28
- Mechanics & Materials 1 FAMU FSU College of Engineering Department of Mechanical Engineering Chapter 14 Chapter 14 Combined Loadings Combined Loadings
-
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
