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1

Problem 11.1

Estimate the bending moment distribution for the cases listed below. Use qualitative reasoning based on relative stiffness. Assume I constant.

(a)

(b)

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2

Problem 11.2Estimate the bending moment distribution. Use qualitative reasoning based on relative stiffness.Assume I constant.

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3

Problem 11.3

Solve problem 11.1 cases (a) and (b) using moment distribution. Compare the approximate and

exact results.

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4

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5

Problem 11.4

Consider the multi-story steel frame shown below. Determine the maximum positive and

negative moments in the beams using the following approaches:

(i) Assume inflection points at .1L from each end of the beams.

(ii) Use a computer software system. Assume g cI 2I

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6

Moment diagram-

max

+

max

M 49 kN-m

M 27.6 kN-m

Problem 11.5

Estimate the axial force, shear force, and bending moment distributions. Assume g cI 2I

(a)

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7

(b)

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8

Assume shear is divided equally between columns.

Problem 11.6

Members AC and FD are continuous. Estimate the bending moment distribution in AC and FD,

and the axial forces in the pin ended members. Compare your results with results generated witha computer software system.

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9

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10

Axial forces & reaction

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11

Moment distribution

Problem 11.7

Repeat problem 11.6 assuming fixed supports at A and F.

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12

Axial forces & reaction

Moment distribution

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13

Problem 11.8

Consider the steel frame shown below. Assume g cI 3 I for all the members. Determine the

moment at each end of each member using

(a) The portal method.

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14

(b) The shear stiffness method.

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16

XX

C 1

FX 2.300E+00

XX

C 2FX 9.398E-01

XX

C 4

FX -9.398E-01

XX

C 3

FX -1.782E+00

XXG3

FX -6.667E+00

XX

C 5

FX 1.367E+00

XXG4

FX -3.040E+00

XX

C 6 FX -1.886E+00

XXG1

FX -7.289E+00

XXG2

FX -2.251E+00

Axial forces( kip)

XX

C 1FY -2.459E+00

XX

C 2FY -1.749E+00

XXG2

FY 9.398E-01

XX

C 4

FY -2.251E+00

XXG1

FY 1.360E+00

XXG3

FY 5.185E-01

XXG4

FY 1.886E+00

XX

C 6 FY -3.040E+00

XX

C 5

FY -3.627E+00

XX

C 3

FY -2.873E+00

Shear forces (kip)

Moment diagram (kip-ft)

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17

(a)

XX

C 1

FX 1.694E+00

XX

C 2FX 5.063E-01

XXG2

FX -4.362E+00

XXG1

FX -7.690E+00

XX

C 3

FX -1.010E+00

XXG3

FX -5.691E+00XX

C 4

FX -2.955E-01

XXG4

FX -3.073E+00

XXG6

FX -1.796E+00

XXG8

FX -4.256E-01

XX

C 1 0

FX -2.582E-01

XX

C 9

FX -8.080E-01

XXG7

FX -2.048E+00

XX

C 7

FX -7.199E-01

XX

C 8

FX -2.166E-01

XXG5

FX -3.822E+00

XX

C 5

FX 8.441E-01

XX

C 6FX 2.640E-01

Axial force (kip)

XX

C 1

FX 1.694E+00FY -2.948E+00

XX

C 2

FX 5.063E-01

FY -6.378E-01

XXG2

FX -4.362E+00FY 5.063E-01

XXG4

FX -3.073E+00FY 2.108E-01

XXG6

FX -1.796E+00FY 4.748E-01

XXG8

FX -4.256E-01FY 2.582E-01

XX

C 1 0

FX -2.582E-01FY -4.256E-01

XXG7

FX -2.048E+00FY 5.498E-01

XX

C 9

FX -8.080E-01FY -2.473E+00

XX

C 7

FX -7.199E-01FY -3.145E+00

XX

C 5

FX 8.441E-01FY -3.146E+00

XXG3

FX -5.691E+00

FY 4.730E-01 XX

C 6

FX 2.640E-01FY -1.277E+00 XX

C 8

FX -2.166E-01FY -1.370E+00XX

C 4

FX -2.955E-01FY -1.289E+00

XXG1

FX -7.690E+00FY 1.188E+00

XX

C 3

FX -1.010E+00FY -3.288E+00

Shear forces (kip)

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18

Bending moment diagram(kip-ft)

(c)

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19

XX

C 2FX 2.003E+00

XX

C 1

FX 4.911E+00

XXG1

FX -7.620E+00

XX

C 3

FX -1.818E+00

XXG3

FX -4.832E+00

XX

C 5

FX 1.808E+00

XXG5

FX -2.149E+00

XX

C 7

FX -4.900E+00

XX

C 8

FX -2.016E+00

XXG6

FX -3.178E+00XX

C 1 0

FX -9.926E-01

XXG7

FX -3.002E+00

XX

C 9FX 9.926E-01

XXG2

FX -4.884E+00

XX

C 4

FX -3.577E-02

XXG4

FX -4.042E+00

XX

C 6

FX 4.871E-02

Axial forces (kip)

XX

C 1

FX 4.911E+00FY -5.496E+00

XXG1

FX -7.620E+00FY 2.908E+00

XX

C 3

FX -1.818E+00FY -6.629E+00

XXG3

FX -4.832E+00

FY 1.125E+00

XX

C 5FX 1.808E+00

FY -6.548E+00

XX

C 7

FX -4.900E+00FY -5.327E+00

XX

C 8

FX -2.016E+00FY -3.178E+00

XXG5

FX -2.149E+00FY 2.884E+00

XXG6

FX -3.178E+00FY 2.016E+00

XXG4

FX -4.042E+00FY 9.750E-01

XX

C 6

FX 4.871E-02FY -3.866E+00

XXG7

FX -3.002E+00FY 9.926E-01

XX

C 9

FX 9.926E-01FY -2.998E+00

XXG2

FX -4.884E+00FY 2.003E+00

XX

C 4

FX -3.577E-02FY -3.841E+00

XX

C 2FX 2.003E+00

FY -3.116E+00

Shear forces (kip)

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20

Moment diagram (kip-ft)

Problem 11.10

For the steel frames shown, estimate the axial force, shear force, and moments for all of the

members. Take 4cI 480 in for all the columns and4

bI 600 in for all the beams. Use the

Stiffness method.(a)

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22

Problem 11.11

Estimate the column axial forces in the bottom story for the distribution of column areas shown.

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23

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24

Problem 11.12

Estimate the column shears for cases (a) and (b). Compare your results with computer basedsolutions.

(a)

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25

(b)

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28

Problem 11.13

Consider the rigid steel frame with bracing shown below. Estimate the column shears and brace

forces. Take6 4 6 4 2

c g bI 40(10) mm I 120(10) mm A 650 mm E 200 GPa . Compare your

results with a computer based solution.

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29

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30

Reactions and brace forces

Column shear

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1

Problem 12.1For the rigid frame shown below, use the direct stiffness method to find the joint displacements,

and reactions for the following conditions.

6 4

2

6 o

I 240(10) mm

A 3800 mm

E 200 Gpa

12(10) / F

Member m n n α

(1) 1 2 090

(2) 2 3 0

Member m α cos sin sin cos 2cos 2sin

(1) 090 0 1 0 0 1

(2) 0 1 0 0 1 0

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2

2 2

3 3 2

2 2

3 3 2m AA

2 2

2 2

3 3

m AB

AE 12EI AE 12EI 6EI( cos sin ) ( )sin cos sin

L L L L L

AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cos

L L L L L

6EI 6EI 4EIsin cosL L L

AE 12EI AE 12EI( cos sin ) ( ) sin cos

L L L L

=

k

k

2

2 2

3 3 2

2 2

6EIsin

L

AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cos

L L L L L

6EI 6EI 2EIsin cos

L L L

2 2

3 3 2

2 2

m BA 3 3 2

2 2

2 2

3 3

m BB

AE 12EI AE 12EI 6EI( cos sin ) ( ) sin cos sin

L L L L L


L L L L L

6EI 6EI 2EIsin cos

L L L

AE 12EI AE 12EI( cos sin ) ( )sin co

L L L L

=

k

k

2

2 2

3 3 2

2 2

6EIs sin

L


L L L L L

6EI 6EI 4EIsin cos

L L L

0 0

0

(1) AA (1) AB (1) AA (1) AB

(1) BA (1) BB (2) AA (2) AB (1) BA (1) BB (2) AA (2) AB

(2) BA (2) BB (2) BA (2) BB

k k 0 k k 00

K = k k 0 k k = k (k +k ) k

0 0 0 0 k k 0 k k

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3

1

11 E 12 I

E 21 22 I

( ) ( - - )

'' ''

E ' ' ' ' '' 'I11 12

___

'' ' ' ' '' ''' ''' '' ''

21 22E I

'P U P|U P U P

= ___ | ___ __ + __ P U U P

|P U P

K K K K

K K

K K

' '11 12

' '21 22

( )

0

(1) BB (2) AA (1) BA (2) AB

(1) AA(1) AB

(2) BA (2) BB

k + k k k

k k

k 0 k

K K

K K

1

1

2

1' ' '

2

3

2

3

3

u

uθ

uθ

θ

U U

1

x 1

y

1' '

x 3

y 3

3

R

R

M

R

R

M

EP

(a) The loading shown

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4

' ' ' '' '

E I I

0

020 kN 0

0

40 kN 45 kN 000 22.5 kN-m

45 kN

22.5 kN-m

P P P U

Using Mathcad:

2

1

2 11 E I

2

u 0.07056

( ) ( - ) -0.148

0.0000193θ

' ' ' 'U P PK

1

x 1

y

1

E 21 I

x 3

y 3

3

R -2.12

R 37.46

M 2.87

-17.87R

2.53R -4.11

M

'' ' ' ''P U PK

(b)

Member 2 experiences to a uniform temperature change throughout its span. The

temperature varies linearly through the depth, from 010 C at the top to 050 C at the

bottom.

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5

um u

62 6

F 6

3

F

62 6

F m 6

3

F m

T T (10 50)T 30 T T T 10 (50) 40

2 2ΔT (200)(240)(10) 40

M EIα (12)(10) ( ) 57.6 kN-mh (10) 0.4

ΔTM EIα 57.6 kN-m

h

(200)(240)(10)F EI α T (12)(10) (30) 17.28 kN

(10)

F EI α T 17.28 kN

l l

' ' ' '' '

E I I

0

017.28 kN

00 0 0

17.28 kN57.6 kN-m

0

57.6 kN-m

P P P U

Using Mathcad:

2

1

2 11 I

2

u -0.0556

( ) ( - ) -0.0556

0.00048θ

' ' 'U PK

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6

1

x 1

y

1

E 21 I

x 3

y 3

3

R -14.1

R 14.1

M 13.5

14.1R

-14.1R 71.1

M

'' ' ' ''P U PK

(c) Support 1 settles 12 mm.

' ' ' '' '

E I I

0

12 mm

00 0 0

0

0

0

P P P U

Using Mathcad:

2

1

2 11 12

2

u -0.342

( ) ( - ) -11.41

0.00294θ

' ' ' ''U UK K

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7

1

x 1

y

1

E 21 22

x 3

y 3

3

R -86.72

R -149.40

M 83.07

86.72R

149.40R -271.11

M

'' ' ' ' ''P U UK K

Problem 12.2

For the rigid frame shown below, use the direct stiffness method to find the joint displacements,

and reactions.4

2

2

I 400 in

A 10 inE 29,000 k/in

Member m n n α

(1) 1 2 - 0116.56

(2) 2 3 0

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8

2 2

3 3 2

2 2

3 3 2m AA

2 2

2 2

3 3

m AB


L L L L L


L L L L L



L L L L

=

k

k

2

2 2

3 3 2

2 2

6EIsin

L


L L L L L

6EI 6EI 2EIsin cos

L L L

2 2

3 3 2

2 2

m BA 3 3 2

2 2

2 2

3 3

m BB


L L L L L


L L L L L

6EI 6EI 2EIsin cos

L L L


L L L L

=

k

k

2

2 2

3 3 2

2 2

6EIs sin

L


L L L L L

6EI 6EI 4EIsin cos

L L L

B A

B A B A

0 0

0

(1) AB (1) AA (1) A (1) A

(1) B (1) B (2) AA (2) AB (1) B (1) B (2) AA (2) AB

(2) BA (2) BB (2) BA (2) BB

k k 0 k k 00

K = k k 0 k k = k (k + k ) k

0 0 0 0 k k 0 k k

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9

A B

BA

' '11 12

' '21 22

0

(1) B (2) AA (1) B (2) AB

(1) A(1) A

(2) BA (2) BB

(k + k ) k k

k k

k 0 k

K K

K K

1

11 E 12 I

E 21 22 I

( ) ( - - )

'' ''

E ' ' ' ' '' 'I11 12

___

'' ' ' ' '' ''' ''' '' ''21 22E I

'P U P|U P U P

= ___ | ___ __ + __ P U U P

|P U P

K K K K

K K K K

1

1

2

1' ' '

2

3

2

3

3

u

uθ

uθ

θ

U U

1

x 1

y

1' '

x 3

y 3

3

R

R

M

R

R

M

EP

' ' ' '' 'E I I

6 kip

8 kip 0 0 0

0

P P P U

Using Mathcad:

2

1

2 11 E

2

u 0.00386

( ) ( ) -0.0062

0.000008θ

' ' 'U PK

1

x 1

y

1

E 21

x 3

y 3

3

R 3.33

R 7.54

M 2.12

-9.33R

0.46R -2.36

M

'' ' 'P UK

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10

Problem 12.3

For the rigid frames shown below, use the direct stiffness method to find the joint displacements,and reactions.

Member m n n α

(1) 1 2 0

(2) 2 3 - 053.13

2 2

3 3 2

2 2

3 3 2m AA

2 2

2 2

3 3

m AB

AE 12EI AE 12EI 6EI( cos sin ) ( )sin cos sinL L L L L


L L L L L

6EI 6EI 4EIsin cos

L L L


L L L L

=

k

k

2

2 2

3 3 2

2 2

6EIsin

L

AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cosL L L L L

6EI 6EI 2EIsin cos

L L L

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11

2 2

3 3 2

2 2

m BA 3 3 2

2 2

2 2

3 3

m BB


L L L L L


L L L L L



L L L L

=

k

k

2

2 2

3 3 2

2 2

6EIs sin

L


L L L L L

6EI 6EI 4EIsin cos

L L L

0 0

0

(1) AA (1) AB (1) AA (1) AB

(1) BA (1) BB (2) AA (2) AB (1) BA (1) BB (2) AA (2) AB

(2) BA (2) BB (2) BA (2) BB

k k 0 k k 00

K = k k 0 k k = k (k +k ) k

0 0 0 0 k k 0 k k

1

11 E 12 I

E 21 22 I

( ) ( - - )

' ' ''E ' ' ' ' '' 'I11 12

___

'' ' ' ' '' ''' ''' '' ''21 22E I

'P U P|U P U P

= ___ | ___ __ + __ P U U P

|P U P

K K K K

K K K K

(a)

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12

1

x 1

2

y

2' ' ' ' ' ' '' '

1 E I I

2

x 33

y 3

R

u 20 kN

R 00 M 0 0

θ 0R

θ 0R

EP P P PU U

Using Mathcad:

Expand the matrix K

(1) AA (1) AB

(1) BA (1) BB (2) AA (2) AB

(2) BA (2) BB

k k 0

K = k (k + k ) k

0 k k

2

2 1

11 E

2

3

u 0.076

0.056 ( ) ( )

θ 0.0000015

-0.000029θ

' ' 'U PK

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13

1

x 1

y

1 E 21

x 3

y 3

R -19.78

R -0.16

M -0.41

-0.22R

0.16R

'' ' 'P UK

(b)

1

x 1

y

2

1' ' ' ' ' ' '' '

2 E I I

x 32

y 3

3

R 0

R 75kNu 0

M 62.5 kN0 0 75 kN

0R θ -62.5 kN

0R

0M

EP P P PU U

' '11 12

' '21 22

0

(1) BB (2) AA (1) BA (2) AB

(1) AA(1) AB

(2) BA (2) BB

(k + k ) k k

k k

k 0 k

K K

K K

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14

Using Mathcad:

2

1

2 11 I

2

u -0.22

( ) ( - ) -0.58

0.0012θ

' ' 'U PK

1

x 1

y

1

E 21 I

x 3

y 3

3

R 57.56

R 86.09

M 82.46 57.56

-57.56R

63.91R 11.49

M

'' ' ' ''P U PK

Problem 12.4

For the rigid frame shown, use partitioning to determine '11K ,'21

K 'EP , and '

IP for the loadings

shown. 6 4I 40(10) mm , A=3000 2mm , and E=200 GPa.

Member m n n α

(1) 1 4 0

(2) 2 4 0180

(3) 3 4 090

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15

2 2

3 3 2

2 23 3 2m AA

2 2

2 2

3 3

m AB


L L L L L

AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cosL L L L L

6EI 6EI 4EIsin cos

L L L


L L L L

=

k

k

2

2 2

3 3 2

2 2

6EIsin

L


L L L L L


2 2

3 3 2

2 2

m BA 3 3 2

2 2

2 2

3 3

m BB


L L L L L


L L L L L

6EI 6EI 2EIsin cos

L L L


L L L L

=

k

k

2

2 2

3 3 2

2 2

6EIs sin

L


L L L L L

6EI 6EI 4EIsin cos

L L L

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16

3 3

3 3

3 3

0 0 0 0 0 0 0

0 0

0 0 0 0

0

(1) AA (1) AB

(2) AA (2) AB

( ) AA ( ) AB

(1) BA (1) BB (2) BA (2) BB ( ) BA ( ) BB

(1) AA (1) AB

(2) AA (2) AB

( ) AA (

k 0 k 0 0

0 k 0 k 0 00 0 0 0K =

0 0 k k 0 0 0 0 0 0 0 0

k 0 k k 0 k k k

k 0 k

0 k 0 k

0 0 k k

2 3 3

( )

) AB

(1) BA ( ) BA ( ) BA (1) BB (2) BB ( ) BB

k k k k k k

3

2 3

3

3

'11

'12

'21

'22

( )

0

(1) BB (2) BB ( ) BB

(1) BA ( ) BA ( ) BA

(1) AB

(2) AB

( ) A B

(1) AA

(2) AA

( ) A A

k k k

k k k

k

k

k

k 0

0 k 0

0 0 k

K

K

K

K

' '

E I

0 0

0 84 kN

16 kN-m 84kN-m

P P

Problem 12.5

For the truss shown, use the direct stiffness method to find the joint displacements, reactions, and

member forces.2A 1200 mm

E 200 GPa

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17

Member m n

n

α L (m)(kN/mm)

AE

L

(1) 1 2 051.34 6.4 37.5

(2) 1 3 0 8 30

(3) 3 2 0128.66 6.4 37.5

3 3

3 3

3 3

3

0 0 0

0 0 0 0

0

( )

( )

(

(1) AA (1) AB (2) AA (2) AB

(1) BA (1) BB ( ) AA ( ) AB

(2) BA (2) BB( ) BA ( ) BB

(1) AA (2) AA (1) AB (2) AB

(1) BA (1) BB ( ) AA ( ) AB

(2) BA ( ) BA (2) B

k k 0 k k 0

K = k k 0 k k

k k 0 0 0 0 k k

k k k k

= k k k k

k k k 3

)

B ( ) BBk

2

( ) ( ) (m)

(m)AA (m)BB ( ) ( ) 2

(m) (m) ( )

2

( ) ( ) (m)

(m)AB (m)BA ( ) ( ) 2

(m) (m) ( )

cos α sinα cosαAE( )

sinα cosα sin αL


sinα cosα sin αL

m m

m m

m

m m

m m

m

k k k

k k k

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18

(m) (m)

F

B ( ) (m) (m) B-+ nn

AEF ( ) cos sin ( ) F

L m

U U

1

11 E 12 I

E 21 22 I

( ) ( - - )

' ' ''E ' ' ' ' '' 'I11 12

___

'' ' ' ' '' ''' ''' '' ''21 22E I

'P U P|U P U P

= ___ | ___ __ + __ P U U P

|P U P

K K K K

K K K K

2' ' ' '

E I I

2

u 30 kN0 0 0

50 kN

' ''P P PU U

Using Mathcad:

1

2 1

11 E

2

x 1

y

E 21

x 3

y 3

u 0.04 ( )

-0.043

R 5

R 6.25

-35R

43.75R

' ' '

'' ' '

U P

P U

K

K

(1)

(2)

(3)

B (1) (1) (1)

B (2) (2) (2)

B (3) (3) (3)

-+

-+

-+

nn

nn

nn

AEF ( ) cos sin ( ) 8

L


L


L

U U

U U

U U

Problem 12.6

For the truss shown, use the direct stiffness method to find the joint displacements, reactions, and

member forces for

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19

2A 2 in

E 29,000 ksi

Member m n n α L (ft) (kip/in.) AE

L

(1) 1 2 053.13 20 241.67

(2) 4 2 0 16 302.1

(3) 3 2 0126.87 22.63 213.58

3 3

3 3

0 0 0 0 0 0 0

0 0

0

0 0 00 0 0

0

( )

(1) AA (1) AB

(2) AB (2) AA(1) BA (1) BB

( ) AB ( ) AA

(2) BB (2) BA ( ) BB ( ) BA

(1) AA (1) AB

(1) BA (1) BB (2) AB (2

k k 0 0 0

0 k 0 k 0 0k k 0 0K =

0 k k 0 0 0 00 0 0 0

k 0 k k k 0

k k 0

k k k 0 k

3 3

3 3

0

0 ( )

) AA

( ) AB ( ) AA

(2) BB ( ) BB ( ) BA (2) BA

0 k k

k k k k

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20

2

( ) ( ) (m)

(m)AA (m)BB ( ) ( ) 2

(m) (m) ( )

2

( ) ( ) (m)

(m)AB (m)BA ( ) ( ) 2

(m) (m) ( )


sinα cosα sin αL

cos α sinα cosαAE

( ) sinα cosα sin αL

m m

m m

m

m m

m m

m

k k k

k k k

(m) (m)

F

B ( ) (m) (m) B-+ nn


L m

U U

1

11 E 12 I

E 21 22 I

( ) ( - - )

'' ''

E ' ' ' ' '' 'I11 12

___

'' ' ' ' '' ''' ''' '' ''21 22E I

'P U P|U P U P

= ___ | ___ __ + __ P U U P

|P U P

K K K K

K K K K

(a) The loading shown

2' ' ' '

E I I

2

u 6 kip0 0 0

10 kip

' ''P P PU U

1

2 1

11 E

2

x 1

y

x 3

E 21y 3

x 4

y 4

u -0.03 ( )

-0.017

R

4.62R 6.16

R 1.37

R -1.37

0R 5.2

R

' ' '

'' ' '

U P

P U

K

K

(1)

(2)

(3)

B (1) (1) (1)

B (2) (2) (2)

B (3) (3) (3)

-+

-+

-+

nn

nn

nn

AEF ( ) cos sin ( ) -7.7

L

AEF ( ) cos sin ( ) -5.2

L

AEF ( ) cos sin ( ) 1.94

L

U U

U U

U U

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21

(b) A support settlement of 0.5 inch at joint 4

2' ' ' '

E I I

2

0

0

u 00 0 0

0

0

.5 in

' ''P P PU U

1

2 1

11 12

2

x 1

y

x 3

E 21 22y 3

x 4

y 4

u 0.0127 ( ) ( - )

-0.268

R

30R 40

R -30

R 30

0R -70

R

' ' ' ''

'' ' ' ' ''

U U

P U U

K K

K K

(1)

(2)

(3)

B (1) (1) (1)

B (2) (2) (2)

B (3) (3) (3)

-+

-+

-+

nn

nn

nn

AEF ( ) cos sin ( ) -50

L


L

AEF ( ) cos sin ( ) -42.4

L

U U

U U

U U

Problem 12.7For the truss shown, use the direct stiffness method to find the joint displacements, reactions, and

member forces due to (a) the loading shown, (b) a temperature decrease of 0C10 for all members,

(c) a support settlement of δ = 12 mm downward at node 4.2

-6 0

A 1200 mm

E 200 GPa

12x10 /C

h 3 m

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22

Member m n n α L (m)

(kN/mm)

AE

L (1) 2 1 045 4.24 56.6

(2) 3 1 0 3 80

(3) 4 1 0135 4.24 56.6

A 3 3

3 3

A 3 3

0 0 0 0

0 0 0 0

0 0 0

0 00 0 0 0 0 0

( )

(1) A (1) AB (2) AA (2) AB ( ) AA ( ) AB

(1) BA (1) BB

(2) BA (2) BB

( ) BA ( ) BB

(1) A (2) AA ( ) AA (1) AB (2) AB ( ) A

k k 0 k k k 0 k

k k 0 0 0 0 0 0K =

k 0 k 0 00 0 0 0

k k 0 0

k k k k k k

3 3

0

0 0

0 0

B

(1) BA (1) BB

(2) BA (2) BB

( ) BA ( ) BB

k k 0

k k

k k

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23

1

11 E 12 I

E 21 22 I

( ) ( - - )

'' ''

E ' ' ' ' '' 'I11 12

___

'' ' ' ' '' ''' ''' '' ''21 22E I

'P U P|U P U P

= ___ | ___ __ + __ P U U P

|P U P

K K K K

K K K K

A 3

3

3

3

( )

0

0 0

0 0

11 (1) A (2) AA ( ) AA

'

12 (1) AB (2) AB ( ) AB

(1) BA

'

21 (2) BA

( ) BA

(1) BB

'

22 (2) BB

( ) BB

' k k k

k k k

k

k

k

k 0

k

k

K

K

K

K

2

( ) ( ) (m)

(m)AA (m)BB ( ) ( ) 2

(m) (m) ( )

2

( ) ( ) (m)

(m)AB (m)BA ( ) ( ) 2

(m) (m) ( )


sinα cosα sin αL


sinα cosα sin αL

m m

m m

m

m m

m m

m

k k k

k k k

(m) (m)

F

B ( ) (m) (m) B-+ nn


L m

U U

(a)

1' ' ' '

E I I

1

u 27 kN0 0 0

45 kN

' ''P P PU U

Using Mathcad:

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1 1

11 E

1

x 2

y 2

x 3

E 21y 3

x 4

y 4

u 0.477 ( ) ( )

0.33

R

-22.82R -22.82

R 0

R -26.36

-4.18R 4.18

R

' ' '

'' ' '

U P

P U

K

K

(1)

(2)

(3)

B (1) (1) (1)

B (2) (2) (2)

B (3) (3) (3)

-+

-+

-+

nn

nn

nn

AE

F ( ) cos sin ( ) 32.27L

AEF ( ) cos sin ( ) 26.36

L

AEF ( ) cos sin ( ) -5.91

L

U U

U U

U U

(b)

F 6F EA α ΔT (200)(1200)(12)(10) (10) 28.8 kNl

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25

1' ' ' '

E I I

1

20.36 kN20.36 kN

u 0 00 0

49.18 kN 28.8 kN

20.36 kN

20.36 kN

' ''P P PU U

Using Mathcad:

1 1

11 E I

1

x 2

y 2

x 3

E 21 Iy 3

x 4

y 4

u 0 ( ) ( - )

-0.51

R

-5.96R

-5.96R

0

R 11.93

5.96R -5.96

R

' ' ' '

'' ' ' ''

U P P

P U P

K

K

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26

(1)

(2)

(3)

B (1) (1) (1)

B (2) (2) (2)

B (3) (3) (3)

-+

-+

-+

nn

nn

nn

AEF ( ) cos sin ( ) 8.43

L

AEF ( ) cos sin ( ) -11.93

L

AEF ( ) cos sin ( ) 8.43L

U U

U U

U U

(c)

1' ' ' '

E I I

1

0

0

u 00 0 0

0

0

12 mm

' ''P P PU U

Using Mathcad:

1 1

11 12

1

x 2

y 2

x 3

E 21 22y 3

x 4

y 4

u 6.0 ( ) ( - )

-2.485

R

-99.4R -99.4

R 0

R 198.8

99.41R -99.41

R

' ' ' ''

'' ' ' ' ''

U U

P U U

K K

K K

(1)

(2)

(3)

B (1) (1) (1)

B (2) (2) (2)

B (3) (3) (3)

-+

-+

-+

nn

nn

nn

AEF ( ) cos sin ( ) 140.59L

AEF ( ) cos sin ( ) -198.82

L

AEF ( ) cos sin ( ) 140.59

L

U U

U U

U U

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27

Problem 12.8

For the truss shown, determine '11K and'21

K .2A 2000 mm

E 200 GPa

Member m n n α L (m) (kN/mm) AE

L

(1) 1 2 0 6 66.67

(2) 4 2 090 5 80

(3) 3 2 0135 7.07 56.58

2

( ) ( ) (m)

(m)AA (m)BB ( ) ( ) 2

(m) (m) ( )

2

( ) ( ) (m)

(m)AB (m)BA ( ) ( ) 2

(m) (m) ( )


sinα cosα sin αL


sinα cosα sin αL

m m

m m

m

m m

m m

m

k k k

k k k

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28

3 3

3 3

3

0 0 0 0 0 0 0

0

0

00 0 0 0 0 0 0

0

(

(1) AA (1) AB

(2) AA (2) AB ( ) AA ( ) AB(1) BA (1) BB

( ) BA ( ) BB

(2) BA (2) BB

(1) AA (1) AB

(1) BA (1) BB (2) AA ( )

k k 0 0 0

0 k 0 k 0 k k k k 0 0K =

0 0 0 0 0 k k 0 0 0 0

k 0 k 0

k k 0

k k k k 3

3 3

)

0

0

AA ( ) AB (2) AB

( ) BA ( ) BB

(2) BA (2) BB

k k

0 k k

k 0 k

3 3

3 3

0

( )

0

0

(1) AA (1) AB

(1) BA (1) BB (2) AA ( ) AA ( ) AB (2) AB

( ) BA ( ) BB

(2) BA (2) BB

k k 0

k k k k k k

0 k k

k 0 k

3

3

( )

11 (1) BB (2) AA ( ) AA

(1) AB'

21 ( ) BA

(2) BA

'k k k

k

k

k

K

K

Problem 12.9

For the truss shown, determine '11K and'22K .

2A 3 in

E 29,000 ksi

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29

Member m n n α L (ft) (kip/in.) AE L

(1) 1 2 037.57 32.8 221

(2) 2 3 - 0116.56 22.36 324.2

(3) 1 3 0 16 453

3 3

3 3

3 3

3

0 0 0

0 0

00

( )

( )

(

(1) AA (1) AB ( ) AA ( ) AB

(1) BA (1) BB (2) AA (2) AB

( ) BA ( ) BB(2) BA (2) BB

(1) AA ( ) AA (1) AB ( ) AB

(1) BA (1) BB (2) AA (2) AB

( ) BA (2) BA (2) B

k k 0 k 0 k

K = k k 0 0 k k 0

k k 0 0 0 k k

k k k k

k k k k

k k k 3

)

B ( ) BBk

Expand the K matrix.

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30

' ' '

' ' '

' ' '

11 12 13

21 ( 22 k) 23

31 32 33

k k k

k k k

k k k

11

'K

Problem 12.10

For the beam shown, use the direct stiffness method to find the joint displacements, reactions,

and member forces for the loading shown.

4I 300 in

E 29,000 ksi

3 2 3 2

AA AB

2 2

3 2 3 2

BA BB

2 2

12EI 6EI 12EI 6EI-

L L L L

6EI 4EI 6EI 2EI-

L L L L

12EI 6EI 12EI 6EI- - -

L L L L

6EI 2EI 6EI 4EI-

L L L L

k k

k k

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31

F

A AB B AA A A

F

B BB B BA A B

= +

= +

P k U k U P

P k U k U P

(1) AA (1) AB (1) AA (1) AB

(1) BB (2) AA (2) AB (1) BB (2) AA (2) AB(1) BA (1) BA

(2) BA (2) BB (2) BA (2) BB

k k 0 k k 00 0 0

K k k 0 0 k k k (k + k ) k

0 k k 0 0 0 0 k k

1

11 E 12 I

E 21 22 I

( ) ( - - )

'' ''

E ' ' ' ' '' 'I11 12

___

'' ' ' ' '' ''' ''' '' ''21 22E I

'P U P|U P U P

= ___ | ___ __ + __ P U U P

|P U P

K K K K

K K K K

y 1

1

' ' ' ' ' '' '

2 E E y 2 I I

y 3

3

'

R 34.6 kip

M 260 kip-ft

θ 0 0 R -185 kip-ft 39.6 kip

5 kipR

75 kip-ftM

P P P PU U

1

11 I

E 21 I

( ) ( - )

' ' '

'' ' ' ''

U P

P U P

K

K

Problem 12.11

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32

For the beam shown, use the direct stiffness method to find the joint displacements, reactions,

and member forces for

a) The loading shown

b) A support settlement of 12 mm at joint 2

6 4I 120(10) mm

E 200 GPa

3 2 3 2

AA AB

2 2

3 2 3 2

BA BB

2 2

12EI 6EI 12EI 6EI-

L L L L

6EI 4EI 6EI 2EI-

L L L L

12EI 6EI 12EI 6EI- - -

L L L L6EI 2EI 6EI 4EI

-L L L L

k k

k k

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33

F

A AB B AA A A

F

B BB B BA A B

= +

= +

P k U k U P

P k U k U P

3 3

3 3

0 0 0 0

0 0

( )

(1) AA (1) AB

(2) AA (2) AB(1) BB(1) BA

( ) AA ( ) AB(2) BA (2) BB

( ) BA ( ) BB

(1) AA (1) AB

(1) BB (2) AA (2) AB(1) BA

(

k k 0 0 0 0 0 0

0 k k 0 0 0k k 0 0K

0 0 k k 0 k k 00 0 0 0

0 0 k k 0 0 0 00 0 0 0

k k 0 0

k k k k 0

0 k 3 3

3 3

( )

2) BA (2) BB ( ) AA ( ) AB

( ) BA ( ) BB

k k k

0 0 k k

1

1

2

2

3

3

θ

=θ

θ

U 2

3

' θ=θ

U 'E 0P

Problem 12.12

Given the system matrix listed below. Determine '11K ,'21K ,

''U , ''I'I ,P P ,

'

EP and ''

EP for the

loading and displacement constraint shown. L= 30 ft, 4I 300 in , and E=29,000 ksi.

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34

3 2 3 2

2 2

3 2 3 3 2

2 2

3 2 3 2

2 2

12EI 6EI 12EI 6EI0 0

L L L L6EI 6EI

- 0L L

12EI 6EI 24EI 12EI 6EI- - 0 -

L L L L L

6EI 6EI0 -L L

12EI 6EI 12EI 6EI0 0 - - -

L L L L

6EI 6E

4EI 2EI0

L L

2EI 8EI 2EI

L L L

2EI 4EI0 I0L L

-L L

K

1

1

2

2

3

3

θ

=θ

θ

U

'' ''

EI11 12

___

' ''' '' ''21 22E I

'P U P|

= ___ | ___ __ + __

|P U P

K K

K K

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35

4EI 2EI0

L L

2EI 8EI 2EI

L L L

2EI 4EI0L L

11

'K

2 2

2 2

2 2

6EI 6EI0

L L

6EI 6EI- 0

L L

6EI 6EI0 - -L L

'

21K

1

2

3

θ 0

θ 0

0.25 inθ

' ''U U

0

10 kip-ft

0

'

EP

I

51.84 kip-ft

0

-51.84 kip-ft

'P I

5.184 kip

13.632 kip

5.184 kip

''P

Problem 12.13

Investigate the effect of varying the spring stiffness on the behavior (moment and deflected

profile) of the structure shown below. Consider a range of values of k v.

6 4

18 kN/mmm

I 120(10) mm , E 200 GPa and k 36 kN/mm

90 kN/mm

v

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37

(1) AA (1) AB (1) AA (1) AB


(2) BA (2) BB (2) BA (2) BB

k k 0 k k 00 0 0

K k k 0 0 k k k (k + k ) k

0 k k 0 0 0 0 k k

2

2

θ0

' ''U U

'E = 14 kN-mP

Problem 12.14 Determine the bending moment and deflection profiles for the following structures.

Take 4I 300 in , 2CableA 3 in , and2E 29,000 k/in .

2

(2) (2) (2)Cable

(2)AA (2)BB (2) 2 2(2) (2) (2)Cable

2

(2) (2) 2)Cable(2)AB (2)BA (2) 2 2

(2) (2) (2)Cable

cos α sinα cosαA E( )

sinα cosα sin αL


sinα cosα sin αL

k k k

k k k

3 2 3 2

AA AB

2 2

3 2 3 2

BA BB

2 2

12EI 6EI 12EI 6EI-

L L L L

6EI 4EI 6EI 2EI-

L L L L

12EI 6EI 12EI 6EI- - -

L L L L

6EI 2EI 6EI 4EI-

L L L L

k k

k k

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38

F

A AB B AA A A

F

B BB B BA A B

= +

= +

P k U k U P

P k U k U P

1

11 E 12 I

E 21 22 I

( ) ( - - )

'' ''E ' ' ' ' '' 'I11 12

___

'' ' ' ' '' ''' ''' '' ''21 22E I

'P U P| U P U P= ___ | ___ __ + __

P U U P|P U P

K K K K

K K K K

Case (a)

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39

3 2 3 2

2 2

3 2 3 2

2 3 2

12EI 6EI 12EI 6EI-

L L L L

6EI 4EI 6EI 2EI-

L L L L

12EI 6EI 12EI 6EI- - -L L L L

6EI 2EI 12EI 6EI-

L L L L

K

y 1

' ' ' ''

E 1 I I

2

R 9 kip

0 M -90 kip-ft 60 kip-ft

21 kipR

''U P P P

2

6EI-

L

11

'K

Using Mathcad:

12 11 Iθ ( ) ( - ) .011 ' ' 'U PK

y 1

1 E 21 I

3

R 13.5

M 105

16.5M

'' ' ' ''P U PK

Case (b)

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40

3 2 3 2

2 2

3 2 3 2

2 3 2

12EI 6EI 12EI 6EI-

L L L L

6EI 4EI 6EI 2EI-

L L L L


6EI 2EI 12EI 6EI-

L L L L

K

y 1

' ' ' ''

2 E 1 I I

2

R 9 kip

M 21 kip 60 kip-ft

90 kip-ftM

'U P P P 0''U

Using Mathcad:

12 11 I( ) (- ) -9.38 ' ' 'U PK

y 1

1 E 21 I

2

R 30

M 375

225M

'' ' ' ''P U PK

Case (c)

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41

3 2 3 2

2 2

3 2 3 2

2 3 2

12EI 6EI 12EI 6EI-

L L L L

6EI 4EI 6EI 2EI-

L L L L


6EI 2EI 12EI 6EI-

L L L L

K

y 12 ' ' ' ''

E I I

2 1

R θ 21 kip 9 kip

-90 kip-ft 60 kip-ftM

'U P P P 0''U

Using Mathcad:

2 1

11 I

2

θ -0.27 ( ) (- )

0.01

' ' 'U PK

y 1

E 21 I

1

R 13.65

109.57M

'' ' ' ''P U PK

Case (d)

(d) Cantilever beam + Cable

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42

(1) (1)

(1) (1)

3 2 3 2

AA AB

2 2

3 2 3 2

BA BB

2 2

12EI 6EI 12EI 6EI-

L L L L

6EI 4EI 6EI 2EI-

L L L L

12EI 6EI 12EI 6EI- - -

L L L L

6EI 2EI 6EI 4EI-

L L L L

k k

k k

2

(2) (2) (2)Cable(2)AA (2)BB (2) 2 2

(2) (2) (2)Cable

2

(2) (2) 2)Cable(2)AB (2)BA (2) 2 2

(2) (2) (2)Cable


sinα cosα sin αL


sinα cosα sin αL

k k k

k k k

(1) AA (1) AB (1) AA (1) AB


(2) BA (2) BB (2) BA (2) BB

k k 0 k k 00 0 0

K k k 0 0 k k k (k + k ) k

0 k k 0 0 0 0 k k

y 1

2 ' ' ' ''

E 1 I I2

y 3

R θ 21 kip 9 kip

M-90 kip-ft 60 kip-ft

R

'U P P P 0''U

Using Mathcad:

1

11 I( ) (- )' ' 'U PK

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E 21 I '' ' ' ''P U PK

Problem 12.15Consider the guyed tower defined in the sketch. The cables have an initial tension of 220 kN.

Determine the horizontal displacements at B, C, the change in cable tension, and the bending

moment distribution in member ABC. Treat the cables as axial elements. Develop a computer based scheme to solve this problem. Take 6 4

tower I 200(10) mm 2

CableA 650 mm ,2

tower A 6000 mm , and the material to be steel.

strudl ' 2012 Problem 12.15 Guyed tower metric GTstrudl input file Connor & Faraji'type plane frameunits kN meter

Joint Coordinates

1 0.0 0.0

2 15.0 60.03 30.0 0.0

4 15.0 0.0

5 15.0 30.0type plane truss

member incidences

1 1 2

2 2 3

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3 3 5

4 1 5

type plane framemember incidences

5 4 5

6 5 2status support joints 1 3 4 joint releases

4 moment z

unit mmconstants

E 200.0

cte 12E-06

Member properties1 2 3 4 AX 650

5 6 AX 6000 IZ 200000000

units m

Loading 1

member tempreture loads

1 2 axial -1823 4 axial -187

Loading 2

joint load2 force x 45.

5 force x 90.

Loading 3

joint load2 force x 45.

5 force x 90.

member tempreture loads1 2 axial -182

3 4 axial -187

stiffness analysislist reactions

list forces

unit mm

list displacements

_____________________________________

Analysis results:

{ 54} > list reactions

--- LOADING - 1 ---

----------------------------------------------------------------------------------------------------------------------------- ------

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45

RESULTANT JOINT LOADS SUPPORTS

JOINT /---------------------FORCE---------------------//--------------------MOMENT--------------------/

X FORCE Y FORCE Z FORCE X MOMENT Y MOMENT Z MOMENT

1 GLOBAL -152.0266571 -410.8855896

3 GLOBAL 152.0266571 -410.8855896

4 GLOBAL 0.0000000 821.7711792 0.0000000

-----------------------------------------------------------------------------------------------------------------------------------

--- LOADING - 2 ---

-----------------------------------------------------------------------------------------------------------------------------------




1 GLOBAL -67.5721207 -180.0000000

3 GLOBAL -67.5721207 180.0000000

4 GLOBAL 0.1442504 0.0000000 0.0000000

----------------------------------------------------------------------------------------------------------------------------- ------

--- LOADING - 3 ---

----------------------------------------------------------------------------------------------------------------------------- ------




1 GLOBAL -219.5987854 -590.8856201

3 GLOBAL 84.4545364 -230.8856049

4 GLOBAL 0.1442504 821.7711792 0.0000000

{ 55} > list forces

ACTIVE UNITS M KN RAD DEGF SEC

----------------------------------------------------------------------------------------------------------------------------- ------

--- LOADING - 1 ---

----------------------------------------------------------------------------------------------------------------------------- ------

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MEMBER FORCES

MEMBER JOINT /-------------------- FORCE --------------------//-------------------- MOMENT --------------------/

AXIAL SHEAR Y SHEAR Z TORSIONAL BENDING Y BENDING Z

1 2 220.2403717

2 3 220.2403717

3 5 220.4998169

4 5 220.4998169

5 4 821.7711792 0.0000000 0.0000000

5 5 -821.7711792 0.0000000 0.0000000

6 5 427.3291016 0.0000000 0.0000000

6 2 -427.3291016 0.0000000 0.0000000

----------------------------------------------------------------------------------------------------------------------------- ------

--- LOADING - 2 ---

----------------------------------------------------------------------------------------------------------------------------- ------

MEMBER FORCES



1 2 92.47249602 3 -92.4724960

3 5 -100.9456024

4 5 100.9456024

5 4 0.0000000 -0.1442504 0.0000000

5 5 0.0000000 0.1442504 -4.3275118

6 5 0.0000000 0.1442504 4.3275118

6 2 0.0000000 -0.1442504 0.0000000

----------------------------------------------------------------------------------------------------------------------------- ------

--- LOADING - 3 ---

----------------------------------------------------------------------------------------------------------------------------- ------

MEMBER FORCES



1 2 312.7128906

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2 3 127.7678833

3 5 119.5542145

4 5 321.4454041

5 4 821.7711792 -0.1442504 0.0000000

5 5 -821.7711792 0.1442504 -4.3275118

6 5 427.3291016 0.1442504 4.3275118

6 2 -427.3291016 -0.1442504 0.0000000

{ 56} > unit mm

{ 57} > list displacements

ACTIVE UNITS MM KN RAD DEGF SEC

-----------------------------------------------------------------------------------------------------------------------------------

--- LOADING - 1 ---

-----------------------------------------------------------------------------------------------------------------------------------

RESULTANT JOINT DISPLACEMENTS SUPPORTS

JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/

X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.

1 GLOBAL 0.0000000 0.0000000

3 GLOBAL 0.0000000 0.0000000

4 GLOBAL 0.0000000 0.0000000 0.0000000

RESULTANT JOINT DISPLACEMENTS FREE JOINTS



2 GLOBAL 0.0000000 -31.2275105 0.0000000

5 GLOBAL 0.0000000 -20.5442810 0.0000000

----------------------------------------------------------------------------------------------------------------------------- ------

--- LOADING - 2 ---

----------------------------------------------------------------------------------------------------------------------------- ------




1 GLOBAL 0.0000000 0.0000000

3 GLOBAL 0.0000000 0.0000000

4 GLOBAL 0.0000000 0.0000000 -0.0014003

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2 GLOBAL 181.3883820 0.0000000 -0.0046460

5 GLOBAL 58.2378464 0.0000000 -0.0030231

-----------------------------------------------------------------------------------------------------------------------------------

--- LOADING - 3 ---

-----------------------------------------------------------------------------------------------------------------------------------




1 GLOBAL 0.0000000 0.0000000

3 GLOBAL 0.0000000 0.0000000

4 GLOBAL 0.0000000 0.0000000 -0.0014003




2 GLOBAL 181.3883820 -31.2275105 -0.00464605 GLOBAL 58.2378464 -20.5442810 -0.0030231

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Problem 12.16

Consider the rigid frame shown above. Investigate how the response changes when 1A is varied.

Use computer software. Vary 1A from2

2 in to 2

10 in . Take 4I 600 in , 2A 5 in , L 200 ft ,4

1I 300 in , 1 k/ftw . Material is steel.

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Analysis Results for 1A 22 in :


ACTIVE UNITS FEET KIP RAD DEGF SEC




1 GLOBAL 0.2750448 26.2271576 0.0000000

4 GLOBAL 0.0000000 26.3188362 0.0000000

5 GLOBAL -0.2750448 147.4540253 -0.0000033

{ 38} > list forces


MEMBER FORCES



1 1 0.2750448 26.2271576 0.0000000

1 2 -0.2750448 40.4394569 -473.7429199

2 2 -75.6282272 33.4004898 421.63412482 3 75.6282272 33.2659035 -417.1477966

3 3 0.0000000 40.3481750 467.6471252

3 4 0.0000000 26.3188343 0.0000000

4 5 105.8844528 -1.4583676 -16.6390419

4 2 -105.8844528 1.4583676 -52.1087914

5 5 105.5302505 -1.4242343 -16.6390438

5 3 -105.5302505 1.4242343 -50.4993286

{ 39} > unit in


ACTIVE UNITS INCH KIP RAD DEGF SEC




1 GLOBAL 0.0000000 0.0000000 -0.0604358

4 GLOBAL 0.4157400 0.0000000 0.0615096

5 GLOBAL 0.0000000 0.0000000 -0.0006194

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2 GLOBAL -0.0015175 -1.4619952 0.0132183

3 GLOBAL 0.4157400 -1.8713145 -0.0138291

____________________________________________________________________

Analysis Results for 1A 210 in :





X FORCE Y FORCE Z FORCE X MOMENT Y MOMENT Z

MOMENT

1 GLOBAL 0.2801770 26.0890694 0.0000000

4 GLOBAL 0.0000000 26.1824589 0.0000000

5 GLOBAL -0.2801770 147.7284851 -0.0000043

{ 38} > list forces


MEMBER FORCES

MEMBER JOINT /-------------------- FORCE --------------------//-------------------- MOMENT --------------------

/

AXIAL SHEAR Y SHEAR Z TORSIONAL BENDING Y BENDING

Z

1 1 0.2801770 26.0890694 0.0000000

1 2 -0.2801770 40.5775414 -482.9487305

2 2 -76.3793488 33.4019623 420.8519592

2 3 76.3793488 33.2644310 -416.2675171

3 3 0.0000000 40.4845543 476.7389832

3 4 0.0000000 26.1824589 0.0000000

4 5 106.5178833 -1.8944246 -27.2068596

4 2 -106.5178833 1.8944246 -62.0968056

5 5 106.1567535 -1.8599567 -27.2068634

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5 3 -106.1567535 1.8599567 -60.4714508

{ 39} > unit in


ACTIVE UNITS INCH KIP RAD DEGF SEC




1 GLOBAL 0.0000000 0.0000000 -0.0581310

4 GLOBAL 0.4198557 0.0000000 0.0592253

5 GLOBAL 0.0000000 0.0000000 -0.0006280




2 GLOBAL -0.0015458 -0.2953888 0.0129835

3 GLOBAL 0.4198557 -0.7126988 -0.0136053

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Problem 12.17

(a) Develop a computer code to automate the generation of the member stiffness matrices

defined by Equation (12.6). Assume A, E, I, L given.(b) Develop a computer code to carry out the operations defined by Equation (12.13).

(c) Develop a computer code to carry out the operation defined by Equation (12.20) and

(12.21).

BB 3 2

2

AE0 0

L

12EI 6EI= 0 -

L L

6EI 4EI0 -

L L

k BA 3 2

2

AE- 0 0

L

12EI 6EI0 - -

L L

6EI 2EI0

L L

k

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AA 3 2

2

AE0 0

L

12EI 6EI= 0

L L

6EI 4EI0 L L

k AB 3 2

2

AE- 0 0

L

12EI 6EI0 -

L L

6EI 2EI0 - L L

k

g

cosα -sinα 0

= sinα cosα 0

0 0 1l

R

cosα sinα 0

= -sinα cosα 0

0 0 1

g l

R

BB BA

T T

BB g BA gg g

T T

AA AA g AB AB gg g

=

=

l l l l

l l l l

k R k R k R k R

k R k R k R k R

B A

B A B A

0 0

0

(1) AB (1) AA (1) A (1) A

(1) B (1) B (2) AA (2) AB (1) B (1) B (2) AA (2) AB

(2) BA (2) BB (2) BA (2) BB

k k 0 k k 00

K = k k 0 k k = k (k + k ) k

0 0 0 0 k k 0 k k

'' ''

EI11 12

___

' ''' '' ''21 22E I

'P U P|

= ___ | ___ __ + __

|P U P

K K

K K

1

11 E 12 I

E 21 22 I

( ) ( - - )

' ' ' ' '' '

'' ' ' ' '' ''

U P U P

P U U P

K K

K K

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Problem 12.18

For the space truss shown, use the direct stiffness method to find displacements at joint1. 2A 3 in

E 29,000 ksi

X-Y Plan

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X-Z Plan

Member l x l y l z l xβ yβ zβ

(1) 12 6 18 22.45 0.53 0.27 0.8

(2) 10 6 18 21.45 0.47 0.28 0.84

(3) 2 12 18 21.72 0.092 0.55 0.83

Using Mathcad:

Lm

X2m

X1m

2

Y2m

Y1m

2

Z2m

Z1m

2

2

( ) ( ) (m)

(m)AA (m)BB ( ) ( ) 2

(m) (m) ( )

2

( ) ( ) (m)

(m)AB (m)BA ( ) ( ) 2

(m) (m) ( )


sinα cosα sin αL


sinα cosα sin αL

m m

m m

m

m m

m m

m

k k k

k k k

E

0

10 kip

20 kip

'P

1

1

1 11 E

1

u .011 in

( ) .066 in

.03 inw

' ' 'U PK

Problem 12.19For the space frame shown, use the direct stiffness method to find displacements at joint B. The

load P is applied perpendicular to the plane ABC. The cross section is circular tube. Take L=4 m,

P=30 kN, E = 200 GPa , and G=77 GPa.

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Cross section

Member m n n 1θ (xg ,x )l 2θ (xg ,x )l 3θ (xg ,x )l

(1) 1 2 π/2 -π/2 0

(2) 2 3 0 π/2 -π/2

i i

11 12 13

21 22 23

31 32 33

i i i

i i i

i

α α α

α α α

α α α

'

iR

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i

lg

ii

0

0

R'R

R'

T

global gl g k R k R

1 AA 1 AB

1 BA 1 BB 2 AA 2 AB

2 BA 2 BB

k k 0

K = k (k + k ) k

0 k k

(1) (1)T

global lg (1)BB lg

1 BBk R k R

( 2) (2)T

global lg (2)AA lg2 AAk R k R

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11 1 BB 2 AA'K = (k +k )

1joint2

2 joint2

3joint2'

1joint2

2 joint2

3joint2

u

u

u

θ

θ

θ

U E

0

0

30

0

0

0

'P

1

11 E

0

023.55

( ) ( ).003

.003

0

' ' 'U K P

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1joint 2 2 joint 2 3 joint2

1joint 2 2 joint 2 3 joint 2

23.55 mm

.003 rad

u u 0 u

θ θ θ 0

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Problem 13.1

(a)

(b) Use a software package to determine:

(i) The maximum values of3

R , 2M , and 1-1M cause by a uniformly distributed dead

load of 2 k/ft.

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(ii) The maximum value of 2M caused by a uniformly distributed live load of 1k/ft.

-

2 maxM 146.7 kip-ft

Problem 13.2

8(28) 32(14)x 9.33 ft

72e

e 14 9.33 4.66 ft 2.33 ft2

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maxM 1703 kip-ft

Moment envelop

maxM 1700 kip-ft

Problem 13.3

Lane load: 10 kN/mw

uniformTruck load:

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(a) 1 2L L 30 m , EI constant

Global moment envelope-Lane

Global moment envelope-Truck

(b) 1L 15 m 2L 30 m , EI constant


Truck

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(c) 1 2L L 30 m , EI constant

Moment diagram-Lane


(d) Compare the global moment envelopes for the structure shown below with the

envelopes generated in part (a). Is there any effect of varying I?


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Problem 13.4

Consider the multi span bridge shown below. Suppose the bridge is expected to experience a

temperature change of ∆T over its entire length. Where would you place a hinge support: at A orat B? Determine the end movement corresponding to your choice of support location.

Case (a)

u α T L

Case (b)

u α T L

Case (b) is better because there is less movement of span.

Problem 13.5

Most design codes limit the deflection due to live loading to some fraction of length, say L/α

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where α is on the order of 500. Generate the global “deflection” envelope for the multi-span

beam and truck loading shown below. Take E = 29,000ksi and I = 60,000 in4.

maxmax req

L L 90(12)2.16 in I (I)

L500 500( )

Problem 13.6

Investigate convergence of the internal forces for the parabolic arch shown as the discretizationis refined. Take the interval as 2.4 m, 1.2 m, and .6 m.

Axial Force

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Bending Moment

Axial Force

Bending Moment

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Axial Force

Bending Moment

Problem 13.7

Suppose a uniform loading is applied to span ABC. Investigate how the response changes as x

varies from L/2 to L. Take h=L/2, 2 2CA 50 in , A 2 in ,2

21k/ft, f 1

L

x

w

.

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Deformation profile (

L

, f 22 x

)

Deformation profile ( L, f 5 x )

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Problem 13.8

Determine the structural response (forces and displacements) of the idealized tied arch shown

below under a uniformly distributed gravity load of 30 kN/m.

Assume 2 6 4 6 2arch Arch hanger A 26000 mm , I 160(10) mm ,A 2(10) mm

Note: roadway girder and arch are pined to together at pints A and B.

Deformation profile

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Axial force

Bending moment

Problem 13.9

Determine the distribution of internal forces and displacements for the cable stayed structureshown below. Member AB acts as counterweight for loading applied on member BC. The 2

members are connected by 9 parallel equally spaced cables. Self weight of members AB and BC

is 16 kN/m and 8 kN/m respectively. Assume 2CableA 50000 mm and E=200GPa. Consider the

following cases:

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(a) 9 4AB BCI I 40(10) mm

Deformation profile

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Axial force

Bending moment

(b) 9 4AB BC BCI 4I I 40(10) mm

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Deformation profile

Bending moment

(c) Uniform Live load of 2 kN/m applied to member BC in addition to self weight.9 4

AB BC BCI 4I I 40(10) mm

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Deformation profile

Bending moment

Problem 13.10

Consider the symmetrical cable structure shown below. Determine a set of cable areas ( 1 10C C )

such that the maximum vertical displacement is less than 375 mm under a uniformly distributed

live load of 10 kN/m. Assume 6 4 3 2girder girder I 400(10) mm ,A 120(10) mm . Take the allowablestress as 700MPa.

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maxδ 371 mm

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Cable A,2mm Tension, kN

C1 100,000 325.5

C2 500,000 62.7

C3 50,000 167

C4 50000 132C5 50000 129

C6 50000 120

C7 50000 113

C8 50000 108

C9 50000 103

C10 50000 104

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Problem 14.1 Consider the plan view of one story rigid frames shown below. Determine the center of twist

corresponding to the brace stiffness patterns shown.(a)

(b)

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(c)

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Problem 14.2

The one story frame shown has an unsymmetrical mass distribution.

Distribution of mass

Stiffness distribution

(a) Determine the center of mass.

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(b) Determine the stiffness parameters1k and

2k such that the center of stiffness coincides

with the center of mass.

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(c) Determine the earthquake floor loads corresponding toa

S =0.3g. Consider both direction,

i.e., N-S and E-W.

Problem 14.3

For the rigid frame shown below, determine (a) The center of twist (T

C ) (b) the seismic floor

loads applied at the center of mass (M

C ) for a N-S earthquake witha

S =0.3g. Assume

properties are equal for each floor.

Elevation

Typical floor plan

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Problem 14.4 Consider the two story rigid frame defined below. Assume the weight of the floor slabs are

equal to floor w . Concentrated masses are located on each floor as indicated.

Plan-Floor 1

Plan –

Floor 2

(a) Determine the position of the center of mass for each floor.

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(b) Assume the structure is subjected to an earthquake acting in the east direction. Determine

the earthquake forces for the individual floors. Assume floor w =1000 kips,

1 2w w 1000 kips , and aS = 0.3g.

(c)

Suppose the story stiffness distribution shown below is used. Describe qualitatively howthe structure will displace when subjected to an earthquake. Consider the stiffness

distribution to be the same for each floor.

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Problem 14.5 Consider the single story multi frame structure shown below. Determine the lateral force in

the frames due to a global load P. Consider both wind and earthquake loading. Assume theslab is rigid.

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Problem 14.6 Consider the stiffness distribution for the one story rigid frame shown below. Determine the

displaced configuration under the action of the loading shown. Assume the slab is rigid.

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Problem 14.7

(a) Determine the center of twist.

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(b) Using Equation (14.47), determine oK .

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Problem 14.8 Consider the plan view of a one story frame shown below. Using the matrix formulation

presented in section 14.3.5, generate the equations of motion for the story. Take 1m 1000 ,

2m 500 , and k=10 kips/in.

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Problem 14.9

Consider the one story plan view shown below.

(a) Locate the center of mass.

(b) Locate the center of twist. Take 1 2 3 4k k k k k .

(c) Take 1 3k k 10 . Suggest values for 2k and 4k such that the center of mass

coincides with the center of twist.

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Problem 14.10 Consider the floor plan shown below. Assume the mass is uniformly distributed over the

floor area. Establish the equations of motion referred to point o.

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Problem 14.11 Consider the roof plan for a one story structure shown below. Assume the shear walls haveequal stiffness and the roof dead load is uniform.

(a) Determine the center of mass and

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