08 2energy deflection
TRANSCRIPT
-
1! External Work and Strain Energy! Principle of Work and Energy! Principle of Virtual Work! Method of Virtual Work:
! Trusses! Beams and Frames
! Castigliano's Theorem! Trusses! Beams and Frames
DEFLECTIONS: ENERGY METHODS
-
2Ue
Eigen work
External Work and Strain Energy
Most energy methods are based on the conservation of energy principle, which statesthat the work done by all the external forces acting on a structure, Ue, is transformed into internal work or strain energy, Ui.
Ue = Ui
L
F
x
F
P
xPF
=
As the magnitude of F is gradually increasedfrom zero to some limiting value F = P, the finalelongation of the bar becomes .
External Work-Force.
Eigen work
FdxdUe =
=x
e FdxU0
=
0
)( dxxPUe
=
=
PxPUe 21)
2(
0
2
-
3P
L
F'
Displacement work
x
F
P Eigen
work
(Ue)Total = (Eigen Work)P + (Eigen Work)F + (Displacement work) P
'
L
F' + P
)'()')('(21))((
21)( ++= PFPU Totale
-
410 mm
L
20 kN
L
x (m)
F
0.01 m
20 kN
mNUe == 100)1020)(01.0(21 3
-
5Displacement work
5 kN
x (m)
F
L
2.5 mm
15 kN
0.0075
Eigen
work
L
15 kN
7.5 mm
L
15 kN
7.5 mm 0.01
20 kN
)1015)(0025.0()105)(0025.0(21)1015)(0075.0(
21 333 ++=W
mN =++= 10050.3725.625.56
-
6 External Work - Moment.
dM
MddUe =
Displacement work
M
M Eigen
work
'
M' + M
=
0
MdUe -----(8-12)
MUe 21
= -----(8-13)
Eigen work
'''21
21)( MMMU Totale ++=
)148()')('(21)( ++= MMU Totale
-
721
=oU
Strain Energy-Axial Force.
L
N
=V
dV ))(21(
=V
dVE
)(21 2
=V
i dVUU 0
=V
dVAN
E2)(
21
=L
AdxAN
E2)(
21
=L
dxEA
N )2
(2
=E
AN
=
-
8 Strain Energy-Bending
21
=oU
x dx
wP
L
=V
dV ))(21(
=V
dVI
MyE
2)(21
M M
dx
d
d=V
dVI
yME
)(21 2
2
2
=L
AdxdAy
IM
E))((
21 2
2
2
=L
dxEIM )
2(
2
=V
i dVUU 0
=V
dVE
)(21 2
IMy
=
I
-
9
=G
dx
cd
J
TT
Strain Energy-Torsion
21
=oU =L
i dxGJTU
2
2
JT =
=V
i dVUU 0
=V
dV)21(
=V
dVG
)(21 2
=V
dVJ
TG
2)(21
=L A
dxdAJT
G))((
21 2
2
2
-
10
VV
dx
dy
AK
Strain Energy-Shear
=G
21
=oU
=V
dV ))(21(
=V
dVG
)(21 2
=V
i dVUU 0
=L A
dxdAIt
QG
V )(2
22
=L
i dxGAVKU2
2
= dVItVQ
G2)(
21
-
11
Principle of Work and EnergyP
L
-PL
M diagram
+ Mx = 0: 0= PxM
PxM =
x
P
xV
M
ie UU =
=L
EIdxMP
0
2
221
=L
EIdxPxP
0
2
2)(
21
L
EIxPP
0
32
621
=
EIPL3
3
=
-
12
dLudVUP +=+ 011 1)21(
Then apply real load P1.
Au
u
L
Principle of Virtual Work
Apply virtual load P' first.
P1
A
P' = 1
1 = u dLReal displacements
Virtual loadings
1 = u dLReal displacements
Virtual loadings
In a similar manner,
u
u
L
dL
ie UU =
1
Real Work
-
13
B
Method of Virtual Work : Truss
External Loading.
N 2
N1
N3 N4
N5N
6
N7 N8 N9
1kN
n 2
n1
n3 n 4 n5
n6
n7 n8 n9
Where:1 = external virtual unit load acting on the truss joint in the stated direction of n = internal virtual normal force in a truss member caused by the external virtual unit load = external joint displacement caused by the real load on the trussN = internal normal force in a truss member caused by the real loadsL = length of a memberA = cross-sectional area of a memberE = modulus of elasticity of a member
P1
P2
B
AEnNL
=1
-
14
Where: = external joint displacement caused by the temperature change = coefficient of thermal expansion of member
T = change in temperature of member
Where: = external joint displacement caused by the fabrication errors
L = difference in length of the member from its intended size as caused by a fabrication error
LTn )(1 =
Ln=1
Temperature
1 = u dL
Fabrication Errors and Camber
1 = u dL
dL
dL
-
15
Example 8-15
The cross-sectional area of each member of the truss shown in the figure isA = 400 mm2 and E = 200 GPa.(a) Determine the vertical displacement of joint C if a 4-kN force is applied to thetruss at C.(b) If no loads act on the truss, what would be the vertical displacement of joint Cif member AB were 5 mm too short?(c) If 4 kN force and fabrication error are both accounted, what would be thevertical displacement of joint C.
A B
C
4 m 4 m
4 kN
3 m
-
16
A B
C4 kN
N(kN)
A B
C
n (kN)
SOLUTION
Virtual Force n. Since the vertical displacement of joint C is to bedetermined, only a vertical 1 kN load is placed at joint C. The n force ineach member is calculated using the method of joint.
1 kN
0.667-0.
833-0.833
2+2
.5 -2.5
1.5 kN1.5 kN
4 kN
0.5 kN0.5 kN
0
Real Force N. The N force in each member is calculated using themethod of joint.
Part (a)
-
17CV = +0.133 mm,
0.667-0.
833-0.833
2+2
.5 -2.5
8
5 5
10.67-10
.41 10.41
A B
C
n (kN)
1 kN
A B
C4 kN
N (kN)
A B
C
L (m)=A B
C
nNL (kN2m)
= AEnNLkN CV ))(1(
)10200)(10400(
67.10)67.1041.1041.10(1
2626
mkNm
mkNAECV
=++=
-
18
Part (b): The member AB were 5 mm too short
5 mm
CV = -3.33 mm,
Part (c): The 4 kN force and fabrication error are both accounted.
CV = 0.133 - 3.33 = -3.20 mm
CV = -3.20 mm,
A B
C
n (kN)
1 kN
0.667-0.
833-0.833
)())(1( LnCV =
)005.0)(667.0( =CV
-
19
Example 8-16
Determine the vertical displacement of joint C of the steel truss shown. Thecross-section area of each member is A = 400 mm2 and E = 200 GPa.
4 m 4 m 4 m
AB C
D
EF
4 m
4 kN4 kN
-
20
4 m 4 m 4 m
AB C
D
EF
4 m
n (kN)
4 m 4 m 4 m
AB C
D
EF
4 m
4 kN4 kNN(kN)
SOLUTION
Virtual Force n. Since the vertical displacement of joint C is to bedetermined, only a vertical 1 kN load is placed at joint C. The n force ineach member is calculated using the method of joint.
Real Force N. The N force in each member is calculated using themethod of joint.
1 kN
0.667-0.
471
-0.47
1
-0.943
0.6670.333
0
.
3
3
3
1
-0.333
4-5.
66 0
-5.66
444 4
-4
0.667 kN0.333 kN
0
4 kN4 kN
0
-
21CV = 1.23 mm,
0.667-0.
471
-0.47
1
-0.943
0.6670.3330
.
3
3
3
1
-0.333
AB C
D
EF
n (kN) 1 kN
4-5.
66 0
-5.66
444 4
-4
AB C
D
EF
4 kN4 kN N(kN)
45.6
6 5.66
5.6644
4 4
4
AB C
D
EF
L(m)
AB C
D
EF
nNL(kN2m)
=
10.6715
.07 0
30.1810.675.33
5
.
3
3
16
5.33
= AEnNLkN CV ))(1(
)10200)(10400(
4.72)]18.3016)67.10(2)33.5(307.15[(1
2626
mkNm
mkNAECV
=++++=
-
22
Example 8-17
Determine the vertical displacement of joint C of the steel truss shown. Due toradiant heating from the wall, members are subjected to a temperature change:member AD is increase +60oC, member DC is increase +40oC and member AC isdecrease -20oC.Also member DC is fabricated 2 mm too short and member AC3 mm too long. Take = 12(10-6) , the cross-section area of each member is A =400 mm2 and E = 200 GPa.
2 mA B
CD
3 m
20 kN
10 kNwall
-
23
2 mAB
CD
3 m
n (kN)
SOLUTION
1 kN
0.667
0
-1.2
01
13.33 kN
23.33 kN
20 kN
23.33
0
-24.04
2020
0.667 kN
0.667 kN
1 kN
Due to loading forces.
CV= 2.44 mm,
2 mAB
CD
3 m
20 kN
10 kN
N (kN)
2
2
3.61
33
AB
CD
L (m)
31.13
0
104.1
2
060
AB
CD
nNL(kN2m)
= AEnNLkN CV ))(1(
)12.10413.3160()200)(400(
1++=CV
-
24
Due to temperature change.
Due to fabrication error.
Total displacement .
1 kN0.667
0
-1.2
01
A B
CD
n (kN)
+40
-20
+60
A B
D
T (oC)
C 2
2
3.61
33
AB
CD
L (m) Fabrication error (mm)
-2
+ 3
A B
D C
LTnkN CV )())(1( =
=++= ,84.3)]61.3)(20)(2.1()2)(40)(667.0()3)(60)(1)[(1012( 6 mmCV
)())(1( LnkN CV =
=+= ,93.4)003.0)(2.1()002.0)(667.0( mmCV
=+= ,35.193.484.344.2)( mmTotalCV
-
25
Method of Virtual Work : Bending
wC
A BC
RBRA
==L
C dxEIMmdm )())((1
Virtual loadings
Real displacements
M M
dx
d
d
dds =
dxEIMdsd =
1
-
26
Method of Virtual Work : Beams and Frames
wC
A BC
RBRA
==L
C dxEIMmdm )())((1
wC
A B
RBRA
C
==L
C dxEIMmdm )())((1
Virtual loadings
Real displacements
Virtual loadings
Real displacements
-
27
Virtual unit load
CA B
wC
A B
Method of Virtual Work : Beams and Frames
C
Real load
1RA RBx1 x2
RBRA
x1 x2
B
RB
x2
v2
m2x1
RA
v1
m1
Vertical Displacement
w
B
x2
RB
V2
M2
x1
RA
V1
M1
=L
C dxEIMm )(1
-
28
Virtual unit couple
CA B
wC
A B
Slope
Real load
x1 x2RBRA
w
B
x2
RB
V2
M2
x1
RA
V1
M1B
RB
x2
v2
m2
RA
v1
m1
RA
x1 x21
C RB
=L
C dxEIMm )(1
-
29
Example 8-18
The beam shown is subjected to a load P at its end. Determine the slope anddisplacement at C. EI is constant.
2a a
AB C
P
C
-
30
Real Moment M
AB
C2a a
P
AB
C2a a
SOLUTION
Virtual Moment m
Displacement at C
1 kN
x1 x2
-a
m
m2 = -x2
x1 x2
-Pa
M
M2 = -Px2
23
21
21
1PxM =
23P
2P
+== a a
LC dxPxxEI
dxPxxEI
dxEI
Mm 2
0 02221
11 ))((1)2
)(2
(11
21
1xm =
=+=+=EI
PaEI
PaEI
PaEI
PxEI
Pxaa
aC 3312
8)
3()
12(
333
0
32
231
-
31
AB
C2a a
P
AB
C2a a
Virtual Moment m Real Moment M
Slope at C
x1 x2
-1
m
x1 x2
-Pa
M
M2 = -Px2
1 kNm
a21
a21
21
1PxM =
23P
2P
M2 = -Px2axm2
11 = a
xm2
11 =
12 =m 12 =m 21
1PxM =
+==aaL
C dxPxEIdxPx
ax
EIdx
EIMmmkN
0221
12
0
1
0
))(1(1)2
()2
(1))(1(
),(67)
2)(1()
38)(
4)(1()
2)(1()
3)(
4)(1(
223
0
22
2
0
31
EIPaPa
EIa
aP
EIPx
EIx
aP
EI
aa
C =+=+=
-
32
C2a a
AB C
P
),(67 2
EIPa
C =
=EI
PaC 3
3
C
Conclusion
-
33
Example 8-19
Determine the slope and displacement of point B of the steel beam shown in thefigure below. Take E = 200 GPa, I = 250(106) mm4.
A5 m
B
3 kN/m
-
34
SOLUTION
Virtual Moment m
A5 m
B
1 kNx
1 kNx
v
m-1x =
xReal Moment M
A5 m
B
3 kN/m
x
3x
2x
V
M= 23 2x
Vertical Displacement at B
-1x = = 23 2x
EImkNx
EIx
EIdxxx
EIdx
EIMmkN
L
B
325
0
45
0
35
0
2
0
375.234)8
3(12
31)2
3)((1))(1( =====
==
=
,69.400469.0
)10250)(10200(
375.234466
3
mmmm
mkN
mkNB
-
35
SOLUTION
Virtual Moment m
A5 m
B
-1 =
xReal Moment M
A5 m
B
3 kN/m
=2
3 2x
Slope at B
x 1 kNm
x
v
m 1 kNm
x
3x
2x
V
M-1 = = 23 2x
EImkNx
EIx
EIdxx
EIdx
EIMmmkN
L
B
325
0
35
0
5
0
22
0
5.62)6
3(12
31)2
3)(1(1))(1( =====
,00125.0)10250)(10200(
5.62466
2
radm
mkN
mkNB =
=
-
36
Example 8-20
Determine the slope and displacement of point B of the steel beam shown in thefigure below. Take E = 200 GPa, I = 60(106) mm4.
A C D
B
5 kN14 kNm
2 m 2 m 3 m
-
37
AC
DB
5 kN14 kNm
2 m 2 m 3 m
AC DB
2 m 2 m 3 m
1 kNVirtual Moment m Real Moment M
0.5 kN0.5 kN
x3x2
6 kN1 kN
111 5.0 xm =
x1
22 5.0 xm =m M
14
x3x2x1
M1 = 14 - x1M2 = 6x2
11 5.0 xm = 22 5.0 xm =M1 = 14 - x1
M2 = 6x2
Displacement at B
=L
B dxEIMmkN
0
))(1(
++=3
03
2
0222
2
0111 )0)(0(
1)6)(5.0(1)14)(5.0(1 dxEI
dxxxEI
dxxxEI
2
0
32
2
0
31
21
2
02
22
2
01
211 )3
3)(1(35.0
27)(1()3(1)5.07(1 x
EIxx
EIdxx
EIdxxx
EI+=+=
==== ,72.100172.0)60)(200(
667.20667.20 mmmEIB
-
38
AC
DB
5 kN14 kNm
2 m 2 m 3 m
AC DB
2 m 2 m 3 m
Virtual Moment m Real Moment M
0.25 kN
x3x2
6 kN1 kN
x1
mM
14
x3x2x1
M1 = 14 - x1M2 = 6x2
1 kNm 0.25 kN
0.5
-0.5
m1 = 0.25x1
m2 = -0.25x2
M1 = 14 - x1M2 = 6x2
m1 = 0.25x1
m2 = -0.25x2
Slope at B
=L
B dxEIMmmkN
0
))(1( ++=3
03
2
0222
2
0111 )0)(0(
1)6)(25.0(1)14)(25.0(1 dxEI
dxxxEI
dxxxEI
+=2
02
22
2
01
211 )5.1(
1)25.05.3(1 dxxEI
dxxxEI
2
0
32
2
0
31
21 )
35.1(1)
325.0
25.3(1 x
EIxx
EI+=
,000194.0)60)(200(
333.2333.2 radEIB
===
-
39
Example 8-21
From the structure shown. Determine the slope and displacement at C. Take E =200 GPa, I = 200(106) mm4.
20 kN
Hinge
30 kNm
AB
C
4 m 3 m
2EI EI
-
40
20 kN
Hinge
30 kNm
A B
C
4 m 3 m
2EI EI
30 kNm
BC
30/3 = 10 kN10 kN20 kN
10 kN30 kN
30 kN30 kN
120 kNmA B
M (kNm) x (m)
-120
30
-
41
Real Moment MDisplacement at B
A BC
4 m 3 m
20 kN 30 kNm
2EI EI
M (kNm) 30
-120 M1 = -30x1
x1 x2
M2 = 10x2
10 kN30 kN
120 kNm
Virtual Moment m
A
BC
4 m 3 m
2EI EI
1 kN
0 kN1 kN
4 kNm
M (kNm) x1 x2
-4 m1 = -x1 m2 = 0
=L ii
iiB dxIE
Mm 02
)30)((4
0
111 += EI
dxxx
4
0)
330(
21 3xEI
=
=
== mEI
008.01040
32323
-
42
Real Moment MSlope at the left of B
A BC
4 m 3 m
20 kN 30 kNm
2EI EI
M (kNm) 30
-120 M1 = -30x1
x1 x2
M2 = 10x2
10 kN30 kN
120 kNm
Virtual Moment m
A
BC
4 m 3 m
2EI EI
0 kN0
1 kNm
M (kNm) x1 x2
m1 = -1 m2 = 0
iL iiBL dxIE
mM= 02)30)(1(4
0
11 += EI
dxx
4
0)
230(
21 2xEI
=
1 kNm
-1 -1
radEI
003.01040
1201203 =
==
-
43
Real Moment MSlope at the right of B
A BC
4 m 3 m
20 kN 30 kNm
2EI EI
M (kNm) 30
-120 M1 = -30x1
x1 x2
M2 = 10x2
10 kN30 kN
120 kNm
Virtual Moment m
A
BC
4 m 3 m
2EI EI
1/3 kN1/3 kN
4/3 kNm
M (kNm) x1 x2
iL iiBR dxIE
mM= ++=3
0
22
24
0
11
1 )10)(3
1(2
)30)(3
(EIdxxx
EIdxxx
3
0
4
0)
910
210(1)
310(
21 32
22
31 xx
EIx
EI++=
1 kNm
-4/3 m1 = -x1/3 m2 = -1 + x2/3-1
radEIEI
0023.0104067.91)3045(167.106 3 =
=++=
-
44
20 kN
Hinge
30 kNm
A B
C
4 m 3 m
2EI EI
B = 8 mm radBR 0023.0=Deflected Curve
radBL 003.0=
-
45
Example 8-22
(a) Determine the slope and the horizontal displacement of point C on the frame.(b) Draw the bending moment diagram and deflected curve. E = 200 GPa
I = 200(106) mm4
A
B C5 m
6 m2 kN/m
4 kN
1.5 EI
EI
-
46
A
B C5 m
6 m2 kN/m
4 kN
x1
x2
1x2
x1
M2= 12 x212 kN
16 kN
12 kN
m2= 1.2 x2
m1= x1
1.2 kN
1 kN
1.2 kN
Real Moment M
M1= 16 x1- x12
M2= 12 x2 m2= 1.2 x2
m1= x1M1= 16 x1- x12
+== 6
0
5
02221
2111 )12)(2.1(
1)16)((5.111 dxxx
EIdxxxx
EIdx
EIMm
LCH
+=5
02
22
6
01
31
21 )4.14(
1)16(5.11 dxx
EIdxxx
EI
+==+=+= ,8.28)200)(200(
1152600552)34.14(1)
4316(
5.11
5
0
32
6
0
41
31 mm
EIEIx
EIxx
EICH
1.5 EIEI
A
C
Virtual Moment m
1.5 EIEI
-
47
A
C
A
B C5 m
6 m2 kN/m
4 kN
x1
x2x2
x1
M2= 12 x212 kN
16 kN
12 kN
m2= 1-x2/5
m1= 0
1/5 kN
0
Real Moment M Virtual Moment m
M1= 16 x1- x12
1 kNm
1/5 kN
M2= 12 x2 m2= 1-x2/5
m1= 0M1= 16 x1- x12
+==6
0
5
022
21
211 )12)(5
1(1)16)(0(5.111 dxxx
EIdxxx
EIdx
EIMm
LC
+=5
02
22
2 )51212(10 dxxx
EI
,00125.0)200)(200(
5050)35
122
12(15
0
32
22 rad
EIxx
EIC+===
=
1.5 EIEI
1.5 EIEI
-
48
+
+
60
60
M , kNm
16
-12-
+V , kN
4
A
B C5 m
6 m2 kN/m
4 kN
12 kN
16 kN
12 kN
CH = 28.87 mm
C = 0.00125 rad ,
-
49
Example 8-23
Determine the slope and the vertical displacement of point C on the frame.Take E = 200 GPa, I = 15(106) mm4.
5 kN
3 m
60o
2 mA
B
C
-
50
Virtual Moment m Real Moment M
3 m
B
C
30o
Displacement at C
3 m
B
C
30ox1
1 kN
x1
1 kN
C
30o
n1
v1
m1 = -0.5x11.5 m1.5 kNm
1 kN
m1 = -0.5x1
x1
5 kN
1.5 m M1 = -2.5x1
x1
5 kN
C
30o
N1
V1
7.5 kNm
M1 = -2.5x1
x22 mA
1.5 kNm
m2 = -1.5
x2
5 kN
2 mA
7.5 kNm
M2 = -7.5
x2
1.5 kNm1 kN
v2n2 m2 = -1.5
x2
7.5 kNm5 kN
N2
V2
M2 = -7.5
=L
CV dxEIMm1 +=
2
0211
3
01 )5.7)(5.1(
1)5.2()5.0(1 dxEI
dxxxEI
)15)(200(75.3375.33)25.11(1)
325.1(1
2
0
3
0
22
31 ==+=
EIx
EIx
EICV= 11.25 mm ,
-
51
Virtual Moment m Real Moment M
3 m
B
C
30ox1
1.5 m
1 kNm
x22 mA
1 kNm
m1 = -1
m2 = -1
2 mA
3 m
B
C
30ox1
5 kN
1.5 m
7.5 kNm
x2
7.5 kNm5 kN
M1 = -2.5x1
M2 =- 7.5
x1
5 kN
C
30o
N1
V1
=L
C dxEIMm1 +=
2
0211
3
0
)5.7)(1(1)5.2()1(1 dxEI
dxxEI
)15)(200(25.2625.26)5.7(1)
25.2(1
2
0
3
0 2
21 ==+=
EIx
EIx
EIC = 0.00875 rad,
Slope at C
1 kNm
x1 C
30o
n1
v1
1 kNm
m1 = -1 M1 = -2.5x1
x2
1 kNm
n2
v2
m2 = -1
x2
7.5 kNm5 kN
N2
V2
M2 = -7.5
-
52
Virtual Strain Energy Caused by Axial Load, Shear, Torsion, and Temperature
Axial Load
Wheren = internal virtual axial load caused by the external virtual unit loadN = internal axial force in the member caused by the real loadsL = length of a memberA = cross-sectional area of a memberE = modulus of elasticity for the material
d
==L
i dxEANndnU )(
-
53
Bending
Wheren = internal virtual moment cased by the external virtual unit loadM = internal moment in the member caused by the real loadsL = length of a memberE = modulus of elasticity for the materialI = moment of inertia of cross-sectional area, computed about the the neutral axis
d
==L
i dxEIMmdmU )(
-
54
Torsion
Where t = internal virtual torque caused by the external virtual unit loadT = internal torque in the member caused by the real loadsG = shear modulus of elasticity for the material J = polar moment of inertia for the cross section, J = c4/2, where c is the
radius of the cross-sectional area
d
==L
i dxGJTtdtU )(
-
55
Shear
Wherev = internal virtual shear in the member, expressed as a function of x and caused by the external virtual unit loadV = internal shear in the member expressed as a function of x and caused by the real loadsK = form factor for the cross-sectional area:
K = 1.2 for rectangular cross sectionsK = 10/9 for circular cross sectionsK 1 for wide-flange and I-beams, where A is the area of the web
G = shear modulus of elasticity for the materialA = cross-sectional area of a member
d
==L
i dxGAKVvdvU )(
-
56
Axial =L
i dxTnU )(
Bending
=L
i dxcTmU )
2
Temperature Displacement :
Where = Differential temperatures:
- between the neutral axis and room temperature, for axial - between two extreme fibers, for bending = Coefficient of thermal expansion
d
d
-
57
Temperature
dx
T1
T2
T2 > T1
dxycTyd )
2()( =
dxcTd )
2()( =
= mdUtemp
=L
temp dxcTmU
0
)2
(
T2
cc
T1T = T2 - T1
cT
2
=
T1
T2
yy
cT
2
y
T2 > T1
T1
O
d
221 TTTm
+= d
MM
-
58
Example 8-24
From the beam below Determine :(a) If P = 60 kN is applied at the mid-span C, what would be the displacement atpoint C. Due to shear and bending moment.(b) If the temperature at the top surface of the beam is 55 oC , the temperature atthe bottom surface is 30 oC and the room temperature is 25 oC.What would be the vertical displacement of the beam at its midpoint C and thethe horizontal deflection of the beam at support B.(c) if (a) and (b) are both accounted, what would be the vertical displacement ofthe beam at its midpoint C.
Take = 12(10-6)/oC. E = 200 GPa, G = 80 GPa, I = 200(106) mm4 and A = 35(103) mm2. The cross-section area is rectangular.
A BC
2 m 2 m
-
59
A B
1 kN x x
A BC
2 m 2 m
PSOLUTION
=L
iibending dxEI
Mm=
2/
0
)2
)(2
(2L
EIdxPxx 2/
0)
34(2
3 LPxEI
=)200)(200(48
)4(6048
33
==EI
PL= 2 mm,
=L
iishear dxGA
VKv=
2/
0
)2
)(21(2
L
GAdxPK
)35000)(80(4)4)(60(2.1
42
2/
0===
GAKPL
GAKPx L
= 0.026 mm,
shearbendingC += = 2 + 0.026 = 2.03 mm,
P/2P/2
Mdiagram
PL/4
x x
xP2 x
P2
P/2
P/2
Vdiagram
Part (a) :
0.5 kN0.5 kN
m diagram 0.5x 0.5x
1
0.5
0.5
vdiagram
-
60
Part (b) : Vertical displacement at CSOLUTION
A B
1 kN x x
m diagram
0.5 kN0.5 kN
0.5x 0.5x1
Troom = 25 oC ,
T1=55oC
T2=30oC
260 mm
Temperature profile
5.422
3055=
+=mT
C = -2.31 mm ,
=L
C dxcTmkN
0 2)())(1(
- Bending
=2
0
)5.0(2
)(2 dxxcT 2
0)
25.0(
)10260()25)(1012(2
2
3
6 x
=
A BC
2 m 2 m
55 oC,
30 oC
260 m
-
61
A B1 kN
Part (b) : Horizontal displacement at B
x
00
Troom = 25 oC ,
T1=55oC
T2=30oC
260 mm
Temperature profile
5.422
3055=
+=mT
BH = 0.84 mm ,
=L
BH dxTnkN )())(1(
- Axial
=4
0
)1()( dxT
4
0))(255.42)(1012( 6 x=
1 kN
1 1n diagram
BH = 0.84 mmCv = 2.31 mm ,
A BC
2 m 2 m
55 oC
30 oC
260 m
Deflected curveA BC
-
62
P
AB
C 55 oC
30 oC
260 m
Part (c) :
C = -2.03 + 2.31 = 0.28 mm,
A B C
C = 2.03 mm
P
=
A B55 oC,
30 oC
C = 2.31 mm
+
BH = 0.84 mm
-
63
Example 8-25
Determine the horizontal displacement of point C on the frame.If thetemperature at top surface of member BC is 30 oC , the temperature at the bottomsurface is 55 oC and the room temperature is 25 oC.Take = 12(10-6)/oC, E = 200GPa, G = 80 GPa, I = 200(106) mm4 and A = 35(103) mm2 for both members.The cross-section area is rectangular. Include the internal strain energy due toaxial load and shear.
A
B C5 m
6 m2 kN/m
4 kN
1.5 EI,1.5AE, 1.5GA
EI,AE,GA260 mm
-
64
B C
Moment, m (kNm)
A
B C
Shear, v (kN)
A
5 m
6 m
A
CB
Virtual load
1
x2
x1
1.2 kN
1 kN
1.2 kN
1.2
11.2
1
1
1-1.2-1.2
6
6
1x1
1.2x2
B C
Axial, n (kN)
A
+
+
-
65
B C
Shear, V (kN)
A
B C
Axial, N (kN)
AA
B C5 m
6 m2 kN/m
4 kN
x1
x2
12 kN
16 kN
12 kN
Real load
12
412
4
16
4
16 - 2x1 -12 -12
60
16x1 - x12
12x2
60 B C
Moment, M (kNm)
A
-
66
Due to Axial
x1
x2
AE
1.5AE
5 m
6 m
B C
Virtual Axial, n (kN)
A1.2
11.2
1B C
Real Axial, N (kN)
A12
412
4
=ii
iiiCH EA
LNnkN ))(1(
AEAE)5)(4)(1(
5.1)6)(12)(2.1(
+=
AEmkN
=26.77
==
=
,0111.0)10(109.1
)10200)(1035000(
6.77 5
2626
mmm
mkNm
mkNCH
-
67
Due to Shear
x1
x2
1.5GA
5 m
6 m
GAB C
Virtual Shear, v (kN)
A1
1-1.2-1.2
B C
Real Shear, V (kN)
A16
4
16 - 2x1 -12 -12
=L
CH dxGAVKkN
0
)())(1(
2
5
01
6
01 )12)(2.1(2.1
5.1)216)(1(2.1 dx
GAdx
GAx
+
=
GAmkNx
GAxx
GA
=+=25
02
6
0
21
14.134)4.14)(2.1()
2216)(
5.12.1(
==
=
,048.0)10(8.4
)1035000)(1080(
4.134 526
26
mmmm
mkN
mkNCH
-
68
Due to Bending
x1
x2
1.5EI
5 m
6 m
EIB C
Virtual Moment, m (kNm)
A
6
6
1x1
1.2x2B C
Real Moment, M (kNm)
A
60
16x1 - x12
12x2
60
=L
CH dxEImMkN
0
))(1(
+=5
0222
6
01
2111 )12)(2.1(
1)16)((5.11 dxxx
EIdxxxx
EI
EImkNx
EIxx
EI
325
0
32
6
0
41
31 1152)
34.14(1)
4316(
5.11
=+=
+==
=
,8.280288.0
)10200)(10200(
115246
26
3
mmmm
mkN
mkNCH
-
69
T1=30oC
T2=55oC
260 mm
Temperature profile
Due to Temperature
CH = 0.0173 m = 17.3 mm ,
Tm= 42.5oC
- Bending
- Axial
CH = 0.00105 m = 1.05 mm ,
A
B C
5 m
x1
x2260 mm
30oC
55oC
Troom = 25oC
B C
m (kNm)
A
6
6
1x1
1.2x2B
C
n (kN)
A1.2
11.2
1+
+
25
03
62
0 )10260()3055)(1012)(2.1(
2)())(1( dxxdx
cTmkN
L
CH
=
=
2
5
0
6
0
)255.42)(1012)(1()())(1( dxdxTnkNL
CH ==
-
70
A
B C
2 kN/m
4 kN
Total Displacement
TempCHBendingCHShearCHAxialCHTotalCH )()()()()( +++=
= 0.01109 + 0.048 + 28.8 + (17.3 + 1.05) = 47.21 mm
CH= 47.21 mm
-
71
P1 P2 Pi + dPi
P
Castiglianos Theorem
Pi + dPi
U
U*iidP
PUdU
=
U = U*
Piiii
dPdPPU
= )(
dU = dU*
Ui = f (P1, P2,, Pn)
iPi P
U
=
P1 P2 Pi
Pi
dPi
(dPi)Pi = dU*
ii
dPPUdU
=
P
-
72
Axial Load
=Li
Pi dxAEN
P)
2(
2
=L i
dxAEN
PN )(
n
Bending )2(
2
=Li
Pi dxEIM
P
= dxEIM
PM
i
)(
m
Shear
=Li
Pi dxGAKV
P)
2(
2
= dxGAV
PVK
i
)(
v
Load Displacement :
Where = external displacement of the truss, beam or frameP = external force applied to the truss, beam or frame in the direction of N = internal axial force in the member caused by both the force P and the loads on the truss, beam or frameM = internal moment in the beam or frame, expressed as a function of x and caused by both the force P and the real loads on the beamV = internal moment in the beam or frame caused by both the force P and the real loads on the beam
-
73
Axial ))((
=Li
Pi dxTNP
= dxT
PN
i
))((
n
Bending
=Li
Pi dxcTM
P))
2((
= dxcT
PM
i
)2
)((
Temperature Displacement :
Where = Differential temperatures:
- between the neutral axis and room temperature, for axial - between two extreme fibers, for bending = Coefficient of thermal expansion
m
-
74
Bending )2(
2
=Li
Mi dxEIM
M
=
L i
dxEIM
MM )(
m
Slope :
Where = external slope of the beam or frameMi = external moment applied to the beam or frame in the direction of M = internal moment in the beam or frame, expressed as a function of x and caused by both the force P and the real loads on the beam
iMi M
U
=
-
75
Castiglianos Theorem : Truss
N 2
N1
N3 N4
N5
N6
N7 N8 N9
P
= iii L
AEN
PN )(
Where: = external joint displacement of the trussP = external force applied to the truss joint in the direction of N = internal force in a member cause by both the force P and the loads on the trussL = length of a memberA = cross-sectional area of a memberE = modulus of elasticity of a member
P1
P2
B
-
76
Example 8-26
Determine the vertical displacement of joint C of the truss shown in the figurebelow. The cross-sectional area of each member of the truss shown in the figureis A = 400 mm2 and E = 200 GPa.
A B
C
4 m 4 m
4 kN
3 m
-
77
=A B
C
LPNN )(
A B
C
N: Virtual Load P
AB
C 4 kN5 m
3 m
4 m 4 m
N: Real Load
SOLUTION
2+2
.5 -2.5
P
0.667P-0.
833P -0.833P
0
=AEL
PNNCV )(
1.5 kN1.5 kN
4 kN
0.5P0.5P
10.656-10
.41 10.41
)10200)(10400(
67.10)67.1041.1041.10(1
2626
mkNm
mkNAECV
=++=
CV = 0.133 mm,
2+2
.5 -2.5
0.667P-0.
833P -0.833P+
-
78
Example 8-27
Determine the vertical displacement of joint C of the steel truss shown. Thecross-section area of each member is A = 400 mm2 and E = 200 GPa.
4 m 4 m 4 m
AB C
D
EF
4 m
4 kN4 kN
-
79
=A
B CD
EF
LPNN )(
AB C
D
EF
N: Virtual Load P
AB C
D
EF
4 kN4 kN
N: Real Load
4 m 4 m4 m
4 m
5.657
m
SOLUTION
P 0.667P0.333P
04-5.
657
0
-5.657
444 4
-4
4 kN4 kN
0 0.667P-0.
471P
-0.47
1P
-0.943P0.667P0.333P 0
.
3
3
3
P
1P
-0.333P
=AEL
PNNCV )(
)10200)(10400(
4.72)]18.3016)67.10(2)33.5(307.15[1
2626
mkNm
mkNAECV
=++++=
CV = 1.23 mm,
4-5.
657
0
-5.657
444 4
-4
0.667P-0.
471P
-0.47
1P
-0.943P0.667P0.333P 0
.
3
3
3
P
1P
-0.333P
10.6715
.07 0
30.1810.675.33
5
.
3
3
16
5.33
+
-
80
Example 8-28
Determine the vertical displacement of joint C of the steel truss shown. Thecross-section area of each member is A = 400 mm2 and E = 200 GPa.
2 mA B
CD
3 m
20 kN
10 kNwall
-
81
2 mAB
CD
3 m
N: Virtual Load P
2 mAB
CD
3 m
20 kN
10 kN
N: Real Load
3.61 m
13.333 kN
23.333 kN
20 kN
23.333
0
-24.03
6
2020
SOLUTION
P
0.667P
0
-1.2P
01P
0.667 P
0.667P
1P
)12.10413.3160(1 ++=AECV
CV= 2.44 mm,
=AEL
PNNCV )(
)10200)(10400(
25.195
2626
mkNm
mkN
=
31.126
0
104.1
24
060
23.333
0
-24.03
6
2020
0.667P
0
-1.2P
01P+
LPNN )(
AB
CD
-
82
wC
A B
Castiglianos Theorem : Beams and Frames
C
=L
dxEIM
PM )(
Px1 x2
RBRA
Displacement
w
B
x2
RB
V2
M2
x1
RA
V1
M1
Where: = external displacement of the point caused by the real loads acting on the beam or frameP = external force applied to the beam or frame in the direction of M = internal moment in beam or frame , expressed as a function of x and cause by
both the force P and the loads on the beam or frame
-
83
w
A B
=L
dxEIM
MM )
'(
Slope
x1 x2RBRA
M
w
B
x2
RB
V2
M2
x1
RA
V1
M1
Where: = external displacement of the point caused by the real loads acting on the beam or frameM = external moment applied to the beam or frame in the direction of M = internal moment in beam or frame , expressed as a function of x and cause by
both the force P and the loads on the beam or frame
-
84
Example 8-29
The beam shown is subjected to a load P at its end. Determine the slope anddisplacement at C. EI is constant.
2a a
AB C
P
C
-
85
AB
C2a a
SOLUTION Displacement at C
=L
C dxEIM
PM )(
+
=aa
dxMP
MEI
dxMP
MEI 0
222
2
011
1 ))((1))((1
+=aa
dxPxxEI
dxPxxEI 0
222
2
01
11 ))((1)2
)(2
(1
,)3
)((1)3
)(4
(133
23
10
2
0 EIPaxP
EIxP
EI
aa
C =+=
x1 x2
-PaM2 = -Px22
11
PxM =
23P
2P
P
Mdiagram
-
86
AB
C2a a
PSlope at C
x1 x2aMP2
5.1 +aMP2
5.0 +
M
A
x1aMP2
5.0 +V1
M1 )25.0( 11 a
MxPx +=
C
P
x2
M
V2
M2 MPx = 2
+
=aa
C dxMMM
EIdxM
MM
EI 022
22
011
1 ))((1))((1
+=aa
dxMPxEI
dxa
MxPxa
xEI 0
22
2
01
11
1 ))(1(1)2
5.0)(2
(1
,6
723
2)2
)((1)3
)(4
(13232
23
10
2
0 EIPa
EIPa
EIPaxP
EIxP
EI
aa
C =+=+=
00
-
87
Example 8-30
Determine the slope and displacement of point B of the steel beam shown in thefigure below. Take E = 200 GPa, I = 250(106) mm4.
A5 m
B
3 kN/m
-
88
SOLUTION
x
A5 m
B
3 kN/m
Displacement at B
=L
B dxEIM
PM )()(
EImkN 32375.234
=
)10250)(10200(
375.234466
3
mmkN
mkN
=
B = 0.00469 m = 4.69mm,
P
=2
3 2xPx
x
3x
2x
V
MP
=5
0
2
)2
3)((1 dxxPxxEI
0
=5
0
3
231 x
EI
)8
3(15
0
4xEI
=
-
89
x
A5 m
B
3 kN/m
slope at B
=L
B dxEIM
MM )
'(
EImkN 325.62
=
)10250)(10200(
5.62466
3
mmkN
mkN
=
B = 0.00125 rad,
=2
3'2xM
=5
0
2
)2
3')(1(1 dxxMEI
0
=5
0
2
231 x
EI
)6
3(15
0
3xEI
=x
3x
2x
V
M M
M
A BDeflected curve
B = 0.00125 rad
B = 4.69mm,
-
90
Example 8-31
Determine the slope and displacement of point B of the steel beam shown in thefigure below. Take E = 200 GPa, I = 60(106) mm4.
A C D
B
5 kN14 kNm
2 m 2 m 3 m
-
91
AC
DB
14 kNm
2 m 2 m 3 m
Px3x2x1
Mdiagram
142
0
2
0)
33)(1()
35.0
27)(1(
32
31
21 x
EIxx
EI+=
)60)(200(667.20667.20
==EI
SOLUTION Displacement at B
=L
B dxEIM
PM )()(
5
111
2
0
1 )22
714()2
(1 dxPxxxEI
+=
++2
02
222 )22
7)(2
(1 dxPxxxEI
5
+=2
02
221
21
2
01 )3(
1)5.07(1 dxxEI
dxxxEI
B = 0.00172 m = 1.72 mm,
227 P
+22
7 P
Vdiagram
)22
7( P+)
227( P
227 22
2PxxM +=
22714 111
PxxM +=
+3
03)0)(0( dx
-
92
AC
DB
14 kNm
2 m 2 m 3 m
5 kNx3x2x1
Mdiagram
SOLUTION Slope at B
=L
B dxEIM
MM
0
)'
(
11
2
0
1 )4
'14()4
(1 dxMxxEI
+=
+3
03)0)(0( dx
+2
02
22
2 )4'6)(
4(1 dxxMxx
EI
0
04
'6 M
Vdiagram
M
4'1 M
)4
'1( M)
4'6( M
1411 )4
'1(14 xMM =
22 )4'6( xMM =
12
1
2
01 )25.05.3(
1 dxxxEI
=
)60)(200(333.2333.2
==EI
B = 0.000194 rad,
2
0
2
0)
35.1(1)
325.0
25.3(1
32
31
21 x
EIxx
EI+=
+2
02
22 )5.1(
1 dxxEI
B = 1.72 mm
B = 0.000194 rad
A C DB
-
93
Example 8-32
Determine the displacement of point B of the steel beam shown in the figurebelow. Take E = 200 GPa, I = 200(106) mm4.
20 kNHinge10 kNm
AB C
4 m 3 m 3 mI 2I
-
94
20 kNP
x2
= (22.5 + P)x3 - (75 + 6P)
= -(2.5 + P)x1
= 10 - 2.5x1
SOLUTION
75 + 6P
22.5 + P2.5 kN
0
P
10 kNm
2.5 kN
0
2.5 kN
10 kNm
2.5 kNV1
M1
x1 P
2.5 kN V2
M2
x2
75 + 6P
22.5 + PV3
M3
x3
x1 x320 kN
10 kNmA
B C4 m 3 m 3 mI 2I
-
95
20 kN
10 kNmA
B C4 m 3 m 3 mI 2I
= (22.5 + P)x3 - (75 + 6P)
= -(2.5 + P)x2
= 10 - 2.5x110 kNm
2.5 kNV1
M1
x175 + 6P
22.5 + PV3
M3
P
2.5 kN V2
M2
x2x3
x2P x3x1
=L
B dxEIM
PM )(
333
3
03 )6755.22()6(2
1 dxPPxxxEI
++
222
3
0211
4
0
)5.2()(2
1)5.210()0(1 dxPxxxEI
dxxEI
+= 0
0 0
+++=3
033
23
3
02
22 )4502105.22(2
1)5.2(2
10 dxxxEI
dxxEI
)200)(200(31531575.30325.11
==+=EIEIEIB = 7.875 mm,
0
-
96
Example 8-33
Determine the displacement of hinge B and the slope to the right of hinge Bof the steel beam shown in the figure below. Take E = 200 GPa, I = 200(106) mm4.
3 m 4 m
20 kNHinge
30 kNm
AB
C2EIEI
5 kN/m
-
97
20 kNSOLUTION
3 m 4 m
30 kNm
AB
C2EIEI
5 kN/mP
M
30 kNm
A B
5 kN/m
15 kN
17.5 kN2.5 kN
B
C
2EI
P
M17.5 kN
P + 17.5
4(P + 17.5) + M
-
98
20 kN
= (P + 17.5)x2 - 4(P+17.5) - M1
21 5.2
2530 xx =
3 m 4 m
30 kNm
AB
C2EIEI
5 kN/m
x1 x2P
M2.5 kN P + 17.5
4(P + 17.5) + M
x1
30 kNm
A
5x1
2.5 kN V1
M1
C2EI4(P + 17.5) + M
P + 17.5x2V3
M3
=L
B dxPMM
EI)(1
++=4
02222 )4)('7045.17(2
10 dxxMPxPxEI
020 20
The displacement of hinge B
==== ,1001.0)200)(200(2
8002800 mmm
EI
-
99
20 kN
= (P + 17.5)x2 - 4(P+17.5) - M1
21 5.2
2530 xx =
3 m 4 m
30 kNm
AB
C2EIEI
5 kN/m
x1 x2P
M2.5 kN P + 17.5
4(P + 17.5) + M
x1
30 kNm
A
5x1
2.5 kN V1
M1
C2EI4(P + 17.5) + M
P + 17.5x2V3
M3
=L
B dxMMM
EI)'
(1
++=4
0222 )1)('7045.17(2
10 dxMPxPxEI
020 20
The slope to the right of hinge B
radEI
31075.3)200)(200(2
3002300 ===
-
100
Example 8-34
Determine the slope and the horizontal displacement of point C on the frame.Take E = 200 GPa, I = 200(106) mm4
A
B C5 m
6 m2 kN/m
4 kN
1.5 EI
EI
-
101
A
B C5 m
6 m
2
k
N
/
m
1.5 EI
EI
12 kN
Horizontal Displacement at C
P
SOLUTION2)5
65
36( xP+=
12 + P
56
536 P
+
56
536 P
+
A
2x112 + P
56
536 P
+
x1V1
M1
C P
56
536 P
+V2
M2
x22
11)12( xxP +=
x1
x2
=L
iiCH dxEI
MP
M )( +++=5
02
2221
2111
6
01 )5
65
36)(5
6(1)12()(5.11 dxPxxx
EIdxxxPxx
EI
+=5
02
22
6
01
31
21 )4.14(
1)16(5.11 dxx
EIdxxx
EI
)200)(200(1152600552)
34.14(1)
4316(
5.11 5
0
6
0
32
41
31 =+=+=
EIEIx
EIxx
EICH = + 28.8 mm ,
44
-
102
A
BC
5 m
6 m
2
k
N
/
m
1.5 EI
EI
12 kN
4 kN
Slope C
5'12 M
x1
x2M
5'12 M
16
2)5'12(' xMM +=
4 N
5'12 M
V2
M2
x2
M
21116 xx =
A
2x116
5'12 M
x1V1
M1
=L
iiC dxEI
MMM
0
)'
( ++=5
02
22
21
211
6
0
)5'12')(
51(1)16()0(
5.11 dxxMxMx
EIdxxx
EI
+=5
02
22
2 )51212(10 dxxx
EI
)200)(200(5050)
3512
212(1
5
0
32
22 ==
=
EIxx
EIC = + 0.00125 rad ,
000