mechanics & materials 1 chapter 14 combined...
TRANSCRIPT
-
-
Mechanics & Materials 1
FAMU FSU College of Engineering
Department of Mechanical Engineering
Chapter 14 Chapter 14
Combined LoadingsCombined Loadings
-
Thin Walled Pressure VesselsThin Walled Pressure Vessels
• Cylindrical or spherical pressure vessels are used in industry as tanks, boilers or containers.
• When under pressure the material is subjected to loadings in all directions.
• In general thin wall refers to an inner radius to wall thickness ratio ri/t ≥ 10, in most cases it is actually > 50.
• If the vessels wall is thin, the stress distribution through thethickness can be assumed to be uniform or constant.
-
Cylindrical Pressure VesselsCylindrical Pressure Vessels
• Consider tubes with an internal pressure and closed ends
• σxx �axial stress due to the pressure on the end walls
• σθθ�“Hoop” stress due to
the pressure acting on the curved surface.
• P is the internal pressure• Look at a FBD of the axial
section
-
Axial Stress
• Consider the axial equilibrium on this structure.
• Which gives the equation for Axial Stress:
( ) 22
2*
20
rPrt
rtrPF
xx
xxx
ππσ
σππ
=
+−==∑
txx 2
Pr2 == σσ
Cylindrical Pressure Vessels: Cylindrical Pressure Vessels: Axial StressAxial Stress
-
Look now at a FBD of the circumferential section
• Equating the forces vertically gives:
• Which simplifies to give the equation for Hoop Stress:
Cylindrical Pressure Vessels: Cylindrical Pressure Vessels: Hoop StressHoop Stress
t
pr== 1σσθθ
rLPLt
LtLRPFy
2)(2
2*2**0
=
+−==∑θθ
θθ
σσ
-
Cylindrical Pressure VesselsCylindrical Pressure Vessels
• Pressure vessels also have stresses created by the weight of the pressurized fluid inside, its own weight, externally applied loads and by an applied torque.
• To analyze this, each loading condition is analyzed individually and the stresses are then combined along their respective axes by the superposition method.
• Pressure vessel with ALL possible loading conditions:
-
Pressure Vessel With ALL Pressure Vessel With ALL Possible Loading Conditions:Possible Loading Conditions:
-
Spherical Pressure VesselsSpherical Pressure Vessels
• Common for storing liquid gases (maximum storage volume for the least volume of material)
• Equating the forces vertically:
• Which simplifies to:
• This is not dependent on the orientation of our FBD, so this stress acts on both surfaces of the element simultaneously.
∑ +−== σππ rtrPFy 2*0 2
t2
Pr=σ
-
Example Example
• A steel gas bottle 2 m long with a diameter of 250 mm and a 3 mm wall thickness is pressurized to 3 MPa.
• Determine the stresses
-
SolutionSolution
-
Problem 14.1Problem 14.1
• The spherical gas tank has an inner radius of r = 1.5m.(If it is subjected to an internal pressure of p=300kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa.
-
mmmtt
tallowz
8.1801875.0 2
)5.1)(10(300)10(12
2
Pr
36
==
=
== σσ
SolutionSolution
-
Example 14.3Example 14.3
• The cap of the cylindrical tank is bolted along the flanges. The tank has an inner diameter of 1.5m and a wall thickness of 18mm.
• If the largest normal stress is not to exceed 150MPa, determine the maximum pressure the tank can sustain .
• Also compute the number of bolts required to attach the cap to the tank if each bolt has a diameter of 20mm.
• The allowable stress for the bolt is 180MPa
-
MPam
N
t
r
60.3000,600,3
018.0
)75.0()150(10
0.759mr here
2
6
1
==
=
==
ρ
ρ
ρσ
1135.112
)02.0)(4/(
)10(73.6361)10(180
73.6361
)5.1)(4/)(10(60.3
2
36
26
⇒=
=
=
=
==
n
n
nA
F
kN
FF
b
ballow
bpr
π
σ
π
Fpr Fb
SolutionSolution
-
Analysis of Stress in Combined LoadingAnalysis of Stress in Combined Loading
• Normal Stress Sources( σ)– Axial Load :- The uniform normal stress distribution due to the normal
force.
– Bending Moment :- Normal stress distribution due to the bending moment.
– Pressure Cylinder :- Biaxial state of stress in the material due to pressure in a thin-walled cylinder.
– Thermal :- Normal stress distribution due to temperature change.
A
F=σ
I
My=σ
t
pr=σ 1t
pr
22=σ
TE∆= ασ
-
• Shear Stress Sources (τ):– Torsion:- Shear-stress distribution that varies linearly from the
central axis to the maximum at the shaft’s outer boundary.
– Transverse Shear Stress :- Shear stress distribution that acts over the cross section.
Analysis of Stress in Combined Analysis of Stress in Combined LoadingLoading
J
T ρτ =
It
VQ=τ
-
Analysis of Stress in Combined Analysis of Stress in Combined LoadingLoading
• After finding each component of the stress we– add (algebraically) all those components which
contribute normal stress to get the total normal stress– add (algebraically) the components which give a shear
stress to get the total shear stress. – We can’t add a shear component to a normal stress
component, because stresses are tensorial quantities which need at least 2 directions rather than one as in the vectors.
-
Analysis of Stress in Combined Analysis of Stress in Combined LoadingLoading
Remember to assign positive or negative signs for the stresses according to
• Axial Loading σ + (tension), σ(-) compression• Bending: σ sign depends on both M sign and c sign, M
sign from Moment diagram, c sign from N.A location• Pressure: Inside pressure σ +, outside pressureσ -• Thermal: Heating: expansion +σ, cooling: contraction: -σ• Torsion: τ sign depends on torque (T) sign (Right hand
rule)• Transverse Shear: τ sign depends on Q sign (N.A location
and section of interest) and shear force sign V, from shear diagram)
-
Combined LoadingCombined Loading• This example has multiple stress
components. The stresses on the rod are axial, shear, two bending, and a torsional stress.
• NOTE THAT ONLY 2 LOADS LEADS TO 5 STRESS COMPONENTS !!
• When found, each stress value is understood as shown in the diagram below.
-
Problem 14.11Problem 14.11
• The offset link supports the loading shown. Determine its required width w if the allowable normal stress is 73MPa. The link thickness is 40mm.
-
σ due to axial force
σ due to bending
wwA
Pa
)10(750
)04.0(
)10(30 33 ===σ
2
3
3
3
205.0)10(4500
))(04.0(121
2205.0)10(30
w
w
w
ww
I
Mcb
+=
+==σ
baallow σσσσ +==max
SolutionSolution
-
mm
mw
ww
www
w
w
w
7.79
0797.0
0225.0373
25.2225.075.073
205.0)10(4500
)10(750)10(73
2
2
2
33
6
==
=−−
++=
++=
SolutionSolution
-
Problem 14.20Problem 14.20
• The bar has a diameter of 40mm. If it is subjected to a force of 800N as shown, determine the stress components that act at points A and B and show results on volume elements located at these points.
-
MPa
I
Mz
A
P
m
AyQ
rA
m
rI
A
A
318.0
0)10(256637.1
400
Bending extension Axial :stress Axial
0Q
)10(3333.5
2
)02.0(
3
)02.0(4
)10(256637.1
)02.0(
)10(1256637.0
)02.0)((4
1
4
1
3
B
36
2
3
22
46
44
=
+=+=
+=
=
=′′=
===
=
==
−
−
−
−
σ
ππ
ππ
ππ
SolutionSolution
-
0
7.21
)10(1256637.0
)02.0(56.138
)10(1256637.0
400
735.0
)04.0)(10(1256637.0
)10)(3333.5(82.692
66
6
6
=−=
−=−=
=
==
−−
−
−
B
B
AA
MPa
I
Mc
A
P
MPa
It
VQ
τ
σ
τ
SolutionSolution
Shear Stress: Only transverse shear
-
Problem 14.25Problem 14.25
• The pliers are made from two steel parts pinned together at A.
• If a smooth bolt is held in the jaws and a gripping force of 10lb is applied at the handles, determine the stress components in the pliers at points B and C.
• The cross section is rectangular having dimensions as shown.
-
lbM
F
lbV
F
c
y
X
30
010(3)-M 0M
5lbN
010sin30-N 0
660.8
010cos30-V 0
=
===
===
==
∑
∑
∑
0
)10(4
)2.0)(2.0)(1.0(
)10(0667.1
)4.0)(2.0(12
1
33
3
3
==
=′′==
=
−
−
B
c
Q
in
AyQ
I
SolutionSolution
xx
yy
BB
0.40.4””
0.20.2””
CC
__yy’’ =.1=.1””
208.0)4.0(2.0 inA ==
-
ksi
psi
I
M
A
N yB
56.5
5562
)10(0666.1
)2.0(30
08.0
53
==
−=+−= −σ
psiIt
VQ
psiI
M
A
N
cc
yc
B
162)2.0)(10(0667.1
)10)(4(660.8
5.6208.0
5
0
3
3
===
−=−=+−=
=
−
−
τ
σ
τ
BB
0.4”0.4”
0.2”
0.2”
__y’=.1”y’=.1”
SolutionSolution
N.AN.A