CAE DS – High Pressure Die Casting Design
Stress Calculation Basics and Examples
Miskolc University
During solidification and cooling stresses appear in the casting. This can be caused:
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by the shrinkage of the casting,
by the formation of temperature differences,
by different wall thicknesses.
If during the elastic deformation inner stresses storage in those parts of the casting
which are not able for elastic deformation, at that time these residual stress decreases
the structural strength of the casting.
In worst case these stored elastic stresses can cause the broke of the casting. And these
castings can change their size during machining and distortion occurs. Because of the
stored elastic stresses the castings can change their size for a long time marginally and
this distortion can endanger the dimensional accuracy of the casting.
The following factors can cause residual stress in the casting:
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originating from different cooling speed of different casting parts,
the form blocks the free shrinkage of the casting,
at multi-phase materials the different expansion coefficient of different
phases, and the anisotropic behavior of them,
originating from the defects of the crystal lattice,
phase transformations occurs which causes the changing of specifical volume.
From other point of view the origin of residual stresses in castings can be attributed to
different factors or conditions that may be present to a greater or lesser extent in all
castings:
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Differences in surface cooling rates
Thermal gradients due to differences in surface cooling rates as compared to
internal cooling rates for the same section can result in residual stresses. In a
casting of uniform metal section, the rate of heat transfer from the exterior
surfaces into the adjacent mold material will normally exceed the rate of heat
flow from the interior surfaces to cores. This tends to cause more rapid solidification and cooling of the exterior portions of the casting. The occurrence f
unequal rates of solid thermal contraction will result in the development of
residual stresses.
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Non-uniform sections
Thermal gradients due to variations in cooling rates between non-uniform
cross-sections in the same casting can result in residual stresses. In castings
containing adjacent thick and thin cross-sections, thermal gradients due to differences in cooling rate during and after solidification are even more
pronounced. The intensity of the resulting internal stresses will be dependent
on the cross-section geometry and the rate of heat transfer from the casting.
Stress Calculation Basics and Examples - 1
CAE DS – High Pressure Die Casting Design
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Mechanical restrain in the mold
Substantial residual stresses can develop, even in castings of uniform crosssection, if the casting design and/or mold material are of such a nature as to
restrict or restrain the normal thermal contraction of the solidifying and cooling metal. During solidification and cooling the vertical portion of the I tends
to undergo contraction. However, in a rigid mold the upper and lower
horizontal portions serve as effective anchors to restrain and restrict the
contraction, thus giving rise to high tensile stresses in the vertical section. Of
course, removal from the mold would relieve the stress in this example, but
patterns of. residual stresses can develop in a casting where the normal
thermal contraction is mechanically restrained during cooling. Sharp corners
and other stress raisers can combine with rigid geometrical shapes to generate
residual stresses of varying intensities.
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Surface peening during cleaning
High energy cleaning with metal shot or grit can peen the surface metal and
induce compressive stresses: These residual stresses are usually of no concern
and may even be beneficial, but in some cases this effect can cause difficulty.
Large, thin sections can be distorted by unequal peening. The removal of
peened surface metal by machining or grinding can cause an imbalance and
result in distortion.
Theoretical basics
The I. term of thermodynamics is
(1.)
.
.
u ρ = F ⋅ ⋅ A+ rρ + h f ⋅ ∇
Where:
u = speed of inner energy changing
ρ = density of mass of the material
F = strain tensor
A = speed tensor of the deformation
r = power density of the heat source;
hf = density of heat flux of the surface
∇ = differential operator of Hamilton.
The change of the inner energy of the body element volume is equal to the sum of the
change of strain energy, the amount of heat arise or disappear in the volume and the
flowed energy through the surface of the volume.
In the (1) equation we can see that the thermodynamical- and mechanical conditions
are not independent from each other. At thermomechanical problems (e.g. hot rolling)
we cannot solve the thermodynamical- and the mechanical problems together on a
connected way. But in case of casting it gives a good approximation if we first solve
the thermodynamical problem and than we determine the distortions (stresses) by the
help of the result temperature distribution.
Stress Calculation Basics and Examples - 2
CAE DS – High Pressure Die Casting Design
Solving of the thermodynamical problem
We can solve the problem by computing the Fourier differential equation with correct
boundary conditions. The differential equation is
∂  ∂T  ∂  ∂T  ∂  ∂T 
∂T
 +  λz
 λx
 +  λy
 + rρ = ρcv
∂x  ∂x  ∂y  ∂y  ∂y  ∂z 
∂t
(2.)
Where:
T-T(x,y,z) = temperature
λx , λ y , λz = thermal conductivity coefficient in x,y,z directions
t = time
cv = coefficient of specific heat of the material.
Calculating imposed stress
This discussion is simplified consideration of the basic relations in the strength of
materials. Its objective is not full coverage of the science of strength of materials, but to
review the basic stress relationships, especially for the metallurgist and foundryman.
The moment of inertia is the basic relation influencing the stress that is imposed by a
load. By formal definition, moment of inertia is the measure of a cross-section's ability
to resist rotation or bending about the axis passing through the center of gravity. In
use, it is the measure of the metal placement to resist beam-type bending stresses,
shaft type torsional stresses, or compressive stresses involving buckling. These three
conditions of loading encompass the majority of service-loading applications.
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The moment of inertia
The moment of inertia has an important mathematical feature. The moment of
inertia in bending is, for all practical purposes, a cube function of one dimension—the height of the section. Since this value is cubed, small changes can
have an important effect on the imposed stress. The moment of inertia enters
directly into radius of gyration, the design property which determines resistance to buckling in compression. Thus, buckling is dependent on a square or
cube function. Again, slight changes in a dimension effect large changes in
imposed stress. A decrease in imposed stress, in turn, is the equivalent of an
increase in the service strength of the metal. Consider the simple rectangles in
figure 1., differing only in their position in relation to the load. Moment of
inertia is determined by:
I=
(3.)
bh * h 2 bh 3
=
12
12
Where:
I
b
h
moment of inertia
base
height.
Stress Calculation Basics and Examples - 3
CAE DS – High Pressure Die Casting Design
Image 1. The moment of inertia of
a simple rectangle around its base
is influenced by its orientation
Because of the cube function, there is a difference of 16 times in moment of inertia due
to the position of the axis of the rectangle of figure 1 in relation to the load. Equation
(1) can be generalized to cover any shape according to Equation (2) and figure 2.
Moment of inertia of any shape about an axis not passing through it can be determined by:
(4.)
I xx = I aa + Ay 2
Where:
Iaa = moment of inertia of the sections about their own centers of gravity,
Ixx = moment of inertia about axis xx,
A = area of sections,
y = distance from center of gravity of the section as a whole to centers of gravity of each
part.
Image 2. Calculations for the
moment of inertia
When most shapes are analyzed, it will be found that the first component of Equation
(2), Iaa, has a negligible value in relation to the second component, Ay2. A general
statement can be made, therefore, that the moment of inertia of most sections increases
closely as the square of the distance of its components from the center of gravity of the
section as a whole. It is these power functions that make moment of inertia important
to load-carrying ability of any section subject to either bending stress or compressive
stress involving buckling or to torsional stress. There are other methods of calculating
moment of inertia of shapes composed of a number of rectangles like those of figure 2.
One is to subtract moments of inertia of solid rectangles, as in figure 1.
Stress Calculation Basics and Examples - 4
CAE DS – High Pressure Die Casting Design
For example, III of the two rectangles of figure 2. would be: b x ( 2y h )3 — b x ( 2y —
h )3. The I-beam section of figure 2. could be calculated as if it were solid, and then
subtract the moments of inertia of the two areas between the flanges on each side of
the web from the moment of inertia of the solid rectangle, accordingly:
(5.)
I xx = b( 2 y + 2h) 3 − [(b − t )(2 y + 2h)]3
Moment of inertia is part of the basic equations determining imposed stress in pounds
per square inch. The term modulus of rupture is used to designate the stress in
pounds per square inch imposed on a beam before there is any plastic deformation.
Modulus of rupture is the maximum fiber stress measured in pounds per square inch
imposed on the metal at the greatest distance from the neutral axis or center of gravity, as in the following equation:
(6.)
s=
Mc
I
Where the numerator is the bending stress and
s = stress in pounds per square inch at the outermost fiber
M = bending moment, load in pounds times distance from concentrated load to point
under study,
c = distance from neutral axis to outer surface
I = moment of inertia.
When any metal is stressed as a beam so that s equals its tensile strength, only the
outermost fibers of the metal in the section are stressed at the tensile strength. Those
outermost fibers are supported by the mass of metal underneath. An approximate
average stress for some sections of uniform cross-section can be calculated by making
the value c of Equation (4) equal to half the distance from the neutral axis to the outer
surface.
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The modulus of rapture
The modulus of rapture, or more correctly maximum fiber stress, has more
meaning and seems particularly important in calculating working stresses for
fatigue. If the fatigue strength of the surface metal is exceeded, a fatigue crack
may start and propagate. The lower stressed metal towards the neutral axis no
longer can be expected to support the outer metal as it does in static loading.
The initial fatigue crack also increases stress concentration by at least several
orders of magnitude.
As the fatigue crack progresses, the outermost fibers are then at the root of the
crack. Both I and c of Equation (4) decrease in magnitude. The values for I decrease as a square function and are in the denominator of the fraction. The
values for c decrease as a first-order function. Stress, therefore, increases at the
base of the advancing crack. In addition, the degree of stress concentration increases with depth of the crack. All these factors cause the fatigue crack to
propagate - and at an increasing rate.
Stress Calculation Basics and Examples - 5
CAE DS – High Pressure Die Casting Design
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Section modulus
Section modulus –S- is a complimentary factor to I that limits I according to:
S=
(7.)
I
c
Where:
I = moment of inertia
c = distance from neutral axis to outer surface.
If moment of inertia –I- is considered alone, a large spider web would be a strong
structure since I increases approximately with the square of the distance from the
center of gravity of the structure as a whole to the outermost fibers. The fallacy is that
the fiber stress would be too great. The most desirable section is one with I and S as
high as is consistent with shear web buckling.
Values for I and S are considered in relation to the two major axes x and y parallel and
perpendicular to the major load. Other axes may have to be considered.
Rigidity and stiffness often are of equal or greater importance than strength. The same
geometric factors that influence strength determine deflection and stiffness, for moment of inertia enters into all the equations. The modulus of elasticity of the metal
must be considered rather than its strength. The basic equation is for a cantilever
beam:
(8.)
PL3
d=
3EI
Where:
d = deflection in inches,
P = load in pounds,
L = length in inches from point of concentrated load to point under study,
E = modulus of elasticity of metal, the ratio of elastic stress to elastic strain,
I = moment of inertia.
Two additional functions are fundamental to design. The radius of gyration measures
the resistance of a section to buckling under compressive stress and the torsional
modulus measures its resistance to twisting. Both are functions of moment of inertia,
as in Equations (7) and (8):
(9.)
r=
I
a
Where:
r = radius of gyration,
I = smaller moment of inertia of the section,
a = area of the section.
Stress Calculation Basics and Examples - 6
CAE DS – High Pressure Die Casting Design
(10.)
Ss =
J
Tr
Where:
Ss = maximum unit shear stress on outer diameter in pounds per square inch (maximum fiber stress),
J = polar movement of inertia,
T = torque in inch/pounds,
R = radius from center of gravity.
The preceding discussion and equations only review the fundamental design calculations that determine applied stress. It is axiomatic that in a successful design the
applied stress must be matched by the ability of the metal to resist the stress. Unfortunately, many cast shapes are too complex for other than very approximate calculations
and the procedures of experimental stress analysis are needed to provide dependable
answers. For critical applications, where reasonably appropriate calculations can be
made, the experimental procedures are desirable for proof or improved design efficiency.
Determination of residual stress with technological specimen
We can determine residual stress by the help of specimens with unequal wall thickness. One of the possible specimens is the stress lattice. In figure 3 we can see different
types of the specimen and in figure 4 we can see the cooling and shrinkage of it.
Image 3. Stress lattice, square- and round
type
If we see the cooling curves and the contractions at the same time we can tell that the
thin bar reaches the limit of elastic deformation at t1 time. By the help of this we can
divide the cooling process to three parts:
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in region t1: both bars has plastic deformation,
in region t2: the thin bar has elastic deformation, the thick bar has plastic deformation,
after t2: both bar has elastic deformation.
Stress Calculation Basics and Examples - 7
CAE DS – High Pressure Die Casting Design
Image 4. Cooling and shrinkage of the
specimen
In the first period of the cooling, till the δkr temperature the bars of the lattice cannot
contract free, so actually the shape of the shrinkage curve is the c0, c1, c2 dashed
curve. So we can see that the bar nr. I. is contracted proportionally to the c0 a 2 distance, and the thicker bar nr. II. to the c0 a 4 . At temperature t2 the thicker bar also
reaches the area of elastic deformation, and in this area the thicker bar contracted
plastically proportionally to the distance a2 c1 − a4 c0 , and the elongation of the thinner
bar doesn’t changes (c0 a3 = c1a1 ). In the end of the cooling, the thinner bar contracted
elastically at c2 c3 amount, and the thicker part elongated elastically at c2 c4 amount.
Because of these tensile stress awakes in the thicker bar and compressive stress
awakes in the thinner bar.
The whole amount of elastic deformations is:
(11.)
ε = c2 c4 + c2 c3 = αδ kt (e
− k t2
− e −k t )
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The amount stresses awaked in bars nr. I. and II. is:
(12.)
ε = c2 c4 + c2 c3 = αδ kt (e
− k t2
− e −k t )
11
Stress Calculation Basics and Examples - 8
CAE DS – High Pressure Die Casting Design
From time t2 the whole system is in the zone of elastic deformation.
(13.)
T1 = Tkrit e − k1t
(14.)
T2 = Tkrit e − k 2t
Where:
k1>k2 ; k = m a
R
a = temperature conductivity
R = modul
m = constant.
(15.)
T2-T1=Tkrit (e
− k 2t
− e − k1t )
Because the shrinkage is proportional to the temperature changing the shrinkage
curve is change similarly:
(16.)
ε 1 = α ⋅ Tkr e − k t
(17.)
ε 2 = α ⋅ Tkr e − k t
(18.)
ε 1 − ε 2 = α ⋅ Tkr (e − k t − e − k t )
1 2
2 2
2 2
1 2
From the common point of c1 the thin bar must contract along c1c3 , and the thick bar
must contract along c 2 c 4 , which is parallel to the original shrinkage curves. Both bars
have the value of c2:
(19.)
-ε1= c 2 c3
(20.)
+ε2= c 2 c4
Stress awakes in the entire cross section:
(21.)
∑ε = c c
(22.)
Σσ=EΣε
2
+ c 2 c3 = α ⋅ Tkr (e − k t − e − k t )
2 2
4
11
For the thin bar:
(23.)
σ 1 = − Eε össz
A2
A2
= −E
α ⋅ Tkr (e − k1t1 − e − k 2 t 2 )
A1 + A2
A1 + A2
Stress Calculation Basics and Examples - 9
CAE DS – High Pressure Die Casting Design
For the thick bar:
A1
A1
=E
α ⋅ Tkr (e − k1t1 − e − k 2 t 2 )
A1 + A2
A1 + A2
(24.)
σ 2 = + Eε össz
(25.)
Tkr = T0e − k2t2
(26.)
T2 =
1 T0
ln
k 2 Tkr
From (25) and (26) we can write the following equation:
(27.)
σ 1, 2

T
A ,A
= ± E 2 1 α ⋅ Tkr c 1 −  kr

A1 + A2
 T0

k1
−1 
 k2 

 

From formula (27.) we can see the following proportionalities:
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the residual stress is proportional to the modulus of elasticity,
the residual stress is inversely proportional to the ratio of the cross sections,
the residual stress is proportional to the coefficient of thermal expansion.
If we examine two points of the geometry we can see the evolution of the stresses. In
figure 5 we can see the examined points, in figure 6 we can see the evolution of the
stresses.
Image 5. The examined points. Left point:
tensile stress; right point: compressive
stress
Stress Calculation Basics and Examples - 10
CAE DS – High Pressure Die Casting Design
stress, N/mm
2
200
150
100
50
0
-50 0
1000
2000
3000
4000
5000
6000
7000
-100
-150
-200
-250
-300
-350
time, s
Image 6. The evolution of the stresses
In figure 7 we can see distorted stress lattice specimen.
Image 7. The distorted stress lattice specimen
If we apply stress lattice specimens with different cross section rates we can see that
the residual stress values are different.
Cross sections
middle bar – side bars
Measured residual stress
32-12mm
114 N/mm2
32-20mm
48 N/mm2
42-20mm
65 N/mm2
42-30mm
42 N/mm2
Stress Calculation Basics and Examples - 11
CAE DS – High Pressure Die Casting Design
Summary
We can tell that by the help of specimens (e.g. Bauer & Shipp stress lattice) it is suitable to determine residual stresses. Because of the geometrical shape stresses awake in
the bars and we can measure them easily. The cross section of the bars can influence if
the specimen is suitable for stress determination. The most precise cross section for
stress determination is with cross section 32 - 12mm. The form material can influence
the size of the casting; the rigid form doesn’t let the metal to expand.
Stress Calculation Basics and Examples - 12