Calculation of equivalent stresses in masonry structures in case of
eccentric loading causing excessive tension
Eccentric loading may result from a load applied off the center of the substructure
or from a concentric load plus a bending moment. For the purpose of determining the
max. stress in the substructure, the moment may be removed by shifting the vertical load
to a fictitious location with an eccentricity e= moment/vertical load. In the analysis of an
eccentrically loaded substructure two separate problems are confronted:
1.
For the purpose of structural design, the stress is assumed to have a planar
distribution. When the load is applied within the kern, common flexural formulae are
applicable.
q
My
Q Mx

x
y
A Iy
Ix
(1)
Where q = stress at a given point (x,y);
Q = vertical load;
A = area
X and y = coordinates of the point at which stress is calculated;
Mx’ My = load Q multiplied by eccentricity parallel to x and y axes, respectively;
Ix’Iy = moment of inertia of the substructure about the x and y axes, respectively.
Equation (1) is valid when one of the following conditions exists:
(a)
The substructure is symmetrical about x and y axes.
(b)
The substructure is symmetrical about x axis and ey = 0
(c)
The substructure is symmetrical about y axis and ex = 0.
For rectangular section, Eq. (1) may be written in a simpler form:
q
e
e 
Q
1  6 l  6 b 
A
L
B
(2)
When el, eb exceed a certain limit, Eq. (2) gives a negative value of q which
indicates tension. If tension portion is to be neglected, above equation does not remain
valid. Therefore, the flexural formulae Eq (2) are applicable only when the load is applied
within a limited area which is known as the kern and is shown shaded in Fig. 1.
L
L/6
B/6
B
Fig. 1
The procedure for determination stresses when the load is applied outside the
kern is simple in principle but laborious. Cases for rectangular and circular sections are
available in some textbooks. For other shapes, the graphical method of successive trials
is probably the simplest for practical solutions.
Let us first understand rectangular section only. Fig 2 shows stresses in the
section due to eccentricity in ‘X’ direction only. As eccentricity is only in X directionneutral axis shall be parallel to face AA of section. Total pressure in shaded portion
should coincide with Q.
A
Q
L
e
B
A
x
Q
B/2 - e
N
e
M
O
qa
N
P
x
Fig 2
i.e. centroid of triangle from right hand edge MP = 1/3
Therefore, (B/2 – e) = 1/3 x
x = 3(B/2 – e)
When x is known, qmax can be determined
Q
= Area of Triangle OMP
L
Q 1
 xqa
L 2
1 B

.3  e q a
=
2 2

Q.B 1  3B  6e 
 
q a
A
2 4 
Q  4B 
qa  
A  3B  6e 
Q 
e 
1 6

A
B

Q 
4B

For e  B
 q max 

6
A  3B  6e 

B

Qmin = 0 at a distance of 3  e)  from edge of footing
2

For e  B
6
 q 
If eccentricity is in two directions, neutral axis shall not be parallel to face of
footing. As such, situation gets complicated, as first step is to calculate neutral axis.
For circular sections
r
Q
e
For e  r  q  1  4 
4
A
r
Q
For e  r 4  q max  k
A
e/r
k
A = r2
r/4
k values are as under
0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.90
2.00 2.20 2.43 2.70 3.10 3.55 4.22 4.92 5.90 7.20 9.20 13.0 80.0
For any general shape, calculations are difficult, but it is possible by using
graphical method. The graphical method, similar to any other method, is based on the
assumption that the pressure varies linearly with the distance to the neutral axis from
zero at the neutral axis to a maximum at the most remote point and on the requirement
of statistical equilibrium that the resultant of the soil pressure should lie on the line of
action of the applied load Q. The procedure is as follows.
N
x
y
Q
l
a
b
o
e
dy
N
Fig.3
qb
Draw a trial neutral axis N-N, Fig. 3 and line ab perpendicular to N-N, starting from
point b which is most remote. The area between point b and N-N is under compression
while the area on the other side of N-N is unstressed. The intensity of stress at a given
point varies in simple proportion with its perpendicular distance from N-N. The
compression area is divided into several narrow strips of uniform width dy, running
parallel to N-N. The unit pressure acting on this strip is equal to (Y/X)qb, where qb is the
unit pressure at point b, and the total pressure is equal to (Y/X)q b l dy. Similar value of
total pressure can be calculated on each strip. If the location of the trial neutral axis N-N
is correct, sum of pressures on all strips will coincide with the point of action of the load
Q. Several such trials will enable the engineer to approach the correct location of the
neutral axis. Once neutral axis is determined then rest is simple.