Paper Information
| Detail | Information |
|---|---|
| Institute | Tribhuvan University, Institute of Engineering — Examination Control Division |
| Exam | Regular, 2081 Bhadra |
| Subject | Seismic Resistance Design of Masonry Structure (Elective I) — CE 72502 |
| Level / Programme / Year–Part | BE / BCE / IV – I |
| Full Marks · Pass Marks · Time | 80 · 32 · 3 hours |
| Instructions | Attempt All questions. IS 1905 code is allowed to use. Assume suitable data if necessary. |
How this paper splits by chapter: Ch 1 — 16 marks (Q1a, Q1b);
Ch 3 — 16 marks (Q2a, Q2b); Ch 2 — 8 marks (Q3a, Q4a);
Ch 4 — 24 marks (Q3b, Q4b, Q5b); Ch 5 — 4 marks (Q5a).
Note that 36 of the 80 marks are pure numericals, all three of which follow the
IS 1905 working-stress sequence.
Question 1
1 (a)
TheoryExplain an earthquake as a lateral force in a building.
[8]
Solution
An earthquake does not apply a force to a building the way wind or a live load does. The ground merely moves; the force is generated inside the building itself as an inertia force.
Earthquake force is an INERTIA force developed because of ground shaking:
FI = mass × acceleration = m × a = (W/g) × a
The imposed acceleration is largest in the horizontal direction, so the resulting inertia force acts essentially horizontally — that is, laterally on the building.
FI = mass × acceleration = m × a = (W/g) × a
The imposed acceleration is largest in the horizontal direction, so the resulting inertia force acts essentially horizontally — that is, laterally on the building.
1. Why a lateral force develops
- When the ground shakes, the foundation moves with the ground.
- The mass of the roof and floors, by inertia (Newton's first law), resists this motion and tends to stay where it was, so the upper part lags behind the base.
- This relative displacement between base and mass distorts the walls and sets up internal forces in them.
- Because the ground reverses direction many times a second, the building swings back and forth and the lateral force reverses in sign every cycle — unlike gravity load, which acts in one direction only.
- If the building were perfectly rigid, every point would move exactly with the ground and the force would be simply m × PGA. Real buildings are flexible, so different parts move by different amounts and the force also depends on the building's own dynamic properties (period, damping, stiffness).
2. Flow of the inertia force — the seismic load path
Mass of roof / floor slab → roof / floor diaphragm → walls parallel to the shaking
(shear walls, acting in-plane) → plinth / bands → foundation → soil
- Walls parallel to the shaking act as shear walls and carry the force efficiently by in-plane shear — they are strong and stiff in that direction.
- Walls perpendicular to the shaking are loaded out-of-plane, i.e. as vertical slabs spanning between floor and roof. Unreinforced masonry is very weak in this mode and overturns — the classic cause of collapse of Nepali masonry houses.
- If any link is missing (no band, no wall-to-roof anchorage, no corner interlocking) the building loses box action and the walls fail individually.
3. Horizontal versus vertical shaking
- Shaking has three components, but the two horizontal components govern design, because buildings have a large factor of safety against vertical load and very little inherent resistance to horizontal load.
- The vertical component is normally taken as about two-thirds of the horizontal. It matters for cantilevers and long spans, and for masonry because it momentarily reduces the compressive pre-load and hence the frictional shear capacity of the wall.
4. Sketch to draw
Draw two panels: (a) building at rest; (b) during shaking — base displaced in the direction of ground motion, roof lagging behind, and the inertia force FI = m a marked at roof level acting opposite to the ground motion. Then add the load-path arrows slab → wall → foundation.
Full treatment with figures: Chapter 1 §1.5.1
(Fig. 1.20 — generation of the lateral inertia force).
1 (b)
TheoryDescribe the earthquake force parameters.
[8]
Solution
The magnitude of the seismic force a building must resist depends on three families of parameters — those of the ground motion, those of the structure, and the code coefficients that combine them.
(A) Ground motion parameters
| Parameter | Why it matters |
|---|---|
| Peak ground acceleration (PGA) | Directly fixes the inertia force F = m a. Expressed in g, m/s² or Gal (1 g = 9.81 m/s² = 981 Gal). |
| Peak ground velocity (PGV) | Best single indicator of damage to medium-period structures; governs the energy input. |
| Peak ground displacement (PGD) | Governs long-period structures, bridges, pipelines and separation gaps. |
| Duration of strong motion | Number of load reversals — masonry degrades cumulatively, so long duration is very damaging. |
| Frequency content / predominant period | Decides whether resonance with the building occurs. |
| Magnitude and focal depth | Control the energy released and how much of it reaches the surface. |
| Epicentral / hypocentral distance | Amplitude attenuates with distance; near-field records contain damaging velocity pulses. |
| Local soil stiffness | Acceleration is amplified in soft soil and the predominant period lengthens. |
| Topographic amplification | Ridges and hill tops amplify base-rock acceleration by resonance. |
(B) Structural parameters
| Parameter | Why it matters |
|---|---|
| Seismic weight / mass W | Force is proportional to mass. Heavy masonry and heavy roofs attract large forces — reducing mass is the cheapest seismic improvement. |
| Stiffness | Governs how the force distributes among the walls, and the drift. |
| Fundamental natural period Ta | Fixes Sa/g read from the spectrum. More flexibility → longer T; more mass → longer T. Range 0.05–2.0 s for 1–20 storeys; T < 0.4 s for low-rise masonry. |
| Damping | Energy dissipation; codes assume 5% of critical. Higher damping reduces response. |
| Ductility | Capacity for inelastic deformation without collapse; permits the design force to be reduced by R. |
| Configuration & regularity | Plan/vertical irregularity, torsion, soft or weak storey and mass irregularity all raise the local demand. |
| Foundation and soil–structure interaction | Modify the motion actually transmitted to the building. |
(C) Code parameters — the design shorthand
VB = AhW
Ah = Z2 ·
IR ·
Sag
Z zone factor (regional seismicity, 0.36 for Zone V) | I importance factor (occupancy, consequence of failure) | R response reduction factor (ductility; I/R ≤ 1.0) | Sa/g spectral acceleration coefficient (period + soil) | W seismic weight
Z zone factor (regional seismicity, 0.36 for Zone V) | I importance factor (occupancy, consequence of failure) | R response reduction factor (ductility; I/R ≤ 1.0) | Sa/g spectral acceleration coefficient (period + soil) | W seismic weight
Full treatment: Chapter 1 §1.5.2; code
coefficients in Chapter 4 §4.3.3.
Question 2
2 (a)
TheoryDiscuss the failure mechanisms of the stone masonry
buildings with mud mortar during the 2080 Jajarkot Earthquake.
[8]
Solution
Background. The 2080 BS (3 November 2023) Jajarkot earthquake was only M 6.4 — a moderate event — yet it destroyed a very large number of houses and killed about 154 people, because the building stock of Jajarkot and Rukum West is almost entirely random rubble stone masonry laid in mud mortar, with heavy mud-and-slate roofs and no seismic features whatever.
| # | Failure mechanism | Why it happened in stone-in-mud construction |
|---|---|---|
| 1 | Delamination of the wall through its thickness | Thick two- or three-leaf walls with a loose rubble core and no through-stones. The leaves have no connection, so the outer leaf peels off and the wall loses most of its effective section. The dominant mechanism in this earthquake. |
| 2 | Out-of-plane overturning of walls | Flexible timber/slate roofs simply resting on the walls, no anchorage, no bands, and heavy walls with a high height-to-thickness ratio. Once a wall is unsupported at the top it topples as a rigid block. |
| 3 | Corner separation and disintegration at junctions | Rounded and irregular stones give no interlocking at the corners, and mud mortar has essentially zero tensile and bond strength — so the corner simply comes apart, leaving each wall unsupported at its ends. |
| 4 | Bulging and buckling of the wall | Random rubble with thick mud joints behaves as loose stacked stone once shaking begins; the wall bulges outwards and buckles. |
| 5 | Diagonal / stepped shear cracking of piers | With weak mortar the diagonal crack steps around the stones through the joints rather than passing through them (bed-joint sliding mode). |
| 6 | Collapse of the heavy mud-and-slate roof | Very large mass high up, unanchored — a large inertia force applied to walls that cannot resist it; the roof then falls in on the occupants. |
| 7 | Gable and parapet collapse | No gable band and almost no vertical load to stabilise them. |
| 8 | Loss of strength of the mud mortar | Rain before and after the event further weakened an already negligible bond. |
| 9 | Total (pancake) collapse | Once the walls disintegrated the floors and roof lost all support — giving a very high fatality rate for a moderate earthquake. |
Sketches to draw
(i) Section through a wall showing delamination of the outer leaf from the rubble core; (ii) elevation showing corner separation with vertical cracks at the junction; (iii) section showing simple out-of-plane overturning about the base with the roof dropping in.
One-line conclusion worth writing: the primary causes were the
mud mortar (no bond), the absence of through-stones (no monolithic wall), the
absence of bands and anchorage (no box action) and the heavy unanchored roof (large
inertia force) — not the size of the earthquake.
Full treatment: Chapter 3 §3.9.2, with the
failure-mode figures in §3.6 and §3.7.
2 (b)
TheoryExplain the strengthening procedure for the damaged
masonry buildings.
[8]
Solution
Answer in two parts — first the procedure (the sequence of activities), then the techniques.
A. The procedure
- Post-earthquake damage assessment. Survey the building from outside, identify and map the cracks, examine the site for geotechnical hazard, inspect from inside and at critical locations, then interpret. Classify the damage on the five-grade EMS scale (G1 negligible → G5 destruction).
- Decide the strategy from the damage grade. G1–G2 → architectural repair plus seismic strengthening; G3 → grouting of cracks + repair + strengthening; G4–G5 → vacate, then either extensive restoration or demolish and rebuild.
- Emergency shoring and propping of unstable walls before any work starts, and removal of dangerous debris, parapets and overhangs.
- Identify the deficiency that caused the damage — no bands, no through-stones, large openings, weak mortar, heavy roof, soft storey — and select measures that address that deficiency.
- Execute in the right order: restore integrity first (crack repair, corner stitching), then add strength and ductility (bands, vertical bars, jacketing), then reduce demand (lighten the roof, reduce openings).
- Quality control and supervision throughout, followed by re-inspection.
B. The techniques
| Purpose | Technique |
|---|---|
| Restore continuity across cracks | Grouting / injection of cement or polymer grout into fine cracks; stitching of wide cracks with steel or reinforced-concrete dowels laid across the crack; re-pointing of eroded joints. |
| Rebuild badly damaged masonry | Dismantle and rebuild the shattered portion in good mortar, toothed into the sound masonry. |
| Stop delamination of stone walls | Insert through-stones (bond stones), or substitutes — S-shaped steel ties, wooden dowels, concrete bond blocks — at about 1.2–1.5 m spacing in both directions. |
| Prevent corner separation | Corner stitching with RC or timber elements connecting the two orthogonal walls at 0.5–0.7 m vertical spacing; add vertical bars at all corners and T-junctions, from foundation to roof. |
| Create box action | Add horizontal bands (RC, steel or timber) at plinth, sill, lintel and eave level in every storey, and a gable band for a sloped roof. |
| Increase in-plane and out-of-plane strength | Splint-and-bandage (vertical splints at corners and jambs plus horizontal bandages at band levels); wire-mesh and cement (ferrocement) jacketing on one or both faces; FRP / textile-reinforced mortar overlays; shotcrete jacketing. |
| Stiffen the diaphragm | Brace the timber floor/roof with diagonals or plywood sheathing; anchor it to the walls with metal anchors. |
| Reduce demand | Replace a heavy roof with a light one; reduce and re-position openings by infilling parts of large openings; remove heavy parapets and projections; reduce the number of storeys. |
| Strengthen the foundation | Underpinning or widening where differential settlement contributed. |
Mitigation table: Chapter 3 §3.7.6 and Fig. 3.18
(containment reinforcement); repair and retrofit techniques belong to Chapter 6.
Question 3
3 (a)
TheoryWhat are the physical properties of bricks?
[4]
Solution
The physical properties are those that can be observed or measured without loading the brick to failure (as opposed to the mechanical properties such as compressive strength and modulus).
| Property | Requirement / typical value |
|---|---|
| Size (Nepal Standard NS 1:2035) | 240 × 115 × 57 mm with a 10 mm mortar joint. Modular rule: length = 2 × width + one vertical joint. (IS 1077 modular brick 190×90×90 mm; common site brick often 230×110×55 mm.) |
| Shape | Regular, truly rectangular, sharp square edges, parallel faces, free from warpage, distortion and bowing. |
| Colour | Deep cherry red or copper colour, uniform throughout. |
| Texture | Even, fine-grained and compact; free from grit, lumps and nodules of lime. |
| Soundness | Emits a clear ringing (metallic) sound when two bricks are struck together; should not break when dropped flat from about 1 m. |
| Hardness | No impression left when scratched with a finger nail. |
| Water absorption | Not more than 15% of its own weight after 24 hours immersion at normal temperature (hand-made bricks with a frog, up to 25%). |
| Efflorescence | Nil to slight — no white salt deposit after immersion and drying. |
| Density / unit weight | About 1800–2000 kg/m³; brick masonry taken as 19–20 kN/m³. |
| Structural quality | Thoroughly burnt, free from cracks, flaws and organic matter. |
Full treatment: Chapter 2 §2.3.1.
3 (b)
NumericalDesign a brick column of height 3 m to support
one end of the roof truss (x–x axis). The axial load on the column is 100 kN.
[12]
Given: Height of column H = 3.0 m; axial load P = 100 kN; column supports one end of a roof truss,
restrained about the x–x axis only.
Assume (data not given): brick size 200 × 100 × 100 mm; minimum crushing strength of unit 10 N/mm²; cement mortar 1:4, i.e. H2 type. Eccentricity e = 0 (axial load).
Assume (data not given): brick size 200 × 100 × 100 mm; minimum crushing strength of unit 10 N/mm²; cement mortar 1:4, i.e. H2 type. Eccentricity e = 0 (axial load).
Effective height in the two directions
Because the truss restrains the column about the x–x axis but not about y–y:
Effective height about x-x = H = 3.0 m = 3000 mm
Effective height about y-y = 2H = 6.0 m = 6000 mm
Slenderness ratio and trial size
For a load-bearing column, IS 1905 limits SR ≤ 12 (cl. 4.6). SR is the greater of the two ratios, so the y–y direction governs:
SR (y-y) = 6000 / l ≤ 12 → l ≥ 500 mm
SR (x-x) = 3000 / b
Adopt l = 500 mm → SR = 6000/500 = 12 ✓
Trial section = 500 mm × b mm
Permissible compressive stress
fp = fck · kA · kSE · ku · kL
f_ck Basic compressive stress (IS 1905 Table 8)
unit 10 N/mm² + H2 mortar → f_ck = 0.96 N/mm²
k_SE Stress reduction factor (IS 1905 Table 9)
SR = 12 , e = 0 → k_SE = 0.84
k_A Area reduction factor (IS 1905 cl. 5.4.1.2)
k_A = 0.70 + 1.5 A , max 1.0 (A ≥ 0.2 m²), min 0.70
first trial, take the mean → k_A = (1.0 + 0.7)/2 = 0.85
k_u Shape modification factor (IS 1905 Table 10)
height/width of unit as laid = 100/100 = 1.0
for unit strength 10 N/mm² → k_u = 1.1
k_L = 1.0 for axial load
f_p = 0.96 × 0.85 × 0.84 × 1.1 × 1.0 = 0.754 N/mm²
Required cross-section
Safe axial load = f_p × cross-sectional area
100 × 10³ = 0.754 × (500 × b)
b = 100000 / (0.754 × 500) = 265.2 mm
Round up to a multiple of half a brick → b = 300 mm
ADOPT COLUMN 500 mm × 300 mm
Check the actual load carrying capacity
Actual slenderness ratio = greater of
6000/500 = 12 and 3000/300 = 10
→ SR = 12 , e = 0 → k_SE = 0.84 (Table 9)
Actual area A = 0.5 × 0.3 = 0.15 m² < 0.2 m²
→ k_A = 0.70 + 1.5 × 0.15 = 0.925
f_ck = 0.96 , k_u = 1.1 , k_L = 1.0 (unchanged)
f_p = 0.96 × 0.925 × 0.84 × 1.1 × 1.0 = 0.8205 N/mm²
Capacity = f_p × A = 0.8205 × (500 × 300) / 1000 = 123.1 kN
123.1 kN > 100 kN → HENCE THE COLUMN IS SAFE.
Adopt a brick column 500 mm × 300 mm, built in 1:4 cement mortar (H2) with bricks of minimum crushing strength 10 N/mm².
Adopt a brick column 500 mm × 300 mm, built in 1:4 cement mortar (H2) with bricks of minimum crushing strength 10 N/mm².
Why 300 mm and not 265 mm? Masonry is a modular material — the dimension must be
buildable from whole and half bricks. With a 200×100×100 unit, 300 mm = 3 × 100 mm and
500 mm = 2×200 + 100 mm, both achievable. Also check that 250 mm would not do:
with b = 250, A = 0.125 m², kA = 0.8875, fp = 0.787 N/mm² and the capacity is
only 98.4 kN < 100 kN — it fails, so 300 mm is the minimum.
Method and factors: Chapter 2 §2.6.7–2.6.8
(IS 1905 Tables 8, 9 and 10 are reproduced there as Figs. 2.17–2.19). Follows the worked example in the
course notes (CH-4, Example 4.7).
Question 4
4 (a)
TheoryDescribe the nature of the masonry structure.
[4]
Solution
Masonry is a composite structural material formed by bonding individual units (brick, stone, block, adobe) with layers of mortar. Its essential character can be summed up in nine points:
| Characteristic | Explanation and structural consequence |
|---|---|
| Composite / two-phase | Not one material but two — stiff strong units bedded in a softer weaker mortar. Behaviour is governed by their interaction and by the unit–mortar bond, not by either alone. |
| Heterogeneous, highly variable | Properties scatter from point to point in the same wall, so design values are conservative “minimum assured” values. |
| Anisotropic (directional) | Bed and head joints form continuous planes of weakness, so strength and stiffness depend on the direction of loading relative to the joints. |
| Strong in compression, very weak in tension and shear | Suits walls, columns, arches and domes; cannot be used for beams or slabs, and cracks as soon as flexural tension appears. |
| Brittle, little ductility | Fails suddenly with little warning and dissipates very little energy — the opposite of what cyclic earthquake loading demands. |
| Heavy — large mass | Large self weight → large inertia force F = m a and heavy foundations. Mass is both its stabiliser (gravity friction) and its enemy (seismic demand). |
| Modular | The unit fixes the module; wall thicknesses, lengths and opening sizes are multiples of the unit dimension plus a joint. |
| Extremely sensitive to workmanship | Full even bed joints, filled perpends, correct bond, corner interlocking and curing govern the real strength far more than the nominal design. |
| Non-engineered by tradition | Most masonry worldwide is still built by artisans without any structural calculation. |
Full treatment: Chapter 2 §2.2.2.
4 (b)
NumericalDesign a shear wall 4 m long and 3 m high to
resist a horizontal earthquake force in its plane under seismic loading. The wall is tied with metal anchors at
the top and bottom supports. Take seismic load 0.2 g acceleration uniformly distributed across the entire
height of the wall.
[12]
Given: L = 4.0 m, H = 3.0 m; horizontal earthquake force acting in the plane of the wall;
wall tied with metal anchors top and bottom; seismic acceleration = 0.2 g, uniformly distributed over the
height.
Assume: wall thickness t = 230 mm (one brick); unit weight of masonry γ = 20 kN/m³; crushing strength of unit 5 N/mm² with M1 mortar (1:5). No floor load on the wall.
Assume: wall thickness t = 230 mm (one brick); unit weight of masonry γ = 20 kN/m³; crushing strength of unit 5 N/mm² with M1 mortar (1:5). No floor load on the wall.
Section properties (bending is in the plane of the wall)
Base area A = L × t = 4000 × 230 = 9.20 × 10⁵ mm²
Section modulus Z = t L² / 6 = 230 × 4000² / 6 = 6.133 × 10⁸ mm³
Note: for in-plane bending the wall bends about its strong axis, so the depth of the section is the wall length L — hence Z = tL²/6, not bt²/6.
Self weight and lateral force
Self weight W = L × H × t × γ
= 4.0 × 3.0 × 0.23 × 20 = 55.20 kN
Lateral force F = 0.2 × W = 0.2 × 55.20 = 11.04 kN
Check for shear stress
Developed shear stress
τ = F / A = 11040 / (9.20 × 10⁵) = 0.012 N/mm²
Compressive stress due to dead load
f_d = W / A = 55200 / (9.20 × 10⁵) = 0.060 N/mm²
Permissible shear stress (IS 1905 cl. 5.4.3)
τ_per = 0.1 + f_d / 6 = 0.1 + 0.060/6 = 0.110 N/mm²
(subject to a maximum of 0.5 N/mm²)
τ = 0.012 < τ_per = 0.110 ✓ SAFE IN SHEAR
Bending moment at the base
The seismic load is uniformly distributed over the height, so its resultant acts at mid-height:
M = F × H/2 = 11040 × 3000/2 = 16.56 × 10⁶ N·mm
Stresses at the base
Axial (compressive) stress
f_ca = W / A = 0.060 N/mm²
Bending (flexural) stress
f_cb = M / Z = 16.56 × 10⁶ / 6.133 × 10⁸ = 0.027 N/mm²
Maximum compressive stress
f_c = f_ca + f_cb = 0.060 + 0.027 = 0.087 N/mm²
Maximum tensile stress
f_t = f_cb − f_ca = 0.027 − 0.060 = −0.033 N/mm²
(negative → still compression, NO TENSION develops)
Slenderness ratio and equivalent eccentricity
Effective thickness t_eff = 230 mm
Effective height h_eff = 3000 mm (anchored top and bottom)
Slenderness ratio SR = h_eff / t_eff = 3000 / 230 = 13.04
Equivalent eccentricity
e = M / W = 16.56 × 10⁶ / 55200 = 300 mm
Eccentricity ratio (in-plane, so divide by the LENGTH)
e / L = 300 / 4000 = 0.075 ≈ between 1/24 and 1/12
Stress reduction factor ks — IS 1905 Table 9
Interpolate for SR = 13.04 :
e/t = 1/24 : SR 12 → 0.83 , SR 14 → 0.76
k = 0.83 − (0.83−0.76)(13.04−12)/(14−12) = 0.794
e/t = 1/12 : SR 12 → 0.81 , SR 14 → 0.74
k = 0.81 − (0.81−0.74)(13.04−12)/(14−12) = 0.774
Now interpolate on e/t between 0.0417 and 0.0833 for e/t = 0.075 :
k_s = 0.794 − [(0.075−0.0417)/(0.0833−0.0417)] × (0.794−0.774)
= 0.794 − 0.80 × 0.020
= 0.778
Basic compressive stress — IS 1905 Table 8
Select crushing strength of unit = 5 N/mm² with M1 mortar (1:5)
Basic compressive stress f_ck = 0.50 N/mm²
Permissible stresses
Permissible stress for axial load
f_ac = k_s × f_ck = 0.778 × 0.50 = 0.389 N/mm²
Permissible combined axial + bending stress
f_acb = 1.25 × k_s × f_ck = 1.25 × 0.389 = 0.486 N/mm²
(25% increase allowed because the excess is due to bending)
Final comparison
Axial : f_ca = 0.060 < f_ac = 0.389 ✓
Combined : f_c = 0.087 < f_acb = 0.486 ✓
Shear : τ = 0.012 < τ_per = 0.110 ✓
Tension : f_t = −0.033 → no tension ✓
All four checks are satisfied → HENCE THE WALL IS SAFE.
Adopt a 230 mm thick brick shear wall, 4 m × 3 m, built with units of crushing strength 5 N/mm² in M1 (1:5) cement mortar, tied to the floor and roof with metal anchors.
Adopt a 230 mm thick brick shear wall, 4 m × 3 m, built with units of crushing strength 5 N/mm² in M1 (1:5) cement mortar, tied to the floor and roof with metal anchors.
Note on the margin. The stresses are only a few percent of the permissible values because a
single isolated wall carries nothing but its own weight. In a real building this wall would also carry the
roof/floor load and the storey shear from the diaphragm, both of which would raise fca and M
substantially. The exam question deliberately isolates the wall so that the method is what is being
tested.
Method: Chapter 4 §4.1.3 and §4.6.6
(10-step shear-wall design). Follows the worked example in the course notes (CH-4, Example 3.2).
Question 5
5 (a)
TheoryExplain reinforced masonry.
[4]
Solution
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 reinforcement, singly or in combination. Design may be by allowable stress design or by
strength design.
- Why it is needed. Brickwork and blockwork, like concrete, have high compressive strength but very low tensile resistance. Steel bars are therefore introduced to carry the tension, exactly as in reinforced concrete.
- Where the steel goes. Advantage is taken of the voids created by particular bonds and/or of the bed joints — vertical bars grouted into the hollow cores of blocks or into pockets, and horizontal bed-joint reinforcement (ladder or truss mesh) laid in the mortar. Bed-joint reinforcement in particular enhances the capacity of walls to resist lateral loading.
- Contrast with unreinforced masonry (URM). URM is designed assuming the flexural tensile stress is resisted by the masonry, and any reinforcement present is neglected in design; design is in the linear elastic range. So “unreinforced masonry” may actually contain steel — the term refers to the design assumption, not to the absence of bars.
- Benefits: genuine flexural and shear capacity, real ductility and energy dissipation, much better out-of-plane performance, and the ability to build taller and with larger openings.
- Applications: shear walls, retaining walls, tanks, footbridges, reinforced masonry columns, beams and RC/reinforced-masonry bands. Prestressed (post-tensioned) masonry adds precompression through vertical tendons.
- History: used in India and Japan since after the First World War, and in America since after the Second.
See Chapter 2 §2.4.2(D); the full treatment is
Chapter 5 (Reinforced Masonry).
5 (b)
NumericalCalculate the centre of mass, centre of
stiffness and eccentricities for the given single-storey masonry building.
[12]
Given (from the figure above): single-storey masonry building, plan 30 m × 10 m.
Grid lines ① at x = 0, ② at x = 14 m, ③ at x = 30 m; grid Ⓐ at y = 10 m (top), Ⓑ at y = 0 (bottom).
Wall stiffnesses:
• walls running in the y-direction (resisting force along Y): k at x = 0 (grid ①), 0.5k at x = 14 m (grid ②), 1.5k at x = 30 m (grid ③)
• walls running in the x-direction (resisting force along X): 1.5k at y = 10 m (grid Ⓐ), 1.2k at y = 0 (grid Ⓑ)
Grid lines ① at x = 0, ② at x = 14 m, ③ at x = 30 m; grid Ⓐ at y = 10 m (top), Ⓑ at y = 0 (bottom).
Wall stiffnesses:
• walls running in the y-direction (resisting force along Y): k at x = 0 (grid ①), 0.5k at x = 14 m (grid ②), 1.5k at x = 30 m (grid ③)
• walls running in the x-direction (resisting force along X): 1.5k at y = 10 m (grid Ⓐ), 1.2k at y = 0 (grid Ⓑ)
Centre of mass
No wall weights are given, so the mass is taken as that of the roof slab, uniformly distributed over the rectangular plan. The centre of mass is therefore the geometric centre of the plan:
X̄_m = L / 2 = 30 / 2 = 15.00 m from grid ①
Ȳ_m = B / 2 = 10 / 2 = 5.00 m from grid Ⓑ
CENTRE OF MASS CM = (15.00 m , 5.00 m)
General formula, to be used whenever wall weights are given:
X̄m = Σ Wi xiΣ Wi
Ȳm = Σ Wi yiΣ Wi
Centre of stiffness (centre of rigidity) — X direction
Only the walls that run in the y-direction resist a force acting along Y, so only they enter X̄r. Out-of-plane wall stiffness is neglected.
| Wall (grid) | Ry | x (m) | Ry · x |
|---|---|---|---|
| ① | 1.0 k | 0 | 0 |
| ② | 0.5 k | 14 | 7.0 k |
| ③ | 1.5 k | 30 | 45.0 k |
| Σ | 3.0 k | 52.0 k |
X̄_r = Σ(R_y · x) / ΣR_y = 52.0 k / 3.0 k = 17.33 m from grid ①
Centre of stiffness — Y direction
Only the walls running in the x-direction enter Ȳr.
| Wall (grid) | Rx | y (m) | Rx · y |
|---|---|---|---|
| Ⓑ (bottom) | 1.2 k | 0 | 0 |
| Ⓐ (top) | 1.5 k | 10 | 15.0 k |
| Σ | 2.7 k | 15.0 k |
Ȳ_r = Σ(R_x · y) / ΣR_x = 15.0 k / 2.7 k = 5.56 m from grid Ⓑ
CENTRE OF STIFFNESS CR = (17.33 m , 5.56 m)
Static (calculated) eccentricities
e_x = X̄_m − X̄_r = 15.00 − 17.33 = −2.33 m
e_y = Ȳ_m − Ȳ_r = 5.00 − 5.56 = −0.56 m
Magnitudes: |e_x| = 2.33 m |e_y| = 0.56 m
(The minus sign means the centre of rigidity lies to the RIGHT of, and
ABOVE, the centre of mass — the building will twist about CR.)
Accidental eccentricity and design eccentricity
A minimum accidental eccentricity of 5% of the plan dimension perpendicular to the direction of the force must be added, to allow for unmodelled variation of mass and stiffness and for rotational ground motion:
For force in the X-direction (dimension perpendicular = 30 m):
e_a,x = 0.05 × 30 = 1.50 m
e_d,x = |e_x| + e_a,x = 2.33 + 1.50 = 3.83 m
For force in the Y-direction (dimension perpendicular = 10 m):
e_a,y = 0.05 × 10 = 0.50 m
e_d,y = |e_y| + e_a,y = 0.56 + 0.50 = 1.06 m
Check the torsional irregularity limit
e_x / L = 2.33 / 30 = 7.8 % < 20–30 % ✓
e_y / B = 0.56 / 10 = 5.6 % < 20–30 % ✓
ANSWER —
Centre of mass CM (15.00 m, 5.00 m);
centre of stiffness CR (17.33 m, 5.56 m);
static eccentricities ex = 2.33 m and ey = 0.56 m;
design eccentricities including 5% accidental —
ed,x = 3.83 m and ed,y = 1.06 m.
Both static eccentricities are well within the 20–30% codal limit, so the building is
not torsionally irregular, although the torsional shear must still be computed and
added to the direct shear in every wall.
Next step in a full analysis (not asked here, but worth one closing line):
the torsional moment is T = Py ed,x and the torsional shear in any wall is
PT = T Ridi/ΣRd², where di is measured from the
centre of rigidity and ΣRd² is taken over all the walls in both
directions. Because the earthquake reverses, the torsional shear is always added to the
direct shear.
Method: Chapter 4 §4.6.1–4.6.4
(centre of mass, centre of rigidity, eccentricity, direct and torsional shear).