Transcript Chapter 09
Chapter 9
Combined Stresses
9-1 Introduction
• Basic types of loading: axial, torsional and
flexural
• Stress formulas:
P
Axial loading - a
A
T
Torsional loading -
J
Flexural loading - f
My
I
9-2 Combined Axial & Flexural Loads
My
f
I
P
a
A
f a
P My
A I
A
20
B
15
P Mc 6M
A ( I bh 2 )
20 103
6(0.45 15 103 0.15 20 103 )
A
(0.05)(0.150)
(0.05)(0.150) 2
(2.67 106 ) (20.00 106 ) 22.67 MPa
20 103
6(0.45 15 103 0.15 20 103 )
B
(0.05)(0.150)
(0.05)(0.150) 2
(2.67 106 ) (20.00 106 ) = 17.33 MPa
F
0:
x
o
2000sin(15 )
Dx 2000cos(15O )
1931.852 lb.
o
2000cos(15 )
C
D
C y 264.598 lb.
2000cos(15o )
Dy 253.04 lb.
y
759.12 lb.ft
1931.852 lb
B
0:
0:
Dy C y 2000sin(15O ) lb.
A
C y 264.598 lb.
D
3
(6)C y ( )2000cos(15O ) (4)2000sin(15O )
12
C y 264.598 lb.
F
517.638 lb
1931.852
M
517.638 lb
Dy 253.04 lb.
Section AB:
3
M (3)C y ( )1931.852 (1)517.638
12
759.12 lb.ft
Normal Stresses
517.638 lb
P Mc 6M
(
2 )
A
I
bh
A
1931.852
759.12 lb.ft
1931.852 lb
C y 264.598 lb.
B
517.638 lb
A
1931.852 6 759.12 12
2 6
2 62
920.1 lb/in 2
B
1931.852 6 759.12 12
2 6
2 62
598.1 lb/in 2
1931.852 6 1012.16 12
2 6
2 62
1173.15 lb/in 2
min
max
1012.16 lb.ft
529.20 lb.ft
BMD
1931.852 6 1012.16 12
2 6
2 62
851.17 lb/in 2
P
0.25P
P
P Mc
A I
P
0.25P 0.05
0.0025
1.5625
8400 P
80 MPa
A 4 D 2 0.0025 0.07854 m 2
I 64
D 4 (0.1) 4 1.5625 106 m 4
80 106
P
29.92 kN
8400
P 2400 [ 12 25 6 25 3]
180,000 180,000 360,000 kg.
5.0
M 180,000 0.5 180,000 3.0
12 (1000 15) 15 5
90,000 540,000 562,500
112,500 kg-m.
4.0
1.5
P Mc 6M
(
2 )
A
I
bh
min
360,000 6 112,500
2
48,333.33
kg/m
1 9
1 9 2
P My
For stiff members the formula
A I
appropriate
is
For long slender members or columns, the effect of
P-d is significant
1
in
2
Fig.(a)
Fig.(b) P
P
1
in
2
1
1
A ( )2 in 2
2
4
1 4
1
1
I 12 ( )
in 4
2
192
max. compressive stress in Fig.(a)
max,( a )
P Mc
P ( 12 P)( 14 )
1
28P
1
A I
(4)
(192 )
max. compressive stress in Fig.(b)
max,(b )
P
P
1 4 P
A
(4)
max,( a ) 28 P
7 :1
max,( a ) 4 P
Hw10
allow
B
D1
D2
Fig. P-908
ค่า z1-z6 ได้จากเลขประจาตัวนิสิต ดังต่อไปนี้
46z1z2z3z4z5z6
D1=(1+z1) in.
D2 = D1(1+z2) in.
หมายเหตุ D2 = D1(1+z2) in.
I1-1=1000(1+z3) in4
Area=10(1+z4) in2
เพื่อให้หน้าตัดมีประสิ ทธิภาพดีในการรับหน่วยแรง
B =10(1+z5) in.
allow=10(1+z6) ksi.
Hw11
L2
L3
L4
b
h
L1
ค่า z1-z6 ได้จากเลขประจาตัวนิสิต ดังต่อไปนี้
L1= (1+z1) in.
L2 = (1+z2) in.
46z1z2z3z4z5z6
L3= (1+z3) in.
L4 = (1+z4) in.
หมายเหตุ h = b(1+z6) in.
b = 0.2(1+z5) in.
h = b(1+z6) in.
เพื่อให้คานมีความลึกไม่นอ้ ยกว่าความกว้างเสมอ
P = (1+z5) kips.
F = (1+z6) kips.
9-3 Kern of Section: Loads Applied off Axes of Symmetry
P My ( Pe)a
A
I
I
a
I
Ae
for b h section
h (bh3 /12)
2
bh e
h
e
6
The maximum eccentricity to avoid tension
e
h
6
That is in designing of masonry or other
structures weak in tension, the resultant load
should fall in the middle third of the section.
The general case:
u
ry2
rx2
ey
P ( Pex ) x ( Pey ) y
A
Iy
Ix
ex
The position of neutral axis (line of zero stress)
P ( Pex ) x ( Pe y ) y
0
2
A
Ary
Arx2
ey
ex
0 1 2 x 2 y
ry
rx
I x Arx2
I y Ary2
P ( Pex ) x ( Pey ) y
Rectangular section: 0 3
3
bh bh /12 hb /12
P ( Pex )(h / 2) ( Pey )(b / 2)
0
bh
bh3 /12
hb3 /12
b h
( , )
2 2
ey
ex
1
h/6 b/6
918 A compressive load P= 12 kips is applied, as in Fig. 9-8a, at a point 1 in. to the
right and 2 in. above the centroid of a rectangular section for which h=10 in. and
b=6 in. Compute the stress at each corner and the location of the neutral axis.
Illustrate the answers with a sketch similar to Fig. 9-8b.
12 kips
6
1
10
2
P ( Pex ) x ( Pey ) y
A
Iy
Ix
Rectangular section:
P ( Pex ) x ( Pey ) y
3
3
bh bh /12 hb /12
A
12
(12 1)(5) (12 2)(3)
0.08 ksi
3
3
6 10 6 10 /12 10 6 /12
B
12
(12 1)(5) (12 2)(3)
0.72 ksi
6 10 6 103 /12 10 63 /12
C
12
(12 1)(5) (12 2)(3)
0.48 ksi
3
3
6 10 6 10 /12 10 6 /12
D 0.32 ksi
12 kips
6
1
10
0
2
12
(12 1)( x) (12 2)( y)
3
6 10 6 10 /12 10 63 /12
3x 2 y
1
25 3
on x axis (y=0) x 25/ 3 8.33
on y axis (x=0) y 3/ 2 1.5
Position of Neutral Axis:
P ( Pex ) x ( Pey ) y
0 3
3
bh bh /12 hb /12
921 Calcualte and sketch the kern of a W360 X 122 section.
ey
ex
Position of Neutral Axis: 0 1 2 x 2 y
ry
rx
ex 257 ey 363
At corner A: 0 1 2
63 2 1532 2
A(
257 363
,
)
2
2
2 632
on x-axis (ey =0): ex
30.89 mm
257
2 1532
on y-axis (ex =0): ey
129.0 mm
363
9-4 Variation of Stress with Inclination of Element
Mc
I
Tc
J
9-5 Stress at A Point
Stress at a point really defines the
uniform stress distributed over a
differential area.
• The most general state of stress at a point may
be represented by 6 components,
x , y , z
normal stresses
xy , yz , zx shearing stresses
(Note : xy yx , yz zy , zx xz )
xx xy xz x xy xz
σ yx yy yz yx y yz
zx zy zz symmetry
zy z
zx
state of stress เมื่อแสดงด้ วยระบบโคออร์ ดเิ นต (xyz)
xy
xz x xy xz
xx
σ yx yy yz yx y yz
zx zy zz symmetry
zx zy z
state of stress เมื่อแสดงด้ วยระบบโคออร์ ดเิ นต (xyz)
• Plane Stress - state of stress in which two faces of
the cubic element are free of stress. For the
illustrated example, the state of stress is defined by
x , y , xy and z zx zy 0.
• State of plane stress occurs in a thin plate subjected
to forces acting in the midplane of the plate.
( n , n )
• State of plane stress also occurs on the free surface
of a structural element or machine component, i.e.,
at any point of the surface not subjected to an
external force.
Plane Stress
y
Two methods to compute the
maximum stresses i.e.,
(1) Analytical approach
(2) Using of Mohr’s circle
y x
z
z
xx
xy x
y
yx
y
x
y
y
yx
xy
x
9-6 Variation of Stress at A Point: Analytical Derivation
A
A cos
A sin
Fn 0 A ( x A cos ) cos ( y A sin ) sin ( xy A cos ) sin
( yx A sin ) cos
Ft 0 A ( x A cos ) sin ( y A sin ) cos ( xy A cos ) cos
( yx A sin ) sin
Fn 0 A ( x A cos ) cos ( y A sin ) sin ( xy A cos ) sin
( yx A sin ) cos
x cos 2 y sin 2 2 xy cos sin
x y x y
2
2
cos 2 xy sin 2
Ft 0 A ( x A cos ) sin ( y A sin ) cos ( xy A cos ) cos
( yx A sin ) sin
x cos sin y sin cos xy cos 2 yx sin 2
x y
2
sin 2 xy cos 2
Note: xy yx , cos 2
1 cos 2
1 cos 2
sin 2
, sin 2
, cos sin
2
2
2
x
y
xy
x y x y
2
x y
2
2
2
2
x y x y
2
x y
2
cos 2 xy sin 2
A cos
sin 2 xy cos 2
x y x y
2
A
A sin
cos 2 xy sin 2
cos 2 xy sin 2
sin 2 xy cos 2
y
yx
xy
x
x
xy
yx
y
cos 2(2 ) cos( 2 ) cos 2
sin 2(2 ) sin( 2 ) sin 2
Eq.(9-5)
Eq.(9-6)
x y x y
2
x y
2
2
A
cos 2 xy sin 2
sin 2 xy cos 2
A cos
A sin
Find maximum or minimum differentiating Eq.(9-5)
w.r.t. and setting the derivative equal to zero
x y
d
2
sin 2 2 xy cos 2 0
d
2
tan 2
2 xy
x y
Find maximum or minimum differentiating Eq.(9-6)
w.r.t. and setting the derivative equal to zero
x y
d
2
cos 2 2 xy sin 2 0
d
2
x y
tan 2 s
2 xy
Eq.(9-5)
Eq.(9-6)
x y x y
2
x y
2
2
A
cos 2 xy sin 2
A cos
sin 2 xy cos 2
A sin
At zero shearing stress 0
0
x y
2
tan 2
sin 2 xy cos 2
2 xy
x y
ซึ่งเป็ นมุมเดียวกับสมการ Eq.(9-7) ดังนั้น ค่า maximum or minimum จะเกิดขึ้นเมื่อ = 0
tan 2
2 xy
xy
sin 21
x y
x y
(
sin 2 2
2
xy
x y
(
2
, cos 21
) 2 xy2
2 (
, cos 2 2
) 2 xy2
x y
x y 2
2
) xy2
y x
x y 2
2 (
2
) xy2
Maximum or minimum (Principal stresses)
2
1 x y
x y 2 2
(
) xy
2
2
2
tan 2
2 xy
x y
max (
2
x y
tan 2 s
2 xy
2
1
มุม และ s ต่างกัน 45O
Maximum or minimum
x y
1
)2 xy2
1 1
2
1
2
s
2
1
P
200
x
0.04 kN/mm2 40 MPa, y 0, xy 0
A 50 100
x y x y
2
2
cos 2(-40O ) xy sin 2(-40O )
40 0 40 0
cos 2(-40O ) 0 sin 2(-40O ) 16.5 MPa
2
2
x y
sin 2 xy cos 2
2
20 0
sin 2(-40O ) 0 cos 2(-40O ) 9.85 MPa
2
8,000 psi
4,000 psi
6,000 psi
x 4,000 psi
y 8,000 psi
xy 6,000 psi
1 x y
x y 2 2 4000 (8000)
4000 (8000) 2
(
)
(
) (6000) 2
xy
2
2
2
2
2
2000 (6000)2 (6000)2 10485.3, 6485.3 psi
x y x y
2
2
cos 2(30O ) xy sin 2(30O )
4000 (8000) 4000 (8000)
cos 2(30O ) (6000) sin 2(30O ) 6,196.15 psi
2
2
x y
2
sin 2 xy cos 2
4000 (8000)
sin 2(30O ) (6000) cos 2(30O ) 2196.15 psi
2
6,196.15 psi
30o
4,000 psi
8,000 psi
2,196.15 psi
6,000 psi
9-7 Variation of Stress at A Point: Mohr’s Circle
Otto Mohr (1882)
Eq.(9-5)
Eq.(9-6)
Eq.(a)2 + Eq.(b)2
x y x y
2
x y
2
2
cos 2 xy sin 2
sin 2 xy cos 2
Rule for Applying Mohr Circle to Combined Stresses
( x , xy )
(0, 0)
( y , xy )
( x , xy )
(0, 0)
C
( y , xy )
( n , n )
n
( x , xy )
R
(0, 0)
n
2
C
( y , xy )
( n , n )
n
( x , xy )
R
(0, 0)
n
2
C
( y , xy )
(C, max )
( x , xy )
( 2 ,0)
R
( 1 ,0)
22 2
1
C
( y , xy )
C (C , 0) (
R (
x y
x y
2
2
, 0)
)2 xy2
1 C R
2 C R
max R
sin 21
xy
tan 21 =
R
or
2 xy
x y
22 180o 21
( x , xy )
( y , xy )
(8000, 6000)
R
( 2 ,0)
(2000, 0)
C (C , 0) (
21
x y
(4000, 6000)
, 0)
2
8000 4000
(
, 0) ( 2000, 0)
2
R (
x y
2
( 1 ,0)
C
)2 xy2 (
4000 8000 2
) 60002 6000 2 psi
2
sin 21
2
1
22.5
1, 2 C R 2000 6000 2 4485.3, 10485.3 psi
1
xy
R
6000
6000 2
1 22.5O
(8000, 6000)
( 2 ,0)
( 30o , 30o )
R
( 1 ,0)
C
60o
(120o ,120o )
(2000, 0)
45o
(4000, 6000)
30 C R cos(15o )
o
2000 6000 2 cos(15o ) 6196.15 psi
30 R sin(15o ) 6000 2 sin(15o ) 2196.15 psi
o
10196.15
120 C R cos(15o )
o
2196.15
30
6196.15
2196.15
2000 6000 2 cos(15o ) 10196.15 psi
120 R sin(15o ) 6000 2 sin(15o ) 2196.15 psi
o
9-8 Absolute Maximum Shearing Stress
2
2
1
2
Rz
1
Rz
Mohr’s circle: Rotation around z-axis
1 2
2
x
1
2
Rx
Mohr’s circle: Rotation around x-axis
Rx
2
2
Ry
1
2
Mohr’s circle: Rotation around y-axis
Ry
1
2
1
2
Rz
1
1 2
x
Ry
2
Rz
Ry
1
2
Rx
2
2
Rx
Absolute maximum shearing stress for plane
stress is equal to the largest of the following three
values
Rz
1 2
2
, Rz
1
2
, Rx
2
2
2
Ry
Rx
Rz
Mohr’s circles for plane stress
1
Absolute maximum shearing stress for general
state of stress is equal to the largest of the
following three values
Rz
1 2
2
, Rz
1 3
2
, Rx
2
2 3
1
2
z
3
Ry
Rx
Rz
Mohr’s circles for general
state of stress
20
Maximum in-plane shearing stress =
1 2
50 20
15 ksi
2
2
Absolute maximum shearing stress is the largest of
1 2
2
1
2
2
2
50
2
20
2
50 20
2
25 ksi,
10 ksi,
15 ksi,
50
x 50
Ex.
20
Maximum in-plane shearing stress =
1 2
50
50 20
35 ksi
2
2
Absolute maximum shearing stress is the largest of
1 2
2
1
2
2
2
50
2
20
2
50 20
2
25 ksi,
10 ksi,
(ksi)
35 ksi,
1 =-50
Ry Rz
Rx
2 =20
(ksi)
Hw17
the figure
( สาหรับข้ อนีใ้ ห้ คานวณ ค่ า absolute maximum shearing stress ด้ วย
โดยกาหนดให้ z = 0
)
ค่า z1-z3 ได้จากเลขประจาตัวนิสิต ดังต่อไปนี้
10( z2 1) MPa
10( z1 1) MPa
10( z3 1) MPa
46xxxz1z2z3
9-9 Application of Mohr’s Circle to Combined Loadings
Combined Loadings
(axial, torsional, flexural)
Combined stresses
Design Criteria, allow ,
allow
(0, )
max
2
Principal stresses and,
Maximum shearing stress
2
1
1
2
1
2
s
1
2
max
max
Mohr’s Circle
1
( , )
Stress Trajectories
max
Tc
J
2
2
1
1
Tc
1
J
Torsional Failure Modes
• Ductile materials generally fail in
shear. Brittle materials are weaker in
tension than shear.
max
Tc
J
1
Tc
J
• A ductile specimen breaks along a
plane of maximum shear
45o
• A brittle specimen breaks along
planes perpendicular to 1
Stress Trajectories
for Torsion
max
Stress Trajectories: lines of
principal stress direction but
of variable stress intensity
Tc
J
1
Tc
J
Stress Trajectories for Beam
My
I
VQ
Ib
(0, )
max
2
Mohr’s Circle
1
( , )
I
D4
J
D 100 mm
80 MPa
100 MPa
64
D4
32
(0.1) 4
64
(0.1) 4
32
1.5625 106 m 4
3.125 106 m 4
M 2500 N.m
Mc (2500 )(0.05)
7
2
8
10
N/m
80 MPa
6
I
1.5625 10
Tc
T (0.05)
1.6 106
1.6T
2
T(
) N/m
MPa
J 3.125 106
80 MPa
(0,
1.6T
1.6T
)
Mohr’s Circle
max
MPa
2
C 40 MPa
(40,0)
1
C
1.6T
max R 402 (
) 2 80 MPa
1.6T
1 C R 40 402 (
)2
100 MPa
(80, 1.6T )
P 2 f T
1.6T 2
402 (
) 60 MPa
P 2 (30)(87.81)
T 87.81 N.m
P 16,551.8 watt
I
r4
4
, J
r4
Mc 4M
3
I
r
Tc 2T
3
I r
2
4M
r3
(0,
2T
r3
2T
r3
)
Mohr’s Circle
max
2
(40,0)
1
C
( 4 Mr3 , 2rT3 )
C 2M /( r 3 )
max R (
2M 2
2T 2
2
2
2
)
(
)
M
T
r3
r3
r3
1 C R
2
2
2
M
M
T
r3
900 12
10.8 kips-in
1000
600 12
M 600 lb-ft
7.2 kips-in
1000
If T 900 lb-ft
2
2
2
M
T
r3
2
8.263
3 7.22 10.82 3 ksi 10 ksi
r
r
max 10 ksi
max 16 ksi
max
2
2
2
M
M
T
r3
2
12.847
2
2
3 7.2 7.2 10.8
ksi 16 ksi
r
r3
1
r 0.938 in.
r 0.929 in.
750 N.m
2500 N
2500 N
3750 N
4000 N
750 N.m
750 N.m
1250 N
2875 N
2500 N
2500 N
1500 N.m
3750 N
4000 N
750 N.m
3625 N
1250 N
1500 N.m
3625 N
2875 N
750 N.m
4000 N
2500 N
2500 N
750 N.m
3750 N
4000 N
750 N.m
1500 N.m
750 N.m
2875 N
3625 N
1m
1250 N
2500 N
2875 N
2m
1m
2m
1500 N.m
3625 N.m
2875 N.m
3625 N
BMzD
2500 N
750 N.m
750 N.m
750 N.m
1500 N.m
1250 N
4m
1500 N.m
3750 N
TMD
2m
BMyD
1250 N.m
3750 N.m
5000 N.m
3625 N.m
750 N.m
2500 N
2500 N
4000 N
B
1250 N
A
D
2875 N.m
E
3750 N
BMzD
C
750 N.m
2875 N
BMyD
1500 N.m
1250 N.m
3750 N.m
3625 N
5000 N.m
| M | M z2 M y2
5000 N.m
4725.2 N.m
My
3834.5 N.m
Mz
A
Cross section of solid shaft
and the resultant moment
B
C
D
E
|M|
1500 N.m
750 N.m
TMD
From Prob. 951 and this problem.
2
2
max 3 M T 2 70 MPa
r
2
2
1 3 M M T 2 120 MPa
r
4M
3625 N.m
r3
(0, )
2T
r3
Mohr’s Circle
max
2875 N.m
BMzD
2
1
At section C
max
1
2
2
2
4725.2
1500
1000 mm 70 MPa
3
r
r 35.6 mm
BMyD
1250 N.m
( , )
3750 N.m
5000 N.m
2
4725.2 4725.22 15002 1000 120 MPa
3
r
r 37.2 mm
5000 N.m
4725.2 N.m
3834.5 N.m
At section D
max
2
50002 7502 1000 mm 70 MPa
3
r
r 35.8 mm
A
B
C
D
E
|M|
1500 N.m
750 N.m
2
1 3 5000 50002 7502 1000 120 MPa
r
r 37.7 mm
TMD
r ≥ 37.7 mm
state of stress on
the element on the
surface of vessel
1 67.5 R
2 67.5 R
x y
2
2
R
xy
2
2
R2 22.52 xy2 32.52
Absolute maximum shearing stress 50 MPa
| 1 2 |
R 50 MPa
2
| 1 | 67.5 R
50 MPa
2
2
| 2 | 67.5 R
50 MPa
2
2
xy2 32.52 22.52 550
xy 23.45 MPa
R 32.5 MPa
Tc
23.45 MPa
J
T (455 mm)
23.45 MPa
9204 9004
32
T 301.8 kN.m
20 mm
120 mm
20 mm
A
40 mm
N.A.
Q (20 40) 40
20 1203
I
=2.88 106 mm 4
12
V 30 kN
=3.2 104 mm3
250 mm
P My
40
7500 20
A I
20 120 2.88 106
68.75 MPa
P 40 kN
M 7500 kN.mm
VQ 30 3.2 104
16.67 MPa
I b 2.88 106 20
250 mm
V 30 kN
P Mc
40
7500 20
A I
20 120 2.88 106
68.75 MPa
P 40 kN
VQ 30 3.2 104
16.67 MPa
I b 2.88 106 20
M 7500 kN.mm
C (C , 0) (
(
x y
2
, 0)
68.75 0
, 0) (34.375, 0)
2
R (
x y
2
)2 xy2
68.75 2
(
) 16.67 2 38.20 MPa
2
Mohr’s Circle at point A
max
2
(68.75,16.67)
C 2
1
(34.375,0)
1 , 2 C R 34.375 38.20
72.578, 3.825 MPa
xy
16.67
R 38.20
12.94O
sin 2
(0,16.67)
72.58
3.83
12.94
3.83
72.58
20 mm
20 mm
120 mm
B
N.A.
40 mm
20 1203
I
=2.88 106 mm 4
12
V 30 kN
Q (20 40) 40
=3.2 104 mm3
300 mm
P My
40
9000 (20)
A I
20 120
2.88 106
45.83 MPa
P 40 kN
M 9000 kN.mm
VQ 30 3.2 104
16.67 MPa
I b 2.88 106 20
300 mm
V 30 kN
45.83 MPa
P 40 kN
16.67 MPa
M 9000 kN.mm
C (C , 0) (
(
x y
2
Mohr’s Circle at point B
, 0)
45.83 0
, 0) (22.915, 0)
2
R (
x y
2
)2 xy2
45.83 2
) 16.67 2 28.34 MPa
2
xy 16.67
sin 2
2 36.03O
R 28.34
(45.83,16.67)
60o
(
(48.81, 11.51)
48.81 MPa
0
30 28.34sin(23.97o ) 11.51 MPa
0
C
(0,16.67)
30 C R cos(60o 36.03o )
22.915 28.34cos(23.97o )
48.81 MPa
(22.915,0)
36.06o
45.83 MPa
11.51 MPa
16.67 MPa
Hw18
L1
L2
1.2D
L3
L4
1.2D
D
ค่า z1-z5 ได้จากเลขประจาตัวนิสิต ดังต่อไปนี้ 46xz1z2z3z4z5
L1= 4(1+z1) in.
L2 = 4(1+z2) in.
L3= 4(1+z3) in.
L4 = 4(1+z4) in.
D = 4(1+z5) in.
Hw19
Also find the maximum shearing stress at point A. Show your results on a
complete sketch of a differential element.
P
L
H
W
ค่า z1-z4 ได้จากเลขประจาตัวนิสิต ดังต่อไปนี้ 46xxz1z2z3z4
L= 0.4(1+z1) m.
P = 4(1+z2) kN
H= 40(1+z3) mm.
W = 40(1+z4) mm
E
G
2(1 )
http://www.kyowa-ei.co.jp/english/products.htm
Strain and deformation of line element
A( II )
A
A
A
A( III )
A
ds
dy
xy dy
y dy
(I )
A
O
O
x dx
x 0, y 0, xy 0
dx
A
A
O
x 0, y 0, xy 0
O
x 0, y 0, xy 0
O
x 0, y 0, xy 0
A
A cos
A sin
x y x y
2
x y
2
2
cos 2 xy sin 2
sin 2 xy cos 2
Eq.(9-5)
Eq.(9-6)
1
2
(800,300)
x 800 106 rad
y 200 106 rad
xy
2
300 106 rad
C 500 106 rad
R 300 2 106 rad
(200, 300)
If we use the stress-strain relation directly the same answer can be obtained
Hw20a จงพิสูจน์ สมการ (9-19) (9-20) ด้วยภาษาของตัวเอง
Hw20b
Hw21
ค่า z1-z3 ได้จากเลขประจาตัวนิสิต ดังต่อไปนี้ 46xxxz1z2z3
a= 100(1+z1)
c= 100(1+z3)
b= -100(1+z2)
ปริมาณทาง Physics สามารถแทนด้ วย Tensor
Order 0 = zero order Tensor (Scalar) – Magnitude (มวล, ความหนาแน่น)
Order 1 = first order Tensor (Vector) – Magnitude, Direction (ความเร็ว, แรง)
Order 2 = second order Tensor – Magnitudes, Directions (stress, strain)
… Higher order ….
ปริมาณทาง Physics ไม่ เปลีย่ นแปลงไปตามระบบโคออร์ ดเิ นตที่ใช้ ในการวัด
temperature
mass
length
mass 2 kg.= ?? lb.
length 5 in. = 12.7 cm.
temperature 50O C = 122O F
ปริมาณทาง Physics ไม่ เปลีย่ นแปลงไปตามระบบโคออร์ ดเิ นตที่ใช้ ในการวัด
แรง Pยังคงมีขนาดและทิศทางเท่ าเดิม ไม่ ว่าจะแสดง component ของเวคเตอร์ ด้วยระบบโคออร์ ดเิ นตอื่น
1
1
0
y
P
x
z
manitude 12 12 2
0.6
0.8
1
y
P
x
z
manitude 0.62 0.82 12 2
สถานะของหน่ วยแรง (state of stress) ยังคงมีคุณสมบัตเิ หมือนเดิม ไม่ ว่าจะแสดงด้ วยระบบโคออร์ ดเิ นตอืน่
1 0.5 0.2
σ 0.5 3 1
0.2 1 4
A
A
A
B
O
O
x0 0, y 0, xy 0
x 0, y 0, xy