Structural Design Analysis Procedures
The structural design analysis used in this project is to design slabs, beams,
columns, and foundation that the structures can withstand through ages.
Design Parameters:
According to Philippine National Standards for Steel Bars for Concrete
Reinforcement under PNS-49:2002, for low-rise buildings, the structural grade used is
Grade 33 with a yield strength (fy) of 230 MPa.
For a normal-weight concrete and lightweight, use minimum f’ c = 17 MPa (Table
419.2.1.1, NSCP 2015)
Table 419.2.1.1 Limits for f’c
Application
General
Concrete
Normal-weight and
lightweight
Normal-weight
Lightweight
Minimum f’c MPa
17
Maximum f’c MPa
None
Special moment
21
None
frames and special
21
35[1]
structural walls
1
The limit is permitted to be exceeded where demonstrated by experimental evidence
that members made with lightweight concrete provide strength and toughness equal to or
exceeding those pf comparable members made with normal weight concrete of the same
strength.
Source: National Structural Code of the Philippines 2015
Strength reduction factor,∅, shall be in accordance with Table 421.2.1 (Sec.
421.2.1, NSCP 2015). Strength reduction factor for moment, axial force, or combined
moment and axial force shall be in accordance with Table 421.2.2.
Table 421.2.1
Strength Reduction Factors, ∅
Action or Structural
Element
Moment, axial force, or
combined moment and
axial force
Shear
∅
Exceptions
0.65 to 0.90
Near ends of pretensioned members where
strands are not fully developed, ∅shall be
in accordance with 421.2.3.
Additional requirements are given in
Section 421.2.4 for structural designed to
resist earthquake effects.
-
0.75
Torsion
0.75
Bearing
0.65
Post-tensioned anchorages
0.85
zones
Brackets and corbels
0.75
Struts, ties, nodal zones,
0.75
and bearing areas designed
with strut-and-ties method
in Section 423
Components of
0.90
connections of precast
members controlled by
yielding of steel elements
in tension
Plain concrete elements
0.60
Anchors in concrete
0.45 to 0,75
elements
Source: National Structural Code of the Philippines 2015
-
-
-
Table 4221.2.2
Strength Reduction Factor, ∅, for Moment, Axial Force, or Combined Moment and
Axial Force
∅
Net tensile
Strain, ϵ t
Classification
ϵ t ≤ ε ty
Compression
controlled
Transition[1]
ϵ ty <ε t <0.005
Type of transverse reinforcement
Spiral conforming to
Other
Sec. 425.7.3
0.75
a
0.65
0.75 + 0.15
(ε t −ε ty )
(0.005−ε ty )
c
ϵ t ≥0.005
b
0.65 + 0.25
(ε t −ε ty )
(0.005−ε ty )
0.90
d
Tension
0.90
e
controlled
[1]
For sections classified as transition, it shall be permitted to use ∅ corresponding to
compression-controlled sections.
Source: National Structural Code of the Philippines 2015
Values of β 1 shall be in accordance with Table 422.2.2.4.3 (Sec.422.2.2.4.3, NSCP
2015).
Table 422.2.2.4.3
Values of β 1 for Equivalent Rectangular Concrete Stress Distribution
f ' c, MPa
17 ≤ f ' c ≤ 28
28 ¿ f ' c < ¿ 55
β1
0.85
0.05(f ' c −28)
0.85
7
f ' c ≥ 55
0.65
Source: National Structural Code of the Philippines 2015
1. Design of One-Way Slab Using Moment Coefficient Method
Design Parameters:
According to ACE Specifications, when
L
≥ 2, it is a one-way slab.
S
Minimum Thickness specified in Sec. 407.3.1.1.1 of NSCP 2015.
a. For simply supported beam
(a)
(b)
(c)
f
t=
f
L
0.40+ y
20
700
(
)
b. For both end continuous
t=
f
L
0.40+ y
28
700
(
)
c. For one-end continuous
t=
f
L
0.40+ y
24
700
(
)
d. For Cantilever Section
t=
f
L
0.40+ y
10
700
(
)
Minimum clear spacing covering = 20 mm (Table 420.6.1.3.1, NSCP 2015).
For parallel non-prestressed reinforcement in a horizontal layer, clear spacing
shall be at least the greatest of 50 mm, db, and (4/3)dagg (Sec. 425.2.1, NSCP 2015)
Maximum spacing of deformed bars shall be the lesser of 3h and 450 mm (Sec.
407.7.2.3, NSCP 2015).
The spacing of deformed shrinkage and temperature reinforcement shall not
exceed the lesser of 5h and 450 mm (Sec. 424.4.3.3, NSCP 2015).
Table 420.6.1.3.1
Specified Concrete Cover for Cast-in-Place Non-Prestressed Concrete Members
Concrete
Member
Reinforcement
Specified Cover,
Exposure
Cast against and
permanently in
contact with ground
Exposed to weather
or in contact with
ground
All
mm
75
All
All
20 mm ∅ through
58 mm ∅ bars
16 mm ∅ bar,
MW200 or MD200
wire, and smaller
Not exposed to
Slabs, joists, and
40 mm ∅ and 58
weather or in
walls
mm ∅ bars
contact with ground
36 mm ∅ bar and
smaller
Beams, columns,
Primary
pedestals, and
reinforcement,
tension ties
stirrups, ties,
spirals, and hoops
Source: National Structural Code of the Philippines 2015
50
40
40
20
40
The ratio of deformed shrinkage and temperature reinforcement area to gross
concrete area shall satisfy the limits in Table 424.4.3.2.
Table 424.4.3.2
Minimum Ratios of Deformed Shrinkage and Temperature Reinforcement Area to
Gross Concrete Area
Reinforcement Type
fy, MPa
Deformed bars
¿ 420
Deformed bars or welded
wire reinforcement
≥ 420
Minimum Reinforcement
Ratio
0.0020
0.0018 x 420
Greater
fy
of:
0.0014
Source: National Structural Code of the Philippines 2015
Number of bars required in “one” meter width of slab = 1000/s. Thickness of slab
should not be less than 75 mm. If the slab is permanently exposed to the ground
minimum clear concrete cover is 70 mm.
Steps in Designing One-Way Slab (USD)
a. Assume thickness of slab using the minimum code requirements.
b. Consider the width to be one meter.
c. Compute the factored load (Wu).
Wu = 1.2D + 1.6L
d. Compute the ultimate moment (Mu).
It shall be permitted to calculate Mu due to gravity loads for continuous beams
and one-way slabs satisfying (d.1) through (d.2).
d.1 Members are prismatic;
d.2. L ≤ 3D; where D refers to the dead load.
Source: Sec. 406.5.1, NSCP 2015
Mu due to gravity loads shall be calculated in accordance with Table 406.5.2 (Sec.
406.5.2).
Table 406.5.2 Approximate Moments for Non-Prestressed Continuous Beams and
One-Way Slabs
Moment
Positive
Location
End Span
Condition
Discontinuous end integral with
support
Discontinuous end unrestrained
Interior Spans
All
Mu
W u l n2
14
W u l n2
11
W u l n2
16
Negative
Interior face of
exterior support
Member built integrally with
supporting spandrel beam
Member built with supporting
column
Exterior face of
first interior
support
Two spans
More than two spans
Face of Other
supports
All
Face of all
supports satisfying
(a) or (b)
a.) Slabs with spans not
exceeding 3m.
b.) Beams where ratio of sum
of column stiffness to beam
stiffness exceeds 8 at each
end of span
Source: National Structural Code of the Philippines 2015
e. Compute the value of the effective depth (d).
d = t – clear covering – ½ ∅
f. Compute “ω” from the equation:
Mu = ∅f’c b d2 ω (1-0.59ω)
g. Compute the value of the steel ratio, ρ.
ρ=
ωf ' c
fy
h. Check the steel ratio. It should be within ρmin & ρmax .
h.1. when ρ> ρ max, increase the slab thickness.
h.2. when ρ< ρ min, decrease slab thickness.
W u l n2
24
W u l n2
16
W u l n2
9
W u l n2
10
W u l n2
11
W u l n2
12
i. Compute the required steel area.
As = ρbd
j. Compute the spacing of main bars.
(
1000
s
π ∅2
=¿ As; s ¿ 3h or 450 mm
4
)( )
k. Compute the required area of temperature bars.
l. Compute the spacing of temperature bars.
(
1000
s
π ∅2
=¿ Ast s ¿ 5h or 450 mm
4
)( )
2. Design of Two-Way Slab
According to ACE Specifications, when
L
<¿ 2, it is a two-way slab.
S
For non-prestressed slabs with beams spanning between supports on all sides, overall slab
thickness h shall satisfy the limits in Table 408.3.1.2.
Table 408.3.1.2 Minimum Thickness of Non-Prestressed Two-Way Slabs with
Beams Spanning Supports on All Sides
α fm[1]
α fm ≤ 0.2
0.22 ¿ α fm ≤ 2.0
α fm >2.0
Minimum h, mm
Sec. 408.3.1.1 applies
Greater
f
l n 0.8 y
of:
1400
36+5 β (α fm−0.2)
125
Greater
fy
l n 0.8
of:
1400
36+ 9 β
(
(
)
)
(a)
(b)[2][3]
(c)
(d)[2][3]
[1]
α fm
90
(e)
is the average value of α f for all beams on edges of a panel and α f shall be
calculated in accordance with Section 408.10.2.7.
[2]
ln is the clear span in the long direction, measured face-to-face of beams (mm).
[3]
β is the ratio of clear span in long to short directions of slab.
Source: National Structural Code of the Philippines 2015
For a panel with beams between supports on all sides, Eq. 408.10.2.7a shall be
satisfied for beams in the two perpendicular directions (Section 408.10.2.7 of NSCP
2015).
α f 1 l 22
≤5.0
0.2 ≤
2
αf 2 l1
(408.10.2.7a)
Where α f 1 and α f 2 are calculated by
αf =
E cb I b
E cs I s
(408.10.2.7b)
Where:
α f = ratio of flexural stiffness of beam section to flexural stiffness of a width of
slab bounded laterally by centerlines of adjacent panels, if any, on each side of the beam.
α fm = average value of α f for all beams on edges of a panel.
α f 1 = α f in direction of l1
α f 2 = α f in the direction of l2.
Ecb = modulus of elasticity of beam, MPa.
Ecs = modulus of elasticity of slab concrete, MPa.
Ib = moment of inertia about centroidal axis of gross section of beam, mm4
Is = moment of inertia of slab about centroidal axis, mm4
Two-Way Slab Design Using the Coefficient Method (Edwards, 2019)
The Coefficient Method is a method of designing two-way slabs that are
supported by edge beams. Previously known as Method 3 of the 1963 ACI Code. This
method makes use of tables of moment coefficients for a variety of slab edge conditions.
The coefficients are based on elastic analysis but also include considerations for inelastic
moment redistribution.
The moments in the middle strips are calculated using formula (1) and (2):
Where:
Ma = Ca w la2
(1)
Mb = Cb w lb2
(2)
Ca = moment coefficient from table
Cb = moment coefficient from table
W = uniform load
la = clear span length in short direction
lb = clear span length in long direction
The panel must be divided into middle strips and edge strips in both the short and
long direction. The width of the middle strip in each direction is equal to ½ the clear span
length. The 2 edge strips are then ¼ the width of the clear span length, Fig. 3.
Figure 3
As expected in two-way slabs, the moments in both directions are larger in the center
portion of the slab than the edges. Therefore, the middle strip must be designed for 1/3 of
the maximum value of the calculated moment.
The ACI Coefficient Tables are designed to give appropriate coefficients based on
the edge conditions of the slab. The floor plan below gives the different edge conditions.
The numbers corresponds to the edge conditions in the following tables:
Case 4: 2 edges continuous, 2 edges discontinuous
Case 8: Figure
3 edges4continuous, 1 edge discontinuous
Case 2: 4 edges continuous
At continuous edges, moments are negative similar to continuous beams at interior
supports.
Table 1 gives the moment coefficients for Negative Moments at Continuous Edges. The
coefficient use depends on the ratio la/lb and the edge conditions of the panel in question.
The maximum negative moment is computed for full Dead Load and Live Load.
Negative moments at discontinuous (free) edges are assumed to be 1/3 of the positive
moment in the same direction.
Table 2 gives the moment coefficients for Positive Moment due to Dead Load. Again,
the coefficient used depends on the ratio of short span to long span as well as the edge
conditions.
Table 3 gives the moment coefficients for Positive Moment due to Live Load. This table
is used in the same manner as Table 2. The reason for the separation of Dead and Live
Load positive moments is due to Live Load placement to achieve maximum effect. For
live load, the maximum positive moment in the panel occurs when the full live load is on
the panel and not on any adjacent panel. This produces rotations at all continuous edges
of the panel which require restraining moments. Dead load across all the panels creates
rotations that cancel each other out (or closely enough).
Placing Reinforcement
The main reinforcement for the two-way edge-supported slab panel should be
placed orthogonally (parallel and perpendicular) to the slab edges. The reinforcement in
the short direction (la) should be placed lower than the reinforcement in the long direction
(lb). Negative reinforcement should be placed perpendicular to the supporting edge
beams.
All other requirements for minimum reinforcement (temperature & shrinkage)
should be observed. For two-way slab systems, the spacing of reinforcement should not
exceed twice (2) the slab thickness (tslab).
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3. Design of Beams and Girders
a. Determine the factored corrected moments using moment distribution method with
sway and non-sway analysis and using the design load combination in Eq. 203-5.
b. The design of beams and girders used in the process is using hand calculation.
V1¿
M 2 −M 1 WL
+
L
2
V2 = WL - V1
Where: M1 = moment at left end
M2 = moment at right end
Figure 5
V1 = shear reaction at left end
V2 = shear reaction at right end
X = distance from left end to the point in which the maximum moment
occurs
W = factored moments of live load, and dead load. The weight in roof
deck is to be added with factored ponding load
L = clear span of a beam or girder
c. Compute the maximum positive or negative moment.
M+ or M- = -M1 – (W)(X2/2) + (V1)(X)
d. Since the beams and girders are continuous, the maximum positive moments,
maximum negative moments, and maximum shear of the continuous beams and girders
should be used.
e. Check the dimension of the beam.
For non-prestressed beams not supporting or attached to partitions or other
construction likely to be damaged by large deflections, overall beam depth h, shall satisfy
the limits in Table 409.3.1.1 (Sec. 409.3.1.1, NSCP 2015).
Table 409.3.1.1
Minimum Depth of Non-Prestressed Beams
Support Condition
Minimum h[1]
Simply Supported
l/16
One end continuous
l/18.5
Both end continuous
l/21
Cantilever
l/8
[1]
Expressions applicable for normal weight concrete and fy = 420 MPa. For other cases,
minimum h shall be modified in accordance with Sections 409.3.1.1 through 409.3.1.1.3,
as appropriate.
Source: National Structural Code of the Philippines 2015
For fy other than 420 MPa, the expressions in Table 409.3.1.1 shall be multiplied
by (0.40 + fy/700) (Sec. 409.3.1.1.1, NSCP 2015).
Suggested width and depth proportion:
½ ≤ b/d ≤ 2/3
For Positive Moment, M+
f. Check the beam or girders if it is rectangular or T-beam section.
From ACI Code:
For effective flange width “b”:
b ≤ center to center spacing of beams
b = base of web, bw + clear distances to the next beam
b = bw + 16hf; hf = thickness of flange
b ≤ ¼ of span length
Use “b” whichever is smaller.
Determine the location of the neutral axis (NA):
f.1 If Muf ¿ Mu: NA will be in the flange
Therefore, design as rectangular beam.
f.2 If Muf ¿ Mu: NA will be in the web
Therefore, design as T-beam.
Note: Mu is the maximum positive moment computed from c.
Muf = ultimate moment capacity of the flange alone.
Mnf = Cf(d –
hf
);
2
Cf = 0.85 f’c hf b
Muf = ∅ Mnf
For Case 1:
Design of Rectangular Beam
f.1.1 Solve for ρ , from the formula of Mu.
ρ=
Mu = ∅f’c b d2ω(1 - 0.59ω);
ω f 'c
fy
d = total depth – effective covering from Table 420.6.1.3.1.
f.1.2 Solve for ρmax and ρmin .
ρmin =
1.4
;
fy
ρmax =0.75 ρb; ρb =0.85 β1
If ρmin < ρ< ρmax , then use ρ.
If ρ< ρ min < ρmax , then use ρmin .
If ρmin < ρmax < ρ, then use ρmax .
f.1.3 Solve for the area of steel requirements.
As = ρbd where ρ is either ρ , ρmin , ρmax .
f.1.4 Determine the number of bars needed.
No. of bars, N = As/A1;
A1 = area of a steel bar.
f.1.5 Check the spacing between longitudinal bars.
f 'c
fy
( )(
600
600+f y
)
Spacing should not be less than 25 mm and should not be less than the bar
diameter (ACI-code 2005). Specified covering for beams should be refer to Table
420.6.1.3.1.
For Case 2:
Design of T-Beam
f.2.1 Solve for Mu2.
Mu2 = ∅ Mn2
-to be resisted by the flange beam
Mn2 = C2 (d – hf/2)
Mn2 = 0.85 f’c hf(b – bw)(d – hf/2)
f.2.2 Solve for Asf.
C2 = T2
0.85 f’c hf (b – bw) = Asf fy
0.85 f ' c h f ( b−b w )
Asf =
fy
f.2.3 Solve for Mu1.
Mu1 = Mu – Mu2
f.2.4. Solve for the restricted steel ratio.
ρ1=0.85
[ √
f 'c
2−M u 1
1− 1−
fy
0.85 ∅ f ' c bw d 2
f.2.5 Check for steel ratio.
]
Determine ρmin and ρmax from f.1.1
f.2.6 Solve for As1.
As1 = ρbd where ρ is either ρ1 , ρmin , ρmax .
f.2.7 Solve for the total steel requirement, As.
As = As1 + Asf
f.2.8 Refer to f.1.4 for the spacing between longitudinal bars.
For Negative Moment, Mg. Determine the area of steel requirement, As, and the number of bars needed for top bars
using the maximum negative moment. Follow the procedures from f.1.1 to f.2.8.
Design of Shear, Vu
Vu
ØVc/2
d
X
h. Solve for the design shear, Vu and nominal shear, Vn.
Vu = V-Wd;
d = effective depth
i. Check if the design needs shear reinforcement.
Vc =
1
f' bd
6√ c
Vu ¿
∅Vc
;
2
Requires web reinforcement
Vu ¿
∅Vc
;
2
Does not require web reinforcement
j. Solve for Vs.
Vs =
Vu
- Vc
∅
k. Solve for the total cross-sectional area of web reinforcement, Av.
Av = 2As;
As = cross-sectional area of stirrups
l. Determine the spacing of stirrups.
S==
Av f y d
Vs
m. Determine the maximum spacing.
m.1 S = d/2 or 600 mm; use whichever is smaller when Vs ¿
1
f' bd
3√ c
m.2 S = d/4 or 300 mm; use whichever is smaller when Vs ¿
1
f' bd
3√ c
m.3 Desirable spacing of stirrups = 100 mm
m.4 The first stirrups should be 50 mm from the face of support.