Dynamics Response Spectrum Analysis

Dynamic Response Spectrum Analysis – Shear Plane Frame

Page 1 of 33

Five-Story Shear Plane Frame Dynamic Response Spectrum Analysis Comparison between hand calculations based on the theory of structural dynamics and ETABS analysis procedure results Problem: Five-story shear plane frame with story-height of 3.0m and single bay of 4.0m • The mathematical model consists from squares columns ( 60 × 60 cm 2 ) with infinitely rigid beams ( I beam = ∞ ). • The entire mass of each story is assumed to be lumped at its level with total value of typical story mass ( m = 100kN . sec 2 / m ). • The material of columns and beams has modulus of elasticity equal to ( E = 2.106 kN / m 2 ). • Assumed damping ratio ( ζ = 0.05 ). • The frame is subjected to dynamic response spectra as defined in UBC-97 with assumed design parameters : ♦ Seismic zone factor ( Z = 0.3 ) ♦ Soil profile type ( S B )

Evaluate the following: (a). Natural vibration frequencies and corresponding vibration mode shapes. (b). Periods corresponding to vibration mode shapes. (c). Response spectrum accelerations corresponding to periods. (d). Maximum modal displacement corresponding to vibration mode shapes. (e). Maximum story-displacement according to modal combination (SRSS). (f). Maximum modal elastic forces (inertia-forces) at story-levels. (g). Maximum modal story-shear forces. (h). Maximum total story-shear forces according to modal combination (SRSS). (i). Modal participation factors. (j). Modal participating mass ratios.

Notes: • The matrix analysis will be done by using MATLAB software (high performance language for technical computing & solve engineering problems). •

Compare hand-calculation results with equivalents obtained from ETABS analysis.

Edited by: Eng.Hussein Rida E-mail: [email protected]


Page 2 of 33

Introduction: A shear frame may be defined as a structure in which there is no rotation of a horizontal section at the level of the floor. In this respect the deflected frame will have many of the features of a cantilever beam that is deflected by shear forces, Hence the name Shear Frame. To accomplish such deflection in frame, we must assume that: (1) the total mass of the structure is concentrated at the levels of the floors; (2) the beams on the floor are infinitely rigid as compared to the columns; and (3) the deformation of the structure is independent of the axial forces present in the columns. These assumptions transform the problem from a structure with an infinite number of degree of freedom (due to the distributed mass) to a structure which has only as many degrees as it has lumped masses at the floor levels. According to previous discussion a five stories frame modeled as a shear frame will have five degrees of freedom, that is, the five horizontal displacements at the floor levels. The second assumption introduces the requirement that the joints between beams and columns are fixed against rotation. The third assumption leads to the condition that the rigid beams will remain horizontal during motion. Determination of Lumped mass matrix: For shear structure; the mass matrix is a diagonal matrix (the nonzero elements are only in the main diagonal) whereas each one of these elements represents the total equivalent entire mass of the story as a concentrated lumped mass at the level of this story with understanding that only horizontal displacement of this mass is possible.

Therefore the lumped mass matrix is given by: 0 0 0 ⎤ ⎡100 0 0 0 0 ⎤ ⎡ m1 0 ⎥ ⎢0 m ⎢ 0 0 0⎥ 0 0 ⎥⎥ 2 ⎢ ⎢ 0 100 0 M =⎢0 0 m3 0 0⎥=⎢ 0 0 100 0 0 ⎥ kN . sec 2 / m ⎥ ⎢ ⎢ ⎥ 0 0 m4 0 ⎥ ⎢ 0 0 0 100 0 ⎥ ⎢0 ⎢⎣ 0 0 0 0 m5 ⎥⎦ ⎢⎣ 0 0 0 0 100⎥⎦ Determination of stiffness matrix: The stiffness matrix of shear frame can be determined by applying a unit displacement to each story alternately and evaluation the resulting story forces. Because the beams are infinitely rigid comparison to columns; then the story forces can easily be determined by adding the side-sway stiffness of the appropriates stories which equal in this case to the total sum of columns stiffness of that stories.

In shear frame as defined previously the stiffness of column with two ends fixed against 12 EI c rotation is given by: Kc = h3 Where ( h ) is the story height, and ( I c ) is the moment of inertia of column's section given by:

a × a3 0.6 × 0.63 Ic = where a = 0.60m ⇒ I c = = 0.0108 m 4 12 12 The stiffness of the story is given by: 24 EI c 24 × 2 × 106 × 0.0108 K i = ∑ K c = 2.K c = = = 19200 kN / m h3 33 Edited by: Eng.Hussein Rida E-mail: [email protected]


Page 3 of 33

The stiffness matrix of the structure is given by: ⎡ K1 ⎢ −K 1 ⎢ K =⎢ 0 ⎢ ⎢ 0 ⎢⎣ 0

− K1 K1 + K 2

0 − K2

0 0

− K2

K 2 + K3

− K3

0

− K3

K3 + K4

0

0

− K4

0⎤ ⎤ ⎡ 1 −1 0 0 ⎥ ⎢− 1 2 − 1 0 0 ⎥⎥ ⎥ ⎢ 0 ⎥ = 19200 ⎢ 0 − 1 2 − 1 0 ⎥ kN / m ⎥ ⎥ ⎢ − K4 ⎥ ⎢ 0 0 − 1 2 − 1⎥ ⎢⎣ 0 0 K 4 + K 5 ⎥⎦ 0 − 1 2 ⎥⎦ 0 0

where K1 , K 2 , K 3 , K 4 and K 5 = K i the entire stiffness of the story. • Natural vibration frequencies and corresponding vibration mode shapes: Based on the dynamics of structures theory, the natural vibration frequencies and corresponding mode shapes can be determined by solve the equation: [ K − ω 2 M ]Φ = 0 This equation is called an eigenvalue problem. The quantities ω 2 are the eigenvalues indicating the square of free vibration frequencies, while the corresponding displacement vectors Φ represent the corresponding mode of vibrating system known as the eigenvectors or mode shapes. Hence a nontrivial solution is possible Φ ≠ 0 only when the determinant K − ω 2 M equal to zero (due to Cramer's rule). Expanding the determinant will give an algebraic equation of the Nth degree in the frequency parameter ω 2 for a system having N degrees of freedom. The N roots of this equation (ω12 , ω 22 , ω32 , ... , ω N2 ) represent the frequencies of the N modes of vibration which are possible in the system. The mode having the lowest frequency is called the first mode or the fundamental mode, the next higher frequency is the second mode, etc.

It is easily to solve this problem by using MATLAB (Mathematical Programming Language), where mathematically we can write: [Φ, Ω] = eig (inv( M ) × K ) Where Ω is the vector of square of frequencies. Edited by: Eng.Hussein Rida E-mail: [email protected]


Page 4 of 33

Type this code in MATLAB editor MATLAB CODE:

>> [ModeShapes,Omega]=eig(inv(M)*K)

The result will be: Square of Frequencies matrix 0 0 0 0 ⎤ ⎡ 15.5547 ⎥ ⎢ 0 132.5335 0 0 0 ⎥ ⎢ ⎥ Ω=⎢ 0 0 329.3511 0 0 ⎥ ⎢ 0 0 0 543.5194 0 ⎥ ⎢ ⎢⎣ 0 0 0 0 707.0414⎥⎦ To get Frequencies Matrix, type the following code: MATLAB CODE:

>> Freq = zeros(5) >> for i=1:5 Freq(i,i)= omega(i,i)^0.5 end

The result will be: ⎡ 3.9439 ⎢ 0 ⎢ ω=⎢ 0 ⎢ ⎢ 0 ⎢⎣ 0

0 0 0 0 ⎤ 11.5123 0 0 0 ⎥⎥ 0 18.1480 0 0 ⎥ rad / sec ⎥ 0 0 23.3135 0 ⎥ 0 0 0 26.5902⎥⎦

Mode shapes matrix:

⎡ 0.0597 ⎢ 0.0549 ⎢ Φ = ⎢ 0.0456 ⎢ ⎢ 0.0326 ⎢⎣ 0.0170 Mode shapes vectors:

0.0549 0.0456 - 0.0326 0.0170 ⎤ 0.0170 - 0.0326 0.0597 - 0.0456⎥⎥ - 0.0326 - 0.0549 - 0.0170 0.0597 ⎥ ⎥ - 0.0597 0.0170 - 0.0456 - 0.0549⎥ - 0.0456 0.0597 0.0549 0.0326 ⎥⎦

⎡0.0597 ⎤ ⎡ 0.0549 ⎤ ⎡ 0.0456 ⎤ ⎡- 0.0326⎤ ⎡ 0.0170 ⎤ ⎢0.0549⎥ ⎢ 0.0170 ⎥ ⎢- 0.0326⎥ ⎢ 0.0597 ⎥ ⎢- 0.0456⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎢ Φ1 = ⎢0.0456⎥ , Φ 2 = ⎢- 0.0326⎥ , Φ 3 = ⎢- 0.0549⎥ , Φ 4 = ⎢- 0.0170⎥ , Φ 5 = ⎢ 0.0597 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢0.0326⎥ ⎢- 0.0597 ⎥ ⎢ 0.0170 ⎥ ⎢- 0.0456⎥ ⎢- 0.0549⎥ ⎢⎣0.0170⎥⎦ ⎢⎣- 0.0456⎥⎦ ⎢⎣ 0.0597 ⎥⎦ ⎢⎣ 0.0549 ⎥⎦ ⎢⎣ 0.0326 ⎥⎦ mode shape-1

mode shape-2

mode shape-3

mode shape-4

mode shape-5

ω1 = 3.9439

ω2 = 11.5123

ω3 = 18.1480

ω4 = 23.3135

ω5 = 26.5902

rad / sec

rad / sec

rad / sec

rad / sec

rad / sec



Page 5 of 33

The five mode shapes for this frame are sketched below:

ω1 = 3.9439

ω2 = 11.5123

ω3 = 18.1480

ω4 = 23.3135

ω5 = 26.5902

rad / sec

rad / sec

rad / sec

rad / sec

rad / sec

Determination of Period Matrix:

The period (T) of motion is given as a function of frequency as following: T =

2π

ω

(sec)

This is mean that each mode shape of vibration has relative period To get the period matrix of the structure; type the following code: MATLAB CODE:

>> Period = zeros(5) >> for i=1:5 Period(i,i) = 2 * pi /freq(i,i) end

The period matrix will be: 0 0 0 0 ⎤ ⎡1.5931 ⎢ 0 0.5458 0 0 0 ⎥⎥ ⎢ T =⎢ 0 0 0.3462 0 0 ⎥ sec ⎢ ⎥ 0 0 0.2695 0 ⎥ ⎢ 0 ⎢⎣ 0 0 0 0 0.2363⎥⎦ st nd rd th th Where the period of 1 , 2 , 3 , 4 and 5 mode shapes are given respectively: (1.5931, 0.5458, 0.3462, 0.2695, 0.2363 sec). Note that the period of the first mode shape is the biggest one (T=1.5931 sec) which is called the fundamental period. The next lesser one is come with second mode shape, etc.



Page 6 of 33

Determination of response spectrum Acceleration Matrix: We can determine the mass acceleration depending on the response spectrum. The response spectrum is a plot of maximum accelerations for all values of periods, in other word for a system has specified period based on its mass and stiffness the response spectrum function gives the maximum acceleration can occur in the entire mass of this system. That is mean; if we know the vibration period of a specific mass we can determine its acceleration depending on response spectrum function.

The response spectrum function depending on the site characteristics, therefore the design codes give the response spectrum as a function of zone and soil profile, where the zone reflects the acceleration occur in the mother bed rock and the soil profile reflect the effect of the soil under structure in decrease or increase the amplitude of the motion. So it is very important to know that for a structure has specified period (T) will vibrate in different accelerations due to the site which the structure located. The determination of the design response spectra as per UBC97 requires two design parameters: Seismic Zone Factor : Z = 0.3 ( for Zone 3)⎫ ⎧Ca = 0.3 ⎫ ⎬⇒⎨ ⎬ Soil profile ( S B ) ⎭ ⎩CV = 0.3⎭

This plot has two characteristics periods CV 0.3 Ts = = = 0.4 sec 2.5Ca 2.5 × 0.3 To = 0.2 × Ts = 0.08 sec



Page 7 of 33

If the period of vibration mode is greater than Ts , then the relative acceleration is given by: C 0.3 0.3 2.943 Sa = V g = g= 9.81 = m / sec 2 T T T T Else, if the period is lesser than Ts and greater than To , then the relative acceleration is given by:

S a = 2.5CV g = 2.5 × 0.3 × 9.81 = 7.3575 m / sec 2

To get the acceleration matrix according to the period of vibration mode shapes, type the following code: MATLAB CODE:

>> Sa = zeros(5) >> for i=1:5 if Period(i,i) > 0.4 Sa(i,i) = 0.3 * 9.81 / Period(i,i) else Sa(i,i) = 0.75 * 9.81 end; end

The acceleration matrix will be: 0 0 0 0 ⎤ ⎡1.8473 ⎢ 0 5.3923 0 0 0 ⎥⎥ ⎢ Sa = ⎢ 0 0 7.3575 0 0 ⎥ m / sec 2 ⎥ ⎢ 0 0 7.3575 0 ⎥ ⎢ 0 ⎢⎣ 0 0 0 0 7.3575⎥⎦ Note: if the structure has a period lesser than or equal to characteristic periods Ts , then the entire mass of this structure will vibrate according to the maximum probable acceleration. This will be lead to create a maximum inertia force in mass. Therefore it is very important to scale the ratio of the structure's stiffness to its mass to get a value of period more than Ts as much as possible, but at the same time we have to avoid getting a more flexible structure Determination of maximum modal displacement:

L Sa m* Ω where: (L) is the matrix of modal excitation factor given by: L = Φ T M {1 } m* is the generalized modal mass matrix given by: m* = Φ T M Φ To get the matrix of modal excitation factor; type the following code:

The maximum modal displacement matrix is given by: U = Φ

MATLAB CODE:

>> LL = ModeShapes' * M * [1;1;1;1;1] >> for i=1:5 L(i,i) = LL(i,1) end



Page 8 of 33

The result will be: 0 ⎡20.9706 ⎢ 0 - 6.6022 ⎢ L=⎢ 0 0 ⎢ 0 ⎢ 0 ⎢⎣ 0 0

0 0 3.4796 0 0

0 0 0 1.9377 0

0 ⎤ 0 ⎥⎥ 0 ⎥ ⎥ 0 ⎥ 0.8853 ⎥⎦

To get the generalized modal mass matrix, type the following code: MATLAB CODE:

>> ModalMass = ModeShapes' * M * ModeShapes

The result will be: ⎡1 ⎢0 ⎢ m * = ⎢0 ⎢ ⎢0 ⎢⎣0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0⎤ 0⎥⎥ 0⎥ ⎥ 0⎥ 1⎥⎦

To get the maximum modal displacement, type the following code: MATLAB CODE:

>> U_Modal = ModeShapes * (L/ModalMass) * (Sa/Omega)

The result will be: ⎡ 0.1487 - 0.0147 0.0035 - 0.0009 0.0002 ⎤ ⎢ 0.1366 - 0.0046 - 0.0025 0.0016 - 0.0004⎥ ⎥ ⎢ U = ⎢ 0.1135 0.0088 - 0.0043 - 0.0004 0.0005 ⎥ m ⎥ ⎢ ⎢ 0.0812 0.0160 0.0013 - 0.0012 - 0.0005⎥ ⎢⎣ 0.0423 0.0122 0.0046 0.0014 0.0003 ⎥⎦ Where the relative displacement vectors due to each mode shape will be as the following: ⎡- 0.0147⎤ ⎡ 0.0035 ⎤ ⎡- 0.0009⎤ ⎡ 0.0002 ⎤ ⎡0.1487 ⎤ ⎢- 0.0046⎥ ⎢- 0.0025⎥ ⎢ 0.0016 ⎥ ⎢- 0.0004⎥ ⎢0.1366⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎥ ⎢ U 1 = ⎢0.1135⎥ , U 2 = ⎢ 0.0088 ⎥ , U 3 = ⎢- 0.0043⎥ , U 4 = ⎢- 0.0004⎥ , U 5 = ⎢ 0.0005 ⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎥ ⎢ ⎢ 0.0160 ⎥ ⎢ 0.0013 ⎥ ⎢- 0.0012⎥ ⎢- 0.0005⎥ ⎢0.0812⎥ ⎢⎣ 0.0122 ⎥⎦ ⎢⎣ 0.0046 ⎥⎦ ⎢⎣ 0.0014 ⎥⎦ ⎢⎣ 0.0003 ⎥⎦ ⎢⎣0.0423⎥⎦ Note that U i is the vector of maximum displacements at story-levels due to relative mode shape Φ i . (we can see clearly that the maximum lateral displacement at 5th story due to 1st, 2nd, 3rd, 4th and 5th mode shapes are given respectively :0.1487, -0.0147, 0.0035, -0.0009, 0.0002 m )



Page 9 of 33

Determination of maximum story-displacement:

Maximum total response cannot be obtained, in general, by merely adding the modal maxima because these maxima usually do not occur at the same time. In most cases, when one mode achieves its maximum response, the other modal responses are less than their individual maxima. Therefore, although the superposition of the modal spectral values obviously provides an upper limit to the total response, it generally over estimates this maximum by a significant amount. A number of different formulas have been proposed to obtain a more reasonable estimate of the maximum response from the spectral values. The simplest and most popular of these is the square root of the sum of the squares (SRSS) of the maximum modal responses. Thus if the maximum modal displacements are given as previous, the SRSS approximation of the maximum total displacements is given by: U max =

n

∑ (U ) i =1

2

i

=

(U1 )2 + (U 2 )2 + (U 3 )2 + (U 4 )2 + (U 5 )2

Where the terms under the radical sign represent the vectors of the maximum modal displacements squared. It is very important to know that the SRSS method is fundamentally sound when the modal frequencies are well separated. However, when the frequencies of major contributing modes are very close together, the SRSS method can give poor results, in which case the more general complete quadratic combination (CQC) method should be used. To get the maximum total displacement matrix, type the following code: MATLAB CODE:

>> for i=1:5 s = 0 for j=1:5 s = s + U_Modal(i,j)^2 end U_Max(i,1) = s^0.5 end

The result will be:

U max


⎡0.1494⎤ ⎢0.1367⎥ ⎥ ⎢ = ⎢0.1139⎥ m ⎥ ⎢ ⎢0.0828⎥ ⎢⎣0.0443⎥⎦


Page 10 of 33

Determination of maximum modal elastic-forces:

The maximum modal elastic forces occur at the story-levels is given by: L f s = M Φ * Sa m To get the matrix of modal elastic forces, type the following code: MATLAB CODE:

>> fs = M * ModeShapes * L /ModalMass * Sa

The result will be: ⎡ 231.2289 ⎢ 212.4961 ⎢ f s = ⎢ 176.5481 ⎢ ⎢ 126.2973 ⎢⎣ 65.8146

- 195.2809 116.6741 - 60.4827 - 83.4652 116.0654 - 140.4308 212.4961 43.4944 162.2452 152.8106

- 46.4792 85.0955 - 24.2207 - 64.9722 78.2015

11.0662 ⎤ - 29.6853 ⎥⎥ 38.8794 ⎥ kN ⎥ - 35.7296 ⎥ 21.2359 ⎥⎦

The relative elastic force vectors due to each mode shape will be as the following: ⎡- 195.2809⎤ ⎡ 116.6741 ⎤ ⎡- 46.4792⎤ ⎡ 11.0662 ⎤ ⎡231.2289⎤ ⎢ - 60.4827 ⎥ ⎢ - 83.4652 ⎥ ⎢ 85.0955 ⎥ ⎢- 29.6853⎥ ⎢ 212.4961⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎢ f s1 = ⎢176.5481 ⎥ , f s 2 = ⎢ 116.0654 ⎥ , f s 3 = ⎢- 140.4308⎥ , f s 4 = ⎢- 24.2207 ⎥ , f s 5 = ⎢ 38.8794 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 212.4961 ⎥ ⎢ 43.4944 ⎥ ⎢- 64.9722 ⎥ ⎢- 35.7296⎥ ⎢126.2973 ⎥ ⎢⎣ 162.2452 ⎥⎦ ⎢⎣ 152.8106 ⎥⎦ ⎢⎣ 78.2015 ⎥⎦ ⎢⎣ 21.2359 ⎥⎦ ⎢⎣ 65.8146 ⎥⎦ where f si is the vector of maximum elastic forces at story-levels due to relative mode shape Φi . (we can see clearly that the maximum elastic force at 5th story due to 1st, 2nd, 3rd, 4th and 5th mode shapes are given respectively :231.2289, -195.2809, 116.6741, -46.4792, 11.0662 kN) Determinations of maximum modal story shear forces:

We can get the shear force acting on a certain story by assembly the elastic-forces acting above the level of this story. Therefore the story-shear force is given by: V j = To assembly the elastic-forces at each story level, type the following code: MATLAB CODE:

>> for i=1:5 for j=1:5 s = 0 for a=1:j s=s+fs(a,i) end V_Modal(j,i)=s end end Edited by: Eng.Hussein Rida E-mail: [email protected]

∑(f ) n

i = j +1

si


The result will be: ⎡ 231.2289 ⎢ 443.7249 ⎢ V = ⎢ 620.2730 ⎢ ⎢ 746.5703 ⎢⎣ 812.3849

- 195.2809 116.6741 - 255.7636 33.2089 - 139.6982 - 107.2219 72.7979 - 63.7274 235.0431 89.0831

- 46.4792 38.6163 14.3956 - 50.5766 27.6249

Page 11 of 33

11.0662 - 18.6190 20.2604 - 15.4692 5.7667

⎤ ⎥ ⎥ ⎥ kN ⎥ ⎥ ⎥⎦

The relative shear force vectors due to each mode shape will be as the following: ⎡ - 195.2809 ⎤ ⎡ 116.6741 ⎤ ⎡- 46.4792⎤ ⎡ 11.0662 ⎤ ⎡231.2289⎤ ⎢- 255.7636⎥ ⎢ 33.2089 ⎥ ⎢ 38.6163 ⎥ ⎢- 18.6190⎥ ⎢443.7249⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎢ V1 = ⎢620.2730⎥ , V2 = ⎢ - 139.6982 ⎥ , V3 = ⎢- 107.2219⎥ , V4 = ⎢ 14.3956 ⎥ , V5 = ⎢ 20.2604 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 72.7979 ⎥ ⎢ - 63.7274 ⎥ ⎢- 50.5766 ⎥ ⎢- 15.4692⎥ ⎢746.5703⎥ ⎢⎣ 235.0431 ⎥⎦ ⎢⎣ 89.0831 ⎥⎦ ⎢⎣ 27.6249 ⎥⎦ ⎢⎣ 5.7667 ⎥⎦ ⎢⎣812.3849 ⎥⎦ where Vi is the vector of shear forces at story-levels due to relative mode shape Φ i . (We can see clearly that the shear force at 1st story due to 1st, 2nd, 3rd, 4th and 5th mode shapes are given respectively: 812.3849, 235.0431, 89.0831, 27.6249, 5.7667 kN) Determinations of maximum total story shear forces:

Similarly to previous (Determination of maximum story-level displacement), the maximum total story shear forces could be approximated from the modal maxima by using SRSS combination method as the following: Vmax =

n

∑ (V ) i =1

2

i

=

(V1 )2 + (V2 )2 + (V3 )2 + (V4 )2 + (V5 )2

To get the maximum total story shear-forces matrix, type the following code: MATLAB CODE:

>> for i=1:5 s = 0 for j=1:5 s = s+V_Modal(i,j)^2 end V_Max(i,1)=s^0.5 end

The result will be:

Vmax


⎡327.8674⎤ ⎢515.0219⎥ ⎥ ⎢ = ⎢645.2662⎥ kN ⎥ ⎢ ⎢754.6690⎥ ⎢⎣850.8506⎥⎦


Page 12 of 33

Determination of modal participation factors:

The modal participation factor representing the interaction between the mode shape and the spatial distribution of the external load. L this factor given by: MPF = * m To get the matrix of modal participation factor, type the following code: MATLAB CODE:

>> MPF = L / ModalMass

The result will be: 0 ⎡20.9706 ⎢ 0 - 6.6022 ⎢ MPF = ⎢ 0 0 ⎢ 0 ⎢ 0 ⎢⎣ 0 0

0 0 3.4796 0 0

0 0 0 1.9377 0

0 ⎤ 0 ⎥⎥ 0 ⎥ ⎥ 0 ⎥ 0.8853 ⎥⎦

Determination of modal participating mass ratio: The modal participating mass ratio represent the part of the total mass which responding to earthquake motion in each mode, therefore this ratio is very important to determine the adequate number of mode shapes which give a reasonable part of vibration mass which will respond to the motion.

The UBC-97 Code declares that we need an adequate number of mode shapes to insure that 90% of the mass at least will respond due to earthquake motion. The modal participating mass ratio given by:

(L

2

)

m* % ∑ mi To get the modal participating mass ratio matrix, type the following code: MPMR =

MATLAB CODE:

>> Segma_M = 0 >> for i=1:5 Segma_M = Segma_M + M(i,i) end >> MPMR = ((L*L/ModalMass)/Segma_M)*100

The result will be: ⎡87.9530 ⎢ 0 ⎢ MPMR = ⎢ 0 ⎢ ⎢ 0 ⎢⎣ 0 Edited by: Eng.Hussein Rida E-mail: [email protected]

0 8.7177 0 0 0

0 0 2.4216 0 0

0 0 0 0.7509 0

0 ⎤ 0 ⎥⎥ 0 ⎥ ⎥ 0 ⎥ 0.1568 ⎥⎦


Page 13 of 33

Note that the part of mass which will respond to the motion for 1st, 2nd, 3rd, 4th and 5th are given respectively: 87.95%, 8.72%, 2.42%, 0.75% and 0.16%). Due to MPMR, we see that the first mode shape is most important one, because 87.95% of mass will respond to ground motion, when only 8.72% of mass will respond in the second mode shape, etc. Note that the only first-two mode shapes will be adequate to insure that more than 90% of the mass will vibrate responding to ground motion.

References: Clough, R., and J. Penzien. 1993. Dynamics of Structures, Second Edition. McGraw- Hill. Paz, M. 1985. Structural Dynamics, theory and computation. Van Nostrand Reinhold. International Conference of Building Official 1997, Uniform Building Code. Whittier, California. CSI Analysis Reference Manual. CSI Computers & Structures, Berkeley, California. ETABS Software Verification Examples, CSI Computers & Structures, Berkeley, California.



Page 14 of 33

MATLAB CODE >> % Define Lumped Mass Matrix >> M = [100 0 0 0 0; 0 100 0 0 0; 0 0 100 0 0; 0 0 0 100 0; 0 0 0 0 100] M = 100 0 0 0 0

0 100 0 0 0

0 0 100 0 0

0 0 0 100 0

0 0 0 0 100

>> % Define Stiffness Matrix >> K = 19200.*[1 -1 0 0 0; -1 2 -1 0 0; 0 -1 2 -1 0; 0 0 -1 2 -1; 0 0 0 -1 2] K = 19200 -19200 0 0 0

-19200 38400 -19200 0 0

0 -19200 38400 -19200 0

0 0 -19200 38400 -19200

>> % ModeShapes & Squared Frequencies >> [ModeShapes,Omega]=eig(inv(M)*K) Omega = 15.5547 0 0 0 0

0 132.5335 0 0 0

0 0 329.3511 0 0

0 0 0 543.5194 0

0 0 0 0 707.0414

0.0456 -0.0326 -0.0549 0.0170 0.0597

-0.0326 0.0597 -0.0170 -0.0456 0.0549

0.0170 -0.0456 0.0597 -0.0549 0.0326

ModeShapes = 0.0597 0.0549 0.0456 0.0326 0.0170

0.0549 0.0170 -0.0326 -0.0597 -0.0456


0 0 0 -19200 38400


Page 15 of 33

MATLAB CODE >> Freq = zeros(5) >> for i=1:5 Freq(i,i)= omega(i,i)^0.5 end Freq = 3.9439 0 0 0 0

0 11.5123 0 0 0

0 0 18.1480 0 0

0 0 0 23.3135 0

0 0 0 0 26.5902

>> %Period Matrix >> Period = zeros(5) >> for i=1:5 Period(i,i) = 2 * pi /freq(i,i) end Period = 1.5931 0 0 0 0

0 0.5458 0 0 0

0 0 0.3462 0 0

0 0 0 0.2695 0

0 0 0 0 0.2363

>> %Acceleration Matrix >> Sa = zeros(5) >> for i=1:5 if Period(i,i) > 0.4 Sa(i,i) = 0.3 * 9.81 / Period(i,i) else Sa(i,i) = 0.75 * 9.81 end; end Sa = 1.8473 0 0 0 0

0 5.3923 0 0 0


0 0 7.3575 0 0

0 0 0 7.3575 0

0 0 0 0 7.3575


Page 16 of 33

MATLAB CODE

>> %Modal Excitation Matrix >> LL = ModeShapes' * M * [1;1;1;1;1] >> for i=1:5 L(i,i) = LL(i,1) end L = 20.9706 0 0 0 0

0 -6.6022 0 0 0

0 0 3.4796 0 0

0 0 0 1.9377 0

0 0 0 0 0.8853

>> % Modal Mass Matrix >> ModalMass = ModeShapes' * M * ModeShapes ModalMass = 1.0000 0.0000 0.0000 0.0000 0.0000

0.0000 1.0000 0.0000 0.0000 0.0000

0.0000 0.0000 1.0000 0.0000 0.0000

0.0000 0.0000 0.0000 1.0000 0.0000

0.0000 0.0000 0.0000 0.0000 1.0000

>> %Modal Displacement >> U_Modal = ModeShapes * (L/ModalMass) * (Sa/Omega) U_Modal = 0.1487 0.1366 0.1135 0.0812 0.0423

-0.0147 -0.0046 0.0088 0.0160 0.0122


0.0035 -0.0025 -0.0043 0.0013 0.0046

-0.0009 0.0016 -0.0004 -0.0012 0.0014

0.0002 -0.0004 0.0005 -0.0005 0.0003


Page 17 of 33

MATLAB CODE >> %Maximum Story Displacement due to SRSS Combination >> for i=1:5 s = 0 for j=1:5 s = s + U_Modal(i,j)^2 end U_Max(i,1) = s^0.5 end U_Max = 0.1494 0.1367 0.1139 0.0828 0.0443 >> %Maximum Modal Elastic Forces Matrix >> fs = M * ModeShapes * L /ModalMass * Sa fs = 231.2289 -195.2809 116.6741 212.4961 -60.4827 -83.4652 176.5481 116.0654 -140.4308 126.2973 212.4961 43.4944 65.8146 162.2452 152.8106

-46.4792 85.0955 -24.2207 -64.9722 78.2015

11.0662 -29.6853 38.8794 -35.7296 21.2359

>> %Maximum Modal Story-Shear Matrix >> for i=1:5 for j=1:5 s = 0 for a=1:j s=s+fs(a,i) end V_Modal(j,i)=s end end V_Modal = 231.2289 -195.2809 116.6741 443.7249 -255.7636 33.2089 620.2730 -139.6982 -107.2219 746.5703 72.7979 -63.7274 812.3849 235.0431 89.0831


-46.4792 38.6163 14.3956 -50.5766 27.6249

11.0662 -18.6190 20.2604 -15.4692 5.7667


Page 18 of 33

MATLAB CODE >> %Maximum Total Story Shear Forces due to SRSS Combination >> for i=1:5 s = 0 for j=1:5 s = s+V_Modal(i,j)^2 end V_Max(i,1)=s^0.5 end V_Max = 327.8674 515.0219 645.2662 754.6690 850.8506 >> %Modal Participation Factor >> MPF = L / ModalMass MPF = 20.9706 0.0000 0.0000 0.0000 0.0000

0.0000 -6.6022 0.0000 0.0000 0.0000

0.0000 0.0000 3.4796 0.0000 0.0000

0.0000 0.0000 0.0000 1.9377 0.0000

0.0000 0.0000 0.0000 0.0000 0.8853

>> %Modal Participating Mass Ratio >> Segma_M = 0 >> for i=1:5 Segma_M = Segma_M + M(i,i) end >> MPMR = ((L*L/ModalMass)/Segma_M)*100 MPMR = 87.9530 0.0000 0.0000 0.0000 0.0000

0.0000 8.7177 0.0000 0.0000 0.0000


0.0000 0.0000 2.4216 0.0000 0.0000

0.0000 0.0000 0.0000 0.7509 0.0000

0.0000 0.0000 0.0000 0.0000 0.1568


Page 19 of 33

ETABS MODEL Build of Mathematical Computer Model:

•

The frame is modeled as five-story consist from two-column line, singly bay system with story-height 3m & length of bay 4m. kN-m-second units are used.



Page 20 of 33

ETABS MODEL

•

Define material properties: (Modulus of elasticity, Self-Mass of Material) Assume that the self-weight of the frame elements is neglected



Page 21 of 33

ETABS MODEL

•

Define Column's Properties:

The column is modeled to have infinite axial area, so that axial deformation is neglected. Also, Zero column shear area is input to trigger the ETABS option of neglecting shear deformation. These deformations are neglected to be consistent with the hand-calculated model with which the result are compared.



Page 22 of 33

ETABS MODEL

•

Define Beam's Properties:

The beam is modeled as a rigid beam to have infinite moment of inertia compared to column, so that axial deformation is neglected. Also, neglecting both shear deformations and axial deformations.



ETABS MODEL

•

Build the model to be as following:


Page 23 of 33


Page 24 of 33

ETABS MODEL

•

Draw Point Object at the mid-span of beams in order to assign lumped mass at the storylevel.



ETABS MODEL

•

Assign Lumped Mass at story-level.


Page 25 of 33


ETABS MODEL

•

Define Mass Source:

•

Assign Diaphragm at Story-Level:


Page 26 of 33


ETABS MODEL

•

Define Response Spectrum Function (UBC97 Design Spectrum):


Page 27 of 33


ETABS MODEL

•

Define & Assign Response Spectrum Case Data:


Page 28 of 33


ETABS MODEL

•

Define Analysis options:

•

Perform analysis.


Page 29 of 33


ETABS MODEL

•

Building Mode-Shapes:


Page 30 of 33


ETABS MODEL

•

Mode-Shapes:


Page 31 of 33


ETABS MODEL

•

Periods and Accelerations:

•

Maximum Story-Displacement according to SRSS combination:

•

Maximum Story-Shear Force according to SRSS combination:


Page 32 of 33


ETABS MODEL

•

Modal Participation Factor:

•

Modal Participating Mass Ratio:


Page 33 of 33

Dynamics Response Spectrum Analysis

Recommend Documents