Chapter 3-1 MECH 303 Chapter 3
3.3 Bending of a simple beam under uniform load
q
O
ql
h/2
ql
h/2
l
x
l
y
 We use the semi-inverse method
 We may assume that y is only a function of y
 y  f y
then we have
 2
 f y
x 2


x2
f  y   xf1  y   f 2  y 
2
f1(y) and f2(y) are arbitrary functions
 Compatibility equation requires
d 4 f 2 y
1 d 4 f  y  2 d 4 f1  y 
d 2 f y
x

x


2
0
2 dy 4
dy
dy 4
dy 2
3-7
MECH 303 Chapter 3
d 4 f1  y 
0
dy 4
d 4 f y
0,
dy 4
d 4 f 2 y
d 2 f y

2
 0,
dy 4
dy 2
We have
f  y   Ay 3  By 2  Cy  D ,
f1  y   Ey 3  Fy 2  Gy. 
f 2 y  
A
10
y5 
B
6
y 4  Hy 3  Ky 2 . 
The  becomes


 

x2
A 5 B 4
Ay 3  By 2  Cy  D  x Ey 3  Fy 2  Gy 
y  y  Hy 3  Ky 2
2
10
6

Stress components will be:
x2
 x  6 Ay  2 B   x6 Ey  2 F   2ay 3  2 By 2  6 Hy  2 K ,
2
 y  Ay 3  By 2  Cy  D ,
 xy   x3Ay 2  2By  C   3Ey 2  2Fy  G
9 constants must be determined by boundary conditions.
 From the condition of symmetry:
E=F=G=0. There are 6 constants left.
 Boundary conditions on the horizontal sides:
3-8
MECH 303 Chapter 3
y
y
h
2
 0,
y
y 
h
2
 q ,
 xy
y 
h
2
0
can determine the constants A. B. C. D:
A
2q
,
h3
C
B=0,
3q
,
2h
D
q
2
 Boundary conditions on the end surfaces:
h
2

x x l
dy  0 ,
h

2
h
2

x x l
ydy  0
h

2

H
ql 2
q
, K=0.

3
10h
h
The resultant of xy must be an upward force equal to ql:
h
2

xy x l
dy  ql
(satisfied)
h

2
Finally we have the solution of the problem:
3-9
MECH 303 Chapter 3

6q 2
y  y2 3 
2
 4 2  


l

x
y

q
 x
3
h
h
5
 h

2

q
y  2 y 


 y   1  1 
2  h 
h 



6q  h 2
 xy   3 x  y 2 
h  4




compared with the solution in mechanics of materials (MECH101):
x 


6q 2
l  x2 y
h3
y 0
 xy  

6q  h 2
x  y 2 
3 
h  4

The obtained solution here is exact only if at the ends ( x  l ) there are
normal forces and shearing forces according to :
X   x  x l
y  y2 3
 q  4 2  
h h
5
Y   xy  x  l

6ql  h 2
  3   y 2 
h  4

3-10
MECH 303 Chapter 3
3.4 Triangular gravity wall (dam or retaining wall)
 The density of wall material :   weight (unit) : g
 The density of liquid :  unit weight : g
 If the height of the wall is considered infinite.
(1) The stress is produced by – gravity, proportional to g
- pressure, proportional to g
(2) Dimensional analysis of stress components ([force] / [length]2)
dimension of g is [force] / [length]3
dimension of g is [force] / [length]3
dimension of x, y is [length] ,   dimensionless
(3) If the stress components can be expressed in the form of polynomials,
they must be combinations of expressions in the form of A gx ,B gy ,C
gx, Dgy, where A.B.C.D. are dimensionless numbers depending upon
only  only. It can be seen that the stress function  must be a polynomial
of the third degree, i.e. ,
  ax 3  bx 2 y  cxy 2  ey 3
Body force X  0 , Y  g
3-11
MECH 303 Chapter 3

 2
 x  2  Xx  2cx  6ey
y



2


 y  2  Yy  6ax  2by  gy 
x


 2
 xy  
 2bx  2cy

xy

The 4 constants are determined by the boundary conditions:
 On the vertical surface x  0 ,  x
x 0
 gy ,  xy
x 0
0
 e   g 6 , c=0
 On the inclined surface x  y tan   ,
l x x  y tan   m xy



m y x  y tan   l xy
x  y tan 
0
x  y tan 
0
( l  cos , m   sin  ) 
b
g
2
cot 2  , a 
g
6
cot  
g
3
cot 3 
 Finally, the stress solution of the problem is
 x  gy

3
2
 y  ( g cot   2g cot  ) x  (g cot   g ) y

2
 xy   yx  gx cot 
 The above results are only exact when the height of the dam is
infinite. For a finite height dam, the stress distribution near the
foundation is approximate. ( Saint-Venant’s principle)
3-12
MECH 303 Chapter 3
3-13