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.
Flexural Response: Uncracked to Nominal Strength
The sequence separates service-response concepts from the code nominal-strength state. Crack shapes and deformation are schematic; the final panel is the state used for the lesson strain-compatibility strength calculation.
RC Beam Flexural Analysis Workflow

Analyze an existing reinforced-concrete flexural section by coupling geometry, equilibrium, and strain compatibility before calculating design strength.

Do not force fs = fy. Solve the neutral axis and each reinforcement-layer strain consistently, revisit any contradicted elastic/yield assumption, then compute Mn and the strain-based phi on the stated NSCP 2015 / ACI 318-14 basis.

RC Beam Flexural Analysis WorkflowAnalyze an existing reinforced-concrete flexural section by coupling geometry, equilibrium, and strain compatibility before calculating design strength.. Do not force fs = fy. Solve the neutral axis and each reinforcement-layer strain consistently, revisit any contradicted elastic/yield assumption, then compute Mn and the strain-based phi on the stated NSCP 2015 / ACI 318-14 basis.. Known section, reinforcement, materials, and Mu → Establish b, h, di, d, dt, cover, and bar fit; Establish b, h, di, d, dt, cover, and bar fit → Set epsilon cu = 0.003 and calculate beta1; Set epsilon cu = 0.003 and calculate beta1 → Choose trial neutral axis c; form a = beta1 c; Choose trial neutral axis c; form a = beta1 c → Calculate layer strains from compatibility; Calculate layer strains from compatibility → Calculate steel stresses; verify yield states; Calculate steel stresses; verify yield states → Does longitudinal force equilibrium close?; Does longitudinal force equilibrium close? — Yes → Take moments of actual resultants to obtain Mn; Does longitudinal force equilibrium close? — No → Update c and repeat compatibility/equilibrium; Update c and repeat compatibility/equilibrium → Calculate layer strains from compatibility; Take moments of actual resultants to obtain Mn → Use extreme tension strain to classify and set phi; Use extreme tension strain to classify and set phi → Report c, a, strains, stresses, Mn, phiMn, and flags

Known section, reinforcement, materials, and Mu → Establish b, h, di, d, dt, cover, and bar fit; Establish b, h, di, d, dt, cover, and bar fit → Set epsilon cu = 0.003 and calculate beta1; Set epsilon cu = 0.003 and calculate beta1 → Choose trial neutral axis c; form a = beta1 c; Choose trial neutral axis c; form a = beta1 c → Calculate layer strains from compatibility; Calculate layer strains from compatibility → Calculate steel stresses; verify yield states; Calculate steel stresses; verify yield states → Does longitudinal force equilibrium close?; Does longitudinal force equilibrium close? — Yes → Take moments of actual resultants to obtain Mn; Does longitudinal force equilibrium close? — No → Update c and repeat compatibility/equilibrium; Update c and repeat compatibility/equilibrium → Calculate layer strains from compatibility; Take moments of actual resultants to obtain Mn → Use extreme tension strain to classify and set phi; Use extreme tension strain to classify and set phi → Report c, a, strains, stresses, Mn, phiMn, and flags

  • Known section, reinforcement, materials, and Mu: terminator
  • Establish b, h, di, d, dt, cover, and bar fit: process
  • Set epsilon cu = 0.003 and calculate beta1: process
  • Choose trial neutral axis c; form a = beta1 c: process
  • Calculate layer strains from compatibility: process
  • Calculate steel stresses; verify yield states: process
  • Does longitudinal force equilibrium close?: decision
  • Update c and repeat compatibility/equilibrium: process
  • Take moments of actual resultants to obtain Mn: process
  • Use extreme tension strain to classify and set phi: process
  • Report c, a, strains, stresses, Mn, phiMn, and flags: terminator

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

Minimum-Reinforcement Exception Must Be Checked Explicitly

On the adopted ACI 318-14 basis, the direct As,min⁡A_{s,\min} requirement for the applicable nonprestressed beam provision need not govern where the reinforcement provided at every section is at least one-third greater than the reinforcement required by analysis. That exception is a separate design check; it does not erase the strain, detailing, development, or serviceability requirements.

The interactive section models in this lesson report the direct As≥As,min⁡A_s\ge A_{s,\min} screen only. They do not calculate or claim the one-third-over-required exception, because doing so requires a complete required-steel design solution for the governing demand and final reinforcement arrangement.

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-

Design from Factored Moment to Actual Reinforcement

A theoretical required steel area is only an intermediate design result. Start from MuM_u and preliminary geometry, estimate AsA_s, select actual bars, calculate the resulting layer depths and steel-area centroid, solve strain compatibility using the provided reinforcement, and then recompute MnM_n, ϵt\epsilon_t, ϕ\phi, and ϕMn\phi M_n.

After the strength solution, verify the minimum tension reinforcement and convert the selected bars into a physical cage: clear cover, stirrup diameter, clear horizontal spacing, clear vertical spacing between layers, and practical bar placement must all be consistent with the effective depth used in the mechanics. Development, cutoff/continuation, anchorage, shear, and serviceability remain separate required checks.

RC Beam Flexural Design Workflow

Design a beam section from factored moment demand through preliminary geometry, reinforcement selection, strain compatibility, strength, detailing, and serviceability.

The loop intentionally continues beyond phiMn >= Mu. Minimum reinforcement—including an applicable adopted-basis exception—physical bar fit, cover/spacing, development/detailing, and serviceability are separate acceptance gates. D-regions or deep-beam behavior require an appropriate strut-and-tie/deep-beam method instead of ordinary plane-section flexure.

RC Beam Flexural Design WorkflowDesign a beam section from factored moment demand through preliminary geometry, reinforcement selection, strain compatibility, strength, detailing, and serviceability.. The loop intentionally continues beyond phiMn >= Mu. Minimum reinforcement—including an applicable adopted-basis exception—physical bar fit, cover/spacing, development/detailing, and serviceability are separate acceptance gates. D-regions or deep-beam behavior require an appropriate strut-and-tie/deep-beam method instead of ordinary plane-section flexure.. Factored flexural demand Mu and design basis → Select b, h, d, and material strengths; Select b, h, d, and material strengths → Estimate required tension reinforcement As; Estimate required tension reinforcement As → Select actual bars and layer geometry; Select actual bars and layer geometry → Solve compatibility and neutral axis c; Solve compatibility and neutral axis c → Are assumed steel states compatible with strains?; Are assumed steel states compatible with strains? — Yes → Calculate Mn, extreme tension strain, phi, and phiMn; Are assumed steel states compatible with strains? — No → Revise steel stresses and resolve equilibrium; Revise steel stresses and resolve equilibrium → Solve compatibility and neutral axis c; Calculate Mn, extreme tension strain, phi, and phiMn → Does phiMn >= Mu?; Does phiMn >= Mu? — Yes → Is the minimum-reinforcement requirement satisfied?; Does phiMn >= Mu? — No → Revise geometry, reinforcement, or materials; Is the minimum-reinforcement requirement satisfied? — Yes → Do bars satisfy cover, spacing, layers, and fit?; Is the minimum-reinforcement requirement satisfied? — No → Revise geometry, reinforcement, or materials; Do bars satisfy cover, spacing, layers, and fit? — Yes → Do development, anchorage, cutoffs, and detailing pass?; Do bars satisfy cover, spacing, layers, and fit? — No → Revise geometry, reinforcement, or materials; Do development, anchorage, cutoffs, and detailing pass? — Yes → Do serviceability checks pass?; Do development, anchorage, cutoffs, and detailing pass? — No → Revise geometry, reinforcement, or materials; Do serviceability checks pass? — Yes → Document final reinforcement and design state; Do serviceability checks pass? — No → Revise geometry, reinforcement, or materials; Revise geometry, reinforcement, or materials — Iterate → Select b, h, d, and material strengths

Factored flexural demand Mu and design basis → Select b, h, d, and material strengths; Select b, h, d, and material strengths → Estimate required tension reinforcement As; Estimate required tension reinforcement As → Select actual bars and layer geometry; Select actual bars and layer geometry → Solve compatibility and neutral axis c; Solve compatibility and neutral axis c → Are assumed steel states compatible with strains?; Are assumed steel states compatible with strains? — Yes → Calculate Mn, extreme tension strain, phi, and phiMn; Are assumed steel states compatible with strains? — No → Revise steel stresses and resolve equilibrium; Revise steel stresses and resolve equilibrium → Solve compatibility and neutral axis c; Calculate Mn, extreme tension strain, phi, and phiMn → Does phiMn >= Mu?; Does phiMn >= Mu? — Yes → Is the minimum-reinforcement requirement satisfied?; Does phiMn >= Mu? — No → Revise geometry, reinforcement, or materials; Is the minimum-reinforcement requirement satisfied? — Yes → Do bars satisfy cover, spacing, layers, and fit?; Is the minimum-reinforcement requirement satisfied? — No → Revise geometry, reinforcement, or materials; Do bars satisfy cover, spacing, layers, and fit? — Yes → Do development, anchorage, cutoffs, and detailing pass?; Do bars satisfy cover, spacing, layers, and fit? — No → Revise geometry, reinforcement, or materials; Do development, anchorage, cutoffs, and detailing pass? — Yes → Do serviceability checks pass?; Do development, anchorage, cutoffs, and detailing pass? — No → Revise geometry, reinforcement, or materials; Do serviceability checks pass? — Yes → Document final reinforcement and design state; Do serviceability checks pass? — No → Revise geometry, reinforcement, or materials; Revise geometry, reinforcement, or materials — Iterate → Select b, h, d, and material strengths

  • Factored flexural demand Mu and design basis: terminator
  • Select b, h, d, and material strengths: process
  • Estimate required tension reinforcement As: process
  • Select actual bars and layer geometry: process
  • Solve compatibility and neutral axis c: process
  • Are assumed steel states compatible with strains?: decision
  • Revise steel stresses and resolve equilibrium: process
  • Calculate Mn, extreme tension strain, phi, and phiMn: process
  • Does phiMn >= Mu?: decision
  • Is the minimum-reinforcement requirement satisfied?: decision
  • Do bars satisfy cover, spacing, layers, and fit?: decision
  • Do development, anchorage, cutoffs, and detailing pass?: decision
  • Do serviceability checks pass?: decision
  • Revise geometry, reinforcement, or materials: process
  • Document final reinforcement and design state: terminator

RC Beam Section — Strain Compatibility Analyzer

Concept and model scope

This nominal-strength analyzer uses the same deterministic solver as the 3D flexure studio. Geometry determines the actual tension-bar locations; equilibrium and linear strain compatibility determine the neutral axis, steel strain/stress, Whitney compression block, internal resultants, Mn, phi, and phiMn.

The model is limited to a singly reinforced rectangular beam with one tension layer on the lesson's NSCP 2015 / adopted ACI 318-14 basis. It verifies steel yield rather than assigning fs = fy in advance.

Bar fit is a simplified one-row geometry screen using a 40 mm longitudinal-bar-surface offset from each side face and clear spacing not less than max(25 mm, bar diameter). It is not a code clear-cover check to outer stirrups. Aggregate-size, transverse-reinforcement geometry, bundling, development, shear, serviceability, and seismic detailing remain outside this component.

One solver drives the reinforcement geometry, neutral axis, strain diagram, compression block, force resultants, and reported strength.

Beam width b

The rectangular compression-block width and available one-row reinforcement width. Increasing b increases the concrete compression area and usually improves bar-fit room.
300 mm

Overall depth h

The total section depth from compression face to tension face. Together with the selected steel-centroid offset it sets effective depth d and therefore the strain geometry and internal lever arm.
500 mm

Tension-steel centroid offset

Distance from the tension face to the centroid of the modeled one-layer tension reinforcement. The solver uses d = h minus this offset. It is not interchangeable with clear concrete cover to the outside of a stirrup.
60 mm

Specified concrete compressive strength

Sets the 0.85f'c Whitney-block stress and the edition-specific beta1 factor. The model follows the lesson's NSCP 2015 / adopted ACI 318-14 basis.
28 MPa

Specified reinforcement yield strength

Sets the idealized yield stress and yield strain epsilon y = fy/Es with Es = 200,000 MPa. The solver assigns fy only after compatible steel strain reaches epsilon y.
420 MPa

Longitudinal tension-bar diameter

Controls the nominal circular area of each modeled tension bar and the simplified one-row clear-spacing screen. Actual projects must use the governing reinforcement standard/bar table where applicable.
Ø20 mm

Number of tension bars

Sets the modeled one-layer tension-steel area As and the horizontal bar positions. Too many bars for the selected width are rejected by the simplified fit screen rather than displayed as acceptable.
3
NA c=65.2 mmCc=395.8 kNT=395.8 kNεc=0.003εt=0.01724strainside bar-surface offset = 40 mm

Nominal-strength state

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
ρ\rho
0.00714
As,minA_{s,min}
440 mm²
MnM_n
163.20 kN·m
ϕ\phi
0.900
ϕMn\phi M_n
146.88 kN·m
C−T closure
0.00e+0 N

This visual is quantitative only for the nominal-strength strain-compatibility state. It does not draw service deflection or crack width, and it does not infer a project's full code compliance from a section-strength calculation.

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.

Common Interior T-Beam Effective Flange-Width Limits

Equivalent SI form of the adopted-basis limits for a monolithic interior T-beam with flange on both sides of the web.

bf≤min⁡(ℓ4,  bw+16hf,  s)b_f\le\min\left(\frac{\ell}{4},\;b_w+16h_f,\;s\right)

Variables

SymbolDescriptionUnit
bfb_fEffective total flange width used for the interior T-beammm
ℓ\ellBeam span length used by the applicable effective-width provisionmm
bwb_wWeb widthmm
hfh_fSlab/flange thicknessmm
ssCenter-to-center spacing of adjacent beams/webs for the stated interior-beam conditionmm

One-Sided Effective Overhang for an Edge L-Beam

Adopted-basis limit for the effective slab overhang on the flange side of a monolithic beam with slab on one side only.

be≤min⁡(ℓ12,  6hf,  sclear2),bf=bw+beb_e\le\min\left(\frac{\ell}{12},\;6h_f,\;\frac{s_{clear}}{2}\right), \qquad b_f=b_w+b_e

Variables

SymbolDescriptionUnit
beb_eEffective one-sided flange overhang beyond the webmm
sclears_{clear}Clear distance to the next web on the flange sidemm
bfb_fTotal effective L-beam flange widthmm

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.

For an L-beam at an edge, flange participation exists on only one side of the web and the applicable effective-width limits must reflect that geometry. For broader nonrectangular sections, equilibrium and compatibility remain the governing mechanics: determine the actual compression region represented by the adopted concrete stress model, locate its resultant, calculate each reinforcement-layer force, and take moments of the true internal resultants rather than forcing the section into a rectangular shortcut.

Rectangular, Multilayer, T-, and L-Beam Section Families
Section geometry changes the compression region and reinforcement strain locations. Effective flange width is code-limited geometry, and multilayer steel requires layer-by-layer compatibility when stresses differ.

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, the deep-beam definition applies to members loaded on one face and supported on the opposite face so that compression struts can develop between the loads and supports, with either clear span ℓn≤4h\ell_n\le4h or a region containing a concentrated load within 2h2h of a support face. Such regions require the applicable deep-beam/D-region provisions and commonly strut-and-tie modeling rather than ordinary slender-beam assumptions.

Side-face skin reinforcement is a separate ordinary-beam detailing requirement. On this adopted basis, beams with overall depth h>900 mmh>900\,\text{mm} require the applicable longitudinal skin-reinforcement detailing along the side faces near the tension zone. That depth trigger does not by itself make a member a deep beam. Conversely, a D-region/deep-beam condition can arise from span/load geometry independently of the skin-reinforcement trigger.

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.

RC Beam Reinforcement Geometry

Concept and model scope

Explore a deterministic reinforcement-cage geometry model for a rectangular beam. The selected dimensions, clear cover, stirrup size/spacing, longitudinal-bar diameter, and bar counts generate the displayed cage and the same clear-spacing checks.

The fit screen uses clear longitudinal-bar spacing not less than max(25 mm, bar diameter). It is deliberately narrower than a complete detailing check: aggregate-size effects, bundling, development, splice, hooks, seismic detailing, and member strength remain outside this viewer.

This component is a detailing/constructability visualizer, not a flexural-capacity solver. Use the strain-compatibility analyzer for c, a, steel stress, Mn, phi, and phiMn.

Adjust section and cage geometry; physically invalid states are reported and the cage is withheld.

Beam width

Overall rectangular section width. It controls the stirrup width and the horizontal room available for longitudinal bars.
300 mm

Beam height

Overall rectangular section depth. It controls the vertical room for the cage; this geometry viewer does not infer flexural capacity from height alone.
500 mm

Displayed beam length

Length of the reinforcement-cage visualization. It determines the modeled stirrup run and longitudinal-bar end offsets, not flexural demand or span-dependent capacity.
3000 mm

Clear cover to outer stirrup

Distance from the concrete surface to the outside face of the modeled stirrup. Longitudinal-bar centerlines are then placed inside that stirrup by the selected stirrup and bar diameters.
40 mm

Number of top longitudinal bars

Number of equally spaced modeled top bars. They are shown for cage geometry; this viewer does not assign them a flexural stress or strength contribution.
2

Number of bottom longitudinal bars

Number of equally spaced modeled bottom bars. The geometry screen rejects a one-row arrangement when the resulting clear spacing is too small.
3

Longitudinal bar diameter

Nominal diameter used to draw the longitudinal bars and calculate the simplified minimum clear-spacing requirement max(25 mm, bar diameter).
20 mm

Stirrup diameter

Nominal diameter of the modeled transverse reinforcement. It shifts the available longitudinal-bar centerline region inward from the clear-cover boundary.
10 mm

Target maximum stirrup spacing

Requested upper spacing for the displayed stirrups. The model divides the available run into an integer number of intervals so the reported actual spacing is no greater than this target.
200 mm
The displayed cage passes this simplified geometry screen.

Top clear spacing 160 mm; bottom clear spacing 70 mm; actual stirrup spacing 194 mm.

Preparing reinforcement scene…

The 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 does not calculate bar force, neutral axis, flexural strength, shear strength, or project compliance. Use the strain-compatibility and other applicable design checks separately.

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. Its one-row horizontal fit screen uses a simplified longitudinal-bar-surface offset rather than modeling outer stirrup cover and aggregate-dependent spacing.

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

The minimum-steel status uses the direct As ≥ As,min screen. The adopted one-third-over-required reinforcement exception is not modeled because this studio does not solve the complete required-steel design problem.

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.

Beam width b

The rectangular compression-block width and simplified horizontal reinforcement-fit screen both use b. Increasing width increases the available compression area and one-row bar-fit room.
300 mm

Overall depth h

Overall concrete depth sets the section geometry. Effective depth d is measured from the compression face to the centroid of the modeled tension reinforcement.
500 mm

Tension-steel centroid from tension face

This is the distance from the tension face to the centroid of the modeled one-layer tension reinforcement, so d = h minus this value. It is not clear cover to a stirrup.
60 mm

Specified concrete compressive strength

The value f'c sets the 0.85f'c Whitney-block stress and the adopted beta1 value. On this lesson basis beta1 decreases above 28 MPa to a minimum of 0.65.
28 MPa

Specified reinforcement yield strength

The model computes yield strain as fy/Es with Es = 200,000 MPa. Steel is assigned fy only after the solved compatible strain reaches that yield strain.
420 MPa

Number of tension bars

The modeled one-row tension-bar count changes steel area As and horizontal bar fit. The component rejects a selected one-row geometry that fails its simplified clear-spacing screen.
3

Displayed simply supported span

Span controls the physical member scene and the equivalent central point-load label P = 4Mu/L. Section flexural capacity itself is independent of displayed span in this model.
4.5 m

Factored moment demand ratio

Mu/phiMn expresses factored moment demand as a fraction of calculated design flexural strength. Values above 100% expose the capacity boundary but do not simulate nonlinear post-failure response.
80%

Schematic deformation factor

This changes only the displayed curvature exaggeration. It is intentionally non-quantitative because cracked service stiffness, loading history, and service deflection are outside this nominal-strength model.
8×

Tension bar diameter

The nominal bar diameter controls each bar area, the one-row bar locations, and the simplified clear-spacing screen. Actual design must use the applicable reinforcement standard and project detailing.
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.
  • The uncracked, first-cracking, cracked-elastic, steel-yield, and nominal-strength stages describe different response regimes; nominal-strength compatibility must not be used as a quantitative service-deflection or crack-width model.
  • T- and L-beam effective flange widths are code-limited. For nonrectangular sections, preserve equilibrium and the actual locations of compression and reinforcement resultants.

References