Determination of Transient Vibration Response of a Portal Frame Simulating Single Span Short Length Bridge

Determination of Transient Vibration Response of a Portal Frame Simulating Single Span Short Length Bridge

Ms. S. J. Modak H. V. Hajare S. S. Kulkarni

Assistant Professor in Civil Engg.,

Professor of Civil Engg.,

Ex-Professor of Structural Engg

Ramdeobaba College of Engineering & Priyadarshini College of Engg, VNIT, Nagpur

Management, Nagpur 440 009 Nagpur

Abstract-The objective of the investigation is to mathematically simulate dynamics and vibrations of a portal frame subjected to a concentrated load moving on its horizontal member with a certain constant velocity. This portal frame is a basic structure of a low length single span bridge. The dynamic analysis of the portal frame is done by matrix method of structural analysis. The vertical members of the portal frame are subjected to time varying axial compressive load. The focus of the paper is to ascertain the transient vibration response of the vertical members treating the members represented by STOF vibratory system subjected to time varying axial compressive load.

Keywords: Bridge, Columns, Portal Frame, Transient Vibrations.

1. INTRODUCTION TO CONSTRUCTION OF A SHORT

   LENGTH BRIDGE

   Figure 1 is a schematic presentation of a short length bridge. The length is so short that the basic structure of a bridge is a simple one span portal frame 01 AB 02. The width of the bridge is also fairly small so that it could be considered as a particular case of a girder bridge [4]. The material of the frame is Mild Steel (M.S.). The philosophy of the analysis is explained through a representative small scale structure with dimensions length of AB = 1005 mm, width = 50 mm and thickness is = 0.01m. The vertical members 01A and 02B are geometry wise identical. The material of 01A & that of 02B is also M.S. A vehicle with total weight W is moving on AB with a constant velocity.

   The objective of the investigation is to estimate vibration response of 01A & 02B.

2. ANALYSIS OF A PORTAL FRAME BY MATRIX

   METHOD

   The specifications of the portal frame under consideration are as follows:

   Material for all the members viz. 01A, AB, 02B is

   1. with modulus of Elasticity E = 2.0 x 108 kN/m2; Lengths of 01A = AB and 02B are equal each

      Figure 1. Schematics of a Portal Frame for a Short Bridge.

      in turn equals to 1m, cross section rectangular 50 mm x 10mm; For member 01A, I = (bd3)/12.0 where b = 0.01m and d = 0.05m. Dimension b is in the plane of the paper where as dimension d is at right angles to the plane of the paper. Geometry wise members 01A & 02B are identical. For member AB the thickness = 0.01m where as width = 0.05mm.

      As far as joints 01, A, B, 02 are concerned the horizontal force components, vertical force components and rotation in vertical plane abut axis, passing through 01 and perpendicular to the plane of the paper are respectively U01, V01 and 01(positive C.W.). Similarly, the same quantities at remaining joints A, B and 02 are concerned are as follows.

      For A : UA, VA, A

      For B : UB, VB B

      And

      For 02 : U02, V02, 2

      For a concentrated load W = 1 kN acting at distance x = x = 0.25m from A, these quantities (i.e. U01, V01, 01, UA, VA, A, UB, VB, B, U02, V02, 02) are estimated adopting Matrix Method of Structural Analysis [3].

      Similar approach for the same load and the same location is applied for the members AB & 02B also. The findings are as shown in Table 1. Similar approach is adopted for the analysis of entire structure when the same load W = 1 kN changed its position x to 0.5m, 0.8m. Te complete findings are shown in Table-1. The graphic variation of V01 and V02 are decided off-course adopting the

      concept of interpolation [5]. These graphic plots are shown in Figure 3.

      1. Determination Of Equivalent Axial Load For 01A And 02B

         This is detailed in this article for the case when x =

         0.25m & W = 1kN.

         Figure 2. Results of Force Analysis of Various Members of Portal Frame

         Effect of V01 & VA is to create an axial compresive load equal to 0.879 x 103 kN

         Effect of U01 & UA is to create a c.c.w. couple =

         0.195 x 103 kN-m

         This stands to reason because of 1 kN load on AB at x = 0.25m a c.w. moment is created on AB of 0.25 kNm by 01A to which 01A should experience an anti clockwise moment.

         rotation of Section B w.r.to 02 is (-0.0299) + (0.0153) = – 0.0146 c.w. which also changes with time and creates time varying bending vibrations in 02B.

         Table 1. Findings of analysis of a portal frame

         This moment 0.195 x 103 kN-m c w on 01A will change its magnitude as the load 1 kN changes its position on AB. Thus, it will induce bending vibrations in 01A. Similarly, the net rotation of cross section at A with respect to cross section at 01 due to A = 0.1263 radians c.c.w. and

         0 = 0.0622 radians.

         c.w. is 0.0641 radians c.c.w. which is also time variant has effect only to the extent of inducing bending vibrations in 01A. F

         On the same lines Fig. 3 shows the Free Body Diagram (FBD) of member 02B of the same portal frame.

         UB = -101.99 N

         VB = 0.0458 kN = 45.8 N

         B = -0.0299 radius U(02) = 101.99 N

         V(02) = -0.0459 kN = 45.8 N

         (02) = 0.0153 radians

         Effect of VB = 45.8 N & V02 4.8 kN is to

         create tension in the member 02B when x = 0.25m and W = 1kN. Similarly, the effect of U(02), +101.99 N and UB = –

         101.99 N is to create c.c.w. moment on 02B. The magnitude of this moment is to change with positionof W. Thus, it is going to create bendign vibrations in the 02B. Similarly, net

         Figure 3. Free Body Diagram (FBD) of member 02B

         As stated earlier, in this paper as emphasis on longitudinal vibrations of 01A and 02B is only considered, bending vibrations is not detailed further.

      2. Influence Line For 01A And 02B

         Figure 4(a) & 4(b) respectively shows the variation of V01 = VA and V02 = VB as the load i.e. the weight of the moving vehicle changes its position on AB. This can be consideed as the influence lines [6] of member AB of the protal frame. This is off-course by the analysis of the concept influence line of a simply supported beam [6]. These figures 4 (a) and 4 (b) shows nonlinear variation of longitudinal load on the members 01A & 02B respectivelly of the portal frame 01AB02 under consideration. In other words as per matrix method of Structural Analysis, one gets nonlinear variation of compresive load on members 01A & 02B of the portal frame as some concentrated constant load changes its position on portion AB of the portal frame. As stated earlier in this paper the single span short length bridge is treated annalogous to a single span portal frame, the loading on columns of the bridge would also be nonlinear. This nonlinearity is in terms of changed position

         of concentrated load. It is obvious that this is also interms of time because the load is nothing but the dead weight of the vehicle which is varying with constant velocity but the position of the vehicle is changing with respect to time . Hence, as per this approach of Structural Analysis [3]. it is concluded that the column reactions change nonlinearly [2] with time as against the linear variation as obtained in the previous analysis [1].

      3. Approach To Estimation Of Vibration Response Figure 4(a) shows influence line of 01A. This variation is having exponetial decay form.

   Figure 4.a &4 b Influences of 01A & 02B

   Thus it can be stated as under, if F(01A) stands for externl longitudinal load acting on the column 01A.

   F(01A) = K1t-n1 . (I)

   Whereas if F(02B) stands for external column load acting on the column 02B, then it is as described in Figure 3(b). Figure 3(b) shows the variation of F(02B) in exponential rising form. Hence, it can be stated as under F(02B) = K2tn2 .. (II)

   The parameters K1, n1, k2, and n2, can be decided by plotting these variations on log-log graph [5]

   If one considers the entire coloumn represented by Single Degree of Freedom (SDoF) system, in which M, K, C represent the mass, elastic stiffness and damping co- eeficient of the structure of the column, then the governing equation of forced vibration phenomena of the column is presented as under.

   + + = 11.. (III)

   For the column 01A where, as that for the column 02B would be

   + + = 22

   The initial conditions are off-course as under, at t = 0; x = 0

   and at t = 0; = 0

   (a). Vibration Response Of 01a

   Equation (III) presents the governing equation of vibrations motion of 01A. In Eq. (III) M, K, C are respectively Mass, Longitudinal Stifness and Damping co- efficient of material of 01A. These are estimated as under:

   M = W/g; W = weight of 01A.

   W = x L x b x d

   = 8 x 103 x 1 x 0.01 x 0.05

   W = 40.0 Kgf

   M = 40.0/9.81

   M = 4.07 Kgf-m-1 sec2

   K = Longitudinal Force per unit longitudinal

   displacment

   K = E x 1/L x bxd

   Upon substitution of pentaniment numerical values; E = 2.0 x 108 KN/m2, L = 1.0m; b = 0.01m; d = 0.05m

   K = 108 Kgf/m

   Let Damping ratio = C/Cc

   For the member 01A, it will be appropriate to assume = 0.01 [7] as the damping is going to be only due to material internal molecular friction or HYSTERISIS DAMPING. Hence, in view of K = 10.08 Kgf/m, M = 4.07 Kgf.m-1 sec2, = 0.01

   C = 403.84 Kgf/m/sec

   In Equation (III) the right hand side is K1t-n1. The parameters K1 & n1 are decided by plotting the values of external longitudinal force to which 01A is subjected with respect to time as depicted in Figure 3(a). This variation if plotted on log-log paper, then it will farily approximately conform to the straight line. The slope of this approximate straight is the value n1 where as from the intersept of this line with ordinate, K1 can be decided [5]. Accordingly, the external longitudinal load on 01A will come out to be

   F = K1t-n1 = (2.05 x 10-3) t-2.5526 . (V)

   Accordingly, now Equation (III) gets transformed

   to

   4.07 + 403.484 + 10.08x = K1t-n1 = (2.05 x 10-3)

   t-2.5526 . (III A)

   upon substitution of numerical values of M, K, C, K1 & n1. This is an ordinary linear diferential equation with constant co-efficients. Hence, adopting the approach of Laplase Transformation [7] and in view of initial conditions t = 0, x

   = 0 & = 0; the solution to this differential equation is obtained which is the vibration response of 01A.

   The details of these calculations are elaborated in Annaxure-I.

   (b) Vibration Response of 02B

   Adopting the procedure by which vibration response of 01A is decided, the vibration response of 02B is obtained. The governing equationof motion of 02B is

   4.07 + 403.484 + 10.08x = K2t-n2 = (3.096 x 10-3)

   t5.19 .. (IV B)

   Equation (IV B) is also an ordinary linear differential equation with constant co-efficients. The initial conditions are at t=0, x=0 and =0. Again the approach of LAPLASE TRANSFORMATION is adopted to arrive at the solution of the differential equation (IV B). This is detailed in Annexure-I(B).

3. POSSIBLE EXTENSION

   A more realistic solution could be by treating the entire column as a multidegree of freedom system represented by ¾ identical masses, stiffness of the springs and damping co-efficient.

   On the same lines an entire column can be considered as a distributed mass, distributed elasticity and distributed damping system. Further all other possible configurations of vehicular trafic should be considered as detailed in the earlier papers of the authors [1 & 2].

   n1 = 2.5526

   Rewriting the Equation (III A) again for the sake of ready ference

   4.07 + 403.484 + 10.08x = K1t-n1 = (2.05 x 10-3) t-2.5526

   . (III.A)

   + 99.136 + (0.245) x 10.08x = (0.5036 x 10-3)t-2.5526

   Performing Laplase Transformation of modified equation (III.A) given above, observing the initial condition tah at t = 0; x = 0 and = 0 , one gets

   2 + 99.136 + 0.245 108

   (0.5036 103)

   = 12.5526

4. CONCLUSION

The paper reports on the possible approach to decide transient vibration response of two columns of a short length bridge which can be simulated by a single span portal frame. As this one is of the very few approaches towards this objective, it is based on large number of oversimplifying assumptions.

References

2 + 99.136 + 0.245 108

(0.5036 103)

= 12.5526

(0.5036 103)1.5526

= [2 + 99.136 + 0.245 108]

Now roots of [2 + 99.136 + 0.245 108]= 0 are

1 and 2 such that

Solving the above quadratic equation one can get

1. S.J.Modak & H. V. Hazare, Vibration Analysis of a Portal Frame subjected to a Moving Concentrated Load,

   (0.5036 103)1.5526

   = +1 +2

   International Journal of Engineering Invention, Vol. I, Issue 9, Nov. 2012, pp 13-19.

2. S.J.Modak, H. V. Hazare & S.S. Kulkarni, Determination of Influence Lines of a Portal Frame Adopting Matrix Method of Structural Analysis, International Index Journal : Indian

   To decide C1:

   = 1

   +1

   + 2

   +2

   Journal of Applied Research, Vol. 3, Issue 7, July 2013, pp 224-226.

3. A.A. Meghre & S. K. Deshmukh, Matrix Method of Structural Analysis, S.B. Patel, Charotar Publishing House,

   (0.5036 103)1.5526

   = +2

   = 1 +1 + 2 +1

   Anand (Gujrat State), India Ed. 2013.

4. FTK Au, Y.S. K. Scheng, Y.K. Cheung, University of

   Now put S = –

   +1 +2

   HongKong, Peoples Republic of China, Prog. Structural Engineering Materials, 2001, 3, pp 299-304

5. H. Schenck Jr., Theory of Engineering Experimentation

   1

   =

   (0.5036 103)1.5526

   1+ 2

   = 1

   McGraw Hill Inc, 1967.

6. S.B. Junarkar, Strength of Materials, Charotar Book Stall,

   1 =

   (0.5036 103)(49.140106)1.5526

   3

   Anand Vallab Vidya Nagar, Gujrat (India).

7. Mario Paz, Structural Dynamics : Theory & Computation, Kluwer Academic Publishers Group, 5 Ed., ISBN No. 1- 4020-7667-3, 2004.

8. K.K.Pujara, Vibrations for Engineers, Dhanpat Rai & Sons, Nai Sarak, Delhi-6, 2012.

10

(0.5036 103)(49.140106)1.5526

= 10 6

= 0.5036 49.140 1.5526 10(0.3156 )

On the same lines C2 can be decided

2 = 0.0512 49.136 1.5526 10(0.3156 )

Annexure I

Vibration Response of 01A and 02B

Then

= 1

+1

+ 2

+2

Governing equation of vibratory motion of 01A is M + c + Kx = K1t-n1 . (III)

Now performing inverse Laplace Transformation,

= 1 1 + 2 2

Substituting for M = 4.07 Kgf-m-1

-sec2

= 1

49.140 106 +

K = 10.08 Kgf/m

C = 403.84 Kgf/m/sec K1 = 2.05 x 10-3

2

49.139 106

(VI)

Equation – (VI) is the vibration response of 01A. Similarly, the vibration response of 02B can be decided.

1 = 0.05125(49.140)1.5526 (10)0.3156

2 = 0.05125(49.140)1.5526 (10)0.3156

1= 49.140 10+6; 2= 49.139 10+6

Table 1. Findings of analysis of a portal frme O1A BO2, For W=1kN & X = 0.25, 0.50, 0.80 M