Paper Information
| Detail | Information |
|---|---|
| Exam | 2076 Chaitra |
| Subject | Seismic Resistant Design of Masonry Structure (Elective I) — CE 72502 |
| Full Marks · Pass Marks · Time | 80 · 32 · 3 hours |
| Instructions | Attempt All questions. NBC 105, IS 1893 and IS 1905 are allowed. Assume suitable data if necessary. |
This is the “band and lintel” paper. Unlike the other years, two of the four
numericals are reinforced-element designs (a seismic lintel band and a reinforced brick masonry lintel)
rather than wall-capacity checks, and one is a full lateral-load analysis. Both reinforced designs are worked
below in full, with sketches.
Question 1
1 (a)
TheoryState and explain briefly the factors affecting the
compressive strength of brick masonry.
[8]
Group the answer into four heads — this structure alone earns most of the marks.
| Group | Factor | How it affects the strength |
|---|---|---|
| A. Properties of the unit |
Crushing strength of the unit | The single most important factor. Masonry strength is roughly proportional to fb0.7, but is only 25–50% of the unit strength. |
| Shape — height/width ratio as laid | A taller unit means fewer bed joints per metre and less lateral tension, so the masonry is stronger. Allowed for by the shape modification factor kp (IS 1905 Table 10). | |
| Frogs, perforations, voids | Reduce the net bearing area and cause stress concentrations. | |
| Water absorption / initial rate of suction | A dry, absorbent brick sucks water out of the mortar and destroys the bond — bricks must be soaked before laying. | |
| Uniformity of size and squareness | Irregular units force thick, uneven joints. | |
| B. Properties of the mortar |
Grade / strength of the mortar | Increases the masonry strength, but much less than proportionately: doubling the mortar strength may gain only 15–25%. Masonry strength ∝ fm0.3. |
| Deformability of the mortar | A softer mortar expands laterally more under load and puts more lateral tension into the unit → earlier vertical splitting. | |
| Workability, water retentivity, mix proportion | Richer than 1:3 shrinks and cracks; leaner than 1:6 is harsh and does not fill the joints. | |
| C. Geometry of the masonry |
Thickness of the mortar joint | Very significant. A 10 mm joint is standard; a 20 mm joint can lose 20–30% of the strength. |
| Slenderness ratio | Higher SR → buckling and secondary moments → permissible stress reduced by ks (Table 9). SR 27 gives only ~43% of the stress allowed at SR 6. | |
| Eccentricity of the load | Produces bending; the permissible stress is reduced through the same factor ks. | |
| Cross-sectional area | Small sections have no redundancy: ka = 0.7 + 1.5A for A < 0.2 m². | |
| D. Workmanship | Filling of the vertical (head) joints | The commonest defect on site; unfilled perpends can lose a third of the strength. |
| Bond pattern and corner interlocking | Continuous vertical joints split a wall into independent slender leaves. | |
| Verticality and uniformity of courses | An out-of-plumb wall introduces an unintended eccentricity. | |
| Curing, age and rate of construction | Building too fast loads green mortar; masonry gains strength for months. |
All of this is condensed by IS 1905 into
fper = fb · ks · ka · kp
where fb covers the unit and mortar (Table 8), ks covers slenderness and
eccentricity (Table 9), ka covers small area, and kp covers the unit shape (Table 10).
Finish with the mechanism: under compression the mortar tries to expand laterally more than the unit; friction restrains it, putting the mortar into triaxial compression and the unit into lateral tension, so failure starts as vertical splitting through the units. This is why the unit strength matters far more than the mortar strength, and why the mortar should be weaker than the unit.
Full treatment: Chapter 2 §2.3.4–2.3.5
(Fig. 2.8).
1 (b)
NumericalDesign a lintel band for a single room one
storey building of 4 m breadth, 6 m length and 3 m ceiling height. The brick masonry wall thickness is 220 mm
and the height of the lintel from the floor is 2.1 m. The roof is an RCC flat slab. The design earthquake
coefficient is 0.36. Neglect the openings on the walls. Show the design details in a sketch.
[12]
Given: single room, one storey; plan 4 m × 6 m (outer); ceiling height 3.0 m; wall t = 220 mm;
lintel band at 2.1 m above the floor; RCC flat slab roof; design earthquake coefficient
αh = 0.36; openings neglected.
Assume: masonry 20 kN/m³; concrete M20, steel Fe 415; band cast integrally with the wall over the full thickness (220 mm wide).
Critical band: the one on the 6 m long wall, which spans between the 4 m cross walls.
Assume: masonry 20 kN/m³; concrete M20, steel Fe 415; band cast integrally with the wall over the full thickness (220 mm wide).
Critical band: the one on the 6 m long wall, which spans between the 4 m cross walls.
What loads act on a lintel band?
A seismic band is a horizontal beam bent about TWO axes at the same time:
• Vertically — by the weight of the masonry it supports (the lintel action).
• Horizontally — by the out-of-plane inertia of the wall strip tributary to it; here the band bends in its own plane, using the wall thickness as the depth.
It must be checked for biaxial bending.
• Vertically — by the weight of the masonry it supports (the lintel action).
• Horizontally — by the out-of-plane inertia of the wall strip tributary to it; here the band bends in its own plane, using the wall thickness as the depth.
It must be checked for biaxial bending.
Vertical load on the band, per metre run
Masonry between the band (2.1 m) and the roof (3.0 m)
height = 3.0 − 2.1 = 0.90 m
w_v = 0.90 × 0.22 × 20 = 3.96 kN/m
(The openings are neglected, so the wall is continuous and
this is the only vertical load the band carries; the roof
slab bears directly on the top of the wall.)
Horizontal (seismic) load on the band, per metre run
Out of plane, the wall spans vertically between the floor, the band and the roof. The band takes half the wall on each side of it:
Tributary height of wall
below the band : 2.10 / 2 = 1.05 m
above the band : 0.90 / 2 = 0.45 m
--------
total tributary height = 1.50 m
Weight of that strip
W_strip = 1.50 × 0.22 × 20 = 6.60 kN/m
Seismic load
w_EQ = α_h × W_strip = 0.36 × 6.60 = 2.376 kN/m
≈ 2.38 kN/m
Design span and bending moments
The band on the 6 m wall spans between the 4 m cross walls.
Clear span = 6.00 − 0.22 = 5.78 m
Design span L (centre to centre) = 6.00 m
The band is continuous and monolithic with the cross walls
→ treat as a fixed / continuous beam : M = w L² / 12
HORIZONTAL bending (the design case for a seismic band)
M_h = 2.376 × 6.00² / 12 = 7.13 kN·m
V_h = 2.376 × 6.00 / 2 = 7.13 kN
VERTICAL bending
M_v = 3.96 × 6.00² / 12 = 11.88 kN·m
V_v = 3.96 × 6.00 / 2 = 11.88 kN
The vertical moment is the larger number, but the band is only 150 mm deep in that
direction and, with no openings, that load is actually carried by arching in the continuous
masonry, not by the band. The band is therefore proportioned for the horizontal
seismic moment, and the vertical bending is covered by providing equal steel on both faces plus the
IS 4326 minimum.
Section of the band
For HORIZONTAL bending the section is :
width b = depth of the band (to be fixed)
depth D = thickness of the wall = 220 mm
Try a band 220 mm wide × 150 mm deep
b = 150 mm , D = 220 mm
effective depth d = 220 − 25 (cover) − 6 (half bar)
= 189 ≈ 190 mm
Limiting moment (M20, Fe 415)
M_u,lim = 0.138 f_ck b d²
= 0.138 × 20 × 150 × 190²
= 14.94 × 10⁶ N·mm = 14.94 kN·m
Factored design moment
M_u = 1.5 × M_h = 1.5 × 7.13 = 10.69 kN·m
10.69 < 14.94 → singly reinforced ✓
Check: with the IS 4326 minimum band depth of 75 mm, Mu,lim = 0.138 × 20 × 75 × 190² = 7.47 kN·m < 10.69 kN·m — a 75 mm band is NOT enough for a 6 m span. Hence 150 mm.
Area of steel
M_u = 0.87 f_y A_st d [ 1 − A_st f_y /(b d f_ck) ]
10.69×10⁶ = 0.87 × 415 × A_st × 190
× [ 1 − A_st × 415/(150 × 190 × 20) ]
= 68 600 A_st [ 1 − A_st/1373.5 ]
Trial A_st = 180 mm² :
68 600 × 180 × (1 − 180/1373.5)
= 12.35×10⁶ × 0.869 = 10.73 kN·m ✓
Required A_st = 180 mm²
Minimum steel A_st,min = 0.85 b d / f_y
= 0.85 × 150 × 190 / 415 = 58 mm² < 180 OK
PROVIDE 2 nos 12 mm φ near EACH face of the wall
= 2 × 113 = 226 mm² per face > 180 ✓
(4 nos 12 mm φ in all)
Shear and links
V_u = 1.5 × 7.13 = 10.69 kN
τ_v = V_u /(b d) = 10 690 /(150 × 190) = 0.375 N/mm²
p_t = 100 × 226 /(150 × 190) = 0.79 %
τ_c (M20, p_t = 0.79%) ≈ 0.57 N/mm²
τ_v = 0.375 < τ_c = 0.57 → nominal links only ✓
PROVIDE 6 mm φ links @ 150 mm c/c throughout
Design details — sketch
Fig. Q1b — Lintel band: section through the wall showing the 220 × 150 mm band with
4 nos 12 mm bars and 6 mm links, and plan showing the band continuous around the whole building.
DESIGN ADOPTED — lintel band 220 mm wide × 150 mm deep, M20 concrete, cast
monolithically at 2.1 m above floor level, reinforced with 4 nos 12 mm φ Fe 415 bars
(2 near each face) and 6 mm φ links at 150 mm centres. The band must be
continuous around the whole building, with L-shaped corner bars lapped
60φ = 720 mm at every corner and junction, and it must never be interrupted at a door
or window.
Detailing rules to write down (IS 4326 cl. 8.4 / NBC 202): minimum band 75 mm deep and
the full wall thickness; minimum 2 longitudinal bars of 8 mm φ with 6 mm links at 150 mm; the band must
be continuous and at the same level all round; at corners provide L-bars of the same diameter with
a lap of 60φ; the lintel band may be combined with the roof band if the two are within 300 mm of each
other; concrete not leaner than M15.
Method: Chapter 4 §4.9;
band function and detailing in Chapter 3 §3.7.6.
Question 2
2 (a)
TheoryExplain the design criteria of walls subjected to
transverse loading.
[8]
“Transverse loading” means load acting perpendicular to the plane of the wall — wind, earthquake inertia, earth pressure or water pressure. IS 1905 lays down the following criteria.
- Slenderness ratio. The SR of the wall must not exceed the limits of Table 7 (27 for a load-bearing wall in cement mortar, 30 for a panel or curtain wall, 10 for a free-standing wall or parapet). This is the first and most important control on out-of-plane behaviour, because the out-of-plane capacity depends primarily on h/t.
- Flexural tensile stress must be within the permissible value. IS 1905 cl. 5.4.2 permits, for cement mortars M1 or richer, 0.07 N/mm² for bending in the vertical direction (tension across the bed joints, i.e. when the wall spans between floor and roof) and 0.14 N/mm² for bending in the horizontal direction (tension across the head joints, i.e. when the wall spans between cross walls). No tension whatever is allowed in an earthquake combination in lime or mud mortar, and no tension is permitted in the direction normal to the bed joint in the case of a DPC or a similar slip plane.
- Precompression must be used first. The correct design approach is to show that the flexural tensile stress fcb is cancelled by the axial compressive stress fca, i.e. fcb − fca ≤ ft,per. If fcb < fca no tension develops at all and the wall is safe by inspection. The critical section for tension is usually at mid-height, where the moment is high but the precompression is only about half its base value.
- Maximum compressive stress. On the compression face, fca + fcb must not exceed the permissible compressive stress, which may be increased by 25% for combined axial and bending stresses.
- Increase in permissible stresses for wind or earthquake. When the transverse load is wind or earthquake, all permissible stresses may be increased by 33⅓% (alternatively, design for 0.75 times the combined load).
- Effective span and the moment coefficient. The moment depends on the support conditions and must be stated: M = pH²/8 for a wall simply supported top and bottom; M = pH²/10 for a wall with partial restraint, e.g. anchored to the roof with metal anchors; M = pH²/12 for full continuity; M = pH²/2 for a wall free at the top (a parapet or free-standing wall — the worst case). If the wall is supported on all four sides it may be designed for two-way bending using the yield-line coefficients.
- Arching action. Where the wall is built tight between rigid supports so that it cannot move away from them, arching action may be relied on to carry the transverse load, and the wall is then checked in compression instead of tension.
- Provision of lateral supports. Cross walls, piers, buttresses, bands and a rigid diaphragm must be provided at a spacing that keeps the effective span — and hence the SR — within limits. Anchoring the wall to the floor and roof is mandatory; an unanchored wall behaves as a free cantilever with five times the moment.
- Stability against overturning and sliding at the base and at the DPC must be checked separately, with a factor of safety of at least 1.5 (or 2.0 for a free-standing wall).
- If the tension cannot be eliminated, the options are: increase the thickness, reduce the span by adding cross walls or piers, add precompression, or provide reinforcement — i.e. change from unreinforced to reinforced masonry.
Full treatment: Chapter 2 §2.6.9 and
Chapter 4 §4.8.
2 (b)
NumericalDesign a reinforced brick masonry lintel
subjected to triangular loading for a window opening of span 1.6 m. The wall thickness is 220 mm and the height
of brickwork above the lintel is 0.9 m. The length of the wall on either side of the lintel is more than half
the span.
[12]
Given: clear span of opening = 1.6 m; wall t = 220 mm; height of brickwork above the lintel
= 0.9 m; the wall on either side extends more than half the span, so arching action can develop.
Assume: masonry 20 kN/m³; concrete/brickwork in the lintel taken as reinforced brickwork with a permissible bending compressive stress σcbc = 1.5 N/mm²; Fe 415 with σst = 230 N/mm²; no floor or roof load falls within the triangle.
Assume: masonry 20 kN/m³; concrete/brickwork in the lintel taken as reinforced brickwork with a permissible bending compressive stress σcbc = 1.5 N/mm²; Fe 415 with σst = 230 N/mm²; no floor or roof load falls within the triangle.
Check that arching action develops
When the wall on either side of the opening extends beyond the springing by at least half the
span, the masonry above the lintel arches over the opening and the lintel carries only the
triangle of masonry enclosed by lines drawn at 45° from the ends of the bearing.
Effective span L = 1.60 m
Height of the 45° load triangle
h_t = L / 2 = 1.60 / 2 = 0.80 m
Height of brickwork actually available above the lintel
= 0.90 m
0.90 m > 0.80 m → the full triangle FITS inside the
available masonry, so the ARCHING ACTION IS COMPLETE and the
loading is genuinely TRIANGULAR. ✓
State the alternative: if the available masonry had been less than ht, the arch could not form and the lintel would have to be designed for the full rectangular load of the masonry above it, plus any floor or roof load landing in that zone.
Load on the lintel
Weight of the load triangle
W = ½ × base × height × thickness × γ
= 0.5 × 1.60 × 0.80 × 0.22 × 20
= 0.5 × 1.60 × 3.52 = 2.816 kN
Assume a lintel section 220 mm wide × 230 mm deep
(3 brick courses + joints)
Self weight of the lintel
= 0.22 × 0.23 × 20 = 1.012 kN/m
over the span = 1.012 × 1.60 = 1.619 kN
Bending moment and shear
For a triangular load of total magnitude W, peaking at mid-span, on a simply supported span L:
Mmax = WL/6 and Vmax = W/2.
For the uniformly distributed self weight: M = wL²/8, V = wL/2.
For the uniformly distributed self weight: M = wL²/8, V = wL/2.
From the triangular masonry load
M₁ = W L / 6 = 2.816 × 1.60 / 6 = 0.751 kN·m
V₁ = W / 2 = 2.816 / 2 = 1.408 kN
From the self weight
M₂ = w L² / 8 = 1.012 × 1.60² / 8 = 0.324 kN·m
V₂ = w L / 2 = 1.012 × 1.60 / 2 = 0.810 kN
-----------
TOTAL M = 0.751 + 0.324 = 1.075 kN·m
TOTAL V = 1.408 + 0.810 = 2.218 kN
Working-stress design constants
σ_cbc = 1.5 N/mm² (reinforced brickwork in bending)
σ_st = 230 N/mm² (Fe 415)
Modular ratio m = 280 / (3 σ_cbc) = 280/(3 × 1.5) = 62.2
Neutral axis depth factor
k = mσ_cbc /(mσ_cbc + σ_st)
= 62.2×1.5 /(62.2×1.5 + 230) = 93.3/323.3 = 0.289
Lever arm factor
j = 1 − k/3 = 1 − 0.289/3 = 0.904
Moment of resistance factor
R = ½ σ_cbc k j = 0.5 × 1.5 × 0.289 × 0.904 = 0.196
Required effective depth
d_req = √( M / (R b) )
= √( 1.075×10⁶ / (0.196 × 220) )
= √( 1.075×10⁶ / 43.1 )
= √ 24 930 = 158 mm
PROVIDE overall depth D = 230 mm (three brick courses)
cover 25 mm , bar 10 mm
d = 230 − 25 − 5 = 200 mm
200 > 158 ✓
Area of tension steel
A_st = M / (σ_st · j · d)
= 1.075×10⁶ / (230 × 0.904 × 200)
= 1.075×10⁶ / 41 584 = 25.9 mm²
This is far below the practical minimum, so provide the
detailing minimum :
PROVIDE 2 nos 10 mm φ Fe 415 = 2 × 78.5 = 157 mm² ✓
(one bar in each of the two outer vertical joints
of the bottom course, fully grouted in 1:3 mortar)
Shear check
τ_v = V / (b j d) = 2218 / (220 × 0.904 × 200)
= 2218 / 39 776 = 0.056 N/mm²
Permissible shear stress in reinforced brickwork ≈ 0.10 N/mm²
0.056 < 0.10 → no shear reinforcement required ✓
Provide nominal 6 mm φ links / MS flats @ 150 mm c/c
to hold the bars in position.
Bearing check
Minimum bearing at each end = 200 mm
End reaction R = V = 2.218 kN
Bearing stress = 2218 / (220 × 200) = 0.050 N/mm²
Permissible bearing (≈ 1.5 × f_per) > 0.5 N/mm² ✓
Overall length of lintel = 1.60 + 2 × 0.20 = 2.00 m
Design details — sketch
Fig. Q2b — Reinforced brick masonry lintel: the 45° dispersion triangle of masonry
carried by the lintel, and the section showing three brick courses with two 10 mm bars grouted into the
bottom vertical joints.
DESIGN ADOPTED — reinforced brick masonry lintel 220 mm wide × 230 mm deep
(three brick courses), overall length 2.00 m with 200 mm bearing at each end, reinforced with
2 nos 10 mm φ Fe 415 bars laid in the two outer vertical joints of the bottom course and
fully grouted with 1:3 cement mortar, with nominal 6 mm φ links at 150 mm centres. The lintel carries the
45° triangle of masonry, W = 2.82 kN, giving M = 1.075 kN·m and a required
effective depth of only 158 mm against the 200 mm provided.
Method: Chapter 4 §4.10 (arching action) and
Chapter 2 §2.5.5.
Question 3
3 (a)
TheoryDiscuss in brief the failure mechanism of masonry
structures and remedial measures.
[10]
Answer this in the form of a table that pairs each failure mechanism with the measure that prevents it — that is what earns full marks.
| Failure mechanism | Cause | Remedial / preventive measure |
|---|---|---|
| Out-of-plane overturning of walls (the dominant killer) | No connection to the floor/roof; no bands; slender wall; flexible diaphragm | Horizontal bands at plinth, sill, lintel, floor and roof; anchor the roof and floor to the walls; reduce the span with cross walls; stiffen the diaphragm |
| Corner separation, vertical cracks at wall junctions | Poor interlocking of courses at the corner; walls built at different times | Proper corner interlocking of alternate courses; L-shaped band bars lapped 60φ at corners; vertical bars at every corner and T-junction |
| Delamination of multi-leaf stone or rubble walls | Two independent leaves with a rubble core, no cross ties | Through-stones or S-shaped/hooked ties every 600 mm vertically and 1.2 m horizontally; grout injection; ferrocement or wire-mesh jacketing |
| In-plane diagonal (X) shear cracking of piers | Storey shear exceeds the diagonal tensile strength; low precompression; weak mortar | Increase the wall length or thickness; restrict the openings; improve the mortar; add splint-and-bandage or a shotcrete/ferrocement overlay; add vertical reinforcement |
| Rocking / toe crushing of slender piers | High aspect ratio, low vertical load | Widen the pier; add vertical steel and confining tie-columns to develop tension at the heel |
| Sliding shear along a bed joint or DPC | Low friction, weak mortar, smooth DPC membrane | Improve the mortar; add dowels / vertical bars across the joint; avoid a smooth slip plane at the DPC |
| Cracking at openings (corners of doors/windows) | Stress concentration; openings too large, too close to a corner, or unbanded | Follow the NBC/IS 4326 opening rules: total width of openings ≤ 50% of the wall length; a pier of at least 600 mm between openings; openings at least 600 mm from a corner; sill and lintel bands and vertical bars at the jambs |
| Gable and parapet collapse | Free-standing cantilever with no top restraint | Gable band along the slope tied to the roof band; limit the parapet height; add vertical bars |
| Roof / floor separation and collapse | Joists simply resting on the wall; heavy tiled or earth roof | Anchor the joists to the wall or band; brace the roof; replace a heavy roof with a light CGI roof (reduces the demand) |
| Soft storey / short column | Open ground storey; partial-height infill; mezzanines | Add infill or shear walls at ground level; full-height or fully separated infill; strengthen the ground-storey columns |
| Pounding between adjacent buildings | Inadequate separation gap | Provide the code separation gap; align the floor levels |
| Torsional failure | Eccentricity between the centre of mass and the centre of rigidity | Symmetric plan; distribute the walls uniformly; add walls to move the centre of rigidity towards the centre of mass |
| Foundation failure / differential settlement | Weak or liquefiable soil, sloping ground | Continuous plinth band; ground improvement; uniform bearing; avoid a split-level foundation |
The four principles behind every one of these measures:
(1) Integrity — tie the whole building into a single box;
(2) Strength — enough wall area, good mortar, no oversized openings;
(3) Ductility — reinforcement and confinement so it cracks instead of collapsing;
(4) Reduced demand — light roof, regular symmetric plan, limited height.
Full treatment with figures: Chapter 3 §3.6 and
§3.7.
3 (b)
NumericalDesign an interior wall of a two storeyed
building to carry a 120 mm thick RCC floor and roof slab with 3 m ceiling height. The wall is unstiffened and
it supports 4 m wide slabs. Live load on the roof and floor is 2 kN/m².
[10]
Given: two storeys; RCC slab 120 mm on both floor and roof; ceiling height 3.0 m; wall
unstiffened; supports 4 m wide slabs on both sides; LL = 2 kN/m² on roof and floor.
Assume: wall thickness t = 230 mm; RCC 25 kN/m³; masonry 20 kN/m³; floor finish 1 kN/m²; interior wall with equal spans on both sides → the slab reactions balance and e = 0. Design the ground-floor wall (worst case), per metre run.
Assume: wall thickness t = 230 mm; RCC 25 kN/m³; masonry 20 kN/m³; floor finish 1 kN/m²; interior wall with equal spans on both sides → the slab reactions balance and e = 0. Design the ground-floor wall (worst case), per metre run.
Slab loads
Slab dead load = 0.120 × 25 = 3.00 kN/m²
Floor finish = 1.00 kN/m²
Live load = 2.00 kN/m²
-----------
Total on roof and on floor = 6.00 kN/m²
Vertical load on the wall, per metre run
Tributary width = 4/2 + 4/2 (slabs on both sides) = 4.00 m
From the ROOF 6.00 × 4.00 = 24.00 kN/m
From the FLOOR 6.00 × 4.00 = 24.00 kN/m
Wall of the upper storey
0.23 × 3.12 × 20 = 14.35 kN/m
Wall of the ground storey (self weight)
0.23 × 3.12 × 20 = 14.35 kN/m
------------
Total at the base, W = 76.70 kN/m
Actual compressive stress
A = 1000 × 230 = 230 000 mm²
f_c = W / A = 76 700 / 230 000 = 0.333 N/mm²
Effective height IS 1905 Table 4
Storey height H = clear 3.00 + slab 0.12 = 3.12 m
RCC slabs bearing on the full thickness top and bottom
→ full lateral restraint
h_eff = 0.75 H = 0.75 × 3120 = 2340 mm
Effective thickness and slenderness ratio
The wall is UNSTIFFENED → no piers, no cross walls
t_eff = t = 230 mm
Long wall, so the length-based ratio does not govern.
SR = h_eff / t_eff = 2340 / 230 = 10.17
Limit for a load-bearing wall in cement mortar = 27 ✓
Modification factors
k_s Table 9 , SR = 10.17 , e = 0
SR 10 → 0.89 , SR 12 → 0.84
k_s = 0.89 − (0.89−0.84)(10.17−10)/(12−10) = 0.885
k_a A = 1.0 × 0.23 = 0.23 m² > 0.2 m² → k_a = 1.0
k_p brick 230×115×57 , h/w = 0.50 < 0.75 → k_p = 1.0
Required basic compressive stress and selection
Required f_b ≥ f_c / (k_s k_a k_p)
= 0.333 / (0.885 × 1.0 × 1.0) = 0.376 N/mm²
From Table 8 :
unit 3.5 N/mm² + M1 f_b = 0.35 ✘ short
unit 5.0 N/mm² + M3 f_b = 0.41 ✓
unit 5.0 N/mm² + M2 (1:6) f_b = 0.44 ✓ ADOPT
unit 5.0 N/mm² + M1 (1:5) f_b = 0.50 ✓ safer
Final check
With unit 5 N/mm² + M2 mortar :
f_per = 0.44 × 0.885 × 1.0 × 1.0 = 0.389 N/mm²
f_c = 0.333 N/mm² < f_per = 0.389 N/mm² ✓
Utilisation = 86 %
With unit 5 N/mm² + M1 mortar :
f_per = 0.50 × 0.885 = 0.443 N/mm²
Utilisation = 75 %
SAFE. Adopt a 230 mm (one brick) thick interior wall built with bricks of
crushing strength 5 N/mm² in M2 (1:6) cement mortar;
fper = 0.389 N/mm² against an actual stress of 0.333 N/mm². Use M1 (1:5) mortar for a
more comfortable margin.
Why e = 0 here. The wall is interior and carries slabs of equal span on both sides,
so the two slab reactions act at equal and opposite eccentricities and cancel. If the spans had been
unequal, the resultant eccentricity would have to be found from
e = (R1e1 − R2e2)/(R1 + R2 + Wwall)
and ks read from the corresponding column of Table 9.
Method: Chapter 2 §2.6.4–2.6.8.
Question 4
4 (a)
NumericalDetermine the seismic coefficient and storey
seismic forces for a single room double storey health post building of 4 m breadth, 6 m length and 3 m ceiling
height located at Kathmandu. The brick masonry wall thickness is 220 mm and the height of the lintel from the
floor is 2.1 m. The floor and roof are RCC flat slab 15 cm thick. Neglect the openings on the walls. The
foundation rests on hard soil.
[12]
Given: two-storey health post, single room; plan 6.0 m × 4.0 m (outer); ceiling height 3.0 m per
storey; wall t = 220 mm; RCC flat slab 150 mm at floor and roof; Kathmandu; hard soil; openings
neglected.
Assume: RCC 25 kN/m³; masonry 20 kN/m³; floor finish 1 kN/m²; LL on floor 2 kN/m²; unreinforced masonry with horizontal bands, R = 3.0.
Assume: RCC 25 kN/m³; masonry 20 kN/m³; floor finish 1 kN/m²; LL on floor 2 kN/m²; unreinforced masonry with horizontal bands, R = 3.0.
Geometry
Storey height h = 3.00 + 0.15 = 3.15 m
Level 1 (first floor) h₁ = 3.15 m
Level 2 (roof) h₂ = 6.30 m
Total height above base H = 6.30 m
Slab plan area = 6.00 × 4.00 = 24.00 m²
Centre-line length of walls
= 2 × [ (6.00 − 0.22) + (4.00 − 0.22) ]
= 2 × (5.78 + 3.78) = 2 × 9.56 = 19.12 m
Wall weight per m² of elevation = 0.22 × 20 = 4.40 kN/m²
Slab loads
Slab dead load = 0.15 × 25 = 3.75 kN/m²
Floor finish = 1.00 kN/m²
-----------
Dead load on roof and on floor = 4.75 kN/m²
Live load on the floor = 2 kN/m² ≤ 3 kN/m²
→ 25 % is taken in the seismic weight
= 0.25 × 2.00 = 0.50 kN/m²
Live load on the ROOF is NOT considered = 0
Seismic weight at the roof level, W₂
Roof slab 4.75 × 24.00 = 114.00 kN
Walls — half the storey height below the roof
19.12 × (3.15/2) × 4.40
= 19.12 × 1.575 × 4.40 = 132.51 kN
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W₂ = 246.51 kN
Seismic weight at the first-floor level, W₁
Floor slab DL 4.75 × 24.00 = 114.00 kN
25 % of LL 0.50 × 24.00 = 12.00 kN
Walls — half a storey above + half a storey below
19.12 × (1.575 + 1.575) × 4.40
= 19.12 × 3.15 × 4.40 = 265.02 kN
-----------
W₁ = 391.02 kN
TOTAL SEISMIC WEIGHT
W = W₁ + W₂ = 391.02 + 246.51 = 637.53 kN
Fundamental natural period IS 1893 cl. 7.6.2
For a building with brick masonry infill / masonry walls:
T = 0.09 h√d
where h = total height and d = base dimension in the direction considered.
Shorter direction d = 4.00 m (critical)
T = 0.09 × 6.30 / √4.00 = 0.567 / 2.00 = 0.284 s
Longer direction d = 6.00 m
T = 0.567 / 2.449 = 0.231 s
Take the CRITICAL value T = 0.284 s
Design parameters
Z Kathmandu is in Seismic Zone V = 0.36
I Health post — important building = 1.5
R Unreinforced masonry with horizontal bands = 3.0
S_a/g HARD soil (Type I), 0.10 ≤ T ≤ 0.40 s
→ S_a/g = 2.5 = 2.5
Design horizontal seismic coefficient
Ah =
Z2 ·
IR ·
Sag
A_h = (0.36/2) × (1.5/3.0) × 2.5
= 0.18 × 0.50 × 2.5
= 0.225
Design base shear
V_B = A_h × W = 0.225 × 637.53 = 143.44 kN
Vertical distribution of the base shear IS 1893 cl. 7.7.1
Qi = VB ·
Wi hi2
Σ Wj hj2
Level W_i (kN) h_i (m) h_i² W_i h_i²
-------------------------------------------------------
1st 391.02 3.15 9.92 3 879.7
Roof 246.51 6.30 39.69 9 783.9
-------------------------------------------------------
Σ W_i h_i² = 13 663.6
Q₁ (first floor) = 143.44 × 3 879.7 / 13 663.6
= 143.44 × 0.2839 = 40.72 kN
Q₂ (roof) = 143.44 × 9 783.9 / 13 663.6
= 143.44 × 0.7161 = 102.72 kN
----------
check : 40.72 + 102.72 = 143.44 kN ✓
Storey shears
Upper storey V₂ = Q₂ = 102.72 kN
Ground storey V₁ = Q₂ + Q₁ = 102.72 + 40.72 = 143.44 kN (= V_B) ✓
ANSWER
Design horizontal seismic coefficient Ah = 0.225
Total seismic weight W = 637.53 kN · Design base shear VB = 143.44 kN
Storey seismic forces: roof Q2 = 102.72 kN, first floor Q1 = 40.72 kN
Storey shears: upper storey 102.72 kN, ground storey 143.44 kN.
Design horizontal seismic coefficient Ah = 0.225
Total seismic weight W = 637.53 kN · Design base shear VB = 143.44 kN
Storey seismic forces: roof Q2 = 102.72 kN, first floor Q1 = 40.72 kN
Storey shears: upper storey 102.72 kN, ground storey 143.44 kN.
Two marks that are easy to lose. (1) Hard soil is stated deliberately — for
Type I soil Sa/g = 2.5 over 0.10–0.40 s, so with T = 0.284 s you sit on the flat plateau;
do not read a sloping branch. (2) Live load on the roof is never included in the seismic weight,
and only 25% of a floor LL ≤ 3 kN/m² is included.
Method: Chapter 4 §4.3 and §4.4
(the 6-step lateral load analysis).
4 (b)
TheoryExplain the following as applied to masonry
structures: (i) Stress reduction factor (ii) Confined masonry.
[4×2]
(i) Stress reduction factor, ks
The stress reduction factor is the factor by which the basic compressive stress
fb of a masonry element must be multiplied to allow for the combined effect of its
slenderness ratio and the eccentricity of the load acting on it.
fper = fb · ks · ka · kp IS 1905 Table 9
fper = fb · ks · ka · kp IS 1905 Table 9
- Why it is needed: the basic stress fb of Table 8 is measured on a short, axially loaded prism. A real wall is slender, so it buckles and develops secondary moments, and the load on it is usually eccentric, so the stress block is not uniform. Both effects reduce the load the wall can carry.
- How it is used: enter Table 9 with the slenderness ratio (rows, SR = 6 to 27) and the eccentricity ratio e/t (columns, 0, 1/24, 1/12, 1/6, 1/4, 1/3) and interpolate linearly between the tabulated values.
- Range: ks = 1.00 at SR = 6 with e = 0, falling to 0.43 at SR = 27 with e = 0 and to only 0.22 at SR = 27 with e/t = 1/6. A slender eccentrically loaded wall may be worth less than a quarter of a short axially loaded one.
- Companion rules: if e ≤ t/24 the eccentricity may be ignored; if t/24 < e ≤ t/6 the stress is reduced through ks; if e > t/6 tension develops, part of the section becomes ineffective, and the stress must be found from fc = 2P/[3L(t/2 − e)].
- Blank entries in Table 9 (e.g. SR 24 with e/t = 1/3) mean that the combination is not permitted.
(ii) Confined masonry
Confined masonry consists of load-bearing masonry wall panels confined on all four sides by
lightly reinforced concrete tie-columns (vertical) and tie-beams (horizontal).
The defining feature is the construction sequence: the masonry wall is built FIRST, with toothed vertical edges, and the tie-columns are cast afterwards against the completed wall. (In an infilled RC frame it is the opposite — frame first, wall afterwards.)
The defining feature is the construction sequence: the masonry wall is built FIRST, with toothed vertical edges, and the tie-columns are cast afterwards against the completed wall. (In an infilled RC frame it is the opposite — frame first, wall afterwards.)
- Components: masonry walls (the main load-carrying element), confining tie-columns and tie-beams, floor and roof slabs acting as diaphragms, a plinth band, and the foundation.
- The ties are NOT a moment frame. They are much smaller than RC frame columns (typically 230 × 230 mm with 4 nos 12 mm bars) and are designed only to confine the masonry, not to carry the gravity load by themselves.
- The confining elements are effective in: (a) enhancing the stability and integrity of the wall under both in-plane and out-of-plane earthquake loads — they hold the cracked masonry together; (b) enhancing the lateral strength and stiffness of the wall; (c) reducing the brittleness of the masonry and hence improving its earthquake performance.
- Where the ties are placed: at every wall intersection, at both sides of every large opening, at free ends of walls, and at a spacing not exceeding about 4 m or 1.5 times the wall height.
- Position in the spectrum: confined masonry is an economical intermediate system between unreinforced masonry and RC frame construction, well suited to 1–4 storey buildings, and it uses the same materials and skills as ordinary masonry.
See Chapter 2 §2.6.7(b) for ks
(Table 9 reproduced as Fig. 2.18) and Chapter 2 §2.4.2(B) for
confined masonry (Figs. 2.10–2.12).