9.3 Forced Vibration of the Shear Beam Under Ground Motion Excitation
375
9.3 Forced Vibration of the Shear Beam Under Ground
Motion Excitation
The equation of motion for the shear beam subjected to ground motion is given by
∂
∂ x
μAG
∂z
∂ x
= ρ A
∂
2 z
∂t 2 + ρ A ¨
y g (t)
(9.23)
where z is the relative displacement with respect to the base and ¨
y g (t) is the ground
acceleration (Fig. 9.2).
Let us assume a solution of the following form
z(x, t) =
Y r (x) ξ r (t)
(9.24)
where ξ r (t) represents the normal coordinates and Y r (x) represents the mode shapes.
Substituting z from Eq. (9.24) into Eq. (9.23), we get
ξ r
d
dx
μAG
dY r
dx
=
ρ AY r ¨
ξ r + ρ A ¨
y g
(9.25)
For the free vibration, it can be shown by substituting v r = Y r sin p r t in
Eq. (8.155), after modifying the equation for a beam having non-uniform cross
section that
d
dx
μAG
dY r
dx
= −ρ A p
2
r Y r
(9.26)
Combining Eqs. (9.25) and (9.26), we get
Fig. 9.2 A shear beam
375
9.3 Forced Vibration of the Shear Beam Under Ground
Motion Excitation
The equation of motion for the shear beam subjected to ground motion is given by
∂
∂ x
μAG
∂z
∂ x
= ρ A
∂
2 z
∂t 2 + ρ A ¨
y g (t)
(9.23)
where z is the relative displacement with respect to the base and ¨
y g (t) is the ground
acceleration (Fig. 9.2).
Let us assume a solution of the following form
z(x, t) =
Y r (x) ξ r (t)
(9.24)
where ξ r (t) represents the normal coordinates and Y r (x) represents the mode shapes.
Substituting z from Eq. (9.24) into Eq. (9.23), we get
ξ r
d
dx
μAG
dY r
dx
=
ρ AY r ¨
ξ r + ρ A ¨
y g
(9.25)
For the free vibration, it can be shown by substituting v r = Y r sin p r t in
Eq. (8.155), after modifying the equation for a beam having non-uniform cross
section that
d
dx
μAG
dY r
dx
= −ρ A p
2
r Y r
(9.26)
Combining Eqs. (9.25) and (9.26), we get
Fig. 9.2 A shear beam