University of Waterloo
Department of Civil Engineering
CIV E 205 – Mechanics of Solids II
Instructor: Tarek Hegazi
Room: CPH 2373 G, Ext. 2174
Email: tarek@uwaterloo.ca
Course Web: www.civil.uwaterloo.ca/tarek/hegazy205.html
Course Notes
University of Waterloo
Civil Engineering
CIV. E. 205 – MECHANICS OF SOLIDS II
COURSE OUTLINE
Instructor:
Office:
Ext.:
E-mail:
Dr. T. Hegazy
Lectures: MWF 9:30 - CPH 3385
CPH – 2373G
2174
tarek@uwaterloo.ca Web: http://www.civil.uwaterloo.ca/tarek/hegazy205.html
T.A.s:
Textbook: - Hibbeler, 2002 “Mechanics of Materials,” 5th Edition, Prentice Hall.
- Course Notes – Download
What is Covered:
1. - Introduction & Review
2. - Analysis of Stress
3. - Analysis of Strain
4. - Stress-Strain Relations
5. - Strain Energy
6. - Theories of Failure
7. - Deflection of Beams
8. - Energy Methods
9. - Principle of Virtual Work
10. - Influence Lines
11. - Reciprocal Theorem
12. - Buckling of Columns
13. - Special Beam Problems (Optional)
Marking:
4 Quizzes @ 10%
40% Held on dates announced in class
Final Examination: 60%
Bridge Competition: Bonus
Notes:
- Each week, a number of suggested problems will be given to serve as background study
for the quizzes. Solutions are not to be handed in.
- Teaching Assistants will provide one-to-one help and will prepare you for quizzes.
- Course notes, solutions to suggested problems, and solutions to quizzes will be posted
on the course web site.
© Dr. Tarek Hegazy
1
Mechanics of Materials II
Mechanics of Materials
Objectives:
-
Solve Problems in a structured systematic manner;
Study the behavior of bodies that are considered
deformable under different loading conditions; &
Analyze and design various machines / systems
1. Basic Concepts
a)
Resultants of Forces
(i.e., one force equivalent to many)
Non-Concurrent
Forces
Concurrent
Forces
P1
P1
P2
P3
P2
Calculating the Resultant Graphically:
Force Polygon Method
Note: R substitutes all its
component forces.
Draw forces subsequently. Each is drawn
parallel to its line of action and length equal
to value of force. Accordingly, resultant is an
arrow that closes the polygon (value &
direction measured graphically).
Calculating the Resultant Analytically: The X and Y components of all forces are summed
X = ∑ (x components of all forces); Y = ∑ (y components of all forces);
Then,
b)
2
2
R = SQRT (X + Y );
tan θ = Y / X
Equilibrium of a system subjected to Forces
(i.e., Resultant of all forces on the system = 0)
Concurrent
Forces
R (resultant) = 0
Two Equilibrium Conditions:
Non-Concurrent
Forces
Three Equilibrium Conditions:
1. ∑ X+ components of all forces = 0
2. +∑ Y components of all forces = 0
1. ∑ X components of all forces = 0
+
2. +∑ Y components of all forces = 0
3. ∑ M (moment at any point) = 0
+
Note: It is possible to put a system in equilibrium by calculating the Resultant of all forces acting on it then
applying an opposite and equal force to that resultant.
© Dr. Tarek Hegazy
1
Mechanics of Materials II
P2
Example: can we set
a system subject to
these two forces
in equilibrium?
P1
R
P1
Step 1: get resultant (R)
Step 2: apply (- R)
P2
Example: Find the resultant and introduce a new force to set the system in Equilibrium
P6
P5
P5
P4
P3
P4
P3
P5
P4
P6
P1
P3
P2
P1
P6
P2
Forces
P1
Resultant
P2
Equilibrium
P6
Example: The Polygon to the right is for a system of forces in Equilibrium.
P7
- The resultant of forces P2, P3, P4, P5, P6, & P7 is ___________
P5
- The resultant of forces P4, P5, P6, P7, P1, & P2 is ___________
P4
P3
P1
P2
Example: Find the forces acting on the cable.
o
45
o
60
F1
=
F2
45o
60o
6 tons
Three force system
Directions of top two
Forces are assumed.
6 tons
© Dr. Tarek Hegazy
2
Mechanics of Materials II
Solution: Graphically using force polygon or analytically using Equilibrium equations.
F2
6 tons
∑X=0
F2 COS 60 – F1 COS 45 = 0
∑Y=0
F2 SIN 60 + F1 SIN 45 -6 = 0
Solving both,
F1
F1 =
t
; F2 =
t
Example: In the following system of parallel forces, determine the reactions Ra, Rb
3
2
1
a
b
1.5
2.5
2
Ra
1.5
Rb
Analytically:
∑ Y components of all forces = 0
∑ M (moment at any point) = 0
+
=>
Ra + Rb – 6 = 0
=>
=>
Ma =
Rb x
+ or, Rb x 7.5 = 18.5
Then, Ra + 2.47 = 6
c)
Ra + Rb = 6
=0
or Rb = 2.47 t
or Ra = 3.53 t
Three-Forces Theorem
Three non-parallel forces in Equilibrium must intersect in a common point. The force polygon of the triangle of
forces shows a continuous loop.
Example: Use the theorem of three forces to determine the reactions
a
Ra
5m
4 tons
b
Rb
© Dr. Tarek Hegazy
4m
3
Using the force polygon,
the reactions are determined.
Mechanics of Materials II
Example: Use the theorem of three forces to determine the reactions on the beam.
Polygon 2
for the
equilibrium
of the
beam
Polygon 1
for the
resultant of
Ra and Rc
6 tons
a
b
o
60
45o
d)
Rc
Ra
Resultant of
Ra and Rc
3m
Rb
3m
1m
Types of Supports
Supports exert reactions in the direction in which they restrain movement.
Roller
Support
(restricts in one
direction only and
allows rotation)
Rubber
Hinged or Pinned
Support (restricts in two
ways and allows rotation)
Fixed Support
(restricts in two directions
and also restricts rotation)
Intermediate Pin or Hinge
(Gives one extra condition)
Gives extra condition
∑ M Right = 0
∑ M Left = 0
Force in
direction of
member
Examples:
_____________
© Dr. Tarek Hegazy
_______
_______________ _______
4
__________
___________
Mechanics of Materials II
e)
© Dr. Tarek Hegazy
Structural Representation of Real Systems
5
Mechanics of Materials II
f)
Stability & Determinacy of Structures
-
A stable structure can resist a general force immediately at the moment of applying the force.
Unstable
Stable
Does not return to original shape if load is released
-
A statically determinate structure is when the reactions can be determined using equilibrium equations.
1. Beams:
if r < c + 3
Unstable
r = unknown support reactions.
if r = c + 3
Statically determinate
c = additional conditions
if r > c + 3
Statically Indeterminate
r = 3 (two at hinge + one at roller)
c = 2 (two intermediate hinges), then,
r < c + 3 Unstable
r = 5 (four at hinges + one at roller)
c = 2 (two intermediate hinges), then,
r = c + 3 Stable & Statically Determinate
r = _______
c = _______
r = _______, then____________________
r = 4 (three at fixed end + one at roller)
c = 0, then
r > c + 3 Stable & Statically Indeterminate
2. Frames:
j = No. of joints
m = No. of membrs
r = unknown support reactions
c = special conditions
if 3m + r < 3j + c
if 3m + r = 3j + c
if 3m + r > 3j + c
j = __; m = __; c = __; r = __
Then,
3m + r =___ , 3j + 2 =___, or
________________________
j = 3; m = 2; c = 0; r = 3
Then,
Stable & Statically determinate
3. Trusses:
j = No. of joints
m = No. of membrs
r = unknown support reactions
if m + r < 2j
if m + r = 2j
if m + r > 2j
j = __; m = __; r = __
Then,
m + r = ___, 2j = ___, or
________________________
© Dr. Tarek Hegazy
Unstable
Statically determinate
Statically Indeterminate
Unstable
Statically determinate
Statically Indeterminate
j = 8; m = 12; r = 3
Then,
m + 3 = 15 < 2j, or Unstable
6
Mechanics of Materials II
2. Analysis of Forces
Forces and their effects at different points:
Distributed Load
Concentrated Load
Rotation (Couple)
M
P
W t/m
P1
L
Effect of a load on
another point
parallel to its axis
=
WxL
L
P1
P
b
M = P1 . L
∑ Ma = 0
+
a
L
P
P
b
a
∑ Ma = 0
+
a
∑ Ma = 0
+
Effect of
loads
P
P
b
M=PxL
Forces:
T
M
V
M
P
Shear = Force In the
X-Section Plane
Bending Moment =
Couple Normal to Plane
Torsion = Couple in
the X-Section Plane
Normal Force =
Perpendicular to X-Section
Example:
Calculate reactions without applying equilibrium equations
P
L/2
© Dr. Tarek Hegazy
P
L/2
M = P.L /2
L/2
7
L/2
Mechanics of Materials II
Example: Determine the forces at section BC
M = 150x5
150 lb
150 lb
150 lb
2”
2”
2”
C
5”
5”
750 lb.in
2”
2”
2”
5”
5”
5”
5”
B
750 lb. in
150 lb
750 lb.in
Example: Determine the forces at section A
z
8”
Note: When the structural system is:__________, then
the free end is a good starting point for the analysis.
10”
x
A
800 lb
800 lb
Mx=800x10
T=800x14
y
14”
A
500 lb
Mz=500x14
500 lb
10”
A
800 lb
T=800x14
Equilibrium equations for each segment:
Mz=500x14
∑Mx=0, ∑My=0, ∑Mz=0
∑Fx=0, ∑Fy=0, ∑Fz=0
500 lb
800 lb
14”
500 lb
Example: Determine the forces at section HK
© Dr. Tarek Hegazy
8
Mechanics of Materials II
Example: Determine the internal forces at section ABCD
3.5
3.5
Example: Determine the internal forces at section C
Example: Determine the forces acting at section D
Example: Determine the internal forces on the section
© Dr. Tarek Hegazy
9
Mechanics of Materials II
3. Internal Loadings on Beams & Frames
Sign convention
for internal forces (N.F., S.F., & B.M.)
+ ive
3.1 Analytical Approach:
Stability & Determinacy – Reactions – Axis – Sections – Signs – N, V, & M Equations – Draw Diagrams
Equilibrium Conditions:
∑ X Components of all forces = 0
+
+
∑ Y Components of all forces = 0
∑ M At any point = 0
+
Extra condition at intermediate Pin:
∑ M right side only = 0 = ∑ M left side only
+
+
Analysis of Shear & Moment Equations:
F.B.D.: Between load changes, make a cut and put 3 (equal & opposite)
internal forces on each side.
P
Segment 1
M
X
Xa
X
V
Cut
V
Ya
Segment 2
M
Yb
Apply Equilibrium Equations to this
segment alone
Apply Equilibrium Equations to this
segment alone
Example: Calculate the Reactions
10 t
Ma
4
3
2m
1m
6t
a
Xa
a
8t
4t
1m
1m
1m
Ya
Step 1: Cantilever beam is stable and statically determinate
Step 2: Looking at the system as a whole (right figure) and applying Equilibrium equations,
∑X
+
+
∑Y
= 0, then Xa – 6 = 0, or Xa = 6 t
= 0, then Ya – 4 – 8 = 0 , or Ya = 12 t
∑ Ma = 0, then Ma – 4 x 1 – 8 x 3 = 0 , or Ma = 28 m.t.
+
© Dr. Tarek Hegazy
10
Check OK
Mechanics of Materials II
3t
Example: Calculate the Reactions
2t/m
Step 1:
m = 5; r = 4; j = 6; c = 1
3m + 4 = 19 = 3j + 1 then,
Frame is stable and statically determinate
c
4m
6m
b
Step 2: Looking at the system as a whole
and applying Equilibrium equations,
2m
a
Xb
Yb
Xa
Ya
∑ Mb = 0, then 3 x 11 + 2 x 8 x 7 + 2 x Xa – 9 x Ya = 0
2m
2m
+
∑ Mc left only = 0 = 3 x 4 + Xa x 6 – Ya x 2 + 2 t/m x 4 x 2 = 0
+
4m
Solving both equations, we get Ya = 16.28 t , and Xa = 0.76 t
+
∑Y
3m
, then
= 0, then Ya + Yb - 3 - 16 = 0 , or Yb = 2.72t
∑ X = 0, then Xa - Xb = 0 then Xb = 0.76 t
+
Check OK
2t/m
Example: Calculate the Reactions
2m
4t
∑ Mb = 0 =
+ =
.
∑ Ma = 0 =
+ =
.
+
∑Y
Ya
. = 0 or Ya = 36 - Yb
=0=
Yb
4m
4m
4m
12 m
Solving these equations, we find
Yb = 20 t
;
Ya = 16 t
; Xa = Xb = 0
Check OK
Example: Calculate the Reactions
50 t
∑ Mb = 0 =
+ = 50x5 - 4 Ya =0 , or Ya = 62.5 t
d
1m
+
∑ Y = 0 = Ya + Yb - 50 = 0 or Yb = -12.5 t
∑ Mc right side only = 0 = 4 Yb - 3 Xb or Xb = - 50/3 t
+
∑ X = 0 = Xa - Xb = 0 or Xa = -50/3 t
+
c
3m
Check OK
Xa
1m
4m
Ya
© Dr. Tarek Hegazy
11
Xb
Yb
Mechanics of Materials II
Examples: Calculate and draw the S.F.D. and the B.M.D.
Reactions:
Reactions:
Sections:
Sections:
Solved examples 6-1 to 6-6
Note on simple beam with distributed load:
© Dr. Tarek Hegazy
12
Mechanics of Materials II
3.2 Graphical Approach:
Stability & Determinacy - Reactions – N, V, & M Relations – Draw Diagrams
Examples on Page 12
Rules:
1- Shear curve is one degree above load curve
2- Moment curve is one degree above shear curve
3- Moment is maximum at point with shear = 0
4- Between any two points: (look at table)
-
Area under load = difference in shear
-
Area under shear = difference in moment
-
Slope of shear curve = - (load trend)
-
Slope of moment curve = shear trend
Solved examples 6-7 to 6-13
© Dr. Tarek Hegazy
13
Mechanics of Materials II
+ ive
Examples: Calculate and draw the S.F.D. and the B.M.D.
From point a to point b:
a
b
-
Load curve =
-
Shear curve =
-
Moment curve =
-
Area under load =
-
Area of shear =
= difference in shear =
= difference in moment =
-
Shear at point of max. Moment =
-
Max. moment can be calculated from shear diagram =
=
-
Slope of shear curve =
-
Slope of moment curve =
From point a to point b:
a
b
-
Load curve =
-
Shear curve =
-
Moment curve =
-
Area under load =
= difference in shear =
-
Area of shear =
= difference in moment =
© Dr. Tarek Hegazy
-
Shear at point of max. Moment =
-
Max. moment can be calculated from shear diagram =
=
-
Slope of shear curve =
-
Slope of moment curve =
14
Mechanics of Materials II
+ ive
Examples: For the problems in page 12, draw the N.F.D., S.F.D., & the B.M.D.
50 t
2t/m
d
1m
4t
c
3m
0
20
4
50/3
50/3
1m
62.5
© Dr. Tarek Hegazy
2
0
4m
4m
12 m
4m
16
12.5
15
Mechanics of Materials II
Examples: Calculate and draw the S.F.D. and the B.M.D.
© Dr. Tarek Hegazy
16
+ ive
Mechanics of Materials II
4. Stresses due to Forces
Internal Forces:
Normal
Moment
Shear
Torsion
V
T
M
M
P
Normal Force =
Perpendicular to X-Section
Stresses
Shear = Force In the
X-Section Plane
Bending Moment =
Couple Normal to Plane
Torsion = Couple in
the X-Section Plane
τ
-
A’
Q = A’ . Y’
In narrow rectangular beams,
τmax
© Dr. Tarek Hegazy
17
= 1.5 V / A
Mechanics of Materials II
Example: Draw the S.F.D. and the B.M.D. for the beam below and then determine the maximum
normal stress due to bending.
Example: Draw the S.F.D. and the B.M.D. for the beam below and then determine the maximum
normal stress to the left and to the right of point D. The beam has a section modulus of 126 in3.
Example: Page 8 of notes - Determine the internal stresses at points B & C
150 lb
2”
150 lb
2”
5”
5”
C
750 lb.in
B
I = 4 x 103 / 12 ; σB = - 150 / A + 750 x 5 / I
© Dr. Tarek Hegazy
= +7.5 psi ;
18
σC = - 150 / A - 750 x 5 / I
= -15.0 psi
Mechanics of Materials II
Example:
Calculate normal stresses at section d
and also at the section just below c.
First, we get the reactions.
C
B
Example: Determine the internal stresses at points A, B, C, & D
VQ / It
= 0.5 MPa
D
A
C
V = 3 KN
T = 3 KN
My = 10.5 KN.m
B
T.c / J
= 15.3 MPa
D
C
A
B
D
M. x / I = 107 MPa
My = 10.5 lb .in
C
B
A
D
A
© Dr. Tarek Hegazy
19
Mechanics of Materials II
Example:
Page 8 of notes
=
Example: The timber used in the beam below has an allowable stresses of 1800 psi (normal)
and 120 psi (shear). Determine the minimum required depth d of the beam.
From the S.F.D. and the B.M.D., maximum values of:
Moment = 7.3 Kip.ft = 90 Kip.in; Shear = 3 Kips
Design based on allowable Normal stress:
Check Shear stress:
>120
(unacceptable)
Redesign based on allowable shear stress:
Solved Problems 6-14 to 6-20, 7-1 to 7-3, 8-4 to 8-6
© Dr. Tarek Hegazy
20
Mechanics of Materials II
5. Transformation of Stresses
- Member under tension only (P) in one direction, i.e., a normal stress. But, let’s consider an inclined plane.
σθ = (P Cos θ) / (A / Cos θ) or
Very important conclusions:
- Under tension only, shear is automatically present at various planes.
- The plane of maximum shear is when Sin 2θ = max or when θ = 45.
- Maximum shear = σx /2 = P / 2A
- It is important to study stress transformation and shear failure.
σθ = σx Cos2 θ
τθ = ½ σx Sin 2θ
Positive Signes
- Member under two dimensional stresses.
y
σy
τxy
σx
Questions:
Is this the maximum stress? If not, then
What is the value of max. normal stress & its orientation? and
What is the value of maximum shear stress & its orientation?
x
y'
x'
τx’y’
σy’
σx’
θ
x
General Equations:
75 MPa
60 MPa
Example:
For the given state of stress, determine the normal and shearing
stresses after an element has been rotated 40 degrees counter-clockwise.
σx = +30 MPa ; σy = -75 MPa ; τxy = +60 MPa ; θ = + 40
30 MPa
y'
90.7
Applying the above equations, we get:
σx’ = +45.7 MPa ; σy’ = -90.7 MPa ; τx’y’ = -41.3 MPa
x'
41.3
45.7
40
x
Solved Examples 9-2 to 9-6
© Dr. Tarek Hegazy
21
Mechanics of Materials II
Important Observations:
1. σx + σy = σx’ + σy’ = Constant
Sum of normal stress is constant (90 degrees apart) for any orientation.
2. The plane in which shear stress
τx’y’ =
0 is when:
=0
or tan 2θ = 2 τxy / (σx
3.
- σy)
or at θ1 , θ2 having 90 degrees apart. These are called principal planes.
σx’ becomes maximum when dσx’ / dθ
= 0, or when differentiating the following equation:
we get,
tan 2 θp = 2 τxy / (σx - σy) or, exactly at the principal planes, which has shear stress = 0.
The value of the principal normal stresses are:
σ max, min = σx + σy
2
4. Since
5.
σx + σy
(σx -2 σy)
2
+
τ2xy
= constant, then, at the principal planes,
τx’y’ is maximum when
max =
σx
is maximum but σy is minimum.
planes, dτ / dθ = 0, or when:
tan 2 θs = - (σx
τx’y’
±
- σy) / 2 τxy
(σx - σy)
2
2
+
and the value of maximum shear stress
τxy is:
τ2xy
6. Similar to single stress situation, maximum is when dτ / dθ = 0, or when: θ =
.
Example:
Check rule 1 for the example in previous page.
In the general equations, even if the original
a value as a function of normal stresses.
© Dr. Tarek Hegazy
τxy
on the element = 0, then still the shear at any plane (τx’y’)has
22
Mechanics of Materials II
Example: Determine the maximum normal and shear stresses at point H.
Forces at the section:
Stresses at Point H:
Principal stresses:
Example: Determine the maximum normal and shear stresses at points H & K.
© Dr. Tarek Hegazy
23
Mechanics of Materials II
Example: Determine the maximum normal and shear stresses at points H & K.
© Dr. Tarek Hegazy
24
Mechanics of Materials II
Circular representation of plane stresses (Mohr’s Circle):
Step 1: Given a state of stress, with
σx
y
and σy having 90 degrees apart.
σy
Step 2: Let’s plot the σx and σy on a horizontal line
Notice, the 90 degrees are now 180 apart.
σx
θ
σy
0
τxy
x
σ
σx
Step 3: Let’s plot the σx and σy on a horizontal line then τxy vertically at points 1 and 2 using signs.
σy
τxy
Y( ? , ? )
τxy
0
σy
Y
?
?
?
?
σx
X
σx
σ
τxy
?
τ
X( ? , ? )
Step 4: Draw a circle from the center to pass by points 1 and 2. Determine σmax , σmin , θp , τmax , θs
?
σy
Y
Y
σx
0
? σy
?
τxy
X
σx
σ
?
X
τ
Solved Examples 9-7 to 9-13
© Dr. Tarek Hegazy
25
Mechanics of Materials II
Example:
For the given state of stress, determine the normal and shearing
stresses after an element has been rotated 40 degrees counter-clockwise.
75 MPa
60 MPa
30 MPa
Y
(-75,60)
X’
75
θ2
θ1
-75
Y’
80o
60
σ
30
Y
X
30
X (30, -60)
From the figure:
Average stress = Center of circle = (30 – 75)/2 = -22.5 ,
R = sqrt (52.52 + 602) = 79.7
o
tan θ1 = 60 / 52.5, then θ1 = 48.8
and
θ2 = 80 - θ1 = 31.2o
Then, points X’ and Y’ have the following coordinates:
σx’ = -22.5 + R cos θ2 = -22.5 + 79.9 * 0.855 = +45.7 MPa
σy’ = -22.5 – R cos θ2 = -90.7 MPa ;
τx’y’ = R sin θ2
= -41.3 MPa
Principal stress values:
σmax, σmin =
Average ± R
= -22.5 ± 79.7 =
57.2, - 102.2
Example:
For the given state of stress, determine: a) principal planes; and b) principal stresses.
60 MPa
σx = -40 MPa; σy = +60 MPa ; τxy = +25 MPa
tan 2θp = 2 τxy / (σx - σy) = 2 x 25 / (-40 -60) = -0.5
Analytically:
25 MPa
40 MPa
or at θp1 = -13.28; θp2 = 76.7
σmax, σmin = Average ± R =
σx + σy
2
±
2
(σx -2 σy) + τ2xy
65.9
= 10 ± 55.9 MPa
x
Graphically: Two points X & Y
13.28
45.9
Center =
R =
σmax, σmin =
Y
(60,25)
Average ± R
Y
2θ
65.9
σ
60
25
13.28
Y
45.9
© Dr. Tarek Hegazy
X
(-40, -25)
26
X
40
Mechanics of Materials II
Example:
You have a Mohr circle of stress as shown below for two
separate points. Draw the stresses on each element and
its orientation from principal planes.
A (50, ? )
R=?
40
A (50, y )
40
B ( ? , -40 )
R = 60
y
x = 50
100
σ
2θ
B’ (100-X , 40 )
160
A’ (150, -y )
R = 60
y=
40
Y
σ
160
100
Y = sqrt (602 – 502) = 33.2
Tan 2θ = y / 50 or 2θ = 33.6o
50
Y
x
40
σ
100
2θ
100-X
Y
B (100 + X, -40 )
40
16.8
160
150
X = sqrt (602 – 402) = 44.7
Tan 2θ = 40 / X or 2θ = 83.6o
41.8
100+X
Special Cases:
1. Case of pure tension
A member under one directional stress
Let’s use Mohr’s circle.
σx = P / A
and
σy = 0
τmax
Y
From Mohr’s circle, maximum shear
X
σx
τmax = σx / 2 = P/ 2A
Maximum shear is at 45 degrees.
y'
X
Y
σx
0
x'
τmax
σy’
σx’
45
2. Case of pure torsion
A member under only torsional stress,
with
σx = 0
and
σy = 0,
Let’s draw Mohr’s circle.
τxy
Y
Y
X
σmax
0
σ
As seen, torsional stress creates normal
Stresses which are maximum at 45 degrees.
X
© Dr. Tarek Hegazy
27
Mechanics of Materials II
Graphical representation of principal planes:
Properties of a circle: an angle 2θ at the center of the circle, corresponds to an angle θ at the circumference.
θ
θ = 90
2θ
2θ = 180
Special case
Y (5,20)
5
20
Y
X
30
+
2θ p
5
30
σ
X
(30, -20)
Plane of principal
normal stress
Plane of maximum
shear stress
Y
Y
σmin = -6.1
σmax = 41.1
2θ p
5
30
30
5
2θs
5
20
θp
θs
30
X
X
17.5
23.6
17.5
© Dr. Tarek Hegazy
28
Mechanics of Materials II
3-Dimensional stress systems: (Absolute maximum shear stress)
Assume
σ1 > σ2 > σ3 are principal normal stresses ( no shear), then let’s draw Mohr’s circle.
τmax
σ2
σ2
σ3
σ1
σ1
σ
σ3
Note : Even if σ3 = 0, 3-D stress analysis becomes essential.
Case 1: both σ1 and σ2 are positive
Then, τmax = σ1 / 2
Case 2: both σ1 and σ2 are negative
Then, τmax = σ2 / 2
τmax
σ2
τmax
σ1
σ2
σ1
Case 3: σ1 and σ2 have opposite signs
Then, τmax = (σ1 - σ1) / 2
τmax
σ2
σ1
Examples
© Dr. Tarek Hegazy
29
Mechanics of Materials II
6. Transformation of Plain Strain
- A structure should be designed so that its material and cross sectional dimensions can resist the maximum
normal and shear stresses imposed on it. Equally important also that the structure does not deform much
under the load, i.e., the ability to resist strains is crucial to the serviceability of structures.
- Normal Strain (due to axial load + bending moment) and Shear Strain (due to transverse shear + torsion).
Normal Strain
Shear Strain
=
+
+
Strain = ε = Unitless = ∆L / L
Positive Signs
(elongation and angle)
Questions:
Is this the maximum strain? If not, then
What is the value of maximum normal strain and the plane in which it exists? and
What is the value of maximum shear strain and the plane in which it exists?
- General equations for strains on a plane at angle θ for a member under two dimensional strain. Notice that all
equations look the same as those of stress transformation, except that τxy is resembled by
General Equations: Given the three constants
,
,
:
then,
Normal strain at any angle θ:
Shear strain at any angle θ:
Principal (Normal) Strain:
Orientation:
Max. Value:
Shear strain at this plane: Zero
Maximum Shear Strain:
Orientation:
Max. Value:
Normal strain at this plane:
Solved Problems 10-1 to 10-8
© Dr. Tarek Hegazy
30
Mechanics of Materials II
- Strains before and after transformation:
Positive Strains
at positive angle θ
Positive Strains
at θ = 0
Negative Strains
at negative angle θ
‘’
‘’
‘
‘
θ
Important Observations:
1. εx
+ εy = εx’ + εy’
= Constant (90 degrees apart) for any orientation.
2. The plane in which shear strain
γx’y’ / 2 =
0 is when:
=0
or tan 2θ = γxy / (εx
3.
or at θ1 , θ2 having 90 degrees apart. These are called principal planes.
εx’ becomes maximum when dεx’ / dθ
tan 2 θp = γxy / (εx
we get,
4.
- εy)
γx’y’ is maximum when dγ / dθ
- εy)
= 0, or when differentiating the following equation:
or, exactly at the principal planes, which has shear strain = 0.
= 0, or when:
tan 2 θs = - (εx
- εy) / γxy
5. Similar to single stress situation θs = 45o from θp.
6. Mohr’s circle of strain:
γ/2
(εy, γ/2)
Y
Y
+
ε
2θ p
X
X
(εx, - γ/2)
εmin ? , εmax ? , γmax ? , θp ? , θs?
© Dr. Tarek Hegazy
31
Mechanics of Materials II
Example:
Given εx’ = -200 x10-6, εy’ = 1000 x10-6, γxy = 900 x10-6. Find the strains associated with x’y’ axes inclined at 30
degrees clockwise. Find principal strains and the maximum shear strain along with the orientation of elements.
Y
Solution
+
First, we sketch the element with the given strains, as follows.
Shorter in X
Longer in Y
+ive shear strain.
Then, we define two points X and Y to draw Mohr’s circle.
γ/2 x10-6
R = Sqrt (60^2 + 450^2) = 750
Y (1000, 450)
R=750
-200
2θ p
Principal Strains:
εmax , εmin = 400 ± 750 = 1150 x10-6 ,
γx’y’ at principal planes = 0
450
ε x10-6
600
400
X
-350 x10
2θp= tan (450 / 600) = 36.8
-1
1000
o
Max Shear Strains:
γmax / 2 = R = 750 x10-6
εx’ = εy’ at Max shear plane =
X
(-200, -450)
400 x10
-6
2θs = 36.8 + 90 = 126.8
o
o
γ/2 x10-6
Y
α = 60 - 36.8 = 23.2
Then
εx’ = 400 - R Cos α = 400 – 750 x Cos 23.2 = -290 x10-6
γx’/ 2 = R Sin α = 750 Sin 23.2 = 295 x10-6
(
,
X’
)
α
o
36.8 400
εy’ = 400 + R Cos α = 400 + 750 x Cos 23.2 = 1090 x10-6
γy’/ 2 = -R Sin α = -750 Sin 23.2 = -295 x10-6
At 30 o Clockwise
R=750
ε x10-6
α
R=750
X
At Principal Planes
Y’
(
Y’
,
)
At Maximum Shear Plane
Y’
+
+
X’
θ=30o
Shorter in X’
Longer in Y’
-ive shear strain (clockwise rotation)
© Dr. Tarek Hegazy
++
+
θp=18.4o
Shorter in X’
Longer in Y’
No Shear strain
32
θs=18.4+ 45o
X’
Longer in X’
Longer in Y’
-ive shear strain
Mechanics of Materials II
-6
Absolute maximum shear strain
Assume
ε1
>
ε2
>
ε3
are principal normal strains (no shear), then let’s draw Mohr’s circle.
γmax/2
γ/2
Note: Even if ε3 = 0, 3-D analysis is essential.
Case 1: both ε1 and ε2 are positive
Then,
ε2
ε3
ε1
ε
γ/2
γmax/2 = ε1 / 2
γmax/2
ε2
γmax/2
= (εmax -
εmin) / 2
Case 2: both ε1 and ε2 are negative
Then,
Case 3: ε1 and ε2 have opposite signs
γmax/2 = ε2 / 2
γmax/2
ε2
ε1
Then,
γmax/2
= (ε1 - ε2) / 2
γ/2 γmax/2
γ/2
ε2
ε1
ε1
Solved Examples: 10-1 to 10-7
© Dr. Tarek Hegazy
33
Mechanics of Materials II
Strain Measurements Using Strain Rosettes:
- 45o strain rosette versus 60o strain rosette
- Cemented on surface
- Its electrical resistance changes when wires are stretched
or compressed with the material being studied
- Resistance changes are measured and interpreted
as changes in deformation
Readings: εa, εb, εc
- Three values to get the state of strain at the point
At:
θa=0, θb=45, θc=90
- Automated condition assessment of bridges
Unknowns: εx, εy, γxy
- Check the strains on older structures
Readings:
At:
εx’ = εx cos2 θ
+
εy Sin2 θ
+
Applying into the general equation:
εx = εa
εy = (2εb + 2εc - εa) / 3
γxy= 2(εb – εc) / Sqrt(3)
εx = εa
εy = εc
γxy= 2εb – (εa + εc)
γxy cos θ . Sin θ
Substitute into either equation 3 times using εa, εb, εc to get the unknowns
εx, εy, γxy
Unknowns:
Applying into the general equation:
or
εa, εb, εc
θa=0, θb=60, θc=120
εx, εy, γxy at the measurement point.
Example:
b
a
Using the strain rosette shown, the measured values at each stain gauge is as follows:
εa = 8 x 10-4 , εb = -6 x 10-4, εc = -4 x 10-4
c
Determine the principal strains at the point.
Solution Using Equations:
γ/2 x10-4
θa = 90, θb = 135, θc = 180
A
Applying into the general strain transformation equation:
R=10
εa = 8 x 10-4 = εx cos2 90 + εy Sin2 90 + γxy cos 90 . Sin 90
εb = -6 x 10-4 = εx cos2 135 + εy Sin2 135 + γxy cos 135 . Sin 135
εc = -4 x 10-4 = εx cos2 180 + εy Sin2 180 + γxy cos 180 . Sin 180
16/2
ε x10-4
-4
8
2
16/2
Then:
εy = εa = 8 x 10-4 ;
εx = εc = -4 x 10-4 ; γxy /2
C
= 16 x 10-4
Using Mohr’s circle, we determine principal strains:
ε1 = 12 x 10-4
;
ε2 = -8 x 10-4
γ/2 x10-4
A
Solution Using Only Mohr’s Circle:
B
Directions (a) and (c) are 90 degrees apart
This means that the center of the circle is the
Average strain = (εa + εc) / 2 = (8 x 10 - 4 x 10 ) /2 = 2 x 10
-4
-4
-6
From the two triangles shown, d = _____,
2
2
Then, R = Sqrt( d + 6 ) = _________
As such,
ε1 = 2 + R = 12 x 10-4
© Dr. Tarek Hegazy
;
2θp
R
R
8
2θp
d=?
-4
ε2 = 2 – R = -8 x 10-4
2
ε x10-4
6
8
C
34
Mechanics of Materials II
Example:
Solution:
Strategy: We draw a Mohr’s circle for strain and on it will find the strains at the
orientations of the strain gauges (45o apart).
Y
+
Longer in X
Shorter in Y
+ive shear strain.
?
?
© Dr. Tarek Hegazy
X
?
35
Mechanics of Materials II
7. Relationship between Stress & Strain
- A stress in one direction causes elongation in its direction and shortening in the other two depending on the
material’s Poisson’s ratio (ν).
Generalized Hooke’s law
εx = σx / E
εy = - ν. σx / E
εz = - ν. σx / E
εx = - ν. σy / E
εy = σy / E
εz = - ν. σy / E
εx =
εy =
εz =
- Assumptions: (1) τ has not correlation with εx and εy; (2) σx and σy have no relation with γxy ; (3) principal strains
occur in directions parallel to principal stresses.
-General Equations:
E.εx = σx - ν (σy + σz);
G. γxy = τxy
E.εy = σy - ν (σx + σz);
G. γyz = τyz
E.εz = σz - ν (σx + σy);
G. γzx = τzx
- Relationship between E, ν , G:
Let’s consider the case of pure torsion, i.e., σx = 0 and
for both stress and strains.
σy = 0,
Let’s draw Mohr’s circles
γxy/2
τ
Y
Y
σmax
0
σ
ε
X
X
Principal stresses are: σ1 = τxy ;
εmax
0
σ2 = - τxy
Principal strains are: ε1 = γxy/2 ;
ε2 = - γxy/2
Now, let’s apply Hook’s Equation, as follows:
E.ε1 = σ1 - ν (σ2) ; then
E.ε1 = E. γxy/2 = τxy - ν (- τxy) =
Then
τxy . (1 + ν) = G. γxy . (1 + ν)
G = E / 2 (1 + ν)
Note: Since most engineering materials has ν = 1/3, then G = 3/8 E
© Dr. Tarek Hegazy
36
,
K = E / 3 (1 - 2 ν)
Bulk Modulus
and K = E
Mechanics of Materials II
Example:
Notice the difference between Mohr’s circles for stress & strain
Example:
Example:
Example:
© Dr. Tarek Hegazy
37
Mechanics of Materials II
Example:
Example: Match each one of the following Mohr’s circle for stress with a Mohr’s circle for strain. Explain.
Example:
A 60 degree strain rosette is put at point A. The strain guage readings are:
εa =
60x10-6
εb =
135x10-6
εc =
264x10
-6
a) Determine the Principal strains at point A and their directions.
b) If the bracket is made of steel for which E = 200 GPa and ν = 0.3,
determine the Principal stresses at point A.
Solution:
Approach: from strain rosette readings, we get strains at the point, then calculate principal strains
and finally convert these principal strains into principal stresses.
© Dr. Tarek Hegazy
38
Mechanics of Materials II
a) First, we determine the normal and shear strains at point A from rosette readings.
Accordingly,
Second, we determine principal strains using Mohr’s circle.
-6
Center of circle = average normal strain = (60 + 246) / 2 = 153x10
-6
-6
Accordingly, Principal strains = 153 ± R = 272 x10 ; 33 x10
b) Using Hooke’s law with σ3 = 0
Another Solution:
Approach: from strain rosette readings, we get strains at the point, then we convert them into stresses
and finally calculate principal stresses.
First, we calculate strains at point A, same as above. Then,
Now, we convert these strains to stresses:
© Dr. Tarek Hegazy
39
Mechanics of Materials II
We now can use Mohr’s circle for stress to determine Principal stresses:
Example
Solution Approach: Since we are given the forces, let’s calculate the
Stresses at point P, then, convert these stresses into strains.
Forces on Section at P.
Stresses at Point P:
© Dr. Tarek Hegazy
Forces at end of beam.
Normal stresses
40
Mechanics of Materials II
Shear Stresses
Strains at Point P:
Example:
Stresses at point A:
Forces at section of Point A:
`
Strains at point A:
© Dr. Tarek Hegazy
41
Mechanics of Materials II
Mohr’s circle of Strain:
Strain Gauge Readings:
Solved Problems 10-9 to 10-11
© Dr. Tarek Hegazy
42
Mechanics of Materials II
8. Theories of Failure
All theories deal with
PRINCIPAL STRESSES
Ductile Material
(Yield Failure)
-
Brittle Material
(Fracture Failure)
Max. normal stress (Rankin’s Theory)
Max. shear stress (Tresca Criterion)
Max. Energy of Distortion (Von Mises Criterion)
Other: Max. principal strain (St. Venant)
Max. normal stress (Rankin’s Theory)
σ2
Failure when:
Principal stresses
σ1
| σ1| > σy / F.S. or
| σ2| > σy / F.S.
where F.S. > 1
σ2
σy
A state of plane stress
Is safe inside the square
And unsafe outside
(assume tension = compression)
-σy
σy
σ1
-σy
Max. shear stress (Tresca Criterion)
A specimen under tension
reached maximum stress
σy, then, the maximum shear
that the material can resist
is σy /2 from Mohr’s
Circle.
Then, failure is when
τ > σy / (2 * F.S.)
Absolute
max. shear
(3-D analysis)
= | σmax – σmin| /2
© Dr. Tarek Hegazy
43
Mechanics of Materials II
Energy of Distortion (Von Mises Criterion)
To be safe,
Ud on element
U = ½ σ.ε
< Ud yield
For the 3-D stress Case:
1 [(σ1 – σ2)2 + (σ2 – σ3)2 + (σ3 – σ1)2 ]
12G
<
2 σ2yield
12G
or Simply,
(σ1 – σ2)2 + (σ2 – σ3)2 + (σ3 – σ1)2
< 2 σ2yield
For the 2-D stress Case: (σ3 = 0)
(σ12 – σ1 σ2 + σ22)
<
σ2yield
Rarely used
Other: Max. principal strain (St. Venant)
Using Hooke’s law
E εmax = σ1 – ν (σ2 + σ3)
<
σyield
Fracture of Brittle Materials
Brittle materials are relatively weak in Tension.
Failure criterion is Maximum Principal Tensile Stress.
Under Tensile force, failure is due to tension.
Under Torsion, failure is still due to tension at an angle.
Element is safe when:
Example: Twist of a piece of chalk.
Solved Examples 10-12 to 10-14
© Dr. Tarek Hegazy
44
Mechanics of Materials II
Example: A steel shaft (45 mm in diameter) is exposed to a tensile yield strength = σyield = 250 MPa.
Determine P at which yield occurs using Von Mises and Tresca critera.
Solution
1) Principal Stresses
T = 1.7 KN.m
σx = P / A = P /π (0.0225)2
τxy = T.c / J = 1.7 x (0.0225) / ½ π (0.0225)4 = 95.01
Mohr's circle:
P?
95.01
σx
Center = σx / 2
R = [(σx/2) 2 + τxy2] ½
σ1 = σx / 2 + R; σ2 = σx / 2 - R
2) Using Von Mises
σ12 + σ1σ2 + σ22 = σ yield 2
(σx / 2 + R)2 + (σx / 2 + R) (σx / 2 - R) + (σx / 2 - R)2 = σyield2
(σx/2)2 + 3 R2 = σyield2 ,
substituting with R,
σx2 + 3 τxy2
substituting with σx & τxy,
X
= σ yield 2 ,
[P /π (0.0225)2] 2 + 3 x (95.01) 2 = σ yield 2 = 2502
then, P = 299.3 KN
3) Using TRESCA
σ1 and σ2 have opposite signs, then
τmax (3-D) = |σ1 - σ2 | / 2 , which reaches failure of τyield = σyield / 2
|(σx/2 + R) - (σx/2 - R)| / 2 = σ yield / 2
R = σ yield / 2
then,
,
(σx/2)2 + τxy2 = (σ yield / 2) 2 ,
substituting with R and squaring both sides,
substituting with σx & τxy,
[P /2 π (0.0225)2] 2 + (95.01) 2 = 1252
then, P = 258.4 KN
Notice the force P under TRESCA (focuses on Shear) is smaller than Von Mises
© Dr. Tarek Hegazy
45
Mechanics of Materials II
Deflection of Beams and Shafts
Beams and shafts deflect under load. For serviceability, we need to make sure deflection is within allowable
values. Also, the shape of the beam under the load (elastic curve) needs to be studied.
Terminology:
-
EI = Flexture rigidity or Bending Stiffness
R = Radius of Curvature
1/R = Curvature
υ
Hookes Law: 1/R = M / EI
The elastic curve:
dθ
R
R
y
ds
Rdθ = ds ≅ dx
or
1/R = dθ/dx
θ+dθ
θ
ds
θ
dx
dυ
Also, dυ/dx = tan θ ≅ θ
Differentiating both sides, then d2υ/dx2 = dθ/dx
Accordingly,
1
R
M
EI
=
=
dθ
dx
=
d2υ
dx2
Notes:
-
Integration of (M/EI) determines the slope of the elastic curve:
-
Double integration of M/EI determines the deflection:
-
Recall relationships between load, shear, and bending moment. Now, we can expand it to:
EI d2υ/dx2 = M(x);
EI d3υ/dx3 = V(x);
EI d4υ/dx4 = -W(x)
Determining the integration constants C1 and C2:
Substituting at points of known
deflection and/or slope, we can
determine the constants of
integration.
© Dr. Tarek Hegazy
46
Mechanics of Materials II
Shape of Elastic Curve:
inflection point at location where moment=0
Calculating Slope & Displacement by Integration:
Step-by-Step
∑ X = 0,
+
+
∑ Y = 0,
∑ M= 0
+
2. Get equation of B.M. at each beam segment with change in load or shape
1. Get beam reactions:
3. Integrate the moment once to get the slope
4. Integrate the moment a second time to get the deflection (elastic curve)
5. Substitute at points of special conditions (boundary conditions) to get the constants C1 & C2
6. Rewrite the slope and deflection equations using the constants
7. Put slope = 0 to determine the location (x) that has maximum deflection
Example:
For the part AB, determine the equation of the elastic curve and maximum deflection if:
P
I = 301x106 mm4, E=200 GPa, P=250 KN, a = 1.2 m, L = 5 m.
Ya
© Dr. Tarek Hegazy
47
L
Yb
a
Mechanics of Materials II
Solution
1. Reactions:
P
∑Ma =0
+
Yb . L - P . (L + a) = 0
or Yb = P (1 + a/L)
+
∑ Y = 0 , then Ya + Yb – P = 0 or
Ya = - P. a/L
P.a/L
P(1 +a/L)
2. Bending moment equations:
Mx
P.a/L
x
0 to L
P
Mx
Fx
Fx
x = 0 to a
V
V
Mx = - P. x
Mx = - P.a.x / L
3. Integrate the moment to get the slope:
8. Applying same steps at the free end:
= - P.a.x2 / 2L + C1
= - P.x2 /2 + C3
…(3)
= - P.x3 / 6 + C3. x + C4
….(4)
4. Integrate a second time to get the (elastic curve)
= - P.a.x3 / 6L + C1. x + C2
5. Substitute at points of known conditions
Slope at B right = Slope at B Left
at support A:
then, C2 = 0
[x = 0 . y = 0]
Slope left = using equation (1), x=L
= -P.a.L/2 + P.a.L/6
also, at support B: [x = L , y = 0]
then, 0 = - P a L3 / 6L + C1. L
or
C1 = P.a.L/6
Slope right = using eq. (3), x=a
= -P.a2 /2 + C3
6. Final equations:
= - P.a.x2 / 2L + P.a.L /6
we get C3
……(1)
= - P.a.x3 / 6L + P.a.L.x /6 ...…(2)
Also, at B:
[x = a , y = 0]
7. Put slope = 0 at maximum deflection
Using Equ. (4), we get C4
0 = - P.a.x2 / 2L + P.a.L /6
get x = 0.577L
Using this value in equation (2), we get
Max deflection = 8 mm Up.
Solved problems 12-1 to 12-4
© Dr. Tarek Hegazy
48
Mechanics of Materials II
Calculating Slope & Displacement by Moment Area Method:
1st Moment Area Theorem:
Recall
M
EI
= dθ
dx
XB
θB/A = M dx
EI
XA
change in
area under
slope
M/EI diagram
Then,
2nd Moment Area Theorem:
tBA = (vertical distance from B on
elastic curve to tangent at A)
= Moment of the area under
M/EI around point B.
=
dB
XB
M dx
dB .
EI
XA
tAB =
Note:
tBA
Case 1: Cantilever
Notice that tangent at point A is horizontal.
-Deflection at any point: _____________
-Slope at any point:
XB
M dx
θB/A =
EI
XA
θA = ___
= θB - θ A = θ B
Case 2: Symmetric Loading – Option 1
Deflection is max at mid beam (C). At this point θC = __
-Deflection at any point: _____________
∆d
A
B
d
tBC
-Slope at any point:
C
θB/A =
Xd
M dx = θD - θC = θD
EI
θC = __
∆C = __
tdc
XC
© Dr. Tarek Hegazy
49
Mechanics of Materials II
Case 3: Unsymmetrical Loading – Option 2
L1
-Deflection at any point: ∆d + tDA = tBA . L1/(L1+L2)
L2
∆d
A
B
d
-Slope at any point to the right:
Xd
M dx = θD - θA with θA being negative = |θD| + |θA|
θD/A =
EI
Xa
tBA / (L1+L2)
L1
A
θA
A
L2
∆d
θD
B
θA
d
θD/A =
-Slope at any point to the left:
tBA
tDA
A
θA
θD
Xd
M dx = θD - θA , both negative = |θA| - |θD|
EI
Xa
tBA
tDA
tBA / (L1+L2)
Case 4: Over-Hanging Beam
L1
θB =
tAB / L1
tCB
A
B
heavy load
∆c
θB
C
= (∆c + |tCB| ) / L2
L2
tAB
Then, ∆c = |θB . L2| - |tCB|
∆
tCA
tBA
∆ = tBA . (L1+L2)/L1
L1
Then, ∆c = |tCA| - | ∆ |
© Dr. Tarek Hegazy
50
C
B
θA
A
L2
heavy load
∆c
Mechanics of Materials II
Case 5: Unsymmetrical Loading – Point of Max. Deflection
x =?
θB = tAB / L
θB/C =
A
tBC
XB
M dx = θB - θC = θB = tAB / L
EI
XC
B
tBA
C
θC = 0
∆C = max
We get x, then ∆C = max = tBC
Note: Equivalence in Bending Moment Diagrams
a
-b
a
=
=
a
+ive
-ive
-b
-a
Method of Superposition:
- Using Standard tables for various beam conditions and types of loads (Appendix C)
- Adding up deflections caused by individual loads
Solved Problems: 12-7 to 12-15
© Dr. Tarek Hegazy
51
Mechanics of Materials II
Example: Determine θC and ∆A
L1
150lb
300lb
600
D
C
A
B
D
4”
6”
24”
250 lb
400 lb
-1800
600
tDC =1/EI [ + (600 x 24 /2) . 2/3 . 24
- (1800 x 24 /2) . 1/3 . 24 ]
M/EI
θC = tDC / 24
= ( |∆A| + | tAC | ) / 4
-1800
600
tDC
∆A
A
θC
tAC = 1/EI [600. 4/2 . 2/3 . 4]
tAC
Example: Determine θA and ∆D
A
B
θA = tBA / L
D
wL/3
wL/6
= Moment of M/EI @ B / L
3rd
degree
= [w.L2 /6EI . L/2 . L/3 - w. L2/6EI . L/4 . L/5] / L
= 7 w.L3 / 360EI
wL2/12
B.M.D.
wL2/6
=
Also, ∆d + tDA = tBA / 2
Then
∆d = | tBA / 2| - | tDA |
3rd
degree
A
θA
© Dr. Tarek Hegazy
- [1/2 .w.L2 /12EI .L/2 .L/6 - ¼ . wL2/48EI. L/10]
B
= 5 w.L4 / 768EI
d
tDA
-wL2/48
= 7 w.L3/720EI
2
L/5 -wL /6
∆d
L/2
tBA
52
Mechanics of Materials II
Using Deflection Calculations to Solve Statically Indeterminate Beams
P
=
+
P
RB ?
∆1 + ∆2 =0
______ statically
indeterminate
P
=
+
MA ?
θ1 + θ2 =0
First, we reduce the beam to a statically determinate, then
We compensate for the change in the deflection behavior.
Example:
12 Kips
Determine the reactions, then draw the S.F.D. & the B.M.D.
3 Kip/ft
C
A
B
12 ft
6 ft
6 ft
12 Kips
=
3 Kip/ft
∆1
A
tAC
+
∆1
B
C
tBC
RB
∆2
∆2 = - RB L3 / 48 EI
∆1 + ∆2 =0
© Dr. Tarek Hegazy
53
Mechanics of Materials II
Example:
Solution:
© Dr. Tarek Hegazy
54
Mechanics of Materials II
Energy Methods
-
For a structural element under load and deformation, External Work Ue = Internal Strain Energy Ui.
-
External work Ue is a function of the load P and deflection ∆.
(deflection is at same point and direction of load)
Ue = ½ P. ∆
-
Also, the Internal strain energy in the structure Ui
is a function of the stress σ and strain ε in the element,
summed over the volume of the structure.
Ui = ½ σ . ε . V = σ
2
σ
.V
2E
Strain Energy
per unit volume
Normal
Ui =
2
σ . dV
ε
Shear
and
Ui =
2E
v
L
2
τ . dV
V=
v 2G
dV =
v
dA
A
when A is constant,
dx
0
V=A
Observe the units.
L
dx
0
Strain Energy calculations for different loading conditions are shown in next page.
Determining Deflections Using Conservation of Energy
Single External load
Deflection in the direction of load:
Ue = Ui
Ue = ½ P. ∆
&
Ui =
Ue = ½ Mo . θ
&
Ui
Ue = ½ P. ∆
&
Ui
Limitations: Applies to single load only. Also, in case 2, only solpe is calculated not deflection.
Also, how to get deflection at a point at which no direct load is applied.
Solved Examples 14-1 to 14-7
© Dr. Tarek Hegazy
55
Mechanics of Materials II
Strain Energy Calculations
Axial
Load
2
L
2
Ui = σ . dV =
N
dA dx
2
2E v
2EA A
0
Example: Truss with varying axial loads on individual members.
(Cross section area A is constant, then V = A . L)
Normal
Stress
L
Bending
Moment
2
Ui = σ /2E
dV
or
Ui =
v
2
M . y
2 E I2
σ = M.y
2
dA
A
dx
0
=I
I
L
M2 dx
Ui =
2EI
0
L
Pure
Shear
2
Ui = ½ τ . γ . dV = τ /2G
A. dx
0
v
τ = V.Q
I.t
Shear
Stress
L
fs V2 dx
Ui =
where, fs = 6/5 - rectangular section
2GA
0
L
Torsion
2
Ui = ½ τ . γ dV = τ /2G . dA . dx
A
v
0
τ = T.c
J
L
T2 dx
Ui =
0
© Dr. Tarek Hegazy
2GJ
56
Mechanics of Materials II
Example: Determine the strain energy due to both shear and bending moment in the following cantilever.
The cross section is a square of length a, with EI being constant.
w
P
Example: Determine deflection at C, neglect shear strain energy.
.
© Dr. Tarek Hegazy
57
2EI
EI
L/2
L/2
Mechanics of Materials II
C
Principle of Virtual Work
Conservation of Virtual Work
∆?
Work-Energy method is not able to determine
the deflection at a point at which no direct load exists.
Solution: Put a virtual load of 1.0 at the desired point of a virtual system. Then apply the principal of
conservation of virtual work, as follows:
Real Beam
1.0
Virtual Beam
∆?
External Virual WorK
½ Virtual load x Real displacement
1.0 x ∆
=
Internal Virtual Energy
=
½ Virtual Stress x Real Strain x Volume
=
σV . εR . V
L
=
σV . σR .
E
L
=
dA
A
dx
0
L
n N A dx = n N dx
A AE 0
0 AE
Axial
Load
L
+
m M dx
0 EI
L
+
Bending
fs v V dx
Shear
GA
0
L
+
Examples:
Real
t T dx
0 GJ
Virtual
Torsion
1.0
1. Determine slope at desired point
Real
Virtual
2. Determine horizontal
deflection at desired point
Solved Examples 14-11 to 14-16
© Dr. Tarek Hegazy
N
M
V
T
1.0
n
m
v
t
58
Mechanics of Materials II
w
Example: Determine the deflection at mid span.
w
Example: Get deflection at A
.
A
L/2
© Dr. Tarek Hegazy
59
L/2
Mechanics of Materials II
Example on Virtual Work - determine the vertical deflection at point A.
A
3 t/m
Determine the horizontal deflection at point A.
A
© Dr. Tarek Hegazy
60
Mechanics of Materials II
2 KN / m
Calculate:
The horizontal displacement at point B.
Vertical displacement at point C.
Slopes at points A, C, and D.
Relative displacement between points E and F.
E
3m
4 KN / m
4 KN
F
C
D
EI = Constant
4m
A
2m
© Dr. Tarek Hegazy
61
B
4m
4m
2m
Mechanics of Materials II
Suggested Problems
Calculate the vertical deflection at point d.
EI = 20,000 m2.t
Calculate:
- The horizontal displacement at point b,
- The vertical displacement at point g
- The slope at point f
EI = 20,000 m2.t
For the shown frame calculate:
-
The vertical deflection at point c
The horizontal deflection at point d
EI = 15,000 m2.t
For the second problem, assume support B is hinged. In this case, draw the S.F.D. and the B.M.D. for the
frame.
© Dr. Tarek Hegazy
62
Mechanics of Materials II
Important Use of the Virtual Work method: SOLVING statically indeterminate structures
w
C
A
B
L/2
L/2
=
w
Reduced
System
∆1
+
∆1 + RB ∆2 = 0
Compensation
∆2
RB .
1.0
Also,
w
L/2
L/2
=
w
Reduced
System
θ1 + MA θ2 = 0
θ1
Compensation
MA .
+
1.0
θ2
Examples:
© Dr. Tarek Hegazy
63
Mechanics of Materials II
Calculating Deflections Using Castigliano’s Theorem
-
Put an external load at the position of required deflection: external load (Q) either horizontal or
vertical to get horizontal or vertical deflection; or an external moment to get slope.
-
Deformation = first derivative of the Strain Energy with respect to the applied load.
∆ = dU / dQ
, & substituting Q = 0
L
L
=
2
δ
δQ
N dx
2EA
0
=
0
L
=
=
δ
δQ
M2 dx
L
δ
δQ
δ M dx
E I δQ
Bending Moment
fs V δ V dx
G A δQ
Shear
δT
G J δQ
Torsion
L
0
L
fs V2 dx
=
2GA
L
δ
δQ 0
Axial Load (Trusses)
M
0
0
=
δ N dx
EA δQ
=
2EI
0
N
L
2
T
2GJ
dx
T
=
0
dx
Example:
Determine the horizontal deflection at point B. Cross-section area= 12 in2
6
E= 30.10 psi. AB = 48 in and BC = 36 in.
Solved Examples
© Dr. Tarek Hegazy
64
Mechanics of Materials II
Example
Example:
© Dr. Tarek Hegazy
65
Mechanics of Materials II
Buckling of Columns
- Slender columns under elastic compression buckle when the load exceeds a critical value.
- Buckling causes column instability.
- Short stocky columns do not buckle.
- W need to study the relation between P, ∆, and shape of buckled column.
- Analysis (Euler 1707 – 1783):
P
P
υ
υ
M
P
M + P. υ = 0
M = d2υ
EI dx2
Recall,
d2υ
dx2
Then,
+ P. υ = 0
EI
υ
Equation of Elastic Curve:
υ = C1 Sin [(P/EI)0.5. x]
υ= 0 at x = L
or when,
or when,
+
C2 Cos [(P/EI)0.5. x]
υ= 0 at x = 0
Sin [(P/EI)0.5. L] = 0
C2 = 0
(P/EI)0.5. L = π, 2π, ….
x
4
© Dr. Tarek Hegazy
66
Mechanics of Materials II
Analysis:
Maximum axial load
before buckling:
P/A should be
within allowable
stresses.
Smaller of the
two directions
x & y.
A
Put,
r = I/A
= radius of gyration
OR
(L/r)2
Note that L/r is the “Slenderness Ratio”
used to classify columns as long, intermediate, or short.
Effect of Column Supports:
;
=
Solved Examples
© Dr. Tarek Hegazy
67
Mechanics of Materials II