Chapter 8 Deflections - موقع المهندس حماده شعبان at both ends of the beam since no deflection is possible at B and C. Notice that segment CD remains un-deformed

July 2011

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Chapter 8

Deflections

،روسثس شسؿ صجز٠ قز

ثثد أدج ظجدمز عد٠د ثضقججس

ثلع١ ثروز٠ ضجفز ؾجج ع

٠صدثد ثصدثلز دتس

. أخرج وج عمج

July 2011






Moment Area Method

دئصذننج beamمننجغ deflections ثننـ ٠slopesىنن ثظننض دث ننر ثطس٠مننز ـ٠ؾننجد ثننـ *

:ث طثس ثضج١ز

(.M-diagram)سظ support reactionsقعخ -1

لنسح أشنىجي ضةنز ٠عن فعنجح عنجفز ون نج أإن ( M-diagram)م دضمعن١ -2

.ج centroidىج ثـ

ثنـ slopes دق١نظ ٠ةنس ع١ن إصؾجنجس ثنـ beamنـ elastic curveسظ شى ثنـ -3

deflections عد ثمجغ ثطح فعجدج ٠ع ثظضضجػ ثعاللجس ثغغ١ز.

أ ( concave up)عن شنى ٠beamعن عن عن ثنـ ve moment+٠سثعن أ -4

–ve moment ثـ ع٠ع عbeam ع شى(concave down )

وره صى ل١ز support fixedعد أ ٠deflectionsؤخر ف ثالعضذجز أ ال ٠ؽد -5

slope عدج صعجzero.

:عد مطز عض د ثةس٠ز ثضج١ز slopeقعجح -6

The angle change between points A and B on the deflected structure,

or the slope at point B relative to the slope at point A, is given by the

area under the M/EI diagram between these two points.

:عد مطز عض د ثةس٠ز ثضج١ز deflectionقعجح -7

The deflection of point B on the deflected structure with respect to a

line drawn tangent to point A on the structure is given by the static

moment of the area under the M/EI diagram between points A and B

taken about an axis through point B.

!إج أ ق١ج ف١جصج أ ضسن غ١سج ٠ق١جج ١جدز عج

July 2011






Example 8-1:

Draw the deflected shape of each of the beams shown in the Fig.

Solution:

In Fig. (a), the roller at A allows free rotation with no deflection while the

fixed wall at b prevents both rotation and deflection. The deflected shape is shown by the bold line. In Fig. (b), no rotation or deflection can occur at A

and B. In Fig. (c), the couple moment will rotate end A. This will cause

deflections at both ends of the beam since no deflection is possible at B and C.

Notice that segment CD remains un-deformed (a straight line) since no

internal load acts within it. In Fig. (d), the pin (internal hinge) at B allows free rotation, and so the slope of the deflection curve will suddenly change at

this point while the beam is constrained by its supports. In Fig (e), the

compound beam deflects as shown. The slope abruptly changes on each side of the pin at B. Finally, in Fig. (f), span BC will deflect concave upwards due

to the load. Since the beam is continuous, the end spans will deflect concave

downwards.

الحوافر تخنه لم جواد وأي.. كاللة تصبه لم امسح وأي

July 2011






Example 8-4:

The beam in Fig. (a) is subjected to load P at its end. Determine the

displacement of C. EI is constant.

Solution:

المسان الطويل داللة

عمي اليد القصيرة

July 2011






.غرور إليها والركون قميل، الدنيا في المقام

4-8 غجيصجدع ف

July 2011






yB

θA

PL

4EI

P/2 -

+

+

P/2

Δ θ A/B = θ A – θ B = θ A

= Area under

between A and B

t A/B = t A – t B = t A = Δ B

= Moment of Area under

between A and B and about A

A

و ٠شس ثععجدر أ٠ج

صىنننن نننن ال . . . ذننننخ

.٠ فج زثء ض ذخ

8-1 Determine the equation of the elastic curve for the beams using x coordinate that’s valid for 0 ≤ x ≤ 12. Specify the slope at A and the beam’s maximum

deflection. EI is constant.

July 2011






P

Pb/L Pa/L

Pb/L

X1

V1

M1 = Pbx1/L

P

Pb/L

X2

M1 = Pbx1/L

V2

Applying the boundary conditions:

. Therefore C2 = 0.

.

Applying the continuity conditions:

إ صى ثألفع ث٢ فغك

.أ ثألس عأز لش

8-2 Determine the equations of the elastic curve using the x1 and x2 coordinates. EI

is constant.

July 2011






2-8صجدع ف ظؤثي

ننج ٠مننك أغذننج ةننسر أننجض لنند ال

٠ىجد ٠ض دج ع ثـغالق

Solving Eqs. (5), (6) and (7) simultaneously yields:

Thus,

OR

And

July 2011






P

3P/2 P/2

P/2

X1

Mx1 = -Px1/2

Vx1

P

X2

Vx2

Mx2 = -Px2

Slops and Elastic Curve:

Boundary conditions:

.

إج أ صضعخ ضقص ع ج صقخ،

أ ظنننضؾذس فعنننه عننن فنننخ نننج

ع١صقص

8-3 Determine the equations of the elastic curve for the beam using the x1 and x2

coordinates. Specify the beam’s deflection. EI is constant.

July 2011






Continuity Conditions:

The slope; Substitute the value of C1 into Eq. (1).

The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs. (2) and

(4), respectively.


.لسز أ صى ثألفع ظضصذـ ثألفع

July 2011






P

P

M1(x) = Px1

X1

P

P

M2(x) = Pa

X2

a

P P

P P


Due to symmetry:

أ صذضع ظج أال : ثقعد

ف أ ٠س ش ص آخس

8-4 Determine the equations of the elastic curve for the beam using the x1 and x2

coordinates. Specify the slop at A and maximum deflection. EI is constant.

July 2011






θ

Continuity Conditions:

Substitute C1 into Eq. (5).


فمد ٠ طنب , ال صؾجدي ثألفك

ثجض ف ثضفس٠ك د١ىج

July 2011






If you fail to plan,

you plan to fail

8-5 Determine the elastic curve for the cantilevered beam, which is subjected to

couple moment Mo. Also compute the maximum slope and maximum deflection of

the beam. EI is constant.

θ


The negative sign indicates clockwise rotation.

.

July 2011






M1 = 0

X1

M2 = Mo

X2 Mo

θ

لننجض ثننثعم١ ال ٠قضنن عنن

.. . .(، فدط أ . . ى إذث )

8-6 Determine the maximum deflection of the beam and the slope at A. Use the

method of double integration. EI is constant.

July 2011






C

θA

2Lw

8EI

- +

+

wL/2

wL/2

Δ θ A/C = θ A – θ C = θ A

= Area under

between A and C

t A/C = t A – t C = t A

= Moment of Area under

between A and C and about A

8-7 Specify the slope at point A and the deflection at point C . EI is constant.

A

صضفق إال إذث ونج ص صصنه ثدزثظن

!ف١ج صقخ أش ١ط ج ٠قذ ث٢خس

July 2011






M(x) = Mo

Mo

x

θ θ θ


Due to symmetry. Occurs at

.

صىن دن١ ثجفعز ثقم١م١ز دثةنج نج

.ج لش دفع ج أش لجدز ع فع

8-8 Determine the elastic curve for the simply supported beam, which is subjected to the couple moments Mo. Also, compute the maximum slope and the maximum

deflection of the beam. EI is constant.

July 2011






Mo

Mo / L Mo / L

Mo / L

V(x)

M(x) = Mo(x) / L

(x)


The slope; Substitute the value of C1 into Eq. (1).

The Elastic Curve: Substitute the values of C1, and C2 into Eqs. (2).

.ثؽع ثض١ص عالضه ثعؾز

8-9 Determine the maximum slope and the maximum deflection for the simply

supported beam that is subjected to the couple moments Mo. Use the method of double integration. EI is constant.

July 2011






P

P

L

PL

M

EI

-PL

EI

tB/A

tan A

tan B

θB/A

θ

θ θ θ

θ

, أظننننس٠ز , د١٠ننننز : ننننضى أنننندثفه عننننز

.ز٠جظ١ز, ١ز , ثلضصجد٠ز , ثؽضجع١ز

8-10 Use the method of moment-area theorems and determine the slope and

deflection at B. EI is constant.

July 2011






IE

LP

IE

LP

IE

LP

AA

CandBbetweenIE

MunderArea

IE

LP

IE

LP

IE

LP

IE

LP

AAA

CandAbetweenIE

MunderArea

B

B

ACACA

2

44

8

5

448

-

2

22

2

2

222

21

/

3

3

P

A C

P

X X X y2 y2

– –

P P

2 P 2 P

V

– IE

M

IE

LP

2

IE

LP

2

3

CAA /

CAtA /

elastic curve

A1 A2

A3

A

8-12 The beam is subjected to the two loads. Use the moment-area theorems and

determine the slope and displacement at points A and B. EI is constant.

م ثعننضق١ أ٠نن : ظننأي ثىنن

فننننن أفنننننال : صمننننن١ا فأؽجدننننن

.ثعجؽص

July 2011






IE

PL

L

IE

LPL

IE

LP

LA

LA

BaboutCandBbetweenIE

MofMoment

t

IE

PL

LIE

LPL

IE

LPL

IE

LP

LLA

LAL

A

AaboutCandAbetweenIE

MofMoment

t

CBB

CAA

48

7

4444

23

2

4

16

7

6

5

44

3

438

23

2

2

4

3

23

2

3

22

32

/

3

222

3

21

/


ثطك، ثةس، : "ثذس عالعز

ف وج طم ف ". ثصش

غ١س ذوـس فمد غ ، وج ةس

، ف غ١س ثعضذجز فمد ظ

. وج صض ف غ١س فـىـس فمد

July 2011






P P P

1.5P 1.5P

∆C tA/C θA/C

tan A

tan C

M

EI 3Pa

2EI

3Pa

2EI

2Pa

EI

θ θ

The displacement at C is.

, إ ثر ٠ضصس عن غ١نس لن

.ى ثر ٠ضصس ع فع أل

8-13 The beam is subjected to the loading as shown. Use the moment-area theorems and determine the slope at A and displacement C. Assume the support

at A is a pin and B is a roller. EI is constant.

July 2011






0.5(P – F)

F

1.5(P + F) tC/A

tan B

tB/A

∆C

1.5 tB/A

M

EI Pa

2EI

- Fa

EI

Require

, ثغ ف ثغسدنز غن

.ثفمس ف ثغ غسدز

8-14 The beam is subjected to the load P as shown. Use the moment-area theorems and determine the magnitude of force F that must be applied at the end

of overhang C so that the displacement at C is zero. EI is constant.

July 2011






P P

P P

V

M

Pa

t B/A

t B/A Area under

between A and B

(between A and B around B)

8-15 The beam is subjected to the load P as shown. If F = P, determine the displacement at D. Use the moment-area theorems. EI is constant.

و ضعرح عذمج لذ ثص جذ ثمنسثزثس

فنننننج نننننن أ ىنننننن طتنننننن١

July 2011






60 kN

60 kN

108 kN.m

60 kN

-108

EI

(V) kN

(M) kN.m A

∆B

∆`

θB area under

between A and B

A1 [

2.43 * 10 -3

rad

ΔB moment of area under

between A and B about B

Δ/ θB * 1.8

ΔA= ΔB + Δ

/

OR

ΔA A1 * (

* 1.8 + 1.8)

[[

8-16 Determine the slope at B and the maximum deflection of the beam. Take E = 200 GPA, I = 200(10

6) mm

4. Use the moment-area theorems.

.ثصد٠ك لش ثع١ك

July 2011






P

P P P

∆E

∆C

P

P

-P

V

M

EI

-Pa

EI

ΔE = ΔC

ΔE = moment of area under

between A and C about A

=

* (

) * (

)

=

( )

2

2ΔC = moment of area under

between A and D about D

=

* (

) * ((

) + a) +

*

* a *(

a )

=

( )

2 +

( ) +

ΔE = ΔC

( )

2 =

( )

2

+

( ) +

دشطخ

ف ثعجدز ٠ضؼ أ

a = 0.125 L دعع ومد٠ ثع صس١*** دعط ثسفجق غ ثضجػ صذع

8-17 At what distance a should the bearing supports at A and B be placed so that

the deflection at its ends? Use the moment-area theorems. The bearings exert

only vertical reactions on shaft. EI is constant.

July 2011






Mo

A + b

Mo

A + b Mo

tA/B tC/A

tan A

tan B

tan C

θA

M

EI

Mo b

EI (a +b)

-Mo a

EI (a +b)

نننن ٠ضلننننف عنننن ثضفى١ننننس

.٠ىه أ صعد ثألثس

8-18 The beam is subjected to the loading shown. Use the moment-area theorems

and determine the slope at B and deflection at C. EI is constant.

θ

The deflection:

July 2011






Mo

Mo

Mo

a Mo

a

t B/A

tan A

tan B

tan C

t C/A

(0.5) t B/A

M

EI

-Mo

EI

-Mo

EI

θ

The slope at A is.

The displacement at C is.

.ثألظد ال ٠أو أظدث

8-19 The shaft is subjected to the loading shown. If the bearings at A and B only exert vertical reactions on the shaft, determine the slope at A and displacement at

C. Use the moment-area theorems. EI is constant.

July 2011






Mo = Pa

P

P

3Pa

tan A

tan B

tan C

θC/A ∆C

M

EI

-Pa

EI -2Pa

EI -3Pa

EI

θ θ

.٠غ١خ عمه إذث دجغش ف عجغفضه غجذج

8-20 Use the moment-area theorems and determine the slope at B and the deflection at C. EI is constant.

July 2011






IE

lw

lIE

lw

BaboutBandA

betweenIE

MunderAreaofMoment

tttt

IE

lw

IE

lwl

BandAbetweenIE

MunderArea

B

BABAB

BABBA

8

4

3

6

6

23

1

4

3

/

3

2

/

B

Use the moment–area method to find the slope and deflection @ B.

فدج ظذخ ثؾجؿ، فى ١عش ثـىججس

!ثةس فه! جؽـ دئىججس ظع١فز

July 2011






∆C

∆/

t C/A

1.33 kN 1.33 kN

V

-1.33 kN

-8

EI

M

EI

A2 A1

tB/A = moment of area under

between A and B about B

= A1 * 2

=

tC/A = moment of area under

between A and C about C

= (A1 * 5) + (A2 * 1.5)

=

=

=

= tC/A – =

θA =

=

θB/A = θB – (– θA) = A1

θB = A1 – θA =

θC/A = θC – (-θA) = A1 +A2

θC = A1 + A2 – θA =

A1 =

*

* 6

=

@ 2m from B

A2 = *

=

@ 1.5m from B

.غجذج ٠ى ثغسز أشد ثألسثض فضىج دجسء

8-21 Use the moment-area theorems and determine the deflection at C and the slope at A, B and C. EI is constant.

July 2011






Mo = Pa

P

P

3Pa

tan A

tan C

θC/A

M

EI

-Pa

EI -2Pa

EI -3Pa

EI

t B/A

θ

θ θ

θ

8-22 Use the moment-area theorems and determine the slope at C and the

deflection at B. EI is constant.

وغ١سث ج ٠قج ث ١جي إ عث ١ز، ى ال

.٠ىج غ١س ث ١جي أ رخ إ أ ىج

.

July 2011






P

P/2 P/2

M

EI PL

4EI

P

P(a + L)

L

P (a)

L M

EI

-Pa

2EI

-Pa

EI

θ θ

θ

Require,

θ θ θ

ننج ثقننع إال فسصننز صمننس

.فأظس إ صمفج, دجده

8-23 Use the moment-area theorems and determine the value of a so that the

slope at A is equal to zero. EI is constant.

July 2011






wa

wa

3wa2

2

tan A

t B/A tan C

tan B

θC/A

M

EI

- wa2

2EI

- 3wa2

2EI

θ θ

The displacement at B is

.ثد ع ثعىس خ١س ثد ع ثمي

8-24 Use the moment-area theorems and determine the slope at C and

displacement at B. EI is constant.

July 2011






+

8-29 Use the moment-area theorems and determine the slope at B and

displacement at C. The member is an A-36 steel structural tee for which I =

32(106) mm

4.

إ ؽصة١ loadضع١ ف ثعؤثي ٠ى صمع١ ثـ

وج دجسظ ف و ؽصء ع فدر ع ؽعج

mm

IEIEIE

AaboutCandAbetweenIE

MofMoment

tttt

rad

IEIE

IEIE

AA

CandBbetweenIE

MunderArea

C

C

CACACA

BCBCB

C

11.1

)1032()10200(12

1085

12

85

3

25

8

56

00172.0

)1032()10200(

1011

56

110

2

11

9

3

2

0

69

3

/

69

3

21

/

1m

+

20 kN

-

1m X X X

18 kN/m

BCAtC /

18 kN/m

18 18

18

18 18

V (kN)

9 A1

+

10

20

10

+

-

10

10

V (kN)

10 A2

M (kN.m)

M (kN.m)

ثن ؽندث فن ظنفسج

.زوح ثمطجز ثصق١ـ

July 2011






Mo

Mo

L

Mo

L

M

EI Mo

2EI

-Mo

2EI

θ

θ

.ثدثز ثىال صأص , ثـصغجء صأص ثقىز

8-26 Use the moment-area theorems and determine the displacement at B and the

slope at A. EI is constant.

July 2011






m

IE

IEIE

IE

IEIE

CaboutCandA

betweenIE

MofMomentt

IEIE

BaboutBandA

betweenIE

MofMomentt

t

t

t

C

AC

AB

AB

AB

ACC

075.0

)10500()10200(

10500,7

500,7

000,55.1

000,15

000,15

53

27505

3

101500

000,5

3

10500,1

5.1

1015

69

3

/

/

/

/

/

60 kN

A C

90 10

60 60

V (kN)

IE

A1500

1

IE

A750

2

300

AB

t/

C

+

-

5

30

30

ACt

/

).( mkNM

B

30

8-27 Use the moment-area theorems and determine the displacement at C. Take E = 200 GPa. I = 500(10

6) mm

4.

( ال)نننن ٠عطننننه، ( ال: )صعنننن لنننني

طننثزا ( ال) ز،ـمجغعننجس ثجصف١نن

.جــــــ ــــدثع الـــضـث

July 2011






Conjugate Beam Method

دئصذج beamsمجغ deflectionsثـ ٠slopesى ثظض دث ر ثطس٠مز ـ٠ؾجد ثـ *

:ث طثس ثضج١ز

.beamـ (M-diagram)سظ support reactionsقعخ -1

ثألصننن ذنننه دضق٠ننن وننن beamثىجفتنننز نننـ conjugate beamسظننن ثنننـ -2

support إ ج ٠ىجفتج فconjugate beam فعخ ثؾدي ثضج:

ف حثقع Mدأفجي عج٠ز ـ فعخ conjugate beamم دضق١ ثـ -3

(.1)ث طر زل

عىنط slope ٠ىن Vعند أ مطنز أخنر مطنع عندج قعنخ slopeـ٠ؾجد ثـ -4

.ؽخ ثعىط دجعىط Vعمجزح ثعجعز إذث وج

٠ى ألع إذث وج Mعد أ مطز أخر مطع عدج قعخ deflectionـ٠ؾجد -5

M ؽخ ألظف إذث وجM ظجخ.

EI

M

غذمنننننننج منننننننث١ ثننننننند٠ج١ىج

ثثة١ننز فننئ ثقننز ثطجننز

ثعننن ز ٠عنننضق١ أ صط١نننس،

ىج ع عد عسفضنج دنر

صط١س ع ذه ثمث١ فئج

July 2011






Example 8-13:

Determine the maximum deflection of the steel beam shown in Fig.(a).

The reactions have been computed. E = 200 GPa, I = 60 (10)6 mm

4.

وس )لجي ع د ثد غجخ

إج شد ثجض (: "هللا ؽ

ف غخ ثع ج ٠س لز

ــ دج عم ـ ".ثضفج عـ

July 2011






Example 8-15:

: عس د ث طجح زظ هللا ع لجي

" وض ظس وج ث ١جز ف ٠د"

July 2011






15-8صجدع ف غجي

لد ٠س ثجض ثؾسؿ ثنر

ف زأظه ىن ال ٠قعن

!!دجأل ثر صعج١

July 2011






ال ٠ى أ ٠غك دفع

ال ٠دزن ع فع

.إال ع١دج أخطجةج

D = 26 mmΔ

EL

60

Use the conjugate beam method to determine the indicated displacement

quantities for each of the indicated structures and loadings. Vertical deflection at

point D. E = 200 Gpa; I = 50(106) mm

4.

:getting reactions due to real loads

:ذ١ وج conjugate beam سظ ع

July 2011






P

P

PL

PL

EI

V`

M`

θ

صنثح فجؽضه أل صى ع

.دالز ع عمه ثسثؽـ

8-28 Use the conjugate beam method and determine the slope and displacement

at A. EI is constant.

July 2011






Mo

Mo(L)

EI

Mo

L

Mo

EI

VC`

MC`

By`

(L)

2

(L)

2

Ay` Mo(L)

2EI

(L)

3

(2L)

3

By`

θ

Segment AB

Segment BC

.ف١ظ ٠غ١خ ثس ٠غد ثألزخ ىج

8-29 Use the conjugate beam method and determine the displacement at C and the slope at A, B, and C. EI is constant.

July 2011






P

2

P

2 P

P

P

P(a)

EI P(a)

EI

3P(a)

2EI

7P(a2)

4EI 7P(a

2)

4EI

7P(a2)

4EI

VC`

MC`

P(a2)

4EI P(a2)

EI P(a

2)

2EI

θ

أظ أ ثة جس فجظدث : لجي ثضذ

. دجس ف عجة ٠ضمخ ♣♣♣♣♣

8-30 Use the conjugate beam method and determine the slope at B and the

displacement at C. EI is constant.

July 2011






4 kN

8 kN

2 kN

10 kN

2.0

EI

2.0

EI

VC`

MC`

18

EI

By`

2 m 1 m

By` Ay`

36

EI

36

EI

3 m 3 m

1 m

θ

.غ١خص ذس أفدث أه و غ١ذج ال

8-31 Use the conjugate beam method and determine the slope and the displacement at the end C of the beam. E = 200 GPa, I = 70(10

6) mm

4.

July 2011






:for real beam

غذمج V ،Mزظ reactionsص فعجح

4طسق ثعجد٠ز ثشسفز ف شجدضس

:for conjugate beam

٠عج loadإ Mص صق٠ IE

Mفعجح

reactions ـconjugate beam ثىجفتز

.ع ثسظ ذجشسر

IE

mkNM

IE

kNV

CC

CC

).(500,7

)(750,1

60 kN

+

-

A

90 kN

B C

60 kN 60 kN

V (kN)

- -

IE

500,1

30 kN 30 kN

).( mkNM

30 kN

IE

750

IE

500,7

IE

500

IE

750,1

30 kN

conjugate beam

8-32 Use the conjugate beam method and determine the slope and the

displacement at C of the beam. Assume A is a pin and B is a roller. EI is constant.

٠ىننه صننف١س ظننجعز ١٠ننج نن لننش

. . . غعجننه والننه أشنن١جء أخننس

!ظضصع دج ثعؾصثس

July 2011






:real beam

ـ V ،Mزظ reactionsجح ـــ فعـص

real beam دجطسق ثعجد٠ز ثشسفز

4دشجدضس

:conjugate beam

٠عج loadإ Mص صق٠ IE

Mفعجح

reactions ـconjugate beam ثىجفتز.

30 kN . m 27 kN

105 kN . m

X X X 27 kN

5 m 5 m

+

27 kN 27 kN

V (kN)

+ +

-

30 kN . m 30 kN . m 3.89 m

X X

X X

1.11

105 kN . m

IE

67.16

IE

150

IE

204

IE

313,1

IE

33.37X 2.5 m 2.5 m X X X

1.296

0.37

3.33

M (kN . m)

mm

IE

mkNM

rad

IE

kNV

CC

CC

41.16

)10400()10200(

10313,1

).(313,1

10467.0

)10400()10200(

1033.37

)(33.37

69

3

3

69

3

8-33 Use the conjugate beam method and determine the slope and deflection at C. E = 200 GPa. I = 400(10

6) mm

4.

فع١ض ده ،صضؾ إذث ال صعسف إ أ٠

.صس٠دال ثطجف ع ثألزؽـ ف ىج

July 2011






6 kN

6 kN

12 kN

9.0

EI

18.0

EI

18.0

EI

(2)( 9.0)(6)

3EI

18.0

EI

VA`

MA`

3 m

3 m

.صىشف ثألخالق ف ظجعز ثشدر

8-34 Use the conjugate beam method and determine the displacement at A.

Assume B is a roller. E = 200 GPa. I = 80(106) mm

4.

July 2011






3wa2

2

wa

wa

a

2

a

2 a

wa2

2EI 3wa

2

2EI

MB`

MB`

VB`

wa3

6EI wa

3

2EI wa

3

2EI

3 a

4 3 a

4

a

6

VC`

MC`

a

2 a

6

wa3

2EI wa

3

2EI

θ

Thus,

ثزث لذ س ٠س ثؾذج

.(غ ٠جدج ) ص

8-35 Use the conjugate beam method and determine the slope at C and the

displacement at B. EI is constant.

July 2011






wL

wL

2

wL

2

wL2

8EI

wL3

24EI

wL3

24EI

MB`

VB`

wL3

24EI

wL3

24EI

5 L

16

θ

.صسفك دفعه ف فجالس ثقص

8-36 Use the conjugate beam method and determine the displacement at B, the

slope at A. Assume the support at A is a pin, and C is a roller. EI is constant.

July 2011






P P

PL

PL

EI

PL

EI

VD`

MD`

2PL2

3EI

PL2

2EI

2 L

3

2PL2

3EI

PL2

2EI

2PL2

3EI

PL2

2EI L

3

2 L

3

2 L

3

θ

ثثصنز دعنند ثضعغننس ثمندزر عنن

. ثض صقدد ثؾجؿ عد

8-37 Use the conjugate beam method and determine the displacement at D, the slope at C. Assume A is a fixed support, and C is a roller. EI is constant.

July 2011






ال صضلع أه

صضم عال

ج ؾسد أه

سرفشش د

8-38 Use the conjugate beam method and determine the slope just to the left and

just to the right of the pin at B. Also, determine the deflection at D. Assume the

beam is a fixed support at A, and that C is a roller. EI is constant.

:beam real

Reactions of conjugate beam:

:slope @ left of B

July 2011






8.38ثؽجدز ثعؤثي صجدع

ال صشننغ دجننه إال دننج صعنننضط١ع،

ثفعنن ننج صعننضط١ع ث٢، عنندج

صعضط١ع أوغس ف١ج ثفع ثص٠د

[[

:slope @ right of B

:@ D deflection

July 2011






-158.7

EI

24

EI

12 kN 24 kN

4 kN/m

4 m 4 m 2 m 3 m

A E B C

D

12 kN

24 kN

4 kN

28 kN

(V) kN

1 m

ME

Ey A6 A5

74

EI 72

EI

Ay

A4

A3

A2

A1

M

EI

179.9

EI 204

EI

-25.4

EI -266.7

EI

V

For the beam shown, determine the slope at B and E. E = 200 GPa, I = 500(106)

mm4.

م سيم IE

M ل س مساحا لما ن

(A4 + A5 يم ن معا )IE

M D ل B من

لد ال صى صع١عج وج صة

July 2011






For conjugate beam:

.ال صضقد إعجج ١ط د٠ ج ٠ عس

ثعجدمزثصفقز صجدع

July 2011






ثصفقز ثعجدمزصجدع

θ

θ

.عندها internal Mعند نقطه نحسة ∆لحساب

أفعننننن غذ١نننننخ فننننن ثعنننننج ننننن

ثذ١طس ف ال ٠عأي س٠ع عن

!شىث د ٠ىضشــــف ذه دفعـــ

Documents

Chapter 8 Deflections - موقع المهندس حماده شعبان at both ends of the beam since no deflection is possible at B and C. Notice that segment CD remains un-deformed