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2078 Bhadra — Regular / Back

BE / BCE · IV/I 80 Marks · 3 hrs
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Paper Information

DetailInformation
Exam2078 Bhadra
SubjectSeismic Resistant Design of Masonry Structure (Elective I) — CE 72502
Full Marks · Pass Marks · Time80 · 32 · 3 hours
InstructionsAttempt All questions. Necessary codes are allowed. Assume suitable data if necessary.
8Questions
24Theory marks
56Numerical marks
4Design numericals
Four 14-mark design numericals in one paper. Questions 3, 4, 5 and 6 are all wall-design problems of exactly the type in Chapter 2 §2.6.8 and Chapter 4 §4.6.6 — solid wall with piers, cavity wall, transverse (wind) wall, and shear wall. If you can do these four you have 56 of the 80 marks.

Question 1

1 (a) TheoryExplain Richter scale and MMI scale. [6]

Richter scale — a MAGNITUDE scale

  • Devised by Charles F. Richter in 1935. It is a quantitative, instrumental measure of the size of an earthquake — of the total energy released at the source.
  • 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: ML = log10A − log10A0.
  • Because it is logarithmic, one unit of magnitude = 10× the wave amplitude and about 31.6 (≈32)× the energy. M7.7 releases ~31× the energy of M6.7 and ~1000× that of M5.7.
  • Energy relation: log10E = 11.8 + 1.5 M (E in ergs).
  • One value per earthquake, independent of the observer's position and of the local soil or buildings. Expressed in Arabic numerals with a decimal.
  • Open-ended and can be negative, but it saturates above about M7 — which is why the non-saturating moment magnitude Mw is now used for large events.
  • Classification: <3 micro/minor, 3–3.9 minor, 4–4.9 light, 5–5.9 moderate, 6–6.9 strong, 7–7.9 major, ≥8 great.

MMI scale — an INTENSITY scale

  • The Modified Mercalli Intensity scale is a qualitative, observational measure of the effects of an earthquake at a particular place — on people, objects, buildings and the ground.
  • It has 12 grades written in Roman numerals, I to XII. No instrument is used; the grade is assigned from damage surveys and human reports.
  • Many values per earthquake — a different value at every location. It is highest near the epicentre and decreases outward.
  • Lines joining points of equal intensity are isoseismals; their map is used to locate the epicentre of historic (pre-instrumental) earthquakes and to prepare seismic hazard maps.
  • It depends on epicentral distance, the local soil, and the type and quality of the buildings.
  • Key grades: VII considerable damage in poorly built masonry; VIII partial collapse of ordinary masonry; X most masonry destroyed with its foundations; XII total damage.
  • Other intensity scales: MSK-64 (also 12 grades, used in Europe and India) and the JMA scale (Japan).

Close with the one-line contrast: the Richter scale measures how big the earthquake was; the MMI scale measures how badly it shook a particular place.

Full treatment including the full comparison table: Chapter 1 §1.2.5–1.2.7.
1 (b) TheoryDescribe in brief with sketch major failure mechanism of masonry building during Gorkha earthquake. [6]

The 2015 Gorkha earthquake (Mw 7.8, 25 April 2015, epicentre Barpak, focal depth ~15 km, followed by an Mw 7.3 aftershock on 12 May) damaged or destroyed about 800,000 buildings, the great majority of them unreinforced masonry. The major failure mechanisms observed were:

  1. Out-of-plane collapse of walls — by far the dominant mechanism, caused by unanchored flexible timber floors and roofs, the absence of bands, and slender walls in the upper storeys.
  2. Corner separation and vertical cracking at wall junctions, from poor interlocking of the courses at the corners.
  3. Delamination and collapse of multi-wythe walls — both burnt-brick-in-cement and un-coursed random rubble walls with CGI roofs, because through-stones were missing.
  4. In-plane diagonal (X) shear cracking of piers between openings, and vertical cracking at the corners of openings.
  5. Failure of gable walls, pulling down the top of the end walls and the roof.
  6. Separation of the roof from the walls, followed by collapse of heavy tiled roofs on the occupants.
  7. Soft / open ground storey failure in shop-front buildings, and short-column failures created by mezzanines.
  8. Total collapse of unreinforced stone-in-mud houses in the rural districts around the epicentre.
  9. Widespread failure of temples, monuments and monasteries, several already weakened by earlier earthquakes; plus lifeline failures (e.g. collapse of a penstock in Sindhupalchowk).
The single most quotable field observation: buildings with even one RC lintel band frequently survived with repairable damage while identical unbanded neighbours collapsed — the clearest available evidence for the value of banding.

Sketch to draw

One isometric of a simple masonry building marked with: separation of walls at corners, diagonal (X) cracking in the in-plane wall, out-of-plane overturning of the transverse wall, separation of the roof from the walls, vertical cracking at openings and gable collapse.

Full treatment with figures: Chapter 3 §3.9.1, with the damage figures in §3.6 and §3.7.

Question 2

2 (a) TheoryDefine masonry structure type based on materials with its advantages and disadvantages. [6]
TypeAdvantagesDisadvantages
Brick masonry
(burnt clay units)		 Uniform shape and size → good bond and thin joints; good strength (3.5–35 N/mm²); easy to form openings and bands; durable, fire and weather resistant; locally manufactured. Needs good clay, fuel and kilns (energy and pollution); quality varies between kilns; heavy; weak in tension.
Stone masonry
(random rubble, coursed rubble, squared, ashlar)		 Very high unit strength; extremely durable; locally available free of cost in hill areas; attractive; good thermal mass. Irregular shapes give poor bond and very thick joints; walls are multi-leaf and delaminate; very heavy; needs skilled masons; the worst-performing type in Nepali earthquakes.
Concrete block masonry Accurate dimensions, fast construction (large units), fewer joints, uniform quality, can be reinforced through the hollow cores. Needs cement and a plant; drying shrinkage cracking; higher initial cost.
Adobe / earthen masonry Cheapest; zero-energy production; excellent thermal comfort; entirely local material and skills. Very low strength, no tensile capacity, degrades on wetting, erodes; extremely vulnerable in earthquakes (NBC 204).
Composite / faced masonry Combines an economical backing (rubble, block) with a good-quality facing (brick, ashlar). The two leaves have different stiffness and separate unless properly bonded through with headers.

Add the general advantages of masonry (high compressive strength, durability, low-cost local materials, no complicated plant, fire and thermal performance, design flexibility, life > 500 years in lime mortar) and its general disadvantages (negligible tensile strength, heavy so large foundations and large inertia forces, brittle, frost and sulphate spalling, efflorescence, slow and weather-dependent construction).

Full treatment: Chapter 2 §2.4.1 and §2.2.4.
2 (b) TheoryDescribe affecting factors and relation between unit and mortar on strength of brick masonry. [6]

1. Relation between unit and mortar — the mechanism

Under vertical compression the mortar is softer and has a higher Poisson's ratio than the unit, so it tries to expand sideways more. Friction and bond at the interface prevent it. The result is that the mortar goes into triaxial compression (which strengthens it) while the unit is put into lateral tension combined with vertical compression. Because units are weak in tension, failure begins as vertical splitting cracks through the UNITS, not by crushing of the mortar.

Consequences:

  • Masonry strength is only about 25–50% of the unit strength.
  • Increasing the mortar strength gives a less than proportionate gain — doubling the mortar strength may raise the masonry strength by only 15–25%.
  • A thinner bed joint gives a stronger wall.
  • The mortar should be weaker than the unit — a mortar stronger than necessary is wasteful, cracks the units rather than the joints, and gives brittle unrepairable failure.
  • Empirically fk = K fb0.7 fm0.3 — unit strength matters more than twice as much as mortar strength.

Sketch: a masonry prism under compression with arrows showing the mortar squeezing outwards, the unit in lateral tension, and vertical splitting cracks through the bricks.

2. Factors affecting the compressive strength — four groups

GroupFactorEffect
A. UnitStrength of the unitThe single most important factor.
Shape / height-to-width ratio as laidTaller unit → fewer bed joints → stronger masonry (factor kp).
Frogs, perforations, voidsReduce net area, cause stress concentration.
Water absorption / suction rateA dry absorbent brick sucks water from the mortar and destroys the bond — bricks must be soaked.
B. MortarGrade and strengthIncreases strength, but far less than proportionately.
Mix proportion and workabilityRicher than 1:3 shrinks and cracks; leaner than 1:6 is harsh and segregates.
Water retentivity, deformabilityA more deformable mortar increases the lateral tension in the unit.
C. GeometryThickness of the mortar jointVery significant — a 20 mm joint can lose 20–30% of the strength.
Slenderness ratioHigher SR → buckling → permissible stress reduced by ks.
Eccentricity of loadingAdds bending; permissible stress reduced by ks.
Cross-sectional areaSmall sections lack redundancy → ka = 0.7 + 1.5A.
D. WorkmanshipBond pattern, corner interlockingContinuous vertical joints split the wall into independent leaves.
Filling of head (vertical) jointsUnfilled perpends are the commonest defect.
Uniformity of bed joints, verticalityUneven joints cause stress concentration; out-of-plumb walls add eccentricity.
Curing, age, rate of constructionGreen mortar overloaded by building too fast.

Design consequence: this is why IS 1905 writes fper = fb ks ka kp.

Full treatment: Chapter 2 §2.3.4 and §2.3.5 (Fig. 2.8).

Question 3

3 NumericalDesign an interior solid wall with pier of a single storey structure supporting a RCC roof of height 6 m. The piers of dimension shown in figure are located at 4 m C/C along the wall. Take load from roof equal to 50 kN/m. [14]
200 mm 100 mm 100 mm t Wall Pier at 4 m c/c
Figure with the questionPlan of the wall and pier. The pier is cruciform — it is 200 mm wide and projects 100 mm from each face of the wall, and the piers repeat at 4 m centre to centre.
Given: single storey, RCC roof; H = 6.0 m; roof load = 50 kN/m.
From the figure: the pier is 200 mm wide and projects 100 mm on each face, so the pier thickness tp = tw + 100 + 100; piers at 4.0 m centres.
Assume: wall thickness tw = 230 mm (one brick); masonry 20 kN/m³; interior wall → e = 0. Design one bay.
Load per bay
Roof load per bay = 4.0 × 50 = 200.00 kN Self weight per bay wall = 4.00 × 0.230 × 6.0 × 20 = 110.40 kN piers = 2 × 0.200 × 0.100 × 6.0 × 20 = 4.80 kN (two projections, one on each face) ---------- Total load per bay, P = 315.20 kN
Cross-sectional area per bay
A = (4000 × 230) + 2 × (200 × 100) = 920 000 + 40 000 = 960 000 mm²
Effective height  IS 1905 Table 4
RCC roof bearing on the full thickness → full restraint h_eff = 0.75 H = 0.75 × 6000 = 4500 mm
Effective thickness  IS 1905 Table 6
S_p / W_p = 4000 / 200 = 20 t_p / t_w = (230 + 100 + 100) / 230 = 430/230 = 1.87 Table 6 : for S_p/W_p = 20 or more the stiffening coefficient is 1.0 for EVERY value of t_p/t_w. → stiffening coefficient = 1.0 → the piers give NO stiffening benefit. t_eff = 1.0 × 230 = 230 mm
Effective length and slenderness ratio
Wall continuous / supported at both ends l_eff = 0.8 L = 0.8 × 4000 = 3200 mm Based on height : 4500 / 230 = 19.57 Based on length : 3200 / 230 = 13.91 ← lesser For a WALL take the smaller → SR = 13.91 Limit for a load-bearing wall in cement mortar = 27 ✓
Modification factors
k_s Table 9 , SR = 13.91 , e = 0 SR 12 → 0.84 , SR 14 → 0.78 k_s = 0.84 − (0.84−0.78)(13.91−12)/(14−12) = 0.783 k_a A per bay = 0.94 m² > 0.2 m² → k_a = 1.0 k_p brick 230×115×57 , h/w = 0.50 < 0.75 → k_p = 1.0
Actual stress and required basic compressive stress
f_c = P / A = 315 200 / 960 000 = 0.328 N/mm² Required f_b ≥ f_c / (k_s k_a k_p) = 0.328 / (0.783 × 1.0 × 1.0) = 0.419 N/mm²
Select unit and mortar  IS 1905 Table 8
unit 5.0 N/mm² + M2 (1:6) f_b = 0.44 ✓ marginal unit 5.0 N/mm² + M1 (1:5) f_b = 0.50 ✓ ADOPT f_per = f_b · k_s · k_a · k_p = 0.50 × 0.783 = 0.392 N/mm²
Final check
f_c = 0.328 N/mm² < f_per = 0.392 N/mm² ✓ Permissible load per bay = 0.392 × 960 000 = 376 320 N = 376.32 kN Applied load per bay = 315.20 kN Utilisation = 315.20 / 376.32 = 84 %
SAFE. Adopt a 230 mm thick interior wall with cruciform piers 200 mm wide projecting 100 mm from each face, at 4 m centres, built with bricks of crushing strength 5 N/mm² in M1 (1:5) cement mortar.
Note: at Sp/Wp = 20 the piers provide no stiffening — they only add 40 000 mm² of bearing area per bay. To gain a stiffening coefficient the spacing would have to be reduced below 4 m.
Method: Chapter 2 §2.6.4–2.6.8.

Question 4

4 NumericalDesign an interior cavity wall of a two storeyed building with a ceiling height of 3 m. The wall is unstiffened and 4 m long. Load from the roof is 10 kN/m and load from the floor is 15 kN/m. [14]
Given: two-storey building; ceiling height 3.0 m per storey; wall unstiffened, 4.0 m long; roof load 10 kN/m; floor load 15 kN/m; cavity wall.
Assume: two leaves of 115 mm (half brick) each with a 50 mm cavity, tied with metal ties → overall 280 mm; RCC slab 120 mm; masonry 20 kN/m³; the floor and roof bear on the inner leaf only. Design the ground-floor wall (worst case), per metre run.
Effective thickness of a cavity wall  IS 1905 cl. 4.5.4
teff = the greater of   (i) 23(t1 + t2)  and  (ii) the thickness of the thicker leaf
(i) (2/3)(115 + 115) = 153.3 mm (ii) thicker leaf = 115.0 mm t_eff = 153.3 mm (the greater)
Vertical load on the loaded (inner) leaf, per metre run
From roof = 10.00 kN/m From floor = 15.00 kN/m Inner leaf above (first storey) = 0.115 × 3.0 × 20 = 6.90 kN/m Inner leaf self weight (ground storey) = 0.115 × 3.0 × 20 = 6.90 kN/m ------------ Total on the inner leaf at the base, W = 38.80 kN/m
Actual compressive stress
Area of the loaded leaf per metre run A = 1000 × 115 = 115 000 mm² f_c = W / A = 38 800 / 115 000 = 0.337 N/mm²
Effective height
H = 3.00 + 0.120/2 = 3.06 m = 3060 mm RCC slabs top and bottom → full restraint h_eff = 0.75 × 3060 = 2295 mm
Effective length
The wall is UNSTIFFENED (no cross walls or piers) l_eff = 1.0 × L = 1.0 × 4000 = 4000 mm
Slenderness ratio
Based on height : 2295 / 153.3 = 14.97 ← lesser Based on length : 4000 / 153.3 = 26.09 SR = 14.97 ≈ 15 (limit 27) ✓
Modification factors
k_s Table 9 , SR = 15 , e = 0 SR 14 → 0.78 , SR 16 → 0.73 k_s = 0.78 − (0.78−0.73)(15−14)/(16−14) = 0.755 k_a A of the loaded leaf = 1.0 × 0.115 = 0.115 m² < 0.2 m² k_a = 0.70 + 1.5 × 0.115 = 0.873 k_p brick 230×115×57 , h/w = 0.50 < 0.75 → k_p = 1.0
Do not forget ka. Each leaf of a cavity wall is only 115 mm thick, so its area per metre run (0.115 m²) is below 0.2 m² and the area reduction factor applies. This is the trap in cavity-wall questions.
Required basic compressive stress and selection
Required f_b ≥ f_c / (k_s k_a k_p) = 0.337 / (0.755 × 0.873 × 1.0) = 0.337 / 0.659 = 0.512 N/mm² From Table 8 : unit 5.0 N/mm² + M1 (1:5) f_b = 0.50 ✘ just short 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
Final check
With unit 7.5 N/mm² + M2 mortar : f_per = 0.59 × 0.755 × 0.873 × 1.0 = 0.389 N/mm² f_c = 0.337 N/mm² < f_per = 0.389 N/mm² ✓ Utilisation = 87 %
SAFE. Adopt a cavity wall of two 115 mm leaves with a 50 mm cavity (overall 280 mm), built with bricks of crushing strength 7.5 N/mm² in M2 (1:6) cement mortar, with metal wall ties at not more than 450 mm vertical and 900 mm horizontal centres so that the two leaves act together.
If the loads were shared between both leaves (i.e. the slab bears on both), the area would be 230 000 mm², ka = 1.0, fc would be about 0.229 N/mm² and a much weaker combination would do. State your assumption clearly — the examiner is looking for the cavity-wall effective-thickness rule and the area reduction factor, not for a particular brick grade.
Method: Chapter 2 §2.4.3 (cavity wall) and §2.6.4–2.6.8.

Question 5

5 NumericalDesign an exterior wall of a single storey workshop of height 4 m. The load on the wall is 30 kN/m from the roof and the wind load is 750 N/m². The wall is connected by metal anchor with the roof structure. [14]
Given: single-storey workshop; H = 4.0 m; vertical load from roof = 30 kN/m; wind pressure p = 750 N/m²; wall tied to the roof with metal anchors.
Assume: t = 230 mm; masonry 20 kN/m³. Design a 1 m run of wall.
Vertical load
Load from roof = 30.00 kN/m Self weight of wall = 0.23 × 20 × 4.0 = 18.40 kN/m ------------ Total at the base, W = 48.40 kN/m
Section properties (per metre run)
A = 1000 × 230 = 230 000 mm² Z = b t² / 6 = 1000 × 230² / 6 = 8.817 × 10⁶ mm³
Bending moment due to wind

The wall is anchored at the floor and the roof, so it is treated as partially constrained at both ends; the moment is critical at the base and at mid-height:

M = p H² / 10 = 750 × 4.0² / 10 = 1200 N·m/m = 1.20 × 10⁶ N·mm/m Flexural stress f_cb = M / Z = 1.20 × 10⁶ / 8.817 × 10⁶ = 0.136 N/mm²
Compressive stress at the BASE
f_ca = W / A = 48 400 / 230 000 = 0.210 N/mm² f_c = f_ca + f_cb = 0.210 + 0.136 = 0.346 N/mm²
Tensile stress at MID-HEIGHT (the critical section for tension)
Load present at mid-height W_mid = 30.00 + 0.23 × 20 × 2.0 = 39.20 kN/m f_ca = 39 200 / 230 000 = 0.170 N/mm² f_t = f_cb − f_ca = 0.136 − 0.170 = −0.034 N/mm² Negative → the section is still wholly in COMPRESSION, NO TENSION develops. ✓
Slenderness ratio and eccentricity
Metal anchors → lateral support with PARTIAL restraint h_eff = H = 4000 mm t_eff = 230 mm SR = 4000 / 230 = 17.39 < 27 ✓ Equivalent eccentricity e = M / W = 1.20 × 10⁶ / 48 400 = 24.8 mm e / t = 24.8 / 230 = 0.108 (between 1/12 and 1/6)
Stress reduction factor  IS 1905 Table 9
At SR = 17.39 : e/t = 1/12 : SR 16 → 0.68 , SR 18 → 0.61 k = 0.68 − 0.07 × (1.39/2) = 0.631 e/t = 1/6 : SR 16 → 0.63 , SR 18 → 0.55 k = 0.63 − 0.08 × (1.39/2) = 0.574 Interpolate on e/t for 0.108 : fraction = (0.108 − 0.0833)/(0.1667 − 0.0833) = 0.296 k_s = 0.631 − 0.296 × (0.631 − 0.574) = 0.614
Permissible stress
k_a A = 1.0 × 0.23 = 0.23 m² > 0.2 m² → k_a = 1.0 k_p h/w = 0.50 < 0.75 → k_p = 1.0 TRIAL 1 : unit 5 N/mm² + M1 mortar , f_b = 0.50 f_per = 0.50 × 0.614 = 0.307 N/mm² with the 25% increase for bending f_per,bend = 1.25 × 0.307 = 0.384 N/mm² f_c = 0.346 < 0.384 ✓ but only 90% utilised TRIAL 2 : unit 7.5 N/mm² + M1 mortar , f_b = 0.74 f_per = 0.74 × 0.614 = 0.454 N/mm² f_per,bend = 1.25 × 0.454 = 0.568 N/mm² f_c = 0.346 < 0.568 ✓ utilisation 61 % ADOPT

Note: because wind is present, IS 1905 also permits the permissible stresses to be increased by one third — not needed here, so the design is conservative.

Summary of checks
Compression at base : 0.346 < 0.568 ✓ Tension at mid-height: none develops ✓ Slenderness : 17.39 < 27 ✓
SAFE. Adopt a 230 mm (one brick) thick exterior wall, 4 m high, built with bricks of crushing strength 7.5 N/mm² in M1 (1:5) cement mortar, anchored to the roof structure with metal anchors at not more than 1 m centres.
Method: Chapter 4 §4.8. Follows the worked example in the course notes (CH-4, Example 3.1).

Question 6

6 NumericalDesign a shear wall of 5 m long and 4 m high to resist a horizontal earthquake force in its plane under seismic loading. The seismic load is 0.3 g acceleration uniformly distributed across the entire height of the wall. The wall is connected with metal anchors at the top and bottom supports. Neglect the load from the floor. [14]
Given: L = 5.0 m; H = 4.0 m; in-plane seismic force = 0.3 g, uniformly distributed over the height; metal anchors top and bottom; floor load neglected.
Assume: t = 230 mm; masonry 20 kN/m³; unit 5 N/mm² with M1 mortar.
Section properties
A = L × t = 5000 × 230 = 1.150 × 10⁶ mm² Z = t L² / 6 = 230 × 5000² / 6 = 9.583 × 10⁸ mm³
Self weight and lateral force
W = 5.0 × 4.0 × 0.23 × 20 = 92.00 kN F = 0.3 × W = 0.3 × 92.00 = 27.60 kN
Shear check  IS 1905 cl. 5.4.3
τ = F / A = 27 600 / 1.150 × 10⁶ = 0.024 N/mm² f_d = W / A = 92 000 / 1.150 × 10⁶ = 0.080 N/mm² τ_per = 0.1 + f_d/6 = 0.1 + 0.080/6 = 0.113 N/mm² τ = 0.024 < τ_per = 0.113 ✓
Bending moment and stresses
Load uniformly distributed over the height → resultant at mid-height : M = F × H/2 = 27 600 × 4000/2 = 55.20 × 10⁶ N·mm f_cb = M / Z = 55.20 × 10⁶ / 9.583 × 10⁸ = 0.058 N/mm² f_ca = W / A = 0.080 N/mm² Max compressive stress f_c = 0.080 + 0.058 = 0.138 N/mm² Max tensile stress f_t = 0.058 − 0.080 = −0.022 N/mm² → still compression, NO tension ✓
Slenderness ratio and eccentricity
h_eff = 4000 mm , t_eff = 230 mm SR = 4000 / 230 = 17.39 e = M / W = 55.20 × 10⁶ / 92 000 = 600 mm e / L = 600 / 5000 = 0.12 (between 1/12 = 0.0833 and 1/6 = 0.1667)
Stress reduction factor  IS 1905 Table 9
At SR = 17.39 (from the interpolation of Q5) : e/t = 1/12 → 0.631 e/t = 1/6 → 0.574 Interpolate for e/t = 0.12 : fraction = (0.12 − 0.0833)/(0.1667 − 0.0833) = 0.44 k_s = 0.631 − 0.44 × (0.631 − 0.574) = 0.606
Permissible stresses and final check
Select unit 5 N/mm² + M1 mortar → f_ck = 0.50 N/mm² k_a = 1.0 (A = 1.15 m² > 0.2 m²) , k_p = 1.0 Axial f_ac = k_s f_ck = 0.606 × 0.50 = 0.303 N/mm² f_ca = 0.080 < 0.303 ✓ Combined f_acb = 1.25 × 0.303 = 0.379 N/mm² f_c = 0.138 < 0.379 ✓ Shear 0.024 < 0.113 ✓ Tension none develops ✓
HENCE SAFE. Adopt a 230 mm thick brick shear wall 5 m × 4 m, built with units of crushing strength 5 N/mm² in M1 (1:5) cement mortar, tied to the supports with metal anchors.
Compare the three shear-wall questions. 2079 used 0.1 g on a 4.2 × 3.3 m wall, 2081 used 0.2 g on 4 × 3 m, and this one 0.3 g on 5 × 4 m. In every case the wall passes comfortably, because an isolated wall carries only its own weight. The marks are for the method, the code references and the four checks — not for the numbers.
Method: Chapter 4 §4.1.3 and §4.6.6.

End of 2078 Bhadra — Solved Past Questions