Example-Modal Analysis Mathcad

ORIGIN := 1

k := 1000⋅ lb

 60 −60 K :=

−60

−120 ⋅

180

  0 −120  1

M :=

⋅

0 1.5 0

  0

K1 := M

300 

0 

0

k⋅ sec

in

2

2

3

−1

⋅ K

 60 −60 K1

1

k

in

2 

0



0

= −40

120

  0 −60

0



−80

sec

-2

150 

evp evp := eige eigenv nval alss( K1)

 21.088  evp

=

96.396

  212.516 

sec

-2

 21.088  Fre Freq := sort sort( evp evp)

Freq

=

96.396

  212.516 

sec

-2

i := 1 .. 3

ωi :=

Freq

ω1 =

4.592sec

ω2 =

9.818sec

ω3 =

:= i

T

i -1

2⋅ π

ωi T

=

1.368sec

T

=

0.64 0.64 sec

T

=

0.431sec

1

-1

2

-1

14.578 14.578 sec

C:\AAA_Files\Teaching\CE7119 \AAA_Mathcads\Example-modal analysis.mcd/Example-modal analysis.mcd

3

CIVL7119/8119

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Find mode shapes: shapes: v := eige eigenv nveecs( K1)

0.739

−0.273 

−0.449

0.694

 0.813 v=

0.527

  0.246 −0.502 −0.666 

Mode shape 1 corresponds with eigenvalue 21.088. 21.088. The first column of the v matrix corresponds with the first evp which was 21.088. 21.088. The second column corresponds to the frequency 96.396 (second row of evp). The third column corresponds to the frequency 212.516.



〈1〉

V1 :=

V2 :=

v

V1



0.649

1, 1

  0.302 

〈2〉



v

v v

V2

1, 2

v v

1, 3

V3

1



= −0.607   −0.679 



〈3〉

V3 :=

=

1

1



= −2.542

  2.44 

 2.531 × 105   2.531 × 105  2 -1   -1 K⋅ V1 = 2.462 × 105 lbft ω1) ⋅ M ⋅ V1 =  2.462 × 105  lbft (     5 5  1.528 × 10   1.528 × 10   1.157 × 106   1.157 × 106  2 -1 ( ω2) ⋅ M⋅ V2 =  −1.053 × 106  lbft-1 K⋅ V =  6 2 1.053 × 10 lbft −   6 6  −1.571 × 10   −1.571 × 10 


CIVL7119/8119

2/7

Check orthogonality condition: T

T

Test1 := V1 M ⋅ V2 Test1

=

Test2 := V1 M⋅ V3

( 6.46 × 10− 12 ) lbsec2 ft-1

Test2

=

( 1.331 × 10− 11 ) lbsec2 ft-1

T

Test3 := V2 M ⋅ V3 Test3

=

( 3.205 × 10− 12 ) lbsec2 ft-1

Sv1 := 1.74⋅

 1  one :=

1

  1 

Sv2 := 1.41⋅

Sv3 := 1.20⋅

M3

= =

sec ft sec

T

T

L2 := V2 ⋅ M⋅ one

T

L3 := V3 ⋅ M⋅ one

T

T

M 3 := V3 ⋅ M ⋅ V3

M2

ft

L1 := V1 ⋅ M⋅ one

M 2 := V2 ⋅ M ⋅ V2

=

sec

T

M 1 := V1 ⋅ M ⋅ V1

M1

ft

( 1.813 ) ( 2.474 )

k⋅sec

2

in k⋅sec

( 22.596 )

k⋅ sec

L1

=

L2

= ( −1.268 )

( 2.577 )

in

2

in k⋅ sec

2

in


L3

=

( 2.066 )

2

k⋅ sec

2

in

k⋅ sec

2

in

CIVL7119/8119

3/7

− 1 Sv1

Y1 := L1⋅ M1

⋅

ω1

− 1 Sv2

Y2 := L2⋅ M2

⋅

ω2

− 1 Sv3

Y3 := L3⋅ M3

⋅

ω3

Y1

=

Y2

= ( −0.074 ) ft

Y3

=

( 0.538 ) ft

( 7.528 × 10− 3 ) ft

u1 := V1⋅ Y1 u2 := V2⋅ Y2 u3 := V3⋅ Y3

 −0.074 

 0.538  u1

=

0.349 ft

u2

  0.163 

=

0.045

  0.05 

ft

 7.528 × 10− 3  u3 =  −0.019  ft  0.018 

SRSS Methods 2

2

2

2

2

2

2

2

2

umax1 :=

(u11) + (u21) + (u31)

umax2 :=

(u12) + (u22) + (u32)

umax3 :=

(u13) + (u23) + (u33)

umax1

=

0.543ft

umax2

=

0.353ft

umax3

=

0.171ft


CIVL7119/8119

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CQC Procedure i := 1 .. 3

ξ :=

0.05

j := 1.. 3

:= i, j

r

ωi

r

i, j

ω j

= 1

2.138 3.175 0.468 1 1.485 3

8⋅ ξ

ρ i , j :=

1 −  u11  U := u12   u13

2

⋅ ( 1 + r i , j) ⋅ ( r i , j)

(r i, j)  2

u2

1

u2

2

u2

3

0.315

2

0.673 1

2

2

+ 4⋅ ξ ⋅ r i , j⋅ ( 1 + r i , j)

 u3  2  u3 3  u3

1

2

 0.538 −0.074 U =  0.349 0.045  0.163 0.05

− 3 

7.528 × 10

−0.019 0.018

 ft 

p := 1.. 3

3

Ucqc

p

:=

 0.542 

3

∑ ∑

U

⋅ ρ m , n⋅ U p , n

p , m

m= 1 n= 1

Ucqc =

0.353 ft

  0.172 

Compare this with SRSS results

Sa1 :=

ω1⋅ Sv1

Sa2 :=

ω2⋅ Sv2

Sa3 :=

ω3⋅ Sv3


CIVL7119/8119

5/7

Modal Forces f1 := M⋅ V1⋅ L1⋅

f2 := M⋅ V2⋅ L2⋅

f3 := M⋅ V3⋅ L3⋅

 136.254 

Sa1

f1

M1

=

132.549 k

  82.257 

 −85.134 

Sa2 f2

M2

Sa3

=

77.464

  115.609 

k

 19.197 

M3

f3

= −73.197

  93.668 

k

Base Shear for Each Mode 3

V01 :=

∑

f1

V01 V01 = 351 351.06 .06 k

f2

V02 V02 = 107 107.938k .938k

f3

V03 V03 = 39. 39.668 668 k

j

j = 1

3

V02 :=

∑

j

j = 1

3

V03 :=

∑

j

j = 1

Total Base Shear

V0 := V0

=

2

V01

+

2

V02

+

2

V03

369. 369.41 414k 4k


CIVL7119/8119

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Total Lateral Force 2

2

Froof :=

(f11) + (f21) + (f31)

F2nd :=

(f12) + (f22) + (f32)

F1st :=

2

2

2

2

(f13) + (f23) + (f33)

2

Froof = 161.807 161.807 k

2

F2nd

=

170.081 170.081 k

2

F1st

=

5

1.7 × 10 lb

Overturning Moments at Floor Levels H1 := 12⋅ ft H2 := 12⋅ ft H3 := 12⋅ ft

OMGround := ( H1

OM1st := ( H1

+

+

H2

+

H3) ⋅ Froof + ( H2

+

H3) ⋅ F2nd

+

H3⋅ F1st

H2) ⋅ Froof + H2⋅ F2nd

OM2nd := H1⋅ F1st

OMGround

OM1st

=

OM2nd


=

=

4

1.195 × 10 k⋅ ft 3

5.924 × 10 k⋅ ft 3

2.04 × 10 k⋅ ft

CIVL7119/8119

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Example-Modal Analysis Mathcad

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