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Chapter 5

Reinforced Masonry

5 Hours · 6 Marks 5 Past Questions · 50 Marks

Syllabus & Marks Distribution

DetailInformation
CourseSeismic Resistant Design of Masonry Structures — ENCE 367 (Elective I)
Chapter5 — Reinforced Masonry
Teaching Hours5 hours
Marks Allocation6 marks in the 60-mark final
Total Course Marks60 marks (final exam) · 45 teaching hours
Weight10% of the final paper
Codes requiredIITK-GSDMA Guidelines for Structural Use of Reinforced Masonry, IS 1905–1987, IS 4326, NBC 202

Syllabus Topics

  • 5.1  Basic concept
  • 5.2  Vertical and horizontal one-way flexural behaviour
  • 5.3  Two-way flexural behaviour
  • 5.4  Flexural and shear strength
  • 5.5  Reinforced masonry column, beam, and bands
The one idea that unlocks this chapter. Everything in Chapters 2 and 4 rested on the assumption that masonry cannot be relied on in tension, so we designed by making sure tension never appeared. Reinforced masonry throws that constraint away: we allow the masonry to crack and put steel where the tension is, exactly as in reinforced concrete. Every provision in this chapter follows from that single change.

Importance & Frequently Asked Topics HIGH

On paper this chapter is only 6 of 60 marks, and the theory part is indeed small — “Explain reinforced masonry” for 4 marks has appeared repeatedly and is almost guaranteed. But the chapter is much more valuable than 6 marks suggests, because the reinforced brick lintel and the seismic band are reinforced-masonry elements, and those have carried 12–18 mark numericals in three of the last six papers. Across the last six papers this chapter accounts for 5 questions and about 50 marks in the 80-mark format.

Marks Distribution by Each Exam

ExamMarksQuestions asked
2075 Chaitra18Q5 — design a reinforced brick masonry lintel over a 1.5 m window opening
2076 Chaitra24Q1b — design a lintel band [12]; Q2b — reinforced brick masonry lintel under triangular loading [12]
2078 Bhadra—Not asked directly
2079 Bhadra4Q5a — what is reinforced masonry? Explain
2080 Bhadra6Q4b — load on the horizontal band at lintel level (band design)
2081 Bhadra4Q5a — explain reinforced masonry

Most Frequently Asked Topics

TopicTimes askedFrequency
Definition and explanation of reinforced masonry (§5.1) 3
Reinforced brick masonry lintel design (§5.5.3) 2
Seismic band — design and detailing (§5.5.4) 2
Flexural and shear strength provisions (§5.4) 1
Reinforced masonry columns and lateral ties (§5.5.2) 1

Topic-wise Importance

SectionTopicImportance
5.1Basic concept, forms of reinforced masonry, permissible stressesVery High
5.2One-way flexural behaviour, vertical vs horizontal spanMedium
5.3Two-way flexural behaviour, yield-line coefficientsMedium
5.4Flexural and shear strength, failure mechanismsHigh
5.5Columns, beams (lintels) and bandsVery High
Exam strategy for this chapter. Prepare §5.1 as a guaranteed 4-marker — learn the formal definition word for word, then the six forms of reinforced masonry, the advantages, and the applications. Then prepare §5.5.3 (lintel) and §5.5.4 (band) as numericals, because that is where the 12–18 mark questions come from. §5.2–5.4 are worth reading for the 6-mark theory slots but are rarely examined as numericals.

5.1  Basic Concept

Past exam questions on this topic
  1. Explain reinforced masonry. 2081 Bhadra · Q5a [4 marks]
  2. What is reinforced masonry? Explain. 2079 Bhadra · Q5a [4 marks]

5.1.1  Definition

Reinforced masonry is masonry designed on the assumption that the flexural tensile stresses cannot be resisted by the masonry and are resisted by reinforcement only. Shear stresses may be resisted by the masonry or by the reinforcement, singly or in combination.

The contrast with unreinforced masonry must be stated carefully, because it is the point of the definition:

BasisUnreinforced masonry (URM)Reinforced masonry (RM)
Design assumption about tension The flexural tensile stress is resisted by the masonry, within a small permissible value (0.07 / 0.14 N/mm²) The masonry is assumed to carry no tension at all; all tension goes to the steel
Reinforcement May actually be present, but is neglected in the calculation Provided by design and fully accounted for
State of the section Designed to remain uncracked; linear elastic throughout The tension zone is allowed to crack; a cracked transformed section is used
Governing check Keep fcb − fca below the permissible tension Provide Ast to carry the whole tensile force T = M/(jd)
DuctilityBrittle — fails suddenly once it cracksDuctile — the steel yields first and absorbs energy
Slenderness limit for a column15Higher, because buckling is restrained by the steel
The examiner's favourite trap. “Unreinforced masonry” does not mean masonry with no steel in it — it describes the design assumption, not the physical contents of the wall. A wall can contain bars and still be designed, quite legitimately, as unreinforced.

5.1.2  Why masonry is reinforced

  1. Masonry is strong in compression but almost useless in tension. Its flexural tensile strength is only 0.1–0.4 N/mm² and the permissible value is 0.07–0.14 N/mm² — roughly a hundredth of its compressive strength, and effectively zero in mud or weak lime mortar.
  2. Earthquake and wind produce reversible tension in walls, both in-plane (at the heel of a shear wall) and out-of-plane (at mid-height of a wall spanning between floor and roof). Unreinforced masonry has no answer to this except mass and precompression.
  3. Unreinforced masonry fails in a brittle manner once flexural or shear cracking starts, with very little warning and almost no energy dissipation — the very worst property in an earthquake.
  4. The scope of URM is limited. It cannot economically produce tall walls, large openings, slender piers, retaining walls or tanks. Reinforcement removes those limits.
  5. Reinforcement supplies ductility, which is what modern seismic codes actually reward: the response reduction factor R rises from about 1.5–2.25 for URM to 3.0–4.0 for reinforced masonry (see §5.5.6), which directly reduces the design earthquake force.

5.1.3  Forms of reinforced masonry — where the steel goes

Six construction forms of reinforced masonry: reinforced hollow blockwork wall, wall with vertical and bed joint reinforcement, pocket type wall, reinforcing pockets formed by Quetta bond, concrete filled cavity wall, beam unit, reinforcement placed in the bed joints of grooved units, and units with bed joint reinforcement
Fig. 5.1 The recognised forms of reinforced masonry. Reinforcement must be located so that it acts compositely with the masonry.Source: IITK-GSDMA Guidelines for Structural Use of Reinforced Masonry, Fig. 2.
FormHow it is builtTypical use
Reinforced hollow blockwork Vertical bars threaded through the aligned hollow cores of concrete blocks; the cores are then grouted The standard modern system for reinforced masonry shear walls
Wall with vertical and bed-joint reinforcement Vertical bars in grouted cavities plus ladder/truss mesh laid in the horizontal mortar beds Walls needing resistance in both directions
Pocket type wall Bars placed in deep pockets formed on one face of the wall and then concreted Retaining walls, tall single-storey walls
Quetta bond A special bonding arrangement that leaves a continuous vertical pocket inside a 1½-brick wall, which is then reinforced and grouted The traditional Indian/Pakistani seismic brickwork — developed after the 1935 Quetta earthquake
Concrete-filled cavity wall Two leaves of masonry with the cavity reinforced and filled with concrete — effectively permanent formwork Heavily loaded walls, basement and retaining walls
Beam unit / bond beam Channel-shaped units laid to form a trough, reinforced and grouted to make a beam within the wall Lintels, bands, and the tie-beams of confined masonry
Bed-joint reinforcement in grooved units Small-diameter wires or mesh laid in the bed joints, sometimes in units grooved to receive them Enhances the capacity of walls to resist lateral loading; crack control

5.1.4  Advantages of reinforced masonry

5.1.5  Applications

Shear walls; load-bearing walls of multi-storey buildings; retaining walls; water tanks and reservoirs; footbridges; reinforced masonry columns, beams and lintels; seismic bands and vertical reinforcement at corners and openings; and, as a strengthening measure, jacketing of existing unreinforced walls. Prestressed (post-tensioned) masonry is the further step, in which vertical tendons apply artificial precompression so that tension never develops at all.

Historical note worth one line in the answer. Reinforced masonry has been used in India and Japan since shortly after the First World War, and in the United States since after the Second. Quetta bond dates from the reconstruction after the 1935 Quetta earthquake.

5.1.6  Materials and permissible stresses

Design follows the working stress (permissible stress) method, with stresses computed on a transformed section using the modular ratio, and stiffness computed on the uncracked section for analysis purposes.

SteelPermissible tensile stress FsPermissible compressive stress
Mild steel to IS 432 (Part I), φ ≤ 20 mm140 MPa130 MPa
Mild steel to IS 432 (Part I), φ > 20 mm130 MPa
HYSD bars to IS 1786 (Fe 415 etc.)230 MPa190 MPa

Size, spacing, cover and anchorage — the detailing rules

ItemRequirement
Maximum bar size25 mm diameter
Minimum bar size5 mm diameter
Bar size vs cell≤ one-half the least clear dimension of the cell, bond beam or collar joint
Bar size vs wall (shear walls)≤ one-tenth of the wall thickness
Clear distance between parallel bars≥ the bar diameter, and ≥ 25 mm
Clear distance in columns and pilasters≥ 1.5 times the bar diameter, and ≥ 35 mm
Cover — bars in filled cavity or pocket≥ 10 mm in mortar; ≥ 15 mm or the bar diameter (whichever is more) in grout
Cover — bars in a bed joint≥ 15 mm from the face of the masonry; ≥ 2 mm of mortar above and below
Corrosion protectionStainless, hot-dip galvanised or epoxy-coated steel; or a 1 mm austenitic stainless coating
Lap spliceStraight lap ≥ 45 bar diameters or 200 mm
Development length   Ld = 0.25 db Fs    but not less than 300 mm

where db = nominal bar diameter (mm) and Fs = permissible stress in the steel (MPa).

It is derived from an allowable bond stress of u = 1.0 MPa: T = πdb u Ld = (πdb2/4)Fs  ⇒  Ld = dbFs/(4u) = 0.25 dbFs.
For mild steel and epoxy-coated bars, increase Ld by 60% (plain bars have no ribs to develop mechanical bond).

Standard hooks may be used to make up any deficiency in development length: a 180° turn plus an extension of at least 4 bar diameters (minimum 64 mm); a 90° turn plus an extension of at least 12 bar diameters; and for stirrups and ties only, a 90° or 135° turn plus an extension of at least 5 bar diameters.

Four lateral load resisting systems: rigid frame action, shear wall action, shear wall to frame interaction, and tube action, each shown with its deflected shape and plan
Fig. 5.2 Where reinforced masonry sits among the lateral-load resisting systems. A reinforced masonry building relies on (b) shear wall action — a flexural cantilever from the foundation — not on frame action.Source: course lecture slides, IOE Pulchowk Campus.
Model answer skeleton — “Explain reinforced masonry” [4 marks]
  1. Definition (write the boxed sentence in §5.1.1 verbatim).
  2. Why — masonry has negligible tensile strength; steel is introduced to carry the tension, exactly as in reinforced concrete.
  3. Where the steel goes — vertical bars in grouted cores/pockets/Quetta bond, and horizontal bed-joint reinforcement; name three or four forms from Fig. 5.1.
  4. Contrast with URM — in URM the tensile stress is assumed resisted by the masonry and any steel present is neglected; the term describes the design assumption, not the absence of bars.
  5. Benefits — flexural and shear capacity, ductility and energy dissipation, better out-of-plane behaviour, taller walls and larger openings, higher R value.
  6. Applications — shear walls, retaining walls, tanks, lintels, bands, columns and beams.

5.2  Vertical and Horizontal One-Way Flexural Behaviour

A wall loaded normal to its plane — by wind, by earth pressure, or by its own out-of-plane inertia in an earthquake — behaves as a slab. If it is supported on only two opposite edges it spans one way, and the direction of that span decides everything about how it must be reinforced.

5.2.1  The two one-way cases

BasisVertical one-way spanHorizontal one-way span
Supported onThe floor below and the roof/floor aboveTwo cross walls, piers or columns
The wall spansVertically, over the storey height HHorizontally, over the spacing L of the cross walls
Bending is aboutA horizontal axisA vertical axis
Cracks formHorizontally, along a bed joint at mid-heightVertically, up the perpends near mid-span
Tension actsNormal to the bed joints — it pulls the bed joint apartParallel to the bed joints — it must break through the units or step round them
Flexural strength of URMWeak — permissible 0.07 N/mm²About twice as strong — permissible 0.14 N/mm²
Reinforcement neededVertical bars in cores, pockets or Quetta-bond cavitiesHorizontal bed-joint reinforcement or bond beams
Helped by gravity load?Yes — precompression directly offsets the flexural tensionNo — vertical load does nothing for horizontal bending
Orthogonal strength ratio   μ = flexural strength normal to the bed jointsflexural strength parallel to the bed joints  ≈ 0.3 to 0.5 for unreinforced masonry.

This is why the two directions are treated separately, and why a wall will always try to span horizontally if it is given the chance.

5.2.2  Design moments for the one-way strip

Take a strip of unit width and analyse it as a beam. The coefficient depends entirely on the edge restraint, and stating it is worth marks:

Support conditionDesign momentWhere it occurs
Simply supported top and bottomM = wH²/8Mid-height
Partial restraint — e.g. anchored to the roof with metal anchorsM = wH²/10Mid-height and support
Fully continuous / built in at both endsM = wH²/12Support (and wH²/24 at mid-span)
Free at the top (parapet, free-standing wall)M = wH²/2Base — the worst case by a factor of 5

The out-of-plane pressure itself is w = the wind pressure, or for earthquake w = Ah × (weight of wall per unit area) = Ah γ t.

5.2.3  Effective span and effective compressive width

The effective span of a one-way member is taken as the lesser of the centre-to-centre distance between supports and the clear span plus the effective depth.

Diagram showing that the effective compressive width of a reinforced masonry section is limited to three times the wall thickness for each vertical bar, with the spacing of reinforcement marked
Fig. 5.3 Effective compressive width for locally concentrated reinforcement. Each vertical bar can only mobilise a limited width of masonry in compression.Source: IITK-GSDMA Guidelines, Fig. 4.
When the reinforcement is concentrated locally so that the section cannot be treated as a flanged member, the width of masonry assumed to work with each bar is limited to the least of:

  (a) the centre-to-centre bar spacing
  (b) 6 times the wall thickness
  (c) 3t — the effective compressive width shown in Fig. 5.3

This is the reinforced-masonry equivalent of the effective flange width of a T-beam: it stops the designer from assuming that one bar can mobilise the whole wall.

5.3  Two-Way Flexural Behaviour

A wall panel supported on three or four edges — by the floor, the roof and one or two cross walls — does not span one way. It bends about both axes at once, and the load finds its way to the supports along the stiffest and strongest path. Because the horizontal direction is roughly twice as strong, a two-way panel sheds a disproportionate share of the load sideways.

5.3.1  Why two-way action matters

5.3.2  Analysis by the yield-line (fracture-line) method

Masonry has no plastic plateau, so the lines are properly called fracture lines, but the algebra is the same as for a concrete slab. The design moment per unit width is written with a coefficient:

M = α · w · L2

where  w = out-of-plane pressure,  L = the shorter span of the panel,
and  α = a bending-moment coefficient that depends on the edge conditions (how many edges are supported, and whether they are simply supported or continuous), the aspect ratio h/L of the panel, and the orthogonal strength ratio μ.
Do not invent the coefficients. The α values are tabulated (BS 5628 Table 9 and its equivalents) for nine standard edge conditions, and they must be read from the table for the actual h/L and μ. In an exam, state the method, quote the formula, name the three things α depends on, and say that the coefficient is taken from the code table — that is what earns the marks.

5.3.3  Design consequences

  1. Reinforce in both directions — vertical bars for the vertical span, bed-joint reinforcement or bond beams for the horizontal span.
  2. The corners attract the largest moments and are where the fracture lines start, so corner reinforcement and proper interlocking of the return walls are essential.
  3. Two-way action only develops if the edges are genuinely restrained — the wall must be properly tied to the diaphragm and bonded into the cross walls. An unanchored wall reverts to a one-way cantilever with five times the moment.
  4. Because it is stronger, two-way action allows thinner walls and larger panels for the same pressure.

5.4  Flexural and Shear Strength

5.4.1  Flexural strength — working stress design of a reinforced section

The section is analysed exactly like a singly reinforced concrete beam, using the cracked transformed section: the masonry below the neutral axis is ignored, and the steel area is transformed by the modular ratio n = Es/Em.

Neutral axis depth factor   k = 11 + Fs / (n Fb)     Lever-arm factor   j = 1 − k/3

Moment of resistance, masonry governing   Mm = ½ Fb k j b d2 = R b d2

Moment of resistance, steel governing   Ms = Fs Ast j d

Required steel   Ast = MFs j d     Required depth   d = √( M / R b )

Where Fb is the permissible bending compressive stress in the masonry (commonly taken as fm/4 for reinforced brickwork) and Fs is the permissible steel stress from §5.1.6. The section should be kept under-reinforced, so that the steel reaches its permissible stress first and the failure is ductile.

5.4.2  Permissible compressive force

Po = ( 0.25 fm An + 0.65 Ast Fs ) ks

An = net area  ·  Ast = area of steel  ·  Fs = permissible steel stress  ·  ks = stress reduction factor, IS 1905 Table 9 (the same table as Chapter 2).

The coefficient 0.25 gives a factor of safety of about 4.0 against crushing of the masonry; the coefficient 0.65 on the steel term was determined from tests. Note that both the masonry and the steel contribute in compression — unlike shear, where they are not added.

Combined axial and flexural compression. For a reinforced member the compressive stress due to the combined action of axial load and bending must not exceed 1.25 Fa, and the compressive stress due to axial load alone must not exceed Fa. The unreinforced interaction equation (P/Pa + M/Ma ≤ 1) does not apply to reinforced sections.

5.4.3  Shear strength

Shear stress in a reinforced flexural member    fv = Vb j d   or approximately   fv = Vb d

If there is tension in any part of a section, that area must be ignored when working out the shear stress — a cracked zone cannot be relied on to carry shear.

Permissible shear stress Fv — flexural members

CasePermissible shear stress FvUpper limit
Without web reinforcement0.083 √fm0.25 MPa
With web reinforcement0.25 √fm0.75 MPa

Permissible shear stress Fv — walls

For walls the permissible shear depends on the aspect ratio, expressed as M/Vd — the ratio of moment to shear times depth, which is essentially the height-to-length ratio of the wall:

Web reinforcementM/VdFv (MPa)Maximum allowable (MPa)
Without< 1 136(4 − M/Vd)√fm 0.4 − 0.2 (M/Vd)
≥ 10.083 √fm0.2
With< 1 124(4 − M/Vd)√fm 0.6 − 0.2 (M/Vd)
≥ 10.125 √fm0.4
Graph of allowable shear stress against M over Vd ratio, showing two curves: reinforcement resists the entire shear, higher; and masonry resists the entire shear, lower; both flatten beyond an M over Vd ratio of 1
Fig. 5.4 Allowable shear for reinforced walls. Squat walls (low M/Vd) are allowed much more shear stress, and the benefit of web reinforcement is greatest exactly where the wall is squat. Beyond M/Vd = 1 both curves flatten out.Source: IITK-GSDMA Guidelines, Fig. C5.
The single most important rule in reinforced masonry shear design — the contributions are NOT added. Unlike reinforced concrete, this document does not permit Vmasonry + Vsteel. Either the masonry resists the entire shear (if fv ≤ Fv), or, if the calculated shear stress exceeds the allowable value, the reinforcement must be designed to carry the FULL shear load on its own. This follows from experiments by Priestley and Bridgeman, and it is why Fig. 5.4 shows two separate curves rather than a sum.
Where the shear reinforcement is designed to carry the shear, the minimum area is

Av,min = V sFs d

V = total applied shear  ·  s = spacing of the shear reinforcement  ·  d = distance from the extreme compression fibre to the centroid of the tension steel.

Maximum spacing of shear reinforcement = the lesser of 0.5d and 1.2 m, so that any 45° shear crack is crossed by at least one bar.

In a cantilever beam the maximum shear at the support must be used. For a member under a uniformly distributed load, the maximum shear may be taken at 0.5d from the face of the support, provided the support reaction causes compression in the end region and there is no concentrated load within that distance.

5.4.4  Behaviour and failure mechanisms of reinforced masonry shear walls

The behaviour under combined shear, axial load and moment depends on the level of axial compression, the amount of horizontal and vertical reinforcement, the wall aspect ratio, and the material properties. There are two families of failure, each with sub-modes (FEMA 306):

FamilyModeWhen it occursCharacter
A. Flexural failure Ductile flexural failure Aspect ratio h/L ≥ 1.0 with a moderate axial stress Tensile yielding of the vertical steel at one end, with cracking and spalling of the masonry at the compression toe; damage concentrated in the plastic hinge zone. This is the preferred mode — ductile and good at dissipating energy.
Flexure / lap splice slip Starter bars from the foundation have insufficient lap length, or the bar is too large for the wall thickness (e.g. 25 mm bars in a 200 mm wall) Vertical cracks appear at the splices, then spalling at the toes; the wall rocks on its foundation. Fairly ductile but causes severe strength degradation and little energy dissipation.
Flexure / out-of-plane instability Very high ductility levels, after large tensile strains have opened the tension zone On load reversal the previously stretched zone goes into compression and buckles out of plane. Seen in laboratory tests but not yet in any post-earthquake field survey. Prevented by ensuring the stability of the compression zone.
B. Shear failure Diagonal tension failure Inadequate or badly anchored horizontal reinforcement The first diagonal cracks widen into a major X-shaped crack pair — sudden and brittle. With adequate, properly anchored horizontal steel the cracks instead become numerous, fine and well distributed, and failure is gradual (“ductile shear failure”), which in a very squat wall may be the only ductile option available.
Sliding shear failure Low gravity load with high seismic shear — at the base of low-rise buildings, or at upper storeys of medium-rise buildings A horizontal crack forms at the base and the wall slides; there may be very little other damage. The frictional resistance is only restored once the vertical bars yield in tension and supply a clamping force. Premature sliding prevents the wall from ever developing its full flexural capacity.
Diagram of a shear or flexural wall under earthquake loading, showing the deflected shape and the plastic hinge forming at the base of the wall
Fig. 5.5 Formation of the plastic hinge at the base of a ductile flexural wall. Design and detailing aim to force the yielding to happen here — and nowhere else — so that the energy is dissipated in a controlled location and the foundation stays undamaged.Source: course lecture slides, IOE Pulchowk Campus.
Capacity design in one sentence. The whole point of reinforced masonry seismic design is to make the ductile mode (flexural yielding at the base) happen before any of the brittle modes (diagonal tension, sliding, lap-splice slip). Ductility is the capacity to deform well beyond yield while retaining most of the load-carrying capacity — it is the property the R factor pays for.

5.5  Reinforced Masonry Column, Beam and Bands

Past exam questions on this topic
  1. Design a lintel band for a single room one storey building of 4 m breadth, 6 m length and 3 m ceiling height … show design details in sketch. 2076 Chaitra · Q1b [12 marks]
  2. Design a reinforced brick masonry lintel subjected to triangular loading for a window opening of span 1.6 m. 2076 Chaitra · Q2b [12 marks]
  3. Design a reinforced brick masonry lintel over a window opening of span 1.5 m … brick masonry characteristic strength 10 N/mm², Fe 415. 2075 Chaitra · Q5 [18 marks]
  4. Calculate the load on the horizontal band at lintel level due to earthquake and dead load. 2080 Bhadra · Q4b [6 marks]

5.5.1  General

A reinforced masonry column is an isolated vertical compression member whose width does not exceed 4 times its thickness; anything longer is a wall. A beam in masonry is usually a lintel spanning an opening. A band is a horizontal reinforced beam cast within the thickness of the wall and running continuously around the whole building.

5.5.2  Reinforced masonry columns

RequirementProvision
Vertical reinforcement ≥ 0.25% and ≤ 4% of the net cross-sectional area of the column
Minimum number of bars4 — so that ties can form a confined core
Lateral ties — diameter≥ 6 mm
Lateral ties — vertical spacing The least of  (i) 16 × diameter of the longitudinal bar,  (ii) 48 × diameter of the tie,  (iii) the least dimension of the column
Tie arrangement Every corner bar and every alternate longitudinal bar must be held by the corner of a tie with an included angle not more than 135°
Clear distance between vertical bars≥ 1.5 × bar diameter and ≥ 35 mm
Minimum eccentricityColumns must be designed for a minimum eccentricity — perfectly axial loading never occurs
Slenderness ratioThe greater of heff/teff in the two principal directions (a column can buckle about either axis; a wall about only one). Limit 15 for an unreinforced load-bearing column, and 12 by IS 1905

Why ties matter: they provide lateral support to the longitudinal bars to prevent them buckling (the critical unsupported length is about 16 bar diameters), and they resist diagonal tension when the column acts in shear. They may be placed in the mortar joints.

5.5.3  Reinforced masonry beams — the reinforced brick lintel

This is the element the exam actually asks you to design. The full worked solutions are on the solved-paper pages; here is the method.

  1. Decide whether arching action develops. If the wall on either side of the opening extends beyond the springing by at least half the span, and the masonry above the lintel is at least as deep as the load triangle, the masonry arches over the opening and the lintel carries only the 45° triangle of masonry (height L/2). Otherwise it carries the full rectangle of masonry above it, plus any floor or roof load landing in that zone.
  2. Compute the load. Triangle: W = ½ × L × (L/2) × t × γ. Add the self weight of the lintel as a UDL.
  3. Compute the moment and shear. For a triangular load of total magnitude W on a simply supported span L: M = WL/6 and V = W/2. For the UDL self weight: M = wL²/8, V = wL/2.
  4. Set the working-stress constants. σcbc = fm/4; σst = 230 MPa for Fe 415; m = 280/(3σcbc); k = mσcbc/(mσcbc + σst); j = 1 − k/3; R = ½σcbckj.
  5. Find the required depth d = √(M/Rb) and provide an overall depth in whole brick courses (typically three courses ≈ 230 mm), then compute the effective depth d = D − cover − φ/2.
  6. Find the steel Ast = M/(σst j d). This is almost always far below the practical minimum, so provide 2 bars of 10–12 mm laid in the vertical joints of the bottom course and fully grouted with 1:3 cement mortar.
  7. Check shear τv = V/(bjd) against about 0.10 N/mm² for reinforced brickwork, and provide nominal 6 mm links at 150 mm centres to hold the bars.
  8. Check bearing and anchorage. Minimum bearing 200 mm each end; the bars run the full length with a 90° hook, giving anchorage well beyond Ld = 0.25 dbFs.
Curtailment rule worth quoting. Reinforcement should not be curtailed in a tension zone, and in all cases at least one quarter of the reinforcement required at mid-span must be carried through to the support and properly anchored there.
Worked examples of this exact design are on 2075 Chaitra Q5 (1.5 m span, 230 mm wall, fm = 10 N/mm², Fe 415) and 2076 Chaitra Q2b (1.6 m span, 220 mm wall). Both come out hugely under-stressed — the section is governed by detailing, not by strength — which is itself a point worth stating in the answer.

5.5.4  Seismic bands

A band is a horizontal reinforced beam cast within the wall thickness and running continuously around the entire building at one level. Bands are the single most cost-effective seismic measure available for masonry construction — in the 2015 Gorkha earthquake, buildings with even one lintel band frequently survived with repairable damage while identical unbanded neighbours collapsed.

The six bands and what each does

BandLocationFunction
Plinth bandAt plinth level, above the foundation Ties the foundation together and distributes differential settlement; the first defence against soil failure
Sill bandAt window sill level Controls cracking at the bottom corners of window openings
Lintel bandAt the top of doors and windows The most important band. Carries the wall above the openings, ties all the walls together at opening head level, and provides the out-of-plane support that stops walls overturning
Floor bandAt each intermediate floor level Ties the walls to the diaphragm and transfers the floor inertia into the shear walls
Roof bandAt roof level Ties the wall tops together and anchors the roof; essential with a flexible or pitched roof
Gable bandAlong the slope of a gable wall Restrains the free-standing triangle of the gable, which otherwise behaves as a cantilever

How a band is loaded — biaxial bending

A band is a horizontal beam bent about TWO axes at once:

• Vertically, by the weight of the masonry it supports (lintel action) — the section is b = wall thickness, D = band depth.
• Horizontally, by the out-of-plane inertia of the wall strip tributary to it — here the band bends in its own plane, so the wall thickness becomes the depth, which is why the band is strong in this direction.

Seismic load per metre run   wEQ = αh × (tributary wall height) × t × γ
Design moment (continuous with the cross walls)   M = wL²/12

Detailing rules — IS 4326 cl. 8.4 / NBC 202

Worked example. 2076 Chaitra Q1b designs a lintel band for a 4 m × 6 m room with αh = 0.36, and shows that the IS 4326 minimum 75 mm band is not sufficient for a 6 m span — a 150 mm deep band with 4 nos 12 mm bars is needed. 2080 Bhadra Q4b computes the band loading for a two-storey health post.

5.5.5  Minimum reinforcement for seismic performance

Elevation of a masonry wall showing the required minimum reinforcement: vertical bars of at least 100 square millimetres at corners, within 400 mm of each side of openings and within 200 mm of wall ends, at 3 m maximum spacing, plus bond beam reinforcement at floor and roof levels and above and below openings extending 500 mm or 40 bar diameters past them
Fig. 5.6 Minimum reinforcement layout for a seismic masonry shear wall. Note the three controlling dimensions: 200 mm from a wall end, 400 mm each side of an opening, and a maximum spacing of 3 m.Source: IITK-GSDMA Guidelines, Fig. C7.
RequirementProvision
Vertical reinforcement At least 100 mm² in cross-sectional area at a maximum spacing of 3 m, and at these critical sections: (a) corners, (b) within 400 mm of each side of an opening, (c) within 200 mm of the end of a wall
Small openingsReinforcement around openings is not needed for openings smaller than 400 mm in either direction — the diagonal strut is not disrupted
Horizontal reinforcement At least 2 bars of 6 mm at not more than 400 mm spacing; or bond-beam reinforcement of at least 100 mm² at not more than 3 m spacing
Where horizontal reinforcement must go At the top and bottom of every wall opening, extending at least 500 mm or 40 bar diameters past the opening; continuously at structurally connected floor and roof levels; and within 400 mm of the top of the wall

5.5.6  The three wall types and their R values

Table listing three wall types: Type A, unreinforced masonry with minimum reinforcement, seismic zones II and III, R equals 2.25; Type B, reinforced masonry with minimum reinforcement, zones IV and V, R equals 3.0; Type C, reinforced masonry with minimum and special reinforcement, zones IV and V, R equals 4.0
Fig. 5.7 Reinforcement requirements and response reduction factors for the three wall types. More reinforcement buys a higher R, and a higher R directly reduces the design base shear (Ah = (Z/2)(I/R)(Sa/g)).Source: IITK-GSDMA Guidelines, Table 3; R values as per IS 1893.
TypeDescriptionZoneRKey requirement
ADetailed unreinforced masonry shear wall — designed as URM but containing the minimum reinforcement of §5.5.5II & III2.25 Improved inelastic response and energy dissipation; can accommodate larger deformation than plain URM. Low to moderate seismic risk only.
BOrdinary reinforced masonry shear wall IV & V3.0 Designed as reinforced masonry and meeting the minimum reinforcement of §5.5.5. Adequate safety in moderate to high seismic zones.
CSpecial reinforced masonry shear wall IV & V4.0 The most restrictive requirements — intended for regions of intense seismic activity. See below.
Additional requirements for a SPECIAL reinforced masonry shear wall (Type C)

1.  Uniformly reinforced in both directions, with the sum of the reinforcement in the two directions ≥ 0.2% of the gross cross-sectional area, and the reinforcement in each direction ≥ 0.07%.
2.  Maximum spacing of horizontal and vertical reinforcement = the lesser of  (a) one-third the length of the shear wall,  (b) one-third the height of the shear wall,  (c) 1.2 m.
3.  The reinforcement in the vertical direction ≥ one-third of the required shear reinforcement.
4.  Shear reinforcement must be anchored around the vertical bars with a 135° or 180° standard hook.

5.5.7  Detailing of reinforced masonry shear walls

Quick Revision

ItemValue / rule
Definition of reinforced masonry Flexural tension resisted by reinforcement only; masonry assumed to carry no tension
Permissible steel stress, HYSD230 MPa tension · 190 MPa compression
Permissible steel stress, MS140 MPa (φ ≤ 20), 130 MPa (φ > 20) tension · 130 MPa compression
Bar size limits5 mm ≤ φ ≤ 25 mm; ≤ ½ the least clear cell dimension; ≤ t/10 in a shear wall
Development lengthLd = 0.25 dbFs ≥ 300 mm; +60% for MS and epoxy-coated
Lap splice≥ 45 db or 200 mm
Cover10 mm in mortar-filled cavity; 15 mm or db in grout; 15 mm from face for bed-joint steel
Orthogonal strength ratio μ0.3–0.5 — horizontal span is about twice as strong as vertical
Permissible flexural tension (URM)0.07 N/mm² vertical bending · 0.14 N/mm² horizontal bending
Two-way momentM = αwL²; α from code tables, depends on edge conditions, h/L and μ
Section constantsk = 1/(1 + Fs/nFb) · j = 1 − k/3 · R = ½Fbkj
Permissible compressive force Po = (0.25 fmAn + 0.65 AstFs) ks
Combined axial + flexure≤ 1.25 Fa; axial alone ≤ Fa
Shear stressfv = V/(bjd); ignore any area in tension
Fv, flexural member0.083√fm ≤ 0.25 MPa (no web steel) · 0.25√fm ≤ 0.75 MPa (with web steel)
Shear contributions are NOT additive Either the masonry takes all the shear, or the steel is designed for the full shear
Shear reinforcementAv,min = Vs/(Fsd); spacing ≤ 0.5d or 1.2 m
Column steel0.25% to 4% of net area, minimum 4 bars
Column ties≥ 6 mm; spacing = least of 16 db, 48 dtie, least column dimension; 135° max included angle
Lintel — arching45° triangle of height L/2; M = WL/6, V = W/2
Band — minimum75 mm deep, full wall thickness, M15, 2 nos 8 mm φ + 6 mm links @ 150 mm, corner lap 60φ
Minimum seismic vertical steel100 mm² at corners, within 400 mm of openings, within 200 mm of wall ends, max spacing 3 m
Special RM wall (Type C)Σρ ≥ 0.2%, each direction ≥ 0.07%, spacing ≤ min(L/3, H/3, 1.2 m), R = 4.0
Boundary elementsRequired where the extreme-fibre compressive stress exceeds 0.2 fck

Chapter Summary

  1. §5.1 — Basic concept. Reinforced masonry is masonry designed on the assumption that flexural tension is carried by reinforcement only. It exists because masonry has essentially no tensile strength and fails in a brittle way. The steel can be placed in grouted hollow cores, in pockets, in Quetta-bond cavities, in filled cavity walls, in beam units, or in the bed joints. The reward is flexural and shear capacity, ductility, better out-of-plane behaviour and a higher R factor. Design is by working stress with permissible steel stresses of 230/190 MPa (HYSD) and 140-130/130 MPa (MS), and Ld = 0.25 dbFs ≥ 300 mm.
  2. §5.2 — One-way flexure. A wall spanning vertically cracks along a bed joint and is weak (0.07 N/mm²), and needs vertical bars; a wall spanning horizontally must break through the units and is about twice as strong (0.14 N/mm²), and needs bed-joint reinforcement. The moment coefficient depends on the restraint: wH²/8, /10, /12, or /2 for a free top. Effective compressive width per bar is limited to 3t (also ≤ the bar spacing and ≤ 6t).
  3. §5.3 — Two-way flexure. A panel supported on three or four edges spans both ways and its design moment falls to a fraction of the one-way value: M = αwL², with α taken from code tables as a function of the edge conditions, the aspect ratio and the orthogonal strength ratio. This is precisely why cross walls, piers and buttresses are so effective, and it produces the diagonal fracture-line crack patterns seen after earthquakes.
  4. §5.4 — Flexural and shear strength. Flexure uses the cracked transformed section with k, j and R exactly as in RC. Compression uses Po = (0.25 fmAn + 0.65 AstFs)ks, with combined stress limited to 1.25 Fa. In shear the crucial rule is that masonry and steel contributions are not added — if fv exceeds Fv, the steel must carry the whole shear. The permissible shear rises sharply for squat walls (low M/Vd). Failure is either flexural (ductile yielding, lap splice slip, out-of-plane instability) or shear (diagonal tension, sliding shear); the ductile flexural mode with a plastic hinge at the base is the one the design must deliberately produce.
  5. §5.5 — Columns, beams and bands. Columns take 0.25–4% vertical steel with at least 4 bars and ties at the least of 16db, 48dtie and the least dimension. Reinforced brick lintels are designed for the 45° arching triangle with M = WL/6, and in practice come out governed by detailing rather than strength. Bands — plinth, sill, lintel, floor, roof and gable — are horizontal beams cast in the wall, bent about two axes at once, and must be continuous all round with 60φ corner laps; they are the cheapest and most effective seismic measure in masonry. Minimum seismic steel is 100 mm² at corners, beside openings and at wall ends at 3 m maximum spacing, and the three wall types A, B and C earn R = 2.25, 3.0 and 4.0 respectively.
How this chapter connects to the rest of the course. Chapter 2 gave the permissible-stress design of unreinforced masonry; Chapter 4 gave the seismic analysis that produces the forces. Chapter 5 is what you do when Chapter 2's tension check fails — instead of thickening the wall indefinitely, you put steel where the tension is. Chapter 6 then applies the same reinforcement ideas to buildings that already exist.

All Past Questions on This Chapter

  1. Explain reinforced masonry. 2081 Bhadra · Q5a [4]  →  solved
  2. Calculate the load on the horizontal band at lintel level due to earthquake and dead load. 2080 Bhadra · Q4b [6]  →  solved
  3. What is reinforced masonry? Explain. 2079 Bhadra · Q5a [4]  →  solved
  4. Design a lintel band for a single room one storey building of 4 m breadth, 6 m length and 3 m ceiling height. Show the design details in a sketch. 2076 Chaitra · Q1b [12]  →  solved
  5. Design a reinforced brick masonry lintel subjected to triangular loading for a window opening of span 1.6 m. 2076 Chaitra · Q2b [12]  →  solved
  6. Design a reinforced brick masonry lintel over a window opening of span 1.5 m. Brick masonry characteristic strength 10 N/mm², Fe 415. 2075 Chaitra · Q5 [18]  →  solved