July 2011
Chapter 8
Deflections
،‫ٌز‬ٍٛ‫ٓ ِق‬٠‫صّجز‬ٚ ‫ِروسثس شسؿ‬
ٖ‫ثد أدٔج‬ٌّٛ‫د ِٓ ث‬٠‫ثِضقجٔجس ظجدمز ٌٍعد‬
ٓ٠‫ز‬ٛ‫ٓ ثٌّرو‬١‫لع‬ٌّٛ‫ ث‬ٍٝ‫ِضجفز ِؾجٔج ً ع‬
‫صدثد‬٠ ‫دتس ثٌصدثلز‬
.ِٕٗ ‫عّمج وٍّج أخرٔج‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
Moment Area Method
‫ دئصذننج‬beam ‫ ٌٕمننجغ‬deflections ‫ ثٌننـ‬ٚ slopes ‫ؾننجد ثٌننـ‬٠‫مننز ـ‬٠‫ّىننٓ ثظننض دثَ ٘ننرٖ ثٌطس‬٠ *
: ‫ز‬١ٌ‫ثس ثٌضج‬ٛ‫ثٌ ط‬
.)M-diagram( ُ‫ٔسظ‬ٚ support reactions ‫ ٔقعخ‬-1
‫نج‬ِٕٙ ً‫ً فعنجح ِعنجفز ون‬ٙ‫عن‬٠ ‫ ألنسح أشنىجي ِٕضةّنز‬ٟ‫) إٌن‬M-diagram( ُ١‫َ دضمعن‬ٛ‫ ٔم‬-2
.‫ج‬ٌٙ centroid ‫ِىجْ ثٌـ‬ٚ
‫ثٌنـ‬ٚ slopes ‫نٗ إصؾج٘نجس ثٌنـ‬١ٍ‫نس ع‬ٙ‫ة‬٠ ‫نظ‬١‫ دق‬beam ‫ ٌٍنـ‬elastic curve ‫ ٔسظُ شىً ثٌنـ‬-3
. ‫ز‬١‫ً ِٕٗ ثظضٕضجػ ثٌعاللجس ثٌّغٍغ‬ٙ‫ع‬٠ٚ ‫ج‬ٙ‫ح فعجد‬ٍٛ‫ عٕد ثٌٕمجغ ثٌّط‬deflections
ْ‫أ‬ٚ )concave up( ً‫ شنى‬ٟ‫ عٍن‬beam ‫ ثٌنـ‬ٟ‫ عٕن‬ٟ‫عّنً عٍن‬٠ +ve moment ْ‫ أ‬ٟ‫سثعن‬٠ -4
)concave down( ً‫ شى‬ٍٟ‫ ع‬beam ‫ ثٌـ‬ٟٕ‫ ع‬ٍٟ‫عًّ ع‬٠ –ve moment
‫ّز‬١‫ْ ل‬ٛ‫ورٌه صى‬ٚ fixed support ٞ‫ عٕد أ‬deflections ‫ؽد‬ٛ٠ ‫ ثالعضذجز أٔٗ ال‬ٟ‫ؤخر ف‬٠ -5
.zero ٞٚ‫ عٕد٘ج صعج‬slope
:‫ز‬١ٌ‫ز ثٌضج‬٠‫ عٕد ٔمطز ٔعض دَ ثٌٕةس‬slope ‫ ٌقعجح‬-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.
:‫ز‬١ٌ‫ز ثٌضج‬٠‫ عٕد ٔمطز ٔعض دَ ثٌٕةس‬deflection ‫ٌقعجح‬- 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.
!‫جدز عٕج‬١ٔ ‫ج٘ج‬١‫ق‬٠ ‫سٔج‬١‫ ٔضسن غ‬ٚ‫جصٕج أ‬١‫ج ف‬١‫إِج أْ ٔق‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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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.
‫ وأي جواد لم تخنه الحوافر‬.. ‫وأي حسام لم تصبه كاللة‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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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
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-4 ‫صجدع فً ِغجي‬
.‫ والركون إليها غرور‬،‫المقام في الدنيا قميل‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
Δ θ A/B = θ A – θ B = θ A
= Area under
between A and B
A
θA
yB
t A/B = t A – t B = t A = Δ B
= Moment of Area under
between A and B and about A
P/2
+
PL
4EI
+
‫ّٕج‬٠‫ٕشس ثٌععجدر أ‬٠ ِّٓ ٓ‫و‬
ٓ‫ال صىننننٓ ِّنننن‬ٚ . . . ‫ذ٘ننننخ‬
.‫ ذ٘خ‬ٝ‫زثءٖ ِض‬ٚ ‫ج‬ٙ‫ ٍف‬٠
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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P/2
July 2011
8-2 Determine the equations of the elastic curve using the x1 and x2 coordinates. EI
is constant.
P
Pb/L
Pa/L
V1
M1 = Pbx1/L
X1
Pb/L
P
Applying the boundary conditions:
.
Therefore C2 = 0.
.
M1 = Pbx1/L
X2
V2
Pb/L
Applying the continuity conditions:
‫ْ فغك‬٢‫إْ ٌُ صىٓ ثألفعً ث‬
.‫لش‬ٚ ‫أْ ثألِس ِعأٌز‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-2 ‫صجدع فً ظؤثي‬
Solving Eqs. (5), (6) and (7) simultaneously yields:
Thus,
OR
And
‫مٍننك أغٍذٕننج ٔةننسر أٔننجض لنند ال‬٠ ‫ِننج‬
‫ ثـغالق‬ٍٝ‫ْ دٕج ع‬ّٛ‫ض‬ٙ٠ ْٚ‫ىجد‬٠
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
Slops and Elastic Curve:
P
3P/2
P/2
P
Mx1 = -Px1/2
X1
P/2
Vx1
Vx2
Mx2 = -Px2
X2
Boundary conditions:
.
،‫ ِج صقخ‬ٍٝ‫إِج أْ صضعخ ٌضقصً ع‬
‫ فنننخ ِنننج‬ٝ‫ ظنننضؾذس ٔفعنننه عٍننن‬ٚ‫أ‬
ٗ١ٍ‫صقصً ع‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
Continuity Conditions:
8-3 ‫صجدع فً ظؤثي‬
The slope; Substitute the value of C1 into Eq. (1).
The Elastic Curve: Substitute the values of C 1, C2, C3, and C4 into Eqs. (2) and
(4), respectively.
.ً‫ْ ثألفعً ظضصذـ ثألفع‬ٛ‫لسز أْ صى‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
P
P
P
P
P
M1(x) = Px1
X1
P
P
a
M2(x) = Pa
X2
P
Boundary conditions:
Due to symmetry:
‫ أْ صذضٍع ظّج أِال‬ٛ٘ :‫ثٌقعد‬
‫س ش ص آخس‬ّٛ٠ ْ‫ أ‬ٟ‫ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-4 ‫صجدع فً ظؤثي‬
Continuity Conditions:
Substitute C1 into Eq. (5).
θ
‫ طنب‬٠ ‫ فمد‬,‫ال صؾجدي ثألفّك‬
‫ٕىّج‬١‫ك د‬٠‫ ثٌضفس‬ٟ‫ثٌٕجض ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
Boundary conditions:
θ
The negative sign indicates clockwise rotation.
If you fail to plan,
you plan to fail
.
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-6 Determine the maximum deflection of the beam and the slope at A. Use the
method of double integration. EI is constant.
M1 = 0
X1
M2 = Mo
Mo
X2
θ
ٝ‫ عٍنن‬ٞٛ‫قضنن‬٠ ‫ٓ ال‬١‫ثعم‬ٛ‫ض ثٌنن‬ِٛ‫لننج‬
.). . . ْ‫ فدط أ‬ٌٛ ، . . ‫(ٌىٓ إذث‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-7 Specify the slope at point A and the deflection at point C . EI is constant.
Δ θ A/C = θ A – θ C = θ A
= Area under
between A and C
A
θA
C
t A/C = t A – t C = t A
= Moment of Area under
between A and C and about A
wL/2
+
wL/2
wL2
8EI
+
ٟ‫ق إال إذث ونجْ ص صصنه ثٌدزثظن‬ٛ‫ٌٓ صضف‬
!ْٚ‫خس‬٢‫قذٗ ث‬٠ ‫ط ِج‬١ٌٚ ‫ّج صقخ أٔش‬١‫ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
Mo
M(x) = Mo
x
Boundary conditions:
θ
θ
Due to symmetry.
θ
Occurs at
ٓ١‫ْ دن‬ٛ‫ز دثةّنج ِنج صىن‬١‫م‬١‫ثٌّٕجفعز ثٌقم‬
.ٍٗ‫ فع‬ٍٝ‫ِج أٔش لجدز ع‬ٚ ٍٗ‫ِج لّش دفع‬
.
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-9 Determine the maximum slope and the maximum deflection for the simply
supported beam that is subjected to the couple moments M o. Use the method of
double integration. EI is constant.
Mo
Boundary conditions:
Mo / L
Mo / L
M(x) = Mo(x) / L
V(x)
(x)
Mo / L
The slope; Substitute the value of C1 into Eq. (1).
The Elastic Curve: Substitute the values of C1, and C2 into Eqs. (2).
.‫ عالِضه ثٌّعؾٍز‬ٛ٘ ‫ص‬١ّ‫ثؽعً ثٌض‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-10 Use the method of moment-area theorems and determine the slope and
deflection at B. EI is constant.
θ
θ
θ
θ
P
θ
PL
L
P
tan A
θB/A
M
EI
tan B
-PL
EI
, ‫ز‬٠‫ أظننننس‬, ‫ننننز‬١ٕ٠‫ د‬:‫عننننز‬ِٕٛ ‫ٌننننضىٓ أ٘نننندثفه‬
.‫ز‬١‫جظ‬٠‫ ز‬, ‫ز‬١ِٕٙ , ‫ز‬٠‫ ثلضصجد‬, ‫ز‬١‫ثؽضّجع‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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tB/A
July 2011
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.
 A / C   A - C   A
 A  Area under
M
between A and C
EI
 A1  A2  A3
P
P
C
A
P L2
P L2
P L2



8EI
4EI
4EI
y2
X
y2
X
X
V
5 P L2

8EI
–
–
P
2P
P
 B  Area under
M
between B and C
EI
A1
 B  A2  A
A2
P L2
P L2


4EI
4EI
M
EI
–
3
P L2

2EI
2P
PL
A3
2 EI
3 PL
2 EI
A  A/C
A  tA/C
elastic curve
ٓ‫نن‬٠‫ أ‬:ً١‫ظننأي ثٌّّىننٓم ثٌّعننضق‬
َ‫ أفنننننال‬ٟ‫ فننننن‬:ٗ‫ُا فأؽجدننننن‬١‫صمننننن‬
.‫ثٌعجؽص‬
‫لع‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫دسظجٌز‬
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July 2011
8-12 ‫صجدع فً ظؤثي‬
 A  tA/C
 Moment of
M
between A and C about A
EI
L
2
3 
 A1      A2   L  
2
3
4 
2
L
L
A3   
 
3
2
2
P L2 L P L2 3
P L2 5

 
 L
 L
8E I 3 4 E I 4
4 EI 6
7 PL3

16 E I

 B  tB / C
 Moment of

M
between B and C about B
EI
L
L 
2
A2     A3    
2
4 
3
P L2 L P L2 L

 

4E I 4 4 E I 4
7 PL3

48 E I

،‫ثٌٕةس‬ٚ ،‫ "ثٌّٕطك‬:‫ثٌذس عالعز‬
ٟ‫ فّٓ وجْ ِٕطمٗ ف‬."‫ثٌصّش‬ٚ
ٖ‫ِٓ وجْ ٔةس‬ٚ ،ٝ‫س ِذ ْوـ ٍس فمد ٌغ‬١‫غ‬
ِٓٚ ،ٝٙ‫س ثعضذجز فمد ظ‬١‫ غ‬ٟ‫ف‬
.ٌٝٙ ‫س فِـىـ ٍس فمد‬١‫ غ‬ٟ‫وجْ صّضٗ ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
θ
θ
P
The displacement at C is.
P
P
tan A
tA/C
1.5P
M
EI
θA/C
∆C
1.5P
tan C
3Pa
2EI
2Pa
EI
3Pa
2EI
, ٞٛ‫نسٖ لن‬١‫ غ‬ٝ‫ٕضصس عٍن‬٠ ٞ‫إْ ثٌر‬
.ٜٛ‫ ٔفعٗ أل‬ٍٝ‫ٕضصس ع‬٠ ٞ‫ٌىٓ ثٌر‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
F
∆C
1.5(P + F) tB/A
0.5(P – F)
M
EI
tan B
tC/A
1.5 tB/A
Pa
2EI
Require
- Fa
EI
ٚ ,ٓ‫غن‬ٚ ‫ ثٌغسدنز‬ٟ‫ ف‬ٕٝ‫ثٌغ‬
.‫غٓ غسدز‬ٌٛ‫ ث‬ٟ‫ثٌفمس ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
P
P
V
P
P
M
t B/A
Pa
Area under between A and B
(between A and B around B)
t B/A
‫وُ ٔضعرح ِعذمج لذً ثص جذ ثٌمنسثزثس‬
ٓ١‫ْ ِ طتنننننن‬ٛ‫فٕنننننج ِننننننٓ أْ ٔىنننننن‬ٛ ٌ
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-16 Determine the slope at B and the maximum deflection of the beam. Take E
= 200 GPA, I = 200(106) mm4. Use the moment-area theorems.
θB
area under
between A and B
60 kN
108 kN.m
A1
[
60 kN
2.43 * 10 -3 rad
ΔB
moment of area under
between A and B about B
60 kN
Δ/
(V) kN
θB * 1.8
A
(M) kN.m
ΔA= ΔB + Δ/
-108
EI
OR
ΔA
∆B
A1 * ( * 1.8 + 1.8)
∆`
[[
.‫ك‬١‫ثٌع‬
‫لش‬ٚ ‫ك‬٠‫ثٌصد‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
ΔE = ΔC
ΔE = moment of area under
between A and C about A
=
*(
)*(
=
(
)2
)
P
P
∆E
∆C
P
P
2ΔC = moment of area under
between A and D about D
=
*(
) * ((
) + a) + *
* a *( a )
P
=
2
(
) +
(
)+
V
ΔE = ΔC
-P
)2=
(
+
(
(
)2
M
EI
)+
ْ‫ٕضؼ أ‬٠ ‫فً ثٌّعجدٌز‬ٚ
a = 0.125 L
‫دشطخ‬
-Pa
EI
ُٙ١ِ‫ُ ثٌّٕعً صس‬٠‫ُ ومد‬ٙ‫دعع‬ٚ ***ُٙ‫دعط ثٌسفجق ِغً ثٌضّجػ صٍذع‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
Mo
A+b
Mo
Mo
A+b
θ
tan A
The deflection:
θA
tA/B
tC/A
tan C
tan B
M
EI
Mo b
EI (a +b)
-Mo a
EI (a +b)
‫ننننس‬١‫لننننف عننننٓ ثٌضفى‬ٛ‫ض‬٠ ٓ‫ِنننن‬
.‫ثس‬ِٛ‫ّىٕه أْ صعدٖ ِٓ ثأل‬٠
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
Mo
Mo
Mo
a
Mo
a
The slope at A is.
θ
tan A
t C/A
The displacement at C is.
tan C
t B/A
(0.5) t B/A
tan B
M
EI
-Mo
EI
-Mo
EI
.‫أوً أظدث‬٠ ‫ثألظد ال‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-20 Use the moment-area theorems and determine the slope at B and the
deflection at C. EI is constant.
θ
θ
3Pa
P
Mo = Pa
P
∆C
tan A
θC/A
tan C
tan B
M
EI
-Pa
EI
-3Pa
EI
-2Pa
EI
.‫خ عمٍه‬١‫غ‬٠ ً‫ عجغفضه غجٌذج‬ٟ‫إذث دجٌغش ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
Use the moment–area method to find the slope and deflection @ B.
 A/ B
 B   A
 B

M
between A and B
EI
Area under
1  w l2 

l  

3  2 E I 

tB / A
B
3
wl
6EI
B
 tB  t A  tB
 Moment of Area under
M
between
EI
A and B about B

w l3
3
 l
6EI
4
w l4

8EI

ُ‫ فى‬،‫فد٘ج ظذخ ثٌٕؾجؿ‬ٚ ‫عش ثـِىجٔجس‬١ٌ
!‫ٌه‬ٛ‫ثٔةس ف‬ٚ !‫فز‬١‫ِٓ ٔجؽـ دئِىجٔجس ظع‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-21 Use the moment-area theorems and determine the deflection at C and the
slope at A, B and C. EI is constant.
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)
=
∆/
t C/A
∆C
1.33 kN
1.33 kN
=
V
=
= tC/A –
=
-1.33 kN
A1
θA =
A2
=
θB/A = θB – (– θA) = A1
θB = A1 – θA =
M
EI
-8
EI
θC/A = θC – (-θA) = A1 +A2
θC = A1 + A2 – θA =
A1 =
=
*
*6
@ 2m from B
A2 = *
=
@ 1.5m from B
.‫ز أشد ثألِسثض فضىج دجٌّسء‬ٚ‫ْ ثٌغس‬ٛ‫ى‬٠ ‫غجٌذج‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-22 Use the moment-area theorems and determine the slope at C and the
deflection at B. EI is constant.
θ
θ
θ
θ
3Pa
P
Mo = Pa
P
θC/A
tan A
tan C
M
EI
-Pa
EI
-3Pa
EI
-2Pa
EI
‫ ٌىٓ ال‬،‫ز‬١ّ٘ٚ ٌُ‫ث‬ٛ‫ ع‬ٌٝ‫جي إ‬١ ٌ‫قٍّٕج ث‬٠ ‫سث ِج‬١‫وغ‬
.ْ‫ ِىج‬ٞ‫ أ‬ٌٝ‫جي أْ ٔر٘خ إ‬١ ٌ‫س ث‬١‫ّىٕٕج ِٓ غ‬٠
.
‫لع‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫دسظجٌز‬
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t B/A
July 2011
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.
θ
θ
P
θ
P/2
Require,
θ
θ
θ
M
EI
P/2
PL
4EI
P
P(a + L)
L
P (a)
L
M
EI
-Pa
2EI
-Pa
EI
‫ِننج ثٌقننع إال فسصننز صمننس‬
.‫ج‬ٙ‫ صٍمف‬ٌٝ‫ فأظس إ‬, ‫دجده‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-24 Use the moment-area theorems and determine the slope at C and
displacement at B. EI is constant.
θ
θ
The displacement at B is
3wa2
2
wa
wa
tan A
tan C
θC/A
t B/A
tan B
M
EI
- wa2
2EI
- 3wa2
2EI
.‫ي‬ٛ‫ ثٌم‬ٍٝ‫س ِٓ ثٌٕدَ ع‬١‫س خ‬ٛ‫ ثٌعى‬ٍٝ‫ثٌٕدَ ع‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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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) mm4.
ٓ١‫ ؽصة‬ٌٝ‫ إ‬load ‫ُ ثٌـ‬١‫ّىٓ صمع‬٠ ‫ً فً ثٌعؤثي‬١ٙ‫ٌضع‬
‫ّج‬ٙ‫ فدر عُ ؽّع‬ٍٝ‫فً وً ؽصء ع‬ٚ ُ‫وّج دجٌسظ‬
C  0
  B / C   B  C   B
20 kN
M
 Area under
between B and C
EI
 A1  A2
 9
2

 

3
E
I

6
5


EI
EI



1  10

1 
 1
2EI


18 kN/m
X 1m
C  t
X
B
18 kN/m
18
( 200  109 )  (32  10 6 )
18
18
+
 0.00172 rad
-
‫ ظنفسٔج‬ٟ‫نُ ؽندث فن‬ٌّٙ‫ِٓ ث‬
.‫ـ‬١‫ح ثٌمطجز ثٌصق‬ٛ‫زو‬
9
C 
M
between A and C about A
EI
6
5
5
2
85


 
EI
8
EI
3 12 E I
V (kN)
18
M (kN.m)
20
10
10
10
+
85  103
6
12  ( 200  10 )  (32  10 )
9
A1
+
t A / C  t A  tC  t A   C

X
A/C
11  103
 C  Moment of
1m
10
10
V (kN)
A2
M (kN.m)
 1.11 mm
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
+
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July 2011
8-26 Use the moment-area theorems and determine the displacement at B and the
slope at A. EI is constant.
Mo
Mo
L
M
EI
θ
Mo
L
Mo
2EI
-Mo
2EI
θ
.‫ ثٌٕدثِز‬ٟ‫ِٓ ثٌىالَ صأص‬ٚ ,‫ ثٌقىّز‬ٟ‫ِٓ ثـصغجء صأص‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-27 Use the moment-area theorems and determine the displacement at C. Take E
= 200 GPa. I = 500(106) mm4.
 C  tC / A   

t
 B/ A
15
10
   1 .5 t B / A
tB / A
M
 Moment of
between
EI
A and B about B

tC / A
1,500
10
5,000


EI
3
EI
M
 Moment of
between
EI
60 kN
A
B
10
30
C
90
5
60
60
+
30
V (kN)
-
30
A1 
1500
EI
A2 
750
EI
M ( kN. m)
A and C about C
1500  10
 750  2



 5 
   5
EI  3
 EI 3

15,000

EI
300
tB / A

tC / A
C
15,000
5,000
 1 .5 
EI
EI
7,500

EI
C 
7,500  103

( 200  109 )  (500  10 6 )
 0.075 m
)‫ (ال‬،‫عطٍننننه‬٠ ٓ‫ (ال) ٌّنننن‬:‫ي‬ٛ‫صعٍننننُ لنننن‬
‫ثزا‬ٛ‫ (ال) ٌٍطنن‬،‫نـز‬١‫جصف‬ٌٙ‫ٌٍّمجغعننجس ث‬
.‫ـــج‬ٙ‫ ٌـــ‬ٟ‫ ال دثعــــ‬ٟ‫ثٌـضـــ‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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‫‪July 2011‬‬
‫‪Conjugate Beam Method‬‬
‫* ‪ّ٠‬ىٓ ثظض دثَ ٘رٖ ثٌطس‪٠‬مز ـ‪٠‬ؾجد ثٌـ ‪ٚ slopes‬ثٌـ ‪ٌٕ deflections‬مجغ ‪ beams‬دئصذج‬
‫ثٌ ط‪ٛ‬ثس ثٌضجٌ‪١‬ز ‪:‬‬
‫‪ٔ- 1‬قعخ ‪ٔٚ support reactions‬سظُ (‪ٌٍ )M-diagram‬ـ ‪.beam‬‬
‫‪ٔ - 2‬سظنننُ ثٌنننـ ‪ conjugate beam‬ثٌّىجفتنننز ٌٍنننـ ‪ beam‬ثألصنننٍ‪ٚ ٟ‬ذٌنننه دضق‪٠ٛ‬نننً ونننً‬
‫‪ support‬إٌ‪ِ ٟ‬ج ‪٠‬ىجفت‪ٙ‬ج ف‪ conjugate beam ٟ‬فعخ ثٌؾد‪ٚ‬ي ثٌضجٌ‪:ٟ‬‬
‫غذمنننننننج ٌمننننننن‪ٛ‬ثٔ‪ ٓ١‬ثٌننننننند‪ٕ٠‬جِ‪١‬ىج‬
‫ثٌ‪ٛٙ‬ثة‪١‬ننز فننئْ ثٌٕقٍننز ثٌطٕجٔننز‬
‫ثٌعننن ّز ‪٠‬عنننضق‪ ً١‬أْ صط‪١‬نننس‪،‬‬
‫ٌىٕ‪ٙ‬ج ‪ِٚ‬ع عدَ ِعسفض‪ٙ‬نج د‪ٙ‬نرٖ‬
‫ثٌم‪ٛ‬ثٔ‪ ٓ١‬فئٔ‪ٙ‬ج ِع ذٌه صط‪١‬س‬
‫‪M‬‬
‫‪ٔ- 3‬م‪ َٛ‬دضقّ‪ ً١‬ثٌـ ‪ conjugate beam‬دأفّجي ِعج‪٠ٚ‬ز ٌـ‬
‫‪EI‬‬
‫ثٌ ط‪ٛ‬ر زلُ (‪.)1‬‬
‫فعخ ‪ M‬ثٌّقع‪ٛ‬ح ف‪ٟ‬‬
‫‪- 4‬ـ‪٠‬ؾجد ثٌـ ‪ slope‬عٕند أ‪ٔ ٞ‬مطنز ٔأخنر ِمطنع عٕند٘ج ‪ٔٚ‬قعنخ ‪٠ٚ V‬ىن‪ slope ْٛ‬عىنط‬
‫عمجزح ثٌعجعز إذث وجْ ‪ِٛ V‬ؽخ ‪ٚ‬ثٌعىط دجٌعىط‪.‬‬
‫‪- 5‬ـ‪٠‬ؾجد ‪ deflection‬عٕد أ‪ٔ ٞ‬مطز ٔأخر ِمطع عٕد٘ج ‪ٔٚ‬قعخ ‪٠ٚ M‬ى‪ ْٛ‬ألعٍ‪ ٟ‬إذث وجْ‬
‫‪ِٛ M‬ؽخ ‪ٚ‬ألظفً إذث وجْ ‪ M‬ظجٌخ‪.‬‬
‫د‪ٕ٠‬جزثْ ٘د‪٠‬ز عٕـد ثٌضٕذ‪ ٗ١‬عٍ‪ ٝ‬وـً خطـأ دّروسثس ثٌّ‪ٛ‬لع دسظجٌز ‪ SMS‬أ‪ ٚ‬دجٌذس‪٠‬د ثالٌىضس‪ٟٔٚ‬‬
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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 mm4.
َ‫ غجٌخ (وس‬ٟ‫ دٓ ثد‬ٍٟ‫لجي ع‬
‫ "إّٔج ش٘د ثٌٕجض‬:)ٗٙ‫ؽ‬ٚ ‫هللا‬
‫ْ ِٓ لٍز‬ٚ‫س‬٠ ‫ غٍخ ثٌعٍُ ِج‬ٟ‫ف‬
."ُ‫ثٔضفج ِٓ ِعـ ٍْ ِـُ دّج عمـٍِـ‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
Example 8-15:
:ٕٗ‫ هللا ع‬ٟ‫لجي عّس دٓ ثٌ طجح زظ‬
"ٖ‫د‬٠ ٟ‫جز ف‬١ ٌ‫"ِٓ وضُ ظسٖ وجْ ث‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-15 ‫صجدع فً ِغجي‬
ٞ‫ ثٌٕجض ثٌؾسؿ ثٌنر‬ٜ‫س‬٠ ‫لد‬
ْٛ‫قعن‬٠ ‫ُ ال‬ٕٙ‫ زأظه ٌىن‬ٟ‫ف‬
!! ٗ١ٔ‫ صعج‬ٞ‫دجألٌُ ثٌر‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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) mm4.
getting reactions due to real loads:
:ٓ١‫ ِذ‬ٛ٘ ‫ وّج‬conjugate beam ُ‫عُ ٔسظ‬
60
EL
ٗ‫غك دٕفع‬٠ ْ‫ّىٓ أ‬٠ ‫ال‬
ٗ‫دزن عٓ ٔفع‬٠ ‫ِٓ ال‬
.‫ج‬ٙ‫أخطجة‬ٚ ‫ج‬ٙ‫د‬ٛ١‫إال ع‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-28 Use the conjugate beam method and determine the slope and displacement
at A. EI is constant.
θ
P
PL
P
M`
V`
PL
EI
‫ثح‬ٛ‫ صن‬ٍٝ‫ْ ع‬ٛ‫فجؽضه ألْ صى‬
.‫ عمٍه ثٌسثؽـ‬ٍٝ‫دالٌز ع‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-29 Use the conjugate beam method and determine the displacement at C and
the slope at A, B, and C. EI is constant.
Segment AB
Mo
Mo
L
Segment BC
Mo
EI
θ
By`
(L)
2
(L)
2
VC`
Mo(L)
EI
(2L)
3
Ay`
(L)
3
Mo(L)
2EI
By`
.‫ ثألزٔخ ٍِىج‬ٚ‫غد‬٠ ‫خ ثٌّٕس‬١‫غ‬٠ ‫ظ‬١‫ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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M C`
July 2011
8-30 Use the conjugate beam method and determine the slope at B and the
displacement at C. EI is constant.
θ
P
2
P
2
P
P
P
3P(a)
2EI
P(a)
EI
P(a)
EI
7P(a2)
4EI
P(a2)
2EI
2
P(a )
EI
7P(a2)
4EI
P(a2)
4EI
VC`
M C`
7P(a2)
4EI
ً‫ أظٍُ أً٘ ثٌةٍُ ِٓ ِجس فجظدث‬ٚ :ٟ‫لجي ثٌّضٕذ‬
.‫ضمٍخ‬٠ ٗ‫ ٔعّجة‬ٟ‫♣♣♣♣♣ ٌّٓ دجس ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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) mm4.
4 kN
8 kN
2 kN
10 kN
2.0
EI
θ
2.0
EI
By`
1m
2m
VC`
M C`
18
EI
36
EI
3m
3m
1m
Ay`
36
EI
By`
.‫خ‬١‫ال ص ذس أفدثً أٔه غ‬ٚ ً‫ذج‬١‫وٓ غ‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
for real beam:
ً ‫ غذمج‬M ،V ُ‫زظ‬ٚ reactions ‫صُ فعجح‬
4 ‫ شجدضس‬ٟ‫فز ف‬ٚ‫ز ثٌّشس‬٠‫ٌٍطسق ثٌعجد‬
60 kN
for conjugate beam:
A
M
‫فعجح‬ٚ
ٞٚ‫عج‬٠ load ٌٝ‫ إ‬M ً٠ٛ‫صُ صق‬
EI
C
B
30 kN
90 kN
60 kN
60 kN
‫ ثٌّىجفتز‬conjugate beam ‫ ٌٍـ‬reactions
+
.‫ ثٌسظُ ِذجشسر‬ٍٝ‫ع‬
V (kN)
30 kN
30 kN
1,750 (kN )
EI
7,500 (kN . m)
 
EI
C  VC  
C  M C
M ( kN . m )
-
30 kN
1, 500
750
7 , 500
EI
EI
EI

500
EI
conjugate beam
1, 750
EI
‫لننش‬ٚ ٓ‫ننج ِنن‬١ِٛ٠ ‫س ظننجعز‬١‫ف‬ٛ‫ّىٕننه صنن‬٠
. . . ٜ‫جء أخننس‬١‫أشنن‬ٚ ‫والِننه‬ٚ ‫غعجِننه‬
!‫ج ثٌّعؾصثس‬ٙ‫ظضصٕع د‬ٚ
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-33 Use the conjugate beam method and determine the slope and deflection at
C. E = 200 GPa. I = 400(106) mm4.
real beam:
‫ ٌٍـ‬M ،V ُ‫زظ‬ٚ reactions ‫صـُ فعـــجح‬
‫فز‬ٚ‫ز ثٌّشس‬٠‫ دجٌطسق ثٌعجد‬real beam
105 kN . m
4 ‫دشجدضس‬
30 kN . m
27 kN
conjugate beam:
X
M
‫فعجح‬ٚ
ٞٚ‫عج‬٠ load ٌٝ‫ إ‬M ً٠ٛ‫صُ صق‬
EI
5m
X
27 kN
27 kN
.‫ ثٌّىجفتز‬conjugate beam ‫ ٌٍـ‬reactions
5m
27 kN
+
V (kN)
30 kN . m
 C  VC
 
X
37.33 (kN )
 
EI
3.89 m
X
-
30 kN . m
+
+
X
X
1.11
37.33  103
6
(200  10 )  (400  10 )
9
X
M (kN . m)
105 kN . m
 0.467  10 3 rad
16.67
EI
150
EI
1, 313
EI
C  M C  
 
1,313 (kN . m)
EI
 1,313  10
204
1.296
3
6
(200  10 )  (400  10 )
9
 16.41 mm
3.33 X X2.5 m X 2.5 m
XE I
37.33
EI
0.37
‫ ده‬ٟٙ‫ٕض‬١‫ فع‬،ٗ‫ٓ صضؾ‬٠‫ أ‬ٌٝ‫إذث ال صعسف إ‬
.ٖ‫د‬٠‫ ِىجْ ال صس‬ٟ‫ ثألزؽـ ف‬ٍٝ‫ثٌّطجف ع‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-34 Use the conjugate beam method and determine the displacement at A.
Assume B is a roller. E = 200 GPa. I = 80(10 6) mm4.
12 kN
6 kN
6 kN
9.0
EI
(2)( 9.0)(6)
3EI
3m
18.0
EI
MA`
3m
18.0
EI
18.0
EI
VA`
.‫ ظجعز ثٌشدر‬ٟ‫صٕىشف ثألخالق ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
8-35 Use the conjugate beam method and determine the slope at C and the
displacement at B. EI is constant.
wa
3wa2
2
a
2
a
Thus,
a
2
wa
θ
3wa2
2EI
wa2
2EI
a
6
3a
4
3a
4
wa3
6EI
3
3
wa
2EI
a
6
wa
2EI
a
2
VC`
M C`
3
wa3
2EI
wa
2EI
VB`
ً‫س ثٌؾذجْ ِِسثزثً لذ‬ّٛ٠
.) ٟٔ‫جدج‬٠ ً‫صٗ ( ِغ‬ِٛ
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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MB`
July 2011
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.
θ
wL
wL
2
wL
2
wL2
8EI
wL3
24EI
wL3
24EI
wL3
24EI
VB`
wL3
24EI
5L
16
MB`
.ْ‫ فجالس ثٌقص‬ٟ‫صسفك دٕفعه ف‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
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July 2011
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.
θ
P
P
PL
PL
EI
PL
EI
VD`
2PL2
3EI
2L
3
M D`
2
PL
2EI
2L
3
2L
3
PL2
2EI
L
3
2PL2
3EI
2PL2
3EI
PL2
2EI
‫ثصنٍز دعنند ثٌضعغننس‬ٌّٛ‫ ث‬ٟ‫ثٌمندزر عٍن‬
.ِٗ‫ صقدد ثٌٕؾجؿ ِٓ عد‬ٟ‫ ثٌض‬ٟ٘
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
Physics I/II, Circuits, English 123/221, Numerical, AutoCAD, Strength, Statics :‫مواد عامة‬
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July 2011
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.
real beam:
Reactions of conjugate beam:
slope @ left of B:
‫لع أٔه‬ٛ‫ال صض‬
‫ٌٓ صضمٓ عّال‬
‫ِج ٌّؾسد أٔه‬
‫فشٍش دٗ ِسر‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
Physics I/II, Circuits, English 123/221, Numerical, AutoCAD, Strength, Statics :‫مواد عامة‬
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Mechanical Design I/II, Structural Analysis I/II, Concrete I/II, Soil, Dynamics, Fluids, System Dynamics :ُ١ّ‫ثد صص‬ِٛ
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July 2011
slope @ right of B:
8.38 ‫صجدع ثؽجدز ثٌعؤثي‬
[[
deflection @ D:
،‫ع‬١‫ال صشننغً دجٌننه إال دّننج صعنننضط‬
‫عٕنندِج‬ٚ ،ْ٢‫ع ث‬١‫ثفعننً ِننج صعننضط‬
‫د‬٠‫ج ثفعً ثٌّص‬ٕٙ١‫ع أوغس ف‬١‫صعضط‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
Physics I/II, Circuits, English 123/221, Numerical, AutoCAD, Strength, Statics :‫مواد عامة‬
C++, Java, MATLAB, Data Structures, Algorithms, Discrete Math, Digital Logic, Concepts :‫صس‬ٛ١‫ثد وّذ‬ِٛ
Mechanical Design I/II, Structural Analysis I/II, Concrete I/II, Soil, Dynamics, Fluids, System Dynamics :ُ١ّ‫ثد صص‬ِٛ
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July 2011
For the beam shown, determine the slope at B and E. E = 200 GPa, I = 500(10 6)
mm4.
24 kN
12 kN
4 kN/m
A
D
B
3m
E
C
4m
2m
4m
24 kN
4 kN
12 kN
(V) kN
1m
28 kN
A3
‫لما ن‬
‫ل س مساحا‬
M
‫م سيم‬
EI
M
D ‫ ل‬B ‫من‬
‫) يم ن معا‬A4 + A5(
EI
A1
72
EI
74
EI
A4
ME
A2
M
EI
Ay
A5
24 A
6
EI
204
EI
Ey
179.9
EI
V
-266.7
EI
-158.7
EI
-25.4
EI
ٓ‫عج وّج صة‬١‫ْ صع‬ٛ‫لد ال صى‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
Physics I/II, Circuits, English 123/221, Numerical, AutoCAD, Strength, Statics :‫مواد عامة‬
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July 2011
‫صجدع ثٌصفقز ثٌعجدمز‬
For conjugate beam:
.ٖ‫ عس‬٠ ‫ٗ ِج‬٠‫ط ٌد‬١ٌ ً‫ إٔعجٔج‬ٜ‫ال صضقد‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
Physics I/II, Circuits, English 123/221, Numerical, AutoCAD, Strength, Statics :‫مواد عامة‬
C++, Java, MATLAB, Data Structures, Algorithms, Discrete Math, Digital Logic, Concepts :‫صس‬ٛ١‫ثد وّذ‬ِٛ
Mechanical Design I/II, Structural Analysis I/II, Concrete I/II, Soil, Dynamics, Fluids, System Dynamics :ُ١ّ‫ثد صص‬ِٛ
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July 2011
‫صجدع ثٌصفقز ثٌعجدمز‬
θ
θ
.‫ عندها‬internal M ‫لحساب ∆ عند نقطه نحسة‬
ٛ‫ ثٌعنننننجٌُ ٘ننننن‬ٟ‫نننننخ فننننن‬١‫أفعنننننً غذ‬
ٓ‫عٗ عن‬٠‫عأي ِس‬٠ ‫ ال‬ٛٙ‫ ف‬ٞ‫طس‬١‫ثٌذ‬
!ٗ‫ىضشــــف ذٌه دٕفعـــ‬٠ ً‫ثٖ د‬ٛ‫شى‬
ٟٔٚ‫د ثالٌىضس‬٠‫ دجٌذس‬ٚ‫ أ‬SMS ‫لع دسظجٌز‬ٌّٛ‫ وـً خطـأ دّروسثس ث‬ٍٝ‫ٗ ع‬١‫ز عٕـد ثٌضٕذ‬٠‫ٕجزثْ ٘د‬٠‫د‬
Physics I/II, Circuits, English 123/221, Numerical, AutoCAD, Strength, Statics :‫مواد عامة‬
C++, Java, MATLAB, Data Structures, Algorithms, Discrete Math, Digital Logic, Concepts :‫صس‬ٛ١‫ثد وّذ‬ِٛ
Mechanical Design I/II, Structural Analysis I/II, Concrete I/II, Soil, Dynamics, Fluids, System Dynamics :ُ١ّ‫ثد صص‬ِٛ
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