2. (A). Column :-
Axial load : When load is acting along the longitudinal
axis of column. It produces compressive stress in column.
Eccentric load : A load whose line of action does not
coincide with the axis of a column. It produces direct
and bending stress in column.
Eccentricity (e) : The horizontal distance between the
longitudinal axis of column and line of action of load.
In axially loaded column e = 0
4. When short column is subjected to axial compressive
force, only direct stress(σo) is produced in the column.
Direct stress = σo =
Eccentric load produces both direct stress (σo) and
bending stress (σb) in column.
Direct stress = σo =
Bending stress = σb = = .y ∴ Z =
I
y
5. Maximum and Minimum Stresses :-
When a column is subjected to eccentric load, the edge of
column towards the eccentricity will be subjected to maximum
stress (σmax) and the opposite edge will be subjected to
maximum stress.
Maximum Stress (σmax) Minimum Stress (σmin)
σmax = direct stress + bending stress
= σo + σb
= +
= (1+ )
σo = direct stress, σb = bending stress, M = Moment = P . e, e = eccentricity
Z = section modulus = , I = moment of Inertia, y = distance of extreme fibre
from c.g. of column.
σmin = direct stress - bending stress
= σo - σb
= -
= (1- )
6. Stress distribution in Column :-
Stress distribution in column as the load (P) moves from
centre of column to the edge of column as shown in fig.
7. Limit of eccentricity (e limit):
The maximum distance of load from the centre of
column, such that if load acts within this distance there
is no tension in the column. The maximum distance is
called Limit of eccentricity.
When load is acting within e limit,
𝜎min will be compressive. (+ve)
When load is acting at the point of e limit,
𝜎min will be zero.
When load is acting beyond e limit,
𝜎min will be tensile (-ve)
8. Example :-
A column of T-section shown in figure is subjected to
a load of 100kN at a point on the centroidal axis, 40mm
below the centroidal x-x axis. Calculate the maximum
stresses induced in the section.
Solution: Data Given,
P = 100 kN
e = 40 mm
Part-1
a1 = 200 x 20 = 4000mm2
y1 = 180 + 10 = 190mm
Part-2
a2 = 180 x 15 = 2700mm2
y2 = 90 = mm
y = = = 149.70 mm
9. Ixx1 = Ig + ah2
=
+ 4000 x (190 – 149.70)2
= 6.629 x 106 mm4
Ixx2 = Ig + ah2
= + 2700 x (149.70 – 90)2
= 16.91 x 106 mm4
Ixx = 6.692 x 106 + 16.91 x 106
= 23.539 x 106mm4
𝑍𝑥𝑥 =
Ixx
𝑦𝑚𝑎𝑥
=
23.539 x 106
149.70
= 157241.15mm3
10. σmax =
P
A
+
M
Z
=
100 𝑥 103
6700
+
100 𝑥 103𝑥 40
157241.15
= 14.92 + 25.43
= 40.35 N/mm2 ( compressive)
σmin =
P
A
−
M
Z
=
100 𝑥 103
6700
−
100 𝑥 103𝑥 40
157241.15
= 14.92 − 25.43
= 10.51 N/mm2(Tensile)
11. (B). Dams and Retaining Wall :-
Maximum and Minimum pressure at the base of Dam :
12. (1) Weight of Dam:
Weight = cross sectional area of dam x density of
dam material
∴ W = (a + b) x
𝐻
2
x ᵟ
where ᵟ = density of dam material in kN/m3
(2) Total water pressure on Dam:
Total water pressure = Area of water pressure diagram
∴ P =
1
2
x wh x h Where, w = density of water
∴ P =
wh2
2
= 1000 kg/m3
= 10 kN/m3
13. (3) Eccentricity (e) :
Total water pressure (p) acts horizontally at height
h
3
from
the base of dam.
Total weight of dam (W) acts vertically downwards.
R is the resultant of P and W.
R = P2 + W2
Resultant (R) cut the base at point K.
distance JK = x =
P
W
x
h
3
distance AJ =
a2+ab+b2
3 (a+b)
∴ d = AJ + JK
∴ eccentricity = e = d −
b
2
14. (4) Maximum and Minimum Pressure :
Maximum Pressure.
𝜎max =
W
b
(1 +
6e
b
)
Minimum Pressure.
𝜎min =
W
b
(1 −
6e
b
)
15. Retaining Wall :
A retaining wall is a structure used to retain soil(earth).
The basic difference between dam and retaining wall is that, a
dam retain water and subjected to water pressure while, a
retaining wall retain earth and subjected to earth pressure.
16. Total earth pressure :
P =
wh2
2
x Ka
∴ P =
wh2
2
x (
1 −sin∅
1+sin∅
)
Where,
Ka = active earth pressure coefficient
=
1 −sin∅
1+sin∅
∅ = Angle of repose of soil
Total earth pressure (P) acts at height
h
3
from the bases of
retaining wall.
18. Consider a chimney or wall having plan dimension b x d and
height h.
Let, 𝛿 = unit weight of chimney or wall
𝛿 =
total weight
volume
Where,
∴ 𝛿 =
W
v
A = base area
∴ w = 𝛿 x V
w = 𝛿 x A x h
∴ direct stress = σo =
W
A
=
δ x A x h
A
σo = 𝛿 x h ….(i)
If there is a uniform horizontal wind pressure (p) acting
on a side of width b,
wind force = P = p x b x h
This force will induce a bending moment on the base.
19. This force will induce a bending moment on the base.
∴ M = P x
h
2
Bending stress caused on the base due to moment,
σb =
𝑀
𝑍
The extreme stresses on the base are,
𝜎max = σo + σb
𝜎min = σo − σb
20. Example :-
A masonry wall, 5m high, is of solid rectangular
section, 3m wide and 1m thick. A horizontal wind pressure of 1.2
kN/m2
acts on the 3m side. Find the maximum and minimum stresses
induced on the base, if unit weight of masonry is 22.4 kN/m3
Solution :-
𝛿 = 22.4 kN/𝑚3
h = 5.0m
b = 3.0m
p = 1.2 kN/m2
𝜎0 = 𝛿 x h
= 22.4 x 5.0
= 112 kN/m2
21. M = P x
h
2
P = total wind force
= p x b x h
= 1.2 x 3 x 5
= 18 kN
M = P x
h
2
= 18 x
5
2
= 45 kN.m
Z =
𝑏𝑑2
6
=
3 x 12
6
= 0.5m3
𝜎𝑏 =
M
Z
=
45
0.5
= 90 kN/m2
𝜎max = σo + σb
= 112 + 90
= 202 kN/m2 (compressive)
𝜎min = σo − σb
= 112 − 90
= 22 kN/m2 (compressive)
