Chapter 13 Homework Figure P135 Solution The Modal Expansion Shown

1

CHAPTER 13

Problems 9.5 and 10.6) excited by horizontal ground

Solution:

u

1

h

Vibration properties (from Problem 10.6):









10.707 1 T





20.707 1 T





Substituting



n, m, and n



in Eq. (13.2.4) gives

The modal expansion of effective earthquake forces is

shown in Fig. P13.1b.

s

h

Part b

Substituting



n and jn



in Eq. (13.2.5) gives floor

displacements due to each mode:

Combining the contributions of the two modes gives

Part c

2

First mode

Second mode

Fig. P13.1c

Substituting Vjn

st in Eq. (13.2.8) gives the modal responses:

Combining the modal responses gives the total responses:

A

Part d

Static analysis of the structure for external floor forces

sn gives the modal static responses Mbn

st and Mn1

st for

M

b

and

M

1, the overturning moments at the base and the first

floor, respectively:

Substituting Mbn

st and Mn1

st in Eq. (13.2.8) gives the modal

responses:

M

t

A

t

M

t

A

t

M

t

A

t

M

t

A

t

11 1

12 2

Combining the modal responses gives the total response:

Figure P13.1c

3

Problem 13.2

The response of the two–story shear frame of Fig. P13.1

(also of Problems 9.5 and 10.6) to El Centro ground

motion is to be computed as a function of time. The

acceleration data are available in Appendix 6 at every

∆t = 0.02 sec.

(a) Determine the SDF system responses Dn(t) and An(t)

using a numerical time-stepping method of your choice

A

each floor, (ii) the story shears, and (iii) the floor and base

overturning moments.

(c) At each instant of time, combine the modal contribu-

tions to each of the response quantities to obtain the total

response; determine the peak value of the total responses.

For selected response quantities plot as a function of time

the modal responses and total response.

Vibration properties (from Problem 10.6):

T

10 321. sec T20 133. sec

M

Modal properties:

Table P13.2a

Mode

M

n Ln

h Lh

n



Part a

The displacements D

t

n()

and pseudo-accelerations

Part b

The modal static responses for the various response

quantities are given in Table P13.2b.

Table P13.2b

Mode n 1 2

un1

st 3

2.23 10

 6 529 10 5

.





st 0.604 –0.104

Step 5c of Section 13.2.4 is implemented to determine

the contribution of the nth mode to selected response

where r

n

st and

A

t

n()

are both known. These results for roof

displacement u

t

, base shear

V

t

, and base overturning

4

Table P13.2c

Floor or

Story Displacement, in. Shear, kips

Overturning

moment,

kip–ft

0.797

0.118

-1

0

1

0.684

0 5 10 15

Mode 1

Time, sec

5

-1

0

-1

0.964

Time, sec

0 5 10 15

Fig. P13.2c

-150

115.28

2.933

-150

115.11

0 5 10 15

Time, sec

2

-2

1959. 25

0 5 10 15

Mode 1

Time, sec

Figure P13.2c

6

Problem 13.3

modes. Verify that Eqs. (13.2.14) and (13.2.17) are

satisfied.

Solution:

From Problem 13.2:

2

1

(0.707) 2 (1)

2

jjj

j

m

Lhmhm h

















j

1



From Eq. (13.2.9a), the effective modal masses are

From Eq. (13.2.9a), the effective modal heights are

m/2

1.457 m0.043 m

Mode 1 Mode 2

Verify Eq. (13.2.14):

2

*

1

1.457 0.043 1.5

n

Mmmm













7

Figure P13.4 shows a two–story frame (the same as that in

Problems 9.6 and 10.10) with flexural rigidity EI for beams

and columns. Determine the dynamic response of this

structure to horizontal ground motion ݑሷ g(t). Express (a) the

Mass and lateral stiffness matrices (from Problem 9.6):

Vibration properties (from Problem 10.10):



Modal properties:

Part a

From Eq. (13.2.5), the floor displacements due to the

first mode are

(a)

The joint rotations associated with u1 are

where T is available in the solution to Problem 10.10:

3

0164 0 411

()

..

ut

R

U



L

O

The floor displacements due to the second mode are

22

TU

W 

TU

W

TU

W

(c)

3

0082

()

.

ut

R

U

R

U

.().()

Dt Dt

R

0647 0353

Part b

The bending moments at the ends of a flexural element

are related to the nodal displacements by

For a first story column, Lh and the nodal displace-

Substituting these ua, ub,



a, and



b and Eqs. (e)–(f) in

Eqs. (g)–(h) gives:

Relate D

t

n()

to

A

t

n()

:

Substituting Eq. (j) in Eq. (i) gives

Similarly,

a b

b

u = 0

a



5

u

= b



6

u

=

a

u = 0

Substituting these ua, ub,



a, and



b and Eqs. (e)–(f) in

Fi

g

ure P13.4b

Problem 13.5

effective earthquake forces, (b) the floor displacement

response in terms of Dn(t), (c) the story shear response in

terms of An(t), and (d) the base overturning moment in













m

u1

1

12

m









m

12 1

011

k



 







k

Vibration properties (from Problem 10.11):



123

0.5 1 0.5

0.866 0 0.866



   

   



   



The first-mode properties are computed from Eq. (13.2.3):

3

22 22

11

(0.5) (0.866) (1) 1.5

jj

m

M

mm m m



 

Similar calculations for the second and third modes give:

20.5

h

Lm



 30.134

h

Lm

Part a

Substituting



n, m, and n



in Eq. (13.2.4) gives

1 0.5 0.6220



   

1 1 0.3333





   

The modal expansion of m1 is shown next:

Part b

Substituting for



n and n



gives

1

( ) 1 0.3333

ut



   

   

10

1

( ) 0.5 0.0447

ut

    

Combining the modal responses gives the floor

displacements:

Part c

Static analysis of the frame for external floor forces sn

gives V

st , i = 1, 2, 3:

The total story shears are

Substituting values of Vjn

st gives

Vt m At A t A t

1123

2 3213 0 1667 0 0121() . () . () . ()

Part d

Static analysis of the frame for external floor forces sn

gives Mbn

st :

Mmh

b10 6220 1 1 0773 2 0 6220 3

st 

.().().()

The base overturning moment response is

Substituting values of Mbn

st gives

11

motion ݑሷ g(t), determine (a) the modal expansion of

effective earthquake forces, (b) the floor displacement

response in terms of Dn(t), (c) the story shear response in



Solution:





Mass and stiffness matrices (from Problem 9.8)







where kEIh83

/ and h = story height

The first mode properties are computed from Eq. (13.2.3):

Mm

11069



.

Similar calculations for the second and third modes give:

22

2

05

L

M.



Part a

Substituting n



, m, and n



in Eq. (13.2.4) gives

0.5 0.701



 



 

The modal expansion of m1 is shown next:

12 ft

EI







12

Part b

44.0

)(

1

tu









25.0

)(

2

1

tu

















)(

212.0

31.0

)(

1

tD

tu



















ut Dt Dt Dt

2123

0 962 0 25 0 212() . () . () . ()

Part c

Part d

sn gives Mbn

st :

The base overturning moment response is:

Substituting values of Mbn

st gives

Static analysis of the frame for external floor forces sn

gives Mn1

st :

The first floor overturning moment response is

Substituting values of Mn1

st gives

s3

m1 s1s2

13

The second floor overturning moment rseponse is

Substituting values of Mn

st

2 gives

14

Problem 13.7

The response of the three-story shear frames of Fig. P13.5

(also of Problems 9.7 and 10.11) to El Centro ground

motion is to be computed as a function

of time. The properties of the frame are h = 12 ft,

m = 100 kips/g, I = 1400 in4, E = 29,000 ksi, and ζn = 5%.

(a) Determine the SDF system responses Dn(t) and An(t)

displacement, (ii) the story shears, and (iii) the base

overturning moment.

(c) At each instant of time combine the modal

contributions to each of the response quantities to obtain

2

100 100 0.2588 kip– sec in.

g 386.4

m 

Vibration properties (from Problem 10.11):

123

0.3418 sec 0.1251 sec 0.0716 secTTT

Modal properties (from Problem 13.5):

Part a

The displacements Dt

n()

and pseudo-acceleration

Part b

The modal static responses for the various response

quantities are given in Table P13.7a (also see Problem

13.5).

Table P13.7a

Mode n 1 2 3

Vm

bn

st 2.3213 0.1667 0.0121

Step 5c of Section 13.2.4 is implemented to determine the

contribution of the nth mode to selected response

quantities:

Part c

responses.

3 1.103 52.22 626.6

-1

0.1096

-1

0.0498

-1

0.7153

-1

0.6065

0 5 10 15

15

-1

0.0365

-1

0.0044

0 5 10 15

-200

11.92

-200

189.29

0 5 10 15

-200

Time, sec

16

-4

0

4

143.1

-4

4 4320. 8

3

Mode 2

0 5 10 15

0

17.5

M

b

,

M

bn

,

17

Problem 13.8

The response of the three-story shear frames of Fig. P13.6

(also of Problems 9.8 and 10.12) to El Centro ground

(a) Determine the SDF system responses Dn(t) and An(t)

(b) For each natural mode, calculate as a function of time

(c) At each instant of time combine the modal

Solution:

System properties:

Vibration properties (from Problem 10.12):

123

0.314 0.5 3.186



Modal properties from (from Problem 13.6):

Part a

The displacements )(tDn and pseudo-acceleration

procedure of Section 5.2 with sec 02.0t. The results

are shown in Figs. P13.8a–b.

Part b

Mode n 1 2 3

st

hmMbn

)()( st tArtr nnn 

where st

n

r and )(tAn are both known. These results for

The modal contributions to each response quantity

are combined at each time instant to obtain Figs. 13.8c–e.

Floor or

Story Displacement, in. Shear

kips.

Overturning

moment, kip–ft.

18

Fig. P13.8a

Figure P13.8a

Figure P13.8b

19

Figure P13.8c

Figure P13.8d