Paper Information
| Detail | Information |
|---|---|
| Exam | 2075 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. Codes are allowed. Assume suitable data if necessary. |
Only six questions, none subdivided. Each is worth 10–18 marks, so every answer must be
long and complete — a two-line answer to Q1 loses 8 marks. Note that Q2 asks for the failure modes
under BOTH static and earthquake load; answering only the seismic modes loses half the marks.
Question 1
1
TheoryDiscuss the design criteria and the design steps
involved in the design of masonry shear walls.
[10]
Part A — Design criteria
A masonry shear wall is the element that carries the in-plane horizontal storey shear down to the foundation. It must satisfy four criteria:
- Shear. The developed shear stress τ = V/A must not exceed the permissible shear stress, which for masonry increases with the precompression: τper = 0.1 + fd/6 ≤ 0.5 N/mm² where fd is the compressive stress due to the dead load only. Note that gravity load helps — a heavily loaded wall has more shear capacity.
- Compression. The maximum combined stress fc = fca + fcb must not exceed the permissible compressive stress. For combined axial and bending stress the permissible value may be increased by 25%, i.e. facb = 1.25 ks fb. The axial stress alone must also satisfy fca ≤ fac = ks fb.
- Tension. The flexural tensile stress ft = fcb − fca must be within the permissible tensile stress (0.07 N/mm² vertical, 0.14 N/mm² horizontal, for M1 or richer cement mortar). Ideally the design should show that no tension develops at all, i.e. that the precompression exceeds the bending stress.
- Stability — overturning and sliding. The restoring moment from the gravity load must exceed the overturning moment with a factor of safety of at least 1.5, and the base friction must exceed the base shear.
To these are added the general design principles of masonry buildings: a symmetric and regular plan; shear walls in both orthogonal directions; walls continuous from foundation to roof; the centre of rigidity as close as possible to the centre of mass to limit torsion; openings restricted in size and position; and a rigid diaphragm to distribute the shear.
Part B — Design steps
| Step | Operation | Key relations |
|---|---|---|
| 1 | Fix the trial geometry and section properties | A = L t · Z = t L²/6 (bending is in the plane, so L is the depth) |
| 2 | Compute the vertical (gravity) load W on the wall | self weight + tributary floor / roof load |
| 3 | Compute the lateral force F on the wall | F = AhW for a single wall, or the share of the storey shear Vi = V · Ri/ΣR plus the torsional share |
| 4 | Check the shear stress | τ = F/A ≤ τper = 0.1 + fd/6 |
| 5 | Compute the in-plane bending moment | M = F·H/2 for a load distributed over the height; M = F·H for a load concentrated at the roof |
| 6 | Compute the axial and flexural stresses | fca = W/A · fcb = M/Z · fc = fca + fcb · ft = fcb − fca |
| 7 | Compute the slenderness ratio and the equivalent eccentricity | SR = heff/teff · e = M/W · e/L |
| 8 | Read the stress reduction factor | ks from IS 1905 Table 9, interpolating on both SR and e/t |
| 9 | Select the unit strength and mortar grade | fb from IS 1905 Table 8; apply ka and kp |
| 10 | Final checks — if any fails, increase t, increase L, or use a stronger unit/mortar and repeat | fca ≤ ksfb · fc ≤ 1.25 ksfb · τ ≤ τper · ft ≤ ft,per |
Remember the reversal. Earthquake load reverses direction, so both ends of the wall must be
checked for maximum compression and for tension; the design must be symmetric.
Question 2
2
TheoryExplain the different modes of failure in brick
masonry structures with sketches for both static load and earthquake load.
[10]
The question has two halves. Give the static (gravity / service) failure
modes first, then the earthquake failure modes. Most candidates write only the seismic
modes and lose half the marks.
A. Failure modes under STATIC (gravity and service) load
| Mode | Cause | Appearance / sketch |
|---|---|---|
| Vertical splitting (the classic compression failure) | The mortar expands laterally more than the unit; the unit is put into lateral tension and splits. | Continuous vertical cracks running through the bricks, roughly parallel to the load. This is the mode on which the basic compressive stress fb is based. |
| Local crushing / bearing failure | A concentrated load (a beam or joist end) applied over a small area. | Crushed and spalled masonry directly under the bearing, with cracks radiating downward. |
| Buckling of a slender wall | High slenderness ratio; secondary moments amplify the deflection. | The wall bows out of plane and fails by combined compression and bending — controlled by ks and the SR limits. |
| Flexural tension failure from eccentricity | e > t/6, so the resultant falls outside the middle third. | Horizontal crack opening on the tension face; only part of the section remains effective, fc = 2P/[3L(t/2 − e)]. |
| Differential settlement cracking | Non-uniform bearing capacity, soft spot, adjacent excavation. | Diagonal cracks widening upward away from the settled portion; sloping stepped cracks over openings. |
| Shrinkage and thermal cracking | Drying shrinkage of the mortar or blocks; thermal movement of a slab over a wall. | Vertical cracks at regular spacing; horizontal crack under the roof slab at the top of the wall. |
| Sliding at the DPC | A smooth membrane forms a slip plane with very low friction. | Horizontal offset at plinth level. No tension whatever may be assumed across a DPC. |
| Arch / vault thrust failure | The horizontal thrust of an arch or vault pushes the supporting wall outwards. | The wall leans outward with a hinge crack; the arch drops. |
| Durability failure | Frost, sulphate attack, efflorescence, dampness. | Spalling of the brick face, powdering mortar, white salt deposits. |
B. Failure modes under EARTHQUAKE load
B1 — In-plane modes (the wall acting as a shear wall):
| Mode | When it governs | Crack pattern |
|---|---|---|
| Rocking (flexural) | Slender pier (high h/L), low vertical load | Horizontal crack at the heel; the pier rotates about the compressed toe; crushing at the toe. Ductile — the best mode to have. |
| Sliding shear | Very low vertical load, weak mortar, low friction | A single horizontal crack along a bed joint with a visible offset. |
| Diagonal shear — bed-joint sliding | Mortar weaker than the unit | Stair-stepped diagonal crack following the joints; the bricks are undamaged. |
| Diagonal tension + toe crushing | Mortar stronger than the unit, high precompression, squat pier | Straight diagonal crack cutting through the bricks; under reversal the two directions form the characteristic “X” crack. Brittle — the dangerous mode. |
B2 — Out-of-plane modes (the wall loaded normal to its plane):
- Simple overturning — the wall rotates about its base as a rigid body and falls outward. Occurs when there is no connection to the return walls or the diaphragm.
- Composed overturning — the falling wall drags a wedge of the return walls with it, so the failure surface is diagonal on the side walls.
- Vertical flexure — the wall is held at the top and bottom but spans too far; a horizontal hinge crack forms at mid-height and the wall folds outward.
- Horizontal flexure — the wall spans horizontally between cross walls; a vertical crack forms at mid-span and the two halves rotate outward.
B3 — System / connection modes: corner separation; delamination of multi-leaf walls; gable and parapet collapse; roof or floor separation; soft-storey and short-column failure; pounding; torsional failure; and liquefaction or settlement of the foundation.
Sketches to draw
Draw two rows of small labelled sketches:
- Static row: a prism under compression with vertical splitting cracks through the bricks; a beam bearing on a wall with local crushing; a slender wall bowing; a wall with settlement cracks widening upward.
- Seismic row: four piers showing rocking, sliding, stair-stepped diagonal and X-shaped diagonal cracking; then a building isometric with the out-of-plane wall overturning, corner separation and the roof pulling away.
Full treatment with all figures: Chapter 3 §3.6
(static in §3.6.2, seismic in §3.6.3–3.6.4) and §3.7.
Question 3
3
TheoryExplain the following and state the influence of it
on the strength of the wall: (i) Slenderness ratio (ii) Shape modification factor.
[12]
(i) Slenderness ratio (SR)
SR = effective height heffeffective thickness teff
or
effective length leffeffective thickness teff
For a WALL, take the LESSER of the two (the wall will buckle in whichever direction is easier — but it is supported in the other, so the smaller ratio governs).
For a COLUMN, take the GREATER of heff/teff in the two principal directions.
For a WALL, take the LESSER of the two (the wall will buckle in whichever direction is easier — but it is supported in the other, so the smaller ratio governs).
For a COLUMN, take the GREATER of heff/teff in the two principal directions.
How the three quantities are obtained:
| Quantity | Condition | Value |
|---|---|---|
| heff Table 4 |
Lateral restraint at both ends, resistance to rotation (RCC slab bearing on the full thickness) | 0.75 H |
| Lateral restraint, no resistance to rotation (timber joists, metal anchors) | 1.00 H | |
| Restraint at one end only | 1.50 H | |
| Free at the top (parapet, free-standing wall) | 2.00 H | |
| leff Table 3 |
Continuous and supported at both ends | 0.8 L |
| Continuous at one end, supported at the other | 0.9 L | |
| Simply supported (braced) at both ends | 1.0 L | |
| Continuous at one end, free at the other / free at one end | 1.5 L or 2.0 L | |
| teff Table 6 |
Solid wall · wall with piers · cavity wall | t · (stiffening coefficient × t) · greater of ⅔(t1+t2) and the thicker leaf |
Influence on the strength of the wall — this is what the question asks for:
- The basic compressive stress fb of Table 8 is measured on a short prism. A tall, slender wall buckles: it deflects laterally, the load becomes eccentric with respect to the deflected shape, and secondary (P–Δ) moments develop that add to the stress. The wall therefore fails at a load far below the short-prism strength.
- IS 1905 accounts for this by the stress reduction factor ks of Table 9, which is a function of SR and the eccentricity ratio: fper = fb ks ka kp.
- The strength falls steeply as SR rises. With e = 0: SR 6 → ks = 1.00; SR 12 → 0.84; SR 18 → 0.67; SR 24 → 0.51; SR 27 → 0.43. Roughly, doubling the slenderness from 12 to 24 halves the permissible stress. With eccentricity present the fall is even steeper — at SR 27 and e/t = 1/6, ks is only 0.22, i.e. barely a fifth of the short-wall value.
- SR interacts with eccentricity: a slender wall is far more sensitive to an eccentric load than a stocky one, because the initial eccentricity is amplified by the lateral deflection.
- Out of plane, the h/t ratio is the single most important parameter governing whether a wall survives an earthquake — hence the code puts an absolute ceiling on SR (Table 7): 27 for a load-bearing wall in cement mortar, 20 or 13 in lime mortar, 30 for a panel or curtain wall, 12 for a column, and only 10 for a free-standing wall or parapet.
- Design consequence: the way to reduce SR is to increase the thickness, or to reduce the effective height or length by adding cross walls, piers, buttresses or bands. Adding piers is usually the most economical, but only if their spacing satisfies Sp/Wp < 20.
(ii) Shape modification factor, kp
The shape modification factor kp is the factor by which the basic compressive stress may be
increased to allow for the fact that the masonry units are taller (or squarer) than the standard
unit on which Table 8 is based. IS 1905 Table 10
How it is read: enter Table 10 with the height-to-width ratio of the unit as laid (rows) and the crushing strength of the unit (columns):
| Height / width of the unit as laid | 5.0 N/mm² | 7.5 | 10.0 | 15.0 and above |
|---|---|---|---|---|
| up to 0.75 | 1.0 | 1.0 | 1.0 | 1.0 |
| 1.0 | 1.2 | 1.1 | 1.1 | 1.0 |
| 1.5 | 1.5 | 1.3 | 1.2 | 1.1 |
| 2.0 to 4.0 | 1.8 | 1.5 | 1.3 | 1.2 |
Influence on the strength of the wall — and why:
- The physical reason. Masonry fails because the softer mortar squeezes out laterally and puts the unit into lateral tension. A taller unit means fewer bed joints per metre of wall height, so there is less mortar to squeeze and the lateral tension induced in the unit is smaller. The wall is therefore stronger for the same unit and mortar.
- Also, a taller unit has a larger height-to-joint ratio, so the proportion of the wall volume that is weak mortar falls, and the restraint the unit provides to the mortar improves.
- Magnitude of the effect. It can be very large — for a low-strength unit (5 N/mm²) laid with h/w = 2 to 4, the permissible stress is increased by 80% (kp = 1.8). This is why hollow concrete blocks and stone blocks are far more efficient than thin bricks.
- The bonus shrinks as the unit gets stronger. At 15 N/mm² the same shape gives only kp = 1.2. A strong unit is already able to resist the lateral tension, so there is less to gain.
- The ordinary Nepali/Indian brick gets nothing. A 230 × 115 × 57 mm brick laid flat has h/w = 57/115 = 0.50, which is less than 0.75, so kp = 1.0. A modular brick 190 × 90 × 90 mm laid flat has h/w = 90/90 = 1.0, so it earns kp = 1.1 to 1.2 — a genuine 10–20% bonus for the same materials.
- Restriction. IS 1905 limits the factor to kp ≤ 1.5 for masonry in which the units are laid in a bond, and it applies only to units laid so that the stated height/width ratio is actually achieved — laying a brick on edge changes the ratio and therefore changes kp.
- Design consequence: kp is the one factor in fper = fbkskakp that can be increased free of cost — simply by choosing a taller unit or by laying the same unit differently.
Full treatment: Chapter 2 §2.6.5 (slenderness) and
§2.6.7(d) (shape modification factor); Tables 6, 7, 9 and 10 are reproduced as
Figs. 2.17–2.19.
Question 4
4
NumericalDesign an interior wall of a two storeyed
building to support a 3 m wide slab of 100 mm thickness. The height of the wall is 3.5 m and it is unstiffened.
Live load on the roof is 2 kN/m², live load on the floor is 3 kN/m², weight of the slab is
2.5 kN/m² and floor finish is 1 kN/m².
[18]
Given: two storeys; slab 100 mm thick, 3 m wide, supported on both sides of the interior wall;
wall height H = 3.5 m, unstiffened; LLroof = 2 kN/m²;
LLfloor = 3 kN/m²; slab weight = 2.5 kN/m²; floor finish = 1 kN/m².
Assume: trial wall thickness t = 230 mm (one brick); masonry 20 kN/m³; interior wall with equal slab spans on both sides, so the slab reactions balance and e = 0. Design per metre run.
Assume: trial wall thickness t = 230 mm (one brick); masonry 20 kN/m³; interior wall with equal slab spans on both sides, so the slab reactions balance and e = 0. Design per metre run.
Slab loads
ROOF FLOOR
Slab weight (given) 2.50 kN/m² 2.50 kN/m²
Floor finish 1.00 kN/m² 1.00 kN/m²
Live load 2.00 kN/m² 3.00 kN/m²
----------- -----------
Total 5.50 kN/m² 6.50 kN/m²
CHECK : 0.100 × 25 = 2.5 kN/m² — the given slab
weight is consistent with a 100 mm RCC slab. ✓
Tributary width
Interior wall → it supports 3 m wide slabs on BOTH sides
Tributary width = 3/2 + 3/2 = 3.00 m
Vertical load on the ground-storey wall, per metre run
From the ROOF 5.50 × 3.00 = 16.50 kN/m
From the FLOOR 6.50 × 3.00 = 19.50 kN/m
Wall of the upper storey
0.23 × 3.50 × 20 = 16.10 kN/m
Wall of the ground storey (self weight)
0.23 × 3.50 × 20 = 16.10 kN/m
-----------
Total at the base, W = 68.20 kN/m
Actual compressive stress
A = 1000 × 230 = 230 000 mm²
f_c = W / A = 68 200 / 230 000 = 0.297 N/mm²
Effective height IS 1905 Table 4
RCC slabs at the top and bottom bear on the full thickness
→ lateral restraint WITH resistance to rotation
h_eff = 0.75 H = 0.75 × 3500 = 2625 mm
Effective thickness and slenderness ratio
The wall is UNSTIFFENED — no piers and no cross walls,
so no stiffening coefficient applies.
t_eff = t = 230 mm
The wall is long, so the length-based ratio does not govern.
SR = h_eff / t_eff = 2625 / 230 = 11.41
Limit for a load-bearing wall in cement mortar = 27 ✓
Eccentricity
Interior wall carrying slabs of EQUAL span on both sides
→ the two slab reactions act at equal and opposite
eccentricities and CANCEL.
e = 0 → use the e = 0 column of Table 9.
(If the spans were unequal, then
e = (R₁e₁ − R₂e₂) / (R₁ + R₂ + W_wall) )
Modification factors
k_s Table 9 , SR = 11.41 , e = 0
SR 10 → 0.89 , SR 12 → 0.84
k_s = 0.89 − (0.89−0.84)(11.41−10)/(12−10)
= 0.89 − 0.05 × 0.705 = 0.855
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
Required f_b ≥ f_c / (k_s k_a k_p)
= 0.297 / (0.855 × 1.0 × 1.0) = 0.347 N/mm²
Select the unit strength and mortar grade IS 1905 Table 8
unit 3.5 N/mm² + M1 (1:5) f_b = 0.35 ✓ but only just
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
ADOPT : brick crushing strength 5.0 N/mm²
mortar grade M2 (cement : sand = 1 : 6)
Final check — ground storey
f_per = f_b · k_s · k_a · k_p
= 0.44 × 0.855 × 1.0 × 1.0 = 0.376 N/mm²
f_c = 0.297 N/mm² < f_per = 0.376 N/mm² ✓
Utilisation = 0.297 / 0.376 = 79 %
Check the upper storey
Load at the base of the UPPER storey
from the roof = 16.50 kN/m
self weight of the upper wall = 16.10 kN/m
-----------
W_upper = 32.60 kN/m
f_c = 32 600 / 230 000 = 0.142 N/mm²
< 0.376 N/mm² ✓
Utilisation = 38 %
A 115 mm wall would just satisfy the stress in the upper
storey (SR = 2625/115 = 22.8 < 27, k_s = 0.54,
k_a = 0.70 + 1.5×0.115 = 0.873, f_per = 0.44×0.54×0.873
= 0.207 > 0.142), but a half-brick wall is NOT recommended
in a seismic region — keep 230 mm throughout.
Summary of the design
Item Ground storey Upper storey
-------------------------------------------------------
Thickness 230 mm 230 mm
Load per metre run 68.20 kN/m 32.60 kN/m
Actual stress f_c 0.297 N/mm² 0.142 N/mm²
Slenderness ratio 11.41 11.41
k_s 0.855 0.855
Permissible stress f_per 0.376 N/mm² 0.376 N/mm²
Utilisation 79 % 38 %
Result SAFE SAFE
SAFE. Adopt a 230 mm (one brick) thick interior wall throughout both
storeys, built with bricks of crushing strength 5 N/mm² in
M2 (1:6) cement mortar; fper = 0.376 N/mm² against a maximum actual stress
of 0.297 N/mm² at the base of the ground storey.
Provide horizontal bands at plinth, sill, lintel, floor and roof level and vertical bars at all corners and junctions in accordance with IS 4326 / NBC 202.
Provide horizontal bands at plinth, sill, lintel, floor and roof level and vertical bars at all corners and junctions in accordance with IS 4326 / NBC 202.
Method: Chapter 2 §2.6.4–2.6.8
(the 12-step design procedure).
Question 5
5
NumericalDesign a reinforced brick masonry lintel over a
window opening of span 1.5 m. The wall thickness is 230 mm and the height of brickwork above the lintel up to
the ceiling is 90 cm. The length of the wall on either side of the lintel is 1.5 m. Take the characteristic
compressive strength of brick masonry as 10 N/mm² and use Fe 415 steel.
[18]
Given: clear span of opening L = 1.5 m; wall t = 230 mm; brickwork above the lintel = 0.90 m;
length of wall on either side = 1.5 m; characteristic compressive strength of brick masonry
fm = 10 N/mm²; Fe 415 steel.
Assume: masonry 20 kN/m³; working-stress design; permissible bending compressive stress in reinforced brickwork σcbc = fm/4 = 2.5 N/mm²; σst = 230 N/mm²; no floor or roof load falls within the load triangle.
Assume: masonry 20 kN/m³; working-stress design; permissible bending compressive stress in reinforced brickwork σcbc = fm/4 = 2.5 N/mm²; σst = 230 N/mm²; no floor or roof load falls within the load triangle.
Does arching action develop?
Arching action develops when the wall on either side of the opening extends beyond the springing by at
least half the span, and when the masonry above the lintel is at least as high as the load triangle.
The lintel then carries only the triangle of masonry bounded by 45° lines from the bearings.
Length of wall on either side = 1.50 m
Half the span = 1.50/2 = 0.75 m
1.50 > 0.75 ✓
Height of the 45° load triangle
h_t = L/2 = 1.50/2 = 0.75 m
Height of brickwork available above the lintel = 0.90 m
0.90 > 0.75 ✓
BOTH conditions are satisfied → the arching action is
COMPLETE and the loading on the lintel is TRIANGULAR.
Load on the lintel
Weight of the load triangle
W = ½ × L × h_t × t × γ
= 0.5 × 1.50 × 0.75 × 0.23 × 20
= 0.5 × 1.50 × 3.45 = 2.588 kN
Trial lintel section : 230 mm wide × 230 mm deep
(three brick courses of 70 mm + joints)
Self weight
w = 0.23 × 0.23 × 20 = 1.058 kN/m
over the span = 1.058 × 1.50 = 1.587 kN
Design bending moment and shear
For a triangular load of total magnitude W peaking at mid-span on a simply supported span L:
Mmax = WL6,
Vmax = W2.
For the UDL self weight: M = wL²/8, V = wL/2.
Triangular masonry load
M₁ = W L / 6 = 2.588 × 1.50 / 6 = 0.647 kN·m
V₁ = W / 2 = 1.294 kN
Self weight (UDL)
M₂ = w L² / 8 = 1.058 × 1.50² / 8 = 0.298 kN·m
V₂ = w L / 2 = 1.058 × 1.50 / 2 = 0.794 kN
-----------
DESIGN MOMENT M = 0.647 + 0.298 = 0.945 kN·m
DESIGN SHEAR V = 1.294 + 0.794 = 2.088 kN
Working-stress design constants
σ_cbc = f_m / 4 = 10 / 4 = 2.50 N/mm²
σ_st = 230 N/mm² (Fe 415, working stress)
Modular ratio
m = 280 /(3 σ_cbc) = 280 /(3 × 2.50) = 37.33
Neutral axis depth factor
k = m σ_cbc /(m σ_cbc + σ_st)
= (37.33 × 2.50) /(37.33 × 2.50 + 230)
= 93.33 / 323.33 = 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 × 2.50 × 0.289 × 0.904 = 0.327
Required effective depth
d_req = √( M /(R · b) )
= √( 0.945 × 10⁶ /(0.327 × 230) )
= √( 0.945 × 10⁶ / 75.2 )
= √ 12 566 = 112 mm
PROVIDE overall depth D = 230 mm (three brick courses)
clear cover 25 mm , bar diameter 10 mm
d = 230 − 25 − 10/2 = 200 mm
d = 200 > d_req = 112 ✓
(an under-reinforced, very safe section)
Area of tension steel
A_st = M /(σ_st · j · d)
= 0.945 × 10⁶ /(230 × 0.904 × 200)
= 0.945 × 10⁶ / 41 584 = 22.7 mm²
This is far below the practical minimum, so the DETAILING
minimum governs.
PROVIDE 2 nos 10 mm φ Fe 415
A_st,prov = 2 × 78.5 = 157 mm² ✓
The two bars are laid in the two outer VERTICAL JOINTS of
the bottom course and the joints are fully grouted with
1 : 3 cement mortar.
Check the compressive stress actually developed
Actual neutral axis depth (transformed section)
b x²/2 = m A_st (d − x)
230 x²/2 = 37.33 × 157 × (200 − x)
115 x² = 5861 (200 − x)
115 x² + 5861 x − 1 172 200 = 0
x = [ −5861 + √(5861² + 4×115×1 172 200) ] / (2×115)
= ( −5861 + √(34.35×10⁶ + 539.2×10⁶) ) / 230
= ( −5861 + 23 949 ) / 230 = 78.6 mm
Lever arm z = d − x/3 = 200 − 26.2 = 173.8 mm
Compressive stress in the masonry
f_c = 2M /(b · x · z) = 2 × 0.945×10⁶ /(230 × 78.6 × 173.8)
= 1.890×10⁶ / 3.142×10⁶ = 0.60 N/mm²
0.60 < σ_cbc = 2.50 ✓
Stress in the steel
f_st = M /(A_st · z) = 0.945×10⁶ /(157 × 173.8) = 34.6 N/mm²
34.6 < σ_st = 230 ✓
Shear check
τ_v = V /(b · j · d) = 2088 /(230 × 0.904 × 200)
= 2088 / 41 584 = 0.050 N/mm²
Permissible shear stress in reinforced brickwork ≈ 0.10 N/mm²
0.050 < 0.10 → NO shear reinforcement required ✓
Provide nominal 6 mm φ links (or 25 × 3 mm MS flats)
@ 150 mm c/c to hold the main bars in position.
Bearing and anchorage
Minimum bearing at each end = 200 mm
End reaction R = V = 2.088 kN
Bearing stress = 2088 /(230 × 200) = 0.045 N/mm²
well below the permissible bearing stress ✓
Overall length of the lintel
= 1.50 + 2 × 0.20 = 1.90 m
Development length (bars grouted in 1:3 mortar,
τ_bd ≈ 0.96 N/mm² for deformed bars)
L_d = φ σ_st /(4 τ_bd) = 10 × 230 /(4 × 0.96) = 599 ≈ 600 mm
The bars run the FULL 1.90 m length with a 90° hook at
each end, so the available anchorage far exceeds 600 mm. ✓
Design summary
Lintel section 230 mm wide × 230 mm deep
(three brick courses)
Overall length 1.90 m (200 mm bearing each end)
Load carried 45° triangle, W = 2.588 kN
Design moment 0.945 kN·m
Effective depth 200 mm (required 112 mm)
Main reinforcement 2 nos 10 mm φ Fe 415, bottom
Links 6 mm φ @ 150 mm c/c
Grout 1 : 3 cement mortar
Masonry f_m = 10 N/mm²
Compressive stress 0.60 < 2.50 N/mm² SAFE
Steel stress 34.6 < 230 N/mm² SAFE
Shear stress 0.050 < 0.10 N/mm² SAFE
DESIGN ADOPTED — reinforced brick masonry lintel 230 mm wide × 230 mm deep
(three brick courses), overall length 1.90 m with 200 mm bearing at each end, reinforced with
2 nos 10 mm φ Fe 415 bars in the bottom course, grouted with 1:3 cement mortar, and
nominal 6 mm φ links at 150 mm centres. All stresses are far below the permissible
values, so the section is governed by detailing requirements rather than by strength.
If arching had NOT developed — that is, if the wall on either side had been shorter
than half the span, or the masonry above shallower than L/2 — the lintel would have had to carry the
full rectangular block of masonry above it (0.90 × 0.23 × 20 = 4.14 kN/m,
giving M = 4.14 × 1.5²/8 = 1.16 kN·m) plus any floor or roof load landing in that
zone. Always state which case applies before computing the load.
Method: Chapter 4 §4.10 (arching action) and
Chapter 2 §2.5.5. Compare
2076 Q2b, the same problem with a 1.6 m span and a 220 mm wall.
Question 6
6
NumericalDetermine the equivalent eccentricity and the
eccentricity ratio of a 230 mm thick and 3 m high wall. The wall carries an axial load of 20 kN/m and an
eccentric load of 12 kN/m acting at 6 mm from the centre of the wall.
[12]
Given: t = 230 mm; H = 3.0 m; axial load P1 = 20 kN/m at the centre line
(e1 = 0); eccentric load P2 = 12 kN/m at e2 = 6 mm from the centre.
Assume: masonry 20 kN/m³; RCC slab bearing top and bottom → full restraint. Work per metre run.
Assume: masonry 20 kN/m³; RCC slab bearing top and bottom → full restraint. Work per metre run.
Principle — what “equivalent eccentricity” means
When several vertical loads act on a wall at different eccentricities, they are replaced by a
single resultant load W = ΣPi acting at the equivalent eccentricity e,
found by taking moments about the centre line of the wall:
e = Σ Pi eiΣ Pi and eccentricity ratio = et
e = Σ Pi eiΣ Pi and eccentricity ratio = et
Resultant load
W = P₁ + P₂ = 20.00 + 12.00 = 32.00 kN/m
Equivalent eccentricity
Taking moments of the loads about the CENTRE LINE of the wall :
W · e = P₁·e₁ + P₂·e₂
32.00 × e = (20.00 × 0) + (12.00 × 6)
32.00 × e = 72.00 kN·mm/m
e = 72.00 / 32.00 = 2.25 mm
Eccentricity ratio
e / t = 2.25 / 230 = 0.00978
= 1 / 102.2 ≈ 1 / 102
Interpret the result IS 1905 cl. 5.4.1
| Range of e | Treatment required by IS 1905 |
|---|---|
| e ≤ t/24 | The eccentricity is so small that it may be IGNORED; ks is read from the e = 0 column of Table 9. |
| t/24 < e ≤ t/6 | The resultant stays within the middle third; no tension develops, but ks must be read from the appropriate e/t column. |
| e > t/6 | The resultant falls outside the middle third; tension develops, part of the section becomes ineffective and fc = 2P/[3L(t/2 − e)] must be used. |
t / 24 = 230 / 24 = 9.58 mm
t / 6 = 230 / 6 = 38.33 mm
e = 2.25 mm < t/24 = 9.58 mm
→ THE ECCENTRICITY IS NEGLIGIBLE AND MAY BE IGNORED.
The wall is designed as if AXIALLY loaded, and k_s is
read from the e = 0 column of Table 9.
Complete the design check of the wall
Self weight of the wall
= 0.23 × 3.00 × 20 = 13.80 kN/m
Total load at the base
W_total = 32.00 + 13.80 = 45.80 kN/m
A = 1000 × 230 = 230 000 mm²
f_c = 45 800 / 230 000 = 0.199 N/mm²
Slenderness ratio and stress reduction factor
h_eff = 0.75 H = 0.75 × 3000 = 2250 mm
t_eff = 230 mm
SR = 2250 / 230 = 9.78 < 27 ✓
k_s Table 9 , SR = 9.78 , e = 0 (eccentricity ignored)
SR 8 → 0.95 , SR 10 → 0.89
k_s = 0.95 − (0.95−0.89)(9.78−8)/(10−8)
= 0.95 − 0.06 × 0.89 = 0.897
k_a A = 0.23 m² > 0.2 m² → k_a = 1.0
k_p h/w = 0.50 < 0.75 → k_p = 1.0
Select the masonry and verify
Required f_b ≥ f_c /(k_s k_a k_p)
= 0.199 / 0.897 = 0.222 N/mm²
From Table 8 :
unit 3.5 N/mm² + L2 / M3 f_b = 0.25 ✓
unit 3.5 N/mm² + M2 (1:6) f_b = 0.35 ✓ ADOPT
f_per = 0.35 × 0.897 × 1.0 × 1.0 = 0.314 N/mm²
f_c = 0.199 N/mm² < f_per = 0.314 N/mm² ✓
Utilisation = 63 %
ANSWER
Resultant load W = 32.00 kN/m
Equivalent eccentricity e = 2.25 mm
Eccentricity ratio e/t = 0.0098 ≈ 1/102
Since e = 2.25 mm is less than t/24 = 9.58 mm, the eccentricity may be neglected and the wall may be designed as axially loaded. With a 230 mm wall of 3.5 N/mm² bricks in M2 mortar, fper = 0.314 N/mm² against fc = 0.199 N/mm² — SAFE.
Resultant load W = 32.00 kN/m
Equivalent eccentricity e = 2.25 mm
Eccentricity ratio e/t = 0.0098 ≈ 1/102
Since e = 2.25 mm is less than t/24 = 9.58 mm, the eccentricity may be neglected and the wall may be designed as axially loaded. With a 230 mm wall of 3.5 N/mm² bricks in M2 mortar, fper = 0.314 N/mm² against fc = 0.199 N/mm² — SAFE.
If the eccentric load acts at 60 mm (a value some versions of this question use), then
e = 12 × 60 / 32 = 22.5 mm and e/t = 22.5/230 = 0.0978 ≈ 1/10.2,
which lies between t/24 and t/6. The eccentricity could then not be ignored: ks would be
interpolated between the 1/12 and 1/6 columns of Table 9, giving about 0.86 instead of 0.897 — a 4%
reduction in the permissible stress. The method is identical; only the interpretation in Step 5 changes.
Method: Chapter 2 §2.6.6 (eccentricity) and
§2.6.7(b). Follows the worked example in the course notes (CH-4, Example 4.10).