Understanding the concepts on Degrees of Freedom

CHALLENGE WEEK 3

ETABS

QUESTION 1: A five storey structure with 5 translational DOF is shown here. Each storey has mass ‘m’. The eigenvalue problem has been solved and the 5 periods of vibration T_n and their corresponding mode shapes, φ_n has been derived.

Answer:

• Mass matrix (m) =

• Modified mass matrix (M) = Φn^T .m.Φn
• For the first mode (n=1) :

[0.334 0.641 0.895 1.078 1.173]

M1 = 3.861m

model property (L) = Φn^T .m.ι

where ι = influence vector (asuuming 100% utilization of sesmic force) =

• For the first mode = [0.334 0.641 0.895 1.078 1.173]

L1 = 4.121m

• Modal participation factor (⌈) = L/M
• For the first mode , ⌈ = L1/M2

(n=1)

• ⌈1 = 4.121m/ 3.861m
• ⌈1 = 1.067
• Similarly

⌈2 = -0.335

⌈3 = 0.177

⌈4 = -0.098

⌈5 = 0.045

• Effective modal mass (M*) = Ln^2/Mn
• For the firat mode, M1* = L1^2/M1

(n=1)

• M1*=(4.121m)^2/3.861m
• M1*= 4.398m
• Similarly,

M2*= 0.436m

M3*= 0.121m

M4*= 0.038m

M4*= 0.008m

• Modal mass aprticipation ratio = Mn*/mj
• where , mj = Total mass of the building = m+m+m+m+m=5m
• Modal mass participation ratio of the first mode (n=1)= 4.398m/5m=0.88
• Simialrly , second mode (n=2) =0.09

third mode (n=3) =0.02

fourth mode (n=4) =0.01

fifth mode (n=5) =0.002

• Adding first two mode mass participation ratios, we get 0.88+0.09=0.97
• To achieve minimum 90% of modal mass participation ratio, first two modes are enough
• Base shear (VB) = Mn*.An
• For the first mode, VB1= 4.398m x 0.27g

(n=1)

VB1 = 1.187mg

• Similarly , VB2=0.327mg

VB3=0.126mg

VB4=0.039mg

VB5=0.008mg

• Average base shear is calculated by the methods of square root of sum of square

√((Vb1^2)+(Vb2^2)+(Vb3^2)+(Vb4^2)+(Vb5^2)

VB=1.15mg