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Chapter 3 Statically Determinate and Indeterminate Systems 3.1 Principles of Solid Mechanics For a stable system, the following requirements must be met: (1) Equilibrium of forces and moments, which directly relates to stresses. (2) Compatibility of strains or displacements, which relates to deformation of geometry. (3) Stress ~ strain relations of materials, which are linked by the physical properties of a
material. 3.2 Statically Determinate Systems 3.2.1 Definition of statically determinate systems A statically determinate system is one that can be solved by force equilibrium only, that is, the number of unknowns in the system, e.g., reactions and internal forces, equals the number of equations available from force equilibrium. Example 3-1 A stepped cylinder carries axial loads, as shown in the figure below. The diameters of bar B and bar C are 25 mm and 50 mm, respectively. Find stresses in the two bars. Solution: For bar B
50=Bf kN
102
225
1050
225 2
3
2 =
=
=
BBf MPa (Tensile)
50 kN
p = 5 MPa
C B
60 mm
2
For bar C
322
1050225
260 =
+ pfC
38320225
26051050
223 =
= Cf N
5.19
250
38320
250 22
=
=
=
CCf
MPa (Tensile)
3.2.2 Typical thin-walled structures (1) Thin-walled pressurized sphere In this structure, stress can be assumed to be uniform through the wall thickness. Radial
stress r is assumed to be zero.
For force equilibrium, RtRp 22 = , then
tpR2
= (Everywhere in the wall, in each direction)
t
p
2R
t
3
(2) Close-ended thin-walled pressurized cylinder In this structure, hoop stress and axial stress l are assumed to be uniform through the wall thickness. Radial stress r is assumed to be zero. Consider force equilibrium in horizontal direction,
RtRp l 22 = , tRp
l 2= (Longitudinal stress)
Consider force equilibrium in vertical direction,
2R
t
p
l l
t
4
ltlRp = 22 , tRp = (Hoop stress or circumferential stress)
l 2= (3) Thin rotating ring Hoop stress is assumed to be uniform through the wall thickness and radial stress r is zero. Consider force equilibrium of a differential element,
=
2sin2 AFr , A Cross-sectional area
The centrifugal force Fr is evaluated as
( ) 2 = RRAFr , Material density
Since is small, 22
sin
. Then 22 R=
R
t
r
r
t
5
3.3 Statically Indeterminate Systems A statically indeterminate system is one that cannot be solved by force equilibrium only, that is, the number of unknowns in the system, e.g., reactions and internal forces, is greater than the number of equations available from force equilibrium. In this case, deformation of geometry has to be involved. Example 3-2 A stepped bar is constrained between two walls with a total deformation , as shown below. Determine the reaction force. Solution: The internal forces of bar a and bar b are equal, that is,
Rff ba ==
ba += bbaa ll +=
bb
ba
a
a lE
lE
+=
b
b
b
b
a
a
a
a
Af
El
Af
El +=
bb
b
aa
a
AR
El
AR
El +=
+=
bb
b
aa
a
AEl
AElR
aabbba
baba
AElAElAAEER +=
R R
la lb
Aa Ab
6
Example 3-3 A load P is applied on a rigid beam which is supported by two rods a and b, as shown in the figure below. Find stresses in the rods. Solution:
lflflP ab 25.1 += Then ba ffP += 25.1 fa and fb cannot be solved by force equilibrium only and deformation of the rods has to be considered.
= lb and = la 2 , therefore ba 2=
la
a = and
lb
b =
a
aa A
f= and b
bb A
f=
aaa E= and bbb E=
b a
l l A B C
l
A B C
a b
P l l
l/2
7
Integrating the above relations, the following expression can be obtained.
bb
b
aa
a
EAf
EAf
2= , Aa and Ab are the cross-sectional areas of rod a and rod b, respectively.
Thus bbb
aa fEAEAP
+= 145.1
bbaa
bbb EAEA
EPAf += 45.1
and bbaa
aaa EAEA
EPAf += 43
bbaa
a
a
aa EAEA
PEAf
+== 43 and
bbaa
b
b
bb EAEA
PEAf
+== 45.1
Example 3-4 Cylinder a and rod b are made of different materials. They are fitted into rigid ends, as shown in the figure below. When temperature changes by T, determine stresses in the cylinder and the rod, respectively. Solution: The length changes of the cylinder and the rod will be different due to the different thermal expansion of the materials. Thus one will be in tension and the other will be in compression.
Cylinder a
Rod b
Cylinder a
Rod b
fa
fa
fb
8
ba ff = fa represents the total internal force in the cylinder and fb represents the total internal force in the rod.
bbaa AA = Since ba ll = and ba = , then ba =
TE aa
aa +=
and TE bb
bb +=
Therefore TE
TE bb
ba
a
a +=+
TAEAT
E bbbaa
aa
a +=+
( )aabb
bbaaba AEAE
AEET +=
( )aabb
abaabb AEAE
AEET +=
If a and b are negative, the cylinder expands more than the rod, then the former is in compression and the latter in tension. Example 3-5 A symmetrical frame consisting of three pin-connected steel bars (E = 200 GPa) is loaded by a force P at the joint, see the figure below. The middle bar is 2 m long and its axial strain is measured to be 0.008. The angle between the inclined bars and the horizontal is = 50. Determine stress in the inclined bars.
9
Solution:
8.261040cos
200040cos
===B
Cll mm
2.167840tan200040tan' === Bll mm ( ) ( ) 1.26232000008.020002.1678008.0 2222' =++=++= BB llll mm
3.128.26101.2623 === Clll mm
0047.08.2610
3.12 ===Cll
940102000047.0 3 === E MPa Example 3-6 A trimetallic bar is uniformly compressed by an axial force P = 2 kN applied through a rigid end plate, as shown in the figure below. The bar consists of a circular steel core surrounded by a brass tube and a copper tube. The steel core has diameter 10 mm, the brass tube has outer diameter 12 mm, and the copper tube has outer diameter 15 mm, with the corresponding elastic constant Es = 200 GPa, Eb = 100 GPa and Ec= 120 GPa. Calculate stresses in the steel, brass and copper, respectively, due to the force P.
50
40 l
l
10
Solution:
cbs fffP ++= , llll cbs === and cbs ==
lEA
flE
lss
s
s
sss ===
, lEA
f
bb
bb = and lEA
f
cc
cc = . Then
ss
bbsb AE
AEff = and ss
ccsc AE
AEff = , ss
ccs
ss
bbss AE
AEfAEAEffP ++=
5.782
10 2 =
= sA mm2
54.342
102
12 22 =
= bA mm2
59.632
122
15 22 =
= cA mm2
+
+=
5.781020059.6310120
5.781020054.341010012000 3
3
3
3
sf , 1172=sf N
84.2575.7810200
54.34101001172 33
==bf N
6.5695.7810200
59.63101201172 33
==cf N
93.145.78
1172 ===s
ss A
f MPa
46.754.3484.257 ===
b
bb A
f MPa
96.859.636.569 ===
c
cc A
f MPa
Example 3-7 A rigid triangular frame is pivoted at C and held by two identical horizontal wires at point A and B, see the figure below. Each wire has axial rigidity 120=EA klb and coefficient of thermal expansion = 6105.12 /F. (1) If a vertical load P = 500 lb acts at point D, what are the tensile forces TA and TB in the wires
at A and B, respectively? (2) If, while the load P is acting, both wires have their temperatures raised by 180F, what are
the forces TA and TB? (3) What further increase in temperature will cause the wire at B to become slack?
11
Solution: 022 = bTbTbP BA PTT BA 22 =+
b
bA
B
2=
, BA 2=
P
TA
TB
Rx
Ry
A
B
12
(1) AE
lTlE
l AAAA === ,
AElTl
El BBBB ===
Therefore AE
lTAE
lT BA 2= , BA TT 2=
( ) PTT BB 222 =+ , 20050052
52 === PTB lb
40020022 === BA TT lb
(2) lTAE
lTAA += and lTAE
lTBB +=
lTAE
lTlTAE
lT BA +=+ 22 TAETT BA = 2 ( ) PTTAET BB 222 =++ ( ) ( ) 92180105.1210120500
52
52 63 === TAEPTB lb
454180105.12101209222 63 =+=+= TAETT BA lb
(3) Set 0=BT , then ( ) 052 == TAEPTB , TAEP = .
Therefore 33.333105.1210120
50063 =
== AEPT F
Further increase in temperature = 333.33F 180F = 153.33F Example 3-8 The shaft assembly shown in the figure below consists of a steel rod A with Youngs modulus of 210 GPa, cross- sectional area of 150 mm2, and coefficient of thermal expansion of 6106 /C, a rigid bearing plate C that is securely fastened to bar A, and a bronze bar B with Youngs modulus of 110 GPa, cross-sectional area of 250 mm2, and coefficient of thermal expansion of 6109 /C, A clearance of 0.5 mm exists between the bearing plate C and bar B before the assembly is loaded. (1) Determine the value of the applied force P to the bearing plate which just closes the gap
between the bearing plate C and bar B, and compute the stresses in bar A and bar B, respectively.
(2) Under the applied load P, the assembly is heated from room temperature (20C) to 100C. Determine the stresses in bar A and bar B, respectively.
13
Solution:
(1) 41025.6800
5.0 ===A
AA l
25.131102101025.6 34 === AAA E MPa 310688.1915025.131 === AA AP N = 19.688 kN 0=B (2) 0= PRR AB AA Rf = and BB Rf = , thus Pff AB =
,TE AA
AA +=
,TE BB
BB +=
A
AA l
= , B
BB l
=
RA
RB
P/2 P/2
14
Since 0=+ BA , 0=+ BBAA ll
0=
++
+ BB
B
BAA
A
A lTE
lTE
0=
++
+ BBBB
BAA
AA
A lTAE
flTAE
f
02008010910110250
8008010610210150
63
63 =
+
+
+
BA ff
26400364.01268.0 =+ BA ff
Since 19688== Pff AB N, then 11785=Af N = 11.785 kN and 31468=Bf N = 31.468 kN
53.78150
11785 ===A
AA A
f MPa (Compressive stress)
88.125250
31468 ===B
BB A
f MPa (Compressive stress)
Example 3-9 A copper tube in the figure below is sealed by two rigid washers. After the nut is tightened by 1/8th of a turn of the thread, the components are in contact with each other. Find the stresses in the bolt and in the copper tube. 207=sE GPa and 120=cE GPa.
200 mm
180 mm
Pitch of thread 1 mm
Rigid washer
Steel bolt 12 mm
Copper tube Do = 25 mm Di = 15 mm
Hamza Mahmood
15
Solution:
sc ff = sscc AA =
Assume that the deformation of copper tube is (compressive), and given that the axial deformation of steel bolt is 125.01
81 = mm (tensile), then the net increment in length of the
bolt is 125.0 .
( ) 310207200
125.0 === s
sssss l
EE
310120180
=== c
ccccc l
EE
( ) 232
1210207200
125.0
=
223
215
22510120
180 , therefore
0448.0= mm
( ) 8310207200
0448.0125.0 3 ==s MPa
30101201800448.0 3 ==c MPa
Example 3-10 In the figure below, a steel pipe (1) is attached to an aluminum pipe (2) at flange B. Both the pipes are attached to rigid supports at A and C, respectively. Pipe (1) has a cross-sectional area of A1 = 3600 mm2, an elastic modulus of E1 = 200 GPa, and an allowable normal stress of 160 MPa. Pipe (2) has a cross-sectional area of A2 = 2000 mm2, an elastic modulus of E2 = 70 GPa, and an allowable normal stress of 120 MPa. Determine the maximum load P that can be applied to flange B without exceeding either allowable stress.
Cylinder a
Rod b
fc
fc
fs
16
Solution:
021 =+ Pff
11
1
1
11 EA
fE
== and 22
2
2
22 EA
fE
==
11
11111 EA
lfl == and 22
22222 EA
lfl ==
Since 021 =+ , 022
22
11
11 =+EAlf
EAlf , then
112
22112 EAl
EAlff = .
Therefore, PEAlEAlff =+
112
22111 or PEAl
EAlf =
+1
112
2211
111 Af = and 222 Af = Let 1601 = MPa,
=
+1
112
2211 EAl
EAlf 53333
112
22111 102.71102003600104.1
10702000108.136001601 =
+
=
+
EAlEAlA N
= 720 kN 720P kN, in this case, 5111 1076.53600160 === Af N = 576 kN,
=== 57672012 fPf 144 kN
722000
1044.1 5
2
22 =
==Af MPa < 120 MPa
221
11221 EAl
EAlff = , then PfEAlEAlf =+ 2
221
1122 or PEAl
EAlf =
+1
221
1122
P/2
P/2
f1 f2
Hamza Mahmood
17
Let 1202 = MPa, then
=
+1
221
1122 EAl
EAlf 63333
221
11222 102.1110702000108.1
102003600104.120001201 =
+
=
+
EAlEAlA N
= 1200 kN 1200P kN, in this case, 5222 104.22000120 === Af N = 240 kN,
960240120021 === fPf kN
2673600
106.9 5
1
11 =
==Af MPa > 160 MPa
Therefore, the maximum load P that can be applied to flange B without exceeding either allowable stress is 720 kN.