1. Stress Analysis
Moment of Inertias
1. Atalet moment of inertia;
2
Ix =  y dA
2. Polar moment of inertia;
2
2
Iy =  x dA
2
Jz =  (x + y )dA
Shape
Ix
Iy
Rectangle
3
bh /12
3
hb /12
Triangle
bh3/36
hb3/36
Circle
πd4/64
πd4/64
J
bh 2 2
b +h
12


 h2 +b2 


 18 
bh 
πd4/32
Stresses
Normal Stresses
Tensile
σ=
Compression
σ=
F
A
Axial
σb =
Bending

σb =
Shear Stresses
Tr
=
J
Torsional
F
3
  = 16T/πd for solid
circular beam
VQ
Q=Ay
 
,
Ib
  max = 4V/3A for
A
Mc
I
32M
3
πd
for solid
circular
beam
solid circular beam
Transverse
(Flexural)
  max = 2V/A for
hollow circular
section
  max = 3V/2A for
rectangular beam
2
Principle stresses
σ1,2
σx + σy -  σx - σy 
2
=
+ 
+  xy

2
 2 
2
Max. and min. shear
stresses
 1,2
- σ -σ 
x
y
2
=+ 
 +  xy
2


tan2φ =
2 xy
σx - σy
2
2
2
2
or σ' = σ x + 3 xy (for biaxial)
σ' = σ1 - σ1σ 2 + σ 2
Von-Mises stresses
Stresses States
ε1 =
ε2 =
Triaxial stress state
ε2 =
σ1
E
σ2
E
σ3
E
-μ
σ2 + σ3
-μ
-μ
E
σ1 + σ 3
E
σ1 + σ 2
E
Stress in Cylinders
Thick-Walled (t/r>1/20) Wessels (internally and externally pressurized
cyclinders):
2
σl =
2
2 2
pia - pob - a b (po - pi )/r
σt =
2
b -a
pi a
2
2
2
2
σr =
2
2 2
pia - pob + a b (po - pi )/r
2
b -a
2
2
2
b -a
2
 If the external pressure is zero (po=0);
 b2 
1+ 2 
2
2 
b -a 
r 
2
σt =

σr =
a pi
At the outer surface;
At the inner surface;
r=a  σ r = -pi σ t = pi
 b2 
1- 2 
2
2 
b -a  r 
2
a pi
2
2
2
2
b +a
b -a
r=b  σ r = 0 σ t = pi
2a
2
2
b -a
If the internal pressure is zero (pi=0);
σ t = -po
 a2 
1+ 2 
2
2 
b -a 
r 
b
2
σ r = -po
 a2 
1- 2 
2
2 
b -a  r 
b
2
2
At the outer surface;
At the inner surface;
2b
r=a  σ r = 0 σ t = -po
2
2
b -a
r=b  σ r = -po σ t = -po
2
a=inside radius of the cylinder b=outside radius of the cylinder
pressure po=external pressure
Thin-Walled Wessels(t/r<1/20):
σt =
pdi
σl =
2t
pdi
4t
Curved Members In Flexure:
r=

A
dA
σ=
My
Ae(r - y)

σo =
Mc o
Aero
,
σi =
Mc i
Aeri
ρ
Press and Shrink Fit:
 σ it = -p
 o =
2
2
2
2
b +a
b -a
2
2
2
2
c +b
c -b

+ μo 

2
2
Eo  c - b

 i = -
bp  c + b
 σ ot = -p
2
2
bp  b + a

- μi 

2
2
Ei  b - a

2
2
bp  c + b
 bp  b 2 + a2 
  =  o -  i =  2 2 + μo  +  2 2 - μi 
Eo  c - b
 Ei  b - a

2
2
  c 2 - b2 b 2 - a 2  
 if Eo = Ei = E; interface pressure = p =


b  2b 2  c 2 - a 2  


E
2
2
2
2
b +a
b -a
pi=internal
2. Deflection Analysis
k=
k=
F
y
, k=spring constant
k=
T
θ
=
GJ
l
,
k=Torsional spring rate
AE
for tension or compression loading
l
Castigliano’s Theorem:
y=
θ=
U
F
Tl
GJ
U
Total energy
F
Force on the deflection point
θ
Angular deflection
Strain Energy
2
Axial Load
U=
Torsional Load
U=
F L
2AE
2
Direct Shear Force
U=
2
Flexural Shear
U= 
T L
2GJ
CF
2AG
2
Bending Moment
2
2GA
F L
U=
M
2EI
dx
C=1.2 for rectangular shape
C=1.11 Circular
C=2.0 for thin walled tubular,
dx , C is constant
Buckling Consideration:
l
Slenderness ratio=   ,
k
1/2
k=
 l   2πEC 

  = 
 k 1  S y 
I
A
Euler column
2

Pcr Cπ E
l l
=
      Critical Unit Load =
2
A
 k   k 1
l/k 
Johnson's Column
or
Pcr =
Cπ2EI
l2
Pcr
l l
  <    Critital Unit Load =
A
 k   k 1

2
 Sy   1   l 
= Sy - 
   
 2π   CE   k 
1. Both ends are rounded-simply supported
2. Both ends are fixed
2
 C=1
 C=4
3. One end fixed, one end rounded and guided
4. One end fixed, one end free
 C=2
 C=1/4
3.Design For Static Strength
Ductile Materials
1. Max. Normal Stress Theory
(MNST):

3. Distortion Energy Theory

If, σ1 > σ 2 > σ 3

σ' =

For biaxial stress state;
If, σ1 > σ 2 > σ 3
2

n=
Sy
σ1
2. Max. Shear Stress Theory
(MSST):
 Yield strength
(Ssy)=Sy/2

 max =
 σ1 - σ 3 
2
for biaxial stress state;
 max =
n=
1
2
S sy
τ max
2
2
σ x + 4τ xy
in
2
2
σ' = σ x + 3τ xy
shear

n=
2
(σ1 - σ 2 ) + (σ 2 - σ 3 ) + (σ 3 - σ1 )
Sy
σ'
2
2
Brittle Materials
1. Max. Normal Stress Theory (MNST):


n=
or
n=
If, σ1 > σ 2 > σ 3
Sut
3. The Modified Mohr Theory
(MMT)

If, σ1 > σ 2 > σ 3

S3 =
σ1
Suc
σ3
Suc
Suc - Sut σ1
Sut
2. The Column Mohr Theory (CMT) or
Internal Friction Theory (IFT):

S3 =
Suc
Suc σ1
Sut σ 3
-1

n=
S3
σ3

n=
σ3
S3
σ3
5. Design for Fatigue Strength
Endurance limit for test specimen (Se’);

For ductile materials:
Se’=0.5 Sut if Sut<1400 MPa
Se’=700 MPa if Sut  1400 MPa

For irons:
Se’=0.4 Sut if Sut<400 MPa
Se’=160 MPa if Sut  400 MPa

For Aliminiums:
Se’=0.4 Sut if Sut<330 MPa
Se’=130 MPa if Sut  330 MPa

For copper alloys:
Se’  0.4 Sut if Sut<280 MPa
Se’  100 MPa if Sut  280 MPa
Se = ka kb kc kd ke Se’
Sf=10c Nb
-1
 0.8Su 
b = - log 

3
 Se 
1
  0.8Su 2 
c = log 

 Se 
ka : surface factor, ka=aSutb

Surface Finish
Factor a
Factor b
Ground
1.58
-0.065
Machined or Cold Drawn
4.51
-0.265
Hot Rolled
57.7
-0.718
As Forged
272
-0.995

kb : size factor;
kb=1 if d  8 mm and kb= 1.189d-0.097 if 8 mm<d  250 mm
bending & torsional loading.
For non-rotating element, k b = 1.189deq
-0.097
for
deq=0.37d
and Se’=0.45Sut
For pure axial loading, kb=1
For combined loading,  =1.11 if Sut  1520 MPa and  =1 if Sut
 1520 MPa for ductile materials.

kc : reliability factor

kd : temperature effects, kd=1 if T  3500 and kd=0.5 if 3500<T  5000

ke : stress concentration factor, ke=1/Kf
Kf=1+q(Kt-1)
Kt : geometric stress concentration factor, q=notch sensitivity.
Modified Goodman
Infinite Life
n=
1
σa
Se
+
σm
Su
 Fa=(Fmax-Fmin)/2
Soderberg
Finite Life
n=
1
σ a σm
+
S f Su
Infinite Life
n=
1
σa
Se
 Fm=(Fmax+Fmin)/2
+
σm
Sy
Finite Life
n=
1
σa
Sf
+
σm
Sy
6. Tolerances and Fits
Clearance fit:
TF=Cmax-Cmin
Cmax=DU-dL
Cmin=DL-dU
Cmax=DU-dL
Imax=dU-DL
Imax=dU-DL
Imin=dL-Du
Transition fit:
TF=Imax+Cmax
Interference fit:
TF=Imax-Imin
Tolerances on the shaft, on the hole and on the fit
TS=dU-dL
TH=DU-DL
TF=TH+TS
7. Design of Power Screws
TR =
Fdm  L + πdmμ 


2  πdm - μL 
TL =
Fdm  πdmμ - L 


2  πdm + μL 
Considering μ = tanρ ;
TR =
Fdm
tan  λ + ρ 
2
If μ  tanλ or  
L
πdm
or
TL =
Fdm
2
tan ρ - λ 
   or TL>0, then screw is self locking.
If the friction between the stationary member and the collar of the screw
is taken into consideration;
TR =
Fdm
2
tan  λ + ρ  +
μc dc F
TL =
2
Fdm
2
tan ρ - λ  +
μc dc F
2
Condition for self locking is : TL>0
Efficiency of screws: ε =
obtain  as, ε =
tanλ
tan  λ + ρ 
To
TR
=
FL
2πTR
when collar friction is negligible, we
Thread Stresses:

Bearing Stresses
σb =


4pF
2
πh d - dr
2
or

σb =
Fp
πdm th
s =
2F
n =
πdr h
6F
πdmNp
πdh
N=h/p
Tensile or Compressive stresses
σx =
F
At =
At
πdt
2
dt =
4
Shear Stresses
 xy =

2F
Bending Stresses
Stresses on the body of screw:

2
For Nut Thread
The maximum bending stress, σ =

p
Shear Stresses
For Screw Thread

where t =
16TR
πdt
3
Combined stresses:
Based on distortion energy theory;
2
σ' = σ x + 3  xy
2
n=
Sy
σ'
Based on maximum shear stress theory;
max =
1
2
2
σ x + 4  xy
2
n=
S sy
max
dr + dm
2
8. Design of Bolted Joints
Feb and Fep are forces shared by the bolt and by the members respectively.
Fe=Feb+Fep
Feb=CFe
Fb=Fi+CFe
Fm=Fi-(1-C)Fe
Stiffness of bolt: k b =

A bE b
L
Fep=(1-C)Fe
stiffness ratio:
Stiffness of members:
1
km
=
1
k1
+
C=
1
k2
Shigley and Mishke approach;
0
For cone angle of  = 30 ,
ki =
1.813Ei d
1
 1.15Li + 0.5d 

 1.15Li + 2.5d 
km
ln  5
=
1
k1
+
1
k2
+ .......... +
1
kn
1.813Ed
If L1=L2=L/2 and materials are same, k m =
 2.885L + 2.5d 

 0.577L + 2.5d 
2ln 
0
For cone angle of  = 45 ,
ki =
πEi d
 5  2Li + 0.5d 

  2Li + 2.5d 
ln 
If L1=L2=L/2 and materials are same, k m =
πEd
 L + 0.5d 

 L + 2.5d 
2ln  5

Wileman approach;
kb
kb  km
+ .......... +
1
kn
k m = EdA i e
(Bid/L)
where Ai and Bi are constants related to the material.
For Steel Ai=0.78715 and Bi=0.62873, for Aliminium Ai=0.79670 and
Bi=0.63816, for Gray cast iron Ai=0.77871 and Bi=0.61616.

Filiz approach;
km =
π
2
Eeqd
π
 d 
 -B1  
 L 
1
1- B 2
e 5
 0.1d 

 L 
where E eq =
2
B1 = 

L1 

L2 
B1 =  1-
E1E 2
E1 + E2
8

Static loading;
S
Fb  S y A t or Fb  Sp A t
1- C nFe  Fi  Sp A t - CnFe
Critical load= Fce =
p
= 0.85S y

Fm  0
n=load factor of safety
Fi
1- C
Dynamic Loading:
σa =
CnFe
2A t
σm =
Fi
At
+ σa
ns =
1
σa
Se
Fi =

A t Su CnFe  Su
 +1
ns
2  Se

+
σm
Su
where: ns is strength factor of safety.
Fi=the maximum value of preload for there is no fatigue failure.
Constraints:

0.6Fp  Fi  0.9Fp

Fi
max
= A t Sut -
where
cFen  Sut

+1

2N  Se

Fp = A t Sp
Fe

(1- c)

3.5d  c b  10d N
 Fi  A t Sp -
cFe
N
9. Design of Riveted Joints
Shearing of Rivets:
=
F
,
A
Secondary Shear Force:
F=Force on each rivet
F''
i =
Mri
N
 ri
2
1
A=
πd
2
4
Bearing (compression) Failure:
=-
F
A
,
A=td,
t=thickness of the
plate
=
F
A
,
A =  w - Nd t
w=width of plate
Primary Shear Force:
F' =
Plate Tension Failure:
N=number of rivets on the selected
cross section
F
N
 Ai
1
10.

Primary Shear Stress
' =

Design of Welded Joints
J = 0.707hJu

I = 0.707hIu
F
A
Secondary Shear Stress
'' =

Mr
J

Bending Stress
σ=
Mc
I
Table 9-3 Minimum weld-metal properties
AWS electrode
Number
E60xx
E70xx
E80xx
E90xx
E100xx
E120xx
Tensile
Strength
Yield Strength
Percent
MPa
340
390
460
530
600
740
Elongation
17-25
22
19
14-17
13-16
14
420
480
530
620
690
830
Table 9-5 Fatigue-strength reduction factors
Type of Weld
Kf
Reinforced butt weld
Toe of transverse fillet weld
End of parallel fillet weld
T-butt joint with sharp corners
1.2
1.5
2.7
2.0
Table 9-1 Torsional Properties of Fillet Welds*
Weld
Throat Area
Location of G
Unit Polar Moment of
Inertia
*G is centroid of weld group; h is weld size; plane of torque couple is in the
plane of the paper; all welds are of the same size.
Table 9-2 Bending Properties of Fillet Welds*
Weld
Throat Area
Location of G
Unit Moment of Inertia
*Iu, unit moment of inertia, is taken about a horizontal axis through G, the
centroid of the weld group; h is weld size; the plane of the bending couple is
normal to the paper; all welds are of the same size