Second Order Theory of Deflections for the
Linear Elastic Isotropic Beams 

Prof. Marcin Kamiński, Ph.D., D.Sc.
Chair of Mechanics of Materials,
Technical University of Ł?dź,
Al. Politechniki 6, 93-590 Ł?dź,
POLANDemail: Marcin.Kaminski@p.lodz.pl,
webpage: kmm.p.lodz.pl/pracownicy/Marcin_Kaminski/index.html 

Abstract: This script has been written to demonstrate a comparison between the first and the second order theories for the elastic isotropic beams deflection. As it is known, the second order theory enables to include directly the influence of the normal forces along the beam on its deflection function. From the mathematical point of view, the differential equation adequate to the deflection function is written on the deformed configuration of this beam, so that the normal forces can play significant role in the beam strength. This script consists of the three parts - the first one describes the classsical deflection determination procedure for the single bay structure clamped at both edges and loaded with the triangular force decreasing from the right to the left end. The extra loading is applied in the form of a concentrated bending moment at the right edge of the beam. The main aim of the second section is to present the procedure related to the second order theory with a compressive force, whereas the third section is devoted to the second order model with tension.  

Classical solution without the normal force  

>

 

>

 

(1.1)

 

we define first the function of the external triangular loading distributed along the beam 

>

 

(1.2)

 

Using the equilibrium of momentum rewritten at the right clamped edge we derive the moment at the left edge and then the cross-sectional momentum equations is derived as  

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(1.3)

 

According to the figure we apply the following boundary conditions:   

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(1.4)

 

which reflects no deflection at the left edge  

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(1.5)

 

because of the zero for the deflection angle at the same point  

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(1.6)

 

which is the second order differential equation for the deflection  

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(1.7)

 

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(1.8)

 

is the third boundary condition - zero deflection at the right clamped edge  

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(1.9)

 

according to the zero of the deflection angle at the same point   

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(1.10)

 

From the differential equation one can derive the reactions in this system, then we have the deflection function  

>

 

(1.11)

 

The deflection as well as the internal forces diagrams can be plotted using the specific values of geometrical, mechanical parameters and loadings and we have   

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	(1.12)

 

>

 

(1.13)

 

which is the final form of a deflection function 

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>

 

 

>

 

 

>

 

 

>

 

 

>

 

(1.14)

 

we look for the location with shear force equal to 0  

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(1.15)

 

being the location of the maximum value for the bending moment.  

Furthermore, we compute the moment at the left clamped edge 

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(1.16)

 

the moment at the right clamped edge  

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(1.17)

 

as well as the shear force at the left hand side  

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(1.18)

 

together with the reaction at the right hand side  

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(1.19)

 

Finally we would like to determine the maximum deflection along the beam which is computed for the zero deflection angle  

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(1.20)

 

and there holds   

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(1.21)

 

2nd order analysis with compressive force  

>

 

>

 

(2.1)

 

we define first the function of the external triangular loading distributed along the beam 

>

 

(2.2)

 

Using the equilibrium of momentum rewritten at the right clamped edge we derive the moment at the left edge and then the cross-sectional momentum equations is derived under the presence of the normal force as  

>

 

(2.3)

 

Let us note that we insert a normal force into this equation, which is an extra component absent in eqn (1.1). According to the figure we apply the following boundary conditions:   

>

 

(2.4)

 

>

 

(2.5)

 

>

 

(2.6)

 

>

 

(2.7)

 

>

 

(2.8)

 

>

 

(2.9)

 

>

 

(2.10)

 

From the differential equation one can derive the reactions in this system, then we have the deflection function  

>

 

(2.11)

 

The deflection as well as the internal forces diagrams can be plotted using the specific values of geometrical, mechanical parameters and loadings and we have   

>

 

 

 

 

 

 

	(2.12)

 

>

 

(2.13)

 

>

 

>

 

 

>

 

 

>

 

 

>

 

 

>

 

(2.14)

 

>

 

(2.15)

 

Furthermore, we compute the moment at the left clamped edge 

>

 

(2.16)

 

the moment at the right clamped edge  

>

 

(2.17)

 

as well as the shear force at the left hand side  

>

 

(2.18)

 

where the right clamped edge equals to  

>

 

(2.19)

 

>

 

(2.20)

 

which returns the maximum deflection for this structure  

>

 

(2.21)

 

2nd order bending of the beam under tension  

>

 

>

 

(3.1)

 

we define first the function of the external triangular loading distributed along the beam 

>

 

(3.2)

 

Using the equilibrium of momentum rewritten at the right clamped edge we derive the moment at the left edge and then the cross-sectional momentum equations is derived under the presence of the normal force as  

>

 

(3.3)

 

According to the figure we apply the following boundary conditions:   

>

 

(3.4)

 

>

 

(3.5)

 

>

 

(3.6)

 

>

 

(3.7)

 

>

 

(3.8)

 

>

 

(3.9)

 

>

 

(3.10)

 

From the differential equation one can derive the reactions in this system, then we have the deflection function  

>

 

(3.11)

 

The deflection as well as the internal forces diagrams can be plotted using the specific values of geometrical, mechanical parameters and loadings and we have   

>

 

 

 

 

 

 

	(3.12)

 

>

 

(3.13)

 

>

 

>

 

 

>

 

 

>

 

 

>

 

 

>

 

(3.14)

 

We look for the maximum of the bending moment function localized at the zero shear force, so that  

>

 

(3.15)

 

Furthermore, we compute the moment at the left clamped edge 

>

 

(3.16)

 

the moment at the right clamped edge  

>

 

(3.17)

 

as well as the shear force at the left hand side  

>

 

(3.18)

 

with the right hand side reaction as  

>

 

(3.19)

 

>

 

(3.20)

 

which returns the maximum deflection for this structure  

>

 

(3.21)

 

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