Analysis and Design of Beams (Flexure)

Learning Objectives

  • Apply equilibrium and strain compatibility to rectangular reinforced-concrete sections at nominal flexural strength.
  • Use the Whitney equivalent rectangular compression block with the correct β1\beta_1 limits.
  • Verify rather than blindly assume whether tension and compression reinforcement has yielded.
  • Calculate MnM_n, ϕ\phi, and ϕMn\phi M_n for singly and doubly reinforced beams.
  • Distinguish minimum reinforcement, the beam minimum tensile-strain requirement, balanced behavior, and the tension-controlled limit.
  • Analyze flanged beams using an effective compression-flange width and the correct compression-block case.
  • Distinguish ordinary beam flexure, side-face skin reinforcement, and deep-beam/strut-and-tie behavior.
  • Check reinforcement geometry, effective depth, extreme-tension depth, and constructability assumptions before accepting a design.

Flexural nominal strength

The moment resistance MnM_n of a section evaluated at the nominal-strength strain state using equilibrium, compatibility, the specified material models, and the applicable reinforced-concrete strength provisions.

NSCP 2015 / Adopted ACI 318-14 Basis

This topic uses the Philippine NSCP 2015 concrete-design context and its adopted ACI 318-14 flexural basis unless a different edition is explicitly identified. The course therefore uses an extreme concrete compression strain of 0.0030.003, the ACI 318-14 rectangular stress block, and the ACI 318-14 strain-based ϕ\phi limits described below. Newer ACI editions must not be mixed silently into these calculations.

Core Flexural Analysis Assumptions

  1. Sections that are plane before bending remain plane after bending, so longitudinal strain varies linearly over the section depth.
  2. Reinforcement strain equals the strain of the surrounding concrete at the same level when adequate bond and anchorage are present.
  3. Internal compression and tension forces satisfy equilibrium.
  4. Concrete tensile stress is neglected when calculating nominal flexural strength after cracking.
  5. The extreme concrete compression strain at nominal flexural strength is taken as ϵcu=0.003\epsilon_{cu}=0.003.
  6. Concrete compression is represented by the equivalent rectangular stress block defined by the adopted code basis.
  7. Reinforcement stress must be obtained from the calculated reinforcement strain; fs=fyf_s=f_y is valid only after yielding has been verified.

Effective depth (dd)

The distance from the extreme compression fiber to the centroid of the longitudinal tension reinforcement.

Extreme tension-reinforcement depth (dtd_t)

The distance from the extreme compression fiber to the center of the reinforcement layer farthest into the tension zone. It is the depth used to evaluate the net tensile strain ϵt\epsilon_t for beam minimum-strain checks, strain classification, and the flexural strength-reduction factor.

Do Not Confuse d with d_t in Multilayer Reinforcement

For one tension-reinforcement layer, d=dtd=d_t, so the distinction is invisible. For multiple tension layers, dd is the depth to the area centroid of the longitudinal tension reinforcement, while dtd_t is the depth to the extreme tension layer. They generally differ. Calculate each layer strain from its own depth. If the layer stresses differ, the internal tension-force resultant does not generally act at dd; calculate the layer forces and their moments individually. If all participating layers develop the same stress, such as the same yielded fyf_y, the force resultant coincides with the steel-area centroid at dd. Never use the area centroid as a substitute for the extreme-layer strain.

Whitney equivalent rectangular stress block

The code-approved equivalent concrete compression block having uniform stress 0.85fc′0.85f'_c over depth a=β1ca=\beta_1c, where cc is the neutral-axis depth measured from the extreme compression fiber.

Equivalent Compression-Block Depth

Relates the equivalent rectangular block depth to the neutral-axis depth.

a=β1ca=\beta_1c

Variables

SymbolDescriptionUnit
aaEquivalent rectangular stress-block depthmm
β1\beta_1Stress-block depth factor-
ccNeutral-axis depth from the extreme compression fibermm

β1\beta_1 for the Adopted ACI 318-14 Basis

For normal-strength ranges used in this course, β1=0.85\beta_1=0.85 for 17≤fc′≤28 MPa17\le f'_c\le28\,\text{MPa}. For fc′>28 MPaf'_c>28\,\text{MPa}, reduce β1\beta_1 by 0.050.05 for each 7 MPa7\,\text{MPa} increase above 28 MPa28\,\text{MPa}, but do not take β1<0.65\beta_1<0.65.

β1=max⁡[0.65,  0.85−0.05(fc′−287)]\beta_1=\max\left[0.65,\;0.85-0.05\left(\frac{f'_c-28}{7}\right)\right]

The reduction applies only above 28 MPa28\,\text{MPa}; do not extrapolate the line below the code range.

Singly Reinforced Rectangular Beams

For a rectangular section with tension steel only, the concrete resultant is

Cc=0.85fc′baC_c=0.85f'_cba,

acting at a/2a/2 from the compression face. The tension force is T=∑AsifsiT=\sum A_{si}f_{si}. For a single yielded tension layer this reduces to T=AsfyT=A_sf_y. Equilibrium requires Cc=TC_c=T. If a yielded-steel shortcut is proposed for multiple tension layers, verify the strain and stress of every participating layer before combining their forces. If any layer remains elastic, use its actual fsi=Esϵsif_{si}=E_s\epsilon_{si} rather than forcing the entire tension reinforcement to fyf_y.

Singly Reinforced Equilibrium When Tension Steel Yields

Closed-form compression-block depth after the assumption fs=fyf_s=f_y has been verified for the participating tension reinforcement.

a=Asfy0.85fc′ba=\frac{A_sf_y}{0.85f'_cb}

Variables

SymbolDescriptionUnit
AsA_sArea of tension reinforcement verified to be at the same yield stressmm2mm^2
fyf_ySpecified reinforcement yield strengthMPa
fc′f'_cSpecified concrete compressive strengthMPa
bbCompression-face width of the rectangular sectionmm

Extreme Tension-Steel Strain from Compatibility

Net tensile strain at nominal strength in the reinforcement layer farthest from the extreme compression fiber.

ϵt=0.003(dt−cc)\epsilon_t=0.003\left(\frac{d_t-c}{c}\right)

Variables

SymbolDescriptionUnit
ϵt\epsilon_tNet tensile strain in the extreme tension reinforcement at nominal strength-
dtd_tDepth from the extreme compression fiber to the extreme tension-reinforcement layermm
ccNeutral-axis depthmm

Layer-by-Layer Compatibility

For any tension layer ii at depth did_i, calculate

ϵsi=0.003(di−cc)\epsilon_{si}=0.003\left(\frac{d_i-c}{c}\right)

and obtain fsif_{si} from the reinforcement stress-strain model. The extreme value used for ϵt\epsilon_t corresponds to dt=max⁡(di)d_t=\max(d_i) on the tension side. If all participating tension layers have yielded to the same fyf_y, their forces may be combined at the steel-area centroid dd for the moment calculation. If their stresses differ, retain the layer forces separately rather than forcing their resultant to act at dd. Neither case changes the requirement to use dtd_t for strain classification.

Nominal Moment of a Yielded Singly Reinforced Rectangular Section

For one tension layer, or multiple tension bars at the same stress, the tension resultant acts at the steel-area centroid d.

Mn=Asfy(d−a2)M_n=A_sf_y\left(d-\frac{a}{2}\right)

Variables

SymbolDescriptionUnit
MnM_nNominal flexural strengthN mm
AsA_sArea of yielded tension reinforcement at the common stress fymm2mm^2
fyf_ySpecified reinforcement yield strengthMPa
ddDepth to the area centroid of the longitudinal tension reinforcementmm
aaEquivalent stress-block depthmm

Do Not Use the Yielded-Steel Formula Before Checking Yield

The equation a=Asfy/(0.85fc′b)a=A_sf_y/(0.85f'_cb) assumes the participating tension steel represented by AsA_s has reached the same fyf_y. For one tension layer, calculate c=a/β1c=a/\beta_1 and its strain immediately. For multiple layers, calculate every ϵsi\epsilon_{si} at its own did_i and verify every layer included in AsfyA_sf_y has yielded. If any layer has ϵsi<ϵy\epsilon_{si}<\epsilon_y, discard the lumped yielded-steel assumption and solve equilibrium with the actual layer stresses. Separately, use the extreme layer at dtd_t to determine ϵt\epsilon_t for the beam minimum-strain check and ϕ\phi classification.

Balanced strain condition

The theoretical condition at which the extreme compression concrete reaches 0.0030.003 at the same time the tension reinforcement reaches its yield strain ϵy\epsilon_y.

Balanced Reinforcement Ratio for the Idealized Yield Model

Useful reference ratio obtained from balanced strain compatibility for ordinary nonprestressed reinforcement.

ρb=0.85β1fc′fy(600600+fy)\rho_b=0.85\beta_1\frac{f'_c}{f_y}\left(\frac{600}{600+f_y}\right)

Variables

SymbolDescriptionUnit
ρb\rho_bBalanced tension-reinforcement ratio-
β1\beta_1Stress-block depth factor-
fc′f'_cSpecified concrete compressive strengthMPa
fyf_ySpecified reinforcement yield strengthMPa

Minimum Reinforcement and Ductility Are Different Checks

Minimum tension reinforcement is intended to prevent an abrupt loss of flexural resistance immediately after concrete cracking. It is not the same as a maximum reinforcement ratio, balanced ratio, or tension-controlled classification. For ordinary nonprestressed rectangular beams on this course basis, evaluate the code minimum area and separately verify the strain limits at nominal strength.

Minimum Tension Reinforcement for Ordinary Nonprestressed Beams

Use the greater of the two ACI/NSCP expressions when the stated beam provision applies.

As,min⁡=max⁡(0.25fc′fybwd,  1.4fybwd)A_{s,\min}=\max\left(\frac{0.25\sqrt{f'_c}}{f_y}b_wd,\;\frac{1.4}{f_y}b_wd\right)

Variables

SymbolDescriptionUnit
As,min⁡A_{s,\min}Minimum required tension-reinforcement areamm2mm^2
bwb_wWeb widthmm
ddEffective depth to the centroid of the longitudinal tension reinforcementmm

Beam Minimum Tensile Strain versus Tension-Controlled Classification

For the adopted ACI 318-14 basis, nonprestressed flexural members with negligible axial compression are generally required to have ϵt≥0.004\epsilon_t\ge0.004 at nominal strength. This is a minimum beam ductility requirement, not the definition of a tension-controlled section.

A section is tension-controlled when ϵt≥0.005\epsilon_t\ge0.005, which permits ϕ=0.90\phi=0.90 for flexure. Between the yield-strain boundary and 0.0050.005, ϕ\phi lies in the transition region. In multilayer reinforcement, these limits apply to the extreme tension reinforcement at dtd_t, not automatically to the centroidal depth dd. Do not use 0.0040.004 and 0.0050.005 interchangeably.

Strength-Reduction Factor for Nonprestressed Flexure on the Adopted Basis

Strain-based phi rule for tied/nonprestressed flexural sections using the actual yield strain.

ϕ={0.65,ϵt≤ϵy0.65+0.25ϵt−ϵy0.005−ϵy,ϵy<ϵt<0.0050.90,ϵt≥0.005\phi= \begin{cases} 0.65, & \epsilon_t\le\epsilon_y\\ 0.65+0.25\dfrac{\epsilon_t-\epsilon_y}{0.005-\epsilon_y}, & \epsilon_y<\epsilon_t<0.005\\ 0.90, & \epsilon_t\ge0.005 \end{cases}

Variables

SymbolDescriptionUnit
ϕ\phiStrength-reduction factor-
ϵt\epsilon_tExtreme tension-steel strain at nominal strength-
ϵy\epsilon_yReinforcement yield strain, fy/Es-

Interactive Section Verification

Use the section analyzer to change fc′f'_c, fyf_y, section dimensions, and reinforcement. Watch the computed yield strain, neutral axis, steel stress, strain classification, MnM_n, and ϕMn\phi M_n. The interactive model uses one tension layer, so its displayed dd is also dtd_t. Invalid bar layouts are flagged rather than treated as legitimate designs.

RC Beam Section Analysis

NSCP 2015 / adopted ACI 318-14 rectangular-section model. Reinforcement stress is solved from the actual strain state.

Beam width bb300 mm
Beam height hh500 mm
Tension-steel centroid from tension face60 mm
fc′f'_c28 MPa
fyf_y420 MPa
Longitudinal bar diameter20 mm
Number of tension bars3
NA c=65.2 mm0.003εt=0.01724

Section results

dd
440.0 mm
AsA_s
942 mm²
β1\beta_1
0.850
cc
65.2 mm
aa
55.4 mm
ϵy\epsilon_y
0.00210
ϵt\epsilon_t
0.01724
fsf_s
420.0 MPa
Steel state
Yielded
Strain class
Tension-controlled
ϕ\phi
0.900
Clear bar spacing
80.0 mm
MnM_n
163.20 kN·m
ϕMn\phi M_n
146.88 kN·m

Scope: singly reinforced rectangular flexure only. The horizontal fit check uses 40 mm side clear cover and a simplified clear-spacing criterion of max(25 mm, bar diameter); aggregate-size, bundling, development, shear, serviceability, and seismic detailing are outside this visualizer.

Doubly reinforced beam

A beam containing both tension and compression longitudinal reinforcement, commonly used when section dimensions are constrained, moment demand is high, moment reversal is expected, or compression steel is otherwise required by structural detailing.

Doubly Reinforced Force Equilibrium

Compression steel does not automatically yield. Its strain is obtained from the same linear strain diagram:

ϵs′=0.003(c−d′)/c\epsilon'_s=0.003(c-d')/c,

and its stress is limited by the reinforcement constitutive model. When compression reinforcement lies inside the equivalent rectangular concrete block, one consistent bookkeeping convention is

Cc=0.85fc′bβ1cC_c=0.85f'_cb\beta_1c

plus the net compression-steel contribution

Cs′=As′(fs′−0.85fc′)C'_s=A'_s(f'_s-0.85f'_c).

The subtraction prevents double-counting the concrete volume displaced by the steel. An alternative convention may exclude the displaced concrete from CcC_c explicitly; either convention is acceptable only if it is applied consistently.

Doubly Reinforced Equilibrium with Compression Steel inside the Stress Block

Consistent equilibrium equation using the net compression-steel contribution for a single equivalent tension-steel stress.

0.85fc′bβ1c+As′(fs′−0.85fc′)=Asfs0.85f'_cb\beta_1c+A'_s(f'_s-0.85f'_c)=A_sf_s

Variables

SymbolDescriptionUnit
As′A'_sCompression-reinforcement areamm2mm^2
fs′f'_sCompression-reinforcement stress from compatibilityMPa
AsA_sTension-reinforcement area represented by the common stress fsmm2mm^2
fsf_sCommon tension-reinforcement stress after compatibility verificationMPa

Doubly Reinforced Nominal Moment with the Net-Steel Convention

Applicable when the tension reinforcement can be represented by one resultant at its area centroid d, such as one layer or equal-stress yielded layers.

Mn=Cc(d−a2)+Cs′(d−d′)M_n=C_c\left(d-\frac{a}{2}\right)+C'_s(d-d')

Variables

SymbolDescriptionUnit
CcC_cConcrete compression resultantN
Cs′C'_sNet compression-steel resultantN
ddDepth to the area centroid of the represented tension reinforcementmm
d′d'Depth from the compression face to compression-steel centroidmm

General Multilayer Tension Reinforcement

If tension layers develop different stresses, do not force them into the preceding AsfsA_sf_s and dd representation. Satisfy equilibrium with T=∑AsifsiT=\sum A_{si}f_{si} and calculate nominal moment from the actual concrete, compression-steel, and individual tension-layer forces and lever arms. The code-defined effective depth dd remains the steel-area centroid, while the force-resultant location is a separate mechanics quantity.

Compression Steel Yield Must Be Verified

Assuming fs′=fyf'_s=f_y can materially distort the neutral-axis solution and moment capacity. Calculate ϵs′\epsilon'_s from the final cc, evaluate fs′=Esϵs′f'_s=E_s\epsilon'_s while the steel is elastic, cap it at fyf_y when yielding occurs, and then re-establish equilibrium. A solved value of cc is not valid if the stress assumptions used to obtain it contradict the resulting strains.

Flanged beam

A beam whose monolithic slab participates as part of the compression flange over an effective width permitted by the code, producing T- or L-shaped compression geometry when the flange is in compression.

T-Beam and L-Beam Analysis

First determine the effective flange width from the applicable NSCP/ACI limits. Then assume the compression block lies within the flange and calculate aa. If a≤hfa\le h_f, analyze the section as a rectangular section of width bfb_f. If a>hfa>h_f, split the concrete compression into the flange overhang and the web so that equilibrium and the centroid of compression are calculated from the actual equivalent block geometry.

Effective flange-width rules depend on beam location, span, flange thickness, and spacing to adjacent webs. They are code limits—not permission to use the entire slab width by default.

Deep beam

A member or region in which the load and support geometry produces significant nonlinear strain distribution and direct compression-strut action, so ordinary beam flexure assumptions are not adequate.

Deep-Beam Behavior Is Not a Skin-Reinforcement Rule

On the adopted basis, a deep beam is identified from span-to-depth and load-to-support geometry, including regions where concentrated loads occur within about twice the member depth from a support and direct strut action develops. Such regions are designed using the applicable deep-beam provisions and commonly the strut-and-tie method.

Side-face skin reinforcement is a separate ordinary-beam detailing requirement. For sufficiently deep flexural members, longitudinal reinforcement is distributed along the side faces near the tension zone to control crack widths. A beam does not become a deep beam merely because its overall depth exceeds the threshold that triggers skin reinforcement, and deep-beam web reinforcement should not be described as ordinary skin reinforcement.

Do Not Apply Plane-Sections Flexure to a Deep-Beam D-Region

Where the strain field is strongly disturbed by nearby concentrated loads, supports, openings, or geometric discontinuities, the ordinary linear strain distribution used for slender beam flexure may be invalid. Use the governing deep-beam or strut-and-tie provisions rather than forcing a familiar Mn=Asfy(d−a/2)M_n=A_sf_y(d-a/2) calculation onto a discontinuity region.

One-Way Joist and Ribbed Systems

A one-way joist system consists of regularly spaced ribs and a monolithic top slab. Qualification for special joist provisions depends on geometric limits such as minimum rib width, maximum depth-to-width ratio, and maximum clear spacing. If those limits are not satisfied, the ribs must be designed under the ordinary provisions applicable to their actual geometry rather than assuming special joist allowances.

Reinforcement Geometry Is Part of the Design

Calculated steel area is not a complete design. Selected bars must fit inside the stirrups and concrete cover with code-compliant clear spacing, practical layer arrangement, and a steel-area centroid consistent with the dd used by code expressions. Compression bars must be laterally supported as required. When bars are arranged in multiple tension layers, recalculate dd, identify dtd_t separately, and calculate each layer strain. If their stresses differ, calculate individual layer forces rather than treating dd as the force-resultant location.

Interactive Reinforcement-Geometry Check

Use the 3D viewer to change beam dimensions, clear cover, stirrup diameter and spacing, longitudinal bar diameter, and bar quantity. The viewer checks simplified clear-spacing and cover geometry and suppresses reinforcement cages that do not physically fit. It is a detailing visualizer, not a substitute for the complete code spacing, aggregate-size, development-length, splice, or seismic-detailing checks.

RC Beam Reinforcement Geometry

Adjust the section and cage geometry. Invalid cover or bar-spacing states are flagged instead of rendered as acceptable reinforcement.

Geometry valid
Beam width300 mm
Beam height500 mm
Beam length3000 mm
Clear cover to stirrup40 mm
Top bars2
Bottom bars3
Longitudinal bar diameter20 mm
Stirrup diameter10 mm
Maximum stirrup spacing200 mm
300 × 500 mm section · 3000 mm long
Cover 40 mm · stirrups Ø10 @ 194 mm actual
Top clear spacing 160 mm · bottom 70 mm
Drag to rotate · scroll/pinch to zoom
Spacing check shown here is a simplified geometric screen using clear spacing ≥ max(bar diameter, 25 mm). Project detailing must also satisfy the governing code provisions, aggregate-size effects, development, splice, and constructability requirements.
The viewer is a reinforcement-geometry teaching model, not a full structural designer. Bar anchorage, hooks, seismic detailing, and member strength are outside this component’s scope.

3D Flexural-Response Studio: Strain Compatibility and Internal Forces

Nominal-Strength 3D Model — NSCP 2015 / Adopted ACI 318-14

The 3D flexural-response studio uses the same one-layer rectangular-beam strain-compatibility basis as the section analyzer above. It solves the neutral axis, Whitney block, actual tension-steel strain/stress, MnM_n, strain-based ϕ\phi, and ϕMn\phi M_n without assuming steel yield before checking ϵt≥ϵy\epsilon_t\ge\epsilon_y. The concrete compression block, neutral-axis plane, tension bars, and CcC_c–TT force couple are generated from that same state. The displayed member curvature and crack pattern are schematic demand cues only; they are not service-deflection or crack-width predictions.

RC Beam Flexure — 3D Strain Compatibility & Internal Force Studio

Concept and model scope

Connect beam geometry, reinforcement, the neutral axis, Whitney compression block, steel strain and yield, internal-force equilibrium, and strain-based flexural strength.

The model is limited to one tension layer in a singly reinforced rectangular beam on the NSCP 2015 / adopted ACI 318-14 lesson basis: εcu = 0.003, the edition-specific β1 rule, actual steel strain/stress verification, the 0.004 ordinary-beam minimum tensile-strain check, the 0.005 tension-controlled threshold, and strain-based φ interpolation.

Rendered member curvature and cracks are schematic demand cues. This nominal-strength studio does not claim service deflection or crack-width prediction.

Modeled singly reinforced flexural state satisfies the displayed lesson checks.

Geometry, minimum steel, adopted minimum tensile strain, and Mu ≤ φMn are satisfied for this one-layer rectangular-beam teaching model.

Preparing RC flexure scene…
Member curvature and crack graphics are schematic, scaled by Mu/φMn and the 8× visual factor. They are not service-deflection or crack-width predictions. Capacity, strain, stress, force, and φ results remain quantitative.

Use the guided sequence to trace equilibrium, strain compatibility, steel yield verification, and the final φMn capacity check.

300 mm
500 mm
60 mm
28 MPa
420 MPa
3
4.5 m
80%
8×
Ø20 mm
Strain compatibility
β1
0.850
c
65.22 mm
a
55.44 mm
εy
0.00210
εt
0.01724
fs
420.0 MPa
Steel state
Yielded
Class
Tension-controlled
Strength / demand
As
942 mm²
As,min
440 mm²
Mn
163.20 kN·m
φ
0.900
φMn
146.88 kN·m
Mu
117.50 kN·m
Equivalent center P
104.4 kN
C−T closure
0.00e+0 N

Professional Flexural Analysis Checklist

  1. Establish the code edition and material strengths before selecting coefficients.
  2. Determine bb, hh, clear cover, transverse reinforcement, the actual tension-layer depths did_i, the steel-area centroid dd, the extreme tension-reinforcement depth dtd_t, and any d′d' from actual geometry.
  3. Calculate β1\beta_1 from fc′f'_c using the adopted edition.
  4. Write equilibrium and compatibility before assuming reinforcement has yielded.
  5. Solve for cc, then compute a=β1ca=\beta_1c and every reinforcement-layer strain and stress from its actual depth.
  6. Revisit the solution if any assumed yielded/elastic steel state is contradicted by the calculated strain.
  7. Compute MnM_n from the actual internal resultants and lever arms. Combine tension layers at dd only when their equal stresses make the force resultant coincide with the area centroid; otherwise retain the layer forces separately.
  8. Determine ϵt\epsilon_t from the extreme tension layer at dtd_t, determine ϵy\epsilon_y, then perform the beam minimum-strain check, strain classification, and ϕ\phi calculation.
  9. Verify ϕMn≥Mu\phi M_n\ge M_u and minimum reinforcement.
  10. Confirm selected bars, spacing, cover, anchorage, development, and serviceability separately.
Key Takeaways
  • Flexural strength is governed by equilibrium plus strain compatibility; assuming every reinforcing bar is at fyf_y is not a valid general method.
  • The adopted stress block uses a=β1ca=\beta_1c, with β1\beta_1 reducing above 28 MPa28\,\text{MPa} to a minimum of 0.650.65.
  • For a single yielded tension layer, the familiar a=Asfy/(0.85fc′b)a=A_sf_y/(0.85f'_cb) and Mn=Asfy(d−a/2)M_n=A_sf_y(d-a/2) apply directly; multiple equal-stress yielded layers may also be combined at their area centroid dd.
  • In multilayer reinforcement, dd is the steel-area centroid defined by the section geometry, while dtd_t locates the extreme tension reinforcement used for ϵt\epsilon_t, minimum-strain checks, strain classification, and ϕ\phi. If layer stresses differ, calculate the force resultant separately from dd.
  • Minimum reinforcement, balanced behavior, ϵt≥0.004\epsilon_t\ge0.004, and tension-controlled ϵt≥0.005\epsilon_t\ge0.005 are distinct concepts on the NSCP 2015 / ACI 318-14 basis.
  • The transition ϕ\phi calculation uses the actual reinforcement yield strain ϵy=fy/Es\epsilon_y=f_y/E_s; it must respond when fyf_y changes.
  • Doubly reinforced analysis must solve compression-steel strain and stress rather than assuming compression-steel yield.
  • Deep-beam/strut-and-tie behavior is distinct from ordinary side-face skin reinforcement for deep flexural members.
  • A reinforcement selection is acceptable only when its actual geometry, layer strains, steel-area centroid, extreme-tension depth, internal force resultants, and ϕMn\phi M_n satisfy the design—not merely when a theoretical required steel area has been computed.