# Galerkin Analysis of Effect of Dead Load on Natural Frequencies of Box Beam

Open Access
Original Contribution

## Keywords

Effect of dead load Box beam Natural frequency Hamilton’s principle Galerkin method

## Establishment of Differential Equation

### Basic Assumption

The cross section and coordinate system which are used for deriving basic are shown in Figs. 1 and 2. Figure 2 represents the formula and force–deformation conditions of the beam when $$\hat{v}$$ represents the deflection of beam under dead loads ($$\hat{p}$$); $$\bar{v}$$ represents the deflection of beam under live loads ($$\bar{p}$$). The deflection state of beam under dead loads is used as the reference state, and $$\bar{v}$$ is derived from this reference state. Fig. 1Cross section Fig. 2Coordinate system and loads distribution
At the assumption of the deformation of beam obeying the hypothesis of Bernoulli–Euler beam equation,the linear strain–displacement relationship can be written as ,
$$\varepsilon_{x} = - yv^{\prime\prime}$$
(1)
where top slab
$$\varepsilon_{xu} = h_{u} v^{\prime\prime}$$
(2)
Bottom slab
$$\varepsilon_{xb} = - h_{b} v^{\prime\prime}$$
(3)
Web slab
$$\varepsilon_{xw} = - yv^{\prime\prime}$$
(4)
The nonlinear strain–displacement relationship can be written as
$$\varepsilon_{x} = - yv^{\prime\prime} + \frac{1}{2}\left( {v^{\prime}} \right)^{2}$$
(5)
where top slab
$$\varepsilon_{xu} = h_{u} v^{\prime\prime} + \frac{1}{2}\left( {v^{\prime}} \right)^{2}$$
(6)
Bottom slab
$$\varepsilon_{xb} = - h_{b} v^{\prime\prime} + \frac{1}{2}\left( {v^{\prime}} \right)^{2}$$
(7)
Web slab
$$\varepsilon_{xw} = - yv^{\prime\prime} + \frac{1}{2}\left( {v^{\prime}} \right)^{2}$$
(8)
where εx is the strain of beam; v is the deflection of beam.

### Total Energy Expression

The strain energy under live loads can be written as
$$U = \hat{U} + \bar{U}$$
(9)
where $$\hat{U}$$ is the strain energy under dead loads ($$\hat{p}$$); $$\bar{U}$$ is the strain energy under live loads ($$\bar{p}$$),
$$\bar{U} = \frac{1}{2}\iiint_{V} {\bar{\sigma }_{x} }\bar{\varepsilon }_{x} {\text{d}}V$$
(10)
$$\hat{U} = \iiint_{V} {\hat{\sigma }_{x} \bar{\varepsilon }_{x} {\text{d}}V}$$
(11)
where the positive stress $$\left( {\hat{\sigma }_{x} ,\bar{\sigma }_{x} } \right)$$ is the bending normal stress produced by dead loads ($$\hat{p}$$) and live loads ($$\bar{p}$$), respectively. It should be noted that the direct stress $$\left( {\hat{\sigma }_{x} } \right)$$ is a constant stress caused by dead loads, but the direct strain $$\left( {\bar{\varepsilon }_{x} } \right)$$ is a strain caused by live loads. The linear strain–displacement relationship provided by Eq. (1) is used for the calculation of $$\bar{U}$$. The nonlinear strain–displacement relationship provided by Eq. (5) is used for the calculation of $$\hat{U}$$. It should be noted that the dead load’s effect of this part is the deterrent effect caused by both the middle surface tension of beam and the bending state stress of dead loads $$\left( {\hat{\sigma }_{x} } \right)$$. It is assumed that the stress–strain relation is linear.
$$\bar{\sigma }_{x} = E\bar{\varepsilon }_{x}$$
(12)
$$\hat{\sigma }_{x} = E\hat{\varepsilon }_{x}$$
(13)

All the strain energy caused by live loads ($$\bar{p}$$) can be expressed as follows:

The strain energy of top slab
$$\bar{U}_{u} = \frac{1}{2}\iiint_{V} {\bar{\sigma }_{xu} }\bar{\varepsilon }_{xu} {\text{d}}V = \frac{1}{2}\iiint_{V} {E\left( { - h_{u} v\prime\prime} \right)^{2} }{\text{d}}V = \frac{1}{2}\int_{0}^{L} {\iint_{A} {Eh_{u}^{2} v^{\prime\prime2} }{\text{d}}A{\text{d}}x} = \frac{1}{2}\int_{0}^{L} {EI_{su} } v^{\prime\prime2} {\text{d}}x$$
(14)
The strain energy of bottom slab
$$\bar{U}_{b} = \frac{1}{2}\int_{0}^{L} {EI_{bu} \bar{v}^{\prime\prime2} {\text{d}}x}$$
(15)
The strain energy of web slab
$$\bar{U}_{w} = \frac{1}{2}\int_{0}^{L} {EI_{w} } \bar{v}^{\prime\prime2} {\text{d}}x$$
(16)
So, the strain energy caused by live loads ($$\bar{p}$$) can be written as
$$\bar{U} = \bar{U}_{u} + \bar{U}_{b} + \bar{U}_{w} = \frac{1}{2}\int_{0}^{L} {E\left( {I_{su} + I_{bu} + I_{w} } \right)} \bar{v}^{\prime\prime2} {\text{d}}x = \frac{1}{2}\int_{0}^{L} {EI} \bar{v}^{\prime\prime2} {\text{d}}x$$
(17)

All kinds of strain energy caused by dead loads ($$\hat{p}$$) can be expressed as follows:

The strain energy of top slab
\begin{aligned} \hat{U}_{u} &= \iiint_{V} {\hat{\sigma }_{x} \bar{\varepsilon }_{x} {\text{d}}V} \\ &= \iiint_{V} {E\left( {h_{u} \hat{v}^{\prime\prime} + \frac{1}{2}\hat{v}^{\prime2} } \right)\left( {h_{u} \bar{v}^{\prime\prime} + \frac{1}{2}\bar{v}^{\prime2} } \right)}{\text{d}}V \\ &= \int_{0}^{L} {\iint_{A} E\left( {h_{u}^{2} \hat{v}^{\prime\prime} \bar{v}^{\prime\prime} + \frac{1}{2}h_{u} \hat{v}^{\prime\prime} \bar{v}^{\prime2} + h_{u} \hat{v}^{\prime2} \bar{v}^{\prime\prime} + \frac{1}{4}\hat{v}^{\prime2} \bar{v}^{\prime2} } \right){\text{d}}z{\text{d}}y{\text{d}}x} \\ &= \int_{0}^{L} {E\left[ {I_{su} \hat{v}^{\prime\prime} \bar{v}^{\prime\prime} - \frac{1}{2}S_{su} \hat{v}^{\prime\prime} \bar{v}^{\prime2} - \frac{1}{2}S_{su}^{prime} \hat{v}^{\prime2} \bar{v}^{\prime\prime} + \frac{{A_{su} }}{4}\hat{v}^{\prime2} \bar{v}^{\prime2} } \right]} {\text{d}}x \\ \end{aligned}
(18)
The strain energy of bottom slab
$$\hat{U}_{b} = \int_{0}^{L} {E\left[ {I_{sb} \hat{v}^{\prime\prime} \bar{v}^{\prime\prime} - \frac{1}{2}S_{sb} \hat{v}^{\prime\prime} \bar{v}^{\prime2} - \frac{1}{2}S_{sb} \hat{v}^{\prime2} \bar{v}^{\prime\prime} + \frac{{A_{sb} }}{4}\hat{v}^{\prime2} \bar{v}^{\prime2} } \right]} {\text{d}}x$$
(19)
The strain energy of web slab
\begin{aligned} \hat{U}_{w} &= \iiint_{V} {E\left( { - y\hat{v}^{\prime\prime} + \frac{1}{2}\hat{v}^{\prime2} } \right)\left( { - y\bar{v}^{\prime\prime} + \frac{1}{2}\bar{v}^{\prime2} } \right)}{\text{d}}V \\ &= \iiint_{V} {E\left( {y^{2} \hat{v}^{\prime\prime} \bar{v}^{\prime\prime} - \frac{1}{2}y\hat{v}^{\prime\prime} \bar{v}^{\prime2} - \frac{1}{2}y\hat{v}^{\prime2} \bar{v}^{\prime\prime} + \frac{1}{4}\hat{v}^{\prime2} \bar{v}^{\prime2} } \right){\text{d}}V} \\ &= \int_{0}^{l} {\iint_{A} {\left[ {Ey^{2} \left( {\hat{v}^{\prime\prime} \bar{v}^{\prime\prime} } \right) - Ey\left( {\frac{1}{2}\hat{v}^{\prime\prime} \bar{v}^{\prime2} + \frac{1}{2}\hat{v}^{\prime2} \bar{v}^{\prime\prime} } \right) + \frac{1}{4}\hat{v}^{\prime2} \bar{v}^{\prime2} } \right]{\text{d}}z{\text{d}}y{\text{d}}x}} \\ &= \int_{0}^{L} {E\left[ {I_{w} \hat{v}^{\prime\prime} \bar{v}^{\prime\prime} - \frac{1}{2}S_{w} \left( {\hat{v}^{\prime\prime} \bar{v}^{\prime2} + \hat{v}^{\prime2} \bar{v}^{\prime\prime} } \right) + \frac{{A_{w} }}{4}\hat{v}^{\prime2} \bar{v}^{\prime2} } \right]} {\text{d}}x \\ \end{aligned}
(20)
So, the strain energy caused by live loads ($$\hat{p}$$) can be given as
\begin{aligned} \hat{U} = \hat{U}_{u} + \hat{U}_{b} + \hat{U}_{w} \\ = \int_{0}^{L} {E\left[ {\left( {I_{su} + I_{sb} + I_{w} } \right)\hat{v}^{\prime\prime} \bar{v}^{\prime\prime} - \frac{1}{2}\left( {S_{su} + S_{sb} + S_{w} } \right)\hat{v}^{\prime\prime} \bar{v}^{\prime2} } \right.} - \frac{1}{2}\left( {S_{su} + S_{sb} + S_{w} } \right)\hat{v}^{\prime2} \bar{v}^{\prime\prime} \left. { + \frac{1}{4}\left( {A_{su} + A_{sb} + A_{w} } \right)\hat{v}^{\prime2} \bar{v}^{\prime2} } \right]{\text{d}}x \\ = \int_{0}^{L} {\left[ {EI\hat{v}^{\prime\prime} \bar{v}^{\prime\prime} + \frac{{EA_{w} }}{4}\hat{v}^{\prime2} \bar{v}^{\prime2} } \right]} {\text{d}}x \\ \end{aligned}
(21)
The total strain energy of the box beam becomes
$$U = \bar{U} + \hat{U} = \int_{0}^{l} {\left[ {\frac{1}{2}EI\left( {\bar{v}^{\prime\prime} } \right)^{2} + \frac{EA}{4}\left( {\hat{v}^{prime} } \right)^{2} \left( {\bar{v}^{\prime}} \right)^{2} + EI\hat{v}^{\prime\prime} \bar{v}^{\prime\prime} } \right]{\text{d}}x}$$
(22)
The potential energy of external loads is
$$V = - \int_{0}^{l} {\left( {\bar{p} + \hat{p}} \right)\bar{v}{\text{d}}x}$$
(23)
With the effect of rotational inertia neglected, the kinetic energy of beam can be expressed as follows:
$$T = - \int_{0}^{l} {\frac{\rho A}{2}\left( {\dot{\bar{w}}} \right)^{2} {\text{d}}x}$$
(24)

## Controlling Differential Equation

According to Hamilton’s principle, as an object move from a location at the time (t1) to another location at the time (t2), to all the distance the object may experience, the distance follows Newton’s law at every moment is the distance makes the value of Lagrangian function average on time become the extreme value [16, 17] ,
$$\delta I = \delta \int_{{t_{1} }}^{{t_{2} }} {\left( {T - U - V} \right){\text{d}}t} = 0$$
(25)
Substituting Eqs. (22), (23), (24) into Eq. (25) gives
\begin{aligned} \delta I & = \int_{{t_{1} }}^{{t_{2} }} {\int_{0}^{L} {\left\{ {\rho A\ddot{\bar{v}} + EI\bar{v}^{\prime\prime\prime\prime} - \frac{EA}{2}\left[ {\left( {\hat{v}^{\prime} } \right)^{2} \bar{v}^{\prime} } \right]^{\prime} + EI\hat{v}^{\prime\prime\prime\prime} - \left( {\bar{p} + \hat{p}} \right)} \right\}\delta v{\text{d}}x{\text{d}}t} } \\ & \quad + \int_{{t_{1} }}^{{t_{2} }} {\left. {\left( {EI\bar{v}^{\prime\prime} + EI\hat{v}^{\prime\prime} } \right)\delta v^{\prime} } \right|_{0}^{L} {\text{d}}t} + \int_{{t_{1} }}^{{t_{2} }} {\left. {\left[ { - EI\bar{v}^{\prime\prime\prime} - EI\hat{v}^{\prime\prime\prime} + \frac{EA}{2}\left( {\hat{v}^{\prime} } \right)^{2} \bar{v}^{\prime} } \right]\delta \bar{v}} \right|_{0}^{L} {\text{d}}t} \\ & = 0 \\ \end{aligned}
(26)
where, as the beam is subjected to dead loads, one can get the following equation,
$$EI\hat{v}^{\prime\prime\prime\prime} - \hat{p} = 0$$
(27)
The boundary conditions are that,
\begin{aligned} \hat{v} = 0 \, , \, or\;\hat{v}^{\prime} = 0,\quad x = 0,L; \hfill \\ \hat{v}^{\prime\prime} = 0 \, ,or\;\hat{v}^{\prime\prime\prime} = 0,\quad x = 0,L. \hfill \\ \end{aligned}
(28)
In Eq. (26), considering the effect of dead loads on bending deformation, the control differential equation of the beam can be written as
$$\rho A\ddot{\bar{v}} + EI\bar{v}^{\prime\prime\prime\prime} - \frac{EA}{2}\left[ {\left( {\hat{v}^{\prime} } \right)^{2} \bar{v}} \right]^{\prime} = \bar{p}$$
(29)
The boundary conditions are that
$$\left\{ {\begin{array}{*{20}l} {\bar{v} = 0, \, or\; \bar{v}^{\prime} = 0,x = 0,L \, ;} \hfill \\ {\bar{v}^{\prime\prime} = 0,or\; EI\bar{v}^{\prime\prime\prime} - \frac{EA}{2}\left( {\hat{v}^{\prime}} \right)^{2} \bar{v}^{\prime} = 0, \, x = 0,L \, .} \hfill \\ \end{array} } \right.$$
(30)

## The Solution of Differential Equation

Considering the effect of dead loads, controlling Eq. (29) can be modified as
$$\rho A\ddot{\bar{v}} + EI\bar{v}^{\prime\prime\prime\prime} - \frac{EA}{2}\left[ {\left( {\hat{v}^{\prime}} \right)^{2} \bar{v}} \right]^{\prime } = \bar{p}$$
(31)
The boundary conditions are
$$\left\{ {\begin{array}{*{20}l} {\bar{v} = 0, \, or \; \bar{v}^{\prime} = 0, \, x = 0,L;} \hfill \\ {\bar{v}^{\prime\prime} = 0,or\; EI\bar{v}^{\prime\prime\prime} - \frac{EA}{2}\left( {\hat{v}^{\prime}} \right)^{2} \bar{v}^{\prime} = 0 \, , \, x = 0,L.} \hfill \\ \end{array} } \right.$$
(32)
The displacement ($$\bar{v}$$) can be expressed as
$$\bar{v}\left( {x,t} \right) = \bar{\varPsi }\left( x \right)\varPhi \left( t \right)$$
(33)
Substituting Eq. (33) into Eq. (31) gives
$$\bar{\varPsi }^{\prime\prime\prime\prime} - \frac{1}{{2r^{2} }}\left[ {2\hat{v}^{\prime}\hat{v}^{\prime\prime} \bar{\varPsi }^{\prime} + \left( {\hat{v}} \right)^{2} \bar{\varPsi }^{\prime\prime} } \right] = k^{4} \bar{\varPsi }$$
(34)
$$\ddot{\varPhi }\left( t \right) + \omega^{2} \varPhi \left( t \right) = 0$$
(35)
Among them,
$$k^{2} = \omega \sqrt {{{\rho A} \mathord{\left/ {\vphantom {{\rho A} {EI}}} \right. \kern-0pt} {EI}}}$$
(36)
Define the parameter $$\bar{v}$$ as
$$\bar{\varPsi }\left( x \right) = \bar{\varPsi }_{n} f_{n} \left( x \right);\;\left( {n = 1,2,3, \ldots \infty } \right)$$
(37)

Among them, the definition of fn is defined as follows:

For simply supported box beam,
$$f_{n} = \sin \frac{n\pi x}{L}$$
(38)
For box beam fixed at both ends,
$$f_{n} = \sin \frac{\pi x}{L}\sin \frac{n\pi x}{L}$$
(39)
For cantilevered box beam,
$$f_{n} = 1 - \cos \frac{{\left( {2n - 1} \right)\pi x}}{2L}$$
(40)
For box beam fixed at one end and simply supported at the other end,
$$f_{n} = \left[ {1 - \cos \frac{{\left( {2n - 1} \right)\pi x}}{L}} \right]\sin \frac{\pi x}{L}$$
(41)
By using Galerkin method, the controlling differential equation for the effect of bending deformation on the natural frequencies of beam can be expressed as
$$\int_{0}^{L} {Q\delta \bar{\varPsi }{\text{d}}x = 0}$$
(42)
$$\delta \bar{\varPsi }_{n} : { }\sum\limits_{m = 1} {\bar{\varPsi }_{m} \left( {A_{mn} - k^{4} B_{mn} } \right)} = 0$$
(43)
As the effect of initial bending deformation is considered, the natural frequencies of beam should be
$$\omega_{i} = k_{i}^{2} \sqrt {{{EI} \mathord{\left/ {\vphantom {{EI} {\rho A}}} \right. \kern-0pt} {\rho A}}} ;\;\left( {n = 1,2,3, \ldots \infty } \right)$$
(44)
Without considering the effect of initial bending deformation, the natural frequencies of beam are
$$\omega_{0n} = k_{0n}^{2} \sqrt {{{EI} \mathord{\left/ {\vphantom {{EI} {\rho A}}} \right. \kern-0pt} {\rho A}}} ;\;\left( {n = 1,2,3, \ldots \infty } \right)$$
(45)

Among them, the parameter k0n is defined as follows:

For simply supported box beam,
$$k_{0n}^{2} = \left( {{{n\pi } \mathord{\left/ {\vphantom {{n\pi } L}} \right. \kern-0pt} L}} \right)^{2}$$
(46)
For box beam fixed at both ends,
$$\cos \left( {k_{0n} L} \right)\cosh \left( {k_{0n} L} \right) = 1$$
(47)
For cantilevered box beam,
$$\cos \left( {k_{0n} L} \right)\cosh \left( {k_{0n} L} \right) = - 1$$
(48)
For box beam fixed at one end and simply supported at the other end,
$$\tan \left( {k_{0n} L} \right) = \, \tanh \left( {k_{0n} L} \right)$$
(49)

## Example and Analysis

The following parameters are adopted for the beam: E = 210 GPa; I0 = 2.5 × 10−4 m4; L0 = 8 m, A0 = 0.01 m2 (the reference value of gyration radius is that: $$r_{0} = \sqrt {I_{0} /A_{0} } = 0.158\;{\text{m}}$$). The dead load intensity is assumed to be: $$\hat{p}_{0} = 6.8\;{\text{kN}}/{\text{m}}$$. The variables ($$\hat{p}$$, l, I, r) can be expressed as follows: $$\hat{p} = \alpha_{{\hat{p}}} \hat{p}_{0}$$, $$l = \alpha_{l} l_{0}$$, I = αII0, r = αrr0. The numerical results change by the above parameters.

The effect of initial bending deformation on the natural frequencies of simple box beam is analyzed. So, in the following expression, ωi represents the circular frequencies of the ith vibration mode effected by initial bending deformation. ωi0 represents the circular frequencies of the ith vibration mode without the effect of initial bending deformation. $$\hat{p}_{0}$$ represents a reference dead load per unit length of the beams. $$\hat{p}$$ represents the practical value of initial load per unit length of the beams. Dimensionless form of coordinate, linear coordinate and logarithmic coordinate are adopted, respectively, for the lateral axis ($$\bar{p}_{0} /\hat{p}$$), while linear coordinate is used for the vertical axis $$\left( {\Delta = \left( {\omega_{i} - \omega_{i0} } \right)/\omega_{i0} \times 100\% } \right)$$.

The relationship between the natural frequencies of the box beam and the change of initial load is shown in Fig. 3. It’s evident that natural frequencies of simple box beams have improved with increase loads. Further, the improvement is more obvious for the first natural frequency. Here the increase of natural frequencies should be regarded as a real increase of the natural frequencies produced by initial load stress. It is different from the natural frequencies those vary with the normal change of beam quality (m) at unit length . Fig. 3Relationship between Δ and $$\bar{p}_{0} /\hat{p}$$ with various constraint conditions
Figures 4, 5 and 6 show the relationship between ∆ (affected by the change of sectional inertia moment αI, span length αL and gyration radius αr, respectively) and $$\bar{p}_{0} /\hat{p}$$ (affected by different constraint conditions). Among them, the gyration radius changes under the condition that the area changes and the sectional inertia moment keep constant. Figure 7 shows comparison of effects of initial loads when the above parameters (αI, αL and αr) of simply supported beam changes. It is demonstrated that the smaller the sectional inertia moment or the gyration radius (the area changes and the sectional inertia moment keeps constant) is or the larger the span length is, the larger effect of dead loads on the natural frequencies of beam will be. Among them, the change of span length makes the largest effect. Fig. 4Relationship between Δ and $$\bar{p}_{0} /\hat{p}$$ for various αI and constraint conditions Fig. 5Relationship between Δ and $$\bar{p}_{0} /\hat{p}$$ for various αL and constraint conditions Fig. 6Relationship between Δ and $$\bar{p}_{0} /\hat{p}$$ for various αr and constraint conditions Fig. 7Relationships between varying deterrent effect and $$\bar{p}_{0} /\hat{p}$$

## Conclusion

The following conclusions can be drawn:
1. 1.

The bending deformation generated by dead loads will produce deterrent effect which causes an increase of the natural frequencies of box beam. This trend will get more obvious when the natural frequencies belong to a lower-order range or larger weight.

2. 2.

The stiffness of box beam has an impact on the deterrent effect which caused by the initial bending deformation. It is concluded that more flexible structures which have a smaller sectional inertia moment, a larger span length or a lower gyration radius, the effect becomes more significant. Among them, the change of span length makes the largest effect.

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## Authors and Affiliations

• Jiawei Zhang
• 1
• Fangyu Luo
• 1
• Tingbin Liu
• 1
• Jianchang Zhao
• 1
• Sudipta Halder
• 2
1. 1.School of Civil EngineeringLanzhou Jiaotong UniversityLanzhouChina
2. 2.Department of Mechanical EngineeringNational Institute of Technology SilcharSilcharIndia

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