΈπ‘₯𝑑 = 2.82 π‘˜π‘ƒπ‘Ž
Interior CHB Wall: Full Grout with Plaster on Both Sides (100-mm thickness),
π‘žπΆπ»π΅πΌπ‘›π‘‘ = 2.69 π‘˜π‘ƒπ‘Ž
Uniformly Distributed Dead Load of CHB on the 2nd Floor Level (𝑀𝐢𝐻𝐡2𝐿 ),
Wall height, h = 3.5 π‘š
Exterior Side:
𝑀𝐢𝐻𝐡2𝐿
= π‘žπΆπ»π΅πΈπ‘₯𝑑 (β„Ž) = 2.82 π‘˜π‘ƒπ‘Ž (3.5 π‘š)
𝑀𝐢𝐻𝐡2𝐿
= 9.87 π‘˜π‘/π‘š
𝐸π‘₯𝑑
𝐸π‘₯𝑑
Interior Side:
𝑀𝐢𝐻𝐡2𝐿
= π‘žπΆπ»π΅πΌπ‘›π‘‘ (β„Ž) = 2.69 π‘˜π‘ƒπ‘Ž (3.5 π‘š)
𝑀𝐢𝐻𝐡2𝐿
= 9.415 π‘˜π‘/π‘š
𝐼𝑛𝑑
𝐼𝑛𝑑
4.1.4.1.d Ground Floor Level (1L)
From Table 204-1 Minimum Densities for Design Loads from Materials
250-mm Concrete Slab
Concrete, Reinforced
Stone, including gravel = 23.6 π‘˜π‘/π‘š3
Masonry, Medium Weight Units = 19.6 π‘˜π‘/π‘š3
From Table 204.2 Minimum Design Dead Loads (kPa)
Ceilings:
Coverings (Roof and Wall):
Floor Fill:
Floor and Floor Finishes:
Ceramic Tile on 25 mm Mortar Bed = 1.10 π‘˜π‘ƒπ‘Ž
Frame Partitions:
Frame Walls:
Windows, glass, frame and sash = 0.38 π‘˜π‘ƒπ‘Ž
Therefore, the Total Uniform Dead Load on the Ground Floor Level (π‘žπ·1𝐿 ),
π‘žπ·1𝐿 = Ceilings + Coverings + Floor Fill + Floor and Floor Finishes
+ Frame Partitions + Frame Walls + Concrete Slab
= 1.10 π‘˜π‘ƒπ‘Ž + 0.38 π‘˜π‘ƒπ‘Ž + 23.6 π‘˜π‘/π‘š3 (0.25 π‘š)
π‘žπ·1𝐿 = 7.38 π‘˜π‘ƒπ‘Ž
Concrete Masonry Units:
Exterior CHB Wall: Full Grout with Plaster on Both Sides (150-mm thickness),
π‘žπΆπ»π΅πΈπ‘₯𝑑 = 2.82 π‘˜π‘ƒπ‘Ž
Interior CHB Wall: Full Grout with Plaster on Both Sides (100-mm thickness),
π‘žπΆπ»π΅πΌπ‘›π‘‘ = 2.69 π‘˜π‘ƒπ‘Ž
Uniformly Distributed Dead Load of CHB on the 2nd Floor Level (𝑀𝐢𝐻𝐡1𝐿 ),
Wall height, h = 4 π‘š
Exterior Side:
𝑀𝐢𝐻𝐡1𝐿
= π‘žπΆπ»π΅πΈπ‘₯𝑑 (β„Ž) = 2.82 π‘˜π‘ƒπ‘Ž (4 π‘š)
𝑀𝐢𝐻𝐡1𝐿
= 11.28 π‘˜π‘/π‘š
𝐸π‘₯𝑑
𝐸π‘₯𝑑
Interior Side:
𝑀𝐢𝐻𝐡1𝐿
= π‘žπΆπ»π΅πΌπ‘›π‘‘ (β„Ž) = 2.69 π‘˜π‘ƒπ‘Ž (4 π‘š)
𝑀𝐢𝐻𝐡1𝐿
= 10.76 π‘˜π‘/π‘š
𝐼𝑛𝑑
𝐼𝑛𝑑
4.1.4.1.1 Partial Summary of Dead Loads on Every Floor Level
Roof Deck (RD)
o Total Uniform Dead Load (π‘žπ·π‘…π· )
π‘žπ·π‘…π· = 6.18 π‘˜π‘ƒπ‘Ž
o Uniformly Distributed Dead Load of Beams (π‘€π΅π‘’π‘Žπ‘š )
π‘€π΅π‘’π‘Žπ‘š = 5.664 π‘˜π‘/π‘š
3rd Floor Level (3L)
o Total Uniform Dead Load (π‘žπ·3𝐿 )
π‘žπ·3𝐿 = 6.93 π‘˜π‘ƒπ‘Ž
o Uniformly Distributed Dead Load of Beams (π‘€π΅π‘’π‘Žπ‘š )
π‘€π΅π‘’π‘Žπ‘š = 5.664 π‘˜π‘/π‘š
o Uniformly Distributed Dead Load of CHB (𝑀𝐢𝐻𝐡3𝐿
𝐸π‘₯𝑑
𝑀𝐢𝐻𝐡3𝐿
= 9.87 π‘˜π‘/π‘š
𝑀𝐢𝐻𝐡3𝐿
= 9.415 π‘˜π‘/π‘š
𝐸π‘₯𝑑
𝐼𝑛𝑑
and 𝑀𝐢𝐻𝐡3𝐿
)
and 𝑀𝐢𝐻𝐡2𝐿
)
and 𝑀𝐢𝐻𝐡1𝐿
)
𝐼𝑛𝑑
2nd Floor Level (2L)
o Total Uniform Dead Load (π‘žπ·2𝐿 )
π‘žπ·2𝐿 = 8.17 π‘˜π‘ƒπ‘Ž
o Uniformly Distributed Dead Load of Beams (π‘€π΅π‘’π‘Žπ‘š )
π‘€π΅π‘’π‘Žπ‘š = 5.664 π‘˜π‘/π‘š
o Uniformly Distributed Dead Load of CHB (𝑀𝐢𝐻𝐡2𝐿
𝐸π‘₯𝑑
𝑀𝐢𝐻𝐡2𝐿
= 9.87 π‘˜π‘/π‘š
𝑀𝐢𝐻𝐡2𝐿
= 9.415 π‘˜π‘/π‘š
𝐸π‘₯𝑑
𝐼𝑛𝑑
𝐼𝑛𝑑
Ground Floor Level (1L)
o Total Uniform Dead Load (π‘žπ·1𝐿 )
π‘žπ·1𝐿 = 7.38 π‘˜π‘ƒπ‘Ž
o Uniformly Distributed Dead Load of Beams (π‘€π΅π‘’π‘Žπ‘š )
π‘€π΅π‘’π‘Žπ‘š = 5.664 π‘˜π‘/π‘š
o Uniformly Distributed Dead Load of CHB (𝑀𝐢𝐻𝐡1𝐿
𝐸π‘₯𝑑
𝑀𝐢𝐻𝐡1𝐿
= 11.28 π‘˜π‘/π‘š
𝑀𝐢𝐻𝐡1𝐿
= 10.76 π‘˜π‘/π‘š
𝐸π‘₯𝑑
𝐼𝑛𝑑
𝐼𝑛𝑑
Floor Level
π‘žπ· (π‘˜π‘ƒπ‘Ž)
π‘€π΅π‘’π‘Žπ‘š (π‘˜π‘/π‘š)
𝑀𝐢𝐻𝐡𝐸π‘₯𝑑 (π‘˜π‘/π‘š)
𝑀𝐢𝐻𝐡𝐼𝑛𝑑 (π‘˜π‘/π‘š)
RD
6.18
5.664
N/A
N/A
3L
6.93
5.664
9.87
9.415
2L
8.17
5.664
9.87
9.415
1L
7.38
5.664
11.28
10.76
Table Figure: Partial Summary of Dead Loads (Incomplete)
Symbols:
RD = Roof Deck
3L = 3rd Floor Level
2L = 2nd Floor Level
1L = Ground Floor Level
π‘žπ· = Total Uniform Dead Load
π‘€π΅π‘’π‘Žπ‘š = Uniformly Distributed Dead Load of Beams
𝑀𝐢𝐻𝐡𝐸π‘₯𝑑 = Uniformly Distributed Dead Load of Exterior CHB
𝑀𝐢𝐻𝐡𝐼𝑛𝑑 = Uniformly Distributed Dead Load of Interior CHB
4.1.4.1.2 Dead Loads on the Frames
4.1.4.1.2.a Longitudinal Exterior Frame (Frame 1)
Total Length = 60 π‘š
Tributary Width (𝑏𝑑 )
𝑏𝑑1 = 2.5 π‘š
Roof Deck (RD)
Total Uniformly Distributed Dead Load on the Roof Deck (𝑀1𝑅𝐷 ),
𝐷
𝑀1𝑅𝐷 = π‘žπ·π‘…π· (𝑏𝑑1 ) + π‘€π΅π‘’π‘Žπ‘š = 6.18 π‘˜π‘ƒπ‘Ž (2.5 π‘š) + 5.664 π‘˜π‘/π‘š
𝐷
𝑀1𝑅𝐷 = 21.114 π‘˜π‘/π‘š
𝐷
3rd Floor Level (3L)
Total Uniformly Distributed Dead Load on the 3rd Floor Level (𝑀13𝐿 ),
𝐷
𝑀13𝐿 = π‘žπ·3𝐿 (𝑏𝑑1 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡3𝐿
𝐷
𝐸π‘₯𝑑
= 6.93 π‘˜π‘ƒπ‘Ž (2.5 π‘š) + 5.664 π‘˜π‘/π‘š + 9.87 π‘˜π‘/π‘š
𝑀13𝐿 = 32.859 π‘˜π‘/π‘š
𝐷
2nd Floor Level (2L)
Total Uniformly Distributed Dead Load on the 2nd Floor Level (𝑀12𝐿 ),
𝐷
𝑀12𝐿 = π‘žπ·2𝐿 (𝑏𝑑1 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡2𝐿
𝐷
𝐸π‘₯𝑑
= 8.17 π‘˜π‘ƒπ‘Ž (2.5 π‘š) + 5.664 π‘˜π‘/π‘š + 9.87 π‘˜π‘/π‘š
𝑀12𝐿 = 35.959 π‘˜π‘/π‘š
𝐷
Ground Floor Level (1L)
Total Uniformly Distributed Dead Load on the Ground Floor Level (𝑀11𝐿 ),
𝐷
𝑀11𝐿 = π‘žπ·1𝐿 (𝑏𝑑1 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡1𝐿
𝐷
𝐸π‘₯𝑑
= 7.38 π‘˜π‘ƒπ‘Ž (2.5 π‘š) + 5.664 π‘˜π‘/π‘š + 11.28 π‘˜π‘/π‘š
𝑀11𝐿 = 35.394 π‘˜π‘/π‘š
𝐷
4.1.4.1.2.b Longitudinal Interior Frame (Frame 2)
Total Length = 60 π‘š
Tributary Width (𝑏𝑑 )
𝑏𝑑2 = 5 π‘š
Roof Deck (RD)
Uniformly Distributed Dead Load on the Roof Deck (𝑀2𝑅𝐷 ),
𝐷
𝑀2𝑅𝐷 = π‘žπ·π‘…π· (𝑏𝑑2 ) + π‘€π΅π‘’π‘Žπ‘š = 6.18 π‘˜π‘ƒπ‘Ž (5 π‘š) + 5.664 π‘˜π‘/π‘š
𝐷
𝑀2𝑅𝐷 = 36.564 π‘˜π‘/π‘š
𝐷
3rd Floor Level (3L)
Total Uniformly Distributed Dead Load on the 3rd Floor Level (𝑀23𝐿 ),
𝐷
𝑀23𝐿 = π‘žπ·3𝐿 (𝑏𝑑2 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡3𝐿
𝐷
𝐼𝑛𝑑
= 6.93 π‘˜π‘ƒπ‘Ž (5 π‘š) + 5.664 π‘˜π‘/π‘š + 9.415 π‘˜π‘/π‘š
𝑀23𝐿 = 49.729 π‘˜π‘/π‘š
𝐷
2nd Floor Level (2L)
Total Uniformly Distributed Dead Load on the 2nd Floor Level (𝑀22𝐿 ),
𝐷
𝑀22𝐿 = π‘žπ·2𝐿 (𝑏𝑑2 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡2𝐿
𝐷
𝐼𝑛𝑑
= 8.17 π‘˜π‘ƒπ‘Ž (5 π‘š) + 5.664 π‘˜π‘/π‘š + 9.415 π‘˜π‘/π‘š
𝑀22𝐿 = 55.929 π‘˜π‘/π‘š
𝐷
Ground Floor Level (1L)
Total Uniformly Distributed Dead Load on the Ground Floor Level (𝑀21𝐿 ),
𝐷
𝑀21𝐿 = π‘žπ·1𝐿 (𝑏𝑑2 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡1𝐿
𝐷
𝐼𝑛𝑑
= 7.38 π‘˜π‘ƒπ‘Ž (5 π‘š) + 5.664 π‘˜π‘/π‘š + 10.76 π‘˜π‘/π‘š
𝑀11𝐿 = 53.324 π‘˜π‘/π‘š
𝐷
4.1.4.1.2.c Transverse Exterior Frame (Frame 3)
Total Length = 30 π‘š
Tributary Width (𝑏𝑑 )
𝑏𝑑3 = 5 π‘š
Roof Deck (RD)
Uniformly Distributed Dead Load on the Roof Deck (𝑀3𝑅𝐷 )
𝐷
𝑀3𝑅𝐷 = π‘žπ·π‘…π· (𝑏𝑑3 ) + π‘€π΅π‘’π‘Žπ‘š = 6.18 π‘˜π‘ƒπ‘Ž (5 π‘š) + 5.664 π‘˜π‘/π‘š
𝐷
𝑀3𝑅𝐷 = 36.564 π‘˜π‘/π‘š
𝐷
3rd Floor Level (3L)
Total Uniformly Distributed Dead Load on the 3rd Floor Level (𝑀33𝐿 )
𝐷
𝑀33𝐿 = π‘žπ·3𝐿 (𝑏𝑑3 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡3𝐿
𝐷
𝐸π‘₯𝑑
= 6.93 π‘˜π‘ƒπ‘Ž (5 π‘š) + 5.664 π‘˜π‘/π‘š + 9.87 π‘˜π‘/π‘š
𝑀33𝐿 = 50.184 π‘˜π‘/π‘š
𝐷
2nd Floor Level (2L)
Total Uniformly Distributed Dead Load on the 2nd Floor Level (𝑀32𝐿 )
𝐷
𝑀32𝐿 = π‘žπ·2𝐿 (𝑏𝑑3 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡2𝐿
𝐷
𝐸π‘₯𝑑
= 8.17 π‘˜π‘ƒπ‘Ž (5 π‘š) + 5.664 π‘˜π‘/π‘š + 9.87 π‘˜π‘/π‘š
𝑀32𝐿 = 56.384 π‘˜π‘/π‘š
𝐷
Ground Floor Level (1L)
Total Uniformly Distributed Dead Load on the Ground Floor Level (𝑀31𝐿 )
𝐷
𝑀31𝐿 = π‘žπ·1𝐿 (𝑏𝑑3 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡1𝐿
𝐷
𝐸π‘₯𝑑
= 7.38 π‘˜π‘ƒπ‘Ž (5 π‘š) + 5.664 π‘˜π‘/π‘š + 11.28 π‘˜π‘/π‘š
𝑀31𝐿 = 53.844 π‘˜π‘/π‘š
𝐷
4.1.4.1.2.d Transverse Interior Frame (Frame 4)
Total Length = 30 π‘š
Tributary Width (𝑏𝑑 )
𝑏𝑑4 = 10 π‘š
Roof Deck (RD)
Uniformly Distributed Dead Load on the Roof Deck (𝑀4𝑅𝐷 )
𝐷
𝑀4𝑅𝐷 = π‘žπ·π‘…π· (𝑏𝑑4 ) + π‘€π΅π‘’π‘Žπ‘š = 6.18 π‘˜π‘ƒπ‘Ž (10 π‘š) + 5.664 π‘˜π‘/π‘š
𝐷
𝑀4𝑅𝐷 = 67.464 π‘˜π‘/π‘š
𝐷
3rd Floor Level (3L)
Total Uniformly Distributed Dead Load on the 3rd Floor Level (𝑀43𝐿 )
𝐷
𝑀43𝐿 = π‘žπ·3𝐿 (𝑏𝑑4 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡3𝐿
𝐷
𝐼𝑛𝑑
= 6.93 π‘˜π‘ƒπ‘Ž (10 π‘š) + 5.664 π‘˜π‘/π‘š + 9.415 π‘˜π‘/π‘š
𝑀43𝐿 = 84.379 π‘˜π‘/π‘š
𝐷
2nd Floor Level (2L)
Total Uniformly Distributed Dead Load on the 2nd Floor Level (𝑀32𝐿 )
𝐷
𝑀42𝐿 = π‘žπ·2𝐿 (𝑏𝑑4 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡2𝐿
𝐷
𝐼𝑛𝑑
= 8.17 π‘˜π‘ƒπ‘Ž (10 π‘š) + 5.664 π‘˜π‘/π‘š + 9.415 π‘˜π‘/π‘š
𝑀42𝐿 = 96.779 π‘˜π‘/π‘š
𝐷
Ground Floor Level (1L)
Total Uniformly Distributed Dead Load on the Ground Floor Level (𝑀41𝐿 )
𝐷
𝑀41𝐿 = π‘žπ·1𝐿 (𝑏𝑑4 ) + π‘€π΅π‘’π‘Žπ‘š + 𝑀𝐢𝐻𝐡1𝐿
𝐷
𝐼𝑛𝑑
= 7.38 π‘˜π‘ƒπ‘Ž (10 π‘š) + 5.664 π‘˜π‘/π‘š + 10.76 π‘˜π‘/π‘š
𝑀41𝐿 = 90.224 π‘˜π‘/π‘š
𝐷
Summary of Dead Loads
Uniformly Distributed Dead Load, 𝑀𝐷 (π‘˜π‘/π‘š)
Floor Level
Longitudinal Frames
Transverse Frames
Frame 1 (Exterior)
Frame 2 (Interior)
Frame 3 (Exterior)
Frame 4 (Interior)
Roof Deck
21.114
36.564
36.564
67.464
3rd Floor
32.859
49.729
50.184
84.379
2nd Floor
35.959
55.929
56.384
96.779
Ground Floor
35.394
53.324
53.844
90.224
Figure: Uniformly Distributed Dead Loads on the Longitudinal Exterior Frame (Frame 1)
Figure: Uniformly Distributed Dead Loads on the Longitudinal Interior Frame (Frame 2)
Figure: Uniformly Distributed Dead Loads on the Transverse Exterior Frame (Frame 3)
Figure: Uniformly Distributed Dead Loads on the Transverse Interior Frame (Frame 4)
4.1.4.2 Live Loads
Live loads refer to the transient forces that move through a building or act on any of its
structural elements. They include the possible or expected weight of people, furniture,
appliances, cars and other vehicles, and equipment. Live load shall be the maximum loads
expected by the intended use or occupancy but in no case shall be less than the load required by
this section.
Roof Deck (RD)
From Table 205-1 Minimum Uniform and Concentrated Live Loads
Residential
Decks = 1.9 π‘˜π‘ƒπ‘Ž
Therefore, the Total Uniform Live Load on the Roof Deck (π‘žπΏπ‘…π· ),
π‘žπΏπ‘…π· = Residential (Deck)
π‘žπΏπ‘…π· = 1.9 π‘˜π‘ƒπ‘Ž
3rd Floor Level (3L)
From Table 205-1 Minimum Uniform and Concentrated Live Loads
Residential
Basic Floor Area = 1.9 π‘˜π‘ƒπ‘Ž
Storage
Light = 6 π‘˜π‘ƒπ‘Ž
Therefore, the Total Uniform Live Load on the 3rd Floor Level (π‘žπΏ3𝐿 ),
π‘žπΏ3𝐿 = Residential Area + Light Storage = 1.9 π‘˜π‘ƒπ‘Ž + 6 π‘˜π‘ƒπ‘Ž
π‘žπΏ3𝐿 = 7.9 π‘˜π‘ƒπ‘Ž
2nd Floor Level (2L)
From Table 205-1 Minimum Uniform and Concentrated Live Loads
Access Floor Systems
Office Use = 2.4 π‘˜π‘ƒπ‘Ž
Bowling Alleys, Poolrooms and Similar Recreational Areas = 3.6 π‘˜π‘ƒπ‘Ž
Theaters, assembly areas and auditorium
Movable Seats = 4.8 π‘˜π‘ƒπ‘Ž
Therefore, the Total Uniform Live Load on the 2nd Floor Level (π‘žπΏ2𝐿 ),
π‘žπΏ2𝐿 = Office Area + Recreational Areas + Assembly Area
= 2.4 π‘˜π‘ƒπ‘Ž + 3.6 π‘˜π‘ƒπ‘Ž + 4.8 π‘˜π‘ƒπ‘Ž
π‘žπΏ2𝐿 = 10.8 π‘˜π‘ƒπ‘Ž
Ground Floor Level (1L)
From Table 205-1 Minimum Uniform and Concentrated Live Loads
Bowling Alleys, Poolrooms and Similar Recreational Areas = 3.6 π‘˜π‘ƒπ‘Ž
Stores
Retail = 4.8 π‘˜π‘ƒπ‘Ž
Storage
Light = 6 π‘˜π‘ƒπ‘Ž
Office
Lobbies and Ground Floor Corridors = 4.8 π‘˜π‘ƒπ‘Ž
Therefore, the Total Uniform Live Load on the Ground Floor Level (π‘žπΏ1𝐿 ),
π‘žπΏ1𝐿 = Recreational Areas + Retail Areas + Light Storage
+ Ground Floor Corridors
π‘žπΏ1𝐿 = 3.6 π‘˜π‘ƒπ‘Ž + 4.8 π‘˜π‘ƒπ‘Ž + 6 π‘˜π‘ƒπ‘Ž + 4.8 π‘˜π‘ƒπ‘Ž
π‘žπΏ1𝐿 = 19.2 π‘˜π‘ƒπ‘Ž
4.1.4.2.1 Summary of Uniform Loadings
Roof Deck (RD)
o Total Uniform Live Load (π‘žπΏπ‘…π· )
π‘žπΏπ‘…π· = 1.9 π‘˜π‘ƒπ‘Ž
3rd Floor Level (3L)
o Total Uniform Live Load (π‘žπΏ3𝐿 )
π‘žπΏ3𝐿 = 7.9 π‘˜π‘ƒπ‘Ž
2nd Floor Level (2L)
o Total Uniform Live Load (π‘žπΏ2𝐿 )
π‘žπΏ2𝐿 = 10.8 π‘˜π‘ƒπ‘Ž
Ground Floor Level (1L)
o Total Uniform Live Load (π‘žπΏ1𝐿 )
π‘žπΏ1𝐿 = 19.2 π‘˜π‘ƒπ‘Ž
Floor Level
π‘žπΏ (π‘˜π‘ƒπ‘Ž)
RD
1.9
3L
7.9
2L
10.8
1L
19.2
Table Figure: Summary of Uniform Live Loads
Symbols:
RD = Roof Deck
3L = 3rd Floor Level
2L = 2nd Floor Level
1L = Ground Floor Level
π‘žπΏ = Total Uniform Live Load
Loadings on the Frames
Longitudinal Exterior Frame (Frame 1)
Total Length = 60 π‘š
Tributary Width (𝑏𝑑 )
𝑏𝑑1 = 2.5 π‘š
Length of the Beams (𝐿𝑏1 )
𝐿𝑏1 = 10 π‘š
Tributary Area of the Beams (𝐴 𝑇1 )
𝐴 𝑇1 = 𝐿𝑏1 (𝑏𝑑1 ) = 10 π‘š (2.5π‘š) = 25 π‘š2
Influence Area of the Beams (𝐴𝐼1 )
𝐴𝐼1 = 2𝐴𝑇1 = 2(25 π‘š2 ) = 50 π‘š2
12
Since 𝐴𝐼1 = 50 π‘š2 > 40 π‘š2 , ∴ 𝐿𝑖𝑣𝑒 π‘™π‘œπ‘Žπ‘‘ π‘Ÿπ‘’π‘‘π‘’π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘
Roof Deck (RD)
π‘žπΏπ‘…π· = πΏπ‘œ = 1.9 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (Section 205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼1
𝐿 = 1.9 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 1.702962 π‘˜π‘ƒπ‘Ž ≈ 1.703 π‘˜π‘ƒπ‘Ž
√50
𝐿 = 1.703 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(1.9 π‘˜π‘ƒπ‘Ž) = 0.95 π‘˜π‘ƒπ‘Ž
∴ π‘ž′𝐿𝑅𝐷 = 𝐿 = 1.703 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the Roof Deck (𝑀1𝑅𝐷 )
𝐿
𝑀1𝑅𝐷 = π‘ž′𝐿𝑅𝐷 (𝑏𝑑1 ) = 1.703 π‘˜π‘ƒπ‘Ž (2.5 π‘š) = 4.257404 π‘˜π‘/π‘š
𝐿
𝑀1𝑅𝐷 = 4.2574 π‘˜π‘/π‘š
𝐿
3rd Floor Level (3L)
π‘žπΏ3𝐿 = πΏπ‘œ = 7.9 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼1
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 7.080735 π‘˜π‘ƒπ‘Ž ≈ 7.0807 π‘˜π‘ƒπ‘Ž
𝐿 = 7.9 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
√50
𝐿 = 7.0807 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(7.9 π‘˜π‘ƒπ‘Ž) = 3.95 π‘˜π‘ƒπ‘Ž
∴ π‘ž′𝐿3𝐿 = 𝐿 = 7.0807 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the 3rd Floor Level (𝑀13𝐿 )
𝐿
𝑀13𝐿 = π‘ž′𝐿3𝐿 (𝑏𝑑1 ) = 7.0807 π‘˜π‘ƒπ‘Ž (2.5 π‘š) = 17.710838 π‘˜π‘/π‘š
𝐿
𝑀13𝐿 = 17.7018 π‘˜π‘/π‘š
𝐿
2nd Floor Level (2L)
π‘žπΏ2𝐿 = πΏπ‘œ = 10.8 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼1
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 9.68 π‘˜π‘ƒπ‘Ž
𝐿 = 10.8 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
√50
𝐿 = 9.68 π‘˜π‘ƒπ‘Ž > 50%πΏπ‘œ = 0.50(10.8 π‘˜π‘ƒπ‘Ž) = 5.4 π‘˜π‘ƒπ‘Ž
∴ π‘ž′𝐿2𝐿 = 𝐿 = 9.68 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the 2nd Floor Level (𝑀12𝐿 )
𝐿
𝑀12𝐿 = π‘ž′𝐿2𝐿 (𝑏𝑑1 ) = 9.68 π‘˜π‘ƒπ‘Ž (2.5 π‘š)
𝐿
𝑀12𝐿 = 24.20 π‘˜π‘/π‘š
𝐿
Ground Floor Level (1L)
π‘žπΏ1𝐿 = πΏπ‘œ = 19.2 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼1
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 17.208875 π‘˜π‘ƒπ‘Ž ≈ 17.209 π‘˜π‘ƒπ‘Ž
𝐿 = 19.2 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
√50
𝐿 = 17.209 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(19.2 π‘˜π‘ƒπ‘Ž) = 9.6 π‘˜π‘ƒπ‘Ž
∴ π‘ž′𝐿1𝐿 = 𝐿 = 17.209 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the Ground Floor Level (𝑀11𝐿 )
𝐿
𝑀11𝐿 = π‘ž′𝐿1𝐿 (𝑏𝑑1 ) = 17.209 π‘˜π‘ƒπ‘Ž (2.5 π‘š) = 43.0221887 π‘˜π‘/π‘š
𝐿
𝑀11𝐿 = 43.022 π‘˜π‘/π‘š
𝐿
Longitudinal Interior Frame (Frame 2)
Total Length = 60 π‘š
Tributary Width (𝑏𝑑2 )
𝑏𝑑2 = 5 π‘š
Length of the Beams (𝐿𝑏2 )
𝐿𝑏2 = 10 π‘š
Tributary Area of the Beams (𝐴 𝑇2 )
𝐴 𝑇2 = 𝐿𝑏2 (𝑏𝑑2 ) = 10 π‘š (5 π‘š) = 50 π‘š2
Influence Area of the Beams (𝐴𝐼2 )
𝐴𝐼2 = 2𝐴𝑇2 = 2 (50 π‘š2 ) = 100 π‘š2
𝑆𝑖𝑛𝑐𝑒 𝐴𝐼2 = 100 π‘š2 > 40 π‘š2 , ∴ 𝐿𝑖𝑣𝑒 π‘™π‘œπ‘Žπ‘‘ π‘Ÿπ‘’π‘‘π‘’π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘
Roof Deck (RD)
π‘žπΏπ‘…π· = πΏπ‘œ = 1.9 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼2
𝐿 = 1.9 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
(πΈπ‘žπ‘›. 205 − 3)
1
√100
)] = 1.3433 π‘˜π‘ƒπ‘Ž
𝐿 = 1.3433 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(1.9 π‘˜π‘ƒπ‘Ž) = 0.95 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′𝐿𝑅𝐷 = 𝐿 = 1.3433 π‘˜π‘ƒπ‘Ž
Uniformly Distributed Live Load on the Roof Deck (𝑀2𝑅𝐷 )
𝐿
𝑀2𝑅𝐷 = π‘ž′′𝐿𝑅𝐷 (𝑏𝑑2 ) = 1.3433 π‘˜π‘ƒπ‘Ž (5 π‘š)
𝐿
𝑀2𝑅𝐷 = 6.7165 π‘˜π‘/π‘š
𝐿
3rd Floor Level (3L)
π‘žπΏ3𝐿 = πΏπ‘œ = 7.9 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼2
𝐿 = 7.9 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
(πΈπ‘žπ‘›. 205 − 3)
1
√100
)] = 5.5853 π‘˜π‘ƒπ‘Ž
𝐿 = 5.5853 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(7.9 π‘˜π‘ƒπ‘Ž) = 3.95 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′𝐿3𝐿 = 𝐿 = 5.5853 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the 3rd Floor Level (𝑀23𝐿 )
𝐿
𝑀23𝐿 = π‘ž′′𝐿3𝐿 (𝑏𝑑2 ) = 5.5853 π‘˜π‘ƒπ‘Ž (5 π‘š)
𝐿
𝑀23𝐿 = 27.9265 π‘˜π‘/π‘š
𝐿
2nd Floor Level (2L)
π‘žπΏ2𝐿 = πΏπ‘œ = 10.8 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼2
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 7.6356 π‘˜π‘ƒπ‘Ž
𝐿 = 10.8 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
√100
𝐿 = 7.6356 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(10.8 π‘˜π‘ƒπ‘Ž) = 5.4 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′𝐿2𝐿 = 𝐿 = 7.6356 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the 2nd Floor Level (𝑀22𝐿 )
𝐿
𝑀22𝐿 = π‘ž′′𝐿2𝐿 (𝑏𝑑2 ) = 7.6356 π‘˜π‘ƒπ‘Ž (5 π‘š)
𝐿
𝑀22𝐿 = 38.178 π‘˜π‘/π‘š
𝐿
Ground Floor Level (1L)
π‘žπΏ1𝐿 = πΏπ‘œ = 19.2 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼2
𝐿 = 19.2 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
(πΈπ‘žπ‘›. 205 − 3)
1
√100
)] = 13.5744 π‘˜π‘ƒπ‘Ž
𝐿 = 13.5744 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(19.2 π‘˜π‘ƒπ‘Ž) = 9.6 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′𝐿1𝐿 = 𝐿 = 13.5744 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the Ground Floor Level (𝑀21𝐿 )
𝐿
𝑀21𝐿 = π‘ž′′𝐿1𝐿 (𝑏𝑑2 ) = 13.5744 π‘˜π‘ƒπ‘Ž (5 π‘š)
𝐿
𝑀21𝐿 = 67.872 π‘˜π‘/π‘š
𝐿
Transverse Exterior Frame (Frame 3)
Total Length = 30 π‘š
Tributary Width (𝑏𝑑3 )
𝑏𝑑3 = 5 π‘š
Length of the Beams (𝐿𝑏3 )
𝐿𝑏3 = 5 π‘š
Tributary Area of the Beams (𝐴 𝑇3 )
𝐴 𝑇3 = 𝐿𝑏3 (𝑏𝑑3 ) = 5 π‘š (5 π‘š) = 25 π‘š2
Influence Area of the Beams (𝐴𝐼3 )
𝐴𝐼3 = 2𝐴𝑇3 = 2 (25 π‘š2 ) = 50 π‘š2
𝑆𝑖𝑛𝑐𝑒 𝐴𝐼2 = 50 π‘š2 > 40 π‘š2 , ∴ 𝐿𝑖𝑣𝑒 π‘™π‘œπ‘Žπ‘‘ π‘Ÿπ‘’π‘‘π‘’π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘
Roof Deck (RD)
π‘žπΏπ‘…π· = πΏπ‘œ = 1.9 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼2
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 1.703 π‘˜π‘ƒπ‘Ž
𝐿 = 1.9 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
√50
𝐿 = 1.703 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(1.9 π‘˜π‘ƒπ‘Ž) = 0.95 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′′𝐿𝑅𝐷 = 𝐿 = 1.703 π‘˜π‘ƒπ‘Ž
Uniformly Distributed Live Load on the Roof Deck (𝑀3𝑅𝐷 )
𝐿
𝑀3𝑅𝐷 = π‘ž′′′𝐿𝑅𝐷 (𝑏𝑑3 ) = 1.703 π‘˜π‘ƒπ‘Ž (5 π‘š) = 8.5148082 π‘˜π‘/π‘š
𝐿
𝑀3𝑅𝐷 = 8.5148 π‘˜π‘/π‘š
𝐿
3rd Floor Level (3L)
π‘žπΏ3𝐿 = πΏπ‘œ = 7.9 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼1
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 7.080735 π‘˜π‘ƒπ‘Ž ≈ 7.0807 π‘˜π‘ƒπ‘Ž
𝐿 = 7.9 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
√50
𝐿 = 7.0807 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(7.9 π‘˜π‘ƒπ‘Ž) = 3.95 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′′𝐿3𝐿 = 𝐿 = 7.0807 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the 3rd Floor Level (𝑀33𝐿 )
𝐿
𝑀33𝐿 = π‘ž′′′𝐿3𝐿 (𝑏𝑑3 ) = 7.0807 π‘˜π‘ƒπ‘Ž (5 π‘š) = 35.403676 π‘˜π‘/π‘š
𝐿
𝑀33𝐿 = 35.4037 π‘˜π‘/π‘š
𝐿
2nd Floor Level (2L)
π‘žπΏ2𝐿 = πΏπ‘œ = 10.8 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼1
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 9.68 π‘˜π‘ƒπ‘Ž
𝐿 = 10.8 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
√50
𝐿 = 9.68 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(10.8 π‘˜π‘ƒπ‘Ž) = 5.4 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′′𝐿2𝐿 = 𝐿 = 9.68 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the 2nd Floor Level (𝑀32𝐿 )
𝐿
𝑀32𝐿 = π‘ž′′′𝐿2𝐿 (𝑏𝑑3 ) = 9.68 π‘˜π‘ƒπ‘Ž (5 π‘š)
𝐿
𝑀32𝐿 = 48.40 π‘˜π‘/π‘š
𝐿
Ground Floor Level (1L)
π‘žπΏ1𝐿 = πΏπ‘œ = 19.2 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼1
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 17.208875 π‘˜π‘ƒπ‘Ž ≈ 17.209 π‘˜π‘ƒπ‘Ž
𝐿 = 19.2 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
√50
𝐿 = 17.209 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(19.2 π‘˜π‘ƒπ‘Ž) = 9.6 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′′𝐿1𝐿 = 𝐿 = 17.209 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the Ground Floor Level (𝑀31𝐿 )
𝐿
𝑀31𝐿 = π‘ž′′′𝐿1𝐿 (𝑏𝑑3 ) = 17.209 π‘˜π‘ƒπ‘Ž (5 π‘š) = 86.044377 π‘˜π‘/π‘š
𝐿
𝑀31𝐿 = 86.0444 π‘˜π‘/π‘š
𝐿
Transverse Interior Frame (Frame 4)
Total Length = 30 π‘š
Tributary Width (𝑏𝑑4 )
𝑏𝑑4 = 10 π‘š
Length of the Beams (𝐿𝑏4 )
𝐿𝑏4 = 5 π‘š
Tributary Area of the Beams (𝐴 𝑇4 )
𝐴 𝑇4 = 𝐿𝑏4 (𝑏𝑑4 ) = 5 π‘š (10 π‘š) = 50 π‘š2
Influence Area of the Beams (𝐴𝐼4 )
𝐴𝐼4 = 2𝐴𝑇4 = 2(50 π‘š2 ) = 100 π‘š2
𝑆𝑖𝑛𝑐𝑒 𝐴𝐼4 = 100 π‘š2 > 40 π‘š2 , ∴ 𝐿𝑖𝑣𝑒 π‘™π‘œπ‘Žπ‘‘ π‘Ÿπ‘’π‘‘π‘’π‘π‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘
Roof Deck (RD)
π‘žπΏπ‘…π· = πΏπ‘œ = 1.9 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (Section 205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼2
𝐿 = 1.9 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
(πΈπ‘žπ‘›. 205 − 3)
1
√100
)] = 1.3433 π‘˜π‘ƒπ‘Ž
𝐿 = 1.3433 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(1.9 π‘˜π‘ƒπ‘Ž) = 0.95 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′′′𝐿𝑅𝐷 = 𝐿 = 1.3433 π‘˜π‘ƒπ‘Ž
Uniformly Distributed Live Load on the Roof Deck (𝑀4𝑅𝐷 )
𝐿
𝑀4𝑅𝐷 = π‘ž′′′′𝐿𝑅𝐷 (𝑏𝑑4 ) = 1.3433 π‘˜π‘ƒπ‘Ž (10 π‘š)
𝐿
𝑀4𝑅𝐷 = 13.433 π‘˜π‘/π‘š
𝐿
3rd Floor Level (3L)
π‘žπΏ3𝐿 = πΏπ‘œ = 7.9 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (Section 205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼2
𝐿 = 7.9 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
(πΈπ‘žπ‘›. 205 − 3)
1
√100
)] = 5.5853 π‘˜π‘ƒπ‘Ž
𝐿 = 5.5853 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(7.9 π‘˜π‘ƒπ‘Ž) = 3.95 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′′′𝐿3𝐿 = 𝐿 = 5.5853 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the 3rd Floor Level (𝑀43𝐿 )
𝐿
𝑀43𝐿 = π‘žπΏ3𝐿 (𝑏𝑑4 ) = 5.5853 π‘˜π‘ƒπ‘Ž (10 π‘š)
𝐿
𝑀43𝐿 = 55.853 π‘˜π‘/π‘š
𝐿
2nd Floor Level (2L)
π‘žπΏ2𝐿 = πΏπ‘œ = 10.8 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (Section 205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼2
(πΈπ‘žπ‘›. 205 − 3)
1
)] = 7.6356 π‘˜π‘ƒπ‘Ž
𝐿 = 10.8 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
√100
𝐿 = 7.6356 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(10.8 π‘˜π‘ƒπ‘Ž) = 5.4 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′′′𝐿2𝐿 = 𝐿 = 7.6356 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the 2nd Floor Level (𝑀42𝐿 )
𝐿
𝑀42𝐿 = π‘ž′′′′𝐿2𝐿 (𝑏𝑑4 ) = 10.8 π‘˜π‘ƒπ‘Ž (10 π‘š)
𝐿
𝑀42𝐿 = 76.356 π‘˜π‘/π‘š
𝐿
Ground Floor Level (1L)
π‘žπΏ1𝐿 = πΏπ‘œ = 19.2 π‘˜π‘ƒπ‘Ž
Alternate Floor Live Load Reduction (205.6),
1
𝐿 = πΏπ‘œ [0.25 + 4.57 (
)]
√𝐴𝐼2
𝐿 = 19.2 π‘˜π‘ƒπ‘Ž [0.25 + 4.57 (
1
√100
(205 − 3)
)] = 13.5744 π‘˜π‘ƒπ‘Ž
𝐿 = 13.5744 π‘˜π‘ƒπ‘Ž > 50% πΏπ‘œ = 0.50(19.2 π‘˜π‘ƒπ‘Ž) = 9.6 π‘˜π‘ƒπ‘Ž
∴ π‘ž′′′′𝐿1𝐿 = 𝐿 = 13.5744 π‘˜π‘ƒπ‘Ž
Total Uniformly Distributed Live Load on the Ground Floor Level (𝑀41𝐿 )
𝐿
𝑀41𝐿 = π‘ž′′′′𝐿1𝐿 (𝑏𝑑4 ) = 13.5744 π‘˜π‘ƒπ‘Ž (10 π‘š)
𝐿
𝑀31𝐿 = 135.744 π‘˜π‘/π‘š
𝐿
Uniformly Distributed Live Load, 𝑀𝐿 (π‘˜π‘/π‘š)
Floor Level
Longitudinal Frames
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
Roof Deck
4.2574
6.7165
8.5148
13.433
3rd Floor
17.7018
27.9265
35.4037
55.853
2nd Floor
24.20
38.178
48.40
76.356
Ground Floor
43.022
67.872
86.0444
135.744
The calculation of live loads on the ground floor are determined however, will not be
considered due to transferring all the ground floor loads to the ground beneath the slab.
Therefore, ground floor live loads can be neglected in structural analysis.
Summary of Live Loads
Uniformly Distributed Live Load, 𝑀𝐿 (π‘˜π‘/π‘š)
Floor Level
Longitudinal Frames
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
Roof Deck
4.2574
6.7165
8.5148
13.433
3rd Floor
17.7018
27.9265
35.4037
55.853
2nd Floor
24.20
38.178
48.40
76.356
Ground Floor
N/A
N/A
N/A
N/A
Note:
Roof Deck: 3rd Floor Level Ratio = 19/79
3rd Floor : 2nd Floor Level Ratio = 79/108
2nd Floor : Ground Floor Level Ratio = 9/16
Longitudinal Exterior Frame (Frame 1)
Figure: Uniformly Distributed Live Load on the Longitudinal Exterior Frame (Frame 1):
Full Live Load
Figure: Uniformly Distributed Live Load on the Longitudinal Exterior Frame (Frame 1):
Live Load Case 1
Figure: Uniformly Distributed Live Load on the Longitudinal Exterior Frame (Frame 1):
Live Load Case 2
Figure: Uniformly Distributed Live Load on the Longitudinal Exterior Frame (Frame 1):
Live Load Case 3
Figure: Uniformly Distributed Live Load on the Longitudinal Exterior Frame (Frame 1):
Live Load Case 4
Figure: Uniformly Distributed Live Load on the Longitudinal Exterior Frame (Frame 1):
Live Load Case 5
Figure: Uniformly Distributed Live Load on the Longitudinal Exterior Frame (Frame 1):
Live Load Case 6
Longitudinal Interior Frame (Frame 2)
Figure: Uniformly Distributed Live Load on the Longitudinal Interior Frame (Frame 2):
Full Live Load
Figure: Uniformly Distributed Live Load on the Longitudinal Interior Frame (Frame 2):
Live Load Case 1
Figure: Uniformly Distributed Live Load on the Longitudinal Interior Frame (Frame 2):
Live Load Case 2
Figure: Uniformly Distributed Live Load on the Longitudinal Interior Frame (Frame 2):
Live Load Case 3
Figure: Uniformly Distributed Live Load on the Longitudinal Interior Frame (Frame 2):
Live Load Case 4
Figure: Uniformly Distributed Live Load on the Longitudinal Interior Frame (Frame 2):
Live Load Case 5
Figure: Uniformly Distributed Live Load on the Longitudinal Interior Frame (Frame 2):
Live Load Case 6
Transverse Exterior Frame (Frame 3)
Figure: Uniformly Distributed Live Load on the Transverse Exterior Frame (Frame 3):
Full Live Load
Figure: Uniformly Distributed Live Load on the Transverse Exterior Frame (Frame 3):
Live Load Case 1
Figure: Uniformly Distributed Live Load on the Transverse Exterior Frame (Frame 3):
Live Load Case 2
Figure: Uniformly Distributed Live Load on the Transverse Exterior Frame (Frame 3):
Live Load Case 3
Figure: Uniformly Distributed Live Load on the Transverse Exterior Frame (Frame 3):
Live Load Case 4
Figure: Uniformly Distributed Live Load on the Transverse Exterior Frame (Frame 3):
Live Load Case 5
Figure: Uniformly Distributed Live Load on the Transverse Exterior Frame (Frame 3):
Live Load Case 6
Transverse Interior Frame (Frame 4)
Figure: Uniformly Distributed Live Load on the Transverse Interior Frame (Frame 4):
Full Live Load
Figure: Uniformly Distributed Live Load on the Transverse Interior Frame (Frame 4):
Live Load Case 1
Figure: Uniformly Distributed Live Load on the Transverse Interior Frame (Frame 4):
Live Load Case 2
Figure: Uniformly Distributed Live Load on the Transverse Interior Frame (Frame 4):
Live Load Case 3
Figure: Uniformly Distributed Live Load on the Transverse Interior Frame (Frame 4):
Live Load Case 4
Figure: Uniformly Distributed Live Load on the Transverse Interior Frame (Frame 4):
Live Load Case 5
Figure: Uniformly Distributed Live Load on the Transverse Interior Frame (Frame 4):
Live Load Case 6
Roof Live Load
From Table 205-3 Minimum Roof Live Loads
Flat Roof = 0.60 π‘˜π‘ƒπ‘Ž
Therefore, the Total Uniform Roof Live Load on the Roof Deck (π‘žπΏπ‘Ÿ 𝑅𝐷 ),
π‘žπΏπ‘Ÿ 𝑅𝐷 = Flat Roof
π‘žπΏπ‘Ÿ 𝑅𝐷 = 0.60 π‘˜π‘ƒπ‘Ž
Longitudinal Exterior Frame (Frame 1)
Tributary Width (𝑏𝑑 )
𝑏𝑑1 = 2.5 π‘š
Total Uniformly Distributed Roof Live Load on the Roof Deck
𝑀1𝑅𝐷
πΏπ‘Ÿ
= π‘žπΏπ‘Ÿ 𝑅𝐷 (𝑏𝑑1 ) = 0.60 π‘˜π‘ƒπ‘Ž (2.5 π‘š)
𝑀1𝑅𝐷
πΏπ‘Ÿ
= 1.5 π‘˜π‘/π‘š
Longitudinal Interior Frame (Frame 2)
Tributary Width (𝑏𝑑2 )
𝑏𝑑2 = 5 π‘š
Uniformly Distributed Roof Live Load on the Roof Deck
𝑀2𝑅𝐷
πΏπ‘Ÿ
= π‘žπΏπ‘Ÿ 𝑅𝐷 (𝑏𝑑2 ) = 0.60 π‘˜π‘ƒπ‘Ž (5 π‘š)
𝑀2𝑅𝐷
πΏπ‘Ÿ
= 3 π‘˜π‘/π‘š
Transverse Exterior Frame (Frame 3)
Tributary Width (𝑏𝑑3 )
𝑏𝑑3 = 5 π‘š
Uniformly Distributed Roof Live Load on the Roof Deck
𝑀3𝑅𝐷
πΏπ‘Ÿ
= π‘žπΏπ‘Ÿ 𝑅𝐷 (𝑏𝑑3 ) = 0.60 π‘˜π‘ƒπ‘Ž (5 π‘š)
𝑀3𝑅𝐷
πΏπ‘Ÿ
= 3 π‘˜π‘/π‘š
Transverse Interior Frame (Frame 4)
Tributary Width (𝑏𝑑4 )
𝑏𝑑4 = 10 π‘š
Uniformly Distributed Roof Live Load on the Roof Deck
𝑀4𝑅𝐷
πΏπ‘Ÿ
= π‘žπΏπ‘Ÿ 𝑅𝐷 (𝑏𝑑4 ) = 0.60 π‘˜π‘ƒπ‘Ž (10 π‘š)
𝑀4𝑅𝐷
πΏπ‘Ÿ
= 6 π‘˜π‘/π‘š
Summary of Roof Live Loads
Uniformly Distributed Roof Live Load, π‘€πΏπ‘Ÿ (π‘˜π‘/π‘š)
Floor Level
Longitudinal Frames
Roof Deck
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
1.5
3
3
6
Figure: Uniformly Distributed Roof Live Load on the Longitudinal Exterior Frame (Frame
1)
Figure: Uniformly Distributed Roof Live Load on the Longitudinal Interior Frame (Frame 2)
Figure: Uniformly Distributed Roof Live Load on the Transverse Exterior Frame (Frame 3)
Figure: Uniformly Distributed Roof Live Load on the Transverse Interior Frame (Frame 4)
Wind Load
Wind loads refer to any pressure or forces that the wind exerts on a building or
structure.
General Requirements (Section 207A):
Use to determine the basic parameters for determining wind loads on Main WindForce Resisting System (MWFRS) for enclosed, partially enclosed, and open building of all
heights. The following steps are taken from Table 207B.2-1.
(1) Risk Category of Building or Other Structure
Occupancy Category: IV (Standard Occupancy)
(π‘‡π‘Žπ‘π‘™π‘’ 103 − 1)
(2) Basic Wind Speed, V (Figure 207A.5 – 1A, B or C)
𝑉 = 250 π‘˜π‘β„Ž =
625 π‘š
π‘š
= 69.4444
9 𝑠
𝑠
(πΉπ‘–π‘”π‘’π‘Ÿπ‘’ 207𝐴. 5 − 1𝐴)
(3) Wind Directionality Factor, 𝐾𝑑 (Section 207A.6)
𝐾𝑑 = 0.85
(π‘‡π‘Žπ‘π‘™π‘’ 207𝐴. 6 − 1)
(4) Exposure Category (Section 207A.7.3)
Exposure Category: B
(π‘†π‘’π‘π‘‘π‘–π‘œπ‘› 207𝐴. 7.3)
(5) Topographic factor, 𝐾𝑧𝑑 (Section 207A.8)
𝐾𝑧𝑑 = 1.0
(π‘†π‘’π‘π‘‘π‘–π‘œπ‘› 207𝐴. 8.2)
(6) Gust Effect Factor (Section 207A.9)
𝐺 = 0.85
(π‘†π‘’π‘π‘‘π‘–π‘œπ‘› 207𝐴. 9.1)
(7) Enclosure Classification (Section 207A.10)
Partially Enclosed Building
(π‘†π‘’π‘π‘‘π‘–π‘œπ‘› 207𝐴. 1.2.1)
(8) Internal Pressure Coefficient, 𝐺𝐢𝑝𝑖 (Section 207A.11)
𝐺𝐢𝑝𝑖 = ± 0.55
(π‘‡π‘Žπ‘π‘™π‘’ 207𝐴. 11 − 1)
Figure 207A.5-1A. Basic Wind Speeds for Occupancy Category III, IV and IV Buildings and
Other Structures
Table 207A.6-1: Wind Directionality Factor, 𝐾𝑑
Table 207.A.11-1: Internal Pressure Coefficient, (𝐺𝐢𝑝𝑖 )
Additional Requirements:
(9) Mean Roof Height, β„Ž
β„Ž = 11 π‘š
(10) Velocity Pressure Exposure Coefficient, 𝐾𝑧 π‘œπ‘Ÿ πΎβ„Ž (Section 207B.3.1)
Height above ground level, z (m)
𝐾𝑧 π‘œπ‘Ÿ πΎβ„Ž (Exposure B)
0 – 4.5
0.57
6.0
0.62
7.5 (3L)
0.66
9.0
0.70
11.0 (RD)
0.74 (Interpolated)
12.0
0.76
Table 207B.3-1: Velocity Pressure Exposure Coefficients, 𝐾𝑧 π‘œπ‘Ÿ πΎβ„Ž
Main Wind Force Resisting System – Part 1
(11) Velocity Pressure, π‘žπ‘§ π‘œπ‘Ÿ π‘žβ„Ž (207B.3.2)
π‘žπ‘§ = 0.613𝐾𝑧 𝐾𝑧𝑑 𝐾𝑑 𝑉 2
(
𝑁
)
π‘š2
From ground level to 4.5-m height,
(πΈπ‘žπ‘›. 207𝐡. 3 − 1)
π‘ž0−4.5
2
625
= 0.613𝐾𝑧 𝐾𝑧𝑑 𝐾𝑑 𝑉 = 0.613(0.57)(1.0)(0.85) (
π‘š/𝑠) = 1432.284433 𝑁/π‘š2
9
2
π‘ž0−4.5 = 1432.2844 𝑁/π‘š2
At 6-m height,
π‘ž6.0
2
625
= 0.613𝐾𝑧 𝐾𝑧𝑑 𝐾𝑑 𝑉 = 0.613(0.62)(1.0)(0.85) (
π‘š/𝑠) = 1557.923418 𝑁/π‘š2
9
2
π‘ž6.0 = 1557.9234 𝑁/π‘š2
At 7.5-m height (3L),
π‘ž7.5
2
625
= 0.613𝐾𝑧 𝐾𝑧𝑑 𝐾𝑑 𝑉 = 0.613(0.66)(1.0)(0.85) (
π‘š/𝑠) = 1658.434606 𝑁/π‘š2
9
2
π‘ž7.5 = 1658.4346 𝑁/π‘š2
At 9-m height,
π‘ž9.0
2
625
= 0.613𝐾𝑧 𝐾𝑧𝑑 𝐾𝑑 𝑉 = 0.613(0.70)(1.0)(0.85) (
π‘š/𝑠) = 1758.945795 𝑁/π‘š2
9
2
π‘ž9.0 = 1758.9458 𝑁/π‘š2
At 11-m height (RD),
2
625
π‘ž11.0 = 0.613𝐾𝑧 𝐾𝑧𝑑 𝐾𝑑 𝑉 2 = 0.613(0.74)(1.0)(0.85) (
π‘š/𝑠) = 1859.456983 𝑁/π‘š2
9
π‘žβ„Ž = π‘ž11.0 = 1859.457 𝑁/π‘š2
(12) External Pressure Coefficient, 𝐢𝑝 π‘œπ‘Ÿ 𝐢𝑛 (Figure 207B.4 -1 )
For walls and flat roof,
Case 1 (Parallel : Normal Dimensions to Wind Direction Ratio):
Figure: 60-m Parallel Dimension and 30-m Normal Dimension to Wind Direction
𝐿1 60 π‘š
=
=2
𝐡1 30 π‘š
Case 2 (Parallel : Normal Dimensions to Wind Direction Ratio):
Figure: 30-m Parallel Dimension and 60-m Normal Dimension to Wind Direction
𝐿2 30 π‘š
=
= 0.50
𝐡2 60 π‘š
Wall Pressure Coefficients, 𝐢𝑝
Surface
L/B
𝐢𝑝
Windward Wall
All Values
0.8
2 (Case 1)
-0.3
0.50 (Case 2)
-0.5
All Values
-0.7
Leeward Wall
Side Wall
(13) Wind Pressure, 𝑝, on Each Building Surface
Figure: General Effect of Wind Pressure on the Longitudinal Frame
Figure: General Effect of Wind Pressure on the Transverse Frame
π‘žπ‘– = π‘žβ„Ž = 1859.457 𝑁/π‘š2
𝑝 = π‘žπΊπΆπ‘ − π‘žπ‘– (𝐺𝐢𝑝𝑖 )
(
𝑁
)
π‘š2
𝑝 = π‘žπΊπΆπ‘ − π‘žπ‘– (𝐺𝐢𝑝𝑖 )
(πΈπ‘žπ‘›. 207𝐡. 4 − 1)
(13.1) Windward Wall
𝑝 = π‘žπ‘§ 𝐺𝐢𝑝 − π‘žπ‘– (𝐺𝐢𝑝𝑖 )
Wind Pressure at 0 – 4.5m height,
𝑝0−4.5 = π‘žπ‘§ 𝐺𝐢𝑝 − π‘žπ‘– (𝐺𝐢𝑝𝑖 ) = 1432.2844(0.85)(0.8) − 1859.457(±0.55)
= −48.74792621 π‘ƒπ‘Ž ≈ −πŸ’πŸ–. πŸ•πŸ’πŸ•πŸ— 𝑷𝒂 ; 1996.654755 π‘ƒπ‘Ž ≈ πŸπŸ—πŸ—πŸ”. πŸ”πŸ“πŸ’πŸ– 𝑷𝒂
Wind Pressure at 6-m height,
𝑝6 = π‘žπ‘§ 𝐺𝐢𝑝 − π‘žπ‘– (𝐺𝐢𝑝𝑖 ) = 1557.9234(0.85)(0.8) − 1859.457(±0.55)
𝑝6 = 36.68658359 π‘ƒπ‘Ž ≈ πŸ‘πŸ”. πŸ”πŸ–πŸ”πŸ” 𝑷𝒂 ; 2082.089265 π‘ƒπ‘Ž ≈ πŸπŸŽπŸ–πŸ. πŸŽπŸ–πŸ—πŸ‘ 𝑷𝒂
Wind Pressure at 7.5-m height,
𝑝7.5 = π‘žπ‘§ 𝐺𝐢𝑝 − π‘žπ‘– (𝐺𝐢𝑝𝑖 ) = 1658.4346(0.85)(0.8) − 1859.457(±0.55)
𝑝7.5 = 105.0341914 π‘ƒπ‘Ž ≈ πŸπŸŽπŸ“. πŸŽπŸ‘πŸ’πŸ 𝑷𝒂 ; 2150.436873 π‘ƒπ‘Ž ≈ πŸπŸπŸ“πŸŽ. πŸ’πŸ‘πŸ”πŸ— 𝑷𝒂
Wind Pressure at 9-m height,
𝑝9 = π‘žπ‘§ 𝐺𝐢𝑝 − π‘žπ‘– (𝐺𝐢𝑝𝑖 ) = 1758.9458(0.85)(0.8) − 1859.457(±0.55)
𝑝9 = πŸπŸ•πŸ‘. πŸ‘πŸ–πŸπŸ– 𝑷𝒂 ; 2218.784481 π‘ƒπ‘Ž ≈ πŸπŸπŸπŸ–. πŸ•πŸ–πŸ’πŸ“ 𝑷𝒂
Wind Pressure at 11-m height,
𝑝11 = π‘žπ‘§ 𝐺𝐢𝑝 − π‘žπ‘– (𝐺𝐢𝑝𝑖 ) = 1859.457(0.85)(0.8) − 1859.457(±0.55)
𝑝11 = 241.7294078 π‘ƒπ‘Ž ≈ πŸπŸ’πŸ. πŸ•πŸπŸ—πŸ’ 𝑷𝒂 ; 2287.132089 π‘ƒπ‘Ž ≈ πŸπŸπŸ–πŸ•. πŸπŸ‘πŸπŸ 𝑷𝒂
z (m)
0 – 4.5
𝑁
π‘žπ‘§ ( 2 )
π‘š
G
1432.2844
0.85
𝐢𝑝
0.8
Wind Pressure, 𝑝 (π‘ƒπ‘Ž)
+ 0.55 (𝐺𝐢𝑝𝑖 )
- 0.55 (𝐺𝐢𝑝𝑖 )
-48.7479
1996.6548
6.0
1557.9234
0.85
0.8
36.6866
2082.0893
7.5 (3L)
1658.4346
0.85
0.8
105.0342
2150.4369
9.0
1758.9458
0.85
0.8
173.3818
2218.7845
11.0 (RD)
1859.457
0.85
0.8
241.7294
2287.1321
Table Figure: Wind Pressure on Windward Wall (All Cases)
(13.2) Leeward Wall
Case 1 (Longitudinal Frame)
Wind Pressure at 11-m height,
𝑝11 = π‘žπ‘§ 𝐺𝐢𝑝 − π‘žπ‘– (𝐺𝐢𝑝𝑖 ) = 1859.457(0.85)(−0.3) − 1859.457(±0.55)
𝑝11 = −1496.862871 π‘ƒπ‘Ž ≈ −πŸπŸ’πŸ—πŸ”. πŸ–πŸ”πŸπŸ— 𝑷𝒂 ; 548.53981 π‘ƒπ‘Ž ≈ πŸ“πŸ’πŸ–. πŸ“πŸ‘πŸ—πŸ– 𝑷𝒂
z (m)
11.0 (RD)
𝑁
π‘ž ( 2)
π‘š
G
1859.457
0.85
𝐢𝑝
-0.3
Wind Pressure, 𝑝 (π‘ƒπ‘Ž)
+ 0.55 (𝐺𝐢𝑝𝑖 )
- 0.55 (𝐺𝐢𝑝𝑖 )
-1496.8629
548.5398
Table Figure: Wind Pressure on Leeward Wall (Case 1)
Case 2 (Transverse Frame)
Wind Pressure at 11-m height,
𝑝11 = π‘žπ‘§ 𝐺𝐢𝑝 − π‘žπ‘– (𝐺𝐢𝑝𝑖 ) = 1859.457(0.85)(−0.5) − 1859.457(±0.55)
𝑝11 = −1812.970558 π‘ƒπ‘Ž ≈ −πŸπŸ–πŸπŸ. πŸ—πŸ•πŸŽπŸ” 𝑷𝒂 ; 232.4321229 π‘ƒπ‘Ž ≈ πŸπŸ‘πŸ. πŸ’πŸ‘πŸπŸ 𝑷𝒂
z (m)
11.0 (RD)
𝑁
π‘ž ( 2)
π‘š
G
1859.457
0.85
𝐢𝑝
-0.5
Wind Pressure, 𝑝 (π‘ƒπ‘Ž)
+ 0.55 (𝐺𝐢𝑝𝑖 )
- 0.55 (𝐺𝐢𝑝𝑖 )
-1812.9706
232.4321
Table Figure: Wind Pressure on Leeward Wall (Case 2)
(13.3) Side Wall (For All Cases)
Wind Pressure at 11-m height,
𝑝11 = π‘žπ‘§ 𝐺𝐢𝑝 − π‘žπ‘– (𝐺𝐢𝑝𝑖 ) = 1859.457(0.85)(−0.7) − 1859.457(±0.55)
𝑝11 = −2129.078246 π‘ƒπ‘Ž ≈ −πŸπŸπŸπŸ—. πŸŽπŸ•πŸ–πŸ 𝑷𝒂 ; −83.67556424 π‘ƒπ‘Ž ≈ −πŸ–πŸ‘. πŸ”πŸ•πŸ“πŸ” 𝑷𝒂
z (m)
11.0 (RD)
𝑁
π‘ž ( 2)
π‘š
G
1859.457
0.85
𝐢𝑝
-0.7
Wind Pressure, 𝑝 (π‘ƒπ‘Ž)
+ 0.55 (𝐺𝐢𝑝𝑖 )
- 0.55 (𝐺𝐢𝑝𝑖 )
- 2129.0782
- 83.6756
Table Figure: Wind Pressure on Side Wall (All Cases)
Table Figure: Wall Pressure and Roof Pressure Coefficients
Calculation of Wind Loads at Floor Levels,
From the computed values of wind pressures, the largest value governs.
∴ 𝑝 = 2287.1321 π‘ƒπ‘Ž = 2.2871 π‘˜π‘ƒπ‘Ž
Case 1 (Longitudinal Frames)
Figure: Governing Wind Pressure on the Windward Wall of Longitudinal Frames
Longitudinal Exterior Frame (Frame 1)
Tributary Width
𝑏𝑑 = 2.5 π‘š
π‘ƒπ‘Šπ‘…π·′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (4.375 π‘š2 ) = 10.00620289 π‘˜π‘ = 10.0062 π‘˜π‘
π‘ƒπ‘Š3𝐿′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (8.75 π‘š2 ) = 20.01240578 π‘˜π‘ = 20.0124 π‘˜π‘
π‘ƒπ‘Š2𝐿′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (9.375 π‘š2 ) = 21.44186334 π‘˜π‘ = 21.4419 π‘˜π‘
Floor-to-Floor
Height (π‘š)
Tributary Area,
𝐴𝑇 (π‘š2 )
Lateral Wind
Load (π‘ƒπ‘Š )
3.5
4.375
10.0062 π‘˜π‘
3.5
8.75
20.0124 π‘˜π‘
2L
4
9.375
21.4419 π‘˜π‘
1L
0
N/A
N/A
Floor
Wind Pressure,
𝑝 (π‘˜π‘ƒπ‘Ž)
RD
3L
2.2871
Table Figure: Computation of Lateral Wind Loads on Frame 1
Figure: Lateral Wind Loads on Longitudinal Exterior Frame (Frame 1)
Longitudinal Interior Frame (Frame 2)
Tributary Width
𝑏𝑑 = 5 π‘š
π‘ƒπ‘Šπ‘…π·′′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (8.75 π‘š2 ) = 20.01240578 π‘˜π‘ = 20.0124 π‘˜π‘
π‘ƒπ‘Š3𝐿 ′′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (17.5 π‘š2 ) = 40.02481156 π‘˜π‘ = 40.0248 π‘˜π‘
π‘ƒπ‘Š2𝐿 ′′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (18.75 π‘š2 ) = 42.88372667 π‘˜π‘ = 42.8837 π‘˜π‘
Floor
Wind Pressure,
𝑝 (π‘˜π‘ƒπ‘Ž)
RD
3L
2L
1L
2.2871
Floor-to-Floor
Height (π‘š)
3.5
Tributary
Area (π‘š2 )
8.75
Lateral Wind
Load (π‘ƒπ‘Š )
3.5
17.5
40.0248 π‘˜π‘
4
18.75
42.8837 π‘˜π‘
0
N/A
N/A
20.0124 π‘˜π‘
Table Figure: Computation of Lateral Wind Loads on Frame 2
Figure: Lateral Wind Loads on Longitudinal Interior Frame (Frame 2)
Case 2 (Transverse Frames)
Figure: Governing Wind Pressure on the Windward Wall of Transverse Frames
Transverse Exterior Frame (Frame 3)
Tributary Width
𝑏𝑑 = 5 π‘š
π‘ƒπ‘Šπ‘…π·′′′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (8.75 π‘š2 ) = 20.01240578 π‘˜π‘ = 20.0124 π‘˜π‘
π‘ƒπ‘Š3𝐿 ′′′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (17.5 π‘š2 ) = 40.02481156 π‘˜π‘ = 40.0248 π‘˜π‘
π‘ƒπ‘Š2𝐿′′′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (18.75 π‘š2 ) = 42.88372667 π‘˜π‘ = 42.8837 π‘˜π‘
Floor-to-Floor
Height (π‘š)
Tributary Area,
𝐴𝑇 (π‘š2 )
Lateral Wind
Load (π‘ƒπ‘Š )
3.5
8.75
20.0124 π‘˜π‘
3.5
17.5
40.0248 π‘˜π‘
2L
4
18.75
42.8837 π‘˜π‘
1L
0
N/A
N/A
Floor
Wind Pressure,
𝑝 (π‘˜π‘ƒπ‘Ž)
RD
3L
2.2871
Table Figure: Computation of Lateral Wind Loads on Frame 3
Figure: Lateral Wind Loads on Transverse Exterior Frame (Frame 3)
Transverse Interior Frame (Frame 4)
Tributary Width
𝑏𝑑 = 10 π‘š
π‘ƒπ‘Šπ‘…π·′′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (17.5 π‘š2 ) = 40.02481156 π‘˜π‘ = 40.0248 π‘˜π‘
π‘ƒπ‘Š3𝐿′′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (35 π‘š2 ) = 80.04962312 π‘˜π‘ = 80.0496 π‘˜π‘
π‘ƒπ‘Š2𝐿 ′′ = 𝑝𝐴𝑇 = 2.2871 π‘˜π‘ƒπ‘Ž (37.5 π‘š2 ) = 85.76745334 π‘˜π‘ = 85.7674 π‘˜π‘
Wind Pressure,
𝑝 (π‘˜π‘ƒπ‘Ž)
Floor
RD
3L
2.2871
2L
1L
Floor-to-Floor
Height (π‘š)
3.5
Tributary
Area (π‘š2 )
17.5
Lateral Wind
Load (π‘ƒπ‘Š )
3.5
35
80.0496 π‘˜π‘
4
37.5
85.7674 π‘˜π‘
0
N/A
N/A
40.0248 π‘˜π‘
Table Figure: Computation of Lateral Wind Loads on Frame 4
Figure: Lateral Wind Loads on Transverse Interior Frame (Frame 4)
Summary of Wind Loads
Lateral Wind Load, π‘ƒπ‘Š (π‘˜π‘)
Floor Level
Longitudinal Frames
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
Roof Deck
10.0062
20.0124
20.0124
40.0248
3rd Floor
20.0124
40.0248
40.0248
80.0496
2nd Floor
21.4419
42.8837
42.8837
85.7674
Ground Floor
N/A
N/A
N/A
N/A
Table Figure: Lateral Wind Loads on the Frames
Reversed Lateral Wind Load, π‘ƒπ‘Š (π‘˜π‘)
Floor Level
Longitudinal Frames
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
Roof Deck
- 10.0062
- 20.0124
- 20.0124
- 40.0248
3rd Floor
- 20.0124
- 40.0248
- 40.0248
- 80.0496
2nd Floor
- 21.4419
- 42.8837
- 42.8837
- 85.7674
Ground Floor
N/A
N/A
N/A
N/A
Table Figure: Reversed Lateral Wind Loads on the Frames
Figure: Reversed Lateral Wind Loads on Longitudinal Exterior Frame (Frame 1)
Figure: Reversed Lateral Wind Loads on Longitudinal Interior Frame (Frame 2)
Figure: Reversed Lateral Wind Loads on Transverse Exterior Frame (Frame 3)
Figure: Reversed Lateral Wind Loads on Transverse Interior Frame (Frame 4)
Note:
No lateral wind load is considered on the ground floor since the joints or supports
located on this level act as fixed supports. The most probable failure may occur due to the
moment overstressing the support to overturn caused by the wind loads acting on the upper
floor levels.
Seismic Load
Seismic loads on the structure during an earthquake result from inertia forces which
were created by ground accelerations. The magnitude of these loads is a function of the
following factors: mass of the building, the dynamic properties of the buildings, intensity,
duration, and frequency content of the ground motion, and soil-structure interaction.
Seismic load refers to the application of an earthquake-generated agitation to a structure or
building. All provisions and specifications are based on NSCP 2015.
Step 1: Determine Risk Category of Building or Other Structure
(π‘‡π‘Žπ‘π‘™π‘’ 103 − 1)
Occupancy Category: IV (Stand Occupancy Structures)
Step 2: Determine the Seismic Importance Factor, I
𝐼 = 1.0
Table 208-1: Seismic Importance Factors
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 1)
Step 3: Determine the Soil Profile Type
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 2)
o Soil Profile Type: 𝑆𝐷
o Soil Profile Name/Generic Description: Stiff Soil Profile
o Shear Wave Velocity, 𝑉𝑠 = 180 π‘‘π‘œ 360 π‘š/𝑠
o SPT, 𝑁 = 15 π‘‘π‘œ 50 π‘π‘™π‘œπ‘€π‘ /360 π‘šπ‘š
o Undrained Shear Strength, π‘†π‘ˆ = 50 π‘‘π‘œ 100 π‘˜π‘ƒπ‘Ž
Table 208-2: Soil Profile Types
Step 4: Determine Seismic Zone Factor, Z
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 3)
Zone 4, 𝑍 = 0.40
Table 208-3: Seismic Zone Factor Z
Step 5: Determine Seismic Source Type
o Seismic Source Description: Faults are not capable of
producing large magnitude earthquakes and that have
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 4)
a relative low rate of seismic activity.
o Seismic Source Definition: Magnitude < 6.5
o Seismic Source Type: C
Step 6: Determine Near-Source Factor, π‘π‘Ž
o Closest Distance to Known Seismic Factor: > 10 km
o Seismic Source Type: C
o Near Source Factor, π‘π‘Ž = 1.0
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 5)
Table 208-5: Near-Source Factor, π‘π‘Ž
Step 7: Determine Near-Source Factor, 𝑁𝑣
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 6)
o Closest Distance to Known Seismic Factor: > 15 km
o Seismic Source Type: C
o Near Source Factor, 𝑁𝑣 = 1.0
Table 208-6: Near-Source Factor, 𝑁𝑣
Step 8: Determine Seismic Coefficient, πΆπ‘Ž
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 7)
o Seismic Zone, 𝑍 = 0.4
o Soil Profile Type: 𝑆𝐷
o Seismic Coefficient, πΆπ‘Ž
πΆπ‘Ž = 0.44π‘π‘Ž
πΆπ‘Ž = 0.44(1) = 0.44
πΆπ‘Ž = 0.44
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 7)
Table 208-7: Seismic Coefficient, πΆπ‘Ž
Step 9: Determine Seismic Coefficient, 𝐢𝑣
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 8)
o Seismic Zone, 𝑍 = 0.4
o Soil Profile Type: 𝑆𝐷
o Seismic Coefficient, 𝐢𝑣
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 8)
𝐢𝑣 = 0.64π‘π‘Ž
𝐢𝑣 = 0.64(1) = 0.64
𝐢𝑣 = 0.64
Table 208-8: Seismic Coefficient, 𝐢𝑣
Step 10: Determining Numerical Coefficients (𝑅, Ω0 , π‘š)
o Basic Seismic – Force Resisting System
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 11𝐴)
•
Moment – Resisting Frame Systems (Ordinary Reinforced Concrete Moment
Frames)
𝑅 = 3.5
Ω0 = 2.8
•
System Limitation and Building Height Limitation by Seismic Zone, π‘š
π‘š = 𝑁𝑃
Table 208-11A: Earthquake-Force-Resisting Structural Systems of Concrete
Step 11: Determine Elastic Fundamental Period of Vibration of the Structure in the Direction
under Consideration
1. Method A (Section 208.5.2.2)
𝑇 = 𝐢𝑑 (β„Žπ‘› )3/4
(πΈπ‘žπ‘›. 208 − 12)
Where:
𝐢𝑑 = 0.0731 for reinforced concrete moment – resisting frames and eccentrically
braced frames
β„Žπ‘› = 11 m (building height)
𝑇 = 𝐢𝑑 (β„Žπ‘› )3/4
𝑇 = 0.0731(11)3/4 = 0.4415317014 𝑠
Summary of Acquired Values
𝐼 = 1.0
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 1)
𝑍 = 0.40
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 3)
π‘π‘Ž = 1.0
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 5)
𝑁𝑣 = 1.0
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 6)
πΆπ‘Ž = 0.44
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 7)
𝐢𝑣 = 0.64
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 8)
𝑅 = 3.5
Ω0 = 2.8
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 11𝐴)
(π‘‡π‘Žπ‘π‘™π‘’ 208 − 11𝐴)
𝑇 = 0.4415317014 𝑠
(πΈπ‘žπ‘›. 208 − 12)
Step 13: Determining Total Design Base Shear in terms of Dead Load, W (Section 208.5.2.1)
o The total design base shear in a given direction shall be determined by the following
equation:
𝑉=
𝐢𝑣 𝐼
π‘Š
𝑅𝑇
𝑉=
0.64(1.0)
π‘Š
3.5(0.4415317014)
𝑉 = 0.414142727 π‘Š
(πΈπ‘žπ‘›. 208 − 8)
o The total design base shear need not exceed the following:
𝑉=
𝑉=
2.5πΆπ‘Ž 𝐼
π‘Š
𝑅
(πΈπ‘žπ‘›. 208 − 9)
2.5(0.44)(1.0)
11
π‘Š=
π‘Š
3.5
35
o The total design base shear shall not be less than the following:
𝑉 = 0.11πΆπ‘Ž 𝐼 π‘Š
(πΈπ‘žπ‘›. 208 − 10)
𝑉 = 0.11(0.44)(1.0) π‘Š
𝑉 = 0.0484 π‘Š
o In addition, for Seismic Zone 4, the total base shear shall also not be less than the
following:
𝑉
=
0.8𝑍𝑁𝑣 𝐼
π‘Š
𝑅
𝑉=
(πΈπ‘žπ‘›. 208 − 11)
0.8(0.4)(1.0)(1.0)
16
π‘Š=
π‘Š
3.5
175
From the computed values of design base shear above, use the largest value.
∴𝑉=
11
π‘Š
35
Longitudinal Exterior Frame (Frame 1)
𝐿 = 60 π‘š
Total Dead Load per Floor Level,
π‘Šπ‘₯ = Uniformly Distributed Dead Load x Total Span Length of the Frame +
+ Total Dead Load due to the Weight of Columns
Total Dead Load on the Roof Deck,
π‘Šπ‘…π·′ = 21.114 π‘˜π‘/π‘š (60 π‘š)
π‘Šπ‘…π·′ = 1266.84 π‘˜π‘
Total Dead Load on the 3rd Floor Level,
π‘Š3𝐿′ = 32.859 π‘˜π‘/π‘š (60 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/π‘š3 )(3.50 π‘š)
π‘Š3𝐿′ = 2341.588 π‘˜π‘
Total Dead Load on the 2nd Floor Level,
π‘Š2𝐿′ = 35.959 π‘˜π‘/π‘š (60 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/π‘š3 )(3.50 π‘š)
π‘Š2𝐿′ = 2527.588 π‘˜π‘
Total Dead Load on the 1st Floor Level,
π‘Š1𝐿′ = 35.394 π‘˜π‘/π‘š (60 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/π‘š3 )(4.0 π‘š)
π‘Š1𝐿′ = 2546.552 π‘˜π‘
Total Dead Load on the Frame 1,
π‘Š1 = π‘Šπ‘…π·′ + π‘Š3𝐿′ + π‘Š2𝐿′ + π‘Š1𝐿′
= 1266.84 π‘˜π‘ + 2341.588 π‘˜π‘ + 2527.588 π‘˜π‘ + 2546.552 π‘˜π‘
π‘Š1 = 8682.568 π‘˜π‘
Floor
Total Dead Load (π‘Šπ‘₯ )
RD
1266.84 π‘˜π‘
3L
2341.588 π‘˜π‘
2L
2527.588 π‘˜π‘
1L
2546.552 π‘˜π‘
Total (π‘Š1 )
8682.568 π‘˜π‘
Table Figure: Total Dead Loads on the Floor Levels on Frame 1
Design Base Shear on Frame 1,
𝑉1 =
11
π‘Š
35 1
𝑉1 =
11
(8682.568 π‘˜π‘) = 2728.807086 π‘˜π‘
35
𝑉1 = 2728.8071 π‘˜π‘
Calculation of Lateral Loads due to Seismic Load on Frame 1,
𝑉 − 𝐹𝑑
𝑆𝑖𝑛𝑐𝑒 𝑇 = 0.4415317014 𝑠 < 0.7 𝑠,
∴ 𝐹𝑑 = 0
𝑉 − 𝐹𝑑 = 2728.8071 π‘˜π‘
𝐹π‘₯ = 𝑃𝐸 =
𝐹𝑅𝐷′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
Σπ‘Šπ‘₯ β„Žπ‘₯
(πΈπ‘žπ‘›. 208 − 17)
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 2728.8071 π‘˜π‘ (0.3349) = 913.9357046 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹𝑅𝐷′ = 913.9357 π‘˜π‘
𝐹3𝐿′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 2728.8071 π‘˜π‘ (0.4221) = 1151.789032 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹3𝐿′ = 1151.789 π‘˜π‘
𝐹2𝐿′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 2728.8071 π‘˜π‘ (0.243) = 663.0823494 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹2𝐿′ = 663.0823 π‘˜π‘
𝐹1𝐿′ =
Level
RD
3L
2L
1L
Total
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 2728.8071 π‘˜π‘ (0) = 0
Σπ‘Šπ‘₯ β„Žπ‘₯
π‘Šπ‘₯ β„Žπ‘₯
π‘Šπ‘₯
Σπ‘Šπ‘₯
β„Žπ‘₯
π‘Šπ‘₯ β„Žπ‘₯
𝑉 − 𝐹𝑑
(π‘˜π‘)
(π‘˜π‘)
(π‘š) (π‘˜π‘ βˆ™ π‘š) Σπ‘Šπ‘₯ β„Žπ‘₯
(π‘˜π‘)
1266.84 1266.84 11
13935.24 0.3349
2341.588 3608.428 7.5 17561.91 0.4221
2728.8071
2527.588 6136.016 4
10110.352 0.243
2546.552 8682.568 0
0
0
41607.502
1
𝑭𝒙
(π’Œπ‘΅)
913.9357
1151.789
663.0823
0
Σ𝐹π‘₯
(π‘˜π‘)
913.9357
2065.7247
2728.8071
Table Figure: Calculation of Lateral Loads due to Seismic Load on Frame 1
Figure: Lateral Seismic Loads on Longitudinal Exterior Frame (Frame 1)
Longitudinal Interior Frame (Frame 2)
𝐿 = 60 π‘š
Total Dead Load per Floor Level,
π‘Šπ‘₯ = Uniformly Distributed Dead Load x Total Span Length of the Frame +
+ Total Dead Load due to the Weight of Columns
Total Dead Load on the Roof Deck,
π‘Šπ‘…π·′′ = 36.564 π‘˜π‘/π‘š (60 π‘š)
π‘Šπ‘…π·′′ = 2193.84 π‘˜π‘
Total Dead Load on the 3rd Floor Level,
π‘Š3𝐿′′ = 49.729 π‘˜π‘/π‘š (60 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/
π‘š3 )(3.50 π‘š)
π‘Š3𝐿′′ = 3353.788 π‘˜π‘
Total Dead Load on the 2nd Floor Level,
π‘Š2𝐿′′ = 55.929 π‘˜π‘/π‘š (60 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/
π‘š3 )(3.50 π‘š)
π‘Š2𝐿′′ = 3725.788 π‘˜π‘
Total Dead Load on the 1st Floor Level,
π‘Š1𝐿′′ = 53.324 π‘˜π‘/π‘š (60 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/π‘š3 )(4.0 π‘š)
π‘Š1𝐿′′ = 3622.352 π‘˜π‘
Total Dead Load on the Frame 2,
π‘Š2 = π‘Šπ‘…π·′′ + π‘Š3𝐿′′ + π‘Š2𝐿′′ + π‘Š1𝐿′′
= 2193.84 π‘˜π‘ + 3353.788 π‘˜π‘ + 3725.788 π‘˜π‘ + 3622.352 π‘˜π‘
π‘Š2 = 12895.768 π‘˜π‘
Floor
Total Dead Load (π‘Šπ‘₯ )
RD
2193.84 π‘˜π‘
3L
3353.788 π‘˜π‘
2L
3725.788 π‘˜π‘
1L
3622.352 π‘˜π‘
Total (π‘Š2 )
12895.768 π‘˜π‘
Table Figure: Total Dead Loads on the Floor Levels of Frame 2
Design Base Shear on Frame 2,
𝑉2 =
𝑉2 =
11
π‘Š
35 2
11
(12895.768 π‘˜π‘) = 4052.955657 π‘˜π‘
35
𝑉2 = 4052.9556 π‘˜π‘
Calculation of Lateral Loads due to Seismic Load on Frame 2,
𝑉 − 𝐹𝑑
𝑆𝑖𝑛𝑐𝑒 𝑇 = 0.4415317014 𝑠 < 0.7 𝑠,
𝑉 − 𝐹𝑑 = 4052.9556 π‘˜π‘
∴ 𝐹𝑑 = 0
𝑃𝐸 = 𝐹π‘₯ =
𝐹𝑅𝐷′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
Σπ‘Šπ‘₯ β„Žπ‘₯
(πΈπ‘žπ‘›. 208 − 17)
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 4052.9556 π‘˜π‘ (0.376) = 1523.737717 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹𝑅𝐷′′ = 1523.7377 π‘˜π‘
𝐹3𝐿′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 4052.9556 π‘˜π‘(0.3919) = 1588.215579 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹3𝐿′′ = 1588.2156 π‘˜π‘
𝐹2𝐿′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 4052.9556 π‘˜π‘(0.2322) = 941.0023606 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹2𝐿′′ = 941.0024 π‘˜π‘
𝐹1𝐿′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 4052.9556 π‘˜π‘(0) = 0
Σπ‘Šπ‘₯ β„Žπ‘₯
π‘Šπ‘₯ β„Žπ‘₯
π‘Šπ‘₯
Σπ‘Šπ‘₯
β„Žπ‘₯
π‘Šπ‘₯ β„Žπ‘₯
𝑉 − 𝐹𝑑
𝑭𝒙
Σ𝐹π‘₯
(π‘˜π‘)
(π‘˜π‘)
(π‘š) (π‘˜π‘ βˆ™ π‘š) Σπ‘Šπ‘₯ β„Žπ‘₯
(π‘˜π‘)
(π’Œπ‘΅)
(π‘˜π‘)
RD
2193.84
2193.84
11
24132.24 0.376
1523.7377 1523.7377
3L 3353.788 5547.628 7.5 25153.41 0.3919
1588.2156 3111.9533
4052.9556
2L 3725.788 9273.416
4
14903.152 0.2322
941.0024 4052.9556
1L 3622.352 12895.768 0
0
0
0
Total
64188.802
1
Level
Table Figure: Calculation of Lateral Loads due to Seismic Load on Frame 2
Figure: Lateral Seismic Loads on Longitudinal Interior Frame (Frame 2)
Transverse Exterior Frame (Frame 3)
𝐿 = 30 π‘š
Total Dead Load per Floor Level,
π‘Šπ‘₯ = Uniformly Distributed Dead Load x Total Span Length of the Frame +
+ Total Dead Load due to the Weight of Columns
Total Dead Load on the Roof Deck,
π‘Šπ‘…π·′′′ = 36.564 π‘˜π‘/π‘š (30 π‘š)
π‘Šπ‘…π·′′′ = 1096.92 π‘˜π‘
Total Dead Load on the 3rd Floor Level,
π‘Š3𝐿′′′ = 50.184 π‘˜π‘/π‘š (30 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/
π‘š3 )(3.50 π‘š)
π‘Š3𝐿′′′ = 1875.568 π‘˜π‘
Total Dead Load on the 2nd Floor Level,
π‘Š2𝐿′′′ = 56.384 π‘˜π‘/π‘š (30 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/
π‘š3 )(3.50 π‘š)
π‘Š2𝐿′′′ = 2061.568 π‘˜π‘
Total Dead Load on the 1st Floor Level,
π‘Š1𝐿′′′ = 53.844 π‘˜π‘/π‘š (30 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/π‘š3 )(4.0 π‘š)
π‘Š1𝐿′′′ = 2038.232 π‘˜π‘
Total Dead Load on the Frame 3,
π‘Š3 = π‘Šπ‘…π·′′′ + π‘Š3𝐿′′′ + π‘Š2𝐿′′′ + π‘Š1𝐿′ ′′
= 1096.92 π‘˜π‘ + 1875.568 π‘˜π‘ + 2061.568 π‘˜π‘ + 2038.232 π‘˜π‘
π‘Š3 = 7072.288 π‘˜π‘
Floor
Total Dead Load (π‘Šπ‘₯ )
RD
1096.92 π‘˜π‘
3L
1875.568 π‘˜π‘
2L
2061.568 π‘˜π‘
1L
2038.232 π‘˜π‘
Total (π‘Š3 )
7072.288 π‘˜π‘
Table Figure: Total Dead Loads on the Floor Levels of Frame 3
Design Base Shear on Frame 3,
𝑉3 =
𝑉3 =
11
π‘Š
35 3
11
(7072.288 π‘˜π‘) = 2222.719086 π‘˜π‘
35
𝑉3 = 2222.7191 π‘˜π‘
Calculation of Lateral Loads due to Seismic Load on Frame 3,
𝑉 − 𝐹𝑑
𝑆𝑖𝑛𝑐𝑒 𝑇 = 0.4415317014 𝑠 < 0.7 𝑠,
∴ 𝐹𝑑 = 0
𝑉 − 𝐹𝑑 = 2222.7191 π‘˜π‘
𝑃𝐸 = 𝐹π‘₯ =
𝐹𝑅𝐷′′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
Σπ‘Šπ‘₯ β„Žπ‘₯
(πΈπ‘žπ‘›. 208 − 17)
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 2222.7191 π‘˜π‘ (0.351) = 780.1121801 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹𝑅𝐷′′′ = 780.1122 π‘˜π‘
𝐹3𝐿′′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 2222.7191 π‘˜π‘(0.4092) = 909. 4597775 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹3𝐿′′′ = 909.4598 π‘˜π‘
𝐹2𝐿′′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 2222.7191 π‘˜π‘(0.2399) = 533.1471283 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹2𝐿′′′ = 533.1471 π‘˜π‘
𝐹1𝐿′′′ =
Level
RD
3L
2L
1L
Total
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 2222.7191 π‘˜π‘(0) = 0
Σπ‘Šπ‘₯ β„Žπ‘₯
π‘Šπ‘₯
Σπ‘Šπ‘₯
β„Žπ‘₯
(π‘˜π‘)
(π‘˜π‘)
(π‘š)
1096.92 1096.92 11
1875.568 2972.488 7.5
2061.568 5034.056 4
2038.232 7072.288 0
π‘Šπ‘₯ β„Žπ‘₯
π‘Šπ‘₯ β„Žπ‘₯
𝑉 − 𝐹𝑑
(π‘˜π‘ βˆ™ π‘š) Σπ‘Šπ‘₯ β„Žπ‘₯
(π‘˜π‘)
12066.12 0.351
14066.76 0.4092
2222.7191
8246.272 0.2399
0
0
34379.152
1
𝑭𝒙
(π’Œπ‘΅)
780.1122
909.4598
533.1471
0
Table Figure: Calculation of Lateral Loads due to Seismic Load on Frame 3
Figure: Lateral Seismic Loads on Transverse Exterior Frame (Frame 3)
Transverse Interior Frame (Frame 4)
𝐿 = 30 π‘š
Σ𝐹π‘₯
(π‘˜π‘)
780.1122
1689.572
2222.7191
Total Dead Load per Floor Level,
π‘Šπ‘₯ = Uniformly Distributed Dead Load x Total Span Length of the Frame +
+ Total Dead Load due to the Weight of Columns
Total Dead Load on the Roof Deck,
π‘Šπ‘…π·′′′′ = 67.464 π‘˜π‘/π‘š (30 π‘š)
π‘Šπ‘…π·′′′′ = 2023.92 π‘˜π‘
Total Dead Load on the 3rd Floor Level,
π‘Š3𝐿′′′′ = 84.379 π‘˜π‘/π‘š (30 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/
π‘š3 )(3.50 π‘š)
π‘Š3𝐿′′′′ = 2901.418 π‘˜π‘
Total Dead Load on the 2nd Floor Level,
π‘Š2𝐿′′′′ = 96.779 π‘˜π‘/π‘š (30 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/
π‘š3 )(3.50 π‘š)
π‘Š2𝐿′′′′ = 3273.418 π‘˜π‘
Total Dead Load on the 1st Floor Level,
π‘Š1𝐿′′′′ = 90.224 π‘˜π‘/π‘š (30 π‘š) + 7(0.80 π‘š)(0.80 π‘š)(23.6 π‘˜π‘/π‘š3 )(4.0 π‘š)
π‘Š1𝐿′′′′ = 3129.632 π‘˜π‘
Total Dead Load on the Frame 4,
π‘Š4 = π‘Šπ‘…π·′′′′ + π‘Š3𝐿′′′′ + π‘Š2𝐿′′′′ + π‘Š1𝐿′′′′
= 2023.92 π‘˜π‘ + 2901.418 π‘˜π‘ + 3273.418 π‘˜π‘ + 3129.632 π‘˜π‘
π‘Š4 = 11328.388 π‘˜π‘
Floor
Total Dead Load (π‘Šπ‘₯ )
RD
2023.92 π‘˜π‘
3L
2901.418 π‘˜π‘
2L
3273.418 π‘˜π‘
1L
3129.632 π‘˜π‘
Total (π‘Š4 )
11328.388 π‘˜π‘
Table Figure: Total Dead Loads on the Floor Levels of Frame 4
Design Base Shear on Frame 4,
𝑉4 =
𝑉4 =
11
π‘Š
35 4
11
(11328.388 π‘˜π‘) = 3560.350514 π‘˜π‘
35
𝑉4 = 3560.3505 π‘˜π‘
Calculation of Lateral Loads due to Seismic Load on Frame 4,
𝑉 − 𝐹𝑑
𝑆𝑖𝑛𝑐𝑒 𝑇 = 0.4415317014 𝑠 < 0.7 𝑠,
∴ 𝐹𝑑 = 0
𝑉 − 𝐹𝑑 = 3560.3505 π‘˜π‘
𝑃𝐸 = 𝐹π‘₯ =
𝐹𝑅𝐷′′′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
Σπ‘Šπ‘₯ β„Žπ‘₯
(πΈπ‘žπ‘›. 208 − 17)
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 3560.3505 π‘˜π‘ (0.3898) = 1387.746523 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹𝑅𝐷′′′′ = 1387.7465 π‘˜π‘
𝐹3𝐿′′′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 3560.3505 π‘˜π‘(0.381) = 1356.424686 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹3𝐿′′′′ = 1356.4247 π‘˜π‘
𝐹2𝐿′′′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 3560.3505 π‘˜π‘(0.2292) = 816.1793044 π‘˜π‘
Σπ‘Šπ‘₯ β„Žπ‘₯
𝐹2𝐿′′′′ = 816.1793044 π‘˜π‘
𝐹1𝐿′′′′ =
(𝑉 − 𝐹𝑑 )π‘Šπ‘₯ β„Žπ‘₯
= 3560.3505 π‘˜π‘(0) = 0
Σπ‘Šπ‘₯ β„Žπ‘₯
π‘Šπ‘₯ β„Žπ‘₯
π‘Šπ‘₯
Σπ‘Šπ‘₯
β„Žπ‘₯
π‘Šπ‘₯ β„Žπ‘₯
𝑉 − 𝐹𝑑
𝑭𝒙
Σ𝐹π‘₯
(π‘˜π‘)
(π‘˜π‘)
(π‘š) (π‘˜π‘ βˆ™ π‘š) Σπ‘Šπ‘₯ β„Žπ‘₯
(π‘˜π‘)
(π’Œπ‘΅)
(π‘˜π‘)
RD
2023.92
2023.92
11 22263.12 0.3898
1387.7465 1387.7465
3L 2901.418 4925.338 7.5 21760.635 0.381
1356.4247 2744.1712
3560.3505
2L 3273.418 8198.756
4 13093.672 0.2292
816.1793 3560.3505
1L 3129.632 11328.388 0
0
0
0
Total
57117.427
1
Level
Table Figure: Calculation of Lateral Loads due to Seismic Load on Frame 4
Figure: Lateral Seismic Loads on Transverse Interior Frame (Frame 4)
Summary of Seismic Loads
Lateral Seismic Load, 𝑃𝐸 (π‘˜π‘)
Floor Level
Longitudinal Frames
Roof Deck
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
913.9357
1523.7377
780.1122
1387.7465
3rd Floor
1151.789
1588.2156
909.4598
1356.4247
2nd Floor
663.0823
941.0024
533.1471
816.1793
Ground Floor
N/A
N/A
N/A
N/A
Table Figure: Lateral Seismic Loads on the Frames
Reversed Lateral Seismic Load, 𝑃′𝐸 (π‘˜π‘)
Floor Level
Longitudinal Frames
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
Roof Deck
- 913.9357
- 1523.7377
- 780.1122
- 1387.7465
3rd Floor
- 1151.789
- 1588.2156
- 909.4598
- 1356.4247
2nd Floor
- 663.0823
- 941.0024
- 533.1471
- 816.1793
Ground Floor
N/A
N/A
N/A
N/A
Table Figure: Reversed Lateral Seismic Loads on the Frames
Figure: Reversed Lateral Seismic Loads on Longitudinal Exterior Frame (Frame 1)
Figure: Reversed Lateral Seismic Loads on Longitudinal Interior Frame (Frame 2)
Figure: Reversed Lateral Seismic Loads on Transverse Exterior Frame (Frame 3)
Figure: Reversed Lateral Seismic Loads on Transverse Interior Frame (Frame 4)
Reversed Lateral Seismic Load, 𝑃′𝐸 (π‘˜π‘)
Floor Level
Longitudinal Frames
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
Roof Deck
- 913.9357
- 1523.7377
- 780.1122
- 1387.7465
3rd Floor
- 1151.789
- 1588.2156
- 909.4598
- 1356.4247
2nd Floor
- 663.0823
- 941.0024
- 533.1471
- 816.1793
Ground Floor
N/A
N/A
N/A
N/A
Note:
No lateral seismic load is considered on the ground floor since the joints or supports
located on this level act as fixed supports.
4.2 Moment Distribution Method
The Moment Distribution Method (MDM) is a primary displacement method of
structural analysis for statically indeterminate beams and frames. In this method, every joint
of the structure to be analyzed is initially assumed fixed so as to develop the Fixed–End
Moments (FEM). Then each fixed joint is locked and unlocked in succession, and the fixedend moments are distributed to adjacent members until equilibrium of joint rotation is
achieved.
The beam dimensions are 400 mm in width by 600 mm in depth with varying lengths
in longitudinal and transverse frames of 10 m and 5 m respectively. The column’s crosssectional dimensions are an 800 – mm square. The structural members’ section properties are
tabulated in the following tables. In accordance with Section 406.6.3.1.1 of the National
Structural Code of the Philippines (NSCP) 2015 as shown in the table below, the columns
and beams shall be taken as 0.7𝐼𝑔 and 0.35𝐼𝑔 respectively. The net moment at each joint is
distributed to the members in proportion to their relative stiffness, 𝐾 = 𝐸𝐼/𝐿. The proportion
of structural members within the fixed joint is defined as the Distribution Factor (𝐷𝐹). In
determining the Distribution Factor (𝐷𝐹), the stiffness factor (𝐾) of the structural member is
divided by the sum of the stiffness factor (Σ𝐾) of adjacent members connected to the same
joint. With the major presence of uniformly distributed loads, 𝑀, applied along the span
length, 𝐿, of the structural members as shown in the figure below, the fixed-end moments at
the fixed supports are computed as 𝑀𝐿2 /12.
The objective of the Moment Distribution Method for the analysis of frames is to
determine the end moments of each frame’s member. With the end moments found, the
behavior of each member can be determined, and the number of unknowns is reduced by a
significant degree. With the acquired end moments, the support (base) reactions can be
computed by applying statics to the structure; equilibrium equations. Shear and Moment
Diagrams are established to determine the other necessary values for the design process of the
structural members.
Table 406.6.3.1.1 (a) Moment of Inertia and Cross – Sectional Area Permitted for Elastic
Analysis at Factored Load Level
Figure: Fixed – End Moments with Uniformly Distributed Load between Fixed Supports
Lateral Loads and Sidesway
Lateral loads such as the wind loads and seismic loads acting parallel to the ground’s
flat surface and floor levels of the frame unlike vertical loads that act downward. These
resulting in sidesway may cause an additional moment for the frame, especially to the
structural members resisting the lateral loads, the columns. In addition, unsymmetrical
vertical loadings and structural properties may contribute to producing additional moments. It
is essential to analyze the frames to resist these lateral loads with the Moment Distribution
Method (MDM). The fixed end moments are assumed to be caused by the storey drift due to
the lateral loads. Each floor level has its relative deflection with respect to the previous floor
level. The storey drift is analyzed one by one and later combined after moment distributions.
Figure: Longitudinal Frame Sidesway Due to Lateral Loads
Figure: Longitudinal Frame Sidesway Due to Reversed Lateral Loads
Figure: Transverse Frame Sidesway Due to Lateral Loads
Figure: Transverse Frame Sidesway Due to Reversed Lateral Loads
The fixed – end moments of a storey due to sidesway parallel to the lateral loads is
computed using the following formula:
Figure: Fixed – End Moments due to Displacement on the Other Fixed Support
πΉπΈπ‘€βˆ† = ±
Sign Convention: πΆπΆπ‘Š (+), πΆπ‘Š (−)
6πΈπΌβˆ†
𝐿2
The displacement due to the sidesway, βˆ†, is determined and considered relatively
equivalent to the other floor levels. The complex solution of the frame sidesway can be
simplified into the following:
π‘†π‘–π‘‘π‘’π‘ π‘€π‘Žπ‘¦ π‘œπ‘› 𝐴𝑙𝑙 πΉπ‘™π‘œπ‘œπ‘Ÿ 𝐿𝑒𝑣𝑒𝑙𝑠
= 𝐴𝑝𝑝𝑙𝑖𝑒𝑑 πΏπ‘œπ‘Žπ‘‘π‘–π‘›π‘”π‘  (π‘π‘œ π‘†π‘–π‘‘π‘’π‘ π‘€π‘Žπ‘¦) + π‘†π‘–π‘‘π‘’π‘ π‘€π‘Žπ‘¦ π‘œπ‘› π‘‘β„Žπ‘’
πΉπ‘–π‘Ÿπ‘ π‘‘ πΉπ‘™π‘œπ‘œπ‘Ÿ 𝐿𝑒𝑣𝑒𝑙 (πΆπ‘Žπ‘ π‘’ 1) + π‘†π‘–π‘‘π‘’π‘ π‘€π‘Žπ‘¦ π‘œπ‘› π‘‘β„Žπ‘’ π‘†π‘’π‘π‘œπ‘›π‘‘ πΉπ‘™π‘œπ‘œπ‘Ÿ
𝐿𝑒𝑣𝑒𝑙 (πΆπ‘Žπ‘ π‘’ 2) + β‹―
Due to the uniformity of displacement in the lateral direction in all floor levels as
shown in the figures below for both longitudinal and transverse frames, the displacement in
different cases are assumed equal.
Figure: Longitudinal Frames Sidesway with Respect to the Original Axis
Figure: Transverse Frames Sidesway with Respect to the Original Axis
Figure: Lateral Loads on Longitudinal Frames (No Sidesway)
Figure: Lateral Loads on Transverse Frames (No Sidesway)
Figure: Longitudinal Frames Sidesway 1 (Case 1)
Figure: Transverse Frames Sidesway 1 (Case 1)
Figure: Longitudinal Frames Sidesway 2 (Case 2)
Figure: Transverse Frames Sidesway 2 (Case 2)
Figure: Longitudinal Frames Sidesway 3 (Case 3)
Figure: Transverse Frames Sidesway 3 (Case 3)
Assume πΉπΈπ‘€βˆ† = 100 π‘˜π‘ βˆ™ π‘š, β†Ί due to sidesway of the 2nd floor Level (Case 1),
hence, the βˆ† and fixed – end moments become,
𝐹𝐸𝑀1𝐿−2𝐿 = 100 π‘˜π‘ βˆ™ π‘š =
βˆ†1𝐿−2𝐿 =
6πΈπΌβˆ†1𝐿−2𝐿
𝐿2
50𝐿2
50 π‘˜π‘ βˆ™ π‘š (4 π‘š)2
π‘˜π‘ βˆ™ π‘š =
3𝐸𝐼
3𝐸𝐼
800 π‘˜π‘ βˆ™ π‘š3
βˆ†1𝐿−2𝐿 =
3𝐸𝐼
𝐹𝐸𝑀2𝐿−3𝐿 = ±
6πΈπΌβˆ†1𝐿−2𝐿
=±
𝐿2
800 π‘˜π‘ βˆ™ π‘š3
)
3𝐸𝐼
(3.5 π‘š)2
6𝐸𝐼 (
𝐹𝐸𝑀2𝐿−3𝐿 =
6400
π‘˜π‘ βˆ™ π‘š, ↻
49
Due to sidesway of the 3rd Floor Level (Case 2),
𝐹𝐸𝑀2𝐿−3𝐿 =
𝐹𝐸𝑀2𝐿−3𝐿 =
6400
π‘˜π‘ βˆ™ π‘š, β†Ί
49
6400
6πΈπΌβˆ†2𝐿−3𝐿 6πΈπΌβˆ†2𝐿−3𝐿
π‘˜π‘ βˆ™ π‘š =
=
(3.5 π‘š)2
49
𝐿2
βˆ†2𝐿−3𝐿 =
𝐹𝐸𝑀3𝐿−𝑅𝐷 = ±
800 π‘˜π‘ βˆ™ π‘š3
3𝐸𝐼
6πΈπΌβˆ†2𝐿−3𝐿
𝐿2
𝐹𝐸𝑀3𝐿−𝑅𝐷 =
800 π‘˜π‘ βˆ™ π‘š3
)
6𝐸𝐼 (
3𝐸𝐼
=±
(3.5 π‘š)2
6400
π‘˜π‘ βˆ™ π‘š, ↻
49
Due to sidesway of the Roof Deck (Case 3),
𝐹𝐸𝑀2𝐿−3𝐿 =
βˆ†3𝑙−𝑅𝐷 =
𝐹𝐸𝑀3𝐿−𝑅𝐷 = ±
800 π‘˜π‘ βˆ™ π‘š3
3𝐸𝐼
6πΈπΌβˆ†3𝑙−𝑅𝐷
=±
𝐿2
𝐹𝐸𝑀3𝐿−𝑅𝐷 =
Vertical Span
1L – 2L
2L – 3L
3L - RD
6400
π‘˜π‘ βˆ™ π‘š
49
800 π‘˜π‘ βˆ™ π‘š3
)
6𝐸𝐼 (
3𝐸𝐼
(3.5 π‘š)2
6400
π‘˜π‘ βˆ™ π‘š, β†Ί
49
Fixed – End Moments Due to Sidesways (π‘˜π‘ βˆ™ π‘š)
Case 1
Case 2
100
N/A
- 6400/49
6400/49
N/A
- 6400/49
Case 3
N/A
N/A
6400/49
Table: Fixed – End Moments Due to Sidesways of Different Floor Levels
Figure: Fixed-End Moments on Frame Sidesway 1 (Case 1)
Figure: Fixed-End Moments on Frame Sidesway 2 (Case 2)
Figure: Fixed-End Moments on Frame Sidesway 3 (Case 3)
The restraining force equations,
𝑅3 = −𝐢′𝑅 ′ 3 − 𝐢′′𝑅 ′′ 3 + 𝐢′′′𝑅′′′3
𝑅2 = −𝐢′𝑅 ′ 2 + 𝐢′′𝑅 ′′ 2 − 𝐢′′′𝑅′′′2
𝑅1 = 𝐢′𝑅 ′1 − 𝐢′′𝑅 ′′1 − 𝐢′′′𝑅′′′1
Where:
𝐢 ′ , 𝐢 ′′ , and 𝐢 ′′′ = Correction Factors
𝑅1 , 𝑅2 , and 𝑅3 = Restraining Forces
𝑅 ′ 𝑛 , 𝑅 ′′ 𝑛 , and 𝑅 ′ ′′𝑛 = External Sidesway-Causing Forces
𝑅3 = 𝑃3 +
|Σ(𝑀𝑅𝐷−3𝐿 + 𝑀3𝐿−𝑅𝐷 )|
3.5
𝑅2 = 𝑃2 + 𝑃3 − 𝑅3 +
|Σ(𝑀2𝐿−3𝐿 + 𝑀3𝐿−2𝐿 )|
3.5
𝑅1 = −𝑃1 − 𝑃2 − 𝑃3 + 𝑅3 + 𝑅2 +
|Σ(𝑀1𝐿−2𝐿 + 𝑀2𝐿−1𝐿 )|
4
Wind Load
No Sidesway
(𝑅𝑛 )
Longitudinal Frames (π‘˜π‘)
Frame 1
(Exterior)
Transverse Frames (π‘˜π‘)
Frame 2 (Interior)
Frame 3 (Exterior)
Frame 4 (Interior)
𝑅3
10.0062
20.0124
20.0124
40.0248
𝑅2
20.0124
40.0248
40.0248
80.0496
𝑅1
21.4419
42.8837
42.8837
85.7674
Reversed Wind Load
No Sidesway
(𝑅𝑛 )
Longitudinal Frames (π‘˜π‘)
Frame 1
(Exterior)
Transverse Frames (π‘˜π‘)
Frame 2 (Interior)
Frame 3 (Exterior)
Frame 4 (Interior)
𝑅3
- 10.0062
- 20.0124
- 20.0124
- 40.0248
𝑅2
- 20.0124
- 40.0248
- 40.0248
- 80.0496
𝑅1
- 21.4419
- 42.8837
- 42.8837
- 85.7674
Seismic Load
Lateral Seismic Load, 𝑃𝐸 (π‘˜π‘)
Floor Level
Longitudinal Frames
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
𝑅3
913.9357
1523.7377
780.1122
1387.7465
𝑅2
1151.789
1588.2156
909.4598
1356.4247
𝑅1
663.0823
941.0024
533.1471
816.1793
Reversed Seismic Load
Lateral Seismic Load, 𝑃𝐸 (π‘˜π‘)
Floor Level
Longitudinal Frames
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
𝑅3
- 913.9357
- 1523.7377
- 780.1122
- 1387.7465
𝑅2
- 1151.789
- 1588.2156
- 909.4598
- 1356.4247
𝑅1
- 663.0823
- 941.0024
- 533.1471
- 816.1793
Both Wind and Seismic Loads
Sidesway 1
(𝑅′𝑛 )
Longitudinal Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Transverse Frames
Frame 3 (Exterior)
Frame 4 (Interior)
𝑅′3
-103.1460088
-104.9062862
𝑅′2
360.9526445
379.6847778
𝑅′1
402.1686298
438.3398994
Sidesway 2
Longitudinal Frames
Transverse Frames
(𝑅′′𝑛 )
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3 (Exterior)
Frame 4 (Interior)
𝑅′′3
185.9344265
211.5955839
𝑅′′2
458.7427194
501.6284677
𝑅′′1
360.9526445
379.6847778
Sidesway 3
Longitudinal Frames
Transverse Frames
(𝑅′′′𝑛 )
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3 (Exterior)
Frame 4 (Interior)
𝑅′′′3
97.40123594
121.3643928
𝑅′′′2
88.53319055
90.2311911
𝑅′′′1
-5.744772867
16.45810658
From the computed tabulated values of end-moments on no sidesway due to applied
loadings and sidesway cases, the restraining forces (𝑅𝑛 , 𝑅′𝑛 , 𝑅′′𝑛 , 𝑅′′′𝑛 ) are acquired to
determine the correction factor to be applied on each case to compensate the neglected
additional moments caused by the sidesway.
𝑅3 = −𝐢′𝑅 ′ 3 − 𝐢′′𝑅 ′′ 3 + 𝐢′′′𝑅′′′3
𝑅2 = −𝐢′𝑅 ′ 2 + 𝐢′′𝑅 ′′ 2 − 𝐢′′′𝑅′′′2
𝑅1 = 𝐢′𝑅 ′1 − 𝐢′′𝑅 ′′1 − 𝐢′′′𝑅′′′1
Correction Factors due to Dead Loads,
𝑅3 = 0
𝑅2 = 0
𝑅1 = 0
Longitudinal Exterior Frame (Frame 1)
0 = −𝐢′(−103.1460088) − 𝐢′′(185.9344265) + 𝐢′′′(97.40123594)
0 = −𝐢′(360.9526445) + 𝐢′′(458.7427194) − 𝐢′′′(88.53319055)
0 = 𝐢′(402.1686298) − 𝐢′′(360.9526445) − 𝐢′′′(−5.744772867)
𝐢′ = 0
𝐢 ′′ = 0
𝐢′′′ = 0
According to the result of correction factors due to dead loads on longitudinal exterior
frame (frame 1) in all sidesway cases, it shows zero correction factors due to uniform and
symmetrical vertical loadings are only applied in which by inspection, there are no sidesways
that occur. Therefore, the correction factors of all other uniform and symmetrical vertical
applied loadings such as the full live load, and roof live load are all zero.
Live Load Case 1
Longitudinal Exterior Frame (Frame 1)
𝑅3 = −0.349931038
𝑅2 = 2.351560731
𝑅1 = −2.292881954
−0.349931038 = −𝐢′(−103.1460088) − 𝐢′′(185.9344265) + 𝐢′′′(97.40123594)
2.351560731 = −𝐢′(360.9526445) + 𝐢′′(458.7427194) − 𝐢′′′(88.53319055)
−2.292881954 = 𝐢′(402.1686298) − 𝐢′′(360.9526445) − 𝐢′′′(−5.744772867)
𝐢 ′ = 0.002510731277
𝐢 ′′ = 0.00933381508
𝐢′′′ = 0.01156632762
Longitudinal Interior Frame (Frame 2)
𝑅3 = −0.552052676
𝑅2 = 3.709833756
𝑅1 = −3.617262231
−0.552052676 = −𝐢′(−103.1460088) − 𝐢′′(185.9344265) + 𝐢′′′(97.40123594)
3.709833756 = −𝐢′(360.9526445) + 𝐢′′(458.7427194) − 𝐢′′′(88.53319055)
−3.617262231 = 𝐢′(402.1686298) − 𝐢′′(360.9526445) − 𝐢′′′(−5.744772867)
𝐢 ′ = 0.003960940023
𝐢 ′′ = 0.01472507146
𝐢′′′ = 0.0182470979
Transverse Exterior Frame (Frame 3)
𝑅3 = −0.28103872
𝑅2 = 2.135398263
𝑅1 = −2.049317416
−0.28103872 = −𝐢′(−103.1460088) − 𝐢′′(185.9344265) + 𝐢′′′(97.40123594)
2.135398263 = −𝐢′(360.9526445) + 𝐢′′(458.7427194) − 𝐢′′′(88.53319055)
−2.049317416 = 𝐢′(402.1686298) − 𝐢′′(360.9526445) − 𝐢′′′(−5.744772867)
𝐢 ′ = 0.003247592302
𝐢 ′′ = 0.009483416387
𝐢′′′ = 0.01177889245
Transverse Interior Frame (Frame 4)
𝑅3 = −0.443367613
𝑅2 = 3.368811475
𝑅1 = −3.233010069
−0.443367613 = −𝐢′(−103.1460088) − 𝐢′′(185.9344265) + 𝐢′′′(97.40123594)
3.368811475 = −𝐢′(360.9526445) + 𝐢′′(458.7427194) − 𝐢′′′(88.53319055)
−3.233010069 = 𝐢′(402.1686298) − 𝐢′′(360.9526445) − 𝐢′′′(−5.744772867)
𝐢 ′ = 0.00512341238
𝐢 ′′ = 0.01496106947
𝐢′′′ = 0.01858242047
According to the result of correction factors due to live load case 1 in all sidesway
cases, it shows near-zero (0 % to 1.9 % max) correction factors due to applied vertical live
pattern loadings are only applied. By interpreting the computation, there are almost negligible
sidesways that occur, thus will result to having imperceptible amount of additional moments
resulted by sidesways. In general, the additional moments caused by sidesways in all live
load cases can be neglected due to the insignificant amount of correction factor to be applied.
Correction Factors due to Wind Loads,
Longitudinal Exterior Frame (Frame 1)
10.00620289 π‘˜π‘ = −𝐢′(−103.1460088) − 𝐢′′(185.9344265) + 𝐢′′′(97.40123594)
20.01240578 π‘˜π‘ = −𝐢′(360.9526445) + 𝐢′′(458.7427194) − 𝐢′′′(88.53319055)
21.44186334 π‘˜π‘ = 𝐢′(402.1686298) − 𝐢′′(360.9526445) − 𝐢′′′(−5.744772867)
𝐢 ′ = 0.763886616
𝐢 ′′ = 0.8049243272
𝐢′′′ = 0.8303538416
Longitudinal Interior Frame (Frame 2)
20.01240578 π‘˜π‘ = −𝐢′(−103.1460088) − 𝐢′′(185.9344265) + 𝐢′′′(97.40123594)
40.02481156 π‘˜π‘ = −𝐢′(360.9526445) + 𝐢′′(458.7427194) − 𝐢′′′(88.53319055)
42.88372667 π‘˜π‘ = 𝐢′(402.1686298) − 𝐢′′(360.9526445) − 𝐢′′′(−5.744772867)
𝐢 ′ = 1.527773232
𝐢 ′′ = 1.609848654
𝐢 ′′′ = 1.660707683
Transverse Exterior Frame (Frame 3)
20.01240578 π‘˜π‘ = −𝐢′(−104.9062862) − 𝐢′′(211.5955839) + 𝐢′′′(121.3643928)
40.02481156 π‘˜π‘ = −𝐢′(379.6847778) + 𝐢′′(501.6284677) − 𝐢′′′(90.2311911)
42.88372667 π‘˜π‘ = 𝐢′(438.3398994) − 𝐢′′(379.6847778) − 𝐢′′′(16.45810658)
𝐢 ′ = 1.168627997
𝐢 ′′ = 1.183419847
𝐢 ′′′ = 1.218004661
Transverse Interior Frame (Frame 4)
40.02481156 π‘˜π‘ = −𝐢′(−104.9062862) − 𝐢′′(211.5955839) + 𝐢′′′(121.3643928)
80.04962312 π‘˜π‘ = −𝐢′(379.6847778) + 𝐢′′(501.6284677) − 𝐢′′′(90.2311911)
85.76745334 π‘˜π‘ = 𝐢′(438.3398994) − 𝐢′′(379.6847778) − 𝐢′′′(16.45810658)
𝐢 ′ = 2.337255994
𝐢 ′′ = 2.366839693
𝐢 ′′′ = 2.436009322
Correction Factor due to Seismic Load,
Longitudinal Exterior Frame (Frame 1)
913.9357046 π‘˜π‘ = −𝐢′(−103.1460088) − 𝐢′′(185.9344265) + 𝐢′′′(97.40123594)
1151.789032 π‘˜π‘ = −𝐢′(360.9526445) + 𝐢′′(458.7427194) − 𝐢′′′(88.53319055)
663.0823494 π‘˜π‘ = 𝐢′(402.1686298) − 𝐢′′(360.9526445) − 𝐢′′′(−5.744772867)
𝐢 ′ = 40.96750615
𝐢 ′′ = 44.62301794
𝐢 ′′′ = 51.18267911
Longitudinal Interior Frame (Frame 2)
1523.737717 π‘˜π‘ = −𝐢′(−103.1460088) − 𝐢′′(185.9344265) + 𝐢′′′(97.40123594)
1588.215579 π‘˜π‘ = −𝐢′(360.9526445) + 𝐢′′(458.7427194) − 𝐢′′′(88.53319055)
941.0023606 π‘˜π‘ = 𝐢′(402.1686298) − 𝐢′′(360.9526445) − 𝐢′′′(−5.744772867)
𝐢 ′ = 60.58800239
𝐢 ′′ = 66.13653042
𝐢 ′′′ = 77.73397197
Transverse Exterior Frame (Frame 3)
780.1121801 π‘˜π‘ = −𝐢′(−104.9062862) − 𝐢′′(211.5955839) + 𝐢′′′(121.3643928)
909.4597775 π‘˜π‘ = −𝐢′(379.6847778) + 𝐢′′(501.6284677) − 𝐢′′′(90.2311911)
533.1471283 π‘˜π‘ = 𝐢′(438.3398994) − 𝐢′′(379.6847778) − 𝐢′′′(16.45810658)
𝐢 ′ = 25.39061373
𝐢 ′′ = 26.57333838
𝐢 ′′′ = 30.81034027
Transverse Interior Frame (Frame 4)
1387.746523 π‘˜π‘ = −𝐢′(−104.9062862) − 𝐢′′(211.5955839) + 𝐢′′′(121.3643928)
1356.424686 π‘˜π‘ = −𝐢′(379.6847778) + 𝐢′′(501.6284677) − 𝐢′′′(90.2311911)
816.1793044 π‘˜π‘ = 𝐢′(438.3398994) − 𝐢′′(379.6847778) − 𝐢′′′(16.45810658)
𝐢 ′ = 40.49899796
𝐢 ′′ = 42.42162996
𝐢 ′′′ = 50.38855692
Actual end moments are computed using the following equation,
𝑀 = π‘€π‘œ + 𝐢 ′ 𝑀′ + 𝐢 ′′ 𝑀′′ + 𝐢′′′𝑀′′′
Summary of Correction Factors
Dead Load, Full Live Load, and Roof Live Load
Longitudinal Frames
Correction
Factors
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
𝐢′
0
0
0
0
𝐢′′
0
0
0
0
𝐢 ′ ′′
0
0
0
0
Table: Correction Factors of Dead, Full Live, and Roof Live Loads
Live Load Cases
Longitudinal Frames
Correction
Factors
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
𝐢′
0 % to 1.9 %
0 % to 1.9 %
0 % to 1.9 %
0 % to 1.9 %
𝐢′′
0 % to 1.9 %
0 % to 1.9 %
0 % to 1.9 %
0 % to 1.9 %
𝐢 ′ ′′
0 % to 1.9 %
0 % to 1.9 %
0 % to 1.9 %
0 % to 1.9 %
Table: Correction Factors of Live Load Cases
Lateral Wind Loads
Longitudinal Frames
Correction
Factors
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
𝐢′
0.76389
1.52777
1.16863
2.33726
𝐢′′
0.80492
1.60985
1.18342
2.36684
𝐢 ′ ′′
0.83035
1.66071
1.218
2.43601
Table: Correction Factors of Lateral Wind Loads
Reversed Lateral Wind Loads
Longitudinal Frames
Correction
Factors
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
𝐢′
- 0.76389
- 1.52777
- 1.16863
- 2.33726
𝐢′′
- 0.80492
- 1.60985
- 1.18342
- 2.36684
𝐢 ′ ′′
- 0.83035
- 1.66071
- 1.218
- 2.43601
Table: Correction Factors of Reversed Lateral Wind Loads
Lateral Seismic Loads
Longitudinal Frames
Correction
Factors
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
𝐢′
40.9675
60.588
25.3906
40.499
𝐢′′
44.623
66.1365
26.5733
42.4216
𝐢 ′ ′′
51.1827
77.734
30.8103
50.3886
Table: Correction Factors of Lateral Seismic Loads
Reversed Lateral Seismic Loads
Longitudinal Frames
Correction
Factors
Transverse Frames
Frame 1
(Exterior)
Frame 2 (Interior)
Frame 3
(Exterior)
Frame 4 (Interior)
𝐢′
- 40.9675
- 60.588
- 25.3906
- 40.499
𝐢′′
- 44.623
- 66.1365
- 26.5733
- 42.4216
𝐢 ′ ′′
- 51.1827
- 77.734
- 30.8103
- 50.3886
Table: Correction Factors of Reversed Lateral Seismic Loads
Load Combination
In accordance with the National Structural Code of the Philippines (NSCP) 2015,
buildings, towers, and other vertical structures shall be designed to resist the load
combinations specified in Sections 203.3, 203.4, and 203.5. The proposed building shall be
designed according to the combinations specified in Section 203.3, Load Combinations Using
Strength Design or Load and Resistance Factor Design (LRFD).
The loads considered in the load combinations are namely:
1. Dead Load (D)
2. Live Load (L) with 7 Cases: Full L, L1, L2, L3, L4, L5, L6
3. Roof Live Load (πΏπ‘Ÿ )
4. Wind Load (W) with 2 Cases: W, W’
5. Seismic Load (E) with 2 cases: E, E’
The possible governing load combinations are considered in this section, and the other
load combinations are omitted due to the absence of other loads that by inspection cannot
exceed the value of the possible governing load combinations. The load combination
formulas are as follows:
1.4𝐷
(πΈπ‘žπ‘›. 203 − 1)
1.2𝐷 + 1.6𝐿 + 0.5πΏπ‘Ÿ
(πΈπ‘žπ‘›. 203 − 2)
1.2𝐷 + 1.6πΏπ‘Ÿ + 𝑓1 𝐿
(πΈπ‘žπ‘›. 203 − 3)
1.2𝐷 + 1.0π‘Š + 𝑓1 𝐿 + 0.5πΏπ‘Ÿ
(πΈπ‘žπ‘›. 203 − 4)
1.2𝐷 + 1.0𝐸 + 𝑓1 𝐿
(πΈπ‘žπ‘›. 203 − 5)
The results from the structural analysis showed a significant difference in shear and
moment among the frames of the proposed building. The longitudinal interior (Frame 2) and
transverse interior (Frame 4) frames are found to produce more shear and moment mainly as
a result of larger tributary or projection areas for applied loadings. Thus, these two frames
shall be considered critical frames for determining the desirable load combinations to be used
for the design of the structural members.
Figure: Load Combination on the Critical Longitudinal Frame
Figure: Load Combination on the Critical Transverse Frame
The maximum end moments of every floor level of two critical frames (frames 2 and
4) are determined for both beams and columns. Based on the computation, the closer supports
to the center frame of both end span flexural and column members experience the maximum
end moments for every floor level.
(Tables of Beam prior to Design)
4.4 Design
Design of Beams
Reinforced concrete beams are known as flexural members that primarily resist
bending moments caused by the applied loadings such as dead load, live load and roof live
load transmitted from the slabs.
The design criteria for beams hint that the beam failure is caused by the bending
resulting in cracks at the outermost tension side especially when the loading moves closer to
the most critical section of the member where it experiences the most bending stress, at the
midspan. In addition, beams are susceptible to vertical and diagonal cracks known as webshear cracks caused by the shear and diagonal tensile stresses. Shear failures of reinforced
concrete beams differ from bending failures in some ways. Shear failures happen
unexpectedly and without warning.
Therefore, beams are designed to fail in bending under loads that are appreciably
smaller than those that will cause shear failures. As a result, those members will fail in a
ductile manner. If the beams are overloaded, crack and droop may produce, but they will not
fall apart as they would if shear failures were feasible.
The formula equations for beam flexural design are as follows:
𝑅𝑛 =
𝜌=
0.85𝑓 ′ 𝑐
𝑓𝑦
𝜌=
πœŒπ‘π‘Žπ‘™ =
𝜌0.005
𝑀𝑒
πœ™π‘π‘‘ 2
(1 − √1 −
2𝑅𝑛
)
0.85𝑓 ′ 𝑐
√𝑓′𝑐
1.4
π‘œπ‘Ÿ
4𝑓𝑦
𝑓𝑦
0.85𝛽1 𝑓′𝑐
600
(
)
𝑓𝑦
600 + 𝑓𝑦
0.31875𝛽1 𝑓 ′ 𝑐
600 + 𝑓𝑦
= 0.375 (
) πœŒπ‘π‘Žπ‘™ =
600
𝑓𝑦
𝐴𝑠 = πœŒπ‘π‘‘
𝑛=
𝐴𝑠
𝐴𝑛
π‘Ž
πœ™π‘€π‘› = πœ™π΄π‘  𝑓𝑦 (𝑑 − )
2
The formula equations for beam shear design are as follows:
𝑉𝑠 =
𝑉𝑒 − πœ™π‘‰π‘
πœ™
𝐴𝑣 = 𝑛𝑠 𝐴𝑠
𝑠=
π‘ π‘šπ‘Žπ‘₯ =
𝐴𝑣 𝑓𝑦𝑑 𝑑
𝑉𝑠
𝐴𝑣 𝑓𝑦𝑑
𝐴𝑣 𝑓𝑦𝑑
π‘œπ‘Ÿ
0.35𝑏𝑀
0.062√𝑓′𝑐 𝑏𝑀
From NSCP 2015 Section 409.7.6.2.2
𝑠=
𝑑
√𝑓′𝑐
≤ 600π‘šπ‘š, 𝑖𝑓 𝑉𝑠 ≤
𝑏𝑀 𝑑
2
3
𝑠=
𝑑
√𝑓 ′ 𝑐
≤ 300π‘šπ‘š, 𝑖𝑓 𝑉𝑠 >
𝑏𝑀 𝑑
4
3
From NSCP 2015 Section 409.7.6.4.3
𝑠 = 16𝑑𝑏 π‘œπ‘Ÿ 48𝑑𝑠 π‘œπ‘Ÿ 300 π‘šπ‘š
From NSCP 2015 Section 418.6.4.4
𝑠=
𝑑
π‘œπ‘Ÿ 6𝑑𝑏 π‘œπ‘Ÿ 150 π‘šπ‘š
4
2
πΆβ„Žπ‘’π‘π‘˜ π‘“π‘œπ‘Ÿ 𝑉𝑒 ≤ πœ™ (𝑉𝑐 + √𝑓 ′ 𝑐 𝑏𝑀 𝑑)
3
General Beam Properties:
Concrete Compressive Strength, 𝑓′𝑐 = 28 π‘€π‘ƒπ‘Ž
Steel Yield Strength, 𝑓𝑦 = 420 π‘€π‘ƒπ‘Ž
Steel Elasticity Modulus, 𝐸𝑠 = 200 πΊπ‘ƒπ‘Ž
Concrete Strain, πœ–π‘ = 0.003
Steel Tension-Controlled Strain, πœ– 𝑇 = 0.005
Steel Yield Strain, πœ–π‘¦ = 0.0021
Beam Width, 𝑏 = 400 π‘šπ‘š
Beam Height, β„Ž = 600 π‘šπ‘š
Effective Depth, 𝑑 = 520 π‘šπ‘š
𝛽1 = 0.85
πœ™π΅ = 0.90
πœ™π‘‰ = 0.75
πœ† = 1.0
Design of Flexural Reinforcements
Longitudinal Interior Frame (Frame 2)
Roof Deck
𝑀𝑒 = 236.9967413929 π‘˜π‘ βˆ™ π‘š
𝑀𝑒
236.9967413929 π‘˜π‘ βˆ™ π‘š(106 )
𝑅𝑛 =
=
= 2.434631219 π‘€π‘ƒπ‘Ž
πœ™π‘π‘‘ 2
0.90(400)(520)2
0.85𝑓 ′ 𝑐
2𝑅𝑛
(1 − √1 −
)
𝜌=
𝑓𝑦
0.85𝑓 ′ 𝑐
=
0.85(28 π‘€π‘ƒπ‘Ž)
2(2.434631219 π‘€π‘ƒπ‘Ž)
[1 − √1 −
]
420 π‘€π‘ƒπ‘Ž
0.85(28 π‘€π‘π‘Ž)
𝜌 = 0.006128095937
Check 𝜌 limits,
πœŒπ‘šπ‘–π‘› =
πœŒπ‘šπ‘–π‘› =
√𝑓′𝑐
√28
=
= 0.0031497
4𝑓𝑦
4(420)
πœŒπ‘šπ‘–π‘› =
𝜌0.005
√𝑓′𝑐
1.4
π‘œπ‘Ÿ
4𝑓𝑦
𝑓𝑦
1.4 1.4
=
= 0.00333
𝑓𝑦
420
0.31875𝛽1 𝑓 ′ 𝑐 0.31875(0.85)(28 π‘€π‘ƒπ‘Ž)
=
=
= 0.0180625
𝑓𝑦
420 π‘€π‘ƒπ‘Ž
𝑆𝑖𝑛𝑐𝑒 πœŒπ‘šπ‘–π‘› < 𝜌 < 𝜌0.005 ,
∴ 𝑂𝐾!
Required Steel Area,
𝐴𝑠 = πœŒπ‘π‘‘ = 0.006128(400 π‘šπ‘š)(520 π‘šπ‘š) = 1274.643955 π‘šπ‘š2
Use 25-mm diameter longitudinal bars,
𝐴20 =
𝑛=
2
πœ‹π‘‘25
πœ‹(25 π‘šπ‘š)2
=
= 156.25πœ‹ π‘šπ‘š2
4
4
𝐴𝑠
1274.643955 π‘šπ‘š2
=
= 2.597 π‘ π‘Žπ‘¦ 3 𝑝𝑐𝑠 (𝐴𝑠 = 468.75πœ‹ π‘šπ‘š2 )
𝐴20
156.25πœ‹ π‘šπ‘š2
Checking the chosen longitudinal bars,
𝑇=𝐢
𝐴𝑠 𝑓𝑦
468.75πœ‹ π‘šπ‘š2 (420 π‘€π‘ƒπ‘Ž) 5625πœ‹
π‘Ž=
=
=
π‘šπ‘š π‘œπ‘Ÿ 64.9686 π‘šπ‘š
0.85𝑓′𝑐 𝑏 0.85 (28 π‘€π‘ƒπ‘Ž)(400 π‘šπ‘š)
272
𝑐=
πœ–π‘‡ =
π‘Ž
64.9686
=
= 76.4336 π‘šπ‘š
𝛽1
0.85
𝑑−𝑐
520 π‘šπ‘š − 76.4336 π‘šπ‘š
(0.003) =
(0.003)
𝑐
76.4336 π‘šπ‘š
πœ– 𝑇 = 0.0174 > 0.005,
∴ π‘‡π‘’π‘›π‘ π‘–π‘œπ‘› − π‘π‘œπ‘›π‘‘π‘Ÿπ‘œπ‘™π‘™π‘’π‘‘ π‘ π‘’π‘π‘‘π‘–π‘œπ‘› (πœ™ = 0.90)
Design Moment Capacity,
π‘Ž
πœ™π‘€π‘› = πœ™π΄π‘  𝑓𝑦 (𝑑 − )
2
= 0.90(468.75πœ‹ π‘šπ‘š2 )(420 π‘€π‘ƒπ‘Ž) (520 π‘šπ‘š −
64.9686 π‘šπ‘š
)
2
πœ™π‘€π‘› = 271,376,077.3 𝑁 βˆ™ π‘šπ‘š
πœ™π‘€π‘› = 271.3761 π‘˜π‘ βˆ™ π‘š > 𝑀𝑒 = 236.9967413929 π‘˜π‘ βˆ™ π‘š
∴ Design Moment Capacity is satisfied.
∴ Use 3 25-mm ∅ Longitudinal Bars at 120-mm Center-to-Center Spacing
(Tables of Design of Beams, Flexural Reinforcement)
Design of Transverse Reinforcements
πœ™ = 0.75
πœ† = 1.0
Concrete shear capacity of the section,
1
πœ™π‘‰π‘ = πœ™ πœ†√𝑓′𝑐 𝑏𝑀 𝑑
6
1
1
πœ™π‘‰π‘ = πœ™ πœ†√𝑓′𝑐 𝑏𝑀 𝑑 = 0.75 (1.0)√28 π‘€π‘ƒπ‘Ž (400 π‘šπ‘š)(520 π‘šπ‘š)
6
6
πœ™π‘‰π‘ = 137579.0682 𝑁 = 137.5791 π‘˜π‘
Maximum concrete strength to resist the shear without vertical reinforcement,
1
1
πœ™π‘‰π‘ = (137.5791 π‘˜π‘) = 68.7895 π‘˜π‘
2
2
Roof Deck of Longitudinal Interior Frame (Frame 2),
From the tabulated values,
𝑉𝑒 π‘šπ‘Žπ‘₯ = 282.42668143068 π‘˜π‘ = 282.4267 π‘˜π‘
𝑉𝑒 @ 𝑑 = 253.242617430682 π‘˜π‘ = 253.243 π‘˜π‘
𝑉𝑒 @ 1π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ = 226.30348143068 π‘˜π‘ = 226.3035 π‘˜π‘
𝑉𝑒 @ 2π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ = 170.18028143068 π‘˜π‘ = 170.1803 π‘˜π‘
𝑉𝑒 @ 3π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ = 114.057081431 π‘˜π‘ = 114.0571 π‘˜π‘
Determining the nominal shear capacity of shear reinforcement,
𝑉𝑠 =
𝑉𝑠 @ 𝑑 =
𝑉𝑒 @ 𝑑 − πœ™π‘‰π‘
πœ™
=
𝑉𝑒 − πœ™π‘‰π‘
πœ™
253.243 π‘˜π‘ − 137.5791 π‘˜π‘
= 154.2181 π‘˜π‘
0.75
𝑉𝑠 @ 1π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ =
=
𝑉𝑒 @ 1π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ − πœ™π‘‰π‘
πœ™
226.3035 π‘˜π‘ − 137.5791 π‘˜π‘
0.75
𝑉𝑠 @ 1π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ = 118.2992 π‘˜π‘
𝑉𝑠 @ 2π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ =
=
𝑉𝑒 @ 2π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ − πœ™π‘‰π‘
πœ™
170.1803 π‘˜π‘ − 137.5791 π‘˜π‘
0.75
𝑉𝑠 @ 2π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ = 43.46828 π‘˜π‘
𝑉𝑠 @ 3π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ =
𝑉𝑒 @ 3π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ − πœ™π‘‰π‘
πœ™
=
114.0571 π‘˜π‘ − 137.5791 π‘˜π‘
0.75
𝑉𝑠 @ 3π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ = −31.3626 π‘˜π‘
The value of 𝑉𝑠 @ 3π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ can be neglected due to its negative value. This
value indicates not needing the shear strength provided by the shear reinforcement as the
concrete shear strength, πœ™π‘‰π‘ , is adequate to carry the factored shear load, 𝑉𝑒 , on that specific
section.
Determining the theoretical stirrups spacing, s, using 10-mm diameter bar,
πœ‹(10 π‘šπ‘š)2
𝐴𝑣 = 𝑛𝑠 𝐴𝑠 = 2 [
] = 50πœ‹ π‘šπ‘š2
4
𝑠=
𝑠@𝑑 =
𝐴𝑣 𝑓𝑦𝑑 𝑑
𝑉𝑠
𝐴𝑣 𝑓𝑦𝑑 𝑑 50πœ‹ π‘šπ‘š(420 π‘€π‘ƒπ‘Ž)(520 π‘šπ‘š)
=
= 222.4525 π‘šπ‘š
𝑉𝑠 @ 𝑑
154.2181 π‘˜π‘ (1000)
𝑠@1π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ =
𝐴𝑣 𝑓𝑦𝑑 𝑑
𝑉𝑠 @ 1π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘
=
50πœ‹ π‘šπ‘š(420 π‘€π‘ƒπ‘Ž)(520 π‘šπ‘š)
118.2992 π‘˜π‘ (1000)
=
50πœ‹ π‘šπ‘š(420 π‘€π‘ƒπ‘Ž)(520 π‘šπ‘š)
43.46828 π‘˜π‘ (1000)
=
50πœ‹ π‘šπ‘š(420 π‘€π‘ƒπ‘Ž)(520 π‘šπ‘š)
−31.3626 π‘˜π‘ (1000)
= 289.9951 π‘šπ‘š
𝑠@2π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ =
𝐴𝑣 𝑓𝑦𝑑 𝑑
𝑉𝑠 @ 2π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘
= 789.2235 π‘šπ‘š
𝑠@3π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ =
𝐴𝑣 𝑓𝑦𝑑 𝑑
𝑉𝑠 @ 3π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘
= −1093.86 π‘šπ‘š
The last computed value, 𝑠@3π‘š π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ , is impractical due to having negative
value of stirrups spacing, thus shall not be neglected.
Determining maximum spacing, s, to provide the minimum area of shear
reinforcement,
π‘ π‘šπ‘Žπ‘₯ =
π‘ π‘šπ‘Žπ‘₯ =
π‘ π‘šπ‘Žπ‘₯ =
𝐴𝑣 𝑓𝑦𝑑
𝐴𝑣 𝑓𝑦𝑑
π‘œπ‘Ÿ
0.35𝑏𝑀
0.062√𝑓′𝑐 𝑏𝑀
𝐴𝑣 𝑓𝑦𝑑
50πœ‹ π‘šπ‘š2 (420 π‘€π‘ƒπ‘Ž)
=
= 471.2389 π‘šπ‘š
0.35𝑏𝑀
0.35(400 π‘šπ‘š)
𝐴𝑣 𝑓𝑦𝑑
0.062√𝑓′𝑐 𝑏𝑀
=
50πœ‹ π‘šπ‘š2 (420 π‘€π‘ƒπ‘Ž)
0.062 √28 π‘€π‘π‘Ž(400 π‘šπ‘š)
= 502.7342 π‘šπ‘š
Checking for maximum spacing according to NSCP 2015 Section 409.7.6.2.2,
𝑠=
𝑑
√𝑓′𝑐
≤ 600π‘šπ‘š, 𝑖𝑓 𝑉𝑠 ≤
𝑏𝑀 𝑑
2
3
𝑑
√𝑓 ′ 𝑐
𝑠 = ≤ 300π‘šπ‘š, 𝑖𝑓 𝑉𝑠 >
𝑏𝑀 𝑑
4
3
√𝑓′𝑐
1 π‘˜π‘
√28 π‘€π‘ƒπ‘Ž
(400 π‘šπ‘š)(520 π‘šπ‘š) (
) = 366.8775 π‘˜π‘
𝑏𝑀 𝑑 =
3
3
1000 𝑁
𝑆𝑖𝑛𝑐𝑒 𝑉𝑠 <
1
√𝑓′𝑐 𝑏𝑀 𝑑, ∴ 𝑒𝑠𝑒 𝑠 = 𝑑/2 π‘œπ‘Ÿ 600 π‘šπ‘š
3
𝑠=
𝑑 520 π‘šπ‘š
=
= 260 π‘šπ‘š
2
2
Checking for maximum spacing according to NSCP 2015 Section 409.7.6.4.3, shall
not exceed the least of:
𝑠 = 16𝑑𝑏 π‘œπ‘Ÿ 48𝑑𝑠 π‘œπ‘Ÿ π‘™π‘’π‘Žπ‘ π‘‘ π‘π‘’π‘Žπ‘š π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›
𝑠 = 16𝑑𝑏 = 16(25 π‘šπ‘š) = 400 π‘šπ‘š
𝑠 = 48𝑑𝑠 = 48(10) = 480 π‘šπ‘š
𝑠 = 𝑏 = 400 π‘šπ‘š
Checking for maximum spacing according to NSCP 2015 Section 418.6.4.4.
The first hoop shall be located not more than 50mm from the face of a supporting
column. Spacing of the hoop shall not exceed the smallest of:
𝑠=
𝑠=
𝑑
π‘œπ‘Ÿ 6𝑑𝑏 π‘œπ‘Ÿ 150 π‘šπ‘š
4
𝑑 520 π‘šπ‘š
=
= 130 π‘šπ‘š
4
4
𝑠 = 6𝑑𝑏 = 6(25 π‘šπ‘š) = 150 π‘šπ‘š
𝑠 = 150 π‘šπ‘š
2
Check for 𝑉𝑒 ≤ πœ™ (𝑉𝑐 + 3 √𝑓 ′ 𝑐 𝑏𝑀 𝑑),
2
2
1 π‘˜π‘
)
πœ™ (𝑉𝑐 + √𝑓′𝑐 𝑏𝑀 𝑑) = 137.5791 π‘˜π‘ + √28 π‘€π‘ƒπ‘Ž (400 π‘šπ‘š)(520 π‘šπ‘š) (
3
3
1000 𝑁
2
πœ™ (𝑉𝑐 + √𝑓′𝑐 𝑏𝑀 𝑑) = 687.8954 π‘˜π‘
3
2
𝑆𝑖𝑛𝑐𝑒 𝑉𝑒 < πœ™ (𝑉𝑐 + √𝑓 ′ 𝑐 𝑏𝑀 𝑑) ,
3
∴ 𝑂𝐾!
∴ Use 10-mm ∅ Spacing 3 at 50 mm, 6 at 130 mm, Rest at 200 mm, Symmetrical at Midspan.
(Tables of Design of Beams, Shear Reinforcement)
Design of Columns
Reinforced concrete columns are used to carry mainly the axial compressive load that
transmits from the upper floor levels and secondarily the bending moment in designing the columns of
the proposed mixed-use building, wherein the ultimate strength column will be considered. From the
structural analysis,
a.) Compute interaction diagram coordinate, 𝐾𝑛 and 𝑅𝑛 .
𝐾𝑛 =
𝑅𝑛 =
𝑃𝑛
𝑓′𝑐 𝐴𝑔
𝑃𝑛 𝑒
𝑒
π‘œπ‘Ÿ π‘œπ‘Ÿ 𝐾𝑛 ( )
𝑓′𝑐 𝐴𝑔 β„Ž
β„Ž
a.) Compute 𝛾, which is the ratio of the center-to-center distance between the bars
and the depth of the column, β„Ž, wherein both values are taken in the direction
of bending. In most cases, the value of 𝛾 falls in between a pair of curves.
𝛾=
π›Ύβ„Ž
β„Ž
c.) Plot the values of 𝐾𝑛 and 𝑅𝑛 on the interaction diagram.
d.) Determine and interpolate the values of πœŒπ‘” from various graphs based on
the direction of bending.
e.) Compute the reinforcing area and select the bars.
𝐴𝑠𝑑 = πœŒπ‘” 𝐴𝑔 π‘œπ‘Ÿ πœŒπ‘” π‘β„Ž
f.) Check the selected bars by using the following equations.
πœŒπ‘” =
𝐴𝑠𝑑
𝐴𝑔
𝑅𝑛 𝑒
=
𝐾𝑛 β„Ž
𝑃′𝑛 =
𝑅𝑛 𝑓′𝑐 𝐴𝑔 β„Ž
≥ 𝑃𝑛
𝑒
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