CODE PROVISIONS

429.4 PERMISSIBLE SERVICE LOAD STRESSES 419.2.3 MODULUS OF RUPTURE OF CONCRETE

429.4.1 Stresses in concrete shall not exceed: 419.2.3.1 Modulus of rupture, fr, for concrete shall be
calculated by:
Flexure ------------------------------------------𝟎. 𝟒𝟓𝒇′𝒄
429.4.2 Tensile stress in reinforcement 𝑓𝑠 shall not 𝒇𝒓 = 𝟎. 𝟔𝟐𝝀√𝒇′𝒄 (419.2.3.1)
exceed where the value of λ is in accordance with Section 419.2.4
For Flexural reinforcement -------------------𝟎. 𝟓𝟎𝒇𝒚 (Lightweight Concrete)
419.2.4 Lightweight Concrete 419.2.4.1 To account for
the properties of lightweight concrete, a modification
429.6 FLEXURE factor λ is used as a multiplier of √𝑓′𝑐 in all applicable
provisions of this Code. 419.2.4.2 The value of λ shall be
For investigation of stresses at service loads, straight-
based on the composition of the aggregate in the concrete
line theory (for flexure) shall be used with the following
mixture in accordance with Table 419.2.4.2 or as
assumptions.
permitted in Section 419.2.4.3.
429.6.1 Strains vary linearly as the distance from the
neutral axis, except for deep flexural members with
overall depth-span ratios greater than 2/5 for continuous
spans and 4/5 for simple spans, a nonlinear distribution
of strain shall be considered.
429.6.2 Stress-strain relationship of concrete is a straight
line under service loads within permissible service load
stresses. 419.2.4.3 If the measured average splitting tensile
strength of lightweight concrete, fct , is used to calculate
429.6.3 In reinforced concrete members, concrete resists λ, laboratory test shall be conducted in accordance with
no tension. 429.6.4 It shall be permitted to take the ASTM C330M to establish the value of fct ,and the
modular ratio 𝑛 = 𝐸𝑠/𝐸𝑐, as the nearest whole number (but corresponding value of fcm and λ shall be calculated by:
NOT less than 6). Except for calculations for deflections,
value of n for lightweight concrete shall be assumed to be 𝒇𝒄𝒕
𝝀= ≤ 𝟏. 𝟎 (419.2.4.3)
the same as for normal weight concrete of the same 𝟎. 𝟓𝟔√𝒇𝒄𝒎
strength.
The concrete mixture tested in order to calculate λ shall
429.6.5 In doubly reinforced flexural members, an be representative of that to be used in the Work.
effective modular ratio of 2𝐸𝑠/𝐸𝑐 shall be used to
transform compression reinforcement for stress where:
computations. Compressive stress in such reinforcement 𝑓𝑐𝑚 = measured average compressive strength of concrete,
shall NOT exceed permissible tensile stress. MPa

419.2.2 MODULUS OF ELASTICITY 424.2.3 CALCULATION OF IMMEDIATE

419.2.2.1 Modulus of elasticity 𝐸𝑐 for concrete shall be DEFLECTIONS
permitted to be taken as 424.2.3.1 Immediate deflections shall be calculated using
methods or formulas for elastic deformations, considering
𝒘𝒄 𝟏.𝟓 𝟎. 𝟎𝟒𝟑√𝒇′𝒄 (𝑖𝑛 𝑀𝑃𝑎) effects of cracking and reinforcement on member
for values of 𝑤𝑐 between 1,440 and 2,560 kg/m³. stiffness.

For normal weight concrete, 𝐸𝑐 shall be permitted to be 424.2.3.2 Effect of variation of cross-sectional properties,
taken as such as haunches, shall be considered when calculating
deflections.
𝑬𝒄 = 𝟒, 𝟕𝟎𝟎√𝒇′𝒄 (𝑖𝑛 𝑀𝑃𝑎)
424.2.3.3 Deflections in two-way slab system shall be
calculated taking into account size and shape of the panel,
conditions of support, and nature of restraints at the panel
edges.
424.2.3.4 Modulus of elasticity, Ec, shall be permitted to 𝜆∆ = Multiplier used for additional deflection due to long-
be calculated in accordance with section 419.2.2. term effects
424.2.3.5 For non-prestressed members, effective 424.2.4.1.2 In the equation above, 𝜌′ shall be calculated at
moment of inertia, Ie, shall be calculated by Eq. mid-span for simple and continuous spans, and at the
424.2.3.5a (show below) unless obtained by a more support for cantilevers.
comprehensive analysis, but 𝐼𝑒 shall not be greater than
𝐼𝑔 . For midspan for simple and continuous span
𝑨′𝒔
𝑰𝒄𝒓 ≤ 𝑰𝒆 ≤ 𝑰𝒈 𝝆′ =
𝒃𝒅
𝑴𝒄𝒓 𝟑 𝑴𝒄𝒓 𝟑 For cantilever
𝑰𝒆 = ( ) 𝑰𝒈 + [𝟏 − ( ) ] 𝑰𝒄𝒓 (424.2.3.6𝑎)
𝑴𝒂 𝑴𝒂
𝑨𝒔
𝝆′ =
where: 𝒃𝒅
𝐼𝑐𝑟 = Moment of inertia of cracked section transformed to 424.2.4.1.3 In the equation above, values of the time
concrete. dependent factor for sustained load, ξ, shall be in
accordance with Table 424.2.4.1.3 (shown below)
𝐼𝑒 = Effective moment of inertia for computation of
deflection.
𝐼𝑔 = Moment of inertia of gross concrete section about the
centroidal axis, neglecting reinforcement.
𝑀𝑐𝑟 = Cracking moment.
𝑀𝑎 = Maximum moment in member at stage deflection is
computed.

Cracking Moment, 𝑴𝒄𝒓 , is calculated by:

𝒇𝒓 𝑰𝒈
𝑴𝒄𝒓 = (424.2.3.5𝑏)
𝒚𝒕
where: FLEXURAL ANALYSIS (WSD)

𝑦𝑡 = Distance from centroidal axis of gross section, A. Uncracked Stage (𝒇𝒕 ≤ 𝒇𝒓 𝒐𝒓 𝑴𝒂 ≤ 𝑴𝒄𝒓 )
neglecting reinforcement.
Rectangular Section
424.2.3.6 For continuous one-way slabs and beams, Ie
shall be permitted to be taken as the average of values
obtained from Eq. 424.2.3.5a for the critical positive and
negative moment sections.
424.2.3.7 For prismatic one-way slabs and beams, Ie shall
be permitted to be taken as the value obtained from Eq.
424.2.3.5a at mid-span for simple and continuous spans,
and at the support for cantilevers.

424.2.4 CALCULATION OF TIME-DEPENDENT

DEFLECTIONS
424.2.4.1.1 Unless values from a more comprehensive 𝒇𝒓 𝑰𝒈
𝑴𝒄𝒓 =
analysis, additional timedependent deflection resulting 𝒚𝒕
from creep and shrinkage of flexural members shall be
calculated as the product of the immediate deflection
caused by the sustained load and the factor 𝜆Δ .
𝝃
𝝀𝚫 = (424.2.4.1.1)
𝟏 + 𝟓𝟎𝝆′
where:
𝜉 = Time-dependent factor for sustained load
B. Cracked Stage (𝒇𝒕 > 𝒇𝒓 𝒐𝒓 𝑴𝒂 > 𝑴𝒄𝒓 ) FLEXURAL DESIGN (WSD)
Transformed Area Method
Rectangular Section

Procedure
Step 1: Calculate the actual expected moment (𝑀𝑎 ) to be
carried by the beam at service level
Location of the neutral axis from extreme compression
fiber Step 2: Solve for the balanced moment capacity

• Singly reinforced beam (SRB): 𝑑𝑓𝑐

𝑥𝑏𝑎𝑙 =
𝑓
1 2 𝑓𝑐 + 𝑛𝑠
𝑏𝑥 = 𝑛𝐴𝑠 (𝑑 − 𝑥)
2
1
• Doubly reinforced beam (DRB): 𝐶𝑏𝑎𝑙 = 𝑓𝑐 (𝑏 ∙ 𝑥𝑏𝑎𝑙 )
2
1 2 1
𝑏𝑥 + (2𝑛 − 1)𝐴′𝑠 (𝑥 − 𝑑′) = 𝑛𝐴𝑠 (𝑑 − 𝑥) 𝑀𝑏𝑎𝑙 = 𝐶𝑏𝑎𝑙 (𝑑 − 𝑥𝑏𝑎𝑙 )
2 3
Cracked Section moment of inertia (𝑰𝒄𝒓 = 𝑰𝑵.𝑨. ) • 𝑀𝑎 ≤ 𝑀𝑏𝑎𝑙 , 𝑑𝑒𝑠𝑖𝑔𝑛 𝑎𝑠 𝑆𝑅𝐵
• 𝑀𝑎 > 𝑀𝑏𝑎𝑙 , 𝑑𝑒𝑠𝑖𝑔𝑛 𝑎𝑠 𝐷𝑅𝐵
• Singly reinforced
Step 3: Design as SRB (𝑀𝑎 ≤ 𝑀𝑏𝑎𝑙 )
𝑏𝑥 3
𝐼𝑐𝑟 = + 𝑛𝐴𝑠 (𝑑 − 𝑥)2
3
• Doubly reinforced

𝑏𝑥 3
𝐼𝑐𝑟 = + (2𝑛 − 1)𝐴′ 𝑠 (𝑥 − 𝑑′ )2 + 𝑛𝐴𝑠 (𝑑 − 𝑥)2
3
Actual stresses

• General flexure formula

𝑀𝑦
𝑓=
𝐼 Locate the neutral axis
• Concrete stress at extreme compression fiber 𝑏𝑥 3 𝑏𝑥 2 (𝑑 − 𝑥) 𝑛𝑀𝑎 (𝑑 − 𝑥)
𝑀𝑎 𝑥 + =
𝑓𝑐 = 3 2 𝑓𝑠
𝐼𝑐𝑟
• Tension steel 𝑀𝑎
𝐴𝑠 =
𝑓𝑠 𝑀𝑎 (𝑑 − 𝑥) 1
= 𝑓𝑠 (𝑑 − 3 𝑥)
𝑛 𝐼𝑐𝑟
• Compression steel Step 4: Design as DRB (𝑀𝑎 > 𝑀𝑏𝑎𝑙 )
𝑓′𝑠 𝑀𝑎 (𝑥 − 𝑑′)
=
2𝑛 𝐼𝑐𝑟
For Tension steel reinforcement (𝐴𝑠 )
i. Solve for 𝐴𝑠1 from balanced condition
𝑀𝑏𝑎𝑙
𝐴𝑠1 =
1
𝑓𝑠 (𝑑 − 3 𝑥𝑏𝑎𝑙 )
ii. Solve for 𝐴𝑠2 from excess of 𝑀𝑎 and 𝑀𝑏𝑎𝑙
(𝑀𝑎 − 𝑀𝑏𝑎𝑙 )
𝐴𝑠2 =
𝑓𝑠 (𝑑 − 𝑑′)
iii. Solve for total area of tension reinforcement
𝐴𝑠 = 𝐴𝑠1 + 𝐴𝑠2
For Compression steel reinforcement (𝐴′𝑠 )
i. Solve for 𝐴′𝑠 from balanced condition and use
𝑀 = (𝑀𝑎 − 𝑀𝑏𝑎𝑙 )
ii. If 𝑓′𝑠 ≥ 𝑓𝑠 , use 𝑓′𝑠 = 𝑓𝑠
2𝑛𝑓𝑐 (𝑥𝑏𝑎𝑙 − 𝑑′)
𝑓′𝑠 =
𝑥𝑏𝑎𝑙
2𝑓𝑠 (𝑥𝑏𝑎𝑙 − 𝑑′)
𝑓′𝑠 =
𝑑 − 𝑥𝑏𝑎𝑙
iii. Solve for 𝐴′𝑠
2𝑛(𝑀𝑎 − 𝑀𝑏𝑎𝑙 )
𝐴′𝑠 =
𝑓′𝑠 (2𝑛 − 1)(𝑑 − 𝑑′)