Bài giảng Strength of materials - Chương 10: Combined loadings - Nguyễn Sỹ Lâm
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.
• Members remain symmetric and bend in
the plane of symmetry.
• The neutral axis of the cross section
coincides with the axis of the couple
• Will now consider situations in which the
bending couples do not act in a plane of
symmetry.
• Cannot assume that the member will bend
in the plane of the couples.
• In general, the neutral axis of the section will
not coincide with the axis of the couple.
10.1 UNSYMMETRIC BENDING
UNSYMMETRIC BENDING
y
0 F dA dA
•
x
x
m
c
or 0 ydA
neutral axis passes through centroid
y
M M y dA
•
z
m
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.
c
σ I
m
or M
I I moment of inertia
z
c
defines stress distribution
• The resultant force and moment
from the distribution of
elementary forces in the section
must satisfy
y
0 M z dA z dA
•
y
x
m
c
or 0 yzdA I product of inertia
yz
F 0 M
M M applied couple
z
couple vector must be directed along
a principal centroidal axis
x
y
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.
M M cos
M M sin
y
z
• Superpose the component stress distributions
My z
Iy
Mz y
Iz
x
• Along the neutral axis,
M y z
Mz y
Iz
M cos
y
M sin z
x 0
Iy
Iz
Iy
y Iz
tan tan
z Iy
10.1 UNSYMMETRIC BENDING
UNSYMMETRIC BENDING
My z
Iy
Mz y
Iz
x
• Along the neutral axis,
M y z
Mz y
Iz
M cos
y
M sin z
x 0
Iy
Iz
Iy
y Iz
tan tan
z Iy
• Equation of Neutral Axis
M y
Iz
M y
Iz
y
z
y
z
M z I y
M z I y
10.1 UNSYMMETRIC BENDING
UNSYMMETRIC BENDING – SUMMARY
Mx
-
-
2
2
Mu M x M y
h
x
z
+
+
y
x
x
y
b
My
-
-
Mu
y
h
x
z
+
+
y
x
y
b
M y
Ix
M y
Iy
Mx
Ix
• Equation of Neutral Axis
y
x
z
y
x
Mx Iy
(N.A.)
10.1 UNSYMMETRIC BENDING
UNSYMMETRIC BENDING – SUMMARY
min
-
-
h
x
Compression
stress
+
+
y
max
b
Mx
x
z
-
-
+
h
x
My
y
Mu
Tension stress
+
M y
Ix
• Equation of Neutral Axis
y
x
y
Mx Iy
(N.A.)
b
M y
Iy
Mx
Ix
(Diagram of Equation : y = ax)
z
y
x
10.1 UNSYMMETRIC BENDING
EXAMPLE 10.01
SOLUTION:
• Resolve the couple vector into
components along the principle
centroidal axes and calculate the
corresponding maximum stresses.
M M cos
M M sin
y
z
• Combine the stresses from the
component stress distributions.
M y
M y
y
z
x
I
I
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.
z
y
• Determine the angle of the neutral
axis.
y I
z
tan tan
z I
y
10.1 UNSYMMETRIC BENDING
EXAMPLE 10.01
• Resolve the couple vector into components and calculate
the corresponding maximum stresses.
M
1600lb in
cos30 1386lb in
sin 30 800lb in
z
M
1600lb in
y
3
4
1
12
I
1.5in3.5in
5.359in
z
3
4
1
12
I
3.5in1.5in
0.9844in
y
The largest tensile stress due to M occursalong AB
z
M y
1386lb in1.75in
z
452.6psi
1
4
I
5.359in
z
The largest tensile stress due to M occursalong AD
z
M z
800lb in0.75in
y
y
609.5psi
2
4
I
0.9844in
• The largest tensile stress due to the combined loading
occurs at A.
452.6609.5
1062psi
max
1
2
max
10.1 UNSYMMETRIC BENDING
EXAMPLE 10.01
• Determine the angle of the neutral axis.
4
I
5.359in
z
tan tan
tan30
4
I
0.9844in
y
3.143
o
72.4
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
x
x
centric
x bending
M
P
y
A I
• Eccentric loading
• Validity requires stresses below proportional
limit, deformations have negligible effect on
geometry, and stresses not evaluated near points
of load application.
F P
M Pd
10.2 ECCENTRIC AXIAL LOADINGS
EXAMPLE 10.02
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.
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
• Find the neutral axis by determining
the location where the normal stress
is zero.
10.2 ECCENTRIC AXIAL LOADINGS
EXAMPLE 10.02
• Normal stress due to a
centric load
2
2
A c
0.25in
2
0.1963in
P
160lb
0
2
A
0.1963in
815psi
• Equivalent centric load
• Normal stress due to
and bending moment
bending moment
4
4
1
1
4
P 160lb
I c
0.25
4
M Pd
160lb0.6in
3
4
3.06810 in
104lb in
Mc
I
104lb in0.25in
m
3
4
.06810 in
8475psi
10.2 ECCENTRIC AXIAL LOADINGS
EXAMPLE 10.02
• Maximum tensile and compressive
• Neutral axis location
stresses
P My
0
0
m
t
0
A
I
9260psi
3 4
8158475
t
P I
3.06810 in
y
815psi
0
A M
105lb in
c
0
m
7660psi
8158475
c
y 0.0240in
0
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.
• Superpose the stress due to a centric
load and the stress due to bending.
• Evaluate the critical loads for the allowable
Given:
tensile and compressive stresses.
3 2
A 310 m
Y 0.038m
• The largest allowable load is the smallest
of the two critical loads.
9
4
I 86810 m
10.2 ECCENTRIC AXIAL LOADINGS
EXAMPLE 10.03
• Determine an equivalent centric and bending loads.
d 0.0380.010 0.028m
P centricload
M Pd 0.028P bending moment
• Superpose stresses due to centric and bending loads
P Mc
P
0.028P0.022
A
377P
A
3
3
9
A
I
310
86810
P Mc
P
0.028P0.022
A
1559P
B
9
A
I
310
86810
• Evaluate critical loads for allowable stresses.
377P 30MPa
P 79.6kN
A
1559P 120MPa P 79.6kN
B
P 77.0 kN
• 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.
P centricforce
M Pa
M Pb
z
y
• By the principle of superposition, the
combined stress distribution is
M z
P M y
y
z
x
A I
I
y
z
• If the neutral axis lies on the section, it may
be found from
M
M
P
A
y
z
y
z
I
I
y
z
10.2 ECCENTRIC AXIAL LOADINGS
SUMMARY
Mx
2
2
Mu M x M y
- -
h
x
z
+
+
x
x
y
y
y
b
My
- +
- +
Mu
y
h
x
z
x
y
b
Nz
+ +
x
h
x
z
+
+
Mu
x
y
y
y
10.2 ECCENTRIC AXIAL LOADINGS
2
2
Mu M x M y
SUMMARY
M y
I y
M x
Ix
Nz
A
z
y
x
M y
I y
M x
Ix
- -
z
y
x
h
x
Mx
+ +
min
y
x
b
Compression
stress
- +
Mu
h
y
x
max
-
+
x
y
My
b
z
Nz
Mu
Tension stress
y
Nz
z
+ +
A
h
x
M y
Nz
Ix
Ix
y
x
+
+
Equation of Neutral Axis (N.A.):
M x I y
A M x
(Diagram of Equation : y = ax +b)
y
10.3 TORSION AND BENDING
CIRCULAR SHAFTS
Mz = M0
P
M0
M z
J
x
z
M z
max c
J
4
1
J c
2
max
exists on the
perimeter of the
circular section
4
4
1
J c c
2
1
2
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