2080 Bhadra – Solved Past Questions

Detail	Information
Exam	Regular, 2080 Bhadra
Subject	Seismic Resistant 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 1905–1987 codes are allowed. Assume suitable data if necessary.

1 (a) TheoryDefine earthquake, its source and influencing factor of shaking in the location. [8]

1. Definition

An earthquake is the sudden shaking or trembling of the earth's surface caused by the rapid release of strain energy accumulated in elastically strained rocks of the earth's crust, usually associated with the sudden rupture (faulting) of rock along a plane of weakness. The released energy radiates in all directions from its source as seismic waves.

2. Source of an earthquake

Focus (hypocentre) — the point inside the earth where the rupture begins and from where the energy is first released. This is the source.
Epicentre — the point on the ground surface vertically above the focus.
Focal depth — vertical depth of the focus. Shallow < 70 km, intermediate 70–300 km, deep 300–700 km. About 90% of foci are shallower than 100 km.
Fault — the fracture along which the two blocks move relative to each other.
Mechanism — plate tectonics loads the fault; when the accumulated strain exceeds the strength of the rock the blocks rupture and “snap back” (elastic rebound theory), releasing the stored energy as seismic waves.
Types by cause: tectonic (>90%), volcanic, collapse, explosion, reservoir-induced.

Draw a section showing focus, epicentre, fault plane, focal depth, epicentral distance and the radiating wave fronts.

3. Factors influencing the shaking at a location

Group them as source – path – site:

Group	Factor	Effect
Source	Magnitude (energy released)	Larger magnitude → larger amplitude and longer duration.
	Focal depth	Shallow focus → much stronger surface shaking for the same magnitude.
	Fault mechanism & rupture directivity	Thrust faults give large vertical components; shaking is amplified in the direction of rupture propagation.
	Length / area of rupture	Longer rupture → longer duration of strong shaking.
Path	Epicentral / hypocentral distance	Amplitude attenuates rapidly with distance.
Path	Geology along the travel path	Reflection and refraction change the amplitude and frequency content.
Site	Local soil type and depth	Soft, deep, unconsolidated soil (e.g. Kathmandu Valley lake deposits) amplifies the shaking and lengthens its period; rock sites shake less.
	Topography	Ridges and hill tops amplify base-rock acceleration by resonance (“ridge effect”).
	Basin geometry	Sediment-filled basins trap surface waves and prolong the shaking.
	Groundwater / liquefiable soil	Saturated loose sand may liquefy, changing the motion and causing settlement.
Structure	Nature of the structure	Damage also depends on the building's own period, mass, stiffness, ductility, configuration and construction quality — resonance greatly increases the response.

Full treatment: Chapter 1 §1.1.1, §1.1.2 and §1.1.9.

1 (b) TheoryDescribe failure mechanism of masonry building during earthquake and suggest its mitigation measures to make earthquake resistant with sketches. [12]

Part 1 — Failure mechanisms

Classification (seven-fold): (1) out-of-plane flexural / shear failure; (2) in-plane shear / flexural failure; (3) separation of walls at junctions; (4) failure of masonry piers between openings; (5) local failures (parapets, gables, chimneys); (6) buckling / delamination of walls; (7) separation of the roof from the walls.

(a) In-plane failure modes — four types

Mode	Mechanism	When it occurs	Crack appearance
Rocking (flexural)	Horizontal crack opens at the heel; the pier rotates as a rigid block about its compressed toe.	Slender piers with low vertical load.	Horizontal cracks top and bottom; toe crushing.
Sliding	The wall slides horizontally along a bed joint.	Low vertical load, weak mortar; at base, sill or DPC.	Single horizontal crack with visible offset.
Diagonal shear — bed-joint sliding	The diagonal crack follows the mortar joints.	Weak mortar, strong units.	Stair-stepped diagonal crack, bricks intact.
Diagonal tension + toe crushing	Principal tension exceeds the tensile strength of the masonry; the crack runs through the units. Reversal gives two crossing cracks.	Strong mortar, weak units, high compression; squat piers.	Straight “X” crack cutting through the bricks.

(b) Out-of-plane failure mechanisms — four types

Mechanism	Description	Cause
Simple overturning	The whole facade rotates outwards about its base as a rigid block.	No connection between facade, transverse walls or floor.
Composed overturning	The facade overturns dragging a wedge of the adjoining transverse walls with it.	Good corner interlocking but no floor/roof tie.
Vertical flexure	Wall restrained at floor and roof bulges and hinges at mid-height.	High h/t; typical of upper storeys.
Horizontal flexure	Long wall spans horizontally between cross walls and cracks vertically at mid-span.	Well tied at the ends but unsupported at the top (flexible diaphragm).

Governing variables: the height-to-thickness (h/t) ratio and the vertical precompression. Walls at the top of buildings and slender walls are the most vulnerable.

(c) Other typical damage

Corner separation (the first damage to appear), delamination of multi-leaf walls, gable-end collapse, roof separation, pounding, and failure of piers between openings.

Part 2 — Mitigation measures (keyed to the failure prevented)

Failure it prevents	Measure
Corner separation	Toothed / stepped joints at corners and T-junctions (corners built first to 600 mm, or toothing in 450 mm lifts); vertical containment reinforcement at corners and junctions from foundation to roof.
Out-of-plane collapse	Horizontal bands at plinth, sill, lintel, roof and gable level; cross walls or buttresses to reduce the span; limit h/t; anchor the wall to the diaphragm.
Roof / floor separation	Positive metal anchors; roof band at eave level.
Loss of box action	Provide a rigid diaphragm (RC slab) or brace the timber floor.
Delamination of stone walls	Through-stones (or S-ties / wooden dowels / concrete bond blocks) at 1.2–1.5 m spacing both ways.
Diagonal shear cracking	Good-quality mortar (never mud) matched to the unit; wide piers; bands above and below openings.
Failure of piers between openings	Openings small, few, central, vertically aligned, ≥ 600 mm from the corner, total ≤ ~50% of the wall length; vertical bars at jambs.
Torsion	Symmetric plan; separate irregular plans into simple rectangles with a seismic gap.
Pounding	Adequate seismic separation; separate the staircase tower.
Excessive inertia force	Light roof; limit the number of storeys; reduce upper-floor mass.
Local / non-structural failure	Avoid or reinforce parapets, gables, chimneys and heavy projections; gable band or light truss gable.
Poor materials / workmanship	Quality control — sound units, correct mortar grade, full joints, soaked bricks, curing, limited daily lift, supervision.

Sketches to draw (the marks are here)

Pier with X-shaped diagonal cracks; and a second pier with a stair-stepped crack — label “strong mortar” / “weak mortar”.
Pier rocking about its toe, and a wall sliding on a bed joint.
The four out-of-plane mechanisms (simple overturning, composed overturning, vertical flexure, horizontal flexure).
A masonry building showing corner separation, diagonal cracking, roof separation, vertical cracking and out-of-plane failure.
Mitigation sketch: building with plinth, sill, lintel, roof and gable bands + vertical bars at corners and jambs.

Full treatment with all the figures: Chapter 3 §3.6 (modes), §3.7 (damage) and §3.7.6 (mitigation).

2 (a) TheoryDefine properties and strength of brick masonry, affecting factors and relation of unit and mortar on strength of masonry. [8]

1. Properties of brick masonry

Type	Property	Typical value
Physical	Size (NS 1:2035)	240 × 115 × 57 mm with a 10 mm joint
	Water absorption	≤ 15% of own weight after 24 h immersion
	Efflorescence	Nil to slight
	Soundness / hardness	Clear ringing sound; no nail impression
	Unit weight of masonry	19–20 kN/m³
Mechanical	Compressive strength of the unit	3.5–40 N/mm² (Nepali kiln brick 5–12)
	Compressive strength of the masonry	only 25–50% of the unit strength
	Tensile / flexural (bond) strength	0.1–0.4 N/mm²; permissible 0.07–0.14 N/mm²; effectively zero in mud mortar
	Shear strength	f_s = 0.1 + f_d/6, max 0.5 N/mm² (IS 1905 cl. 5.4.3)
	Modulus of elasticity E_m	≈ 550–1100 f_m (often 750 f_m)
	Poisson's ratio / modulus of rigidity	0.15–0.20 / G = 0.4 E_m

2. Relation between unit and mortar on the strength of masonry

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 — and these are the marks:

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 (less soft material → less lateral tension).
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.

Empirical form: f_k = K · f_b^α · f_m^β with α ≈ 0.7 and β ≈ 0.3
→ unit strength matters more than twice as much as mortar strength.

3. Factors affecting the compressive strength — four groups

Group	Factors
A. The unit	Strength of the unit (most important); shape / height-to-width ratio as laid (shape modification factor k_p); frogs, perforations and voids; water absorption and suction rate.
B. The mortar	Grade and strength; mix proportion and workability; water retentivity and deformability.
C. Geometry	Thickness of the mortar joint (a 20 mm joint can lose 20–30% of the strength); slenderness ratio (k_s); eccentricity of loading (k_s); cross-sectional area (k_a).
D. Workmanship	Bond pattern and corner interlocking; filling of vertical (head) joints; uniformity of bed joints and verticality; curing and age; rate of construction.

Design consequence: this is exactly why IS 1905 writes f_per = f_b k_s k_a k_p.

Full treatment: Chapter 2 §2.3.1, §2.3.4 and §2.3.5 (with Fig. 2.8 — the lateral-tension mechanism).

2 (b) NumericalCalculate the effective height, effective length, effective thickness and slenderness ratio of walls A, C, G and P of the plan below. The building is of single storey with one brick thick wall, 3 m height of floor, RCC roof 100 mm thick, height of door 2.1 m and height of window 1.2 m. [12]

Plan of a single storey masonry building showing walls A and B along the top each 4.40 m long, walls C and E at the sides 3.0 m, internal wall D, and the lower band of short walls F, G, J, K, M, N and the piers P, with all dimensions in metres — Figure with the questionPlan referred to in the question. All dimensions are in metres. Walls A (4.40 m) and B (4.40 m) form the top wall; C and E (3.0 m) are the side walls; G (1.0 m) are the short piers in the lower wall; P (0.6 m) are the piers in the bottom wall at 3.8 m clear spacing.

Given: single storey; wall thickness = one brick = 230 mm; clear floor height = 3.0 m; RCC roof slab = 100 mm; door height = 2.1 m; window height = 1.2 m.
Read off the plan above: wall A = 4.40 m long (top wall, continuous at one end and supported at the other); wall C = 3.0 m (left wall); G = 1.0 m of brickwork between two openings in the lower wall; P = 0.6 m pier in the bottom wall, with 3.8 m clear on either side.

Which are walls and which are columns? A length of masonry is treated as a wall if its length exceeds 4t and as a column otherwise. Here 4t = 4 × 0.23 = 0.92 m, so A and C are walls, while G and P are so short and so poorly supported that they must be treated as columns — which is exactly why the examiner picked those four.

Actual height of the wall

H = clear height of floor + half the thickness of the roof slab = 3.00 + 0.100/2 = 3.05 m = 3050 mm

Is each element a WALL or a COLUMN?

IS 1905 definition: an isolated vertical load-bearing member whose width does not exceed 4 t is a column; if it exceeds 4 t it is a wall.

4 t = 4 × 0.23 = 0.92 m Wall A L = 4.4 m > 0.92 m → WALL Wall C L = 3.0 m > 0.92 m → WALL Brickwork G L ≈ 1.0 m — has openings on BOTH sides, so it has no lateral support on either side → treated as a COLUMN Brickwork P L ≈ 0.6 m < 0.92 m → COLUMN

Effective height IS 1905 Table 4

The RCC roof bears on the full thickness of the wall, so the walls have full restraint at the top and at the bottom → h_eff = 0.75 H.

For masonry between openings acting as a pier, with full restraint at the top (cl. 4.5.3):

perpendicular to the plane of the wall: h_eff = 0.75 H + 0.25 H₁
parallel to the plane of the wall: h_eff = H
(H₁ = height of the taller adjacent opening)

Wall A h_eff = 0.75 × 3050 = 2287.5 mm Wall C h_eff = 0.75 × 3050 = 2287.5 mm Brickwork G (openings both sides, taller = DOOR 2.1 m) ⊥ to plane : h_eff = 0.75×3050 + 0.25×2100 = 2812.5 mm ‖ to plane : h_eff = H = 3050.0 mm Brickwork P (adjacent opening = WINDOW 1.2 m) ⊥ to plane : h_eff = 0.75×3050 + 0.25×1200 = 2587.5 mm ‖ to plane : h_eff = H = 3050.0 mm

Effective length IS 1905 Table 3

Wall	End condition	Coefficient	l_eff
A	Continuous at one end, discontinuous at the other but supported by a cross wall	0.9 L	0.9 × 4.4 = 3.96 m
C	Discontinuous on one side, continuous on the other, supported by cross walls on both sides	0.9 L	0.9 × 3.0 = 2.70 m
G	No support on either side (openings both sides)	Not applicable — the slenderness ratio of this element is governed by its height.
P	Discontinuous and supported on one end	2.0 L	2.0 × 0.6 = 1.20 m

Effective thickness IS 1905 Table 6

Walls stiffened by cross walls get t_eff = (stiffening coefficient) × t_w. Taking t_p = 3 t_w (thickness of the cross wall not given) so t_p/t_w = 3, and W_p = t_w = 0.23 m:

Wall A : S_p/W_p = 4.4 / 0.23 = 19.13 Table 6 at t_p/t_w = 3 : S_p/W_p 15 → 1.2 , 20 → 1.0 coeff = 1.2 − (1.2−1.0)(19.13−15)/(20−15) = 1.035 t_eff = 1.035 × 230 = 238 mm Wall C : S_p/W_p = 3.0 / 0.23 = 13.04 Table 6 at t_p/t_w = 3 : S_p/W_p 10 → 1.4 , 15 → 1.2 coeff = 1.4 − (1.4−1.2)(13.04−10)/(15−10) = 1.278 t_eff = 1.278 × 230 = 294 mm Brickwork G and P : no cross-wall stiffening t_eff = actual thickness = 230 mm (in the direction parallel to the plane, the effective thickness is the LENGTH of the element)

Note: the stiffening coefficient may only be used when the slenderness ratio is based on the effective height; no modification is made when it is based on the effective length (cl. 4.5.4).

Slenderness ratio

Wall — SR = lesser of h_efft_eff and l_efft_eff, limit 27
Column — SR = greater of the ratios of effective height to the respective effective thickness in the two principal directions, limit 12 (cl. 4.6)

Wall A h_eff/t_eff = 2287.5 / 238 = 9.61 ← lesser l_eff/t_eff = 3960.0 / 238 = 16.64 SR = 9.61 < 27 ✓ Wall C h_eff/t_eff = 2287.5 / 294 = 7.78 ← lesser l_eff/t_eff = 2700.0 / 294 = 9.18 SR = 7.78 < 27 ✓ Brickwork G (column) ⊥ : 2812.5 / 230 = 12.23 ← greater ‖ : 3050.0 / 1000 = 3.05 SR = 12.23 > 12 ✘ Brickwork P (column) ⊥ : 2587.5 / 230 = 11.25 ← greater ‖ : 3050.0 / 600 = 5.08 SR = 11.25 < 12 ✓

Summary table (write this out in the exam)

Element	Type	h_eff (mm)	l_eff (mm)	t_eff (mm)	SR	Limit	Remark
A	Wall	2287.5	3960	238	9.61	27	Safe
C	Wall	2287.5	2700	294	7.78	27	Safe
G	Column	2812.5 (⊥) 3050 (‖)	—	230 / 1000	12.23	12	Marginally exceeds
P	Column	2587.5 (⊥) 3050 (‖)	1200	230 / 600	11.25	12	Safe

Walls A, C and brickwork P are within their slenderness limits. Brickwork G, at SR = 12.23, marginally exceeds the limit of 12 for a column.
Remedy: thicken that pier to 1½ brick (350 mm), or reduce the door height, or widen the pier so that it is no longer an isolated column, or provide vertical reinforcement in it.

Method and all three tables: Chapter 2 §2.6.4 and §2.6.5. Follows the worked example in the course notes (CH-4, Example 4.9), which uses the same lettered-wall plan.

3 (a) NumericalDesign an exterior solid wall with pier of the ground floor of a double storey brick masonry building with 120 mm thick RCC floor and roof. The heights of ground and first floor are 3 m. The piers are located 5 m centre to centre and dimensions 200 mm × 100 mm as shown in the figure. [10]

Figure with the questionPlan of the wall and pier. The pier is 200 mm wide and projects 100 mm from each face of the wall; the piers repeat at 5 m centre to centre.

Given: double storey; RCC floor and roof 120 mm thick; storey height 3.0 m each; piers at 5 m centre to centre.
From the figure: pier 200 mm wide, projecting 100 mm from each face.
Assumed (not given — “assume suitable data”): wall thickness t = 230 mm (one brick); slab spans 3.0 m onto the wall (tributary width for an exterior wall = 1.5 m); RCC 25 kN/m³; floor finish 1.0 kN/m²; LL on roof 1.5 kN/m², on floor 2.0 kN/m²; masonry 20 kN/m³.

Loads on a 1 m run of the ground-floor wall

Slab dead load = 0.12 × 25 = 3.00 kN/m² Floor finish = 1.00 kN/m² From ROOF : (3.00 + 1.00 + 1.50) × 1.5 = 8.25 kN/m From FLOOR : (3.00 + 1.00 + 2.00) × 1.5 = 9.00 kN/m Wall above (first floor) = 0.23 × 3.0 × 20 = 13.80 kN/m Wall self weight (ground floor, at base) = 13.80 kN/m ----------- Total vertical load at the base, W = 44.85 kN/m

Effective height

H = 3.00 + 0.120/2 = 3.06 m = 3060 mm RCC slab bears on the full wall thickness → full restraint h_eff = 0.75 H = 0.75 × 3060 = 2295 mm

Effective thickness — do the piers help?

S_p = 5000 mm (centre-to-centre spacing of piers) W_p = 200 mm (width of pier in the direction of the wall) t_p = 230 + 100 = 330 mm , t_w = 230 mm → t_p/t_w = 1.43 S_p / W_p = 5000 / 200 = 25 IS 1905 Table 6 : for S_p/W_p ≥ 20 the stiffening coefficient is 1.0 for EVERY value of t_p/t_w. → stiffening coefficient = 1.0 → t_eff = 1.0 × 230 = 230 mm

Important conclusion — state it explicitly. At S_p/W_p = 25 the piers are too far apart to provide any stiffening at all. They add cross-sectional area locally but do not reduce the slenderness ratio of the wall. To gain a benefit the spacing would have to come down to about 4 m (S_p/W_p = 20) or less.

Slenderness ratio

SR = h_eff / t_eff = 2295 / 230 = 9.98 ≈ 10 Limit for a load-bearing wall in cement mortar = 27 ✓

Eccentricity

The RCC slab bears on part of the wall thickness, so for an exterior wall the load is eccentric. Adopt the usual conservative value (IS 1905 Appendix A):

e / t = 1/6 (resultant just at the edge of the middle third)

Actual compressive stress

Area per metre run A = 1000 × 230 = 230000 mm² f_c = W / A = 44850 / 230000 = 0.195 N/mm²

Modification factors

k_s Stress reduction factor (Table 9) SR = 10 , e/t = 1/6 → k_s = 0.85 k_a Area reduction factor (cl. 5.4.1.2) A per metre = 1.0 × 0.23 = 0.23 m² > 0.2 m² → k_a = 1.0 k_p Shape modification factor (Table 10) brick 230×115×57, h/w = 57/115 = 0.50 < 0.75 → k_p = 1.0

Required basic compressive stress and choice of unit + mortar

Required f_b ≥ f_c / (k_s · k_a · k_p) = 0.195 / (0.85 × 1.0 × 1.0) = 0.229 N/mm² From IS 1905 Table 8, select a comfortable combination: unit crushing strength 5 N/mm² + M2 mortar (1:6) → f_b = 0.44 N/mm² Permissible stress f_per = f_b · k_s · k_a · k_p = 0.44 × 0.85 × 1.0 × 1.0 = 0.374 N/mm²

Final check

f_c = 0.195 N/mm² < f_per = 0.374 N/mm² ✓ Utilisation = 0.195 / 0.374 = 52 %

SAFE. Adopt a 230 mm (one brick) thick exterior wall with 200 mm × 100 mm piers at 5 m centres, built with bricks of crushing strength 5 N/mm² in M2 (1:6) cement mortar.
Note for the answer script: because S_p/W_p = 25 the piers give no stiffening benefit; they are provided for architectural reasons and for local bearing under the floor beams.

Method: Chapter 2 §2.6.4–2.6.8 (12-step wall design; Table 6 stiffening coefficients; Tables 8, 9, 10).

3 (b) NumericalDesign a wall of a basketball sports building of height 6 m. The roof is of 150 mm thick RCC slab and take wind load 850 N/m². The wall is connected by metal anchor with the roof structure. [10]

Given: wall height H = 6.0 m; roof = 150 mm RCC slab; wind pressure p = 850 N/m²; wall connected to the roof by metal anchors (lateral support, partial restraint).
Assumed: roof slab spans 4 m onto the wall (tributary width 2.0 m); RCC 25 kN/m³; finish 1.0 kN/m²; roof LL 1.5 kN/m²; masonry 20 kN/m³. Design a 1 m run of wall.

Load from the roof (per metre run)

Slab dead load = 0.15 × 25 = 3.75 kN/m² Finish = 1.00 kN/m² Live load on roof = 1.50 kN/m² ------------ 6.25 kN/m² Roof load on wall = 6.25 × 2.0 = 12.50 kN/m

FIRST TRIAL — assume t = 230 mm (one brick)

Self weight of wall = 0.23 × 20 × 6.0 = 27.60 kN/m Total load at base W = 12.50 + 27.60 = 40.10 kN/m A = 1000 × 230 = 2.30 × 10⁵ mm² Z = b t²/6 = 1000 × 230²/6 = 8.817 × 10⁶ mm³ Bending moment (wall anchored top and bottom) M = p H² / 10 = 850 × 6.0² / 10 = 3060 N·m/m = 3.06 × 10⁶ N·mm/m Flexural stress f_cb = M / Z = 3.06 × 10⁶ / 8.817 × 10⁶ = 0.347 N/mm² TENSION CHECK at MID-HEIGHT (least precompression): W_mid = 12.50 + 0.23 × 20 × 3.0 = 26.30 kN/m f_ca = 26300 / 230000 = 0.114 N/mm² f_t = f_cb − f_ca = 0.347 − 0.114 = +0.233 N/mm² TENSION

f_t = 0.233 N/mm² far exceeds the permissible flexural tensile stress of masonry (0.07–0.14 N/mm², IS 1905 Table 4). A 230 mm wall is NOT adequate — increase the thickness.

SECOND TRIAL — t = 350 mm (one and a half brick)

Self weight = 0.35 × 20 × 6.0 = 42.00 kN/m Total load at base W = 12.50 + 42.00 = 54.50 kN/m A = 1000 × 350 = 3.50 × 10⁵ mm² Z = 1000 × 350² / 6 = 2.0417 × 10⁷ mm³ M = 850 × 6.0² / 10 = 3.06 × 10⁶ N·mm/m f_cb = M / Z = 3.06 × 10⁶ / 2.0417 × 10⁷ = 0.150 N/mm²

Check the tensile stress at mid-height

W_mid = 12.50 + 0.35 × 20 × 3.0 = 33.50 kN/m f_ca = 33500 / 350000 = 0.096 N/mm² f_t = f_cb − f_ca = 0.150 − 0.096 = +0.054 N/mm² Permissible flexural tension (bending in the vertical direction, tension normal to bed joints, M1 mortar) = 0.07 N/mm² 0.054 < 0.07 ✓

Check the compressive stress at the base

f_ca = W / A = 54500 / 350000 = 0.156 N/mm² f_c = f_ca + f_cb = 0.156 + 0.150 = 0.306 N/mm²

Slenderness ratio and eccentricity

Metal anchors give lateral support with PARTIAL restraint → h_eff = H = 6000 mm t_eff = 350 mm (no cross walls given) SR = 6000 / 350 = 17.14 < 27 ✓ Equivalent eccentricity e = M / W = 3.06 × 10⁶ / 54500 = 56.1 mm e / t = 56.1 / 350 = 0.160 ≈ 1/6

Permissible stress

k_s Table 9 , SR = 17.14 , e/t = 1/6 SR 16 → 0.63 , SR 18 → 0.55 k_s = 0.63 − (0.63−0.55)(17.14−16)/(18−16) = 0.584 k_a A = 1.0 × 0.35 = 0.35 m² > 0.2 m² → k_a = 1.0 k_p h/w of brick = 57/115 = 0.50 < 0.75 → k_p = 1.0 Select unit 7.5 N/mm² with M1 mortar (1:5) Table 8 → f_b = 0.74 N/mm² f_per = f_b · k_s · k_a · k_p = 0.74 × 0.584 = 0.432 N/mm² Excess is due to bending → 25% increase allowed f_per,bending = 1.25 × 0.432 = 0.540 N/mm² (Wind is present, so a further 1/3 increase is also permissible — not needed here.)

Final comparison

Compression : f_c = 0.306 < 0.540 ✓ Tension : f_t = 0.054 < 0.070 ✓ Slenderness : SR = 17.14 < 27 ✓

SAFE. Adopt a 350 mm (one and a half brick) thick wall, 6 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.
Recommended additions: because H = 6 m is tall and the tension check is the governing criterion, provide pilasters or buttresses at 4–5 m centres (which would reduce the effective span and hence M), and an RC band at eave level tied to the anchors.

Exam technique. Always start with the smallest sensible thickness, show that it fails, and then increase. Showing one failed trial and one successful trial demonstrates that you understand which check governs — here it is the flexural tension at mid-height, not the compression at the base.

Method: Chapter 4 §4.8 (wall subjected to out-of-plane bending) and Chapter 2 §2.6.9. Follows the worked example in the course notes (CH-4, Example 3.1).

4 (a) NumericalCalculate the floor level shear and earthquake force for a double storey health post building located at Surkhet with plan dimension 4.5 m long and 3.5 m wide with 120 mm thick RCC slab. The wall thickness is one and half brick with 1:4 cement sand mortar and each floor height is 3 m. The soil of the location is gravel mixed. Neglect the door and openings. [14]

Given: double storey health post at Surkhet; plan 4.5 m × 3.5 m; RCC slab 120 mm; wall = 1½ brick = 350 mm in 1:4 CM (H2 mortar); storey height 3.0 m each; soil gravel mixed → medium (Type II) soil; doors and openings neglected.
Assumed: RCC 25 kN/m³; masonry 20 kN/m³; floor finish 1.0 kN/m²; live load on the floor 3.0 kN/m².

Component weights

Plan area = 4.5 × 3.5 = 15.75 m² Slab = 0.12 × 25 × 15.75 = 47.25 kN (per floor) Finish = 1.00 × 15.75 = 15.75 kN (per floor) Wall perimeter (centre-line) = 2 × (4.5 + 3.5) = 16.00 m Wall weight per metre of height = 16.00 × 0.35 × 20 = 112.0 kN per m height Wall weight of one storey = 112.0 × 3.0 = 336.0 kN

Seismic weight of each floor IS 1893 cl. 7.3

W = DL + n% of LL. For LL ≤ 3 kN/m², take 25%. Live load on the roof is NOT considered (cl. 7.3.2). Each wall is lumped half to the floor above and half to the floor below.

LEVEL 1 (first floor, h₁ = 3.0 m) Slab + finish = 47.25 + 15.75 = 63.00 kN 25% of floor LL = 0.25 × 3.0 × 15.75 = 11.81 kN Half wall below = 336.0 / 2 = 168.00 kN Half wall above = 336.0 / 2 = 168.00 kN ---------- W₁ = 410.81 kN LEVEL 2 (roof, h₂ = 6.0 m) Slab + finish = 47.25 + 15.75 = 63.00 kN Roof live load = 0.00 kN Half wall below = 336.0 / 2 = 168.00 kN ---------- W₂ = 231.00 kN TOTAL SEISMIC WEIGHT W = 410.81 + 231.00 = 641.81 kN

Fundamental natural period IS 1893 cl. 7.6.2

T = 0.09 h / √d (masonry, treated as infilled frame) h = total height = 6.0 m Along the SHORTER direction, d = 3.5 m T = 0.09 × 6.0 / √3.5 = 0.54 / 1.871 = 0.289 s Along the LONGER direction, d = 4.5 m T = 0.09 × 6.0 / √4.5 = 0.54 / 2.121 = 0.255 s Both lie in the range 0.10 ≤ T ≤ 0.55 s, so for MEDIUM (Type II) soil the spectrum is on the plateau: S_a / g = 2.50

Design horizontal seismic coefficient

Z Zone factor — Surkhet is in the highest seismic zone (equivalent to IS 1893 Zone V) = 0.36 I Importance factor — a HEALTH POST is a post- earthquake essential facility = 1.5 R Response reduction factor — load-bearing unreinforced masonry with horizontal bands = 3.0 check I/R = 1.5/3.0 = 0.5 ≤ 1.0 ✓ A_h = (Z/2) × (I/R) × (S_a/g) = (0.36/2) × (1.5/3.0) × 2.50 = 0.18 × 0.50 × 2.50 = 0.225

Design base shear IS 1893 cl. 7.5.3

V_B = A_h × W = 0.225 × 641.81 = 144.41 kN

Vertical distribution IS 1893 cl. 7.7.1

Q_i = V_B · W_i h_i²Σ W_j h_j²

Level	W_i (kN)	h_i (m)	h_i²	W_ih_i²	W_ih_i² / Σ	Q_i (kN)
2 — Roof	231.00	6.0	36	8316.0	0.6922	99.96
1 — First floor	410.81	3.0	9	3697.3	0.3078	44.45
Σ	641.81			12013.3	1.0000	144.41

Storey (floor level) shear

The storey shear at any level is the sum of all the lateral forces above that level:

Level	Lateral force Q_i (kN)	Storey shear V_i (kN)
Roof (level 2)	99.96	99.96
First floor (level 1)	44.45	144.41
Base	—	144.41 = V_B ✓

ANSWER — seismic weight W = 641.81 kN; A_h = 0.225; base shear V_B = 144.41 kN.
Earthquake forces: Q_roof = 99.96 kN and Q_{first floor} = 44.45 kN.
Floor level shears: V_roof = 99.96 kN, V_{first floor} = V_base = 144.41 kN.

Marks are lost on three things. (1) Including the roof live load — it must be excluded. (2) Forgetting to split the wall weight half up / half down. (3) Forgetting the importance factor — a health post is a post-earthquake essential facility, so I = 1.5, not 1.0. With I = 1.0 the base shear would be only 96.3 kN.

Method: Chapter 4 §4.3 (seismic weight, base shear, vertical distribution, with the worked pattern in §4.3.4).

4 (b) NumericalFor the same building of question 4(a), which is to be constructed with horizontal and vertical bands, calculate the load on the horizontal band at lintel level due to earthquake and dead load. (Use the appropriate data of 4a.) [6]

Carried over from 4(a): A_h = 0.225; wall thickness 0.35 m; masonry 20 kN/m³; storey height 3.0 m; plan 4.5 m × 3.5 m.
Assumed: lintel level at 2.1 m above floor level (standard door height); the band spans the longer wall, i.e. between cross walls L = 4.5 m.

Read the question carefully. The band carries two different loads acting in two perpendicular planes:
• a VERTICAL dead load — the masonry above it, exactly like a lintel;
• a HORIZONTAL earthquake load — the out-of-plane inertia of the strip of wall tributary to it, the band acting as a horizontal beam spanning between cross walls.
They must be computed separately and the band checked for biaxial bending.

Vertical (dead) load on the band

Check whether arching action relieves the band. The masonry above the band would arch over a triangle of apex height 0.866 L:

0.866 × L = 0.866 × 4.5 = 3.90 m Height of masonry actually above the lintel band = storey height − lintel level = 3.00 − 2.10 = 0.90 m 0.90 m < 3.90 m → the arch CANNOT form fully, so the band carries the WHOLE rectangle of wall above it. w_DL (masonry) = 0.90 × 0.35 × 20 = 6.30 kN/m Add the slab reaction if the slab bears on this wall = (0.12×25 + 1.0) × (3.5/2) = 7.00 kN/m ---------- Total vertical load w_v = 13.30 kN/m

Horizontal (earthquake) load on the band

The band supports the out-of-plane inertia of the strip of wall midway to the supports above and below:

Tributary height of wall for the band = half the wall above the band + half the wall below = (3.00 − 2.10)/2 + 2.10/2 = 0.45 + 1.05 = 1.50 m Weight of that wall strip per metre length = 1.50 × 0.35 × 20 = 10.50 kN/m Earthquake (inertia) load on the band w_EQ = A_h × 10.50 = 0.225 × 10.50 = 2.36 kN/m

Design moments in the band

The band is continuous over the cross walls, so take M = wL²/12:

Span between cross walls L = 4.5 m VERTICAL bending (dead load) M_v = w_v L² / 12 = 13.30 × 4.5² / 12 = 22.44 kN·m HORIZONTAL bending (earthquake) M_h = w_EQ L² / 12 = 2.36 × 4.5² / 12 = 3.98 kN·m Corresponding shears V_v = w_v L / 2 = 13.30 × 4.5 / 2 = 29.93 kN V_h = w_EQ L / 2 = 2.36 × 4.5 / 2 = 5.31 kN

ANSWER — load on the horizontal band at lintel level:
• vertical dead load w_v = 13.30 kN/m (6.30 kN/m from the masonry above + 7.00 kN/m slab reaction), giving M_v = 22.44 kN·m;
• horizontal earthquake load w_EQ = 2.36 kN/m, giving M_h = 3.98 kN·m.
The RC lintel band must therefore be designed for biaxial bending (M_v together with M_h), with the reinforcement of the door and window lintels provided in addition to the band steel.

Typical band detail to quote. For a two-storey building of this size, NBC 202 / IS 4326 would call for a lintel band of the full wall thickness × 75 mm deep with 2 – 10 mm bars top and bottom and 6 mm links at 150 mm centres, continuous around the whole building and properly lapped at the corners, together with vertical bars at all corners, T-junctions and opening jambs.

Band function and detailing: Chapter 2 §2.5.4 and Chapter 3 §3.8(g); arching action over openings: Chapter 4 §4.10.1.

2080 Bhadra — Regular

Paper Information

Question 1

1. Definition

2. Source of an earthquake

3. Factors influencing the shaking at a location

Part 1 — Failure mechanisms

(a) In-plane failure modes — four types

(b) Out-of-plane failure mechanisms — four types

(c) Other typical damage

Part 2 — Mitigation measures (keyed to the failure prevented)

Sketches to draw (the marks are here)

Question 2

1. Properties of brick masonry

2. Relation between unit and mortar on the strength of masonry

3. Factors affecting the compressive strength — four groups

Question 3

Question 4