Calculation of the Natural Frequency of Steel and Wooden Glued- Laminated Beams of Similar Load Bearing Capacities.

A Project for

Math 222-Differential Equations

Spring 2004 FVCC

By

Marc Pitman

**Abstract:** Two simply supported beams with symmetric
overhang, one made of steel and the other of glued-laminated wood, are designed
to carry a uniformly distributed load of 300 lb/ft. Normal functions and boundary conditions are
applied to develop a system of 12 equations in 12 unknowns. These equations are solved using Matlab. The
fourth order differential equation, used to model the transverse vibration of
beams and the general solution to this equation are discussed in appendices B
and C.

**Contents: **

** **

** **Beam Design:

Shear
and Moment calculations

Steel
Beam Design

** **Glued-Laminated Wood Beam Design

Solving a System of

Normal Equations

Boundary Conditions

Determinant Solution using Matlab

Calculating Natural Frequency

Appendix A References

Appendix B Differential
Equation for Vibration

Appendix C
General Solution to the Differential Equation

**Beam
Design:** Design steel and
glued-laminated wood beams that are 50 feet long, simply supported with 5 foot
overhangs, and carrying a uniformly distributed load of 300 lb/ft.

Figure 1 Beam geometry and load schematic.

First calculate reactions at supports.

Figure 2 Equivalent point load and dimensions.

Calculation of reactions:

_{}

_{}

Shear diagram:

Figure 3 Shear diagram

Maximum shear occurs at the supports and is 7 kips.

Moment Diagram:

The left end is a free end so the moment is zero.

The moment between the left end and the left reaction, R1, (in kip-ft) is:

_{}

The moment to the right of R1 and left of R2 is:

_{}

Maximum moment occurs when the derivative of m(x) is equal to 0.

_{}

Figure 4 Moment diagram

The maximum shear is 6.0kips and the maximum bending moment is 56.25 kip-ft.

Steel beam design:

Using the __Manual of Steel Construction Allowable Stress
Design__ [3] a W shape (wide flange) beam constructed of steel with a yield
stress, F_{y} = 36 ksi and an unbraced length,
L_{c} = 5 ft.

The max bending stress, F_{b} is:

_{}

The required section modulus, S_{x} is:

_{}

From the Allowable Stress Design Selection Table a W 14 x 22
has an S_{x} = 29.0 in^{3}.^{ }The W 14 x 22 is chosen. L_{u}
for this shape is 5.6 ft. M_{u}
is 66 Kip-ft.

This beam has the following properties:

Area = 6.49 in^{2}

^{ }

Weight 22 lb/ft

I_{xx}=199 in^{4}

I_{yy}=7.00 in^{4}

^{ }

*E* = 2.9 x 10^{7} psi (modulus of elasticity)

*G* = 11.5 x 10^{6} psi (shear modulus)

*r* = 490
lbm/ft^{3} (density)

^{ }

^{ }

^{ }

^{ }^{}

^{ }

^{Figure.}^{ 5 W-Shape steel beam selected for
problem.}^{}

Wood beam design:

For Glued Laminated Timber, The American

From The NDS [4], using a dry service condition the only factor for the bending stress will be the volume factor:

_{}

Where:

*K _{L}
= *1.0 for a single span uniformly distributed load.

*L* =
40’ the length of the span between supports.

*x *= 10 for all wood species
other than southern pine.

_{}

The maximum bending stress is the allowable bending stress multiplied by the volume factor:

*F _{b'}*

For the design *F _{b }*is:

_{}

For a 5-1/8" x 21" glued-laminated beam
constructed of outer and inner laminations of Douglas fir, the allowable
bending stress, *F _{b}* is 2200.
Multiplying this by the volume factor,

5-1/8" x 21" glued-laminated beam can be used.

Area = 107.625" *E* = 1.6 x 10^{6} psi

**Figure 6.**** Selected dimension for wood beam**

I = (b x d^{2})/12 = 188.34 in^{4}** **

*r* = 35.1
lbm/ft^{3} (density)** **

**Solution of system of equations for k:**

The fourth order differential equation that models transverse vibration along a beam is:

_{} (See appendix A)

The general solution to this differential equation has the form:

_{} (See appendix B)

and for the transverse vibration of beams:

_{}

Where:

_{}= beam natural frequency,

_{}= beam modulus of elasticity,

_{} = beam moment of
inertia,

_{}= beam mass density, and

_{}= beam cross-sectional area

*C*_{1}, *C*_{2}, *C*_{3},
and *C*_{4 }are determined from boundary conditions at the ends of
a beam.

Because the beam designed in this project has symmetrical
overhangs, it will be divided into three sections. Each section will have a separate coordinate
system for measuring the distance *x* with the origin for each section
being at the left end of each section.

In general the geometry of such a beam is as shown below:

Figure 7 Beam geometry and coordinate system

The following normal functions and corresponding first, second, and third derivatives are shown for each section of the beam:

Left overhang:

Center span:

Right overhang:

The following boundary conditions are applied:

At the ends of the beam both moment and shear have to be zero. At the supports deflection is zero, slope and moment are continuous.

At *x*_{1}=0
*d *^{2}*X*_{1}/*dx*_{1}^{2}=0 Eq. 1

*d** *^{3}*X*_{1}/*dx*_{1}^{3}=0 Eq. 2

At *x*_{1}=d=5, *x*_{2}=0
*X*_{1}=0 Eq.
3

*X*_{2}=0 Eq.
4

*dX*_{1}/*dx*_{1}_{ }^{-
}*dX*_{2}/*dx*_{2}=0 Eq. 5

*d** *^{2}*X*_{1}/*dx*_{1}^{2}
- *d *^{2}*X*_{2}/*dx*_{2}^{2}=0 Eq. 6

At *x*_{2}=*S*=40, *x*_{3}=0
*X*_{2}=0 Eq.
7

*X*_{3}=0 Eq.
8

*dX*_{2}/*dx*_{2}_{ }^{–
}*dX*_{3}/*dx*_{3}=0 Eq. 9

*d** *^{2}*X*_{2}/*dx*_{2}^{2}
- *d *^{2}*X*_{3}/*dx*_{3}^{2}=0 Eq. 10

At *x*_{3}=d=5 *d
*^{2}*X*_{3}/*dx*_{3}^{2}=0 Eq. 11

*d** *^{3}*X*_{3}/*dx*_{3}^{3}=0 Eq. 12

Applying these 12 boundary conditions yields the following system of equations:

The system of equations has a nontrivial solution when the determinant of the 12 X 12 matrix = 0.

The determinant of this system of homogenous equations is calculated for a range of expected values of k and plotted using Matlab and the following .m file:

L=50;

S=40;

d=(L-S)/2;

D_vect=[];

K_vect=[];

%initial value of k (minimum expected value)

k=.063;

%increment by j

j=.00001;

%number of iterations required to reach maximum expected value of k

n=2000;

for i=1:n

k=k+j;

K_vect=[K_vect,k];

%defines each equation

a1=[-k^2 0 k^2 0 0 0 0 0 0 0 0 0];

a2=[0 -k^3 0 k^3 0 0 0 0 0 0 0 0];

a3=[cos(d*k) sin(d*k) cosh(d*k) sinh(d*k) 0 0 0 0 0 0 0 0];

a4=[0 0 0 0 1 0 1 0 0 0 0 0];

a5=[-k*sin(d*k) k*cos(d*k) k*sinh(d*k) k*cosh(d*k) 0 -k 0 -k 0 0 0 0];

a6=[-k^2*cos(d*k) -k^2*sin(d*k) k^2*cosh(d*k) k^2*sinh(d*k) k^2 0 -k^2 0 0 0 0 0];

a7=[0 0 0 0 cos(S*k) sin(S*k) cosh(S*k) sinh(S*k) 0 0 0 0];

a8=[0 0 0 0 0 0 0 0 1 0 1 0];

a9=[0 0 0 0 -k*sin(S*k) k*cos(S*k) k*sinh(S*k) k*cosh(S*k) 0 -k 0 -k];

a10=[0 0 0 0 -k^2*cos(S*k) -k^2*sin(S*k) k^2*cosh(S*k) k^2*sinh(S*k) k^2 0 -k^2 0];

a11=[0 0 0 0 0 0 0 0 -k^2*cos(d*k) -k^2*sin(d*k) k^2*cosh(d*k) k^2*sinh(d*k)];

a12=[0 0 0 0 0 0 0 0 k^3*sin(d*k) -k^3*cos(d*k) k^3*sinh(d*k) k^3*cosh(d*k)];

%defines the 12 X 12 Matrix

A=[a1;a2;a3;a4;a5;a6;a7;a8;a9;a10;a11;a12];

D=det(A);

D_vect=[D_vect,D];

end

plot(K_vect,D_vect)

grid on

xlabel('k')

ylabel('Det(A)')

shg

eq=polyfit(K_vect,D_vect,6);

k=max(roots(eq))

hold on

plot(k,0,'ro')

text(k,0,' (k,0)','HorizontalAlignment','Left','VerticalAlignment','Bottom')

title('k vs. determinant of normal functions')

The output and graph plotted by the above .m file are:

k = 0.0780

Figure 8 k vs. det(A)

The determinant of the matrix is zero near a value of *k*
= 0.078.

This closely matches the results found in fig.2 of Murphy
[1]. *K*_{1 }shown on that
figure is calculated as follows:

_{}

Using the value for *k* as determined above for the
given beam geometry *K*_{1 }is:

_{}

*S/L *for this beam is 40/50 = 0.8.

The values for *K*_{1} and *S/L* match the
curve very well as shown in the figure below:

The natural frequency, *f *can be calculated from the
known properties of the beams designed for this project using the following
equation:

_{}

Where *k* = 0.0780 and

For the steel W14 x 22 beam:

A = 6.49 in^{2}

*r* = 490
lbm/ft^{3} =0.2836 lbm/in^{3}

*E* = 2.9 x 10^{7} psi

I_{xx}=199 in^{4}

_{}

^{ }

For a 5-1/8" x 21" glued-laminated beam

A= 107.625"

*r* = 35.1
lbm/ft^{3} =0.0203 lbm/in^{3}

*E* = 1.6 x 10^{6} psi

I = 188.34 in^{4}** **

_{}

The natural frequency of the steel beam is about 4.77 times greater than that of the wood beam.

**Appendix A**

**References**

**Derivation of the
Differential Equation of Transverse Vibration [2]**

From this free body diagram and the equation of equilibrium of the
vertical forces according to

Eq. 1

The sum of moments about any point of the element yields:

Eq. 2
_{}

Substituting Eq. 2 into Eq. 1:

Eq. 3 _{}

Since the beam is assumed to have a plane of symmetry and the vibrations occur in that plane, the known differential equation of the deflected curve is:

Eq. 4 _{}

Combining Eq. 4 and Eq. 3:

Eq. 5 _{}

Using the method of separation of variables:

Eq. 6 _{}

Substituting Eq. 6 into Eq. 5 and letting_{}:

Eq. 7 _{}

The term on the left side of Eq. 7 is dependent only on *x*
and the right side only on *t*. To
be equal to each other the sides must both be equal to the same constant, such as _{}. Then the left hand
side of Eq. 7 can be written as:

Eq. 8 _{}

Where: _{}

Or if _{}:

Eq. 9 _{}

**General Solution
to the Differential Equation of Transverse Vibration**

The differential equation:

** **_{}

Has the characteristic equation:

_{}

Which factors as:

_{}

Which gives the eigenvalues:

_{}

Which yields the general solution:

_{}

Or:

_{}