Structural Engineering II

7/18/2019 Structural Engineering II

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1

Approximate Lateral Load Analysis by Portal Method

Por ta l F r a m e

Portal frames, used in several Civil Engineering structures like buildings, factories, bridges have the primary

purpose of transferring horizontal loads applied at their tops to their foundations. Structural reuirements

usually necessitate the use of statically indeterminate layout for portal frames, and appro!imate solutions are

often used in their analyses.

"i# "ii# "iii# "iv#

P ort al Fr a m e St r u ct u r e s

$ s s u m p ti ons f or t he $ pp r o ! i m a t e So l u ti on

%n order to analyze a structure using the euations of statics only, the number of independent force

components must be eual to the number of independent euations of statics.

%f there are n more independent force components in the structure than there are independent euations of

statics, the structure is statically indeterminate to the nth

degree. &herefore to obtain an appro!imate solutionof the structure based on statics only, it 'ill be necessary to make n additional independent assumptions. $solution based on statics 'ill not be possible by making fe'er than n assumptions, 'hile more than nassumptions 'ill not in general be consistent.

&hus, the first step in the appro!imate analysis of structures is to find its degree of statical indeterminacy

"dosi# and then to make appropriate number of assumptions.

For e!ample, the dosi of portal frames sho'n in "i#, "ii#, "iii# and "iv# are 1, (, ) and 1 respectively. *ased on

the type of frame, the follo'ing assumptions can be made for portal structures 'ith a vertical axisof symmetry that are loaded horizontally at the top

1. &he horizontal support reactions are eual

). &here is a point of inflection at the center of the unsupported height of each fi!ed based column

$ssumption 1 is used if dosi is an odd number "i.e., + 1 or (# and $ssumption ) is used if dosi 1.

Some additional assumptions can be made in order to solve the structure appro!imately for different loading

and support conditions.

(. orizontal body forces not applied at the top of a column can be divided into t'o forces "i.e.,

applied at the top and bottom of the column# based on simple supports-. For hinged and fi!ed supports, the horizontal reactions for fi!ed supports can be assumed to be four

times the horizontal reactions for hinged supports



/0

./

./

0

(.((

0

)

E!a m p l e

ra' the a!ial force, shear force and bending moment diagrams of the frames loaded as sho'n belo'.

12 k* C

12 k* C

12

$ 10

F E

0

$ 10

Soluti on

"i# "ii#

"i# For this frame, dosi + ( ( 3 - ( - + 14 i.e., $ssumption 1 $ + + 125) + 0 k

6$ + 2 12 12 7 10 + 2 7 + ./ k

Fy + 2 7$ 3 7 + 2 7$ + ./ k

12 k 02 02

.

0

0 k 0 k

./ k ./ k

8eactions $F "k# SF "k# *6 "k9ft#

"ii# dosi + ( ( 3 ( - + (

$ssumption 1 $ + + 125) + 0 k, $ssumption ) *6E + *6F + 2

*6F + 2 $ 0 3 6$ + 2 6$ + )0 k9ft4 Similarly *6E + 2 6 + )0 k9ft

6$ + 2 )0 )0 3 12 12 7 10 + 2 7 + (.(( k Fy + 2 7$ 3 7 + 2 7$ + (.(( k

12 k

0 k

)0 k9ft

0

(.(((.(( 0

0 k

)0 k9ft

)0 )0

)0 )0

(.(( k (.(( k




12 0 7 10 + 2 7 + (.(( k

7 + (.(( k

)0)0

(.((0

).0

).0

/.0

"iii#* :

12 k

E

0

C

F 12

* : C12 k

E F$

$ $

6$ /.0 /.06

7$

7

7$

/.0

7

/.0

dosi + ( - 3 ( 0 1 + )4 $ssumption 1 and ) *6E + *6F + 2

*6E + 2 "bottom# $ 0 3 6$ + 2 6$ + 0$4 Similarly *6F + 2 6 + 0

$lso *6E + 2 "free body of E*CF#

Fy + 2 7$ 3 7 + 2 7$ +

*6: + 2 "bet'een E and :# 7$ /.0 $ 0 + 2 $ + 0 k 6$ + 0$ + )0 k9ft

F! + 2 "entire structure# $ 3 3 12 + 2 0 3 3 12 + 2 + 0 k 6 + 0 + )0 k9ft

12 k

0 k

)0 k9ft

0

(.(( (.(( 0

0 k)0

)0 k9ft

(.(( k (.(( k


"iv# dosi + ( 0 3 ; ( + $ssumptions needed to solve the structure

$ssumption 1 and ) $< *< C + 1< )< 1 $ + 125- + ).0 k, * + 0 k, C + ).0 k

6$ + 6C + ).0 0 + 1).0 k9ft, 6* + 0 0 + )0 k9ft

&he other - assumptions are the assumed internal hinge locations at midpoints of columns and one beam

12 k 1).0

)0

1).0

).0 k

=$

1).0 k9ft

0 k

)0 k9ft

=*

).0 k

=C

1).0 k9ft

02

1).0 )01).0

8eactions *6 "k9ft#

1./ 1./

).0 1./ 1./

SF "k# $F "k#



E F : 7$E + ( 15"# + k , 7*F + 1)

k , 7C: + 1)

k , 7 +

k

*ending moments are$ * C 6%6 + ( 125) + 10

k, 6>? + (2

k, 6@A + (2

k, 6BP + 10

k

10 12 10 6E% + 0 125) + )0k, 6F> + 02

k, 6:@ + 02

k, 6B + )0

k

6$E + 125) + (2k, 6F> + 2

k, 6:@ + 2

k, 6B + (2

k

(

12 12 0

1) 1)

910 9; 9(

-2 -2 -2

1 11

9

) 1 91 9)

/.(( (./ 9(./ 9/.((

10.0 /./0 9/./0 910.0

) D 1 2 + ) 2

1 )

Analysis of Multi-storied Structures by Portal Method

$ppro!imate methods of analyzing multi9storied structures are important because such structures are

statically highly indeterminate. &he number of assumptions that must be made to permit an analysis by

statics alone is eual to the degree of statical indeterminacy of the structure.

$ s s u m p ti ons

&he assumptions used in the appro!imate analysis of portal frames can be e!tended for the lateral load

analysis of multi9storied structures. &he Portal Method thus formulated is based on three assumptions

1. &he shear force in an interior column is t'ice the shear force in an e!terior column.). &here is a point of inflection at the center of each column.

(. &here is a point of inflection at the center of each beam.

$ssumption 1 is based on assuming the interior columns to be formed by columns of t'o adacent bays or

portals. $ssumption ) and ( are based on observing the deflected shape of the structure.

E!a m p l e

se the Portal 6ethod to dra' the a!ial force, shear force and bending moment diagrams of the three9storied

frame structure loaded as sho'n belo'.

1k

6 ? A P

1)k

% > @ B

Column shear forces are at the ratio of 1<)<)<1.Shear force in "7# columns %6, >?, @A, BP are

G1 15"1 3 ) 3 ) 3 1# +H (k , G1 )5"1 3 ) 3 ) 3 1# +H

k ,

k , (

krespectively. Similarly,

k k k k 7E% + (2 15"# + 0 , 7F> + 12 , 7:@ + 12 , 7B + 0 4 and

k

&he rest of the calculations follo' from the free9body diagrams

( 10 (2

0 )0 02

( /)

(2 10

02 )0

/) (

9129

9)

90 9( 91

Column SF "k# Column *6 "k9ft# *eam $F "k#

10 10 10

9) 9( 9)

90.(( 90.((

9.1( 91).) 9.1(

*eam *6 "k9ft# *eam SF "k# Column $F "k#

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(

0

- D 1 2 + - 2

F

Problems on Lateral Load Analysis by Portal Method

1. &he figure belo' sho's the shear forces "kips# in the interior columns of a t'o9storied frame. se the

Portal 6ethod to calculate the corresponding"i# applied loads P1 and P), "ii# column bending moments, "iii# beam a!ial forces.

P)

12

P1

1)

)2 )2

). &he figure belo' sho's the applied loads "F1, F)# and shear force "7EF# in column EF of a t'o9storied

frame. %f F) + 12 k, and 7EF + 0 k, use the Portal 6ethod to calculate the"i# applied load F1, "ii# ma!imum column bending moments.

F1$

12* EF)

7EF 1)

C F

)2 10 10

(. For the structure sho'n in Iuestion ), use the Portal 6ethod to calculate the lateral loads F1, F) if the

a!ial forces in beams $ and *E are 12 kips and 10 kips respectively.

-. For the structure sho'n belo', use the Portal 6ethod to

"i# dra' the bending moment diagrams of the top floor beams $* and *C

"i# calculate the applied load F1 if the ma!imum bending moment in column E is (2 k9ft.

$ * CF) + 12 k

12 E F

1

1-

: %

10 10

0. &he figure belo' sho's the e!terior column shear forces "kips# in a four9storied fame.Calculate "i# the applied loads, "ii# beam shear forces.

0

12

10

)2

)2 12



-).- (0.( -).-

0(. -.-.

9.1 912./ 9.1

91-.) 9.; 9(.2-

) D 1 2 + ) 2

1 )

Analysis of Multi-storied Structures by Cantilever Method

$lthough the results using the Portal Method are reasonable in most cases, the method suffers due to the

lack of consideration given to the variation of structural response due to the difference bet'een sectional

properties of various members. &he Cantilever Method attempts to rectify this limitation by considering the

cross9sectional areas of columns in distributing the a!ial forces in various columns of a story.

$ s s u m p ti ons&he Cantilever Method is based on three assumptions

1. &he a!ial force in each column of a storey is proportional to its horizontal distance from the

centroidal a!is of all the columns of the storey.

). &here is a point of inflection at the center of each column.

(. &here is a point of inflection at the center of each beam.

$ssumption 1 is based on assuming that the a!ial stresses can be obtained by a method analogous to that

used for determining the distribution of normal stresses on a transverse section of a cantilever beam.

$ssumption ) and ( are based on observing the deflected shape of the structure.

E!a m p l e

se the Cantilever 6ethod to dra' the a!ial force, shear force and bending moment diagrams of the three 9storied frame structure loaded as sho'n belo'.

1k

6 ? A P

1)k

% > @ B

&he dotted line is the column centerline "at all floors#

Column a!ial forces are at the ratio of )2< 0< 0< )2.

$!ial force in "P# columns %6, >?, @A, BP are

G1 0 )25J)2)

3 0)

3 " 0#)

3 " )2#)K + H ).1)

k , G1 0

k 05")2 3 0 3 " 0# 3 " )2# K + H 2.0( , 2.0( , ).1)

E F :

$ * C

respectively.

1.;-k , PB + /./

k 4 and

10 0 0 10 P$E + ; )25"02# + 1.(k , P*F + -.2;

k , PC: + -.2;

k ,

P + 1.(k


10.; 1(.) 10.;

).1) 2.0( 92.0( 9).1) 9).1) 9).0 9).1)

/./ 1.;- 91.;- 9/./ 90.09/.2

90.-

1.( -.2; 9-.2; 91.(

Column $F "k# *eam SF "k# *eam *6 "k9ft#

10.;

).0

(.1

);.1

-.0

;.;

);.1

-.0

;.;

10.;

).0

(.1

(.1

0.(2

.(0

0.)

;./2

11.0

0.)

;./2

11.0

(.1

0.(2

.(0

9;.9.22 9).1)

9-.;0 9(.22 91.20

Column *6 "k9ft#Column SF "k#

*eam $F "k#

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90.; 912.1/ 90.

(;.) 9(.1 00.1 (.) 9(;

90./( 91).(/ 90.10

) D 1 2 + ) 2

1 )

1k

Results from ‘Exact Structural Analysis

6 ? A P

1)k

% > @ B

k

E F :

$ * C

$ll members have eual cross9sections

10 12 10

1;.- 91.( 1.- 1.) 91;.(

/.1; 0.1( 90.1 9/.1- 9-.2 9.( 9-./

(.1

-/.-

9((.;

9-2.;

-(.)

02.

((. 9(/.;

-2. 9-/.1

1(.2/ 9;.-1 9;.- 91(.22


1;.-

9;.0

).

9)).;

901.0

(-./

9).0

02.0

9-.1

92.;

(-.

9).-

02.0

9-.1

92./

1;.(

9;.-

).0

9)).

901.2

/.0; 12.-0 12.-1 /.0( 9(.00 9).;/ 9).-2

Column *6 "k9ft# Column SF "k# *eam $F "k#

%nterior columns have t'ice the area of e!terior columns

.1

1).) 1).) 91).;0 911.0/


1.0

9(.

-(.(

92.-

-).-

9/;.

1.-

9(.-

0./( 1).(/ 1).)1 0.; 9-. 9).0 91.(1


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6 ? A P

% > @ B

E

$

F

*

:

C

( D 1 2 +

( 2

) D 1 2 + ) 2

1 )

C o l u m n s ' i t h

a r e a ) $

( D 1 2 +

( 2

F

F

Problems on Lateral Load Analysis by Cantilever Method

1. &he figure belo' sho's the a!ial forces "kips# in the e!terior columns of a t'o9storied frame.

%f the cross9sectional area of column $*C is t'ice the area of the other columns, use the Cantilever 6ethod to calculate the corresponding applied loads P1 and P).

$P)

( 12

*

P1

1) 1)

C

)2 )2

). For the structure sho'n belo', use the Cantilever 6ethod to calculate the lateral loads F1, F) if the shear

forces in beams $* and E are 12 kips and 10 kips respectively. $ssume all the columns have the samecross9sectional area.

$ * C)

E F :

1

10

% > @

)2 12 )2

(. se the Cantilever 6ethod to dra' the a!ial force, shear force and bending moment diagrams of the

three9storied structure loaded as sho'n belo'.

1k

1)k

k

,

10 12 10

-. Figure "a# belo' sho's the e!terior column a!ial forces "kips# in a three9storied fame.

se the Cantilever 6ethod to calculate "i# the applied loads, "ii# beam bending moments, "iii# column

bending moments. $ssume all the columns to have eual cross9sectional areas.

0

12

100 / )

)2 12 )2 12

Fig. "a# Fig. "b#

0. Figure "b# above sho's the column shear forces "kips# in a three9storied fame.

Calculate the column *6, beam *6, beam SF and column $F.$lso check if they satisfy the conditions for Cantilever 6ethod "for eual column areas#.

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Approximate !ertical Load Analysis

$ pp r o! i m a ti on b a s e d o n t h e Boc a ti o n of i n g es

%f a beam $* is subected to a uniformly distributed vertical load of ' per unit length GFig. "a#H, both the

oints $ and * 'ill rotate as sho'n in Fig. "b#, because although the oints $ and * are partly restrained

against rotation, the restraint is not complete. ad the oints $ and * been completely fi!ed against rotation

GFig. "c#H the points of inflection 'ould be located at a distance 2.)1B from each end. %f, on the other hand,

the oints $ and * are hinged GFig. "d#H, the points of zero moment 'ould be at the end of the beam. For the

actual case of partial fi!ity, the points of inflection can be assumed to be some'here bet'een 2.)1 B and 2

from the end of the beam. For approximate analysis, they are often assumed to be located at one-tenth (0.

!" of the span len#th from each end $oint.

$ *$ C * $ C * $ *

Bk 1B k )B

2.)1B 2.)1B

"a# "b# "c# "d#

epending on the support conditions "i.e., hinge ended, fi!ed ended or continuous#, a beam in general can be

statically indeterminate up to a degree of three. &herefore, to make it statically determinate, the follo'ing

three assumptions are often made in the vertical load analysis of a beam

1. &he a!ial force in the beam is zero

). Points of inflection occur at the distance 2.1 B measured along the span from the left and right support.

* end i ng 6o m ent and S he a r Fo r ce f r om $ pp r o! i m a t e $ na l y s i s

*ased on the appro!imations mentioned "i.e., points of inflection at a distance 2.1 B from the ends#, the

ma!imum positive bending moment in the beam is calculated to be

6"3# + '"2.B#)5 + 2.2 'B

), at the midspan of the beam

&he ma!imum negative bending moment is

6" # + 'B)5 2.2 'B

)+ 2.2-0 'B

), at the oints $ and * of the beam

&he shear forces are ma!imum "positive or negative# at the oints $ and * and are calculated to be

7$ + 'B5), and 7* + 'B5)

6o m e n t and S h ear 7 a l u es us i ng $C % C oe ff ic ien ts6a!imum allo'able BB5B + (, ma!imum allo'able adacent span difference + )2L

1. Positive 6oments

"i# For End Spans"a# %f discontinuous end is unrestrained, 6 "3# + 'B

)511

"b# %f discontinuous end is restrained, 6 "3# + 'B)51-

"ii# For %nterior Spans, 6"3# + 'B)51

). ?egative 6oments"i# $t the e!terior face of first interior supports

"a# &'o spans, 6" # + 'B

)

5;"b# 6ore than t'o spans, 6" # + 'B

)512

"ii# $t the other faces of interior supports, 6" # + 'B)511

"iii# For spans not e!ceeding 12 , of 'here columns are much stiffer than beams, 6" # + 'B)51)

"iv# $t the interior faces of e!terior supports

"a# %f the support is a beam, 6" # + 'B)5)-

"b# %f the support is a column, 6" # + 'B)51

(. Shear Forces

"i# %n end members at first interior support, 7 + 1.10'B5)

"ii# $t all other supports, 7 + 'B5)

G'here B + clear span for 6"3# and 7, and average of t'o adacent clear spans for 6" #H



) D 1

2 + ) 2

1 )

E!a m p l e

$nalyze the three9storied frame structure loaded as sho'n belo' using the appro!imate location of hinges to

dra' the a!ial force, shear force and bending moment diagrams of the beams and columns.

1 k5

6 ?

1 k5

% >

1 k5E F

$ *

&he ma!imum positive and negative beam moments andA P shear forces are as follo's.

For the 10 beam, 6"3# + 2.2 1 10)

+ 1 k9ft

@ B 6" # + 2.2-0 1 10)

+ 12.1( k9ft7" # + 1 105) + /.0 k

: For the 12 beam, 6"3# + 2.2 1 12)

+ k9ft

6" # + 2.2-0 1 12)

+ -.0 k9ftC 7" # + 1 125) + 0 k

10 12 10 $!ial Force P in all the beams + 2


9/.0 91).0

/.0 0.2 /.0 910.2 9)0.2 1.2.2

9)).0 9(/.0 912.1( 9-.0

*eam SF "k# Column $F "k# *eam *6 "k9ft#

0.2

90.0(

(.2-

(.()

91.0) 2.;1

91.0) 92.1 91.0)

2.- 2.1; 2.-

2.- 2.10 2.-

9).(2 1.(

92.0 2.(0


sing the $C% coefficients "for pattern loading#

/.0 0.2.(

9/.0

910.2

91(.(

9)/.)01.2/

.)0

9.( 9/.0 9)).0 9-2.91-.2

910.( 91-.1;

*eam SF "k# Column $F "k# *eam *6 "k9ft#

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Approximate Analysis of "rid#e Portal and Mill "ent

* ri d g e Po r t a l s a n d 6 i ll * e n t s

Portals for bridges or bents for mill buildings are often arranged in a manner to include a truss bet'een t'o

fle!ural members. %n such structures, the fle!ural members are continuous from the foundation to the top and

are designed to carry bending moment, shear force as 'ell as a!ial force. &he other members that constitute

the truss at the top of the structure are considered pin9connected and to carry a!ial force only.

*ridge Portal 6ill *ent

Such a structure can be statically indeterminate to the first or third degree, depending on 'hether the

supports are assumed hinged of fi!ed. &herefore, the same three assumptions made earlier for portal frames

can be made for the appro!imate analysis of these structures also4 i.e., for a load applied at the top

1. &he horizontal support reactions are eual

). &here is a point of inflection at the center of the unsupported height of each fi!ed based column

E!a m p l e

%n the bridge portal loaded as sho'n belo', dra' the bending moment diagrams of columns $* and C.

*

1 k5ftE

$

6$

C $ssuming the total load to be applied eually "i.e., )05) +

1).0 k and 1).0 k# at $ and *, the horizontal reactions are12

$ + 1).0 3 1).05) + 1./0 k, * + 1).05) + .)0 k $lso, *6 + 2 at the midpoint of the free height4 i.e., at 105)

F + /.0 from the bottom.

10 6$ 3 1./0 /.0 /.0 /.05) + 2 6$ + 11).0 k9ft

6 3 .)0 /.0 + 2 6 + -. k9ft

6$ + 2 11).0 -. 3 )0 1).0 7 )2 + 2$

6

7$ 7

)2

7 + /. k

Fy + 2 7$ 3 7 + 2 7$ + /. k

8 * 8 C

8 E 0.)0 8 F -.

1./0 k 11).0 k9ft 11).0

.)0 k -. k9ft -.

/. k *6 "k9ft# of $* /. k *6 "k9ft# of C



: %

(.;

Approximate Analysis of Statically $ndeterminate %russes

&'o appro!imate methods are commonly used for the analysis of statically indeterminate trusses. &he

methods are based on t'o basic assumptions

6ethod 1< iagonal members take eual share of the sectional shear force

6ethod )< iagonal members can take tension only "i.e., they cannot take any compression#

E!a m p l e

Calculate member forces :C, *, :, *C of the statically indeterminate truss sho'n belo' assuming

"i# iagonal members take eual share of the sectional shear force,

"ii# iagonal members can take tension only.

0 k 12 k 12 k 12 k 0 k

!

)2 kF >

$

* C !

10

E)2 k

)(./0 k - D ) 2 + 2 1.)0 k

F! + 2 E! 3 )2 + 2 E! + )2

6E + 2 )2 10 3 "0 3 12 3 12 3 12 3 0# -2 3 $y 2 + 2 $y + )(./0 k

Fy + 2 $y 3 Ey 3 0 3 12 3 12 3 12 3 0 + 2 Ey + 1.)0 k

"i# $t section !9!,

F! + 2 F: 3 F*C 3 F* cos (.; 3 F:C cos (.; 3 )2 + 2

F: 3 F*C 3 2. F* 3 2. F:C 3 )2 + 2

Fy + 2 F* sin (.; F:C sin (.; 3 0 3 12 )(./0 + 2

2. F* 2. F:C + ./0

$ssuming diagonal members to take eual share of the sectional shear force2. F* + 2. F:C + ./05) + -.(/0 F* + /.); k, F:C + /.); k

6* + 2 )(./0 )2 3 0 )2 3 )2 10 2. /.); 10 3 F: 10 + 2 F: + 12.( k

F! + 2 F: 3 F*C 3 2. F* 3 2. F:C 3 )2 + 2 F*C + (2.( k

"ii# $ssuming the diagonal member :C to fail in compression "i.e., to be non9e!istent#

$t section !9!,

F! + 2 F: 3 F*C 3 F* cos (.; 3 )2 + 2 F: 3 F*C 3 2. F* 3 )2 + 2

Fy + 2 F* sin (.; 3 0 3 12 )(./0 + 2 F* + 1-.0 k

6* + 2 )(./0 )2 3 0 )2 3 )2 10 3 F: 10 + 2 F: + 0 k

F! + 2 F: 3 F*C 3 2. F* 3 )2 + 2 F*C + (./ k

?ote< &he actual values from :8$SP "assuming identical member sections# are

F* + -. k, F:C + ;./1 k, F: + 1).// k, F*C + ).;2 k

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' k i p p e r f t

Problems on Approximate Analysis of "rid#e Portal& Mill "ent and %russ

1. %n the mill bent sho'n belo', use the portal method to calculate the a!ial forces in members *: and E

and dra' the shear force and bending moment diagrams of $*C and EF.

12 k

C

-0

*1 k5ft

$

- D) 2 + 2

12

-0 )2

E

-2

F

). %n the mill bent sho'n belo',

"i# se the Portal 6ethod to dra' the bending moment diagram of the member @B6."ii# Calculate the forces in E: and F, assuming them to take eual share of the sectional shear.

12 k

12 k C

E % ?

12

F : >6

12

* B

12

$ @

- D1 2 + -2

(. %n the bridge portal sho'n belo', compression in member : is 12 kips. se the Portal 6ethod to

"i# calculate the load ' per unit length, assuming the diagonal members to share the sectional shear

force eually.

"ii# dra' the *6 and SF of the member F: for the value of ' calculated in "i#.

C

)0

* E:

02

$ F) D 0 2 + 122

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C

E F %

*

$ - D 1 2 + -2>

-. %n the structure sho'n belo',"i# se the Portal 6ethod to calculate the reactions at support $, : and dra' the *6 of $*C.

"ii# Calculate the forces in members C, *E, CF, assuming diagonal members to take tension only.

C12 k

*

0 k

12

E %

12

F

)2

$ :

) D) 2 + -2

0. %n the bridge portal sho'n belo',

"i# se the Portal 6ethod to calculate the reactions at support $ and force in member *E."ii# Calculate the forces in members :% and F, assuming diagonal members to take tension only.

1 k5ft

: 6

B

@

12 k

12

12

12

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6et hod of 7i rtual Mork

'eflection Calculation by the Method of !irtual (or)

$nother 'ay of representing the euilibrium euations is by energy methods, 'hich is based on the la' of

conservation of energy. $ccording to the principle of virtual 'ork, if a system in euilibrium is subected to

virtual displacements, the virtual 'ork done by the e!ternal forces " ME# is eual to the virtual 'ork done

by the internal forces " M%#

ME + M% N...NNNNNNN"1#

'here the symbol is used to indicate Ovirtual. &his term is used to indicate hypothetical incrementsof displacements and 'orks that are assumed to happen in order to formulate the problem.

Consider the body loaded as sho'n in Fig. 1. nder the given

loading conditions, the point $ deflects an amount in the direction*

sho'n in the Figure. 6oreover the same load causes the element * dB'ithin the body to e!tend an amount dB in the direction sho'n.

%f a virtual unit load "i.e., a load of magnitude #, 'hen applied in$

u

the direction of , causes a virtual internal force u in the element *in the direction of dB, the virtual 'ork done by the e!ternal forces

ME + . N...NNNNNNN")#Fig. 1

'hile the virtual 'ork done by the virtual internal force "u# on * is + u. dB N...NNNNNNN"(#

&he total internal virtual 'ork done is M% + u. dB N...NNNNNNN"-#

'here the symbol indicates the summation over the lengths of all the elements 'ithin the body.

%n this formulation, the terms in italic indicate virtual loads or internal forces.

&he principle of virtual 'ork GE. "1#H . + u. dB + u. dB N...NNNNNNN"0#

%t is to be noted here that the term above can indicate the deflection or rotation of the body, depending on

'hich the virtual load (" can be a unit force or a unit moment applied in the direction of .

e fl e c ti on o f & r uss d ue t o E!t e r nal Boa d s

&he above principle can be applied to calculate the deflection of a truss due to a!ial deformation of itsmembers. &his a!ial deformation can be caused be caused by e!ternal loads on the truss, temperature change

or misfit of member length. &he a!ial deformation due to e!ternal loads is caused by the internal forces

'ithin the truss members, the resulting e!tension of a truss member being

dB + ?2B5E$ N...NNNNNNN"#

'here ?2, B, E and $ stand for the a!ial force "due to e!ternal loads#, length, modulus of elasticity and

cross9sectional area of a truss member. &he internal force u due to the unit virtual load is often e!pressed by

% , from 'hich the euation for truss deflection GE. "0#H becomes + % . ?2B5E$ NNN.NN"/#

E!a m p l e

Calculate the vertical deflection of the point * of the truss $*CEF due to the e!ternal loads applied

G:iven< E$5B + 022 kip5ft, for all the truss membersH.

F E

)212 k

$ * C $

/.2/

90

F12

2 9/.2/

*90

E12

2 91-.1-

C

12 k

$

F

0.&0& -

-0.'*

E 0 0

0.&0&0

0

-0.' C

( D )2 + 2 ?2 "k# %

sing member forces ?2 and % from the above analyses, + ?2 % B5E$ NNN.NN"/#

%gnoring zero force members,

*,v + J"/.2/# "0.&0& # 3 "Q/.2/# "0.&0& # 3 "Q0# "Q0.'# 3 "Q0# "Q0.'#K5022 + 2.21 ft



e fl e c ti on o f & r uss d ue t o & e m pe r a tu r e Chan g e a n d 6 i s f it

%n addition to e!ternal loads, a truss oint may deflect due to change in member lengths "i.e., become longer

or shorter than its original length# caused by change in temperature or geometrical misfit of any truss

member "being longer or shorter than its specified length#.

%n E. "0#4 i.e., + u. dB N...NNNNNNN"0#

the tem dB "elongation of a truss member# can also be due to temperature change or fabrication defect of any

truss member.

&he change in length due to increase in the temperature & is + & B N...NNNNNNN"#

'here + Coefficient of thermal e!pansion4 i.e., change of length of a member of unit length due to unit

change of temperature, & + Change of temperature of a member of length B.

$dding to it a geometric misfit "due to fabrication defect# of B, the total elongation of a truss member

dB + ?2B5E$ 3 & B 3 B N...NNNNNNN";#

from 'hich the euation for truss deflection GE. "0#H becomes

+ % dB + % "?2B5E$ 3 & B 3 B# N...NNNNNNN"12#

E!a m p l eCalculate the vertical deflection of oint * of the truss $*CEF sho'n belo' due to

"i# temperature rise of (2 F in the bottom cord members $* and *C,

"ii# fabrication defects resulting in vertical members *F and CE to be made 2.)0 shorter than specified

G:iven< Coefficient of thermal e!pansion + 0.0 129

5 F, for all the truss membersH.

F E 0 0

)292.)0

92.)00.&0& - 0.&0&

00

$*

C 2.2(; 2.2(; -0.' -0.'

( D )2 + 2 dB "in# %

"i# For members $* and *C, + 0.0 129

5 F, & + (2 F, B + )2 ft + )-2 in

dB + & B + "0.0 129

# "(2# ")-2# + 2.2(; in

%gnoring zero force members, *,v + "2.2(;# "-0.'# 3 "2.2(;# "-0.'# + 2.2(; in

"ii# For members *F and CE, dB + 2.)0 in

%gnoring zero force members, *,v + " 2.)0# "-# 3 " 2.)0# "0# + 2.)0 in

Suppo r t S e ttl e m ent

Settlement of supports due to consolidation or instability of the subsoil5foundation is a maor reason of deflection of structures. &here is a fundamental difference bet'een the effect of support settlement onstatically determinate and indeterminate structures. Mhile it causes deflection due to geometrical changes

only in statically determinate structures, it induces internal stresses in statically indeterminate structures"'hich may even be more significant than the forces due to e!ternal loads#.

&he effect of support settlement on statically indeterminate structures is dealt separately but the follo'ing

figure sho's the deflected shape of the truss $*CEF sho'n above due to settlement of support C.



1) 1)

Problems on %russ 'eflection by the Method of !irtual (or)

$ssume E$5B + 022 k5ft, + 0.0 129

5 F for the follo'ing trusses.

*1. Calculate E,h due to $

"i# &he e!ternal load

12 k

"ii# & + 02 F for C and CE. 1)

C E1) 1)

). Calculate *9C"rel# due to 10 k 10 k

"i# &he e!ternal loads

"ii# B + 2.0 for C and CE. C E

$ *

(. Calculate $9C"rel# due to 12 k C

"i# &he e!ternal loads

"ii# & + 02 F for $* and $.

(2 2

*

2 (2

1)

1/.()

1/.()

$ 12 k

-. Calculate C,v and C,h due to the e!ternal loads.

12k $ * C

-0F

E )2k

)0

- D) 0 + 122

0. Calculate *,v and "along $# due to the e!ternal loads.

12 k

)2 k

1)

$ C

*

1) 1)

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)

'eflection due to *lexural 'eformations

Fle!ural deformation is the main source of deflection in many civil engineering structures, like beams, slabs

and frames4 i.e., those designed primarily against bending moment. %t is often much more significant than

other causes of deflection like a!ial, shear and torsional deformation. From E. "0# of the previous section,

the principle of virtual 'ork + u. dB NNNNNNN"0#

'here the term above can indicate the deflection or rotation of the body, u is the virtual internal force in

an element 'ithin the body, 'hich deforms by an amount dB in the direction of u.

e fl e c ti on o f * ea m 5 Fr a m e due t o E ! t e r n a l B o ads

For fle!ural deformation, u is be the virtual internal moment m in the element 'hile dB is the rotation d

caused by e!ternal forces4 i.e., dB + d + curvature ds + "m25E%# ds.

+ m m25E% ds NNNNNNN"11#

'here m2 is the bending moment caused by e!ternal forces and E% is called the fle!ural rigidity of the

member. ere, the integration sign is used instead of summation because the bending moments vary'ithin the length of each member "unlike the trusses, 'here a!ial forces do not vary 'ithin the members#.

% n t e g r a ti on & a b l e

%n order to facilitate the integration sho'n in E. "11#, the follo'ing table is used bet'een functions f 1 and

f ), both of 'hich can be uniform or vary linearly or parabolically along the length "B# of a member.

$nte#ration of Product of *unctions +$ , f f . dS/

f ) f 1 $

B

*

B

$

B

$ *

B

$ C *

B

a

B

$aB *aB5) $aB5) "$3*#aB5) G$3-C3*HaB5

b

B

$bB5) *bB5( $bB5 G$3)*HbB5 G)C3*HbB5

a

B

$aB5) *aB5 $aB5( G)$3*HaB5 G$3)CHaB5

a b

B

$"a3b#B5) *"a3)b#B5 $")a3b#B5 G$")a3b#3*"a3)b#HB5G$a3*b3

)C"a3b#HB5

E!a m p le < Calculate the tip rotation and deflection of the beam sho'n belo' G:iven< E% + constH.

P2

$

B

m2

sing the m2 diagram along 'ith m1 for unit anticlock'ise

moment at $

*

P2B

$ + " 1# " P2B5E%# B5) + P2B 5)E% m11

sing the m2 diagram and m1 for unit up'ard load at $

v$ + "B# " P2B5E%# B5( + P2B(5(E%

B

m1



R12

v2 "k#

'eflection due to Combined *lexural& Shear and Axial 'eformations

* C12 k

12

E$ + -22 12(

k, :$

+ 1)0 12(

k, E% + -2 12(

k9ft)

Calculate the

"i# horizontal deflection at C " C,h#

"ii# vertical deflection at " ,v#.

$

12

R12

!2 "k# m2 "k #

For horizontal deflection at C " C,h#<

1

1

122

!1 v1 m1 " #

,v + "!1 !25E$# dS 3 "v1 v2 5:$# dS 3 "m1 m2 5E%# dS

R12

+ 12" 12#"1#5"-22 12(# 312" 12#"1#5"1)0 12

(#3125("122#" R 12#5"-2 12

(# + " R 2.)0 R 2.2 R (.((# 12

(

+ 2.2-- ft

For vertical deflection at " ,v#<

12

R1

1 R1

!1 v1 m1 " #

,v + "!1 !25E$# dS 3 "v1 v2 5:$# dS 3 "m1 m2 5E%# dS

+ 2 3 2 3 125) "12#"122#5"-2 12(# + 2 3 2 3 1)0 12

(+ 2.1)0 ft



Problems on 'eflection of "eams0*rames usin# Method of !irtual (or) +1nit Load Method/

$ssume E$ + -22 12(

k, :$

+ 1)0 12(

k, E% + -2 12(

k9ft)

* ea m s

1. 122 k ). 12 k )E% 12 k

$ *

$ C C$, $T

12 - D 0C, $T

12 k 1 k5

(. -.$ C *

:uided 8oller $ C *

12 12 12 12C, $T

*, $T

0. 1 k5

$ * C * is an %nternal inge*, *"B# , *"8#T

12 12 )2

Fra mes

. 12 k 12 k

/.

* C $*

C

1 k5

C, CT

)2 )2

$, $T

$

12 12 12 0 0

. 1 k5

*C

12 , $T

$

12 )2

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&he

Analysis of Statically $ndeterminate Structures by *lexibility Method



*lexibility Method for .-de#ree $ndeterminate %russes

C 12 k

E$5B + constant + 1222 k5ft

"?ote< E$ constant#

dosi + 1 3 - R ) - + )

12 &he horizontal reaction * and member force F*

are taken as the t'o redundants.

$ *

12

"2# 12 k "2# " 2./2/#

"2# "1-.1-# " 12# "2#

"2#

"2# "2#

"1#

" 2./2/#

"1#

"1#

" 2./2/#

" 2./2/#

Case 2 "* + 2, F* + 2#GForces ?2 "k#H Case 1 "* + 1#GForces ?1H

Case ) "F* + 1#GForces ?)H

1,2 + J?1 ?2 5"E$5B#K + J2 2 3 2 2 3 2 1-.1- 3 2 "R12# 3 1 2K51222 + 2 ft

),2 + J?) ?2 5"E$5B#K + J1-.1- 1 3 "R12# "R2./2/#K51222 + )1.)1 129(

ft

1,1 + J?1 ?1 5"E$5B#K + J2)

3 2)

3 2)

3 2)

3 1)K51222 + 1 12

9(ft5k

1,) + ),1 + J?1 ?) 5"E$5B#K + J1 "R2./2/#K51222 + R2./2/ 129(

ft5k

),) + J?) ?) 5"E$5B#K + J- "R2./2/#)

3 ) 1)K51222 + - 12

9(ft5k

"1 129(

# * 3 "R2./2/ 129(

# F* + 2

"R2./2/ 129(

# * 3 "- 129(

# F* + R)1.)1 129(

* + R -.); k, and F* + R .2 k

? + ?2 3 ?1 * 3 ?) F*

F$* + 2 31 "R -.);# 3 "R2./2/# "R.2# + 2, F*C + R12 3 2 3 "R2./2/# "R.2# + R0./1 k

FC + 2 32 3 "R2./2/# "R.2# + -.);, F$ + 2 32 3 "R2./2/# "R.2# + -.); k

F$C + 1-.1- 3 2 3"1# "R.2# + .2 k, F* + R .2 k



Problems on *lexibility Method for %russes +from past exams/

1. $ *12 k

$lso solve if C moves 2.12 to right 1)

C E

1) 1)

). 12 k

$lso solve if * settles 2.12 12

$ * C

12 12

(.10 k

C E

$ *

1) 1)

-.C

1)

12 k

(2 2*

1/.()

2 (2

$ - D12 + -2

1/.()

0. 12 k

$lso solve if C settles 2.12

)2 k

1)

$ * C

1) 1)

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$ *2 12 k

C

-0

(2

:

.$

* 12 k

-(.(

(2

E(2 2 2

C

02 02

/. $

*

C

).;

0/./

(2

%

2

E

02

(2:

2

02

2

2

12 k

F

02

0/./

.

1/.()

(2

;.

.

E

%

D( 2 + 12 F

C

* 12 k

Support E moves 2.12 right'ards

Support $ moves 2.12 right'ards

$ (2 2 2 (2 E

: F

12.

( D 12 2 + (22

$ * C

)0

12 k

)0

E12 k

-0:

-0F ,

- D ) 0 + 122

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Solutions for Problems on *lexibility Method for %russes

1. $ *

12 k

1)

$lso support C moves 2.12 right'ards

$ssume E$5B + Constant + 022 k5ft

dosi + / 3 - ) 0 + 1

C E

1) 1)

"2# 12 k

"2# "2#

"/.2/# " /.2/#

1

" 0# " 0# " 1# " 1#

?2 "k# "UC + 2# ?1 "UC + 1#

1,2 + J" 0# " 1#3 " 0# " 1#K5022 + 2.2) , 1,1 + J" 1# )3 " 1#)K5022 + 2.22-

2.22- UC 3 2.2) + 2.12 UC + )2 k

? + ?2 3 UC ?1 P$* + 2, P$C + 2, P*C + /.2/ k, P* + 2, P*E + /.2/ k, PC + )0 k, P*E + )0 k

).

12 k

12

$lso support * settles 2.12$ssume E$5B + Constant + 022 k5ft

dosi + 03 - ) - + 1

$ * C

12 12

12 k

"2# "2./2/# "2./2/#

"/.2/# " /.2/# " 1#

"0# "0# " 2.0# " 2.0#

1

?2 "k# "=* + 2# ?1 "=* + 1#

1,2 + J"0# " 2.0#3"0# " 2.0#3 "/.2/# "2./2/#3 " /.2/# "2./2/#K5022 + 2.21

1,1 + J" 2.0#)3 " 2.0#)3 "2./2/#)3 " 1#)3 "2./2/#)K5022 + 2.220

2.220 =* 2.21 + 2.12 =* + 1 k

? + ?2 3 =* ?1 P$* + 1- k, P*C + 1- k, P$ + 0. k, P* + 1 k, PC + 1;.2 k



(. 10 k 10 k

C E "2# "2# " 2./2/# "2#

1) "2# "10# " 2./2/# " 2./2/#"2# "2# "2# "1# "1# "2#

$ * "2#

1) 1) " 2./2/#

?2 "k# "P*C + 2# ?1 "P*C + 1#

E$5B + Constant + 022 k5ft, dosi + 3( ) 0 + 1

1,2 + J"10# " 2./2/#K5022 + 2.2)1) , 1,1 + J- " 2./2/#)3) "1#

)K5022 + 2.22

2.22 P*C 2.2)1) + 2 P*C + ).0 k

? + ?2 3 P*C ?1

P$* + 1./ k, P$C + 1./ k, PC + 1./ k, P* + 10.1( k, P*C + ).0 k, P$ + ).0 k,

PE + 2, P*E + 2

-. C

12 k

2

* (2

2 (2

E$5B + Constant + 1222 k5ft

1/.() dosi + 3( ) - + 1

$

- D 12 + -212 k

1/.()

".# "2# " 0# " 2.# " 2.0#

"1#

").0# "1#

"2# "2# " 2.0# " 2.#

?2 "k# "P$C + 2# ?1 "P$C + 1#

1,2 + J".# " 2.#3" 0# " 2.0#3").0# "1#K51222 + 2.22)0

1,1 + J) " 2.0#)3) " 2.#

)3) "1#

)K51222 + 2.22-

2.22- P$C 2.22)0 + 2 P$C + 2.( k

? + ?2 3 P$C ?1

P$* + 2.(1 k, P*C + .1) k, PC + 0.(1 k, P$ + 2.0- k, P$C + 2.( k, P* + (.1( k

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0. 12 k

)2 k

12

$lso support C settles 2.12$ssume E$5B + Constant + 022 k5ft

dosi + 030 ) - + )

$ * C

12 12

12 k

)2 k

" ).)# "(2# "2# "2# "2# "2# " 1.-1# " 1.-1# 1")#

"2# "2# "1# 1 "2# "1# "1#

?2 "k# ?1 "U* + 1# ?) "=C + 1#

1,2 + 2, ),2 + J" ).)# " 1.-1# 3 "(2# ")#K5022 + 2.)2

1,1 + "1#)5022 + 2.22), 1,) + ),1 + "1# "1#5022 + 2.22), ),) + J) " 1.-1#

)3 ")#

)3) "1#

)K5022 + 2.2)

2.22) U* 3 2.22) =C 3 2 + 2

2.22) U* 3 2.2) =C 3 2.)2 + 2.12

U* + 1./ k, =C + 1./ k

? + ?2 3 U* ?1 3 =C ?)

P$* + 2, P*C + 1./ k, P$ + -./1 k, P* + (.(( k, PC + )(.0/ k



*lexibility Method for -de#ree $ndeterminate "eams

E!a m pl e 1

PE% + constant

$ * C dosi + ( 1 3 - R ( ) + 1&ake 8 $ as the redundant

B5) B5) 1,2 3 8 $ 1,1 + $ + 2 NNNN..N"i#

m2 1,2 + "B5)#5 G)B 3 B5)H "RPB5)#5E% + R 0PB(5-E%

B5)

PB5)

B 1,1 + m1 m1 dS5E%

+ B5( "B# "B#5E% + B(5(E%

m1

"i# R 0PB(5-E% 3 B

(5(E% + 2 8 $ + 0P51

0PB5() 6 + m2 3 8 $ m1 + m2 3 "0P51# m1

6 6$ + 2, 6* + 0PB5(), 6C + RPB5) 3 0PB51 + R(PB51

(PB51

E!ampl e )

1 k5ft E% + constant

$ * C dosi + ( ) 3 - R ( ( + 1

&ake 8 * as the redundant

12 12 1,2 3 8 * 1,1 + * + 2 NNNN..N"i#

(/.002

m2 "k #1,2 + ) G) (/.0 3 02H "R0 125#5E% + R)2(.((5E%

1,1 + m1 m1 dS5E%m1 " # + ) 125( "R0# "R0#5E% + 1./5E%

R 0 "i# R)2(.((5E% 3 1./ 8 *5E% + 2

8 * + 1).0 k .)0 .)0

6 + m2 3 8 * m1 + m2 3 1).0 m1

6 "k # 6$ + 2, 6* + 02 R ).0 + R 1).0 k , 6C + 2

R 1).0



*lexibility Method for .-de#ree $ndeterminate "eams

E!ampl e (

12 k 12 k

$ * C E

dosi + ( ) 3 0 R ( ( + )

E% + 1 E% + 1 1,2 3 8 $ 1,1 3 8 C 1,) + $ + 2 NN"i#

0 0 0 0 ),2 3 8 $ ),1 3 8 C ),) + C + 2 NN"ii#

m2 "k # 1,2 + m1 m2 dS5E%

2 2 + J125 "R122# "(230#

R02 R122 3 05G"R122#"(23)2# 3 "R)22#"-2310#HK5E%

R)22 + R1;1./5E%

12 10 )2 ),2 + m) m2 dS5E%

0 + J05 "0# "R)22R02#m1 " # 3 05 G"0#"R)22R)22#3"12#"R-22R122#HK5E% + R /05E%

1,1 + m1 m1 dS5E% + )25( ")2# ")2#5E%

+ )./5E%

0 12 1,) + ),1 + m1 m) dS5E%

m) " # + J05"0#"(2312#

305 G"0#"(23)2#3"12#"-2310#HK5E%+ ((.((5E%

),) + m) m)dS5E% + 125( "12# "12#5E%

+ (((.((5E%

$voiding the factors E%

"i# )./ 8 $ 3 ((.(( 8 C + 1;1./

"ii# ((.(( 8 $ 3 (((.(( 8 C + /0

8 $ + G1;1./ (((.(( R ((.(( /0H5G)./ (((.(( R ((.(()H + (.(; k

and 8 C + G)./ /0 R ((.(( 1;1./H5G)./ (((.(( R ((.(()H + 1).1- k

6 + m2 3 8 $ m1 3 8 C m) + m2 3 (.(; m1 3 1).1- m)

6$ + 2, 6* + 2 3 (.(; 03 2 1).1- + 1.;0 k , 6C + R02 3 (.(; 12 32 1).1- + R1.12 k ,

6 + R122 3 (.(; 10 3 0 1).1- + 11.00 k , 6E + R)22 3 (.(; )2 3 12 1).1- + R12.2 k

1.;0

11.00

6 "k #

R1.12

R12.2



Analysis for Support Settlement

E!a m p l e -

Support * settles 2.12

$ * C E% + -2 12(

k9ft)

dosi + ( ) 3 - R ( ( + 1

1,2 3 8 * 1,1 + * + R2.12 NN"i#

12 12 1,2 + 2

m1 " # 1,1 + m1 m1 dS5E%

+ ) 125( "R0# "R0#5E% + 1./5-2 12(

R0 "i# 1./ 8 *5-2 12(

3 2 + R2.12

8 * + R -22251./ + R)- k 1)2

6 + m2 3 8 * m1 + R)- m1 Gin k H

6 "k #

E!a m p l e 0

Support C settles 2.12

E% + -2 12(

k9ft)

$ * C E dosi + )

1,2 3 8 $ 1,1 3 8 C 1,) + $ + 2.N..N."i#

12 12 ),2 3 8 $ ),1 3 8 C ),) + C + R2.12.N"ii#

1,2 + 2, ),2 + 2

1,1 + m1 m1 dS5E% + )./5E%, 1,) + ),1 + ((.((5E%, ),) + (((.((5E%

"i# ")./5E%# 8 $ 3 "((.((5E%# 8 C + 2

"ii# "((.((5E%# 8 $ 3 "(((.((5E%# 8 C + R2.12

8 $ + 1/.1- k, and 8 C + R0-. k

6 + m2 3 8 $ m1 3 8 C m) + 1/.1- m1 R 0-. m) Gin k H

6$ + 2, 6* + 1/.1- 0 R 2 0-. + 0./2, 6C + 1/.1- 12 R 2 0-. + 1/1.-2,

6 + 1/.1- 10 R 0 0-. + R1/.)2, 6E + 1/.1- )2 R 12 0-. + R )20.2

1/1.-2

6 "k #

R)20.2



R12

Combined *lexural& Shear and Axial 'eformations

* C12 k

12

E$ + -22 12(

k, :$

+ 1)0 12(

k

E% + -2 12(

k9ft)

dosi + ( ( 3 - R ( - + 1

&he vertical reaction at "7# is taken

as the redundant.

$

12

R12

122

!2 "k# v2 "k# m2 "k #

12

R1

1 R1

!1 v1 m1 " #

1,2 + "!1 !25E$# dS 3 "v1 v2 5:$# dS 3 "m1 m2 5E%# dS

+ 2 3 2 3 125) "122#"12#5"-2 12(# + 2.1)0 ft

1,1 + "!1 !15E$# dS 3 "v1 v1 5:$# dS 3 "m1 m1 5E%# dS

+ ) 12 "1 1#5"-22 12(#312 "1 1#5"1)0 12(#3G12 "12 12#312 "12 12#5(H5"-2 12(#+ 2.20 12

R(3 2.2 12

R(3((.(( 12

R(+ ((.- 12

R(

7 + R2.1)05((.- 12 R(

+ R(./- k



Problems on *lexibility Method +"eams0*rames/

$ssume E$ + -22 12(

k, :$

+ 1)0 12(

k, E% + -2 12(

k9ft)

* ea m s

1. 122 k ). 12 k 12 k

12 - D0

(. 12 k -.

:uided 8ollers

1 k5

12 12 12 12

0. 12 k .

1 k5

0 0 12 12 12 12

/. 12 k 12 k . Support settles 2.12

0 0 12 0 0 ( D12

Fr a m es

12 k 12 k

;. 12.

1 k5

)2 )2

12 12 12 0 0

mailto:4@5

mailto:3@10

mailto:3@10

mailto:4@5

mailto:3@10



)0

Solution of Problems on *lexibility Method +"eams0*rames/

1. dosi + ( 3 - R + 14 i.e., assume 8 $ as the redundant12

m2 "k # m1 " #

R122 8 $ + 1

1,1 + "m1 m15E%# dS + 125( "12 12#5"-2 12(# + .(( 12

R(ft5k

8 $ + 2.1)05".(( 12 R(

# + 10 k

02

6 "k #

). dosi + 3 - R ; + 14 i.e., assume 8 * as the redundant

R122

02 02

m2 "k # R 0

m1 " #

8 * + 1

1,2 + "m2 m15E%# dS + ) J05( "02# "R).0# 305) "02# "R).0 R0#K5"-2 12(# + R2.20/( ft

1,1 + "m1 m15E%# dS + ) J125( "R0# "R0#K5"-2 12(# + -.1/ 12

R(ft5k

8 * + 2.20/(5"-.1/ 12 R(

# + 1(./0 k

(. dosi + ( 3 - R + 14 i.e., assume 6$ as the redundant

10.(

R 1./0

6 "k #

122

m2 "k #

6$ + 1

m1 " #

1

1,2 + "m2 m15E%# dS + J125) "122# "R1# 312 "122# "R1#K5"-2 12(# + R2.2(/0 ft

1,1 + "m1 m15E%# dS + )2 "R1# "R1#5"-2 12(# + 2.0 12

R(ft5k

6$ + 2.2(/05"2.0 12 R(

# + /0 k9ft6 "k #

-. dosi + 3 - R ; + 14 i.e., assume 8 * as the redundant

(/.002

R /0

m2 "k # R 0m1 " #

1,2 + "m2 m15E%# dS + ) J125 ") (/.0 3 02# "R0#K5"-2 12(# + R2.20)1 ft

1,1 + -.1/ 12 R(

ft5k "as in Problem )#

8 * + 1

8 * + 2.20)15"-.1/ 12 R(

# + 1).0 k .)0 .)0

6 "k #

R 1).0



)0

R

0. dosi + ; 3 0 R 1) R 1 + 14 i.e., assume 8 C as the redundant

)0

m2 "k #

12

m1 " #

02122

8 C + 1

1,2 + "m2 m15E%# dS + 125 "R02R) 122# "12#5"-2 12(

# + R2.12-) ft1,1 + "m1 m15E%# dS + 125( "12# "12#5"-2 12

(# + .(( 12

R(ft5k

8 * + 2.12-)5".(( 12 R(

# + 1).0 k

. dosi + 3 0 R ; + )4 i.e., assume 8 * and 6C as the redundants

(/.002

)0

6 "k #

R 02

2.01

m2 "k # 0m1 " # m)

8 * + 1

1,2 + R2.20)1 ft, 1,1 + -.1/ 12 R(

ft5k "as in Problem -#

),2 + "m2 m)5E%# dS + )25 ") 0232# "1#5"-2 12(# + .(( 12

R(rad

6C + 1

1,) + ),1 + "m1 m)5E%# dS + J125( "R0#"2.0# 3125 "R0#"13) 2.0#K5"-2 12(# + R2.)0 12

9(rad5k

),) + "m) m)5E%# dS + )25( "1#"1#5"-2 12(# + 2.1/ 12

9(rad5k9ft

-.1/ 8 * R 2.)0 6C + 0).14 and R2.)0 8 * 3 2.1/ 6C + R.((

8 * + 11.-( k, 6C + R/.1- k9ft

/.1-1.(-

R 12./1

6 "k #

R /.1-

/. dosi + 3 0 R ; + )4 i.e., assume 8 * and 8 C as the redundants

02 02

m2 "k # ./ m1 " #./

m) " #

1,2 + "m2 m15E%# dS

8 * + 1 8 C + 1

+ J05("02#"(.((#305)"02#"(.((3./#3105)"02#"./31./#305("02#"1./#K5"-2 12(#+ 11;. 12

R(ft

1,1 + "m1 m15E%# dS + J125( " ./# " ./# 3 )25( " ./# " ./#K5"-2 12(

# + 11.11 12 R(

ft5k 1,) + ),1 + "m1 m)5E%# dS + G125(" ./#" (.((# 3 125J" ./#" ) (.(( ./#

3" (.((#" ) ./ (.((#K3125(" ./#" (.((#H5"-2 12(# + ;./) 12

R(ft5k

),2 + 1,2 + 11;. 12 R(

ft, ),) + 1,1 + 11.11 12 R(

ft5k

8 * + 8 C + 0./0 k "i.e., up'ard#

)1.)0

6 "k #

/.0



. dosi + 3 0 R ; + )4 i.e., assume 8 * and 8 C as the redundants

m2 is zero here, but m1 and m) are same as in Problem /.

1,2 + "m2 m15E%# dS + 2, ),2 + "m2 m)5E%# dS + 2

1,1 + 11.11 12 R(

ft5k, 1,) + ),1 + ;./) 12 R(

ft5k, ),) + 1,1 + 11.11 12 R(

ft5k

11.11 12 R(

8 * 3 ;./) 12 R(

8 C + 2.12

;./) 12 R(

8 * 3 11.11 12 R(

8 C + 2 8 * + (.- k "i.e., do'n'ard#, 8 C + ((. k "i.e., up'ard#

1--

6 "k #

;

;. dosi + 3 - R ; + 14 i.e., assume 8 C as the redundant

12122

102

12 !2 "k# )2 v2 "k#

1

1

(22

)2

m2 "k #

!1 v1 m1 " #

1,2 + "!2 !15E$# dS 3 "v2 v15:$# dS 3 "m2 m15E%# dS

+ )2" 12#"1#5"-22 12(# 3 12"12#" 1#5"1)0 12

(#

J125"122#") )2312#3)25"1223- 1023(22#")2#K5"-2 12(# + " 2.0 2. 1/0# 12

9(+ 1./ ft

1,1 + "!1 !15E$#dS 3 "v1 v15:$#dS 3 "m1 m15E%#dS + )2"1#"1#5"-22 12

(# 3 )2" 1#" 1#5"1)0 12

(# 3

J)25( ")2# ")2# 3 )2 ")2# ")2#K5"-2 12(# + "2.20 3 2.1 3 )./# 12

9(+ 2.); ft5k

8 * + 1./5"2.);# + /.2( k

).;/-2.(

/2.(1

/.2(

;.(/

).;/ U "k# 7 "k#)2

10;.(/6 "k #



12. dosi + 3 - R ; + 14 i.e., assume 8 $ as the redundant

)00

0

0

!2 "k# v2 "k#m2 "k #

1 120

1

)

!1 v1 m1

" #

1,2 + "!2 !15E$# dS 3 "v2 v15:$# dS 3 "m2 m15E%# dS

+ )2" 0#")#5"-22 12(# 3 0J"0#" 1#3" 0#" 1#K5"1)0 12

(# 3J05")0#") 0312#305(")0#"0#K5"-2 12

(#

+ " 2.0 3 2 3 10.(# 129(

+ 10.1( 129(

ft

1,1 + "!1 !15E$# dS 3 "v1 v15:$# dS 3 "m1 m15E%# dS

+ )2")#")#5"-22 12(# 3 J12"1#"1#312" 1#" 1#K5"1)0 12

(# 3 )J125("12#"12#K5"-2 12

(#

+ "2.) 3 2.1 3 1./# 129(

+ 1/.2( 129(

ft5k

8 $ + 10.1( 129(

5"1/.2( 129(

# + 2.; k

2.;

0.;

-.11

)2.00

./ .;

U "k# 7 "k#6 "k #



%he Moment 'istribution Method

*ixed End Reactions for 2ne-dimensional Prismatic Members under %ypical Loadin#s

PB5 P PB5 'B)51) ' 'B

)51)

B5) B5) B

P5) P5) 'B5) 'B5)

Pab)5B

)P Pa

) b5B

)'B

)5(2

'

'B)5)2

a b B

Pb)"(a3b#5B

(Pa

)"a3(b#5B

(('B5)2 /'B5)2

65- 6 65-

P

B5) B5) Pb5B a b Pa5B

(65)B (65)B

-E% 5B )E% 5B E% 5B)

E% 5B)

E% 5B)

E% 5B)

1)E% 5B(

1)E% 5B(

B B



End Rotation and Rotational Stiffness of *ixed Ended Prismatic Members

6$ 6*

$ E% + Constant *

7$ 7*

6$5E% 6*5E%

B B

sing the 6oment $rea &heorems bet'een $ and *

1st

&heorem "6$5E% 3 6*5E%# B5) +

6$ 3 6* + ) E% 5B NNNNNNNN..NN..."1#

)nd

&heorem 6$5E% B5) B5( 3 6*5E% B5) )B5( + 2 "6$5 3 6*5(# B)

+ 2

6* + 6$5) ..NNNNNNN..NNN")#

"1# 6$5) + ) E% 5B 6$ + - E% 5B NNNNNNN..NNN.."(#

and ")# 6* + ) E% 5B NNNNNNN..NNN.."-#

&he term -E%5B is called the rotational stiffness and the ratio " 6*56$ +# 2.0 the ca r r y o v er f ac t o r of the

member $*.

&aking 6* + 2 and 6$ + 2, 7$ and 7* can be derived to be E%5B)and E%5B

).

?ote that for anti9clock'ise rotation , the moments 6$ and 6* are both anti9clock'ise but have different

signs in the *6.

End 'eflection and Shear Stiffness of *ixed Ended Prismatic Members

6$ 6*

$ E% + Constant *

7$ 7*

6$5E% 6*5E%

B B

sing the 6oment $rea &heorems bet'een $ and *

1st

&heorem "6$5E% 3 6*5E%# B5) + 2 6* + 6$ NNNNNNNN..NN..."1#

)nd

&heorem 6$5E% B5) B5( 3 6*5E% B5) )B5( + 6$ 3 )6* + E% 5 B)

...NNN")#

"1#, ")# 6$ + E% 5 B)

6$ + E% 5 B)

NNNNNNN..NNN.."(#

and ")# 6* + E% 5 B) NNNNNNN..NNN.."-#

&aking 6* + 2 and 6$ + 2, 7$ and 7* can be derived to be 1)E% 5B(and 1)E% 5B

(.

& he t e r m 1) E % 5B( i s c a ll ed the s h e ar s tif f ne s s of the member $*.

?ote that 6$ and 6* are both anti9clock'ise here but have different signs in the *6.



Rotation of a 3oint and Moment 'istribution *actors +M'*/

*

$ @ A*

@ A$ 6A

A @ AC C

@ AE

E

@ A

Fle!ural members A$, A*, ACN...are oined at oint A and have rotational stiffnesses of @ A$, @ A*,@ ACNN.respectively4 i.e., for unit rotation of the oint A they reuire moments @ A$, @ A*, @ ACNN.respectively to be applied at A.

%f a moment 6A applied at oint A causes it to rotate by an angle , the follo'ing moments are needed to

rotate members A$, A*, ACN...6A$ + @ A$ NNN.."1#

6A* + @ A* NNN..")#

6AC + @ AC NNN.."(#

NNNNNNNNN

$dding "1#, ")#, "(#N. 6A$ 3 6A* 3 6AC 3 NN.+ @ A$ 3 @ A* 3 @ AC 3NN NNN.."-#

Since 6A + 6A$ 3 6A* 3 6AC 3 NNN.

6A + "@ A$ 3 @ A* 3 @ AC 3NN# + @ A G@ A + @ A$ 3 @ A* 3 @ AC 3NNH

+ 6A5"@ A# NNN.."0#

"1# 6A$ + G@ A$5@ AH 6A NNN.."#

")# 6A* + G@ A*5@ AH 6A NNN.."/#

"(# 6AC + G@ AC5@ AH 6A NNN.."#NNNNNNNNN

&he factors G@ A$5@ AH, G@ A*5@ AH, G@ AC5@ AHNNN..are the moment distribution factors "6F# of members

A$, A*, ACNN..respectively. &herefore the distributed moments in members are proportional to their respective 6Fs.

E!ampl e

12 k

2.2 1).0 1).0 (./0 /.0 1.)0 0.2

2.-2 0.2 0.2

10

E% + Constant

0 0

).0 ).0

GBoad and 6FH GFE6H Gist 6H G&otal end 6H



Problems on Moment 'istribution

$ssume E% + constant + -2 12(

k9ft)

* ea m s

1. 1 k5 Support $ settles 2.20

$

)2

*

). 0 k 1 k5 (. $

$ )2

0 k

C

)2 0

12 )2

$ and * are guided roller supports

E%$* + ) E%

-.1 k5

*

$

12 k

C

* in an %nternal inge

E%E + ) E%

E

12 0 0 0 10

0. 122 k 12 k . 1 k5

12

:uided 8oller

Support $ settles 2.20 >oint $ rotates 1V anticlock'ise

120 0

1 k5 1 k5 1 k5 1 k5

/. .

Fr a m es

- D 12 0 D 12

Support settles 2.12;. 12 k 12.

0

0

12 12

12

1 k5 12

12 12

mailto:4@10

mailto:5@10

mailto:4@10

mailto:5@10



1. 12 k

1 k5ft

$ * C E

F

- -

).

$ * C

E% + Constant

1) C is an %nternal inge

12 k

; E% + -2,222 k9ft)

Support E moves 2.20 do'n'ards

E

1) 10

(.

$ * C

E% + -2,222 k9ft

)

*oth and E settle by 2.)0

E

1 ;

-.

E

E% + -2,222 k9ft)

12

*oth $ and * settle by 1

$ C

12

*

12 12



(;.; 2.02 0(.(( 0(.((2.)0

((.(( .0 ((.(- 1./ -0.2

1(.1( ).)0 -(./.0/ (.)

2.11 1.)(

1./ (.) -0.2

Moment 'istribution for *rames +from past exams/

1. 12 k 1 k5ft0 k 0 k 2 2.2

$ * C )2.2 0.(( 0.((10 k

1) E% + Constant2.-2

- 12.1( 10.)2 /.2

;./ 1).;(E

0.2/

-

)2.2

;./

-2 -2

12.1( 1).;(

*6 "k #

0.2/

12 k

).

$ * C

;E% + -2,222 k9ft

)

Support E moves 2.20 do'n'ards E% 5B)

+ -2222 2.20510)

+0(.(( k E

1) 10

1)2 2.02 2.(/0

1;.;0

).20

)1. 1.2)

)).;2

1)2 2.11 -0.2

(;.;

*6 "k #

1;.;- )).;2



/

/.01 )2.2) ).; 1(.(-

0.- ).20 )./-

2.1) 2.(( 2.-0 2.))

2.2( 2.2; )2.(0 (-.0; -.1/

0.-; 0.-; 12.1/

*6 "k #

)./- 12.1/

(.

$ * C

E% + -2,222 k9ft)

*oth and E settle by 2.)0

E

1 ; (-.0;

2./(

2.)/ 2.)-

2.((

2.-( 0.-;

1-.)-

)2.(0

1./( 1./(

9-.1/

-. E-5/

)22

(5/

122

)22

)22

0-.;) ).;/ 0.1; 0/.1- -).3).))

12 0-.;) 2

$ C 1;.1) 2.0- 1.- 12.(; ).0-511

E% + -2,222 k9ft)

*oth $ and * settle by 1(511 -511

/./; 12.(;

12 12 12 2.-2 2.0-

.1; 12.;(*

0-.;) 0.-/

0-.;)

*6 "k #

1;.1)

.1; 12.;(

0.-/



7 7

4ualitative $nfluence Lines and Maximum *orces

"8#1. For the beam sho'n belo', dra' the influence lines of 8 $, 8 *, 7*"B#

, 7* , 6$, 6*.

$ * C

12 0

8 $ 8 *

"B#*

"8#*

6$ 6*

). For the beam sho'n belo', B + 1 k5 , moving BB + 2.0 k5 "B#, 0 k "concentrated#.

Calculate the ma!imum values of 8 $, 8 *, 6E, 6* and 6F GEach span is 12 longH.

$ * C $ * C

E F : "8 $# E F : "8 *#

$ * C $ * C

E F : "6E# E F : "6*#

1.0 k5 0 k

1 k5 1 k5

"6F# Boad arrangement for 6F"ma!#

Final end moments 2 1.)0 1.)0

6F"ma!# + 1.)0 3 1.0 12)5 3 0 125- + 10 k



1.0 k5 ) k5 ) k5 1.0 k5

2 1 (5/ -5/ 2.0 2.0

1./0 1./ 1./

").2#1.2-

" 2.;#

" 2.2#

1./ 1.

").1.2-

" 1.1;# 2.22.10 "2.(2#

" 2.2;#

1).0 1).0

2# ").2#1.2-

"2.(2# 2.10

1./0 1./0 1.0 1.0 1-. 1-. 11.(1

E

(. For the beam sho'n belo', dra' the ualitative influence lines for

"i# *ending moments 6C, 6, 6E, 6F

"ii# Support reactions 8 *, 8 , 8 E, 8 F"8# "B# "8# "B# "8#"iii# Shear forces 7* , 7 , 7 , 7E , 7E ,7F

%f the beam is subected to a uniformly distributed B + 1.0 k5ft and moving BB + 2.0 k5ft "uniformly

distributed# and 0 k "concentrated#, calculate the ma!imum values of "8#"i# positive 6C, "ii# positive 8 and "iii# positive 7E G:iven< E% + constantH.

$ * C E F

0 0 0 12 12

%B of 6C %B of 8 %B of 7"8#

%B of 6 %B of 8 F %B of 7"B#

"i# 6a!imum positive value of 6C< "ii# 6a!imum positive value of 8 <

0 k 0 k

1.0 k5 ) k5 1.0 k5 ) k5 B

2 1 (5/ -5/ 2.0 2.0 F

1./0 )).;) )).;) 1).0 1).0 1./ 1./ FE6

" -.1/#

).2

"0.(# "/.1-# (.0/

1.;( " (./# " (./# 1.;(

"2.(# "1.12# 2.00 2.1- " 2.)/# " 2.)# 2.1-

"2.2# "2.2#1./0 1./0 1./0 1./0 1).0) 1).0) 1./- 7

"B#+ ) 125)3"1./0 1.0#512+ ;./ k

7

"8#+ ) 125)3"1.0 1-.#512 + 12.1/ k

6a!imum value of 6C 6a!imum 8 + 7

"8#7

"B#30 +)-.;0 k

+ 1./0 3 ) 12)5 3 0 125- + 1./0 k



sing v"B# + 2 in "(# 2 + 8 $B(5 3 C1 B 3 1 8 $B

(5 3 C1 B + 1 NNNNNNN.."#

sing "B# + 2 in ")# 2 + 8 $B)5) 3 C1 8 $B

)5) 3 C1 + 2 NNNNNNN.."/#

S h e a r F o r c e

* e n d i n g 6

o m e n t 5 B

8 e a c t i o n

4uantitative $nfluence Lines for $ndeterminate Structures

$ *

E% + Constant + 1 "assume#

B

From 6oment9Curvature 8elationship, E% d)v5d!

)+ 6"!# + 8 $!

%n this case, d

)

v5d!

)

+ 6"!# + 8 $ ! NNNNNNN.."1#dv5d! + "!# + 8 $ !)5) 3 C1 NNNNNNN..")#

v"!# + 8 $!(5 3 C1! 3 C) NNNNNNN.."(#

&here are three unkno'ns in these euations4 i.e., 8 $, C1 and C)

For the given beam, there are three kno'n boundary conditions from 'hich these three unkno'ns can becalculated.

&he boundary conditions are, v"2# + 1, v"B# + 2 and "B# + 2 NNNNNNN.."-#

sing v"2# + 1 in "(# 1 + 2 3 2 3 C) C) + 1 NNNNNNN.."0#

Solving "# and "/#, 8 $ + (5B(and C1 + (5)B NNNNNNN.."#

v"!# + "!5B#(5) ("!5B#5) 3 1 ...NNNNNN..";#

1.22

81 8)

2./0

2.02

B2.)0

2.22

2.22 2.)0 2.02 2./0 1.22

?on9dimensional istance, !5B Fig.

1< %nfluence Bines for 8eactions

Ance the euation of %B for 8 $ is determined, the euations of %B for shear force and bending moment at any

section can also be derived.

1.22

2.02

71 7) 7(

2.)2

2.12

61 6) 6( 6-

2.22

92.02

91.22

2.22 2.)0 2.02 2./0 1.22

?on9dimensional istance,!5B

Fig. )< %nfluence Bines for Shear

Forces

2.22

92.12

92.)2

2.22 2.)0 2.02 2./0 1.22

?on9dimensional istance, !5B

Fig. (< %nfluence Bines for *ending 6

oments



Short 4uestions and Explanations

Fle!ibility 6ethod for ) &russes vs. ) Frames1. nkno'ns< Forces only vs. Forces 3 6oments

). ?o. of nkno'ns< dosi + m 3 r ) vs. dosi + (m 3 r (

(. 6ember Properties< E, $ vs. E, :, $, $ , %-. eformations considered< $!ial vs. $!ial, Shear, Fle!ural

0. Forces Calculated< 6ember $!ial Forces vs. 6ember $!ial, Shear Forces, *6s

. Structural isplacements< eflections vs. eflections 3 8otations

$lso learn9 Bateral Boad $nalysis by Portal vs. Cantilever 6ethod

9 7ertical Boad $nalysis by $C% Coefficients vs. $ppro!imate inge locations

9 ifference bet'een appro!imate 6ethods for &russ $nalysis

9 Fle!ibility 6ethod vs. 6oment istribution 6ethod

*riefly e!plain 'hy

9 it is often useful to perform appro!imate analysis of statically indeterminate structures

9 dosi of ( truss + m 3 r ( and dosi of ( frame + m 3 r h9 a!ial deformations are sometimes neglected for structural analysis of beams5frames but not trusses

9 support settlement is to be considered5avoided in designing statically indeterminate structures9 unit load method is often used in the structural analysis by Fle!ibility 6ethod

9 a guided roller can be used in modeling one9half of a symmetric structure9 the terms moment distribution factor and carry over factor in the 6oment istribution 6ethod

9 the influence lines of statically determinate structures are straight 'hile the influence lines of

statically indeterminate structures are curved

Comment on9 t'o basic characteristics of the Fle!ibility 6atri! of a structure

9 the main advantage and limitation of the 6oment istribution 6ethod9 advantage of using modified stiffness in the 6oment istribution 6ethod

9 the applicability of Oualitative and Ouantitative influence lines



A !

5on-coplanar *orces and Analysis of Space %russ

? o n 9 cop l anar F o r ce$ vector in space may be defined or located by any three mutually perpendicular reference a!es A!, Ay and

Az "Fig. 1#. &his vector may be resolved into three components parallel to the three reference a!es.

%f the force AC "of magnitude F# makes angles , and y'ith the three reference a!es A!, Ay and Az, then the

components of the force along these a!es are given by

F! + F cos NN.NNN.NNN.NNN"i#

Fy + F cos N...NNN.NNNN..NN."ii# C

Fz + F cos NN.NN..NNNN..NN"iii#

G"i#)

3 "ii#)

3 "iii#)H F + GF!

)3 Fy

)3 Fz

)H N.N...N."iv#

"i# cos + F!5 GF!)

3 Fy)

3 Fz)H NN.NN.NN"v#

"ii# cos + Fy5 GF!)

3 Fy)

3 Fz)H NNN.N..N..."vi#

"iii# cos + Fz5 GF!)

3 Fy)

3 Fz)H NNN.N..N."vii#

z&he values of cos , cos and cos given by Es. "v#, "vi#

and "vii# are called the direction cosines of the vector F. Fig. 1< ?on9coplanar Force and Components

Space &ru s s$lthough simplified t'o9dimensional structural models are uite common, all civil engineering structuresare actually three9dimensional. $mong them, electric to'ers, offshore rigs, rooftops of large open spaceslike industries or auditoriums are common e!amples of three9dimensional or space truss. &he members of aspace truss are non9coplanar and therefore their a!ial forces can be modeled as non9coplanar forces.

Since there is only one force per member and three euilibrium euations per oint of a space truss, the

degree of statical indeterminacy "dosi# of such a structure is given by

dosi + m 3 r ( NNNNNNNNNNN"viii#

&he three euilibrium euations per oint of a space truss are related to forces in the three perpendicular a!es

!, y and z

F! + 2, Fy + 2 and Fz + 2 NN..NNNNNNNNN"i!#

o'ever the other three euilibrium euations related to moments4 i.e.,

6! + 2, 6y + 2 and 6z + 2 NN..NNNNNNN.NN"!#

may also be needed to calculate the support reactions of the truss. ere, it is pertinent to mention that the

moment of a force about an a!is is zero if the force is parallel to the a!is "'hen it does not produce any

rotational tendency about that a!is# or intersects it "'hen the perpendicular distance from the a!is is zero#.



6ember B! By Bz C! Cy Cz

$* (2 2 2 1.22 2.22 2.22

*C 2 2 )2 2.22 2.22 1.22* (2 2 )2 2.( 2.22 2.0

$ 2 2 )2 2.22 2.22 1.22

$E 10 )0 12 2.-; 2.1 2.()

*E 10 )0 12 2.-; 2.1 2.()

CE 10 )0 12 2.-; 2.1 2.()

E 10 )0 12 2.-; 2.1 2.()

E!ample < Calculate the support reactions and member forces of the truss sho'n belo'.

E

)0

$

)2 k

*

%gnoring the zero force member C

dosi + m 3 r ( + 3 / ( 0 + 2

&he structure is statically determinate.

C

12

12

y

10 1012 k

!

z

6C + 2 =$ )2 12 )2 + 2 =$ + 12 k N.NNNN"1#

6*C + 2 =$ (2 3 = (2 3 )2 )0 + 2 = + )./ k NNNNN.")#

6y"# + 2 )2 12 3 WC (2 + 2 WC + ./ k NNNN.N"(#Fy + 2 =$ 3 =C 3 = 12 + 2 =C + )./ k N...NNN.."-#

F! + 2 U 3 UC 3 )2 + 2 NNNNN."0#

Fz + 2 W 3 WC + 2

W + WC + ./ k Gusing "(#H NNNNN.."#

Euilibrium of >oint $ "unkno'ns F$*, F$ and F$E#<

F! + 2 F$* 3 2.-; F$E + 2 NNNN..N"/#

Fy + 2 2.1 F$E 3 12 + 2 F$E + 1).(( k N..NNN...."#

F$* + 2.-; F$E + .22 k Gusing "/#H N..NNN....";#

Fz + 2 F$ 2.() F$E + 2 F$ + -.22 k Gusing "#H N....NNN"12#

Euilibrium of >oint * "unkno'ns F*C, F* and F*E#<

F! + 2 F*$ 2.( F* 2.-; F*E + 2 NNNNN"11#

Fy + 2 2.1 F*E 12 + 2 F*E + 1).(( k N...NNN."1)#

F* + " F*$ 2.-; F*E#52.( + 1-.-) k Gusing "11#H N...NNN."1(#

Fz + 2 F*C 2.0 F* 2.() F*E + 2 F*C + -.22 k Gusing "1)#, "1(#H N.N..N.N"1-#

Euilibrium of >oint C "unkno'ns UC and FCE#<

F! + 2 UC 2.-; FCE + 2 NNNNN."10#

Fy + 2 )./ 3 2.1 FCE + 2 FCE + (). k NNNN.N"1#

UC + 2.-; FCE + 1 k Gusing "1#H N...NN..N"1/#

Fz + 2 ./ 3 2.() FCE 3 FC* + 2 FC* + -.22 k GverifiedH

Euilibrium of >oint "unkno'ns U and FE#<

F! + 2 U 3 2.-; FE 3 2.( F* + 2 NNNNN..."1#

Fy + 2 )./ 3 2.1 FE + 2 FE + (). k N..NNNN."1;#

U + -.22 Gusing "1(#, "1;#H NN...NNN")2#

UC + )2 U + 1.22 Gusing "0#H N...NNNN")1#

Fz + 2 ./ 3 2.() FE 3 F$ 3 2.0 F* + 2 ./ 3 12./ 3 -.22 .22 + 2

2 + 2 GverifiedH



Problems on the Analysis of Space %russes

1. Calculate the member forces of the space truss loaded as sho'n belo'.

ainge Support

d

)212 k

)2

y bc, d b

az

)2

!

)2 -2! c

). Calculate the horizontal "along ! a!is# deflection of oint E and vertical "along y a!is# deflection of oint

* of the space truss analyzed in class G:iven< E$5B + constant + 022 k5ftH.

(. Calculate the support reactions and member forces of the space truss loaded as sho'n belo'. $lso

calculate the vertical "along y a!is# deflection of the oint d G:iven< E$5B + constant + 022 k5ftH.

12 k

y zd

! !

)2 U b

0

Wa Wc

0a

bc

=a = b =c

12 12

-. Calculate the support reactions, member forces and also the horizontal "along ! a!is# deflection of the

oint a of the space truss loaded as sho'n belo' G:iven< E$5B + constant + 022 k5ftH.

12 ka

y W b

bU b

!

)2

z

c

0 !a 12 k

12

b, e c, d e d Ud

= b, =e =c, =d

0 10



'eflection of 6rids due to Combined *lexural and %orsional 'eformations

:rids are )9dimensional "coplanar# structures 'ith one deflection and t'o rotations at each node.

%f the structure is in the !9z plane, the deflection is out9of9plane "along the y a!is# 'hile the rotations are

about t'o in9plane a!es "! and z a!is#. :rids are loaded perpendicular to the structural plane and have three

forces per member4 i.e., shear force, bending moment and torsion.

E!ampl e

E Calculate $,v and C,v ifE

0

00 0

E% + -2 12(

k9ft)

:> + (2 12(

k9ft)

12 k

12 k

$ * C$ * C

&op 7ie' %sometric 7ie'

12 k

12 k

102

02

Calculation of $,v

m2 "k # t2 "k #

02 0212

0

0

1

$,v + "m1 m25E%# dS 3 "t1 t25:># dS

m1 " # t1 " #

0

+ JG0 0 "R02#5(H 3 0 G0 "R)02#312 "R(02#H5K5"-2 12(# 3J12 "R0# "R02#K5"(2 12

(#

+ R12;.( 12 R(

3 (.(( 12 R(

+ R).20 12 R(

ft12

Calculation of C,v

0

0 0

1m1 " # t1 " #

C,v + JG0 0 "R02#5(H3G0 0 "R02#5(H30 G0 "R)02#312 "R(02#H5K5"-2 12(# 3 J12 "0# "R02#K5"(2 12

(#

+ R1)2.2 12 R(

R (.(( 12 R(

+ R)2-.1( 12 R(

ft



Problems on 'eflection of 6rids usin# Method of !irtual (or) +1nit Load Method/

$ssume E% + -2 12(

k9ft), :> + (2 12

(k9ft

)

1.$

y

1 k5ft

, X x,, X z ,T 12 x

12 k

z

* C

0 0

).$

E, *T12

0 0

(.

, *T

$

0

* C0

12 k 0

* C

012 k

E

12 k

E

12 k

12

-.$

1 k5ft

C, X x,C, X z ,CT 12

*

C

0 0



'eflection of 6rids usin# Method of !irtual (or)

1.12

$

* C

02

).0102

0 0

02

m2 "k9ft# t2 "k9ft#m1 "ft#

For t1 "ft#

+ G"Q02#"0#05( 3 J)"Q).0# 3 "Q102#K"12#"12#5H5"-2 12(# 3 G"Q02#"0#12H5"(2 12

(#

+ Q2.1)0 Q2.2(( + Q2.)2( ft

).

10

$

0

* C

02 02

)020

02 0

0

E

m2 "k9ft# t2 "k9ft# m1 "ft# t1 "ft#

For E

E + G"Q02#"0#05( 3 "Q02#"0#05( 3 J"Q02# ")0# 3 "Q)02# "(0#K"12#5H5"-2 12(# 3 G"02#"Q0#0H5"(2 12

(#

+ Q2.-(/0 Q2.2-1/ + Q2.-/;) ft



1

(.12

$102

C*02

E

122

m2 "k9ft#

102

122

t "k9ft#

0

12

12

For 2 m1 "ft# t1 "ft#

+ G"Q122#"12#125( 3 "Q02#"0#05( 3 J"Q02# ")2# 3 "Q102# ")0#K"0#5H5"-2 12(# 3

G"122 3 102#"Q12#0H5"(2 12(#

+ Q2.1;)/ Q2.-1/ + Q2.2;- ft

-. $

1).0

1).0

*

C

Q).0

m2 "k9ft# t2 "k9ft# m1 "ft# t "ft#

For C

+ G"1).0#"Q).0# 05( )H5"-2 12(#

+ Q).2- 129(

ft



*lexibility Method for 6rids +Combined *lexural and %orsional 'eformations/

:rids are )9dimensional "coplanar# structures 'ith one deflection and t'o rotations at each node.

%f the structure is in the !9z plane, the deflection is out9of9plane "along the y a!is# 'hile the rotations are

about t'o in9plane a!es "! and z a!is#.

:rids are loaded perpendicular to the structural plane and have three forces per member4 i.e., shear force,

bending moment and torsion.

E!a m p l e

E

E

0 E% + -2 12(

k9ft)

12 k

:> + (2 12(

k9ft)

dosi + ( ( 3 - R ( - + 1

0 12 k $ * C $ * C

0 0

&op 7ie' %sometric 7ie'

12 k

102

12 k 02

1,2

02 02

Case 2 m2 "k # t2 "k #

12

0

1,1

0

1 0

Case 1 m1 " # t1 " #

1,2 + "m1 m25E%# dS 3 "t1 t25:># dS

+ JG0 0 "R02#5(H 3 0 G0 "R)02#312 "R(02#H5K5"-2 12(# 3J12 "R0# "R02#K5"(2 12

(#

+ R12;.( 12 R(

3 (.(( 12 R(

+ R).20 12 R(

ft

1,1 + "m1 m15E%# dS 3 "t1 t15:># dS

+J0 "0 0#5(312 "12 12#5(K5"-2 12(# 3J12 "R0# "R0#K5"(2 12

(#

+ ;.( 12 R(

3 .(( 12 R(

+ 1/./1 12 R(

ft5k

7$ + ).20 12 R(

51/./1 12 R(

+ 1.-/ k



Problems on *lexibility Method for 6rids

1.E

12 k

0

0

$ * C

0 0

12 k

$ssume "for all members#

E% + -2 12(

k9ft)

:> + (2 12(

k9ft)

dosi + ( ( 3 0 R ( - + )

%t 'ill be convenient to take =$ and =C as the) redundants

02

Case 2 "=$+2, =C+2# m2 "k # t2 "k #12

0

0

1 0

Case 1 "=$+ 1, =C+ 2# m1 " # t1 " #

12

0

0

0

1

Case ) "=$+ 2, =C+ 1# m) " # t) " #

1,2 + "m1 m25E%# dS 3 "t1 t25:># dS + J0 "R02#")230#5K5"-2 12(# 32 + R).2- 12

R(ft + ),2

1,1 + "m1)

5E%# dS 3 "t1)

5:># dS +J0 "0 0#5(312 "12 12#5(K5"-2 12(

# 3J12 "R0#)

K5"(2 12(

#+ ;.( 12

R(3 .(( 12

R(+ 1/./1 12

R(ft5k + ),)

1,) + ),1 + "m1 m)5E%# dS 3 "t1 t)5:># dS + J12 "12 12#5(K5"-2 12(# R .(( 12

R(+ 2

=$ + =C + ).2- 12 R(

51/./1 12 R(

+ 1.-/ k

=E + 12 R =$ R =C + /.2 k



).

1 k5

12

$ * C 1).0 02

m2 "k #

0 0

&his problem is very similar to Problem 1, the main difference being m2 sho'n above.

1,2 + "m1 m25E%# dS 3 "t1 t25:># dS

+ J12 "2 2R- 1).0 0R02 12#5K5"-2 12(# 32 + R(1.)0 12

R(ft + ),2

1,1, 1,), ),1, ),) remaining the same, =$ + =C + (1.)0 12 R(

51/./1 12 R(

+ 1./ k

= + 12 R =$ R =C + .- k

(.

12 12 k 12 k

$ * C122

m2 "k #

0 0

ere, 1,2 + "m1 m25E%# dS 3 "t1 t25:># dS

+ J12 "R122 12#5(K5"-2 12(# 32 + R(.(( 12

R(ft + ),2

1,1, 1,), ),1, ),) remaining the same, =$ + =C + (.(( 12 R(

51/./1 12 R(

+ -./1 k = + 2.0 k

-.

12 k E 12 k

1

$ * C

0 0 Case 2 "=$+ 2# Case 1 "=$+ 1#

-2 0

m2 "k # t2 "k # m1 "k # t1 "k #

ere, 1,2 + "m1m25E%# dS3 "t1t25:>#dS + ) J -2 "R#5(K5"-2 12(# 32 + R -)./ 12

R(ft

)5E%# dS3 "t 5:>#dS + ) J0 "0# 5(3 "R# 5(K5"-2 12 # 32 + 12.) 12 ft

1,1 + "m1 1)

) ) ( –3

=$ + -)./ 12 R(

512.) 12 R(

+ -.2) k, =C + -.2) k, = + 2.; k, =E + 2.; k













































































































































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Structural Engineering II