Giáo trình Strength of Materials - Chapter 10: Combined loadings

CHAPTER 10: COMBINED LOADINGS 
10.1 Unsymmetric Bending 
10.2 Eccentric Axial Loadings 
10.3 Torsion and Bending 
10.4 General Combined Loadings 
10.1 UNSYMMETRIC BENDING 
UNSYMMETRIC BENDING 
• Analysis of pure bending has been limited 
to members subjected to bending couples 
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. 
10.1 UNSYMMETRIC BENDING 
UNSYMMETRIC BENDING 
Wish to determine the conditions under 
which the neutral axis of a cross section 
of arbitrary shape coincides with the 
axis of the couple as shown. 
• 
 couple vector must be directed along 
a principal centroidal axis 
inertiaofproductIdAyz
dA
c
y
zdAzM
yz
mxy

 






0or
0 
• The resultant force and moment 
from the distribution of 
elementary forces in the section 
must satisfy 
coupleappliedMMMF zyx  0
• 
 neutral axis passes through centroid 

 






dAy
dA
c
y
dAF mxx
0or 
0 
• 
 defines stress distribution 
inertiaofmomentII
c
Iσ
dA
c
y
yMM
z
m
mz

 






Mor 

10.1 UNSYMMETRIC BENDING 
UNSYMMETRIC BENDING 
Superposition is applied to determine stresses in 
the most general case of unsymmetric bending. 
• Resolve the couple vector into components along 
the principle centroidal axes. 
 sincos MMMM yz 
• Superpose the component stress distributions 
y
y
z
z
x
I
zM
I
yM

• Along the neutral axis, 
   



tantan
sincos
0
y
z
yzy
y
z
z
x
I
I
z
y
I
zM
I
yM
I
zM
I
yM


10.1 UNSYMMETRIC BENDING 
UNSYMMETRIC BENDING 
y
y
z
z
x
I
zM
I
yM

• Along the neutral axis, 
   



tantan
sincos
0
y
z
yzy
y
z
z
x
I
I
z
y
I
zM
I
yM
I
zM
I
yM


z
I
I
M
M
y
y
z
z
y

• Equation of Neutral Axis 
z
I
I
M
M
y
y
z
z
y

10.1 UNSYMMETRIC BENDING 
UNSYMMETRIC BENDING – SUMMARY 
x
I
I
M
M
y
y
x
x
y
• Equation of Neutral Axis 
(N.A.) 
x
I
M
y
I
M
y
y
x
x
z 
My 
x 
y 
Mx 
z 
x 
y 
z 
h 
h 
b 
x 
y 
+ 
- - 
+ 
b 
x 
y 
+ 
- - 
+ 
x 
y Mu 
22
yxu MMM 
10.1 UNSYMMETRIC BENDING 
UNSYMMETRIC BENDING – SUMMARY 
x
I
I
M
M
y
y
x
x
y
• Equation of Neutral Axis 
(N.A.) 
x
I
M
y
I
M
y
y
x
x
z 
h 
h 
b 
x 
y 
- 
- + 
+ 
b 
x 
y 
+ 
- - 
+ 
 x 
y Mu 
Mx 
My 
max
min
z 
Tension stress 
Compression 
stress 
(Diagram of Equation : y = ax) 
10.1 UNSYMMETRIC BENDING 
EXAMPLE 10.01 
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. 
 sincos MMMM yz 
• Combine the stresses from the 
component stress distributions. 
y
y
z
z
x
I
yM
I
yM

• Determine the angle of the neutral 
axis. 
 tantan
y
z
I
I
z
y

10.1 UNSYMMETRIC BENDING 
EXAMPLE 10.01 
• Resolve the couple vector into components and calculate 
the corresponding maximum stresses. 
 
 
  
  
  
  
psi5.609
in9844.0
in75.0inlb800
 along occurs todue stress nsilelargest te The
psi6.452
in359.5
in75.1inlb1386
 along occurs todue stress nsilelargest te The
in9844.0in5.1in5.3
in359.5in5.3in5.1
inlb80030sininlb1600
inlb138630cosinlb1600
42
41
43
12
1
43
12
1










y
y
z
z
z
z
y
z
y
z
I
zM
ADM
I
yM
ABM
I
I
M
M


• The largest tensile stress due to the combined loading 
occurs at A. 
5.6096.45221max   psi1062max 
10.1 UNSYMMETRIC BENDING 
EXAMPLE 10.01 
• Determine the angle of the neutral axis. 
143.3
30tan
in9844.0
in359.5
tantan
4
4

 
y
z
I
I
o4.72
10.2 ECCENTRIC AXIAL LOADINGS 
ECCENTRIC AXIAL LOADING IN A PLANE OF SYMMETRY 
• 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 
   
bendingcentric xxx
 
• Eccentric loading 
PdM
PF


• Validity requires stresses below proportional 
limit, deformations have negligible effect on 
geometry, and stresses not evaluated near points 
of load application. 
y
I
M
A
P

10.2 ECCENTRIC AXIAL LOADINGS 
EXAMPLE 10.02 
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. 
10.2 ECCENTRIC AXIAL LOADINGS 
EXAMPLE 10.02 
• Equivalent centric load 
and bending moment 
  
inlb104
in6.0lb160
lb160



PdM
P
 
psi815
in1963.0
lb160
in1963.0
in25.0
20
2
22




A
P
cA


• Normal stress due to a 
centric load 
 
  
psi8475
in10068.
in25.0inlb104
in10068.3
25.0
43
43
4
4
14
4
1








I
Mc
cI
m

• Normal stress due to 
bending moment 
10.2 ECCENTRIC AXIAL LOADINGS 
EXAMPLE 10.02 
• Maximum tensile and compressive 
stresses 
8475815
8475815
0
0




mc
mt


psi9260t
psi7660c
• Neutral axis location 
 
inlb105
in10068.3
psi815
0
43
0
0





M
I
A
P
y
I
My
A
P
in0240.00 y
10.2 ECCENTRIC AXIAL LOADINGS 
EXAMPLE 10.03 
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. 
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. 
Given: 
49
23
m10868
m038.0
m103





I
Y
A
• Superpose the stress due to a centric 
load and the stress due to bending. 
10.2 ECCENTRIC AXIAL LOADINGS 
EXAMPLE 10.03 
• Determine an equivalent centric and bending loads. 
moment bending 028.0
load centric
m028.0010.0038.0



PPdM
P
d
• Evaluate critical loads for allowable stresses. 
kN6.79MPa1201559
kN6.79MPa30377


PP
PP
B
A


• Superpose stresses due to centric and bending loads 
  
  
P
PP
I
Mc
A
P
P
PP
I
Mc
A
P
A
B
A
A
1559
10868
022.0028.0
103
377
10868
022.0028.0
103
93
93














kN 0.77P• The largest allowable load 
10.2 ECCENTRIC AXIAL LOADINGS 
GENERAL CASE OF ECCENTRIC AXIAL LOADING 
• Consider a straight member subject to equal 
and opposite eccentric forces. 
• The eccentric force is equivalent to the system 
of a centric force and two couples. 
PbMPaM
P
zy 
 force centric 
• If the neutral axis lies on the section, it may 
be found from 
A
P
z
I
M
y
I
M
y
y
z
z 
• By the principle of superposition, the 
combined stress distribution is 
y
y
z
z
x
I
zM
I
yM
A
P

10.2 ECCENTRIC AXIAL LOADINGS 
SUMMARY 
x 
y 
Mx 
z 
x y 
z 
h 
b 
x 
y 
- 
- + 
+ 
h 
b 
x 
y 
+ 
- - 
+ 
x 
y Mu 
22
yxu MMM 
x y 
z 
My 
Nz 
h x 
y 
+ 
+ + 
+ 
x 
y Mu 
10.2 ECCENTRIC AXIAL LOADINGS 
SUMMARY 
22
yxu MMM 
h 
b 
x 
y 
- 
- + 
+ 
My 
Nz 
h x 
y 
+ 
+ + 
+ 
x 
y Mu 
h 
b 
x 
y 
+ 
- - 
+ 
Mx 
x 
y Mu 
max
min
z 
A
N
x
I
M
y
I
M z
y
y
x
x
z 
x
I
M
y
I
M
y
y
x
x
z 
A
N z
z 
Equation of Neutral Axis (N.A.): 
(Diagram of Equation : y = ax +b) 
x
xz
y
x
x
y
M
I
A
N
x
I
I
M
M
y 
Tension stress 
Compression 
stress 
10.3 TORSION AND BENDING 
CIRCULAR SHAFTS 
4
2
1 cJ 
 414221 ccJ  
P 
M0 
z 
x 
Mz = M0 
c
J
M zmax

J
M z
max exists on the 
perimeter of the 
circular section 
10.3 TORSION AND BENDING 
CIRCULAR SHAFTS 
Mz = M0 
u 
v 
Mu 
max
min
max
max
z 
max
max
max
min
   2max
2
max 4
   2max
2
max 3
or 
   2max
2
min 4
   2max
2
min 3
or 
10.3 TORSION AND BENDING 
RECTANGULAR SECTIONS 
OR 
A 
B 
C 
D 
E 
F 
G 
H 
C 
)(
max
yx MM 
G 
)(
min
yx MM 
E 
)(
min
)(
max
yx MM  
A 
)(
max
)(
min
yx MM  
H 
)(
min
xM
1
B 
max
)(
max
yM
D 
)(
max
xM
1
F 
)(
min
yM
max
Mx 
My 
Mz 
10.3 TORSION AND BENDING 
RECTANGULAR SECTIONS 
OR 
 
n
S y
TB
22
4
2
1
2
13 


 
n
S y
TB
33
3
2
1
2
14 


10.4 GENERAL COMBINED LOADINGS 
RECTANGULAR SECTIONS 
A 
B 
C 
D 
E 
F 
G 
H 
C 
)()(
max
zyx N
z
MM
 

G 
)()(
min
zyx N
z
MM
 

E 
)()(
min
)(
max
zyx N
z
MM
 
A 
)()(
max
)(
min
zyx N
z
MM
 
H 
)()(
min
zx N
z
M
 
1
B 
max
)()(
max
zy N
z
M
 
D 
)()(
max
zx N
z
M
 
1
F 
)()(
min
zy N
z
M
 
max
Mx 
My 
Mz 
Nz 
10.4 GENERAL COMBINED LOADINGS 
RECTANGULAR SECTIONS 
OR 
 
n
S y
TB
22
4
2
1
2
13 


 
n
S y
TB
33
3
2
1
2
14 

