Beam Bending and Stress Analysis

Bending and extension of beams
The elastic field problem is solved by satisfying a set of 15 equations in 15 unknowns.
Out of these 9 are differential equations. The geometric shapes and boundary
conditions for most practical structures are so complex that exact theory of elasticity
solutions are either imposible to obtain or too complex to be useful in the field.
Approximate solutions using strength of materials techniques are accurate enough
for engineering purposes and are commonly used.
Stress Resultants
A beam is defined to be a member capable of resisting bending moments and whose
length is large compared to it's cross sectional area. It is assumed that the beam
length is along the x axis and that the cross sectional area A of the beam varies
smoothly in x. In beam theory, it is convenient to work with force and moment
resultants of the stresses that act upon the cross section of the beam.
Consider a beam with
externally applied loadings as
shown below. The externally
applied forces per unit length
are px, py and pz and moments
per unit length are mx,my and
mz. The two loadings are
shown on two separate figures
below. Note that externally
forces and moments are
assumed to be positive if they
have the same sense as their
respective coordinate
directions.
Bending and extension of beams file:///home/kamle/Desktop/Old_HDD/courses_combin...
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Now suppose that a cutting
plane is passed through the
beam at a distance x from the
origin and a free body diagram
is constructed as shown. In
general, six internal resultants
can be constructed at the
intersection of the x axis with
the cross section defined by
the cutting plane.
On element area dA, Force=dF̂
dF̂= σxxdA îx + σxydA îy +
σxzdA îz
Total force on face = ∫ (σxxdA
îx + σxydA îy + σxzdA îz)
or, F̂=(P îx+ Vy îy+ Vz îz)
Due to element are dA, the
moment dM̂ is
dM̂= r̂ × dF̂
The total moment on the face
is the integral of this value.
M̂=(Mx îx+ My îy+ Mz îz)
The six stress resultants can
now be written as:
Axial Force P = ∫ σxxdA
Shear Force Vy = ∫ σxydA
Shear Force Vz = ∫ σxzdA
Torsional Moment Mx = ∫ (yσxz-zσxy)dA
Bending Moment My = ∫ zσxxdA
Bending Moment Mz = ∫ (-yσxx)dA
The face on which these internal resultants act is called a positive face because the x
coordinate axis is coming out of it. Internal resultants are defined to be positive when

they act in their respective coordinate axis directions on a positive face.
Linear Differential equations of equilibrium
For the beam shown in the above figure subjected to externally applied forces per
unit length px, py and pz and moments per unit length mx,my and mz, the differential
equations of equilibrium can be written as:
dP/dx = -px
dVy/dx = -py
dVz/dx = -pz
dMx/dx = -mx
dMy/dx = -my + Vz(x)
dMz/dx = -mz - Vy(x)
The above equations satisfy the equilibrium equations only in an average sense over
the cross section. The above equations can be solved if the boundary conditions are
known.
Stresses due to extension and bending
Normal stresses σxx are developed in the x coordinate direction due to the axial
loading P and bending moments My and Mz. We will be making the following
assumptions (called Bernoulli Euler assumptions):
The transverse
components of normal
stress σyy and σzz are
assumed to be negligible
compared to the axial
stress σxx.
Cross sections are
assumed to remain
planar and normal to the

centroidal axis of
deformation.
It can be seen that the axial displacement u(x,y,z) is the sum of the contributions due
to the axial extension u0=u(x,0,0), rotation due to bending about the z and y axes
respectively:
u(x,y,z)=u0 - θz(x)y + θy(x)z
The axial strain εxx is given by:
εxx= ∂u/∂x = du0/dx -y dθz/dx + z dθy/dx
Consider a beam whose material properties are hetrogenous. Substitution in linear
elastic stress strain equations gives:
σxx= E (du0/dx -y dθz/dx + z dθy/dx)
Substitution into resultant equations gives:

P = ∫E (du0/dx -y dθz/dx + z dθy/dx) dA
My = ∫E (du0/dx -y dθz/dx + z dθy/dx) z dA
Mz = - ∫E (du0/dx -y dθz/dx + z dθy/dx) ydA
Now define the modulus weighted section properties as follows:
A* = dA
ȳ* = y dA
z̄* = z dA
I*
yy = z2 dA
I*
yz = yz dA
I*
zz = y2 dA
In the case of a homogeneous beam, the reference modulus E1 may be chosen to be
the same as actual material modulus. In heterogeneous beams, the modulus weighted
centroidal axes(x*,y*,z*) should be used. If the modulus weighted centroidal axes are
chosen, then ȳ* and z̄* defined above will be identically zero. With such a choice, the
equations become:
du0/dx = P/ (E1 A*)
dθz/dx = (MzI*
yy+MyI*
yz)/ (E1(I*
yyI*
zz- I*
yz
2))
dθy/dx = (MyI*
zz+MzI*
yz)/ (E1(I*
yyI*
zz- I*
yz
2))
Substituting these results into the strain equations gives:
εxx = P/ (E1 A*) -y (MzI*
yy+MyI*
yz)/ (E1(I*
yyI*
zz- I*
yz
2))

+z (MyI*
zz+MzI*
yz)/ (E1(I*
yyI*
zz- I*
yz
2))
σxx = E P/ (E1 A*) -y E(MzI*
yy+MyI*
yz)/ (E1(I*
yyI*
zz- I*
yz
2))
+z E(MyI*
zz+MzI*
yz)/ (E1(I*
yyI*
zz- I*
yz
2))
Determination of modulus weighted section properties
Consider a beam composed of n discrete portions of homogeneous make up. In this
case the equations for A* etc. reduce to:
A* = Ei/E1 dA
or, A* = dA
or, A*= Ai
Also,
ȳ* = y dA
or, ȳ'* = ȳi'Ai
Similarly z̄* = z dA
or, z̄'* = z̄i'Ai
where ȳi' and z̄i' are the coordinates of the ith portion taken with respect to the
arbitrary y' and z' axes.
The modulus weighted moments of inertia may be determined about the modulus
weighted centroid, or alternatively, we can determine these properties about the
arbitrary y' and z' axes and then use the parallel axis theorem.

I*
yy = I*
y'y' - (z̄'*)2 A*
I*
yz = I*
y'z' - (ȳ'*)(z̄*') A*
I*
zz = I*
z'z' - (ȳ'*)2 A*
Example Problem 1
The figure below shows an
unsymmetrical beam section
composed of four stringers
a,b,c,d of equal area 0.1 sq.
inch connected by a thin web.
Determine stress and total load
on each stringer if the
moments applied on the
section are My=10,000 lb-in
and Mz=5000 lb-in.
Example Problem 2
An airfoil is idealized as prismatic
with average cross section shown
below. All stringers are made of
structural steel (E= 30 X 106 psi) with
dimensions shown below. All skins
are Aluminum(E=10 X 106 psi).
determine the modulus weighted
section properties assuming E1= 10 X
106 psi.

Example Problem 3
The idealized airfoil described
in the previous example is
subjected to the loading
shown. Determine the stresses
in the stringers at x=0.
Deformations due to bending and extension
It is possible to determine u(x,y,z), v(x,y,z) and w(x,y,z) at various points in the beam.
It is of practical importance to determine the three displacement components along
the centroidal axes i.e. u(x,0,0),v(x,0,0) and w(x,0,0).The axial component u0= u(x,0,0)
can be ontained as:
du0/dx = P/(E1 A*)
To obtain v0 and w0, we use the relations :
dθz/dx = d2v0/dx2
dθy/dx = d2w0/dx2
The above equations containing the variables u0,v0,w0 can be solved by integrating
directly after applying displacement boundary conditions to account for integration
constants.

Beam Bending and Stress Analysis

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