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  106 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 21 
 x  y
 x  y
(
sin 2 2 
2
 xy
 x  y
(
2
, cos 21 
) 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)
22 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 21 
 xy
tan 21 =
R
or
2 xy
 x  y
22  180o  21

( x , xy )
( y ,  xy )

(8000, 6000)
R
( 2 ,0)
(2000, 0)
C  (C , 0)  (
21
 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 21 
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  106  m 4
 3.125  106  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  106 



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,  1.6T )
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 106 rad
 y  200 106 rad
 xy
2
 300  106 rad

C  500 106 rad
R  300 2 106 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 