Equilibrium Equations for a Beam
The equilibrium equations can be derived very quickly by simply considering a
differential element and summing forces and moments. This derivation emphasizes
the connection between 3D elasticity and beam theory.
Three-dimensional equilibrium equations
 x  xy  xz


  gx  0
x
y
z
 y  xy  yz


 gy  0
y
x
z
 z  yz  xz


  gz  0
z
y
x
and each of these stresses and body forces are functions of (x,y,z)
Now let’s exploit the assumptions to obtain simplified relations. Consider the following
free body diagram of a differential element of dimensions dx * w * h. For comparison the
differential element used for deriving three-dimensional equations is also shown. Only
positive-face tractions are shown.
3-D Continuum
y
Beam
y
yy
yy
yx
yz
yx
xy
zy
xx
zx xz
dy
xy
h
dz
xx
gy
gx
x
w
x
zz
dx
z
dx
z
Recall that we assumed that some of the stresses are zero for a beam. This results in most
of the tractions being zero for the beam differential element.
The beam differential element extends the full height and width of the beam… not a
differential distance. The only differential distance is along the x-axis,
Since the beam differential element is of finite height, equilibrium requires that both the
summation of forces and moments =0. (=>three equations) This can be expressed as
1) Force equilibrium in the x-direction
 g x h wdx  
h/2
h / 2

 xx nx wdy
h/ 2
 xx nx wdy
h / 2
left
0
right
2) Force equilibrium in the y-direction
 g y h wdx  
h/2
h / 2
 xy nx wdy
left

h/2
h / 2
 xy nx wdy
Mz = r ´ F
3) Moment equilibrium about the z - axis
ò y ˆj ´ (s
n
xx x
) iˆ wdy
+ ò y ˆj ´ (s xx nx ) iˆ wdy
left
dx ˆ
i´
2
h/2
ò
- h/2
0
right
s xy nx ˆj wdy
+
left
+
right
dx ˆ
i´
2
h/2
ò
- h/2
s xy nx ˆj wdy
=0
right
Remember: nx = -1 on the left face and 1 on the right face.
ˆj ´ iˆ = - kˆ
iˆ ´ ˆj = kˆ and
Hence, the moment equilibrium equation simplifies to
òy s
xx
-
wdy
left
-
òy s
dx
2
xx
wdy
right
h/2
ò
- h/2
s xy wdy
+
left
dx
2
h/2
ò
- h/2
s xy wdy
=0
right
Also, it is convenient to define some terms (F,M,V) referred to as stress resultants, as
follows. Note that these characterize stresses inside the body (note that there is no normal
vector component in these formulas since these are quantities defined inside the body).
Axial force: F =
òs
xx
wdy
Moment:
M = - ò y s xx wdy
Shear:
V=
òs
xy
wdy
Now we can express the three equilibrium equations as
2)
 g x h w dx   FL  FR  0
 g y h w dx  VL  VR  0
3)
 M L  M R  VL
1)
where the subscripts => left and right
dx
dx
 VR
0
2
2
ends
Express the stress resultants as follows
FL = F
VL = V
ML = M
and
dF
FR = F +
dx
dx
FR = F + dF
dV
or VR = V + dV
VR = V +
dx
dx
M R = M + dM
dM
MR = M +
dx
dx
Substitute these expressions into the equilibrium equations above. After dividing through
each equation by dx, we obtain
dF
  gx h w  0
dx
dV
 gy h w  0
dx
dM
V  0
dx
Generally we will replace  g x h w with f x and define it to be the axial force per unit
length. Similarly, we will define f y   g y h w to be the transverse force per unit length.
Also, we will often combine the last two equations to obtain

d2M
 fy  0
dx 2
Equilibrium Equations in Terms of Displacements
The stresses depend on the strains and the strains on the displacements, as shown below.
 u

 xx  E( xx  T)  E   T 
 x

 du

d2 v
 E  0  y 2  T 
dx
 dx

Hence, the equilibrium equations can be expressed in terms of displacements. The first
step is to express F and M in terms of displacements. Substitute the expression for stress
into the equations for F and M given earlier
Axial Force
 du

d2v
F   E  0  y 2  T  wdy
dx
 dx

du
d2v
  Ewdy 0   Eywdy 2   Twdy
dx
dx
2
du
d v
F= EA 0  B 2  FT
dx
dx
where EA   Ewdy
B   Eywdy
FT   Twdy
Moment
 du

d2v
M    yE  0  y 2  T  wdy
dx
 dx

du
d2v
   yEwdy 0   y 2 Ewdy 2   yETwdy
dx
dx
2
du
d v
 B 0  EI 2  M T
dx
dx
where
B   Eywdy
EI   y 2 Ewdy
M T   yETwdy
Now we can use these expressions for F and M in the equilibrium equations.
To keep things simple for this course, let's assume that B = 0 (This is always the
case if the beam is symmetric about the x-axis. It is also the case if we shift the location
of the x-axis so that  Eywdy  0 .)
Simplified equations for a symmetric beam
du 0
d 

 FT   f x  0
 EA
dx 
dx

Axial force equilibrium
Transverse force equilibrium


d2  d2v
EI 2 +M T   f y  0
2 
dx  dx

d2  d2 v  d2MT
 fy  0
 EI

dx 2  dx 2  dx 2
Note: The moment equilibrium equation was used to express V in terms of M in the
transverse equilibrium equation, so we have 2 rather than 3 equations.
or 
V cannot be expressed in terms of displacements through the constitutive relations, since
we assumed the transverse shear strain to be zero. (This is a well recognized
inconsistency.)
dM
However, moment equilibrium equation tells us that V  
. Hence
dx
d  d 2v 
V    EI 2 
dx  dx 
===============================================================
The equilibrium equations can be derived very quickly by simply considering a
differential element and summing forces and moments. The derivation above emphasizes
the connection between 3D elasticity and beam theory.