• Unsymmetric Bending(不对称弯曲)
• Tension & Bending(拉弯组合)
• Eccentric Compression(偏心压缩)
• Core of Cross-sections(截面核心区域)
• Core of Rectangular Cross-sections(矩形截面核心区域)
• Core of Circular Cross-sections(圆形截面核心区域)
• Tension & Torsion(拉扭组合)
• Bending & Torsion(弯扭组合)
• Tension, Bending & Torsion(拉弯扭组合)
Contents
2
• A circular bar subjected to a single type of load
3
Introduction
x z zM y I
*
Sxy z zF S I b
x pT I
• The principle of superposition is used to determine the
resultant stress & strain
• Prismatic bars are frequently subjected to several loads
simultaneously
• Conditions for the principle of superposition
- Linear elasticity & small deformation
4
Introduction
- No interaction between variously loads
Unsymmetric Bending
• Analysis of pure bending has been limited
to members subjected to bending moments
acting in a plane of symmetry.
• Will now consider situations in which the
bending couples do not act in a plane of
symmetry.
• In general, the neutral axis of the section will
not coincide with the axis of the couple.
• Cannot assume that the member will bend
in the plane of the couples.
• The neutral axis of the cross section
coincides with the axis of the couple
• Members remain symmetric and bend in
the plane of symmetry.
5
Wish to determine the conditions under
which the neutral axis of a cross section
of arbitrary shape coincides with the
axis of the moment as shown.
• Moment vector must be directed
along a principal centroidal axis
0
or 0 product of inertia
zy
z
yz
M yM z dA
I
yz dA I
• The resultant force and moment
from the distribution of
elementary forces in the section
must satisfy
0 applied couplex y zF M M M
• If neutral axis passes through centroid
0
or 0
zx x
z
M yF dA dA
I
y dA
• Stress distribution
zz
z
M yM M y dA
I
• Superposition is applied to determine
stresses in the most general case of
unsymmetric bending.
Unsymmetric Bending
6
, .x z z y z zM y I w M EI
• Construct a coordinate system.
• In x-y plane: positive Mz results in compression for y < 0.
• Bending stress and deflection:
• The same sign conventions can be used for bending in x-z plane.
• Positive My results in compression for z < 0.
• Bending stress and deflection: , .x y y z y yM z I w M EI
• Sign Convention
7
z
x
y
b
h
x
zF
,y z
yFa
Unsymmetric Bending
, ,
, , 0
zz y x
yz zx x x
yz y
y z x
y
M yM F x a a x L
M zI M yM z I I
M F x x LI
8
z
x
y
b
h
x
zF
,y z
yFa
Unsymmetric Bending
• Equation for Neutral axis:
00
0
0
0
tan
tan
ta
0
n
yzx
z y
y y
z z
z
y
y z
y
z
y y
z z
M zM y
I I
I Ix az
y I x I
I Ix a
I x
M
M
I
F
F
I a
I
• The neutral axis passes through the centroid of the cross-section.
• The maximum stresses occur at the two farthest points from the neutral axis.
• With the exception of Iy = Iz, i.e. for circular/square cross-section, the bending
stress and deflection cannot be calculated from the resultant moment.
• Bending in a single plane occurs if and only if the orientation of neural axis stays
the same for every cross-section, i.e. Mz/My=constant for a prismatic beam.
• In general, superposition should be resorted to determine both the bending stress
and the deflection. 9
0 0,y z
yM
zM
yF
zF
z
y
Unsymmetric Bending
A 1600 lb-in couple is applied to a
rectangular wooden beam in a plane
forming an angle of 30 deg. with the
vertical. Determine (a) the maximum
stress in the beam, (b) the angle that the
neutral axis forms with the horizontal
plane.
SOLUTION:
• Resolve the couple vector into
components along the principle
centroidal axes and calculate the
corresponding maximum stresses.
• Combine the stresses from the
component stress distributions.
• Determine the angle of the neutral axis.
Sample Problem
10
• Resolve the couple vector into components and
calculate the corresponding maximum stresses.
3 4112
3 4112
1 4
1600lb in cos30 1386lb in
1600lb in sin 30 800lb in
1.5in 3.5in 5.359in
3.5in 1.5in 0.9844in
The largest tensile stress due to occurs along
1386lb in 1.75in452.6ps
5.359in
z
y
z
y
z
z
z
M
M
I
I
M AB
M y
I
2 4
i
The largest tensile stress due to occurs along
800lb in 0.75in609.5psi
0.9844in
z
y
y
M AD
M z
I
• The largest tensile stress due to the combined
loading occurs at A.
max 1 2 452.6 609.5 1062psi
z
y
11
• Determine the angle of the neutral axis.
0 0 000 04
0
0
1386lb in 800lb in0 258.63 812.68
5.359in 0.9844
812.68tan 3.142
258.63
72.4
yzx
z y
o
M z y zM yy z
I I
y
z
12
z
y
0 0,y z
• Solution
2. Critical cross-section
• For the I-32a beam shown, L = 4 m, [σ] = 160 MPa, F = 80 kN, α =
5°. Analyze the strength condition.
; ;2 4 4
y zz y
F L F LLx M M
3. Neutral axis & maximum stresses
4 4
y yz zx
y z y z
M z F LM y F Lz y
I I I I
1. Decompose the load
Sample Problem
13
cosyF F
sinzF F z
y
z
y
F
2L 2L
xF
4. Strength check:
max 217.8MPa 160MPay z
y z
M M
W W
6. If α = 0, max 115.6z
z
MMPa
W
22
zy www 5. Deflections:
cosyF F
sinzF F z
y
Remark: The fact that the neutral axis is not necessarily perpendicular
to the resultant moment can greatly affect the stresses in a beam,
especially if the ratio of the principal moments of inertia is very large.
Under these conditions the stresses in the beam are very sensitive to
slight changes in the direction of the load and to irregularities in the
alignment of the beam itself.
14
• Stress due to eccentric loading found by
superposing the uniform stress due to a centric
load and linear stress distribution due a pure
bending moment
centric bendingx x x
z
z
M yP
A I
• Validity requires stresses below proportional
limit, deformations have negligible effect on
geometry, and stresses not evaluated near points
of load application.
Tension & Bending (Eccentric Tension)
15
, F P M Pd
• Strength condition:
max max, .
An open-link chain is obtained by
bending low-carbon steel rods into the
shape shown. For 160 lb load, determine
(a) maximum tensile and compressive
stresses, (b) distance between section
centroid and neutral axis
SOLUTION:
• Find the equivalent centric load and
bending moment
• Superpose the uniform stress due to
the centric load and the linear stress
due to the bending moment.
• Evaluate the maximum tensile and
compressive stresses at the inner
and outer edges, respectively, of the
superposed stress distribution.
• Find the neutral axis by determining
the location where the normal stress
is zero.
Sample Problem
16
• Equivalent centric load
and bending moment
160lb
160lb 0.65in
104lb in
z
P
M Pd
22
2
2
0.25in
0.1963in
160lb
0.1963in
815psi
x
A c
P
A
• Normal stress due to a
centric load
441 14 4
3 4
3 4
0.25
3.068 10 in
104lb in 0.25in
3.068 10 in
8475psi
z
zx
z
I c
M c
I
• Normal stress due to
bending moment
17
• Maximum tensile and compressive
stresses
max
max
815 8475 9260psi
815 8475 7660psi
• Neutral axis
0
3 4
0 2
0
160lb 3.068 10 in
0.1963in 104lb in
0.0240in
z
z
z
z
M yP
A I
IPy
A M
18
x
z
y
e A(yF,zF)
Fc
x
z
y
e A(yF,zF)
F
c m mz
myx
z
y
e A(yF,zF)
F
• Bars subjected to axial forces deviated a distance from axis.
1. Find the equivalent system of forces at the centroid.
Eccentric Compression
19
mz
my
x
z
y
e A(yF,zF)
F
B(y,z)
y
zO D1
D2.
2. Normal stress on cross-sections.
Nx
F F
A A
z Fx
z z
M y F y y
I I
y Fx
y y
M z F z z
I I
F F
z y
F F y y F z z
A I I
FN = -F,
My = -F · zF,
Mz = -F · yF,
2 2
2 2 1 F F
y y z z
z y
y y z zFI Ai I Ai
A i i
,
• Note the negative sign in bending stresses.
20
y
z
ay
az
A(yF,zF)
3. Equation of neutral axis.
0
2
0 0 z
y zF
ia y
y ;
F
y
yzz
iza
2
000
4. Strength condition:
• The intercepts of neutral axis:
2 2
0 0
2 2
0 1
1 0
F F
z y
F F
z y
y y z zF
A i i
y y z z
i i
21
View from top
max max, .
The largest allowable stresses for the cast
iron link are 30 MPa in tension and 120
MPa in compression. Determine the largest
force P which can be applied to the link and
the neutral axis.
SOLUTION:
• Determine an equivalent centric load and
bending moment.
• Evaluate the critical loads for the allowable
tensile and compressive stresses.
• The largest allowable load is the smallest
of the two critical loads.
• Superpose the stress due to a centric
load and the stress due to bending.
Sample Problem
22
• Determine an equivalent centric and bending loads.
0.038 0.010 0.028m
0.028 bending momentz
d
M Pd P
• Evaluate critical loads for allowable stresses.
377 30MPa 79.6kN
1559 120MPa 77.0kN
A
B
P P
P P
kN 0.77P• The largest allowable load
• Superpose stresses due to centric and bending loads
3 9
3 9
0.028 0.022377
3 10 868 10
0.028 0.0381559
3 10 868 10
z AA
z
z BB
z
PM cP PP
A I
PM cP PP
A I
From geometry,
23
0
0
2 9 3
0 0
2
0 0
868 10 3 10 = 10.33 mm
0.028
zy z
F
y
z yF
ia y
y
ia z
z
• Neutral axis
y
B
A
Cz
y5
10
• Find the maximum normal stress for the bar shown.• Solution:
1. Equivalent System of forces
FN = F =10 kN
Mz = 10×103×5×10-2 = 500 Nm
My = 10×103×2.5×10-2 = 250 Nm
2. Maximum normal stress
max 14 MPay AN z A
A
z y
M zF M y
A I I
10max
MPay Bz BN
B
z y
M zM yF
A I I
; 2
F
zy
y
ia
F
y
zz
ia
2
Intercepts:
5 cm
10 cm
Sample Problem
24
View from left
z
y
O
1
2
3
①
②
③
ay1
az1
• Intercepts of neutral axis:
• Portion of a cross-section within which eccentric
loading results in only compressive stresses.
22
22
;
yzy z
F F
yzF F
y z
iia a
y z
iiy z
a a
;
Core of Cross-sections
25
z
y
h
b
①
②
③
④A
B C
D
4
3 1
6b
6h
2
,12
22 h
A
Ii zz
12
22 b
A
Ii
y
y
• Take side AB as neutral axis
211
ba,a zy
1 1
1 1
22
06
yzF F
y z
ii by z
a a ,
• Take side BC as neutral axis
06 22
FF zh
y ,
• Through corner B, there exists
infinite number of neutral axes
we can take.
2
,zF
y
iy
a
z
y
Fa
iz
2
2 2
2 2
1 0
1 02 2
12 126 6
1 0
F B F B
z y
F F
F F
y y z z
i i
y zh bh b
y z
h b
Core of Rectangular Cross-sections
26
z
y
O
d
①
1
8
d
• Due to axial symmetry, only one neutral axis is
necessary to determine the radius of the core
area:
①:
2
22
1 1
,2
4
40
82
y z
y z
yzF F
y z
da a
di i
dii dy z
da a
,
• Note: no matter whatever shape of a cross-section area, the
corresponding core area is always solid and non-concave.
Core of Circular Cross-sections
27
• Simultaneous tensile and torsional loading of a circular shaft.
2
3
4
16
Nx
x
P
F P
A dT T
W d
28
T
P
T
P
• Solution
Tension & Torsion
yl
aA B
Cz
F
A1
A2
τ
Pz
z
W
T
W
M ,
τ
σ
A1
• Critical section: AA1
σ
A2
A
][4 22
][3 22
3
4
r
r
• Critical points on A: A1 & A2
2 2 2 2
3
2 2 2 2
4
1( ) 4( ) [ ]
1( ) 3( ) 0.75 [ ]
zr z
z P z
zr z
z P z
M TM T
W W W
M TM T
W W W
Bending & Torsion
29
View from left View from right1A
2A
• For the circular shaft shown, L = 50 cm, F = 8 kN, a = 37.5 cm, [σ] =
100 MPa. Find the minimum diameter of shaft AB based on
maximum shearing stress criteria.
yl
aA B
Cz
F
F
M
2 2
2 2 2 2
3
2 2
5 3
3
14 4 [ ]
5 10
3279.8
zr z
z P z
z
z
z
M TM T
W W W
M TW m
Wd mm
• Solution:
2 2
2 2 2 2
4
2 2
13 3 0.75 [ ]
0.75
zr z
z P z
z
z
M TM T
W W W
M TW
What about the maximum
distortion energy criteria?
max3 kNm, 4 kNmT M FL
Sample Problem
30
• Two forces P = 18 kN and F = 15 kN are applied to the shaft with a
radius of R = 20 mm as shown. Determine the maximum normal and
shearing stresses developed in the shaft.
Tension, Bending & Torsion – an Exercise
31
• Unsymmetric Bending(不对称弯曲)
• Tension & Bending(拉弯组合)
• Eccentric Compression(偏心压缩)
• Core of Cross-sections(截面核心区域)
• Core of Rectangular Cross-sections(矩形截面核心区域)
• Core of Circular Cross-sections(圆形截面核心区域)
• Tension & Torsion(拉扭组合)
• Bending & Torsion(弯扭组合)
• Tension, Bending & Torsion(拉弯扭组合)
Contents
32