Shear and Torsion in Beams

Shear and Torsion in Beams

$Learning Objectives$

Explain how flexure and shear produce principal tensile stresses and inclined cracking.
Distinguish flexure-shear cracking from web-shear cracking without assuming every inclined crack forms at exactly $45^\circ$ .
Calculate $V_c$ , $V_s$ , $V_n$ , and $\phi V_n$ using a consistent NSCP 2015 / ACI 318-14 basis.
Determine when minimum shear reinforcement is required and check both strength and maximum stirrup spacing.
Recognize the concrete-web strength limit beyond which increasing stirrup area alone is not an acceptable solution.
Distinguish equilibrium torsion from compatibility torsion and apply the threshold-torsion concept correctly.
Explain the closed transverse and distributed longitudinal reinforcement required after torsional cracking.
Check the combined shear-torsion section-size limit for solid rectangular beams.

$Course Code Basis and Design Philosophy$

This lesson uses NSCP 2015 provisions that adopt the ACI 318-14 structural-concrete framework. Unless stated otherwise, equations are written in SI units with $f'_c$ and steel stresses in MPa, section dimensions in mm, force in N, and torque in N-mm. The shear and torsion strength-reduction factor used here is $\phi=0.75$ . Nominal strength and design strength are kept distinct: $V_n$ is nominal resistance, while $\phi V_n$ is the design resistance compared with factored demand $V_u$ .

$Diagonal tension$

Diagonal tension is the tensile principal-stress condition produced by the combined normal stress from flexure and shear stress in a beam web. Because concrete has low tensile strength, inclined cracking forms approximately perpendicular to the tensile principal-stress direction.

$Inclined-Crack Mechanisms$

Flexure-shear crack: begins as a flexural crack at the tension face where bending tension is significant. As it grows into a region where shear stress is important, the crack turns and propagates on an inclined path toward the load or compression zone.
Web-shear crack: can initiate within the web before a flexural crack reaches that location when the principal tensile stress caused by shear is sufficiently large while flexural tension is comparatively small. It is common in high-shear regions and in some prestressed or deep-member situations, but it is not limited to one member type.
Diagonal-tension failure: describes brittle loss of shear resistance associated with inclined cracking when the concrete-and-transverse-reinforcement mechanism cannot carry the demand.

Inclined cracks are often idealized near $45^\circ$ for elementary truss models and detailing rules, but the actual crack angle depends on the stress field, reinforcement, geometry, and loading. Deep beams and other disturbed regions can develop pronounced arching or strut-and-tie action and should not be reduced to ordinary slender-beam behavior.

$Nominal and Design Shear Strength$

For a nonprestressed beam designed with ordinary transverse reinforcement, nominal shear resistance is the sum of the concrete and transverse-steel contributions. The factored demand is checked against the reduced nominal resistance.

Nominal Shear Strength

Concrete and transverse reinforcement contribute to nominal shear resistance.

V_n=V_c+V_s

Variables

Symbol	Description	Unit
$V_n$	Nominal shear strength	N
$V_c$	Nominal shear strength attributed to concrete mechanisms	N
$V_s$	Nominal shear strength provided by shear reinforcement	N

Shear Strength Requirement

Factored shear demand must not exceed design shear strength.

V_u\leq \phi V_n=\phi\left(V_c+V_s\right)

Variables

Symbol	Description	Unit
$V_u$	Factored shear at the section being checked	N
$\phi$	Strength-reduction factor for shear; $0.75$ for the course basis	-

$V_c$

The simplified NSCP 2015 / ACI 318-14 expression below is appropriate for the ordinary nonprestressed beam cases used in this lesson. It represents the combined empirical effect of the uncracked compression zone, aggregate interlock, and dowel action rather than a literal single resisting force. For lightweight concrete, the applicable $\lambda$ must be selected from the governing code rather than assumed from density alone.

Simplified Concrete Shear Strength

SI expression for the ordinary nonprestressed beam cases used in the worked examples.

V_c=0.17\lambda\sqrt{f'_c}\,b_wd

Variables

Symbol	Description	Unit
$f'_c$	Specified concrete compressive strength	MPa
$b_w$	Web width	mm
$d$	Effective depth from extreme compression fiber to centroid of longitudinal tension reinforcement	mm
$\lambda$	Lightweight-concrete modification factor; $1.0$ for normal-weight concrete	-

$V_c$

For eligible nonprestressed members, the adopted ACI 318-14 basis also permits the more detailed expression

V_c=\left(0.16\lambda\sqrt{f'_c}+17\rho_w\frac{V_ud}{M_u}\right)b_wd \leq0.29\lambda\sqrt{f'_c}\,b_wd

with the code limit on $V_ud/M_u$ observed. Do not combine this alternative with coefficients from later ACI editions. The examples and simulator below intentionally use the simplified $0.17\lambda\sqrt{f'_c}b_wd$ expression so one code basis is carried consistently through the calculations.

$Shear Reinforcement Contribution$

For vertical stirrups, $A_v$ is the total cross-sectional area of the stirrup legs crossing a potential shear crack within one spacing $s$ . A two-leg stirrup therefore uses twice the area of one stirrup bar. On the adopted ACI 318-14 basis, the value of $f_{yt}$ used to calculate shear and torsion reinforcement strength is limited to $420\text{ MPa}$ ; a higher specified steel grade must not be inserted into these equations without applying the governing code limit.

Vertical-Stirrup Shear Strength

Nominal shear contribution of vertical transverse reinforcement.

V_s=\frac{A_v f_{yt}d}{s}

Variables

Symbol	Description	Unit
$A_v$	Area of shear reinforcement within spacing $s$	$mm^2$
$f_{yt}$	Transverse reinforcement yield strength used in design; not greater than $420$ MPa on this ACI 318-14 basis	MPa
$s$	Longitudinal center-to-center stirrup spacing	mm

Strength-Required Shear Reinforcement Ratio

Required $A_v/s$ when transverse steel must supply $V_{s,req}$ .

V_{s,\mathrm{req}}=\max\left(0,\frac{V_u}{\phi}-V_c\right),\qquad \left(\frac{A_v}{s}\right)_{\mathrm{strength}}=\frac{V_{s,\mathrm{req}}}{f_{yt}d}

Variables

Symbol	Description	Unit
$V_{s,\mathrm{req}}$	Nominal steel shear contribution required by strength	N

$When Minimum Shear Reinforcement Is Required$

For a general nonprestressed beam, minimum shear reinforcement is required where $V_u>0.5\phi V_c$ . If $V_u>\phi V_c$ , the amount required by strength is also calculated and the larger of the strength requirement and the minimum-reinforcement requirement governs. ACI 318-14 Table 9.6.3.1 contains explicit exceptions for certain shallow beams, members integral with slabs, qualifying steel-fiber members, and one-way joist construction; those exceptions must be checked before treating the $0.5\phi V_c$ trigger as universal.

Minimum Shear Reinforcement

Minimum $A_v/s$ for the general nonprestressed-beam case in SI units.

\left(\frac{A_v}{s}\right)_{\min} = \max\left( \frac{0.062\sqrt{f'_c}\,b_w}{f_{yt}}, \frac{0.35b_w}{f_{yt}} \right)

Variables

Symbol	Description	Unit
$(A_v/s)_{\min}$	Minimum transverse-reinforcement area per unit beam length	$mm^2/mm$

$Complete Vertical-Stirrup Design Check$

STEP-BY-STEP

Calculate $V_c$ and $\phi V_c$ using the selected code equation.
Determine whether the general minimum-reinforcement trigger $V_u>0.5\phi V_c$ applies and check any applicable code exception.
Calculate $V_{s,\mathrm{req}}=\max(0,V_u/\phi-V_c)$ .
Calculate both $(A_v/s)_{\mathrm{strength}}$ and, when applicable, $(A_v/s)_{\min}$ ; use the larger requirement.
Select a stirrup size, number of legs, and spacing such that the provided $A_v/s$ is not less than the governing requirement.
Recalculate the provided $V_s=A_vf_{yt}d/s$ , use that selected strength state to determine the applicable maximum-spacing regime, and verify the selected spacing against the resulting $s_{\max}$ .
Recalculate $V_n=V_c+V_s$ and verify $\phi V_n\geq V_u$ .
Check the concrete-web strength ceiling. If the required shear exceeds the permitted section strength, increase the section rather than merely adding more stirrup steel.

$Maximum Spacing and Web-Strength Limits$

For vertical shear reinforcement in a nonprestressed beam on the adopted ACI 318-14 basis:

For the selected provided reinforcement, if $V_s\leq0.33\sqrt{f'_c}\,b_wd$ , use $s\leq\min(d/2,600\text{ mm})$ .
If the selected provided $V_s>0.33\sqrt{f'_c}\,b_wd$ , use the tighter $s\leq\min(d/4,300\text{ mm})$ .
A trial layout should therefore be rechecked after bar size, leg count, and spacing are selected; an intentionally overprovided cage can enter the tighter spacing regime even when the theoretical $V_{s,\mathrm{req}}$ is below the threshold.
The section-size limit is equivalent to requiring that the steel contribution needed for ordinary shear design not exceed $0.66\sqrt{f'_c}\,b_wd$ ; equivalently, $V_u\leq\phi\left(V_c+0.66\sqrt{f'_c}\,b_wd\right)$ for the simplified beam case.

Shear Design Workflow — NSCP 2015 / ACI 318-14

From factored shear demand through concrete contribution, section-size limits, required transverse reinforcement, spacing, strength, and detailing.

Support shear Vu300 kN

Beam span L (visualization)6.0 m

Concrete strength f'c28 MPa

Specified transverse steel fyt420 MPa

Web width bw300 mm

Overall depth h600 mm

Effective depth d540 mm

Selected spacing s100 mm

Stirrup bar / shear-effective legs2 legs

Crack teaching modelflexure-shear

Selected stirrup layout satisfies the displayed shear checks.

145.7 kN

φVc

109.3 kN

Required Vs

254.3 kN

Provided Vs

356.3 kN

Design fyt

420 MPa

Required Av/s

1.121 mm²/mm

Provided Av/s

1.571 mm²/mm

Maximum spacing

135 mm

Section φVn ceiling

533.6 kN

Geometry identity. The solver uses

A_v

= 157.1 mm² from 2 rendered vertical legs of Ø10 mm within each 100 mm spacing. Reducing spacing physically increases the number of rendered stirrups and increases the same solver quantity

A_v/s

Loading chart...

General-beam minimum shear reinforcement is active because

V_u>0.5\phi V_c

. Explicit ACI 318-14 exceptions still require project-specific checking.

$Shear friction$

Shear friction is a design model for transfer of shear across a defined potential sliding plane, such as a construction joint or an interface. Reinforcement crossing the plane develops clamping force as relative slip tends to open the rough interface, allowing friction and aggregate interlock to resist sliding.

Basic Shear-Friction Model

Nominal resistance for the simplified shear-friction cases discussed in this lesson.

V_n=A_{vf}f_y\mu

Variables

Symbol	Description	Unit
$A_{vf}$	Area of reinforcement crossing the shear plane	$mm^2$
$\mu$	Code friction coefficient for the interface condition	-

$Shear-Friction Interface Conditions$

For the adopted course basis, common values include $\mu=1.4\lambda$ for concrete placed monolithically, $1.0\lambda$ for concrete placed against hardened concrete intentionally roughened to the specified amplitude, and $0.6\lambda$ for concrete placed against hardened concrete not intentionally roughened. The complete shear-friction design also includes code limits on nominal shear stress, reinforcement anchorage, and interface preparation; the single equation above should not be used as an unrestricted capacity formula.

$Brackets, Corbels, and Disturbed Regions$

Corbels and brackets have short shear spans and highly nonlinear strain fields. Their load path is better represented by strut-and-tie action and, where a defined interface governs, shear-friction concepts. A small $a/d$ is a warning that ordinary slender-beam flexure/shear assumptions are inappropriate; do not diagnose every deep-member failure using a single $45^\circ$ crack model.

$Equilibrium torsion$

Equilibrium torsion is required for static equilibrium of the structure. If that torque is not resisted, the intended load path cannot exist, so the member must be designed for the equilibrium torsion demand.

$Compatibility torsion$

Compatibility torsion results from deformation compatibility in an indeterminate framing system. After torsional cracking reduces stiffness, forces may redistribute if equilibrium and deformation compatibility can still be maintained by the surrounding structure.

$Threshold and Cracking Torsion$

Torsion need not be included in member design when the factored torque does not exceed the reduced threshold torque. For compatibility torsion, the code permits redistribution only when the structural system can support it; equilibrium torsion cannot simply be discarded. The threshold expression is one-quarter of the corresponding cracking-torque expression on this ACI 318-14 basis.

Threshold Torsion

Threshold torque for the nonprestressed member case used here; compression is positive in the axial-load term.

T_{th}=0.083\lambda\sqrt{f'_c}\left(\frac{A_{cp}^2}{p_{cp}}\right) \sqrt{1+\frac{N_u}{0.33A_g\lambda\sqrt{f'_c}}}

Variables

Symbol	Description	Unit
$T_{th}$	Nominal threshold torsion	$N\cdot mm$
$A_{cp}$	Area enclosed by the outside perimeter of the concrete section	$mm^2$
$p_{cp}$	Outside perimeter of the concrete section	mm
$A_g$	Gross concrete area	$mm^2$
$N_u$	Factored axial force, positive in compression for this expression	N

Cracking Torsion for Compatibility Redistribution

Corresponding cracking-torque expression on the same code basis.

T_{cr}=0.33\lambda\sqrt{f'_c}\left(\frac{A_{cp}^2}{p_{cp}}\right) \sqrt{1+\frac{N_u}{0.33A_g\lambda\sqrt{f'_c}}}

Variables

Symbol	Description	Unit
$T_{cr}$	Nominal torsional cracking torque	$N\cdot mm$

$Threshold check is a factored comparison$

With $\phi=0.75$ , torsion may be neglected only when $T_u\leq\phi T_{th}$ for the applicable member case. If $T_u$ exceeds that level, torsion is included in design. Do not compare factored $T_u$ directly with unreduced $T_{th}$ .

$Compatibility-Torsion Redistribution Is a System Decision$

When torsion is caused by compatibility in a statically indeterminate system, ACI 318-14 permits redistribution after torsional cracking only when the surrounding system can accept the redistributed forces and still satisfy equilibrium and compatibility. For the eligible nonprestressed compatibility-torsion case represented in this lesson, if the elastic analysis gives $T_u>\phi T_{cr}$ and the redistribution conditions are satisfied, the member section may be designed for a reduced torque not less than $\phi T_{cr}$ . This is not available for equilibrium torsion, and a section-level calculation cannot prove that the rest of the structure can carry the redistributed actions.

Compatibility-Torsion Design Torque

Section design torque after an explicitly justified compatibility redistribution on the adopted ACI 318-14 basis.

T_{u,\mathrm{design}}= \begin{cases} 0, & T_u\leq\phi T_{th}\\ \phi T_{cr}, & \text{eligible compatibility torsion with justified redistribution and }T_u>\phi T_{cr}\\ T_u, & \text{otherwise} \end{cases}

Torsion Decision and Design Workflow

Threshold screening, equilibrium-versus-compatibility classification, permitted redistribution, reinforcement design, section limit, and detailing.

Start: Tu and complete structural load path → Tu ≤ ϕTth?; Tu ≤ ϕTth? — Yes → Confirm torsion may be omitted; Confirm torsion may be omitted → Torsion reinforcement not required; Tu ≤ ϕTth? — No → Compatible?; Compatible? — Yes → Eligible?; Compatible? — No → Design for unreduced factored Tu; Eligible? — Yes → Redistribute globally and design section for Tu = φTcr; Eligible? — No → Design for unreduced factored Tu; Redistribute globally and design section for Tu = φTcr → Establish Acp, pcp, Aoh, ph, Ao and closed-hoop geometry; Design for unreduced factored Tu → Establish Acp, pcp, Aoh, ph, Ao and closed-hoop geometry; Establish Acp, pcp, Aoh, ph, Ao and closed-hoop geometry → Confirm threshold/cracking basis and chosen θ; Confirm threshold/cracking basis and chosen θ → Design closed transverse torsion reinforcement At/s; Design closed transverse torsion reinforcement At/s → Design distributed longitudinal torsion reinforcement Al; Design distributed longitudinal torsion reinforcement Al → Section OK?; Section OK? — No → Resize section; recheck; Resize section; recheck — Iterate → Start: Tu and complete structural load path; Section OK? — Yes → Check closed hoops, spacing, perimeter bars, corner bars, anchorage, extension; Check closed hoops, spacing, perimeter bars, corner bars, anchorage, extension → Torsion design accepted

Start: Tu and complete structural load path: terminator
Tu ≤ ϕTth?: decision
Confirm torsion may be omitted: process. Omit torsion reinforcement only when the adopted code threshold and all governing conditions permit it; continue all other required member checks.
Torsion reinforcement not required: terminator. This outcome closes only the torsion-reinforcement workflow; it does not waive shear, flexure, serviceability, or other member checks.
Compatible?: decision. Classify torsion using the structural system and load path. A No branch represents equilibrium torsion and proceeds with unreduced factored Tu.
Eligible?: decision. Use redistribution only when all code conditions are met and the relevant factored torsional demand exceeds ϕTcr.
Redistribute globally and design section for Tu = φTcr: process
Design for unreduced factored Tu: process
Establish Acp, pcp, Aoh, ph, Ao and closed-hoop geometry: process
Confirm threshold/cracking basis and chosen θ: process
Design closed transverse torsion reinforcement At/s: process
Design distributed longitudinal torsion reinforcement Al: process
Section OK?: decision. Check the combined shear–torsion section-size limit using the adopted code equations and the selected section.
Resize section; recheck: process. After a section or load-path revision, restart threshold screening because the torsional threshold and design basis may change.
Check closed hoops, spacing, perimeter bars, corner bars, anchorage, extension: process
Torsion design accepted: terminator

$Post-Cracking Torsion Model$

After torsional cracking, a reinforced concrete beam is idealized as a thin-walled space truss. Diagonal concrete compression fields form the struts, closed transverse reinforcement forms transverse tension ties, and longitudinal bars distributed around the perimeter form longitudinal tension ties. The concrete core inside the effective shear-flow tube is not treated as an independent solid torsion-resisting core.

Nominal Torsional Strength

Space-truss expression for closed transverse torsional reinforcement.

T_n=2A_o\frac{A_t f_{yt}}{s}\cot\theta

Variables

Symbol	Description	Unit
$A_o$	Gross area enclosed by the shear flow path; commonly taken as $0.85A_{oh}$ for design	$mm^2$
$A_{oh}$	Area enclosed by the centerline of the outermost closed transverse torsional reinforcement	$mm^2$
$A_t$	Area of one leg of closed transverse reinforcement resisting torsion within spacing $s$	$mm^2$
$\theta$	Angle of the compression diagonals used by the code truss model	degree

Longitudinal Torsion Reinforcement

Required longitudinal reinforcement associated with the selected torsion truss angle.

A_l=\frac{A_t}{s}p_h\frac{f_{yt}}{f_y}\cot^2\theta

Variables

Symbol	Description	Unit
$A_l$	Total longitudinal reinforcement required for torsion	$mm^2$
$p_h$	Perimeter of the centerline of the outermost closed torsional stirrup	mm

Combined Transverse Reinforcement for Shear and Torsion

One physical closed transverse cage must satisfy the combined shear and torsion requirement.

\left(\frac{A_v+2A_t}{s}\right)_{\mathrm{strength}} = \left(\frac{A_v}{s}\right)_{\mathrm{shear}} + 2\left(\frac{A_t}{s}\right)_{\mathrm{torsion}}

\left(\frac{A_v+2A_t}{s}\right)_{\min} = \max\left( \frac{0.062\sqrt{f'_c}\,b_w}{f_{yt}}, \frac{0.35b_w}{f_{yt}} \right)

The selected cage must provide at least the governing combined requirement while also satisfying the independent shear and torsion spacing limits. $A_v$ counts the effective legs crossing a shear crack; $A_t$ is the area of one outer closed-hoop leg used in the torsion truss.

Minimum Longitudinal Torsion Reinforcement

Minimum perimeter longitudinal torsion reinforcement used by the simulation on the adopted ACI 318-14 basis.

A_{l,\min}= \min\left[ \frac{0.42\sqrt{f'_c}A_{cp}}{f_y} -\frac{A_t}{s}p_h\frac{f_{yt}}{f_y}, \; \frac{0.42\sqrt{f'_c}A_{cp}}{f_y} -\frac{0.175b_w}{f_{yt}}p_h\frac{f_{yt}}{f_y} \right]

Use a nonnegative required area and combine this minimum with the longitudinal steel required by torsional strength. The bars must be distributed around the perimeter rather than concentrated only at the flexural tension face.

$Torsional Reinforcement and Detailing$

Use closed stirrups or hoops capable of developing a complete closed transverse tension path; an open U-stirrup is not a torsional tie.
Distribute longitudinal torsion reinforcement around the perimeter of the closed stirrup, with at least one longitudinal bar at each corner.
Keep longitudinal torsion-bar spacing around the perimeter within the adopted code limit, commonly $300\text{ mm}$ for this basis.
Limit transverse torsion-reinforcement spacing to the smaller of $p_h/8$ and $300\text{ mm}$ , while also satisfying any more restrictive shear-spacing requirement.
Design shear and torsion together: the same physical closed cage participates in both mechanisms, so required $A_v/s$ , $2A_t/s$ , minimum combined transverse steel, and the governing spacing limit must be checked together.
Verify that the provided longitudinal perimeter steel is not less than the governing $A_l$ requirement; calculating $A_l$ without detailing it is not an acceptable design state.
Treat the simulator's geometry gate as an overlap check, not a complete constructability approval. Project detailing must still satisfy the governing minimum clear spacing, cover, aggregate-size compatibility, development/anchorage, and congestion requirements.

Combined Shear-Torsion Section-Size Check

Compression-strut limit for a solid section on the adopted ACI 318-14 SI basis.

\sqrt{\left(\frac{V_u}{b_wd}\right)^2+ \left(\frac{T_up_h}{1.7A_{oh}^2}\right)^2} \leq \phi\left(\frac{V_c}{b_wd}+0.66\sqrt{f'_c}\right)

Variables

Symbol	Description	Unit
$A_{oh}$	Area enclosed by the outermost closed torsion stirrup centerline	$mm^2$
$p_h$	Perimeter of that stirrup centerline	mm

$Do not cure a failed section-size check with more steel$

If the combined shear-torsion stress exceeds the right-hand-side compression-strut limit, increasing $A_v$ or $A_t$ does not remove the concrete crushing mechanism. Increase the section dimensions or otherwise change the structural demand/load path.

Combined Shear–Torsion Acceptance Workflow

Coordinates section-size, shared transverse-cage, longitudinal torsion, spacing, strength, and detailing checks.

Use one physical closed transverse cage for the combined shear and torsion demand, then verify its most restrictive spacing limit, shear and torsion strengths, provided longitudinal torsion steel, and detailing. A failed section-size check requires resizing or a justified demand/load-path revision before steel design proceeds.

3D Shear–Torsion Studio: Closed Cage, Force Flow, and Combined Action

Selected transverse reinforcement satisfies the displayed shear and torsion checks.

The transverse cage, longitudinal torsion steel, spacing checks, and combined concrete-strut limit all pass for the displayed section state.

Preparing shear–torsion scene…

ConcreteClosed hoopsLongitudinal barsShear cracksTorsion diagonalsSection flow

Use the guided sequence to connect shear cracking, the torsion space-truss cage, and the combined section-size compression-strut limit.

300 kN

20.0 kN·m

28 MPa

420 MPa

300 mm

600 mm

540 mm

40 mm

100 mm

45°

5.0 m

2 legs

1608 mm²

Shear checks

Vc: 145.7 kN
φVn provided: 376.5 kN
Required Av/s: 1.121 mm²/mm
Provided Av/s: 1.571 mm²/mm
Shear smax: 135 mm
Provided Vs: 356.3 kN
Design fyt: 420 MPa
Section φVn ceiling: 533.6 kN

Torsion / combined checks

φTth: 5.93 kN·m
φTn provided: 45.04 kN·m
Design Tu: 20.00 kN·m
Torsion active?: Yes
Torsion smax: 180 mm
Required Av/s + 2At/s: 1.819 mm²/mm
Provided Av/s + 2At/s: 3.142 mm²/mm
Required Al: 502 mm²
Provided Al: 1608 mm²
Longitudinal max perimeter spacing: 242 mm
Longitudinal min center spacing: 92 mm
Shear-leg center spacing: 210 mm
Governing smax: 135 mm
Combined demand: 2.369 MPa
Combined limit: 3.294 MPa

Equilibrium torsion cannot simply be discarded when it is required by the structural load path. The 3D crack/strut field is a deterministic mechanism visualization, not a nonlinear crack-width or finite-element solution.

Key Takeaways

$V_n=V_c+V_s$ is nominal strength; the design comparison is $V_u\leq\phi V_n$ .
For a general nonprestressed beam, minimum shear reinforcement is triggered above $0.5\phi V_c$ , subject to the explicit ACI 318-14 exceptions.
A complete stirrup design checks required $A_v/s$ , minimum $A_v/s$ , maximum spacing, provided $\phi V_n$ , and the concrete-web strength ceiling.
Flexure-shear and web-shear cracks describe different initiation mechanisms; neither should be reduced to a universal fixed $45^\circ$ crack narrative.
Torsion may be neglected only below $\phi T_{th}$ . Equilibrium torsion must preserve the load path, while compatibility torsion may redistribute only when the surrounding system can carry the redistributed forces.
Post-cracking torsion requires a closed transverse cage plus longitudinal reinforcement distributed around the perimeter, and combined shear-torsion compression-strut limits still govern section adequacy.

$References$

Association of Structural Engineers of the Philippines (ASEP) — National Structural Code of the Philippines, C101-15, Volume I, Seventh Edition (2015) — Official publisher/source metadata for the NSCP 2015 teaching basis; consult the licensed code for complete provisions, exceptions, and detailing requirements.
ACI 318 Building Code Portal — American Concrete Institute
ACI 318-14 equilibrium-torsion worked design example — StructurePoint

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Learning Objectives

Course Code Basis and Design Philosophy

Diagonal tension

Inclined-Crack Mechanisms

Nominal and Design Shear Strength

Nominal Shear Strength

Shear Strength Requirement

Concrete Contribution VcV_cVc​

Simplified Concrete Shear Strength

Optional detailed VcV_cVc​ expression

Shear Reinforcement Contribution

Vertical-Stirrup Shear Strength

Strength-Required Shear Reinforcement Ratio

When Minimum Shear Reinforcement Is Required

Minimum Shear Reinforcement

Complete Vertical-Stirrup Design Check

Maximum Spacing and Web-Strength Limits

Shear friction

Basic Shear-Friction Model

Shear-Friction Interface Conditions

Brackets, Corbels, and Disturbed Regions

Equilibrium torsion

Compatibility torsion

Threshold and Cracking Torsion

Threshold Torsion

Cracking Torsion for Compatibility Redistribution

Threshold check is a factored comparison

Compatibility-Torsion Redistribution Is a System Decision

Compatibility-Torsion Design Torque

Post-Cracking Torsion Model

Nominal Torsional Strength

Longitudinal Torsion Reinforcement

Combined Transverse Reinforcement for Shear and Torsion

Minimum Longitudinal Torsion Reinforcement

Combined Shear-Torsion Section-Size Check

Do not cure a failed section-size check with more steel

3D Shear–Torsion Studio: Closed Cage, Force Flow, and Combined Action

References

$Learning Objectives$

$Course Code Basis and Design Philosophy$

$Diagonal tension$

$Inclined-Crack Mechanisms$

$Nominal and Design Shear Strength$

$V_c$

$V_c$

$Shear Reinforcement Contribution$

$When Minimum Shear Reinforcement Is Required$

$Complete Vertical-Stirrup Design Check$

$Maximum Spacing and Web-Strength Limits$

$Shear friction$

$Shear-Friction Interface Conditions$

$Brackets, Corbels, and Disturbed Regions$

$Equilibrium torsion$

$Compatibility torsion$

$Threshold and Cracking Torsion$

$Threshold check is a factored comparison$

$Compatibility-Torsion Redistribution Is a System Decision$

$Post-Cracking Torsion Model$

$Do not cure a failed section-size check with more steel$

$References$