Mechanical Metallurgy - A VISION TO A MISSION · 2017. 4. 29. · GENERAL INFORMATION Course Number...
Transcript of Mechanical Metallurgy - A VISION TO A MISSION · 2017. 4. 29. · GENERAL INFORMATION Course Number...
-
Mechanical MetallurgyMechanical Metallurgy
INME 6016
Pablo G. Caceres‐Valencia
Professor in Materials Science,
B.Sc., Ph.D. (UK)
-
GENERAL INFORMATION
Course Number INME 6016
Course Title Mechanical Metallurgy
Credit Hours 3 (Lecture: 3hours)
Instructor Dr. Pablo G. Caceres‐Valencia
Office Luchetti Building L‐212
Phone Extension 2358
Office Hours Mo and Wed 1:00pm to 5:00pm
e‐mail [email protected]
Web‐site http://academic.uprm.edu/pcaceres
mailto:[email protected]://academic.uprm.edu/pcaceres/
-
AssessmentThe course will be assessed in the following manner:
Partial Exam 30%
Final Exam 30%
Quizzes (3)* 33%
Attendance and Class Participation 7%(*) A total of three quizzes will be performed.
(**) Class Attendance (after the second absence ‐ 1 point will be deducted for each non‐authorized absence). The participation in class will be taken into account.
AttendanceAttendance and participation in the lectures are mandatory and will be considered in the grading. Students should bring calculators, rulers, pen and pencils to be used during the lectures. Students are expected to keep up with the assigned reading and be prepared for the pop‐quizzes or to answer questions on these readings during lecture.
-
TexbooksG.E. Dieter; Mechanical Metallurgy; Mc Graw HillM.A. Meyers and K.K. Chawla; Mechanical Metallurgy: Principles and Applications; Prentice‐Hall I will also post my lecture notes in the web: http://academic.uprm.edu/pcaceres
TENTATIVES DATESJan/9‐11
Basic PrinciplesJan/14‐18
Stress‐StrainJan/21‐25
Basic Elasticity
Jan/28‐Feb/01 Basic Elasticity
Feb/04‐08 Single Crystals
Feb/11‐15Dislocation Theory
March/3‐7 Strengthening Mechanisms
Mar/24‐28 Fracture
April /14‐18 Mechanical Properties
Feb/18‐22 Dislocation Theory 1st Exam
Feb/25‐29 Strengthening Mechanisms
March/10‐14 Strengthening Mechanisms
Mar/17‐21–No Class ‐ Holy Week
April/1‐4Fracture
April/7‐11 Mechanical Properties Exam 2
April/22‐25 –Mechanical Properties
Apr/28‐May02‐Mechanical Properties
-
Content
•Stress and Strain Relationships for Elastic Behavior•Elements of the Theory of Elasticity•Plastic Deformation of Single Crystals•Dislocation Theory•Strengthening Mechanisms•Fracture•Mechanical Properties
-
The Concept of Stress
0AF
=σ
Uniaxial tensile stress: A force F is applied perpendicular to the area (A). Before the application of the force, the cross section area was AOEngineering stress or nominal stress: Force divided by the original area.
True stress: Force divided by the instantaneous area A
FT =σ
00
0
ll
lll Δ
=−
=εEngineering Strain or Nominal Strain: Change of length divided by the original length
True Strain: The rate of instantaneous increase in the instantaneous gauge length. ∫ ⎟⎟
⎠
⎞⎜⎜⎝
⎛==
o
iT
d lnε
-
⎟⎟⎠
⎞⎜⎜⎝
⎛ Δ+⇒⎟⎟
⎠
⎞⎜⎜⎝
⎛ Δ+=
⎟⎟⎠
⎞⎜⎜⎝
⎛== ∫
oo
o
o
oT
o
iT
d
lnln
ln
ε
ε
Relationship between engineering and true stress and strain
i
o
oo
o
iiT A
AAF
AA
AF
AF ** ===σ
iioo AA =
)1(1 ε+=+Δ=+Δ==oo
o
o
i
i
o llAA )1( εσ +=
oT A
F
We will assume that the volume remains constant.
)1( εσσ +=T
)1ln( εε +=T
-
Primary Types of Loading
(a) Tension (b) Compression(c) Shear(d) Torsion(e) Flexion
-
Hooke’s LawWhen strains are small, most of materials are linear elastic.
σ
ε
E
Normal: σ = Ε ε
Young’s modulus
EAlFl
o
o
⋅⋅
=Δ
Springs: the spring rate
o
o
lEA
lFk ⋅=Δ
=
-
Torsion Loading resulting from the twist of a shaft.
lr
zrZ
θδδθγθ
⋅≅⋅=,
lrG
zrGG zz
θδδθγτ θθ
⋅⋅≅⋅⋅=⋅= ,,
G = Shear Modulus of Elasticity
Shear strain
Twist Moment or Torque
∫∫ ⋅⋅
=⋅⋅=A zA z
Arl
GArT δθδτ θθ2,,
∫ ⋅= A ArJ δ2Area Polar Moment
of InertiaJGlT
lJGT
⋅⋅
=⋅⋅
= θθ or
JrT
z⋅
≅,θτThus: JrT o
Max⋅
≅τAngular spring rate:
lGJTka
⋅==
θ
-
From equilibrium principles:τxy = τyx , τxz = τzx , τzy = τyz
Stress Components Normal Stresses σx, σy, σzShear Stresses τxy , τyx , τxz , τzx , τzy , τyz
Normal stress (σ) : the subscript identifies the face on which the stress acts. Tension is positive and compression is negative.Shear stress (τ) : it has two subscripts. The first subscript denotes the face on which the stress acts. The second subscript denotes the direction on that face. A shear stress is positive if it acts on a positive face and positive direction or if it acts in a negative face and negative direction.
-
A shear strain in an element is positive when the angle between two positive faces (or two negative faces) is reduced, and is negative if the angle is increased.
Sign Conventions for Shear Stress and StrainThe Shear Stress will be considered positive when a pair of shear stress acting on opposite sides of the element produce a counterclockwise (ccw) torque (couple).
-
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
==
zyzxz
zyyxy
zxyxx
ij
στττστττσ
σσ
For static equilibrium τxy = τyx , τxz = τzx , τzy = τyz resulting inSix independent scalar quantities. These six scalars can be arranged in a 3x3 matrix, giving us a stress tensor.
The sign convention for the stress elements is that a positive force on a positive face or a negative force on a negative face is positive. All others are negative.The stress state is a second order tensor since it is a quantityassociated with two directions (two subscripts direction of the surface normal and direction of the stress).
-
F1F3
F2 Cube with a Face area Ao
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
==
000
0
0
23
31
oo
oo
ij AF
AF
AF
AF
σσ
A property of a symmetric tensor is that there exists an orthogonal set of axes 1, 2 and 3 (called principal axes) with respect to which the tensor elements are all zero except for those in the diagonal.
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
==
zyzxz
zyyxy
zxyxx
ij
στττστττσ
σσ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡==
3
2
1'
000000
'σ
σσ
σ ijσ
-
Plane Stress or Biaxial Stress : When the material is in plane stress in the plane xy, only the x and y faces of the element are subject to stresses, and all the stresses act parallel to the x and y axes.
Stresses on Inclined SectionsKnowing the normal and shear stresses acting in the element denoted by the xy axis, we will calculate the normal and shear stresses acting in the element denoted by the axis x1y1.
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
00000
yyxy
yxxx
σττσ
-
x1y1
x
y
θ
θ
σy
σx
σx1
τx1y1
τyx
τxy
Equilibrium of forces: Acting in x1
θθθτ
θθθσθτθσ
θσ
CosSinA
CosSinASinAAA OYXOYOXYOXOX ⋅+⋅+⋅+⋅=⋅ cossincoscos1
AoAO/Cosθ
AOSinθ/Cosθ
θ
-
Eliminating Ao , secθ = 1/cosθand τxy=τyx
θθτθσθσσ cossin2sincos 221 XYYXX ++=
Acting in y1
τx1y1Aosecθ = − σxAosinθ + τxyAocosθ + σyAotanθcosθ − τyxAotanθsinθ
Eliminating Ao , secθ = 1/cosθ and τxy=τyx
θθτθσθσσ cossin2cossin 221 XYYXY −+=
( )θθτθθσθθστ 2211 sincoscossincossin −+⋅⋅+⋅⋅−= xyyxyx
-
Transformation Equations for Plane StressUsing the following trigonometric identities:Cos2θ = ½ (1+ cos 2θ) Sin2θ = ½ (1- cos 2θ) Sin θ cos θ = ½ sin 2θ
These equations are known as the transformation equations for plane stress.
θτθσσ
τ
θτθσσσσ
σ
2cos2sin2
2sin2cos22
11
1
xyyx
yx
xyyxyx
x
+−
−=
+−
++
=
-
Case 1: Uniaxial stress
Special Cases
⎟⎠⎞
⎜⎝⎛⋅−=
⎟⎠⎞
⎜⎝⎛ +⋅=
===
22
221
0 0
1,1
1
xy
θστ
θσσ
ττσ
Sin
Cos
xyx
xx
yxy
Case 2 : Pure Shear
θττ
θτσ
σσ
2
2
0
1,1
1
Cos
Sin
xyyx
xyx
yx
⋅=
⋅=
==
θτθσσ
τ
θτθσσσσ
σ
2cos2sin2
2sin2cos22
11
1
xyyx
yx
xyyxyx
x
+−
−=
+−
++
=
-
Case 3: Biaxial stress
θσσ
τ
θσσσσ
σ
τ
22
222
0
1,1
1
Sin
Cos
yxyx
yxyxx
xy
⋅−
−=
⋅−
++
=
=
-
Example: An element in plane stress is subjected to stresses σx=16000psi, σy=6000psi, and τxy=τyx= 4000psi (as shown in figure below). Determine the stresses acting on an element inclined at an angle θ=45o (counterclockwise - ccw).
Solution: We will use the following transformation equations:
θτθσσ
τ
θτθσσσσ
σ
2cos2sin2
2sin2cos22
11
1
xyyx
yx
xyyxyx
x
+−
−=
+−
++
=
-
Numerical substitution½ (σx + σy) = ½ (16000 + 6000) = 11000psi½ (σx - σy) = ½ (16000 – 6000) = 5000psi τxy = 4000psisin 2θ = sin 90o = 1 cos 2θ = cos 90o = 0 Then σx1 = 11000psi + 5000psi (0) + 4000psi (1) = 15000psiτx1y1 = - (5000psi) (1) + (4000psi) (0) = - 5000psiσx1 + σy1 = σx + σy then σy1 = 16000 + 6000 – 15000 = 7000psi
psipsipsi
Max 403,6597,4403,17
2
1
===
τσσ
-
Example: A plane stress condition exists at a point on the surface of a loaded structure such as shown below. Determine the stresses acting on an element that is oriented at a clockwise (cw) angle of 15owith respect to the original element.
Solution:We will use the following transformation equations:
θτθσσ
τ
θτθσσσσ
σ
2cos2sin2
2sin2cos22
11
1
xyyx
yx
xyyxyx
x
+−
−=
+−
++
=
-
Numerical substitution½ (σx + σy) = ½ (- 46 + 12) = - 17MPa½ (σx - σy) = ½ (- 46 – 12) = - 29MPa τxy = - 19MPasin 2θ = sin (- 30o) = - 0.5 cos 2θ = cos (- 30o) = 0.866 then σx1 = - 17MPa + ( - 29MPa)(0.8660) + (-19MPa)(- 0.5) = - 32.6MPaτx1y1 = - (- 29MPa) (- 0.5) + (- 19MPa) (0.8660) = - 31.0MPaσx1 + σy1 = σx + σy then σy1 = - 46MPa + 12MPa – (- 32.6MPa) = - 1.4MPa
-
Example : A rectangular plate of dimensions 3.0 in x 5.0 in is formed by welding two triangular plates (see figure). The plate is subjected to a tensile stress of 600psi in the long direction and a compressive stress of 250psi in the short direction. Determine the normal stress σw acting perpendicular to the line or the weld and the shear stress τw acting parallel to the weld. (Assume σw is positive when it acts in tension and τw is positive when it acts counterclockwise against the weld).
-
SolutionBiaxial stress weld joint σx = 600psi σy = -250psi τxy = 0From the figure tanθ = 3 / 5 θ = arctan(3/5) = arctan(0.6) = 30.96o
We will use the following transformation equations:
Numerical substitution½ (σx + σy) = 175psi ½ (σx - σy) = 425psi τxy = 0psisin 2θ = sin 61.92o cos 2θ = cos 61.92o
θτθσσ
τ
θτθσσσσ
σ
2cos2sin2
2sin2cos22
11
1
xyyx
yx
xyyxyx
x
+−
−=
+−
++
=
Then σx1 = 375psi τx1y1 = - 375psiσx1 + σy1 = σx + σythen σy1 = 600 + (- 250) – 375 = -25psi
-375psi
Θ = 30.96o
375psi25psi
x
y
-
Stresses acting on the weld
σw
θτw
25psi
375psiθ = 30.96o
σw = -25psi and τw = 375psi
-
Principal Stresses and Maximum Shear StressesThe sum of the normal stresses acting on perpendicular faces of plane stress elements is constant and independent of the angle θ.
YXYX σσσσ +=+ 11As we change the angle θ there will be maximum and minimum normal and shear stresses that are needed for design purposes.The maximum and minimum normal stresses are known as the principal stresses. These stresses are found by taking the derivative of σx1 with respect to θ and setting equal to zero.
θτθσσσσ
σ 2sin2cos221 xy
yxyxx +
−+
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
=+−−=
2
2tan
02cos22sin)(1
yx
xyP
xyyxx
σστ
θ
θτθσσδθ
δσ
-
The subscript p indicates that the angle θp defines the orientation of the principal planes. The angle θp has two values that differ by 90o. They are known as the principal angles.
For one of these angles σx1 is a maximum principal stress and for the other a minimum. The principal stresses occur in mutually perpendicular planes.
⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
2
2tanyx
xyP σσ
τθ
2
)(2cos 2sin
RRyx
Pxy
P
σσθ
τθ
−==
P
1
stress maximum for the
2sin2cos22
θθ
θτθσσσσ
σ
=
+−
++
= xyyxyx
x
-
( )
( )
( ) ( )22
12
2
22
1
22
1
22
2 But
21
2
1122222
xyyxyx
xyyx
xyyxyx
xyyxyxxy
xyyxyxyx
R
R
RRRR
τσσσσ
στσσ
τσσσσ
σ
τσσσστ
τσσσσσσ
σ
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −+
+=+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −=
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −⎟⎠⎞
⎜⎝⎛+
+=
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛ −+
+=⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −−+
+=
Principal stresses:
The plus sign gives the algebraically larger principal stress and the minus sign the algebraically smaller principal stress.
( )
( )22
2
22
1
22
22
xyyxyx
xyyxyx
τσσσσ
σ
τσσσσ
σ
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −−⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −+⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
-
Maximum Shear Stress
The location of the angle for the maximum shear stress is obtained by taking the derivative of τx1y1 with respect to θ and setting it equal to zero.
θτθσσ
τ 2cos2sin211 xy
yxyx +
−−=
xy
yx
S
xyyxyx
τ
σσ
θ
θτθσσδθ
δτ
22tan
02sin22cos)(11
−
−=
=−−−=
( )( )
( )( )oP
oP
Po
Po
P
P
S
S
PPxy
yxS
902cos902sin
290cos290sin
2sin2cos
2cos2sin
2cot2tan1
2)(
2tan
−−
=−−−
=−=
−=−=−
−=
θθ
θθ
θθ
θθ
θθτ
σσθ
Therefore, 2θs-2θp=-90o or θs= θp +/- 45o
The planes for maximum shear stress occurs at 45o to the principal planes. The plane of the maximum positive shear stress τmax is defined by the angle θS1 for which the following equations apply:
0P1s111 45 and 2
)(2sin 2cos −=
−−== θθ
σσθ
τθ
RRyx
sxy
s
-
The corresponding maximum shear is given by the equationAnother expression for the maximum shear stress
The normal stresses associated with the maximum shear stress are equal to
( )22
2 xyyx
MAX τσσ
τ +⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
( )2
21 σστ−
=MAX
( )2
yxAVER
σσσ
+=
Equations of a Circle
( )2
21
1
1
2sin2cos2
2sin2cos2
2sin2cos22
⎥⎦
⎤⎢⎣
⎡+
−=−
+−
=−
+−
++
=
θτθσσ
σσ
θτθσσ
σσ
θτθσσσσ
σ
xyyx
AVERx
xyyx
AVERx
xyyxyx
xGeneral equation
Consider
Equation (1)
-
Equation (2) ( )2
211
11
2cos2sin2
2cos2sin2
⎥⎦
⎤⎢⎣
⎡+
−−=
+−
−=
θτθσσ
τ
θτθσσ
τ
xyyx
yx
xyyx
yx
Equation (1) + Equation (2)
( ) ( )22
211
21 2cos2sin2
2sin2cos2 ⎥⎦
⎤⎢⎣
⎡+
−−+⎥
⎦
⎤⎢⎣
⎡+
−=+− θτθ
σσθτθ
σστσσ xy
yxxy
yxyxAVERx
( ) ( )
( ) ( )
( ) 222
22222
22222
2
2cos2sin2
22cos2sin2
2cos2sin2
2cos2sin2
22sin2cos2
2sin2cos2
RSUM xyyx
xyyx
xyyx
xyyx
xyyx
xyyx
xyyx
=+⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −−=⎥
⎦
⎤⎢⎣
⎡+
−−
⎟⎟⎠
⎞⎜⎜⎝
⎛ −++⎟⎟
⎠
⎞⎜⎜⎝
⎛ −=⎥
⎦
⎤⎢⎣
⎡+
−
τσσ
θθτσσ
θτθσσ
θτθσσ
θθτσσ
θτθσσ
θτθσσ
( ) ( ) 221121 RyxAVERx =+− τσσ
-
τxy2θP
σ
τ
C
( )22
2 xyyxR τ
σσ+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −=
⎟⎟⎠
⎞⎜⎜⎝
⎛ +
2yx σσ
The radius of the Mohr circle is the magnitude R.
Mohr Circle
( )22
2 xyyxR τ
σσ+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −=
The center of the Mohr circle is the magnitude
( )2
yxAVER
σσσ
+=
( ) ( ) ( )22
211
21 2
xyyx
yxAVERx τσσ
τσσ +⎟⎟⎠
⎞⎜⎜⎝
⎛ −=+−
-
Alternative sign conversion for shear stresses:
(a) clockwise shear stress,(b) counterclockwise shear
stress, and (c) axes for Mohr’s circle.
Note that clockwise shear stresses are plotted upward and counterclockwise shear stresses are plotted downward.
a) We can plot the normal stress σx1 positive to the right and the shear stress τx1y1 positive downwards, i.e. the angle 2θ will be positive when counterclockwise or
b) We can plot the normal stress σx1 positive to the right and the shear stress τx1y1 positive upwards, i.e. the angle 2θ will be positive when clockwise.
Both forms are mathematically correct. We use (a)
Forms of Mohr’s Circle
-
Two forms of Mohr’s circle: (a) τx1y1 is positive downward and the angle 2θis positive counterclockwise, and(b) τx1y1 is positive upward and the angle 2θ is positive clockwise. (Note: The first form is used here)
-
Construction of Mohr’s circle for plane stress.
-
Example: At a point on the surface of a pressurized cylinder, the material is subjected to biaxial stresses σx = 90MPa and σy = 20MPa as shown in the element below.Using the Mohr circle, determine the stresses acting on an element inclined at an angle θ = 30o (Sketch a properly oriented element).
Solution (σx = 90MPa, σy = 20MPa and τxy = 0MPa)Because the shear stress is zero, these are the principal stresses.Construction of the Mohr’s circleThe center of the circle is σaver = ½ (σx + σy) = ½ (90 + 20) = 55MPaThe radius of the circle is R = SQR[((σx – σy)/2)2 + (τxy)2] R= (90 – 20)/2 = 35MPa.
-
Stresses on an element inclined at θ = 30o
By inspection of the circle, the coordinates of point D areσx1 = σaver + R cos 60o = 55MPa + 35MPa (Cos 60o) = 72.5MPaτx1y1 = - R sin 60o = - 35MPa (Sin 60o) = - 30.3MPaIn a similar manner we can find the stresses represented by point D’, which correspond to an angle θ = 120o ( 2θ = 240o)σy1 = σaver - R cos 60o = 55MPa - 35MPa (Cos 60o) = 37.5MPaτx1y1 = R sin 60o = 35MPa (Sin 60o) = 30.3MPa
-
Example: An element in plane stress at the surface of a large machine is subjected to stresses σx = 15000psi, σy = 5000psi and τxy = 4000psi, as shown in the figure.
Using the Mohr’s circle determine the following:a) The stresses acting on an element inclined at
an angle θ = 40ob) The principal stresses andc) The maximum shear stresses.
SolutionConstruction of Mohr’s circle:Center of the circle (Point C): σaver = ½ (σx + σy) = ½ (15000 + 5000) = 10000psiRadius of the circle: R = SQR[((σx – σy)/2)2 + (τxy)2] R = SQR[((15000 – 5000)/2)2 + (4000)2] = 6403psi.Point A, representing the stresses on the x face of the element (θ = 0o) has the coordinates σx1 = 15000psi and τx1y1 = 4000psiPoint B, representing the stresses on the y face of the element (θ = 90o) has the coordinates σy1 = 5000psi and τy1x1 = - 4000psiThe circle is now drawn through points A and B with center C and radius R
-
Stresses on an element inclined at θ = 40o
These are given by the coordinates of point D which is at an angle 2θ = 80o from point A. By inspection the angle ACP1 for the principal stresses (point P1) is tan ACP1 = 4000/5000 = 0.8 or 38.66o.Then, the angle P1CD is 80o – 38.66o = 41.34o
-
Knowing this angle, we can calculate the coordinates of point D (by inspection)σx1 = σaver + R cos 41.34o = 10000psi + 6403psi (Cos 41.34o) = 14810psiτx1y1 = - R sin 41.34o = - 6403psi (Sin 41.34o) = - 4230psiIn an analogous manner, we can find the stresses represented by point D’, which correspond to a plane inclined at an angle θ = 40o + 90o = 130o
σy1 = σaver - R cos 41.34o = 10000psi - 6403psi (Cos 41.34o) = 5190psiτx1y1 = R sin 41.34o = 6403psi (Sin 41.34o) = 4230psiAnd of course, the sum of the normal stresses is 14810psi + 5190psi = 15000psi + 5000psi
-
Principal StressesThe principal stresses are represented by points P1 and P2 on Mohr’s circle.σ1 = 10000psi + 6400psi = 16400psiσ2 = 10000psi – 6400psi = 3600psiThe angle it was found to be 2θ = 38.66oor θ = 19.3o
Maximum Shear StressesThese are represented by point S1 and S2 in Mohr’s circle. The angle ACS1 from point A to point S1 is 2 θS1 = 51.34o. This angle is negative because is measured clockwise on the circle. Then the corresponding θS1 value is – 25.7o.
-
Example:At a point on the surface of a generator shaft the stresses are σx = -50MPa, σy = 10MPa and τxy = - 40MPa as shown in the figure. Using Mohr’s circle determine the following:
(a) Stresses acting on an element inclined at an angle θ = 45o,
(b) The principal stresses and(c) The maximum shear stressesSolutionConstruction of Mohr’s circle:Center of the circle (Point C): σaver = ½ (σx + σy) = ½ ((-50) + 10) = - 20MPaRadius of the circle: R = SQR[((σx – σy)/2)2 + (τxy)2] R = SQR[((- 50 – 10)/2)2 + (- 40)2] = 50MPa.Point A, representing the stresses on the x face of the element (θ = 0o) has the coordinates σx1 = -50MPa and τx1y1 = - 40MPaPoint B, representing the stresses on the y face of the element (θ = 90o) has the coordinates σy1 = 10MPa and τy1x1 = 40MPaThe circle is now drawn through points A and B with center C and radius R.
-
Stresses on an element inclined at θ = 45oThese stresses are given by the coordinates of point D (2θ = 90oof point A). To calculate its magnitude we need to determine the angles ACP2 and P2CD.
tan ACP2=40/30=4/3 ACP2=53.13o P2CD = 90o – ACP2 = 90o – 53.13o = 36.87oThen, the coordinates of point D areσx1 = σaver + R cos 36.87o = - 20MPa – 50MPa (Cos 36.87o) = - 60MPaτx1y1 = R sin 36.87o = 50MPa (Sin 36.87o) = 30MPaIn an analogous manner, the stresses represented by point D’, which correspond to a plane inclined at an angle θ = 135o or 2θ = 270oσy1 = -20MPa + 50MPa (Cos 36.87o) = 20MPa τx1y1 = -30MPaAnd of course, the sum of the normal stresses is -50MPa+10MPa = -60MPa +20MPa
-
Principal StressesThey are represented by points P1 and P2 on Mohr’s circle.σ1 = - 20MPa + 50MPa = 30MPa σ2 = -20MPa – 50MPa = - 70MPaThe angle ACP1 is 2θP1 = 180o + 53.13o= 233.13o or θP1 = 116.6oThe angle ACP2 is 2θP2 = 53.13o or θP2= 26.6o
Maximum Shear StressesThese are represented by point S1 and S2in Mohr’s circle. The angle ACS1 is 2θS1 = 90o + 53.13o = 143.13o or θ = 71.6o . The magnitude of the maximum shear stress is 50MPa and the normal stresses corresponding to point S1 is -20MPa.