A three degree-of-freedom model of a railroad bridge is shown in Figure P7.84. The bridge is com

step 1 Hi everyone here it is given bridge plus extend 200 meters log. The structure is free to slide h

A three degree-of-freedom model of a railroad bridge is...

Key Concepts

Rigid Body Dynamics

Rigid body dynamics focuses on the motion of bodies in which deformations are negligible. For each rigid span of the structure, inertia properties such as moments of inertia and mass distributions play a key role in determining the dynamic response. The idealization of bridge spans as rigid bodies simplifies the analysis because the internal deformations within each span are ignored, and only the overall motion (such as rotations about pinned supports) is considered.

Multi-Degree-of-Freedom Systems

This concept involves analyzing systems in which multiple independent movements (degrees of freedom) occur simultaneously. In structural dynamics, complex structures such as a railroad bridge can be modeled as a collection of interconnected rigid bodies, each with its own degree of freedom. This allows the engineer to capture the coupled dynamic behavior of various parts of the structure, leading to a more accurate prediction of its response under dynamic loads.

Generalized Coordinates

Generalized coordinates are chosen parameters that uniquely define the configuration of a dynamical system relative to its constraints. In the context of multi-body dynamics, using angular displacements to represent the state of each span simplifies the formulation of the problem since rotational motions can be directly related to the system’s kinetic and potential energies. The use of such coordinates reduces the problem’s complexity by minimizing redundant or non-independent variables.

Lagrangian Dynamics

Lagrangian dynamics is a powerful method for deriving the equations of motion of a system by formulating the Lagrangian as the difference between the kinetic energy and potential energy. This energy-based approach is particularly useful for systems with complex constraints and multiple degrees of freedom, as it converts the physical problem into a set of differential equations. It allows the systematic incorporation of generalized coordinates and their associated velocities, leading to a formulation that directly yields the motion equations.

Derivation of Equations of Motion

The derivation of the equations of motion entails combining the kinetic and potential energy contributions into the Lagrangian and then applying Euler-Lagrange equations. This process converts the energy expressions into a set of ordinary differential equations that govern the dynamic behavior of the system. It involves accurately computing the kinetic energy from the rotational movement of each span and the potential energy from gravitational or elastic restoring forces, thus resulting in differential equations that capture the interplay of inertia and stiffness in the dynamic response.

Transcript

00:01 Hi everyone here it is given bridge plus extend 200 meters log.

00:06 The structure is free to slide horizontally to permit thermal expansion.

00:12 The structural components are connected by the pit joint and their barred are of mass having negligible as compared to the car of mass 1 360 at the center.

00:40 Calculate force of tension in each structure component here normal reaction will be n -a your normal reaction is n -e say f -a -b here f -a -b here f -v -d this is f -b -d f -b -t this is f -a -c here weight of the car this is 40 this is 40 so this will be 40 now let us see here for the bridge as a whole if we find the torque about a for complete bridge, torque about a if you find, let us see it, an a into 0, weight of the car in 200 pillars, and e into 200 must be 0.

03:32 So from here normal reaction at e point you will get 1360 9 .8 in 200 divided by 200.

03:47 So this value you will get 666 .6 or it can be written as 6 .6 .6.

04:17 Just a moment i have put the wrong decimal kilo -nuter.

04:30 So this is your one of the answer.

04:36 Summation of f5 must be zero so it will give you an a minus weight of the car 1330 newton plus n e must be zero so from here normal reaction and a you will get 6 .66 kiloons now at pin a if we draw the force at pin a see here, this is the pin a, this is the force f, a, this is the force f, a, c.

05:40 This angle is given 40 degree and this is normal reaction.

05:59 Summission of forces along y -axis must be 0.

06:07 So you may write minus f -av, sign of 40 plus n -a, must be so from here f a b will be n a upon sign 40 and a upon sign of 40 n a is 6 .60 kilo newton so this value will be 10 .4 kilo newton and nature is compressive now if we find submission of fx to be zero then it could be f ac minus f .a .v.

07:35 Cos 40.

07:38 That must be zero.

07:41 If you will get fac is equal to fab cost 40.

07:50 So you will get fac to be 7 .94 kiloons and it will be a tension.

08:09 Now at pin v we draw the forces at pin v you may see just a moment this is fv fvc and fav...

Please give Ace some feedback