Kindly use the unit (N/m) and (mm) where ever applicable. Question 5: SDOF Free

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Kindly use the unit (N/m) and (mm) where ever applicable.

Question 5: SDOF Free Vibration Analysis A solid slab of mass 10,000 kg is supported by 3 columns as shown in Fig.QS below. All 3 columns are rigidly fixed at the top and at the base except for the middle column which is pinned at the base. The columns are of steel section UB 305x 102x25, having elastic modulus E-205 GPa and moment of inertia lyy damping 4.364x10 m. Answer the following questions, assuming there is no 8 sec. same arks) massa loooe ky L a velocity time t Fig. Q5 arks) a.) If it is observed that the mass has a undamped free vibration period of T- 1.278 sec. What would be the height, Lo of the columns, assuming that they are all of the same height? Round off the answer to the nearest meter. n the (7 marks) b.) If it is further observed that at time t 0, the mass attains an amplitude (i.e. max displacement)- 100 mm, what would be the displacement function y(t) and velocity function v()? (5 marks) c.) What is the max velocity, Vmax (in m/s) it could attain and what is the minimum time t (3 marks) required from the start to attain (1/2) of vmas? (5 marks) d.) What is the maximum force acting on the mass? e.) What is the maximum shear force and maximum bending stress fmax incurred in the midle column of section UB 305x102x257 Express fax in units of MPa. (5 marks)

Explanation / Answer

a)

Mass (M) of slab = 10000 kg

Stiffness of fixed columns = 12EI/Lc3

Stiffness of pinned columns = 3EI/Lc3

Total lateral stiffness (K) of the system = 12EI/Lc3 + 3EI/Lc3 + 12EI/Lc3 = 27EI/Lc3

Time period (T) = 1.278 sec = 2 (M/K)0.5

=> (M/K)0.5 = 1.278/2

=> M/K = (1.278/2)2

=> K = M/(1.278/2)2

=> 27EI/Lc3 = M/(1.278/2)2

=> Lc3 = 27EI (1.278/2)2 / M

=> Lc = (27EI (1.278/2)2 / M)1/3 = (27 x 205 x 109 x 4.364 x 10-5 x (1.278/2)2 / 10000)1/3

=> Lc = 10 m

b)

The equation of motion:

M + K x = 0

Let the solution of the equation is x(t) = A cos(wt) + B sin (wt)

(t) = -Aw sin(wt) + Bw cos(wt)

(t) = -Aw2 cos(wt) – Bw2 sin(wt)

where w = (K/M)0.5 = 2/T = 4.916 rad/s

At t = 0, x = 100mm = 0.1m

0.1 = A + 0

=> A = 0.1

Since, velocity is zero when displacement is maximum, so at t = 0, (t) = 0

0 = 0 + Bw

=> B = 0

So, x(t) = 0.1 cos(4.916t) m [Displacement function]

(t) = -Aw sin(wt) = -0.492 sin(4.916t) m/s [Velocity function]

c)

vmax = Max |(t)| = Max |-0.492 sin(4.916t)| = 0.492 m/s

Now,

0.5 vmax = vmax sin(4.916t)

=> sin(4.916t) = 0.5

=> 4.916 t = sin-1 (0.5)

=> t = sin-1 (0.5) / 4.916 = 0.1065 sec

So, 0.5 vmax will be attained in 0.1065 sec.

d)

(t) = -Aw2 cos(wt) = -2.417 cos(4.916t) m/s2 [Acceleration function]

amax = Max | (t)| = Max |-2.417 cos(4.916t)| = 2.417 m/s2

Maximum force (Fmax) acting on the mass = M amax = 10000 x 2.417 = 24171.19 N = 24.17 kN

e)

The maximum force acting on the mass will be resisted horizontally by the three columns at their bases.

The horizontal reactions acting in each column will be in proportion to the column stiffness.

Horizontal reaction acting on the middle column

= Fmax x (3EI/Lc3 / (12EI/Lc3 + 3EI/Lc3 + 12EI/Lc3))

= Fmax x (3/27)

= Fmax / 9

= 2.686 kN

So, maximum shear force acting in the middle column = 2.686 kN

Maximum bending moment (BM) acting in the middle column = 2.686 x 10 = 26.86 kNm

Maximum bending stress acting in the middle column

= BM y / Iyy

= 26.86 x (0.3051/2) / (4.364 x 10-5)

= 93882.14 kPa

= 93.882 MPa

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