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Solved Past Questions

2079 Bhadra — Regular

BE / BCE · IV/I 80 Marks · 3 hrs

Paper Information

DetailInformation
ExamRegular, 2079 Bhadra
SubjectSeismic Resistant Design of Masonry Structure (Elective I) — CE 72502
Full Marks · Pass Marks · Time80 · 32 · 3 hours
InstructionsAttempt All questions. IS 1905–1987 codes are allowed. Assume suitable data if necessary.
9Questions
44Theory marks
36Numerical marks
3Numericals

Question 1

1 (a) TheoryDefine elastic rebound theory. Explain the origin of earthquakes in Himalayas. [8]

1. Elastic rebound theory

Proposed by H. F. Reid after studying the 1906 San Francisco earthquake. Rocks on either side of a fault are slowly deformed (bent) by tectonic forces and store elastic strain energy. When the accumulated strain exceeds the strength of the rock (or the frictional resistance of the fault), the rock suddenly ruptures and the two blocks “snap back” — rebound — to a new, unstrained position. The stored elastic energy is released in a few seconds as heat, fracturing and seismic waves; those waves are the earthquake.

Draw the four stages: (a) original unstrained state, with a fence or road crossing the fault straight; (b) strain build-up over decades — the fence bends; (c) rupture when the strength is exceeded — the fence snaps; (d) rebound — the two halves are straight again but permanently offset. The cycle then restarts.

2. Origin of earthquakes in the Himalayas

The Himalaya is the classic continent–continent convergent margin:

  1. About 225 million years ago the Indian plate separated from Gondwanaland and drifted northwards at roughly 10 cm/yr, the intervening Tethys oceanic crust subducting beneath Eurasia.
  2. Around 55 million years ago the Tethys closed and India collided with Asia. Because both are continental crust of the same low density, neither could subduct beneath the other.
  3. The collision pressure could only be relieved by crustal thickening and thrusting upward, raising the Himalayan range. This is why the Himalaya has great earthquakes but no volcanoes.
  4. Convergence continues today; about 2 cm/yr of the ~4–5 cm/yr India–Eurasia convergence is absorbed across the Himalaya, so Nepal shortens by roughly 2 m per century.
  5. The shortening is taken up on a system of north-dipping thrusts that all sole into one gently north-dipping detachment, the Main Himalayan Thrust (MHT): MFT (Main Frontal Thrust, southernmost and active), MBT (Main Boundary Thrust), MCT (Main Central Thrust) and STDS (South Tibetan Detachment System, a normal fault to the north).
  6. The shallow part of the MHT is locked by friction between earthquakes. Strain accumulates for decades to centuries and is then released suddenly as a great earthquake — exactly the elastic rebound mechanism. The 2015 Gorkha earthquake (Mw 7.8) was a rupture of the locked MHT that propagated eastward from Barpak, Gorkha, at a focal depth of about 15 km.
  7. Seismic gap: no great earthquake has ruptured the western Nepal segment of the MHT since 1505, so more than five centuries of strain is stored there. The 2015 event did not release it.

Draw a simplified north–south section: Indian plate underthrusting, the Himalayan wedge, the MHT with the MFT / MBT / MCT splays branching off it, and the typical focus at 10–20 km depth.

Full treatment with figures: Chapter 1 §1.1.6 and §1.1.8 (Figs. 1.4 and 1.7).
1 (b) TheoryExplain the magnitude and intensity of the earthquake. Compare the MMI scale with the Richter scale. [8]

1. Magnitude

Magnitude is a quantitative, instrumental measure of the size of an earthquake — of the total energy released at the source. It is a single number for a given earthquake, independent of where it is measured.

  • Richter scale (local magnitude ML), 1935: based on the logarithm (base 10) of the maximum trace amplitude recorded by a standard Wood–Anderson seismograph, corrected to a standard epicentral distance of 100 km.
  • An increase of one unit of magnitude = 10 times larger wave amplitude and about 31.6 (≈32) times more energy. So M7.7 releases ~31 times the energy of M6.7 and ~1000 times that of M5.7.
  • Open-ended, can be negative, but it saturates above about M7.
  • Moment magnitude Mw is derived from the seismic moment M0 = µAD; it does not saturate and is now used for all large earthquakes (Gorkha 2015 = Mw 7.8).
ML = log10 A − log10 A0      Energy–magnitude (Gutenberg–Richter):   log10 E = 11.8 + 1.5 M  (E in ergs)

2. Intensity

Intensity is a qualitative, observational measure of the actual effects of an earthquake at a particular place — on people, objects, buildings and the ground. Because the effects change from place to place, one earthquake has many intensity values but only one magnitude. It is highest near the epicentre and decreases outward; lines joining points of equal intensity are isoseismals, used to locate the epicentre of historic (pre-instrumental) earthquakes and to prepare seismic hazard maps. The scale used is the Modified Mercalli Intensity (MMI) scale with 12 grades, I to XII (others: MSK-64, JMA).

3. Comparison — Richter vs MMI (this table carries most of the marks)

BasisMagnitude — Richter scaleIntensity — MMI scale
What it measuresEnergy released at the source (size of the earthquake)Effect / severity of shaking at a particular place
NatureQuantitative and objectiveQualitative and subjective (observational)
How obtainedMeasured instrumentally from a seismogramAssessed from damage surveys and human reports; no instrument
Number of valuesOne value per earthquakeMany values — different at every location
Depends on distance?NoYes — decreases with epicentral distance
Depends on soil / building type?NoYes — strongly
NotationArabic numerals with decimals (e.g. 7.8)Roman numerals I to XII (e.g. IX)
Scale typeLogarithmic, open-endedDescriptive, bounded I–XII
Represented byA single point — the epicentreAn isoseismal map
Engineering useSource characterisation, energy, recurrence relationsDamage assessment, seismic zoning, vulnerability studies
Applicable to historic earthquakes?No — needs instruments (post-1900)Yes — can be inferred from records

Close with the approximate correlation: M 5–5.9 → MMI V–VI; M 6–6.9 → VI–VII; M 7–7.9 → VII–IX; M 8–8.9 → IX–X.

Full treatment: Chapter 1 §1.2.5–1.2.7.

Question 2

2 (a) TheoryExplain load-bearing masonry walls, infill masonry, and confined masonry. [8]

These are the three structural systems in which masonry is used. The distinguishing features are the load path and whether the wall is a structural element.

(A) Load-bearing masonry

  • The masonry walls themselves are the main supporting elements — there is no separate frame.
  • Load path: roof / floor slab → main (load-bearing) walls → plinth → foundation → soil. Both gravity and lateral loads travel through the walls.
  • Walls parallel to the shaking act as shear walls (in-plane); walls perpendicular are loaded out-of-plane.
  • Economical for 1–3 storeys; thickness increases rapidly with height. Unreinforced load-bearing masonry is brittle, so earthquake resistance must be supplied by bands, vertical bars, corner bonding and box action.

(B) Confined masonry

Load-bearing masonry wall panels confined on all four sides by lightly reinforced concrete tie-columns (vertical ties) and tie-beams (horizontal ties).

The defining construction sequence: the masonry wall is built FIRST, with toothed vertical edges, and the tie-columns are then cast against the completed wall.
  • Tie-columns (“practical columns”) are of much smaller cross-section than RC frame columns. They are not a moment frame — the masonry, not the ties, carries the load.
  • The confining members are effective in: (1) enhancing the stability and integrity of the wall for in-plane and out-of-plane earthquake loads — they contain the damaged masonry; (2) enhancing the lateral strength; and (3) reducing the brittleness, thereby improving earthquake performance.
  • Components: masonry walls, confining elements, floor/roof slabs acting as diaphragms, plinth band and foundation.
  • It has emerged over the last 100 years as an economical alternative between unreinforced masonry and RC frame construction.

(C) Infill masonry

  • Masonry panels enclosed by RC or steel frame members on all four sides. The frame, not the wall, is the designed structure.
  • Load path: slab → beam → column → foundation. The infill is nominally non-structural.
  • In reality it greatly increases the lateral stiffness of the frame, and under lateral load acts as a diagonal compression strut, so the infilled frame can be analysed as a truss.
  • Dangers: a sudden loss of stiffness when the infill cracks can create a soft-storey mechanism; irregular distribution in elevation (the open ground storey) concentrates drift in one storey; partial-height infill produces the short-column effect.
  • Conversely, continuous well-distributed infill has been shown to reduce the vulnerability of an RC structure.

Comparison table

BasisLoad-bearingConfinedInfill (RC frame)
Vertical load pathWallsWalls (ties assist)Beams and columns
Lateral resisting elementWalls in-planeConfined wall panelsFrame + infill strut
Construction sequenceWalls onlyWall first, then tie-columnsFrame first, then wall
Is the wall structural?YesYesNominally no
Seismic performancePoor unless bandedGood — damage containedRisk of soft storey / short column
Suitable height1–3 storeysUp to 3–4 storeysMulti-storey
Full treatment with figures: Chapter 2 §2.4.2 (Figs. 2.9–2.12).
2 (b) TheoryDifferentiate between in-plane and out-of-plane behaviour of masonry structures. Explain the different failure modes of masonry structures with a neat sketch. [8]

Part 1 — In-plane vs out-of-plane behaviour

BasisIn-plane behaviourOut-of-plane behaviour
Direction of loadParallel to the plane of the wall (along its length)Perpendicular to the plane of the wall
Name of the wallShear wall — the “strong direction”Cross / transverse wall — the “weak direction”
Structural actionDeep cantilever / coupled piers carrying storey shear by in-plane shear and bendingVertical slab spanning between floor and roof, and horizontally between cross walls
Governing stressDiagonal tension, bed-joint shear, toe compressionFlexural tension normal to the bed joints — the weakest property of masonry
Stiffness & strengthHigh — one or two orders of magnitude stifferVery low
Governing parametersAspect ratio L/heff, vertical precompression, mortar and unit strength, openingsh/t ratio and vertical stress; span between cross walls
Failure modesRocking, sliding, diagonal shear, toe crushing — the X crackSimple overturning, composed overturning, vertical flexure, horizontal flexure
ConsequenceSevere cracking but the wall usually stays standingCollapse — the wall falls out and the floor loses its support
The key sentence: because the earthquake shakes the building in both horizontal directions and reverses continually, every wall is an in-plane wall at one instant and an out-of-plane wall the next. A wall is only safe if it is adequate in both roles — and it is the out-of-plane role that governs survival.

Part 2 — Failure modes

In-plane (four modes):

ModeMechanismCrack appearance
Rocking (flexural)Horizontal crack at the heel, pier rotates about its compressed toeHorizontal cracks top and bottom, toe crushing
SlidingHorizontal sliding along a bed jointSingle horizontal crack with offset
Diagonal shear — bed-joint slidingCrack follows the mortar joints (weak mortar)Stair-stepped diagonal crack, bricks intact
Diagonal tension + toe crushingCrack runs through the units (strong mortar, high compression)Straight “X” crack cutting through the bricks

Out-of-plane (four mechanisms): simple overturning; composed overturning (dragging a wedge of the return walls); vertical flexure (hinge at mid-height); horizontal flexure (vertical crack at mid-span between cross walls).

Other typical damage: corner separation, delamination of multi-leaf walls, failure of piers between openings, gable and parapet collapse, roof separation, pounding.

Sketch to draw: one building isometric showing the in-plane wall with X cracks, the out-of-plane wall overturning, the diaphragm, and corner separation at the junction — all labelled.

Full treatment with all figures: Chapter 3 §3.4 and §3.6 (Figs. 3.7–3.11).

Question 3

3 (a) TheoryDefine repair, restoration, and retrofitting. [4]
TermDefinitionAim & typical work
Repair Cosmetic and functional works carried out on a damaged building to bring it back to a usable condition, without altering its structural strength. Aim: restore serviceability and appearance only. Work: patching and re-plastering, filling fine cracks, repairing doors, windows, wiring, plumbing, whitewashing, rebuilding non-structural partitions, re-fixing roof tiles. The strength is NOT restored.
Restoration
(rehabilitation)
Structural works that bring the damaged building back to its original strength — i.e. to the condition it was in before the earthquake. Aim: recover the lost strength and stiffness. Work: grouting and injection of cracks, stitching of wide cracks with dowels, re-pointing eroded joints, dismantling and rebuilding shattered portions, replacing damaged lintels. No more than the original capacity is provided.
Retrofitting
(seismic strengthening)
Structural works that upgrade the building beyond its original strength, so that it can resist a future earthquake better than it resisted the last one. Aim: increase strength, stiffness, ductility and integrity to meet current code requirements. Work: adding horizontal bands and vertical bars, splint-and-bandage, wire-mesh / ferrocement or FRP jacketing, through-stones, corner stitching, stiffening the diaphragm, reducing openings, lightening the roof, adding buttresses or shear walls. Applied to undamaged buildings too.
The one-line distinction to write: Repair → looks better; Restoration → as strong as it was; Retrofitting → stronger than it ever was.
See Chapter 3 §3.7.6; full treatment is Chapter 6 §6.1.
3 (b) NumericalDesign an interior wall of a single-storeyed workshop of height 5.4 m supporting an RCC roof. The bottom of the wall rests over a foundation block. Assume roof load equal to 45 kN/m. [12]
230 115 115 230 4830 4830 All dimensions in mm — plan of the wall with piers
Figure with the questionPlan of the workshop wall. The wall is 230 mm thick; the piers are 230 mm wide and are offset 115 mm from each face (so the pier is 460 mm thick), repeating at 4830 mm centre to centre.
Given: single-storey workshop; wall height H = 5.4 m; RCC roof; bottom of wall on a foundation block; roof load = 45 kN/m.
From the figure above: wall thickness tw = 230 mm; pier offsets 115 mm on each side (so tp = 230 + 115 + 115 = 460 mm); pier width Wp = 230 mm; piers at 4830 mm centre to centre.
Assume: brick crushing strength 10 N/mm² with M1 mortar (1:5); masonry 20 kN/m³. Design one bay (one pier spacing).
Basic compressive stress  IS 1905 Table 8
Unit 10 N/mm² + M1 mortar → f_ck = 0.96 N/mm²
Load per bay
Total load from roof per bay = 4830 × 45 / 1000 = 217.34 kN Self weight of wall per bay = [ (4.830 × 0.230) + (2 × 0.230 × 0.115) ] × 5.4 × 20 = [ 1.1109 + 0.0529 ] × 108 = 1.1638 × 108 = 125.69 kN ---------- Total load per bay, P = 343.04 kN
Cross-sectional area per bay
A = (4830 × 230) + (2 × 230 × 115) = 1 110 900 + 52 900 = 1 163 800 mm²
Effective height  IS 1905 Table 4
RCC roof at the top, foundation block at the bottom → full lateral restraint at both ends h_eff = 0.75 H = 0.75 × 5400 = 4050 mm
Effective thickness  IS 1905 Table 6
S_p / W_p = 4830 / 230 = 21 t_p / t_w = 460 / 230 = 2 From Table 6 : for S_p/W_p ≥ 20 the stiffening coefficient is 1.0 for every t_p/t_w. → stiffening coefficient = 1.0 → THERE IS NO STIFFENING EFFECT from the piers. t_eff = 1.0 × 230 = 230 mm
Slenderness ratio

Take the lesser of the two ratios; effective length for a wall continuous/supported at both ends = 0.8 L:

Based on effective HEIGHT SR = 4050 / 230 = 17.6 ≈ 18 Based on effective LENGTH l_eff = 0.8 × 4830 = 3864 mm SR = 3864 / 230 = 16.8 ≈ 17 For a wall, take the SMALLER → SR = 17 Limit for a load-bearing wall in cement mortar = 27 ✓
Stress reduction factor  IS 1905 Table 9
SR = 17 , e = 0 (axial roof load, interior wall) SR 16 → 0.73 , SR 18 → 0.67 k_s = 0.73 − (0.73−0.67)(17−16)/(18−16) = 0.70
Permissible stress and load
k_a A per bay = 1.1638 m² > 0.2 m² → k_a = 1.0 k_p brick 230×115×57 , h/w = 0.50 < 0.75 → k_p = 1.0 σ_al = k_s × f_ck = 0.70 × 0.96 = 0.672 N/mm² Permissible load per bay = σ_al × A = 0.672 × 1 163 800 = 782 074 N = 782.07 kN
Final check
Applied load per bay P = 343.04 kN Permissible load per bay = 782.07 kN 343.04 < 782.07 ✓ Utilisation = 343.04 / 782.07 = 44 %
THE DESIGNED WALL IS SAFE. Adopt a 230 mm thick interior wall with 115 mm offsets each side forming 460 mm thick, 230 mm wide piers at 4830 mm centres, built with bricks of crushing strength 10 N/mm² in M1 (1:5) cement mortar.
The point of this question is Sp/Wp = 21 ≥ 20. State clearly that the piers give no stiffening benefit at this spacing — they only add local area. Many candidates wrongly apply a coefficient of 1.2 or 1.4 here.
Method: Chapter 2 §2.6.4–2.6.8. Follows the worked example in the course notes (CH-4, Example 4.2) — identical data.

Question 4

4 (a) TheoryWhat are the steps involved in retrofitting strategies? [4]
  1. Data collection and condition survey. Collect the as-built drawings; survey the building from outside and inside; record the structural system, materials, wall thicknesses, openings, roof type and the site conditions; note existing cracks and past interventions.
  2. Seismic vulnerability assessment. Rapid visual screening first (e.g. FEMA 154 / NBC type checklists), then a detailed evaluation — material testing (NDT), damage grade classification (EMS G1–G5), and analysis of the existing capacity.
  3. Identify the deficiencies. List exactly what is wrong: no bands, no through-stones, large or badly placed openings, weak mortar, heavy roof, soft storey, torsional irregularity, poor corner bonding, inadequate diaphragm.
  4. Define the performance objective. Life safety, immediate occupancy or collapse prevention, and the corresponding target earthquake level.
  5. Select the retrofitting strategy. Choose between increasing strength/stiffness (jacketing, new shear walls, bands), increasing ductility and integrity (vertical bars, corner stitching, confinement), reducing demand (lightening the roof, removing storeys, base isolation), or a combination.
  6. Analyse and design the intervention, checking that the retrofitted structure meets the performance objective and that the new elements are compatible with the old.
  7. Cost–benefit and feasibility check against the option of demolition and reconstruction; consider heritage value, disruption to occupants and available skills.
  8. Prepare drawings, specifications and a construction sequence (shoring first, then integrity, then strength, then demand reduction).
  9. Execute with strict quality control and supervision, followed by inspection and monitoring.
See Chapter 3 §3.7.6; full treatment is Chapter 6.
4 (b) NumericalA wall 20 cm thick, using modular bricks, carries at the top a load of 80 kN/m having a resultant eccentricity ratio of e = 1/12. The wall is 5 m long between cross walls and is of 3.4 m clear height between RCC slabs at top and bottom. What should be the strength of the brick and the grade of mortar for the masonry? [12]
Given: t = 200 mm (modular bricks 190 × 90 × 90 mm); load at the top Wtop = 80 kN/m with eccentricity ratio e/t = 1/12; wall length between cross walls L = 5.0 m; clear height between RCC slabs h = 3.4 m.
Assume: masonry 20 kN/m³; joints not raked.
Total vertical load at the base of the wall (per metre run)
Load at the top = 80.00 kN/m Self weight of the wall = 0.20 × 3.4 × 20 = 13.60 kN/m ------------ Total at the base, W = 93.60 kN/m
Actual compressive stress
Area per metre run A = 1000 × 200 = 200 000 mm² f_c = W / A = 93 600 / 200 000 = 0.468 N/mm²
Effective height  IS 1905 Table 4
RCC slabs at the top and bottom bear on the full thickness → full restraint at both ends h_eff = 0.75 h = 0.75 × 3400 = 2550 mm
Effective length  IS 1905 Table 3
Wall supported by cross walls at both ends l_eff = 0.8 L = 0.8 × 5000 = 4000 mm
Effective thickness  IS 1905 Table 6
S_p / W_p = 5000 / 200 = 25 ≥ 20 → stiffening coefficient = 1.0 t_eff = 1.0 × 200 = 200 mm
Slenderness ratio
h_eff / t_eff = 2550 / 200 = 12.75 ← lesser l_eff / t_eff = 4000 / 200 = 20.00 SR = 12.75 (limit 27 for a load-bearing wall) ✓
Stress reduction factor  IS 1905 Table 9
SR = 12.75 , e/t = 1/12 Column e/t = 1/12 : SR 12 → 0.81 , SR 14 → 0.74 k_s = 0.81 − (0.81 − 0.74)(12.75 − 12)/(14 − 12) = 0.81 − 0.07 × 0.375 = 0.784
Area and shape modification factors
k_a A per metre = 1.0 × 0.20 = 0.20 m² k_a = 0.70 + 1.5 × 0.20 = 1.00 k_p Modular brick 190 × 90 × 90 laid flat height / width = 90 / 90 = 1.00 > 0.75 From Table 10, for a unit of 7.5 N/mm² with h/w = 1.0 → k_p = 1.1
Required basic compressive stress
f_c ≤ f_b · k_s · k_a · k_p Required f_b ≥ f_c / (k_s · k_a · k_p) = 0.468 / (0.784 × 1.00 × 1.1) = 0.543 N/mm²
Select the unit strength and mortar grade  IS 1905 Table 8
Look for a combination with f_b ≥ 0.543 N/mm² : unit 5.0 N/mm² + M1 (1:5) f_b = 0.50 ✘ too low unit 7.5 N/mm² + M3 f_b = 0.56 ✓ marginal unit 7.5 N/mm² + M2 (1:6) f_b = 0.59 ✓ ADOPT unit 7.5 N/mm² + M1 (1:5) f_b = 0.74 ✓ safer ADOPT : brick crushing strength 7.5 N/mm² mortar grade M2 (cement : sand = 1 : 6)
Verification
f_per = f_b · k_s · k_a · k_p = 0.59 × 0.784 × 1.00 × 1.1 = 0.509 N/mm² f_c = 0.468 N/mm² < f_per = 0.509 N/mm² ✓ With M1 mortar instead : f_per = 0.74 × 0.784 × 1.00 × 1.1 = 0.638 N/mm² utilisation = 0.468 / 0.638 = 73 %
ANSWER — use bricks of crushing strength 7.5 N/mm² with M2 grade mortar (cement : sand = 1 : 6); fper = 0.509 N/mm² against an actual stress of 0.468 N/mm².
For a more comfortable margin (and because the mortar grade must in any case be matched to a unit of 5–14.9 N/mm²), M1 mortar (1:5) with the same 7.5 N/mm² brick is recommended — utilisation then falls to 73%.
Watch the shape modification factor. With modular bricks laid flat the height-to-width ratio is 90/90 = 1.0, which is greater than 0.75, so kp = 1.1 — a genuine 10% bonus. With conventional 230×115×57 bricks the ratio is 0.50 and kp = 1.0. The question specifically says “using modular bricks”, so this factor is being tested.
Method: Chapter 2 §2.6.5–2.6.8 (Tables 8, 9, 10 reproduced as Figs. 2.17–2.19).

Question 5

5 (a) TheoryWhat is reinforced masonry? Explain. [4]
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. Design may be by allowable stress design or by strength design.
  • Why: brickwork and blockwork have high compressive strength but very low tensile resistance, so steel is introduced to carry the tension — exactly as in reinforced concrete.
  • Where the steel goes: vertical bars grouted into the voids created by particular bonds or into the hollow cores of blocks, and horizontal bed-joint reinforcement (ladder or truss mesh) laid in the mortar. Bed-joint reinforcement enhances the capacity of walls to resist lateral loading.
  • Contrast with URM: unreinforced masonry is designed assuming the flexural tensile stress is resisted by the masonry, and any reinforcement present is neglected; the design is in the linear elastic range. So “unreinforced masonry” may actually contain steel — the term describes the design assumption, not the absence of bars.
  • Benefits: real flexural and shear capacity, genuine ductility and energy dissipation, much better out-of-plane performance, taller walls and larger openings possible.
  • Applications: shear walls, retaining walls, tanks, footbridges, reinforced masonry columns, beams and bands. Prestressed (post-tensioned) masonry adds precompression through vertical tendons.
  • History: used in India and Japan since after the First World War, in America since after the Second.
See Chapter 2 §2.4.2(D); full treatment is Chapter 5.
5 (b) NumericalDesign a shear wall 4.2 m long and 3.3 m high to resist a horizontal earthquake force in its plane. The wall is tied with metal anchors at the top and bottom supports. Assume the seismic load to be uniformly distributed across the height of the wall. Earthquake acceleration = 0.1 g. [12]
Given: L = 4.2 m; H = 3.3 m; in-plane horizontal earthquake force; wall tied with metal anchors top and bottom; seismic acceleration = 0.1 g, uniformly distributed over the height.
Assume: t = 230 mm; masonry 20 kN/m³; unit crushing strength 5 N/mm² with M1 mortar (1:5).
Section properties
Base area A = 4200 × 230 = 9.66 × 10⁵ mm² Section modulus Z = t L²/6 = 230 × 4200²/6 = 6.76 × 10⁸ mm³
Self weight and lateral force
W = 4.2 × 3.3 × 0.23 × 20 = 63.76 kN F = 0.1 × W = 0.1 × 63.76 = 6.37 kN
Shear check  IS 1905 cl. 5.4.3
Developed shear stress τ = F / A = 6370 / (9.66 × 10⁵) = 0.006 N/mm² Compressive stress due to dead load f_d = W / A = 63 760 / (9.66 × 10⁵) = 0.066 N/mm² Permissible shear stress τ_per = 0.1 + f_d/6 = 0.1 + 63 760/(6 × 9.66×10⁵) = 0.1 + 0.011 = 0.11 N/mm² τ = 0.006 < τ_per = 0.11 ✓
Bending moment and flexural stress

The seismic load is uniformly distributed over the height, so the resultant acts at mid-height:

M = F × H/2 = 6370 × 3300 / 2 = 10.51 × 10⁶ N·mm f_cb = M / Z = 10.51 × 10⁶ / 6.76 × 10⁸ = 0.015 N/mm² f_ca = W / A = 0.066 N/mm² Maximum compressive stress f_c = f_ca + f_cb = 0.066 + 0.015 = 0.081 N/mm² Maximum tensile stress f_t = f_cb − f_ca = 0.015 − 0.066 = −0.051 N/mm² (negative → still compression, NO tension)
Slenderness ratio and equivalent eccentricity
t_eff = 230 mm , h_eff = 3300 mm SR = h_eff / t_eff = 3300 / 230 = 14.35 e = M / W = 10.51 × 10⁶ / 63 760 = 164.83 ≈ 165 mm e / L = 165 / 4200 = 1 / 25.24 ≈ 1/24
Stress reduction factor  IS 1905 Table 9
SR = 14.35 , e/t = 1/24 SR 14 → 0.76 , SR 16 → 0.71 k_s = 0.76 − (0.76 − 0.71)(14.35 − 14)/(16 − 14) = 0.76 − 0.05 × 0.175 = 0.751
Basic compressive stress  IS 1905 Table 8
Select crushing strength of unit = 5 N/mm², M1 mortar f_ck = 0.50 N/mm²
Permissible stresses and final check
Permissible stress due to axial load f_ac = k_s × f_ck = 0.751 × 0.50 = 0.375 N/mm² f_ca = 0.066 < 0.375 ✓ Permissible combined axial + bending stress f_acb = 1.25 × k_s × f_ck = 1.25 × 0.375 = 0.469 N/mm² f_c = 0.081 < 0.469 ✓ Shear : 0.006 < 0.11 ✓ Tension : none develops ✓
HENCE SAFE. Adopt a 230 mm thick brick shear wall 4.2 m × 3.3 m, built with units of crushing strength 5 N/mm² in M1 (1:5) cement mortar, tied to the supports with metal anchors.
Method: Chapter 4 §4.6.6 (the 10-step shear-wall method). Follows the worked example in the course notes (CH-4, Example 3.2) — identical data.