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
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.
Shear and Moment calculations
Steel Beam Design
Glued-Laminated Wood Beam Design
Solving a System of
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:
Figure 3 Shear diagram
Maximum shear occurs at the supports and is 7 kips.
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  a W shape (wide flange) beam constructed of steel with a yield stress, Fy = 36 ksi and an unbraced length, Lc = 5 ft.
The max bending stress, Fb is:
The required section modulus, Sx is:
From the Allowable Stress Design Selection Table a W 14 x 22 has an Sx = 29.0 in3. The W 14 x 22 is chosen. Lu for this shape is 5.6 ft. Mu is 66 Kip-ft.
This beam has the following properties:
Area = 6.49 in2
Weight 22 lb/ft
E = 2.9 x 107 psi (modulus of elasticity)
G = 11.5 x 106 psi (shear modulus)
r = 490 lbm/ft3 (density)
Figure. 5 W-Shape steel beam selected for problem.
Wood beam design:
For Glued Laminated Timber, The American
From The NDS , using a dry service condition the only factor for the bending stress will be the volume factor:
KL = 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:
Fb' = Fb x Cv
For the design Fb is:
For a 5-1/8" x 21" glued-laminated beam constructed of outer and inner laminations of Douglas fir, the allowable bending stress, Fb is 2200. Multiplying this by the volume factor, Cv = 0.882 gives a maximum allowable bending stress, Fb' = 1940 lb/in3. The
5-1/8" x 21" glued-laminated beam can be used.
Area = 107.625" E = 1.6 x 106 psi
Figure 6. Selected dimension for wood beam
I = (b x d2)/12 = 188.34 in4
r = 35.1
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:
= beam natural frequency,
= beam modulus of elasticity,
= beam moment of inertia,
= beam mass density, and
= beam cross-sectional area
C1, C2, C3, and C4 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:
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 x1=0 d 2X1/dx12=0 Eq. 1
d 3X1/dx13=0 Eq. 2
At x1=d=5, x2=0 X1=0 Eq. 3
X2=0 Eq. 4
dX1/dx1 - dX2/dx2=0 Eq. 5
d 2X1/dx12 - d 2X2/dx22=0 Eq. 6
At x2=S=40, x3=0 X2=0 Eq. 7
X3=0 Eq. 8
dX2/dx2 – dX3/dx3=0 Eq. 9
d 2X2/dx22 - d 2X3/dx32=0 Eq. 10
At x3=d=5 d 2X3/dx32=0 Eq. 11
d 3X3/dx33=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:
%initial value of k (minimum expected value)
%increment by j
%number of iterations required to reach maximum expected value of 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
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 . K1 shown on that figure is calculated as follows:
Using the value for k as determined above for the given beam geometry K1 is:
S/L for this beam is 40/50 = 0.8.
The values for K1 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 in2
r = 490 lbm/ft3 =0.2836 lbm/in3
E = 2.9 x 107 psi
For a 5-1/8" x 21" glued-laminated beam
r = 35.1 lbm/ft3 =0.0203 lbm/in3
E = 1.6 x 106 psi
I = 188.34 in4
The natural frequency of the steel beam is about 4.77 times greater than that of the wood beam.
Derivation of the Differential Equation of Transverse Vibration 
From this free body diagram and the equation of equilibrium of the
vertical forces according to
The sum of moments about any point of the element yields:
Substituting Eq. 2 into Eq. 1:
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:
Combining Eq. 4 and Eq. 3:
Using the method of separation of variables:
Substituting Eq. 6 into Eq. 5 and letting:
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:
Or if :
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: