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A state of plane stress exists at a point Q with ππ§ = ππ§π₯ = ππ§π¦ = 0. The state of plane stress is defined by the
stress components ππ₯ , ππ¦ , ππ₯π¦ associated with the material shown:
If the material is rotated through an angle π about the z axis, the stress components change to ππ₯β², ππ¦
β², ππ₯π¦β²
which can be expressed in terms of ππ₯ , ππ¦ , ππ₯π¦ and π
Consider a prismatic element with faces respectively perpendicular to the x, y and xβ axes:
If the area of the oblique face is ΞA, the areas of the vertical and horizontal faces are equal to ΞA cosΞΈ, and ΞA sinΞΈrespectively.
The mechanical equilibrium along the xβ and yβ axes require that
πΉπ₯β² = 0, ππ₯β²Ξπ΄ β ππ₯ Ξπ΄ cos π cos π β
ππ₯π¦ π΄ cos π sin π β ππ¦ Ξπ΄ sin π sin π β
ππ₯π¦ π΄ sin π cos π = 0
πΉπ¦β² = 0, ππ₯β²π¦β²Ξπ΄ + ππ₯ Ξπ΄ cos π sin π β
ππ₯π¦ π΄ cos π cos π β ππ¦ Ξπ΄ sin π cos π +
ππ₯π¦ π΄ sin π sin π = 0
The first equation is solved for ππ₯β² and the second for ππ₯β²π¦β² as
ππ₯β² = ππ₯ cos
2 π + ππ¦ sin2 π + 2ππ₯π¦ sin π cos π
ππ₯β²π¦β² = β ππ₯ β ππ¦ sin π cos π + ππ₯π¦ cos2 π β sin2 π
After simplifications using trigonometric substitutions we obtain the normal and shear stresses on the rotated material as
ππ₯β² =
ππ₯ + ππ¦
2+ππ₯ β ππ¦
2cos 2π + ππ₯π¦ sin 2π
ππ¦β² =
ππ₯ + ππ¦
2βππ₯ β ππ¦
2cos 2π β ππ₯π¦ sin 2π
ππ₯β²π¦β² = βππ₯ β ππ¦
2sin 2π + ππ₯π¦ cos 2π
The expression for the normal stress ππ¦β² is obtained by replacing ΞΈ by the angle ΞΈ+90 that the yβ axis forms with
the x axis.
Adding the two normal stresses we see that
ππ₯β² + ππ¦
β² = ππ₯ + ππ¦
In the case of plane stress, the sum of the normal stresses exerted on a cubic material is independent of the orientation of the material since ππ§ = ππ§β² = 0
The equations obtained for the normal shear stresses in the rotated material under plane stress condition are the parametric equations of a circle
If we plot a point M in the rectangular axes with the coordinates (ππ₯β², ππ₯β²π¦β²) for any given value of the parameter
ΞΈ, all the other possible points will lie on a circle.
ππ₯β² =
ππ₯ + ππ¦
2+ππ₯ β ππ¦
2cos 2π + ππ₯π¦ sin 2π
ππ₯β²π¦β² = βππ₯ β ππ¦
2sin 2π + ππ₯π¦ cos 2π
The angle ΞΈ in the equations can be eliminated by algebraic simplifications and addition of the two equations:
ππ₯β² β
ππ₯ + ππ¦
2
2
+ ππ₯β²π¦β²2 =
ππ₯ β ππ¦
2
2
+ ππ₯π¦2
Where ππ₯+ππ¦
2= πππ£π and
ππ₯βππ¦
2
2+ ππ₯π¦
2 = π 2
So ππ₯β² β πππ£π
2 + ππ₯β²π¦β²2 = π 2
Which is the equation of a circle of radius R centered at the point C of coordinates (πππ£π , 0)
The point A where the circle intersects the horizontal axis is the maximum value of the normal stress ππ₯β² and the
other intersection point B is the minimum value. Both points correspond to a zero value of shear stress ππ₯β²π¦β².
These are the principle stresses.
Since ππππ₯ = πππ£π + π ππππ = πππ£π β π
ππππ₯,πππ =ππ₯ + ππ¦
2Β±
ππ₯ β ππ¦
2
2
+ ππ₯π¦2
The rotation angles that produce the principal stresses with no shear stress is obtained from the equation of shear stress
ππ₯β²π¦β² = βππ₯ β ππ¦
2sin 2π + ππ₯π¦ cos 2π = 0
tan 2ππ =2ππ₯π¦
ππ₯ β ππ¦
This equation gives two ππ values that are 90 apart. Either of them can be
used to determine the orientation of the corresponding rotated plane. These planes are the principal planes of stress at point Q
The points D and E are located on the vertical diameter of the circle corresponding to the largest numerical value of the shear stress ππ₯β²π¦β².
These points have the same normal stresses of πππ£π. So the rotation that produces the maximum shear stresses can be obtained from the normal stress equations.
ππ₯β² = πππ£π =
ππ₯ + ππ¦
2+ππ₯ β ππ¦
2cos 2π + ππ₯π¦ sin 2π =
ππ₯ + ππ¦
2
tan 2ππ = βππ₯ β ππ¦
2ππ₯π¦
This equation gives two ππ values that are 90 apart. Either of them can be used to determine the orientation of corresponding rotated plane that produces the maximum shear stress which is equal to
ππππ₯ = π =ππ₯ β ππ¦
2
2
+ ππ₯π¦2
The normal stress corresponding to the condition of maximum shear stress is
ππ₯β² = πππ£π =
ππ₯ + ππ¦
2
Also
tan 2ππ = βππ₯ β ππ¦
2ππ₯π¦= β tan 2ππ
β1= β
2ππ₯π¦
ππ₯ β ππ¦
β1
This means that the angles ππ and ππ are 45 apart
So the planes of maximum shear stress are oriented at 45 to the principal planes
Example β Determine the principal planes, principle stresses, maximum shear stress and the corresponding normal stress for the state of plane stress shown
Yield criteria for ductile materials under plane stress
When a ductile material is under uniaxial stress, the value of the normal stress ππ₯ which will cause the material to yield can be obtained simply from a stress-strain diagram obtained by a tensile test.
The material will deform plastically when ππ₯ > ππ
On the other hand when a material is in a state of plane stress, the material will yield when the maximum value of the shear stress exceeds the corresponding value of the shear stress in a tensile-test specimen as it starts to yield.
Maximum shear stress criterion is based on the observation that yield in ductile materials is caused by slippage of the material along oblique surfaces and is due primarily to shear stresses.
In the plane stress condition the material can be represented as a point under principal stresses ππ , ππ
Recall that the maximum value of shear stress at a point under a centric axial load is equal to half the value of the corresponding normal axial stress.
Thus at yielding
ππππ₯ =1
2ππ
Also for plane stress condition if the principle stresses are both positive or both negative, the maximum value of
the shear stress is equal to 1
2ππππ₯
Therefore ππ > ππ or ππ > ππ
If the maximum stress is positive and the minimum stress negative, the maximum value of the shear stress is
equal to 1
2ππππ₯ β ππππ
Therefore ππ β ππ > ππ
These relations produce a hexagon in the xy plane, called Trescaβs hexagon. Any given state of stress will be represented in the figure by a point.
Maximum distortion energy criterion is based on the determination of the distortion energy in a given material, which is the energy consumed by a change in the shape of the material.
Also called von Mises criterion, it states that a material will yield when the maximum value of the distortion energy per unit volume exceeds the distortion energy per unit volume required to cause yield in a tensile test specimen.
The distortion energy in an isotropic material under plane stress is
ππ =1
6πΊππ
2 β ππππ + ππ2
In the case of a tensile test specimen yielding at ππ
ππ =1
6πΊππ2
Thus the maximum distortion energy criterion indicates that the material yields when ππ > ππ:
ππ2 β ππππ + ππ
2 > ππ2
This equation produces an ellipse in the principal stress plane
Prediction of yielding under multiaxial loading according to the maximum shear stress and von Mises criterion iinvolves the analysis of the octahedral planes
Figure octahedral plane
There are eight octahedral planes making equal angles with the principal stress directionsThe shearing stress on these planes is given by
ππππ‘ =1
3π1 β π2
2 + π2 β π32 + π3 β π1
2
Or with nonprinciple stresses:
ππππ‘ =1
3ππ₯ β ππ¦
2+ ππ¦ β ππ§
2+ ππ§ β ππ₯
2 + 6 ππ¦π§2 + ππ§π₯
2 + ππ₯π¦2
The shear strain acting on an octahedral plane is given by
πΎπππ‘ =2
3π1 β π2
2 + π2 β π32 + π3 β π1
2
Or
πΎπππ‘ =2
3ππ₯ β ππ¦
2+ ππ¦ β ππ§
2+ ππ§ β ππ₯
2 +3
2πΎπ¦π§
2 + πΎπ§π₯2 + πΎπ₯π¦
2
Von Mises yield criterion is visualized as a circular cylinder in the stress space
Fig stress space
The axis of the cylinder passes through the origin of the coordinates for unyielded material
It is inclined equal amount to the three axes and represents pure hydrostatic stress for elastic deformations.
The von Mises yield criterion is given by
π1 β π22 + π2 β π3
2 + π3 β π12 = 2ππ¦
Or
ππ₯ β ππ¦2+ ππ¦ β ππ§
2+ ππ§ β ππ₯
2 + 6 ππ¦π§2 + ππ§π₯
2 + ππ₯π¦2 = 2ππ¦
In terms of effective stress the criterion is
ππππ =1
2π1 β π2
2 + π2 β π32 + π3 β π1
2 = ππ¦
ππππ =2
2ππ₯ β ππ¦
2+ ππ¦ β ππ§
2+ ππ§ β ππ₯
2 + 6 ππ¦π§2 + ππ§π₯
2 + ππ₯π¦2 = ππ¦
For plane states of stress, the yield condition is the interaction of the cylinder with the principal stress plane, which is a yield ellipse
The effective stress is the uniaxial stress that is equally distant from or located on the yield surface
Yield criteria for deformation of metals under plane stress
The data for the mild steel and Cr-V steel which behave in a ductile manner agree well with the octahedral shear stress (von Mises) criterion
Data for cast iron which behaves in a brittle manner, agrees better with the maximum principal stress criterion:π1 = ππ¦
Brittle materials fail suddenly in a tensile test by rupture without any prior yielding.
When a brittle material is under uniaxial tensile stress, the value of the normal stress which causes it to fail is equal to the ultimate strength of the material as determined from a tensile test.
When a brittle material is under plane stress, the principal stresses are compared to the ultimate strength obtained from the uniaxial tensile test.
Maximum normal stress criterion states that a brittle material will fail when the maximum normal stress exceeds the ultimate strength obtained from the uniaxial tensile test:
ππ > ππ or ππ > ππ
This criterion forms a square area centered on the xy plane. The criterion is based on the assumption that the ultimate strength of materials under tension and compression are equal, which is an overestimation for most materials as the presence of cracks and flaws often weaken the material under tension
A rod of length L=500 mm and cross-sectional area A=60 mm2 is made of an elastoplastic material having a modulus of elasticity E= 200 GPa in its elastic range and a yield point ππ= 300 MPa. The rod is subjected to an axial load until it is stretched 7 mm and the load is removed. What is the resulting permanent set?
A nylon thread is to be subjected to an 11-N tension. What is the required diameter of the thread if its modulus is 3.1 GPa, the maximum allowable normal stress is 40 MPa and the length of the thread should not increase by more than 1%.
An aluminum test specimen that is in the shape of a bar of length 300mm*60mm*15mm is subjected to two equal and opposite centric axial forces of magnitude P. What is the maximum allowable value of P and the corresponding total elongation of the specimen if E= 70 GPa and ππππππ€ππππ= 200 MPa.
The cylindrical rod AB has a length L= 2m and a 32mm diameter. It is made of a mild steel which is elastoplastic with E= 200 GPa and ππ= 250 MPa. A force P is applied to the rod until its end A has moved down relative to the its top side that is connected to a support. What is the maximum value of the force P and the permanent set of the rod after the force has been removed if the elongation is a) 3mm, b) 6mm?
A steel rod of length L and uniform cross section of area A is attached to rigid supports and is unstressed at a temperature of 20 C. The steel is assumed to be elastoplastic with E= 200 GPa and ππ= 250 MPa. What is the stress in the rod after the temperature has been raised to 150 C if the coefficient of thermal expansion is 11.7*10-
6/C?
A cylindrical block of brass, that is 160 mm high and 120 mm in diameter is lowered into the ocean to a depth of 7500 m where the pressure is 75 MPa. Determine a) the change in height of the block, b) the change in its diameter, c) the change in its volume if the E= 105 GPa and π= 0.35. What is the pressure which should be applied d) to its top and bottom faces only, and e) to its cylindrical surface only to cause the same change of volume as the hydrostatic pressure?
What is the change in volume of a steel block of dimensions 80*60*40 mm when it is subjected to the hydrostatic pressure p= 180 MPa? What are the changes in the length of each side? E= 200 GPa, π= 0.29.
In a standard tensile test an aluminum rod of 20 mm diameter is subjected to a tensile force of magnitude P= 30 kN. What is the a) elongation of the rod in a 150 mm gage length and b) change in diameter of the rod if E= 70 GPa and π= 0.35. Also consider c) the dilatation (the change in volume) of the rod.
In a standard tensile test a 20 mm diameter rod made of an experimental polymer is subjected to a tensile force of magnitude P= 6 kN. What are the modulus of elasticity, the modulus of rigidity and the poissonβs ratio of the material if the elongation is 14 mm and decrease in diameter is 0.85 mm?
A 1.2 m concrete post is reinforced by four steel bars, each of 19 mm diameter. What are the normal stresses in the steel and concrete when a 700 kN centric force is applied to the post if the moduli for steel and concrete are Es= 200 GPa, Ec= 25 GPa ?
What are the normal stresses induced in the steel and in the concrete by a temperature rise of 45 C if the coefficients of thermal expansion are Ξ±s= 11.7*10-6/C and Ξ±c= 9.9*10-6/C ?
The aluminum rod AD is fitted with a jacket which is used to apply a hydrostatic pressure of 40 MPa to the 300 mm portion BC of the rod. What is a) the change in the total length AD of the rod, b) the change in the diameter at the middle of the rod if the E= 70 GPa and π= 0.36. Also determine the forces which should be applied to the ends A and D of the rod c) if the axial strain in portion BC is to remain zero as the hydrostatic pressure is applied and d) if the total length AD of the rod is to remain unchanged.
The assembly shown consists of an aluminum shell that is fully bonded to a steel core and is unstressed at a temperature of 20. Compressive centric forces of 180 kN are applied at both ends of the assembly by means of rigid end plates. What are a) the normal stresses in the steel core and the aluminum shell and b) the deformation of the assembly ? The moduli of aluminum and steel are Ea= 70 GPa, Es= 200 GPa.
Considering only axial deformations, determine the stress in the aluminum shell when the temperature reaches 180 C. The coefficients of thermal expansion are Ξ±a= 23.6*10-6/C and Ξ±s= 11.7*10-6/C.
What is the stress in the aluminum shell if the core is made of brass with Eb= 105 GPa, Ξ±b= 20.9*10-6/C ?
A rectangular block of a material with a modulus of rigidity (shear modulus) G= 600 MPa is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. What are the a) average strain in the material and b)the force P on the upper plate if the upper plate moves through 0.8 mm under the action of the force?
An 800 mm long cylindrical rod of cross-sectional area Ar= 45 mm2 is placed inside a tube of the same length and of cross-sectional area At= 60 mm2. The ends of the rod and tube are attached to a rigid support on one side, and to a rigid plate on the other, as shown in the figure. The rod and tube are both assumed to be elastoplastic with moduli of elasticity Er= 200 GPa and Et= 100 GPa and yield strengths (Οr)Y= 200 MPa and (Οt)Y= 250 MPa. A) What is the load-deflection diagram of the rod-tube assembly when a load P is applied to the plate. B) What is the elongation of the assembly if the load P applied is increased from zero to 19.5 kN and decreased back to zero. C) What is the permanent set after the load has been removed.
In many situations it is known that the normal stress in a given direction is zero in the case of a thin plate that is under plane stress condition. Show that the following expressions for stresses and strain are correct for the strains in the x and y directions that are determined experimentally.
ππ₯ = πΈππ₯ + πππ¦
1 β π2
ππ¦ = πΈππ¦ + πππ₯
1 β π2
ππ§ = βπ
1 β πππ₯ + ππ¦
In many situations physical constraints prevent strain from occurring in a given direction. For example Ξ΅z= 0 in the case of plane strain condition where longitudinal movement of the long prism is prevented at every point. Plane sections perpendicular to the longitudinal axis remain plane and the same distance apart. Show that for this situation the stresses and strains are expressed as:
ππ§ = π ππ₯ + ππ¦
ππ₯ =1
πΈ1 β π2 ππ₯ β π 1 + π ππ¦
ππ¦ =1
πΈ1 β π2 ππ¦ β π 1 + π ππ₯
A 20 mm square was scribed on the side of a large steel pressure vessel. The plane stress condition of the material after pressurization is shown. What is a) the change in the lengths of the sides and the percent change in the slope of the diagonal DB if E= 200 GPa and π= 0.3 ?
A circle of diameter d= 200 mm is scribed on an unstressed aluminum plate of thickness t= 18 mm. Forces acting in the plane of the plate later cause normal stresses ππ₯= 85 MPa and ππ§= 150 MPa. What are the changes in a) the length of diameter AB, b) the length of diameter CD, c) the thickness of the plate, d) the volume of the plate if E= 70 GPa and π=1/3?
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