The vertical stresses in a soil mass due to a vertical line load can be obtained using Boussinesq's solution.

1. SUBJECT :- SOIL MECHANICS
VERTICAL STRESSES DUE TO
LINE LOAD AND STRIP LOAD

2. VERTICAL STRESSES DUE TO LINE LOAD
 The vertical stresses in a soil mass
due to a vertical line load can be
obtained using boussinesq’s
solution.
 Let the vertical line load be of
intensity q’ per unit length, along y
axis acting on the surface of a
semi-infinite soil mass, as show in
figure

3.  Consider a small length dy of the line load. The load can be taken as a point load
of (q’ * dy) and boussinesq’s solution can be applied to determine the vertical
stress at p(x, y, z) from equation
dσz =
3Q
2π
∗
1
Z2 ∗ [
1
1+(
r
z
)2
]
5
2
 dσz =
3 q′∗dy
2π
*
Z3
( r2+ z2 )
5
2
 The vertical stress at p due to the line load extending from -∞ to +∞ is obtained
by the integration.
σz =
3 q′ 𝑍3
2π +∞
−∞ 𝑑𝑦
(𝑥2+𝑦2+𝑧2)5 2

4.  σz =
3 q′ 𝑍3
2π
2 0
−∞ 𝑑𝑦
(𝑥2+𝑦2+𝑧2)5 2
 Substituting (𝑥2 + 𝑦2 = 𝑢2) we get,
 σz =
3 q′ 𝑍3
2π
2 0
−∞ 𝑑𝑦
(𝑢2+𝑧2)5 2
 Let, y = u tan θ, then dy = y sec2
θ * dθ and limits of integration being changed
from θ = 0 to θ =
π
2
σz =
3 q′ 𝑍3
2π
2 0
π 2 𝑈 sec2 θ ∗𝑑θ
𝑢5 sec5 θ
σz =
3 q′ 𝑍3
2π
2
𝑢4 0
π 2
cos3
θ ∗ 𝑑θ

5.  Let, sin θ = t
 cos θ * dθ = dt and limits of integration being changed from t = 0 to t = 1,
now equation is written as,
 σz =
3 q′ Z3
2π
2
u4 0
1
(1 − t2) ∗ dt
σz =
3 q′ Z3
2π
2
u4 t −
t3
3
σz =
3 q′ Z3
2π
∗
2
u4 ∗
2
3
σz =
2 q′ Z3
πu4

6.  σz =
2 q′ Z3
π( 𝑥2+𝑦2)2
𝛔 𝐳 =
𝟐 𝐪′
𝛑𝐳
[
𝟏
𝟏+(
𝐱
𝐳
) 𝟐
] 𝟐

7. VERTICAL STRESSES UNDER A STRIP LOAD
 The expression for the stresses
developed in a soil mass due to a
strip load acting at the soil
surface can be obtained by using
the expressions for the stresses
developed in the soil mass due to
line load.

8.  Case-1 :- Point p below the center of the strip :-
 Let, a uniform load of intensity q per unit area be acting on a strip of infinite length
and constant width B=2b at the surface of soil mass. Let us consider the load acting
on a small elementary width dx at a distance c from the center of the load. This small
load of (qdx) can be considered as a line load of intensity q’.
 In this case the vertical stress at point p can be obtained from equation of line load
σz =
2(qdx)
πz
[
1
1+(
x
z
)2
]2
 q = load intensity in KN/m2
 q’ = line load intensity in KN/m

9.  σz =
2 q
πz −b
b 1
[ 1+(
x
z
)2 ]2
dx
 let,
x
2
= tan 𝑢 dx = z * sec2 𝑢 * du
σz =
2 q
πz
* 2 0
0 𝑧 sec2 𝑢 ∗𝑑𝑢
[ 1+ tan2 𝑢 ]2
Where, θ = tan−1 𝑏
𝑧
= angle made by extremities of the strip at P.
 σz =
4q
π
* 0
0
cos2 𝑢 ∗ 𝑑𝑢
σz =
4q
π
* 0
0 1+cos 2𝑢
2
𝑑𝑢
𝛔 𝐳 =
𝐪
𝛑
(2θ+𝐬𝐢𝐧 𝟐𝛉)

10.  Case-2 :- point P not below the
center of the strip
 Fig show the case when the point
P is not below the center of strip.
The extremities of the strip make
angle 𝛽1 𝑎𝑛𝑑 𝛽2 at point P. As in
the previous case, the load (q . dx)
acting on a small length dx can be
considered as a line load of
intensity q’ .the vertical stress at P
is given by
d𝜎𝑧 =
1
1+(
𝑥
𝑧
)2
2

11.  Equation (a) can be simplified by making the following substitution.
X = z tan 𝛽 or dx = z . 𝑠𝑒𝑐2 𝛽. 𝑑𝛽
d𝜎𝑧 =
2 𝑞( 𝑧 sec2 𝛽 .𝑑𝛽)
𝜋𝑧
[
1
1+ 𝑡𝑎𝑛2 𝛽
]2 1 + tan2 𝛽 = 𝑠𝑒𝑐2 𝛽
or d𝜎𝑧 =
2𝑞
𝜋
𝑐𝑜𝑠2 𝛽 . 𝑑𝛽
𝜎𝑧 =
2𝑞
𝜋 𝛽1
𝛽2 ( 1+𝑐𝑜𝑠2𝛽)
2
d𝛽

12. or 𝜎𝑧 =
𝑞
𝜋 𝛽1
𝛽2
(1 + 𝑐𝑜𝑠2β)d𝛽 1+cos 2𝛽 = 2 𝑐𝑜𝑠2 𝛽
=
𝑞
𝜋
𝛽 +
𝑠𝑖𝑛2𝛽
2
𝛽2
𝛽1
or
𝑞
𝜋
[ 𝛽2 − 𝛽1 + (sin 𝛽2 cos 𝛽2 − sin 𝛽1 cos 𝛽1)]

14. Substituting,
𝛽2 - 𝛽1 = 2𝜃
𝜎𝑧 =
𝑞
𝜋
[2𝜃 + (sin 𝛽2 . cos 𝛽2 − sin 𝛽1 . cos 𝛽1)]
If (𝛽1 + 𝛽2) = 2∅ 𝑡ℎ𝑒𝑛 𝑖𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑠ℎ𝑜𝑤𝑛 𝑖𝑛 𝑓𝑖𝑔
Therefore equation (b) becomes,
𝝈 𝒛 =
𝒒
𝝅
[ 2𝜽 + 𝒔𝒊𝒏 𝟐𝜽 . 𝒄𝒐𝒔 𝟐∅ ]