5-4-Staticallysozen

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5-4 Statically Indeterminate Frame

In Section 1-5, we distinguished between statically determinate and statically indeterminate

trusses: the axial forces in the statically determinate trusses could canbe calculated considering

the equilibrium of forces only,butwhereasthe calculation of the axial forces in the statically

indeterminate trusses requiresd, in addition to statics, considerationing of compatibility of

deformation compatibilitys as well. In this section, we shall define study a statically

indeterminate frame, where for which the calculation of the internal forces axial forces, shear

forces and bending moment! requires consideration of both the force equilibrium of forces and

the deformation compatibility. of deformations. "egrettably, many frames are statically

indeterminate. #e need to learn to determinecalculatetheir response to load.$ompared withto

the calculation process for a determinate structure, that for an indeterminate structure is longer,

yebut straightforward.

"ecall %ig. 1-5-&, where the axial forces in statically indeterminate trusses changed when we

changed the axial stiffness EA for the section the 'oung(s modulus multiplied by the cross-

sectional area!. Similarly, the internal forces in an indeterminate frame changes if we change the

bending stiffnessEIfor the section the 'oung(s modulus multiplied by the moment of inertia!.

In %ig. 5-)-1*d, we obtained a bending moment diagram for a portal frame with all of its

members ha+ing the same stiffness. If we assume that the stiffness of column $ is extremely

large as shown in %ig. 5--1, which should be the correct bending moment diagram among

included in %igs. 5--a through dis the correct one/

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hori2ontal force of a gi+en magnitude, R, at reaction %ig. 5--a!6 and determine the

hori2ontal displacement u" it causes at reaction %ig. 5--c!. he +alue of R is not

important but it may simplify arithmetic if it is chosento be unity.!

Step : he condition we must satisfy is that the hori2ontal displacement at reaction must

be 2ero under the influence of the +ertical loadFand the hori2ontal loadR. #e determine the

hori2ontal reaction at which must be equal to the hori2ontal reaction at ! from

u%7 u" 5..1!

#e shall go through the process described abo+e in detail in the following paragraphs.

Step 1: #ith the hori2ontal restraint released at support 8, the portal frame has no hori2ontal

reaction. s we ha+e obser+ed earlier Section 5-)!, the beam responds as a simply supported

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beam. 4i+en that the slope is 2ero at mid-span, the rotation at each end is

19B C

FL

EI = = 5..!

he columns are not subected to bending moment. hey do not bend. he reforen, the hori2ontal

deflections contributionsed ofby the column rotationss to the hori2ontal deflections are

left Bu L= and right Cu L= because the column height is equal to the beam spanL!. he totaldisplacement is

)

% left right;

B C

FL

u u u L LEI

= + = + = 5..)!

Step : #e apply a hori2ontal forceRat support 8. he resulting bending-moment distribution is

shown in %ig. 5--b. #e note that the bending deformation of the column on the right with an

extremely large bending stiffnessEI! is negligible. he slopes at the two ends of the beam are

the same

B CRL

EI = = 5..!

hereforen, the total hori2ontal displacement is the sum of the contribution of the column on the

left

) )

left

5

) 9B

RL RLu L

EI EI= + = 5..5!

and that of the +ery stiff column on the right

)

right

CRLu LEI

= = 5..9!

resulting in

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)

" left right

)

RLu u u

EI= + = 5..0!

o get the proper hori2ontal reaction at 8, we equate the determined displacements

) )

; )

FL RL

EI EI= 5..;!

leading to

FR).

)= 5..&!

3ow we ha+e the hori2ontal reaction for the indeterminate frame in terms of the force %.

8ecause we


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In this case, the displacement of the roller caused by the hori2ontal force R%ig. 5--a! is:

)

RD B C

RLu L L

EI = + = 5..11!

o get the proper hori2ontal reaction at , we equate this displacement with that caused by the

+ertical force Dq. 5--)!.

) )

;

RL FL

EI EI= 5..1!

which leads to

FR;

1= 5..1)!

herefore, the bending moment is as shown in %ig. 5--c. he deformed deflected shape is

shown in %ig. 5--5. 3ote that the columns do not deformdeflect. he eflected

shapedeformationand thebending moment of the beam are, therefore, the same as those of a

beam of which ends are with both ends fixed as shown in %ig. 5--9.

%ig. 5--5 wo rigid columns %ig. 5--9 8eam of which ends are fixed

Example 5-4-1.$onstruct the bending moment diagram of the structure shown in %ig. 5--0a and calculate the hori2ontal displacement of the beam, ux.

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%ig. 5--0 symmetric frame with a rigid beam

Solution: 8ecause the beam is rigid and does not deform, the top of each column does cannot

rotate 7 *! as shown in %ig. 5--0b!. herefore, the columns are deformed as if they are

cantile+er beams withof length L with their free ends at the supports. 8ecause the bending

stiffness EI! and the lateral displacement ux! of the two columns are the same, the shear force

in each column is also the same. hus, w#e conclude that the shear force in each column isFC

and the bending moment diagram is as shown in %ig. 5--0c. 3ote that the deflection of a

cantile+er beam with a load ofFC is:

EI

FL

EI

LFux 9) !.C

))

== 5..1)!

his is the lateral displacement of the beam shown in %ig. 5--0b. In 4='->, modify 'oung(s

modulus of the columns to 10; 3Cmmso thatEI7 1*93.mmand ma


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%ig. 5--; n asymmetric unsymmetrical frame with a rigid beam

Solution: 8ecause the beam is rigid, the columns may be treated as cantile+er beamswith theirfixed ends at the beam le+el. EetR1andRdenote the reactions of the supports as shown in %ig.

5--;b and note that the deformation ux! of each column is the same. hat leads us to:

!.))

).

)1

EI

LR

EI

LRu

x == 5..1!

or 1. .RR = . 3oting that .1 RRF += , we obtain )C1 FR = and )C.. FR = . he shear force in

the right column is twice that in the left column because the bending stiffness of the right

column is twice that of the left column. If we substitute )C1 FR = into Dq. 5..1!, we get:

EI

FL

EI

LFu

x&)

!)C ))== 5..15!

'ou should chec< this result using 4='->.

Example 5-4-3. $onstruct the bending moment diagram of the structure with distributedload shown in %ig. 5--&.

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%ig. 5--& frame with distributed load

Solution: #e can sol+e this example replacing one of the pin supports by a roller support as we

did before. Fowe+er, we note that the columns do not deform. hat obser+ation leads to a faster

solution. 8ecause the ends of the beam do not rotate, the beam may be considered as a beam

with fixed ends %ig. 5--1*a!.

%ig. 5--1* Solution

3ote that the +ertical reaction at each support is wLC because of symmetry. Integrating dV/dx7

w, we obtain %ig. 5--1*b or:

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agreement is also obser+ed for the case of a concentrated load. he bending moment in an

equi+alent simple-beam is called Astatic moment.B%igure 5--11a shows a frame subected to a

uniform load of wand a hori2ontal load ofFH. If you call the positi+e moment at mid-spanM1

andthe two negati+e moments at the ends of the beamM*andM, you will find

* 1

;

M M wLM

+ = or

*

1 ;

M M wLM

++ =

because the moment distribution in %ig. 5--11a can be decomposed into those in %igs. 5--11b

and c. %or a beam with a concentrated load at the middle%ig. 5--1!

* 1

M M FL

M

+

= or * 1

M M FL

M

+

+ =

he left-hand term, for any loading distribution, is equal to the static moment or the moment at

mid-span of a simply supported beam for the same loading distribution. his result is useful to

remember because it can be used to chec< the reliability of solutions for statically indeterminate

beams.

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Exercise 1: a


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5-4-Staticallysozen

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