AXIAL LOAD
!
!
!
Principle of Superposition
Axial - force / Stress diagram
Strain
!
!
!
!
!
Mechanical
Thermal
Force - deformation relationship
Deformation
Compatibility conditions
1
Principle of Superposition
P=P1+ P2
The following two conditions must be
valid if the principle of superposition is
to be applied.
=
d
P1
2. The loading must not significantly change the original geometry or
configuration of the member.
d
+
1. The loading must be linearly
related to the stress or displacement that
is to be determined.
P2
d
2
Strain
P
• Mechanical Strain
A
σ (MPa)
LAB
400
350
B
LBC
300
200
150
100
50
0
0.00
0.0000
φBC
C
250
σAB
φAB
P
σpl
εAB
0.10
ε (mm/mm)
0.20
0.30
0.40
0.0010 0.0020 0.0030 0.0040
ε AB =
Upper scale
Lower scale
δ AB
LAB
3
Strain
• Thermal Strain
A
εΤ = δΤ/L
LAB
B
LBC
C
(∆Τ)ΑΒ
αAB
αBC
(εΤ)AB
(∆T)AB
α=
(ε T ) AB
(∆T ) AB
∆T (Co)
(mm/mm)/oC
(ε T ) AB = α ( ∆T ) AB
4
• Hookean’s Force - deformation relationship
P
σ (MPa)
A
LAB
400
350
B
LBC
300
φBC
C
250
σAB
φAB
200
P
150
σpl
100
50
0
0.0000
εAB
ε (mm/mm)
0.0010 0.0020 0.0030 0.0040
σ
δ AB = ε AB LAB = AB LAB
E
PLAB
=
AE
5
• Temperature - deformation relationship
δAB
εΤ = δΤ/L
LAB
(∆Τ)ΑΒ
A
B
LBC
C
αAB
αBC
(εΤ)AB
(∆T)AB
α=
(ε T ) AB
(∆T ) AB
∆T (Co)
(mm/mm)/oC
(δ T ) AB = α ( ∆T ) AB LAB
6
• Total Strains
P1
LAB
(∆Τ)ΑΒ
A
P2
φAB
B
LBC
φBC
C
P3
(ε Total ) AB = (ε T ) AB + (ε F ) AB
= α (∆T ) AB +
σ AB
E
(δ Total ) AB = (δ T ) AB + (δ F ) AB
= α (∆T ) AB LAB +
PAB LAB
AE
7
Elastic Deformation
x
δ (x)
dx
P0
P0
L
x
σ
P0
σ
E=
ε
ε (mm/mm)
P(x)
σ=εE
P
A(x)
P (x )
dδ
=E
A( x )
dx
P(x)
x
σ
σ(x) =
dδ =
P ( x )dx
A ( x )E
L
P(x)
A(x)
dδ = ∫
0
x
P( x)dx
A( x) E
8
• Axial-force diagram
RA
A
P1/2
B
P1/2 C
ACD = (π/4)(do2 - di2 )
LAB
LBC
di
P2/2
P3
D
P2/2
AAC= (π/4)(di2 )
LCD
do
PCD = RA + P1 + P2 = P3
P
PAB=RA
PBC = RA + P1
+
+
x
σCD
σ
σAB
σBC
+
+
x
9
• Deformation diagram
RA
A
B
P1/2 C
ACD = (π/4)(do2 - di2 )
LAB
LBC
P3
D
P2/2
AAC= (π/4)(di2 )
LCD
do
PCD = RA + P1 + P2 = P3
P
PAB=RA
δx/A (mm)
di
P2/2
P1/2
PBC = RA + P1
+
x
δC
δB
+
δ B = δ A + δ B/ A = o +
PAB LAB
AAB E AB
δC = δC/ A = δ B + δC/B = δ B +
PBC LBC
ABC E BC
δ D = δ D/ A = δ C + δ D/C = δ C +
PCD LCD
ACD ECD
x (mm)
δD
10
Example 1
The rod is made from a solid steel section AB and aluminum section BC. If it is
fixed to a rigid support at A. A solid aluminum having an inner diameter of 30
mm and outer diameter of 60 mm. Determine (a) draw the quantitative stress and
strain diagram of the bar (b) the horizontal displacement of end D and the
displacement of C relative to B. Take Est = 200 GPa, Eal = 70 GPa
100 kN
di = 30 mm
375 kN
400 kN
B
A
500 mm
100 kN C 375 kN
400 mm
600 mm
D
do = 60 mm
11
100 kN
150 kN
di = 30 mm
375 kN
400 kN
B
A
500 mm
D
100 kN C 375 kN
400 mm
do = 60 mm
600 mm
P (kN)
σ AB =
350
150
+
-
x (mm)
-400
• Stress diagram
350 kN
π
( )( 0 . 06 2 )
4
= 159 . 2 MPa
159.2
σ CD =
+
-188.6
π
( )(0.06) 2
4
= 53.05 MPa
σ BC =
σ (MPa)
53.05
150 kN
x (mm)
− 400 kN
π
( )( 0 . 06 2 − 0 . 03 2 )
4
= −188 . 6 MPa
12
100 kN
150 kN
di = 30 mm
375 kN
400 kN
D
B 100 kN C 375 kN
Est = 200 GPa
Eal = 80 GPa
A
500 mm
400 mm
do = 60 mm
600 mm
P (kN)
150
δB/A
350
-
δx/A (mm)
x (mm)
-400
δC /B
( 350 )( 0 . 4 )
π
( 0 . 06 2 ))( 200 × 10 6 )
4
= 0 . 248 mm
(
0.133+0.248 = 0.381
0.133
π
( 0 . 06 2 ))( 200 × 10 6 )
4
= 0 . 133 mm
(
+
• Displacement diagram
(150 )( 0 . 5 )
( −400 )( 0 . 6 )
δ D /C =
π
( 0 . 06 2 − 0 . 03 2 ))( 70 × 10 6 )
4
= −1 . 617 mm
σD = 0.381-1.617 = -1.236
(
13
Example 2
The assembly shown consists of an aluminum tube AB having a cross-sectional
area of 400 mm2. A steel rod having a diameter of 10 mm is attached to a rigid
collar and passes through the tube. If a tensile load of 80 kN is applied to the rod,
determine the displacement of the end C of the rod. Take Est = 200 GPa, Eal = 70
GPa.
400 mm
B
A
C
80 kN
600 mm
14
400 mm
B
A
80 kN
C
600 mm
- Rod BC
- Aluminum tube AB
80 kN
B
A
0.4 m
δB/A =
=
PAB = 80 kN
A = 400 mm2
PAB LAB
Atube E al
80 kN
= −1 . 143 × 10 −3 m = 1 . 143 × 10 −3 m →
d = 10 mm
C
80 kN
0.6 m
δC /B =
( −80 kN)( 0 . 4 m)
( 400 × 10 −6 m 2 )( 70 × 10 6 kN / m 2 )
B
=
PBC LBC
Arod E st
( 80 kN)( 0 . 6 m )
[π ( 0 . 005 m) 2 ]( 200 × 10 6 kN / m 2 )
= 3 . 056 × 10 −3
m →
- Displacement of the end C
+
( →
)
δ C = δ B + δ C / B = 0 . 001143 + 0 . 003056 = 0 . 00420 m = 4 . 20 m →
15
Example 3
A rigid beam AB rests on the two short posts shown. AC is is made of steel and
has a diameter of 20 mm, and BD is made of aluminum and has a diameter of 40
mm. Determine the displacement of point F on AB if vertical load of 90 kN is
applied over this point. Take Est = 200 GPa, Eal = 70 GPa.
90 kN
200 mm
400 mm
B
A
F
300 mm
C
D
16
90 kN
200 mm
- Compatibility Equation
400 mm
600 mm
B
A
400 mm
F
60 kN
A
30 kN
0.286 mm
dAC = 20 mm
C
Est = 200 GPa
60 kN
δA =
dBD = 40 mm
Eal = 70 GPa.
300 mm
D
30 kN
PAC LAC
( −60 kN)( 0 . 3 m )
=
AAC E st
[π ( 0 . 01 m ) 2 ]( 200 × 10 6 kN / m 2 )]
= −2 . 86 × 10 −4 m = 0 . 286 mm ↓
δB =
PBD LBD
( −30 kN)( 0 . 3 m )
=
ABD E al [π ( 0 . 02 m ) 2 ]( 70 × 10 6 kN / m 2 )
= −1 . 02 × 10 −4 m = 0 . 102 mm ↓
B
0.102 mm
F
δF
y
0.286 − 0.102
=
400
600
y = (0.286 − 0.102)(
400
)
600
y = 0 . 123 mm
δ F = 0 . 102 + y
= 0 . 102 + 0 . 123
= 0 . 225 mm ↓
17
Example 4a
The rod is made from a solid steel section AB and aluminum section BC. If it is
fixed to a rigid support at A. A solid aluminum having an inner diameter of 30
mm and outer diameter of 60 mm. Draw the quantitative stress diagram of the
rod. Temperature in member AB and CD are increase 20 oC and 10 oC
respectively prior to the loads are applied. Take Est = 200 GPa, Eal = 70 GPa, αst
= 12x10-6 /oC and αal = 23x10-6 /oC.
steel
375 kN
di = 30 mm
aluminum
400 kN
A
+ 20 oC
900 mm
B 375 kN
+ 10 oC C
600 mm
do = 60 mm
18
steel
350 kN
α = 12x10-6 /oC
A
P (kN)
aluminum
C
400 kN
375 kN
(∆T)st = 20 oC
Est = 200 GPa
900 mm
(∆T)al = 10 oC
B 375 kN E = 80 GPa
al
600 mm
AAB = (π/4)(0.062 )
= 2.827x10-3 m2
di = 30 mm
do = 60 mm
ABC = (π/4)(0.062 - 0.032 )
= 2.121x10-3 m2
350
+
x (mm)
σ (MPa)
σ=
P
A
350 × 10 N
( 2 . 827 × 10 −3 m 2 )
= 123 . 8 MPa
3
σ st =
123.8
-400
+
x (mm)
-
− 400 × 10 3 N
( 2 . 121 × 10 −3 m 2 )
= 188 . 6 MPa
-188.6 σ =
al
19
steel
350 kN
α = 12x10-6 /oC
A
P (kN)
aluminum
C
400 kN
375 kN
(∆T)st = 20 oC
Est = 200 GPa
900 mm
AAB = (π/4)(0.062 )
= 2.827x10-3 m2
(∆T)al = 10 oC
B 375 kN E = 80 GPa
al
600 mm
di = 30 mm
do = 60 mm
ABC = (π/4)(0.062 - 0.032 )
= 2.121x10-3 m2
350
+
-
δx/A (mm)
δ=
x (mm)
-400
(350)(0.9)
−6
δ B/ A =
+
(
12
×
10
)(20)(0.9)
−3
6
(2.827 ×10 )(200 ×10 )
= 0.557 mm + 0.216 mm = 0.773 mm
0.773
PL
+ α (∆T ) L
AE
x (mm)
0.773-1.479 = -0.706
−6
+ ( 23 × 10 )(10 )( 0 . 6 )
( −400 )( 0 . 6 )
( 2 . 121 × 10 −3 )( 70 × 10 6 )
= −1 . 617 mm + 0 . 138 mm = −1 . 479 mm
δC /B =
20
Example 4b
The rod is made from a solid steel section AB and aluminum section BC. If it is
fixed to a rigid support at A. A solid aluminum having an inner diameter of 30
mm and outer diameter of 60 mm. Temperature in member AB and CD are
increase 20 oC and 10 oC respectively. Determine the maximum load P the rod
can take at end C, if the allowable normal stress in steel is (σst)allow = 120 MPa
and the in aluminum is (σal)allow = 180 MPa . Take Est = 200 GPa, Eal = 70 GPa,
αst = 12x10-6 /oC and αal = 23x10-6 /oC.
steel
375 kN
aluminum
di = 30 mm
P
+20 oC
A
900 mm
B 375 kN
+10 oC C
600 mm
do = 60 mm
21
RA
375 kN
steel
aluminum
P
di = 30 mm
+20 oC
+10 oC C
αst = 12x10-6 /oC
375
kN
B
Est = 200 GPa
A
Eal = 80 GPa
αal = 23x10-6 /oC
900 mm
600 mm
(σst)allow = 120 MPa
do = 60 mm
+
(σal)allow = 180 MPa
→ ΣFx = 0 : − RA + 750 − P = 0,
RA = 750 − P
δC = 0
AAB = (π/4)(0.062 )
ABC = (π/4)(0.062 - 0.032 )
P (kN)
= 2.827x10-3 m2
= 2.121x10-3 m2
750-P
+
(σ st ) allow :
750 − P
= 120 × 10 3 ,
−3
( 2 . 827 × 10 )
-P
P1 = 410 . 76 kN
(σ al ) allow :
−P
= 180 × 10 3 ,
−3
( 2 . 121 × 10 )
P2 = 381 . 78 kN
δC :
x (mm)
( 750 − P )( 0 . 9)
( −P )( 0 . 6 )
−6
+
(
12
×
10
)(
20
)(
0
.
9
)
+
( 2 . 827 × 10 −3 )( 200 × 10 6 )
( 2 . 121 × 10 −3 )( 70 × 10 6 )
+ ( 23 × 10 −6 )(10 )( 0 . 6 ) = 0 ,
P3 = 274 . 80 kN
22
Principle of Superposition
P=P1+ P2
The following two conditions must be
valid if the principle of superposition is
to be applied.
=
d
P1
1. The loading must be linearly
related to the stress or displacement that
is to be determined.
2. The loading must not significantly change the original geometry or
configuration of the member.
d
+
P
P2
d
23
Statically Indeterminate Axially Loaded Members
Compatibility Conditions
• End to end bars
P
δD/A = δB/A + δC/B+ δDC = 0
D
A
B P
C
• Co-axial bars
material#1, core
B
A
material#2, sleeve
• Parallel bars
P
(δB)1 = (δB)2
F
E
P
A
B
C
δB
D
LAB
δB
δC
δD
=
δC
LAC
=
δD
LAD
24
Statically Indeterminate Axially Loaded Members: End to End Bars
d2
d1
RA
P1/2
B P1/2
A
Steel
LAB
P (kN)
d1
P2/2
P2/2
D
C
Aluminum
Steel
LBC
LCD
RA+ P1
RA
RD
Est
Eal
RA+ P1 - P2 = RD
x
- Equilibrium Equation : Σ Fx = 0
- RA- P1 + P2 + RD = 0
-----(1)
- Compatibility Equation :
δD/A = δB/A + δC/B+ δD/C = 0
R A LAB ( R A + P1 ) LBC RD LCD
+
+
=0
AAB Est
ABC Eal
ACD Est
− − − − − (2)
25
Example 5
A solid steel rod shown in the figure, having an inner diameter of
30 mm and outer diameter of 60 mm. Determine the reactions at supports. Draw
the quantitative normal stress and deformation diagrams of the rod. Take E = 200
GPa.
100 kN
B
A
500 mm
100 kN C 375 kN
400 mm
di = 30 mm
375 kN
600 mm
D
do = 60 mm
26
100 kN
di = 30 mm
375 kN
RA
RD
B
A
500 mm
P (kN)
100 kN C 375 kN
400 mm
AAC = (π/4)(0.062 )
= 2.827x10-3 m2
RA + 200
RA
+
D
600 mm
do = 60 mm
ACD = (π/4)(0.062 - 0.032 )
= 2.121x10-3 m2
RA -550 = RD
x (mm)
- Equilibrium equation:
RA -550 = RD
-----(1)
- Compatibility equation:
δD/A = δB/A + δC/B + δD/C = 0
R A ( 0 . 5)
(R A + 200 )( 0 . 4 )
(R A − 550 )( 0 . 6 )
+
+
= 0 − −( 2 )
( 2 . 827 × 10 −3 )( 200 × 10 6 ) ( 2 . 827 × 10 −3 )( 200 × 10 6 ) ( 2 . 121 × 10 −3 )( 200 × 10 6 )
RA = 211.73 kN
RD = RA - 550 = 211.73 -550 = -338.27 kN
27
100 kN
211.73 kN
di = 30 mm
375 kN
338.27 kN
A
500 mm
P (kN)
400 mm
AAC = (π/4)(0.062 )
= 2.827x10-3 m2
211.73
D
100 kN C 375 kN
B
600 mm
do = 60 mm
ACD = (π/4)(0.062 - 0.032 )
= 2.121x10-3 m2
411.73
+
x (mm)
-338.27
σ (MPa)
σ=
P
A
74.9
σ AB
145.64
+
( 211 . 73 kN)
=
(2.827 × 10 -3 m 2 )
= 74 . 9 MPa
σ BC =
( 411 . 73 kN)
(2.827 × 10 - 3 m 2 )
= 145 . 64 MPa
-159.49
x (mm)
σ CD =
( −338 . 27 kN)
(2.121 × 10 - 3 m 2 )
= −159 . 49 MPa
28
100 kN
211.73 kN
di = 30 mm
375 kN
338.27 kN
A
500 mm
P (kN)
D
100 kN C 375 kN
B
400 mm
600 mm
do = 60 mm
ACD = (π/4)(0.062 - 0.032 )
= 2.121x10-3 m2
AAC = (π/4)(0.062 )
= 2.827x10-3 m2
411.73
211.73
+
-
x (mm)
-338.27
δx/A (mm)
0.19
0.48
x (mm)
δD/A =
( 211 . 73 )( 0 . 5 )
( 411 . 73 )( 0 . 4 )
( −338 . 27 )( 0 . 6 )
+
+
( 2 . 827 × 10 −3 )( 200 × 10 6 ) ( 2 . 827 × 10 −3 )( 200 × 10 6 ) ( 2 . 121 × 10 −3 )( 200 × 10 6 )
= 0 . 19 + 0 . 29 − 0 . 48 = 0 mm
29
Statically Indeterminate Axially Loaded Members: Coaxial Bars
RA = F1+ F2
Axial
Force
material#1 core, d1
material#1 core, d1
P
B
A
material#2, sleeve
L
material#2 post, d2
(F1+ F2) = P
x
P
F1
F2/2
- Equilibrium Equation :
- (F1+ F2) + P = 0
F2/2
B
------(1)
- Compatibility Equation :
(δB/A)1 = (δB/A)2
F1 L
FL
= 2
A1 E1 A2 E2
− − − − − ( 2)
30
Example 6
The rod is made from a solid steel section AB and a tubular portion made of
steel and having a aluminum core section BC. If it is fixed to a rigid support
at A. Take Est = 200 GPa, Eal = 80 GPa
(a) Compute the normal stress in steel and aluminum.
(b) Determine the displacement of the end C of the rod.
175 kN
B 175 kN
A
1.0 m
0.6 m
200 kN
Aluminum core
dal = 30 mm
C
steel dst = 50 mm
31
(a) Compute the normal stress in steel and aluminum.
175 kN
150 kN
A
200 kN
B 175 kN
0.6 m
1.0 m
P (kN)
Aluminum core
di = 30 mm
C
do = 50 mm
200
+
-
x (mm)
-150
- Compatibility equation
(
- Equilibrium equation
(δC/B)st = (δC/B)al
Pst + Pal = 200
PL
PL
) st = (
) al
AE
AE
4 . 444 Pal + Pal = 200
Pst (0.6)
Pal (0.6)
=
− − − (1)
2
2
6
2
6
(π / 4)(0.05 − 0.03 )(200 × 10 ) (π / 4)(0.03 )(80 × 10 )
Pst = ( 4 . 444 )Pal
− − − (2)
Pal = 36 . 75 kN
Pst = 163 . 25 kN
32
175 kN
150 kN
A
1.0 m
P (kN)
200 kN
B 175 kN
0.6 m
C
do = 50 mm
200
+
-150
- Normal stress
Aluminum core
di = 30 mm
x (mm)
Pal = 36.75 kN
Pst = 163.25 kN
• (σ BC ) st =
(PBC ) st
(163 . 25 )
=
= 129 . 91 MPa ⇐
2
2
Ast
(π / 4 )( 0 . 05 − 0 . 03 )
• (σ BC ) al =
(PBC ) al
( 36 . 75 )
=
= 51 . 28 MPa ⇐
Aal
(π / 4 )( 0 . 03 2 )
33
(b) Determine the displacement of the end C of the rod.
175 kN
150 kN
A
1.0 m
P (kN)
200 kN
B 175 kN
0.6 m
C
do = 50 mm
200
+
-150
Aluminum core
di = 30 mm
x (mm)
Pal = 36.75 kN
Pst = 163.25 kN
δC/A = δB/A + δC/B
•δC / A =
( −150 )(1 . 0 )
( 36 . 75 )( 0 . 6 )
+
= −0 . 115 mm ⇐
(π / 4 )( 0 . 05 2 )( 200 × 10 6 ) (π / 4 )( 0 . 03 2 )( 80 × 10 6 )
34
Statically Indeterminate Axially Loaded Members: Parallel Bars
- Equilibrium Equation
F
E
- Compatibility Equation
F
E
P
B
A
Ay
A
C
FBE
B
D
A
B
D
δB
ΣFy = 0:
Ay + FBE + FCF - P = 0 -----(1)
+ ΣMA = 0:
FBE(LAB) + FCF(LAC) + P(LAD) = 0 -----(2)
=
δC
LAC
D
δC
P
LAB
+
C
δB
FCF
C
P
=
δD
δD
LAD
FBE LBE
FCF
=
ABE EBE ACF ECF
− − − − − (3)
35
Example 8
The three A-36 steel bar shown are pin-connected to a rigid member. If the
applied load on the member is 15 kN, determine the fore developed in each bar.
Bars AB and EF each have a cross-sectional area of 25 mm2 , and bar CD has a
cross-sectional area of 15 mm2
B
D
F
0.5 m
C
A
0.2 m
0.2 m
E
0.4 m
15 kN
36
B
D
F
0.5 m
C
A
0.2 m
FAB
E
0.2 m
15 kN
0.4 m
FCD
FEF
C
A
0.2 m
0.2 m
E
0.4 m
15 kN
- Equilibrium Equation
+
ΣFy = 0:
+ ΣMC = 0:
FAB + FCD + FEF - 15 kN = 0 -----(1)
-FAB(0.4 m) + 15 kN(0.2 m) + FEF(0.4 m) = 0 -----(2)
37
B
D
F
0.5 m
C
A
0.2 m
0.2 m
15 kN
A
E
δA
A´
- Compatibility Equation
0.4 m
C
δC C ´
E
E´
δE
δ A −δE δC −δE
=
0.8 m
0.4 m
δC =
1
1
δA + δE
2
2
FCD ( 0 . 5 m )
1 FAB ( 0 . 5 m )
1 FEF ( 0 . 5 m )
[
]
[
=
+
]
2
2
2
2 ( 25 mm )E st
2 ( 25 mm )E st
(15 mm )E st
Solving Eqs. 1-3 : FAB = 9.52 kN,
FCD = 3.46 kN,
− − − − − ( 3)
FEF = 2.02 kN,
38
Thermal Stress
δ/L
∆T
∆(δ/ L)
α
L
δ
L
L
δ T = ∫ ε T dx = ∫ α (∆T )dx
0
∆T
α = ∆(δ/ L)/(∆T)
Temp(oC)
(mm/mm)/oC
0
δΤ = α(∆T)L
εT = α(∆T)
39
∆Τ
RA
A
P (kN)
L
RB
B
d
RA = RB
x (mm)
δB/A = 0 =
R AL
+ α ( ∆ T )L
AE
R A = −α ( ∆ T ) AE = R B
40
• Compatibility condition with thermal strain
d1
RA
Steel
LAB
P (kN)
RA
d1
P2/2
P1/2
P2/2
B P1/2
A
∆
d2
D
C
Aluminum
Steel
LBC
LCD
RA+ P1
RD
Eal
αst
RA+ P1 - P2 = RD
Eal
αal
x
δ = δForce + δ∆Τ = Σ [(εForce L) + (ε∆Τ L)] = Σ[(PL/ AE)] + Σ [α(∆Τ) L]
Compatibility equation:
δD/A = ∆
δB/A + δC/B + δD/C = ∆
41
Example 7
A solid steel rod shown in the figure, having an inner diameter of
30 mm and outer diameter of 60 mm. Draw the quantitative normal
stress and the Elongation of the rod. Temperature in member AB and
CD are increase 30 oC. Take E = 200 GPa, α = 12x10-6 /oC.
100 kN
B
A
500 mm
375 kN
100 kN C 375 kN
400 mm
0.5 mm
600 mm
di = 30 mm
D
do = 60 mm
42
+30
RA
100 kN
oC
375 kN
+30
di = 30 mm
oC
RD
B
A
500 mm
100 kN C 375 kN
400 mm
AAC = (π/4)(0.062 )
= 2.827x10-3 m2
- Equilibrium equation:
P (kN)
RA
D
600 mm
ACD = (π/4)(0.062 - 0.032 )
= 2.121x10-3 m2
do = 60 mm
RA + 200
+
RA -550 = RD
x (mm)
- Compatibility equation:
0 . 5 × 10 −3 =
δD/A = 0.5x10-3 m = δForce + δ∆Τ = Σ[(PL/ AE) + α(∆Τ)L]
R A ( 0 . 5)
(R A + 200 )( 0 . 4 )
−6
+
×
+
+
(
12
10
)(
30
)(
0
.
5
)
( 2 . 827 × 10 −3 )( 200 × 10 6 )
( 2 . 827 × 10 −3 )( 200 × 10 6 )
+
(R A − 550 )( 0 . 6 )
−6
+
(
12
×
10
)( +30 )( 0 . 6 )
−3
6
( 2 . 121 × 10 )( 200 × 10 )
RA = 246.32 kN ,
RD = RA - 550 = 246.32 -550 = -303.68 kN
43
100 kN
246.32 kN
di = 30 mm
375 kN
303.68 kN
A
500 mm
P (kN)
400 mm
AAC = (π/4)(0.062 )
= 2.827x10-3 m2
246.32
D
100 kN C 375 kN
B
600 mm
ACD = (π/4)(0.062 - 0.032 )
= 2.121x10-3 m2
do = 60 mm
446.32
+
-
x (mm)
-303.68
σ (MPa)
σ =
P
A
87.13
σ AB =
157.88
σ BC =
( 446 . 32 kN)
( 2 . 827 × 10 −3 m 2 )
+
( 246 . 32 kN)
( 2 . 827 × 10 −3 m 2 )
= 87 . 13 MPa
-143.18
x (mm)
σ CD =
( −303 . 68 kN)
( 2 . 121 × 10 −3 m 2 )
= −143 . 18 MPa
44
100 kN
246.32 kN
di = 30 mm
375 kN
303.68 kN
B
A
500 mm
D
100 kN C 375 kN
400 mm
600 mm
ACD = (π/4)(0.062 - 0.032 )
= 2.121x10-3 m2
AAC = (π/4)(0.062 )
= 2.827x10-3 m2
P (kN)
446.32
246.32
+
-
do = 60 mm
x (mm)
-303.68
δ (mm)
0.4
0.72
0.50
x (mm)
δ D/ A
( 246.32)(0.5) ×103
( 446.32)(0.4) ×103
−6
=
+ (12 ×10 )(30)(.500) +
−3
6
(2.827 ×10 )(200 ×10 )
(2.827 ×10 −3 )(200 ×106 )
( −303.68)(0.6) ×103
+
+ (12 ×10 −6 )(30)(.600)
−3
6
( 2.121×10 )(200 ×10 )
= 0.218 + 0.18 + 0.316 − 0.43 + 0.216 = 0.5 mm
45
Example 8
From the frame shown, determine:
(a) Reactions at all support
(b) Normal stress in steel and aluminum
(c) Displacement at A and B
Take Est = 200 GPa, Eal = 70 GPa, αst = 12x10-6 /oC and αal = 23x10-6 /oC.
800 kN
0.75 m
A
0.90 m
B
+20oC
Steel
d = 60 mm
0.5 m
+10oC
Aluminum
di = 30 mm
C
0.6 m
do = 60 mm
0.5 m
46
800 kN
0.75 m
A
B
Steel
d = 60 mm
Fst
δst
A
Cy
+10oC
Aluminum 0.6 m
di = 30 mm
+20oC
0.90 m
C
• Compatibility
A´
0.5 m
δ st δ al
=
1 0.5
δ st = 2δ al
δ st =
π
0.5 m
− − − (2)
Fst (0.9)
( (0.06 2 ))(200 ×106 )
4
+ (12 ×10 −6 )(20)(0.9)
0.5 m
(a) Reactions at all support
• Equilibrium equation
δ al =
+ ΣM C = 0 : 800(0.75) + 0.5 Fal + 1Fst = 0
Fal = −1200 − 2 Fst
C
B
Fal do = 60 mm
0.5 m
B´ δ
al
− − − (1)
π
Fal (0.6)
( (0.062 − 0.032 ))(70 ×106 )
4
+ (23 ×10−6 )(10)(0.6)
Substitute δst and δal in eq.(2) from eq.(1) and (2) can solve
Fst = -542.78 kN (C)
Fal = -114.44 kN (C)
+↑
ΣFy = 0 :
542 . 78 + 114 . 44 − 800 − C y = 0 ,
C y = −142 . 78 kN, ↑
47
800 kN
(b) Normal stress in steel and aluminum
0.75 m
A
0.90 m
B
+20oC
Steel
C
+10oC
142.78 kN
Aluminum 0.6 m
di = 30 mm
δ al
do = 60 mm
114.44 kN
d = 60 mm
542.78 kN
0.5 m
− 542 . 78 kN
= −191 . 97 MPa (C )
π
( 0 . 06 2 )
4
− 114 . 44 kN
=
= −53 . 97 MPa(C )
π
( 0 . 06 2 − 0 . 03 2 )
4
σ st =
0.5 m
(c) Displacement at A and B
− (542 . 78 )( 0 . 9)
δ st =
Est = 200 GPa
Eal = 70 GPa
αst = 12x10-6 /oC
αal = 23x10-6 /oC.
+ (12 × 10 −6 )( 20 )( 0 . 9)
π
( 0 . 06 2 ))( 200 × 10 6 )
4
= −0 . 864 mm + 0 . 216 mm = −0 . 648 mm
(
( −114 . 44 kN)( 0 . 6 )
δ al =
+ ( 23 × 10 −6 )(10 )( 0 . 6 )
π
( 0 . 06 2 − 0 . 03 2 ))( 70 × 10 6 )
4
= −0 . 463 mm + 0 . 138 mm = −0 . 325 mm
(
48