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NookanCranny
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Homework Statement
"If a horizontal force of F = 299kN is applied to the right at the girder level by a hydraulic jack and then the force is removed suddenly, find the equation of motion of the girder. Consider 5% critical damping. Find the
displacement and velocity of the girder at 0.2 seconds after release of the force."
This is of a braced single storey building. (bottom left is braced to top right etc)
I = 7.6*10^6mm^4
E = 200 GPa
Length of Girder = 6m
Height of Columns = 3.5m
The braced bars are 25mm diameter. This meant that the Area of them came to 490.874mm^2 as Area = (pi*d^2) / 4
mass of frame = 4000kg.
stiffness (girder and two columns) = 11404.775N/mm
natural frequency = 53.4 rad/s
Homework Equations
m*(acceleration) + c*velocity +k*displacement = F(x) -> (Inertia Force + Viscous force + Stiffness force = F(x))
critical damping constant = 2*mass*omega
damped natural frequency = natural frequency * sqrt (1-ξ^2)
natural frequency = sqrt (stiffness / mass)
The Attempt at a Solution
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I calculated Stiffness in both the columns and the rigid girder. This came to 11404.775. I solved for the columns by using (12EI/L^3) multiplied by 2 as there are two columns. For the rigid girder, I had to take into account theta.
I then calculated natural frequency of vibration after finding stiffness and dividing it out by the mass. After square rooting the fraction, it came to 53.4 rad/sec.
I then calculated the critical damping constant, this came to 427200kg/sec as critical damping constant Cc = 2*mass*natural frequency.
C actually was less than Cc, meaning it is under critically damped. ξ = 0.05.
I then went back to m*(acceleration) + c*velocity +k*displacement = F(x)
I then made all of it equal to zero, deriving from a characteristic equation and then solving for the roots using the quadratic formula. All this did was bring me back to an equation I already knew from my notes, as shown with attached image.
I am wondering, since it is applied instantaneously, does m*acceleration + c*velocity + k*displacement still equal to 0? or does it equal to 299kN?? The 299 is throwing me off because I was told if the frame is free moving, then it should equal to zero.
Also I'm not sure about writing the equation of motion when I am not given the initial velocity or displacement constraints.
Apologies if this question sounds a bit silly - I haven't come across one of these questions in my tutes!