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Chapter 7 · 8 hours

Seismic Design and Strengthening of Masonry Buildings

Practice questions

Practice questions and answers

6 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 6 marks

Explain the seismic behaviour of unreinforced masonry (URM) buildings. How does reinforced masonry perform better? Differentiate between them.

Answer

Seismic behaviour of unreinforced masonry

  • Heavy and brittle: the mass attracts large inertia force (F=maF = ma) while the masonry has low tensile and shear strength, so cracks form early.
  • Load path: seismic forces from the roof pass through the floor diaphragm to the walls parallel to the force (shear walls); walls normal to the shaking bend out of plane.
  • Typical damage: diagonal cracks in piers, corner separation, out-of-plane collapse of walls and gables, and delamination of wall leaves.
  • Weak connections between walls and floors, and no continuous ties, lead to collapse of walls one by one rather than box action.
  • URM has very small ductility; the response reduction factor RR used in design is small (about 1.5).

Seismic behaviour of reinforced masonry

  • Vertical bars (in grouted cores) and horizontal bars (in bond beams) carry the tension caused by bending and shear, giving good strength and ductility.
  • Walls act as a box, with continuous bond beams tying walls and floors, and out-of-plane strength is higher.
  • Cracks are fine and distributed; energy is dissipated by yielding of steel, and the wall retains vertical load capacity after cracking.
  • Because of its ductility, a larger RR (about 3) is allowed, so the design force is smaller.

Comparison

PointUnreinforced masonryReinforced masonry
Tensile strengthVery lowProvided by steel
FailureBrittle shear / out-of-planeDuctile flexure (if designed)
Energy dissipationPoorGood
DuctilityAbout 1–1.5Moderate to high
Design RRLowHigher
Post-cracking capacityDrops quicklyRetained
CostLowerHigher (steel, grout, labour)
Height permittedLow (1–3 storeys)More storeys

URM can be made safer by confining with RC bands, tie columns and vertical bars at corners.

  • Practice · 6 marks

State and explain the principles to be followed in the planning and construction of masonry buildings in earthquake-prone areas.

Answer

Masonry buildings should behave as a strong, tied box so that all walls support each other during shaking.

1. Simple, symmetric and regular plan

  • Rectangular, symmetrical plans are best. Avoid L, T and U shapes (or separate them by seismic gaps), because they twist (torsion) and have stress concentrations at re-entrant corners.
  • Keep the length-to-breadth ratio not more than about 3, and place walls so that the centre of mass and the centre of rigidity are close.

2. Limited height and storey height

Keep the number of storeys and storey height low (commonly not more than 3.0–3.4 m per storey). Heavy roofs (stone slabs, thick mud) should be avoided; use light roof materials.

3. Walls

  • Provide adequate wall density in both directions, with the cross-walls spaced closely to limit out-of-plane span.
  • Keep the thickness related to height (the h/th/t ratio small) and avoid slender walls.
  • Use good-quality units and cement or cement–lime mortar, and fill all joints; use through stones in stone walls.

4. Openings

Keep openings small, symmetrical and away from corners. As a guide (IS 4326): the total width of openings should not exceed about 50 percent of the wall length in a single-storey building (less for higher buildings); pier width should be not less than about 340 mm and openings should be about 600 mm from inside corners. Openings in adjacent storeys should be vertically aligned.

5. Bands and vertical steel

Provide horizontal bands (plinth, lintel, roof, gable) and vertical bars at corners, junctions and door/window jambs to tie the structure.

6. Floors and roofs

Use rigid diaphragms (RC slabs) or well-braced timber floors firmly anchored to the walls to distribute force by box action.

7. Foundation

Use a uniform type of foundation on firm soil, with plinth beams; avoid foundations at different depths.

8. Quality of workmanship

Fill the joints fully, toothed or bonded wall junctions, curing, and avoid cutting chases in walls that weaken them.

  • Practice · 8 marks

A two-storey load-bearing masonry building (plan 8 m × 6 m, height 6 m, each storey 3 m) is in Zone IV (Z=0.24Z = 0.24) on medium soil. Seismic weights are 450 kN at the roof and 600 kN at the first floor. Importance factor I=1.0I = 1.0, response reduction factor R=1.5R = 1.5. Use the seismic coefficient method of IS 1893: Ah=Z2IRSagA_h = \dfrac{Z}{2}\dfrac{I}{R}\dfrac{S_a}{g}, with T=0.09h/dT = 0.09h/\sqrt{d} taking dd the plan dimension in the direction of force (8 m). Take Sa/g=2.5S_a/g = 2.5 for the period found. Find the design base shear and its distribution to floors.

Answer

Step 1: Fundamental period

T=0.09 hd=0.09×68=0.19 sT = \frac{0.09\,h}{\sqrt{d}} = \frac{0.09 \times 6}{\sqrt{8}} = 0.19\ \text{s}

For medium soil with 0.1≤T≤0.550.1 \le T \le 0.55 s, the spectral value Sa/g=2.5S_a/g = 2.5 (plateau).

Step 2: Design horizontal coefficient

Ah=0.242×1.01.5×2.5=0.12×0.667×2.5=0.20A_h = \frac{0.24}{2}\times\frac{1.0}{1.5}\times 2.5 = 0.12 \times 0.667 \times 2.5 = 0.20

Step 3: Base shear

Seismic weight W=450+600=1050W = 450 + 600 = 1050 kN.

VB=AhW=0.20×1050=210 kNV_B = A_h W = 0.20 \times 1050 = 210\ \text{kN}

Step 4: Distribution of base shear

Qi=VBWihi2∑Wjhj2Q_i = V_B \frac{W_i h_i^2}{\sum W_j h_j^2}
LevelWiW_i (kN)hih_i (m)Wihi2W_i h_i^2QiQ_i (kN)Storey shear (kN)
Roof450616 200210×16200/21600=157.5210 \times 16200/21600 = 157.5157.5
First floor60035 400210×5400/21600=52.5210 \times 5400/21600 = 52.5210.0
Sum105021 600210.0
        Q2 = 157.5 kN
   --->  =================  roof (6 m)
   |
        Q1 = 52.5 kN
   --->  =================  floor (3 m)
   |
   ^ base shear VB = 210 kN

Remarks

  • The design force is 20 percent of the weight. NBC 105 uses a similar approach: base shear = Cd(T) WC_d(T)\,W, where Cd(T)C_d(T) depends on zone factor, soil type, importance factor, ductility and overstrength factors.
  • The shear is then shared among the walls in the direction of force in proportion to their stiffness (rigid diaphragm) or their tributary area (flexible diaphragm), and each wall is checked for shear and overturning.

Answer: T=0.19T = 0.19 s, Ah=0.20A_h = 0.20, base shear VB=210V_B = 210 kN; lateral forces: 157.5 kN at roof and 52.5 kN at first floor.

  • Practice · 6 marks

A brick masonry pier 230 mm thick and 1.5 m long, 3.0 m high, has lateral and rotational restraint at top and bottom. It carries an axial load of 150 kN and an in-plane seismic moment of 35 kN·m at the base. The basic compressive stress is fb=1.2f_b = 1.2 N/mm². Check the pier using the working stress method allowing a 33 percent increase in permissible stresses for earthquake. Use ksk_s (SR 8 → 0.95, SR 10 → 0.89). State the permissible tension is zero.

Answer

Step 1: Section properties

A=230×1500=3.45×105 mm2Z=230×150026=8.625×107 mm3\begin{aligned} A &= 230 \times 1500 = 3.45\times10^5\ \text{mm}^2 \\ Z &= \frac{230 \times 1500^2}{6} = 8.625\times10^{7}\ \text{mm}^3 \end{aligned}

Step 2: Stresses

NA=150×1033.45×105=0.435 N/mm2MZ=35×1068.625×107=0.406 N/mm2\begin{aligned} \frac{N}{A} &= \frac{150\times10^3}{3.45\times10^5} = 0.435\ \text{N/mm}^2 \\ \frac{M}{Z} &= \frac{35\times10^6}{8.625\times10^7} = 0.406\ \text{N/mm}^2 \end{aligned} σmax=0.435+0.406=0.841 N/mm2 (compression)σmin=0.435−0.406=+0.029 N/mm2 (compression)\begin{aligned} \sigma_{max} &= 0.435 + 0.406 = 0.841\ \text{N/mm}^2 \ (\text{compression}) \\ \sigma_{min} &= 0.435 - 0.406 = +0.029\ \text{N/mm}^2 \ (\text{compression}) \end{aligned}

There is no tension (the resultant lies within the middle third).

Step 3: Permissible stress

Heff=0.75×3000=2250H_{eff} = 0.75 \times 3000 = 2250 mm, SR=2250/230=9.78SR = 2250/230 = 9.78 and ks=0.89+0.22/2×0.06=0.897k_s = 0.89 + 0.22/2 \times 0.06 = 0.897.

A=0.345A = 0.345 m² >0.2> 0.2, so ka=1.0k_a = 1.0.

fc=1.2×0.897=1.076 N/mm2,fc,seismic=1.33×1.076=1.43 N/mm2f_c = 1.2 \times 0.897 = 1.076\ \text{N/mm}^2, \qquad f_{c,seismic} = 1.33 \times 1.076 = 1.43\ \text{N/mm}^2

Step 4: Checks

  • Maximum compression: 0.841<1.430.841 < 1.43 N/mm². Safe.
  • Gravity alone: 0.435<1.0760.435 < 1.076 N/mm². Safe.
  • Tension: none.

The moment that just causes zero tension is M0=NZA=NL6=150×0.25=37.5M_0 = \dfrac{N Z}{A} = N\dfrac{L}{6} = 150 \times 0.25 = 37.5 kN·m, so the applied 35 kN·m is 93 percent of this limit. If the moment is larger, the pier will crack and tension must be taken by vertical bars (reinforced masonry), or the axial load must be increased or the pier lengthened.

Answer: σmax=0.841\sigma_{max} = 0.841 N/mm² < 1.43 N/mm² and σmin=+0.029\sigma_{min} = +0.029 N/mm² (no tension); the pier is safe. Zero-tension limit moment = 37.5 kN·m.

  • Practice · 5 marks

Write short notes on seismic bands and vertical reinforcement in masonry buildings, giving their purposes, locations and typical details.

Answer

Horizontal seismic bands

A band is a continuous horizontal beam (RC or timber) running through all load-bearing walls at one level, forming a ring.

Purposes

  • Tie the walls together so the building acts as a box.
  • Reduce the out-of-plane span of the wall (supports it horizontally).
  • Prevent cracks from growing by confining the masonry, and distribute the roof load.

Types and locations

  1. Plinth band at the plinth level, also reduces differential settlement.
  2. Lintel band at the lintel level of doors and windows (can double as lintel).
  3. Roof / floor band below the roof or floor slab (not needed when an RC slab is cast over all walls).
  4. Gable band along the top of gable walls.

Typical details (as per IS 4326 / NBC 202; check the code)

  • RC band depth at least 75 mm, full wall thickness, concrete grade not less than M15 (1:2:4).
  • At least 2 longitudinal bars of 8 mm diameter, for spans up to about 5 m, with 6 mm stirrups at about 150 mm; bars are lapped and anchored at the corners.
  • Timber bands may be used in low-strength construction as two longitudinals connected by cross-members.

Vertical reinforcement

  • Provided at corners, wall junctions and the sides of door and window openings.
  • Bars (typically 10–12 mm diameter, one or two as per the number of storeys and zone) start in the foundation, pass through every band and extend into the roof band; they are embedded in concrete or mortar (in the hole made in the masonry).
  • Purpose: increase flexural strength and ductility, tie walls at the corners, and prevent collapse of wall portions.
   roof band ================      <- tie everything
        |   |         |   |
        |   | window  |   |
        |   |         |   |  <- vertical bar at jamb
 lintel =======================
        | pier |     | pier |
 plinth band =================
        foundation (bars anchored)

Together, bands and vertical bars form a confined or reinforced masonry system that increases the ductility and avoids brittle collapse.

  • Practice · 6 marks

Describe the methods used for the seismic strengthening (retrofitting) of existing masonry buildings.

Answer

Strengthening of existing masonry aims at (i) restoring strength after damage (repair), and (ii) improving the strength, ductility and integrity of the building (retrofitting).

1. Repair of cracks

  • Grouting of cement–sand or epoxy grout into cracks and voids in rubble walls restores continuity.
  • Stitching (staples or bars across cracks) and re-pointing of joints with stronger mortar.

2. Jacketing of walls

  • Ferrocement or wire-mesh jacketing: welded wire mesh fixed on one or both faces with anchors through the wall, covered by 25–40 mm cement mortar. It increases shear and flexural capacity and out-of-plane strength.
  • Reinforced concrete jacket or FRP (fibre-reinforced polymer) strips and sheets on the face for greater strength with little added weight.

3. Adding bands and vertical elements

  • External or internal RC bands (plinth, lintel, roof) and vertical bars at corners and jambs, added to existing walls to give box action.
  • Corner stitching with steel or timber ties.

4. Connection improvement

  • Anchoring floor and roof to walls with steel angles or through bolts; tie rods or cables across rooms to tie parallel walls.
  • Adding cross-walls or buttresses to limit out-of-plane bending.

5. Modifying the structure

  • Replacing heavy roofs with lighter ones.
  • Closing unnecessary or weakening openings by infilling, or lining openings with frames.
  • Adding new shear walls or RC frames to share the load.

6. Foundation improvement

Underpinning, plinth beams or foundation grouting.

MethodImprovesDrawback
GroutingIntegrity, strengthNeeds skilled labour
Mesh jacketingShear, out-of-planeAdded thickness, cost
Bands / tiesBox actionNeeds access
FRPStrength, ductilityExpensive

The retrofitting scheme should follow a prior assessment: survey, material tests (see non-destructive tests), and analysis of the weakest link.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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