DIRECT AND BENDING

STRESSES
CHAPTER VII

ECCENTRICITY
Prof. Yassin Sallam
1
 DIRECT AND BENDING
STRESSES
Direct stress come plays at
CHAPTER VII

every section of the body. If

a body subjected to a
bending stress comes into ECCENTRIC LOADING
play. It is thus obvious that A load whose line of action
if a member, subjected to does not coincide with the
an axial loading along with axis of a column or strut is
a transverse bending. known as an eccentric load
The magnitude and nature
of these stresses may be
easily found out from the
magnitude and nature of the
load and the moment. 2
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
a. COLUMNS WITH ECCENTRIC LOADING
Consider a column subjected to an eccentric loading. This eccentric load
CHAPTER VII

may be analyses as shown

❑ The given load P P

P P
acting at an eccentricity e
e

of e Fig.(a)

❑ We introduce, along
the axis of the column
two equal and opposite =
forces P Fig.(b)

(a)
(b)

3
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
❑ The forces column acting may be split up into three forces:
❑ One of these forces will be acting along the axis of the column.
CHAPTER VII

This force will cause a direct stress,  o = P Fig.(c)

A
❑ The other two forces will form a couple, Fig. (d).
❑ The moment of this couple will be equal to M = P.e
❑ This couple will cause a bending stress:
P P
e

P
P

Direct Bending
(c) (d)

P M . y Pey
o = b = =
A I I 4
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
2. SYMMETRICAL COLUMN WITH ECCENTRIC
LOADING ABUT ONE AXIS
Consider a column ABCD, subjected to an eccentric load about one axis/ yy ,
CHAPTER VII

as shown.
eP
P : Load acting on the column
e : Eccentricity of the load
b : Width of the column section
d : Depth of the column section

•Area of column section: A = b.d

b.d 3
•Moment of inertia of the column section/ yy: I=
12
I bd 3 / 12 bd 2
•Modulus of section: Z= = =
y d /2 6 A B
P
•Direct Stress on the column due to the load: o = e d

A C b D
•Moment due to load: M = P.e  min
 max
• Bending stress at any point of the column section
• at a distance y from yy axis. y y 1 M
b = M = Pe =M = 5
Prof. Yassin Sallam I I Z Z
 DIRECT AND BENDING
)
STRESSES
y=
d M . ymax P.e. ymax
•Bending stress at the extreme ( 2) b = =
d I I
M( )
CHAPTER VII

b = 2 = 6M = 6 P.e = 6 P.e
bd 3 bd 2 bd 2 A.d
12
Total stress at the extreme fiber:  = − o   b
✓(+) or (-) signs will depend upon the position P Pey
 =− 
of the fiber with respect to eccentric load. A I
P M
✓The total stress along the width of the column  =− 
A Z
will vary a straight line law. P 6e
 = − (1 + ) P 6 Pe
•The maximum Stress: max  =− 
A d A Ad
P 6e P 6e
•The minimum Stress:  min = − (1 − )  = − (1  )
A d A d
If  max  min
1 o   b Same nature Compressive
2 o =  b 2 b 0
6
3 o  b Partly compressive Partly tensile
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
Load Example (1)
180 kN
A rectangular strut is 150 mm and 120 mm thick. It curries a e
CHAPTER VII

load of 10 mm in a plane bisecting, the thickness. Find the

maximum and the minimum intensity of stress in the section.
Solution:
Width b = 150 mm
Depth d = 120 mm
Load P = 180 kN = 180x103 N
Eccentricity e = 10 mm

Area of section column:

A = bd = 150 x120 e 120 mm

= 18 000 mm 2 150 mm

7
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
•The maximum Stress:
P 6e
 max = − (1 + )
CHAPTER VII

A d
180 x103 6 x10
 max = − (1 + ) e 120 mm
18 x103 120
 max = −10 (1 + 0.5) = 15 MPa 150 mm
Compression
 min = 5MPa
•The minimum Stress:  max = 15MPa
P 6e
 min = − (1 − )
A d
180 x10 3
6 x10
 min =− (1 − )
18 x103 120
 min = −10(1 − 0.5) = 5 MPa Compression

There is no tension 8

Prof. Yassin Sallam

 DIRECT AND BENDING
STRESSES
120 kN
Example (2) 50 mm

A rectangular column 200 mm wide and 150 mm thick

CHAPTER VII

is carrying a vertical load of 120 kN at an eccentricity of

50 mm in a plane bisecting the thickness.
Determine the maximum and the minimum intensity of
stress in the section.

Solution:
Width b = 200 mm
Depth d = 150 mm
Load P = 120 kN = 120x103 N
e 150 mm

Eccentricity e = 50 mm 200 mm

Section area A = bd = 200 x150 = 30 000 mm 2 9

Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
•The maximum Stress:
P 6e
CHAPTER VII

 max = − (1 + )
A d
120 x103 6 x50
 max =− (1 + ) e 150 mm
30 x10 3
150
 max = −4(1 + 2) = 12 MPa 200 mm
Compression
•The minimum Stress:  min = 4MPa
 max
P 6e
 min = − (1 − ) = 12MPa
A d
120 x103 6 x50
 min =− (1 − )
30 x10 3
150
 min = −4 x(1 − 2) = 4 MPa Tension
10
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
Example (3) P kN
e
In a tension specimen 13 mm in diameter, the line of pull
CHAPTER VII

is parallel to the axis of the specimen but is displaced

from it. Determine the distance of line of pull from the
axis, when the maximum stress is 15 percent greater
than the mean stress on a section normal to this axis.

Solution:
Diameter d = 13 mm
Max. Stress  max = 1.15  av .................(1)
Area of section column: 13 mm
e
d 2 2
(13)
A= = = 132.7 mm 2
4 4
11
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
•Modulus of section:
 
Z= d3 = x(13)3
CHAPTER VII

32 32
Z = 215.7 mm3
•Moment due to eccentricity load:
M = P.e
By definition the mean (average) stress:
P P
 av =  o = =− N / mm 2
A 132.7
•Maximum stress:
 max = − o −  b
 max = − av −  b
P P.e
=− −
A Z
P P .e
 max = − − ............(2)
132.7 215.7 12
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
 max = 1.15  av ....................(1)
P P.e
 max = − − ............(2)
CHAPTER VII

From equations (1) and (2) 132.7 215.7

P P.e
 max= − −
132.7 215.7
P P P.e
1.15 x =− −
132.7 132.7 215.7
115 10 e
=− −
13270 1327 215.7  min
115 10  max
e = 215.7( − )
13270 1327
e = 215.7(0.0087 − 0.0075)

e = 215.7 x 0.00112 = 0.25 mm

13
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
Example (4) 2 MN
A hollow rectangular masonry pier is 1.2 m x 0.8 m wide 100 mm

and 150 mm thick. A vertical load of 2 MN is transmitted in

CHAPTER VII

the vertical plane bisecting 1.2 m side and at an eccentricity

of 100 mm from the geometry axis of the section.
Calculate the maximum and the minimum stress intensities
in the section.
Solution:
Outer depth D = 1.2 m = 1.2 x10 3 mm
Outer width B = 0.8 m = 0.8 x103 mm
Thickness t = 150 mm d =? b=?
Load P = 2 MN = 2 x106 N 0 .8 m

Eccentricity e = 100 mm
Area of section column:
A = DB − db 1 .2 m

A = [(1.2 x0.8) − (0.9 x0.5)]x106

A = (0.96 − 0.45) x10 6 = 0.51x10 6 mm 2 14
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
•Modulus of section:
1
CHAPTER VII

Z = ( BD 2 − bd 2 )
6
1
Z= [(1.2 x103 )(0.8 x103 )2 ] − [(0.9 x103 ) x(0.5 x103 )2 ]
6
1
Z = 768 x106 − 225 x106 
6
Z = 90.5 x106 mm3

•Moment due to eccentricity load:

M = P.e
M = 2 x106 x100
M = 200 x106 N .mm 15
Prof. Yassin Sallam
 DIRECT AND BENDING
STRESSES
• Maximum stress the section:
P M
 max = −( + )
CHAPTER VII
A Z 6
2 x10 200 x106
 max = −( + )
6
0.51x10 90.5 x10 6

 max = −(3.92 + 2.21)

 max = −6.13 N / mm 2 1 .2 m

= 6.13 MPa Compression

• Minimum stress the section:
P M  min
 min = −( − ) = 1.71 MPa
 max
A Z = 6.13MPa
2 x106
200 x10 6

 min = −( − )
0.51x106 90.5 x106
 min = −3.92 + 2.21
 min = −1.71 N / mm2 = 1.71 MPa Compression
16
Prof. Yassin Sallam There is no tension