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STRUCTURA ACADEMIC · LESSON AREA

In-Plane Shear, Stability and Wall-System Behaviour

Chapter 05 · Masonry Design to Eurocode 6

Approved course
Illustrative masonry course visual showing brick and block cavity-wall materials; not a construction detail.
Original course visual generated for STRUCTURA Academic. Use the reviewed lesson diagrams—not this editorial image—for technical interpretation.
StandardEN 1996-1-1 and EN 1990 teaching references
Source1 source file
Review stateApproved · 2026-08-19
LEARNING OUTCOMES

After this chapter, you should be able to

  • Distinguish in-plane and out-of-plane action.
  • Trace lateral action from diaphragm to foundation.
  • Review shear and sliding resistance.
  • Assess return-wall assumptions.
  • Recognise when full stability analysis is required.

5.1 Purpose and system warningSource §5.1

Masonry walls act as a building system. Some carry gravity actions, some form lateral load paths, and returns or cross-walls may provide restraint. Material capacity alone is insufficient if diaphragms, junctions, foundations or openings interrupt the assumed path.

5.2 NotationSource §5.2

Shear and stability notation
SymbolMeaningUnit
V_Ed / V_RdDesign in-plane shear action / resistancekN or kN/m
f_vk / f_vdCharacteristic / design masonry shear strengthN/mm²
σ_dDesign vertical compressive stressN/mm²
M_Ed / M_RdDesign overturning or lateral moment / flexural resistancekNm/m
M_RdsCracked-section stability moment resistancekNm/m
t, h, lThickness, height and lengthmm or m

5.3 Out-of-plane versus in-plane actionSource §5.3

Out-of-plane panel action
Wind bends the wall across its thickness; support, span and flexural strength govern.
In-plane shear-wall action
Horizontal force travels along the wall to foundations; shear, sliding and overturning govern.
Figure 5.R — regenerated building load-path diagram showing diaphragm transfer to wall lines and foundations.Approved original STRUCTURA academic diagram

5.4 Load path and wall-system behaviourSource §5.4

A typical path is external pressure or frame action → floor or roof diaphragm → shear wall or return → foundation. Every link must exist and be capable of transferring the action.

  • Confirm sufficient wall continuity.
  • Confirm diaphragm transfer to the wall line.
  • Verify returns and cross-walls are bonded or tied.
  • Check openings, chases and movement joints.
  • Confirm foundation restraint and overturning resistance.

5.5 In-plane shear and slidingSource §5.5

Teaching shear check
VEd ≤ VRd ; VRd = fvd t l

Uses the effective wall length and a reviewed design shear strength.

Teaching sliding check
VEd ≤ μf NEd

The friction assumption, DPC type and reliable vertical compression at the interface are critical.

5.6 Returns, cross-walls and stiffeningSource §5.6

Potentially useful
Fully bonded or properly tied return; compatible loading and movement; junction unlikely to crack.
Use conservatively
Openings close to return; unbonded/touching walls; movement joints or chases interrupting the path.

5.7 Free-standing and boundary-wall stabilitySource §5.7

A free-standing wall acts as a cantilever from its base. The source presents a flexural-strength route and a cracked-section equilibrium route.

Boundary-wall flexural resistance
MRd = (fxk1/γM + σd) Z ; MEd = WEd h2/2

Limited compression contributes in the source teaching route; verify the applicable design basis.

Cracked-section stability concept
bci = Nid/fd ; z = t/2 − bci/2 ; MRds = Nid z

The cracked-section stability moment may be more restrictive than the flexural-strength result.

5.8 When masonry should not be the assumed primary stability systemSource §5.8

Conditions requiring more complete analysis
ConditionWhy the simplified wall-line model is incomplete
Large openingsInterrupt effective wall length and force transfer.
Flexible diaphragmMay not distribute action to the selected wall.
Poor junctionsUnbonded or cracked returns do not provide full restraint.
Incomplete foundation modelOverturning and sliding cannot be confirmed.

5.9 Shear and Stability CalculatorSource §5.9

APPROVED ACADEMIC CALCULATOR

Shear and Stability Calculator

Provide source-supported teaching checks for in-plane shear, sliding and boundary-wall flexural/cracked stability while clearly labelling the result as a concept check.

Inputs
Wall length/thickness · f_vd · V_Ed · N_Ed · Friction coefficient · Boundary height/thickness · Material strengths · Unit weight · Partial factors
Outputs
Shear resistance/utilisation · Sliding utilisation · Boundary-wall flexural capacity · Characteristic pressure capacity · Cracked stability moment
Status states
Concept check · Fail · Invalid input
Validation
Approved against the supplied worked-example results; project-specific verification remains required
APPROVED ACADEMIC CALCULATOR · WE-06

Shear and Stability Calculator

Approved educational implementation reproducing the supplied source example.

Inputs
CONCEPT CHECK

Simplified shear and sliding checks pass; full building stability analysis is still required.

Shear resistance
129.00 kN
Shear utilisation
0.39
Sliding utilisation
0.63
Boundary-wall M_Rd
0.253 kNm/m
Characteristic wind capacity
0.23 kN/m²
Cracked stability moment
0.141 kNm/m
Calculation trail
  1. V_Rd = f_vd t l = 129.00 kN
  2. Sliding resistance = μ_f N_Ed = 80.00 kN
  3. M_Rd = (f_xd1 + σ_d)Z = 0.253 kNm/m
  4. M_Rds = N_id z = 0.141 kNm/m

5.10 WE-06 — Boundary-wall stabilitySource §5.10

WORKED EXAMPLE

WE-06 · MAS-WE-06 · Rev. C

Compare 105 mm and 220 mm thick, 1.20 m high free-standing walls using flexural capacity, then check cracked-section stability for the 105 mm wall.

  1. Design strengths

    fd = 2.33/3.0 = 0.776; fxd1 = 0.30/2.70

    fxd1 = 0.111 N/mm2
  2. Self-weight stress

    σd = 1.20(22) = 26.4 kN/m2

    0.0264 N/mm2
  3. 105 mm modulus

    Z = 1000(1052)/6

    1.838 × 106 mm3/m
  4. 105 mm resistance

    MRd = (0.111 + 0.0264)(1.838 × 106)

    0.253 kNm/m
  5. 105 mm pressure capacity

    WEd,cap = 2(0.253)/1.22; divide by 1.5

    Wk,cap ≈ 0.23 kN/m2 · LOW
  6. 220 mm comparison

    Z = 8.067 × 106; MRd = 1.109 kNm/m

    Wk,cap ≈ 1.03 kN/m2
  7. Cracked-section force

    Nid = 1.2(0.105)(22) = 2.77 kN/m; bci = 2.77/0.776

    bci = 3.57 mm
  8. Cracked moment

    z = 50.7 mm; MRds = 2.77(0.0507)

    0.140 kNm/m · more restrictive

Result. The 105 mm wall has very low lateral resistance and is governed by cracked-section stability. Increasing thickness to 220 mm greatly improves capacity; final design must include wind exposure, piers, foundations, movement joints and workmanship.

5.11 Chapter summarySource §5.11

Key points

  • In-plane shear and out-of-plane bending are different behaviours.
  • Diaphragms, junctions and foundations determine whether a wall system works.
  • Free-standing walls are often governed by lateral stability.
  • Simplified results must be labelled concept checks.
Source teaching summary — ULS wall-system combinations
CaseG_kQ_kW_k
Compression side · dead + imposed + wind1.351.500.75
Compression side · alternative1.351.051.50
Tension side · dead + wind1.00—1.50
Teaching summary derived by the source from EN 1990 Eq. 6.10 and lecture notes; not an official numbered Eurocode table.

Source references recorded by the supplied chapter

  • EN 1996-1-1 ULS, shear, flexural and stability concepts
  • EN 1990 Equation 6.10 teaching combinations
  • Masonry Note Parts 1, 2 and 4
  • Updated interactive book Chapter 05